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Notations, (The number after the item indicates the page where the notation is defined.), , SET THEORY, , SPECIAL SETS, , >i[I Si, <i[I Si, [a], |s|, , intersection of sets Si , i [ I, union of sets Si , i [ I, {x [ S | x , a}, equivalence class of S containing a, 18, number of elements in the set of S, , Z, , integers, additive groups of integers, ring of integers, rational numbers, field of rational numbers, multiplicative group of positive rational numbers, set of nonzero elements of F, real numbers, field of real numbers, multiplicative group of positive real numbers, complex numbers, , Q, Q1, F*, R, R1, C, , FUNCTIONS, AND ARITHMETIC, , ALGEBRAIC SYSTEMS, , f 21, t|s, tBs, gcd(a, b), lcm(a, b), |a 1 b|, f(a), f: A → B, gf, ab, D4, Dn, e, Zn, det A, U(n), , Rn, SL(2, F), GL(2, F), g21, 2g, UGU, UgU, H#G, H,G, kal, Z(G), , inverse of the function f, t divides s, 3, t does not divide s, 3, greatest common divisor of the integers a and b, 4, least common multiple of the integers a and b, 6, 2a2 b2, 13, image of a under f, 20, mapping of A to B, 20, composite function, 21, group of symmetries of a square, dihedral group of, order 8, 33, dihedral group of order 2n, 34, identity element, 43, group {0, 1, . . . , n 2 1} under addition modulo n, 44, the determinant of A, 45, group of units modulo n (that is, the set of integers, less than n and relatively prime to n under multiplication modulo n), 46, {(a1, a2, . . . , an) U a1, a2, . . . , an [ R}, 47, group of 2 3 2 matrices over F with, determinant 1, 48, 2 3 2 matrices of nonzero determinants with coefficients from the field F (the general linear group), 48, multiplicative inverse of g, 51, additive inverse of g, 52, order of the group G, 60, order of the element g, 60, subgroup inclusion, 61, subgroup H 2 G, 61, {an U n [ Z}, cyclic group generated by a, 65, {a [ G U ax 5 xa for all x in G}, the center of G, 66, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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C(a), kS l, C(H), f(n), N(H), cl(a), Gn, Sn, An, G<G, fa, Aut(G), Inn(G), aH, aHa21, UG:HU, HK, stabG(i), orbG(i), G1 % G2 % ? ? ? % Gn, Uk(n), G9, HvG, G/H, H3K, H1 3 H2 3 ? ? ? 3 Hn, Ker f, f21(g9), f21(K ), Z[x], M2(Z), R1 % R2 % ? ? ? % Rn, nZ, Z[i], U(R), char R, kal, ka1, a2, . . . , anl, R/A, A1B, AB, Ann(A), N(A), F(x), R[x], , {g [ G U ga 5 ag}, the centralizer of a in G, 68, subgroup generated by the set S, 71, {x [ G U xh 5 hx for all h [ H}, the centralizer, of H, 72, Euler phi function of n, 84, {x [ G U xHx21 5 H} 5 {x [ G U Hx 5 xH}, the, normalizer of H in G, 95, conjugacy class of a, 95, {gn U g [ G}, 96, group of one-to-one functions from, {1, 2, ? ? ? , n} to itself, 101, alternating group of degree n, 110, G and G are isomorphic, 128, mapping given by fa(x) 5 axa21 for all x, 135, group of automorphisms of the group G, 136, group of inner automorphisms of G, 136, {ah U h [ H}, 144, {aha21 | h [ H}, 144, the index of H in G, 148, {hk U h [ H, k [ K}, 150, {f [ G U f(i) 5 i}, the stabilizer of i under the permutation group G, 151, {f(i) U f [ G}, the orbit of i under the, permutation group G, 151, external direct product of groups G1, G2, . . . , Gn, 162, {x [ U(n) U x mod k 5 1}, 166, commutator subgroup, 181, H is a normal subgroup of G, 185, factor group, 187, internal direct product of H and K, 196, internal direct product of H1, . . . , Hn, 197, kernel of the homomorphism f, 208, inverse image of g9 under f, 210, inverse image of K under f, 211, ring of polynomials with integer coefficients, 246, ring of all 2 3 2 matrices with integer entries, 246, direct sum of rings, 247, ring of multiples of n, 249, ring of Gaussian integers, 249, group of units of the ring R, 251, characteristic of R, 258, principal ideal generated by a, 268, ideal generated by a1, a2, . . . , an, 268, factor ring, 268, sum of ideals A and B, 275, product of ideals A and B, 275, annihilator of A, 277, nil radical of A, 277, field of quotients of F[x], 291, ring of polynomials over R, 298, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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deg f (x), Fp(x), M2(Q), kv1, v2, . . . , vnl, F(a1, a2, . . . , an), f 9(x), [E:F], GF( pn), GF( pn)*, cl(a), Pr(G), np, W(S), ka1, a2, . . . , an U w1 5 w2 5 . . . 5 wtl, Q4, Q6, D`, fix(f), Cay(S:G), k * (a, b, . . . , c), (n, k), Fn, d(u, v), wt(u), Gal(E/F), EH, Fn(x), C⊥, , degree of the polynomial, 300, pth cyclotomic polynomial, 316, ring of 2 3 2 matrices over Q, 352, subspace spanned by v1, v2, . . . , vn, 353, extension of F by a1, a2, . . . , an, 363, the derivative of f (x), 368, degree of E over F, 378, Galois field of order pn, 389, nonzero elements of GF( pn), 390, {xax21 U x [ G}, the conjugacy class of a, 409, probability that two elements from G commute, 411, the number of Sylow p-subgroups of a group, 416, set of all words from S, 446, group with generators a1, a2, . . . , an and relations w1, 5 w2 5 . . . 5 wt , 449, quarternions, 453, dicyclic group of order 12, 453, infinite dihedral group, 454, {i [ S U f(i) 5 i}, elements fixed by f, 497, Cayley digraph of the group G with generating set S,, 506, concatenation of k copies of (a, b, . . . , c), 514, linear code, k-dimensional subspace of Fn, 531, F % F % ? ? ? % F, direct product of n copies of the, field F, 531, Hamming distance between vectors u and v, 532, the number of nonzero components of the vector u, (the Hamming weight of u), 532, the automorphism group of E fixing F, 554, fixed field of H, 554, nth cyclotomic polynomial, 571, dual code of a code C, 582, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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This is an electronic version of the print textbook. Due to electronic rights restrictions,, some third party content may be suppressed. Editorial review has deemed that any suppressed, content does not materially affect the overall learning experience. The publisher reserves the right, to remove content from this title at any time if subsequent rights restrictions require it. For, valuable information on pricing, previous editions, changes to current editions, and alternate, formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for, materials in your areas of interest., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7RP\ZRQGHUIXOJUDQGVRQ-RH\<DSHO, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Contents, , Preface xi, PART 1, , Integers and Equivalence Relations 1, 0 Preliminaries 3, Properties of Integers 3 | Modular Arithmetic 6 |, Complex Numbers 13 | Mathematical Induction 14 |, Equivalence Relations 17 | Functions (Mappings) 20, Exercises 23, , PART 2, , Groups 29, 1 Introduction to Groups 31, Symmetries of a Square 31 | The Dihedral Groups 34, Exercises 37, Biography of Niels Abel 41, , 2 Groups 42, Definition and Examples of Groups 42 | Elementary, Properties of Groups 50 | Historical Note 53, Exercises 54, , 3 Finite Groups; Subgroups 60, Terminology and Notation 60, Examples of Subgroups 65, Exercises 68, , |, , Subgroup Tests 61, , |, , 4 Cyclic Groups 77, Properties of Cyclic Groups 77, of Cyclic Groups 82, , |, , Classification of Subgroups, , iv, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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v, , &RQWHQWV, , Exercises 87, Biography of James Joseph Sylvester 93, Supplementary Exercises for Chapters 1–4 95, , 5 Permutation Groups 99, Definition and Notation 99 | Cycle Notation 102 | Properties of, Permutations 104 | A Check-Digit Scheme Based on D5 115, Exercises 118, Biography of Augustin Cauchy 126, , 6 Isomorphisms 127, Motivation 127 | Definition and Examples 127 |, Cayley’s Theorem 131 | Properties of Isomorphisms 133, Automorphisms 134, Exercises 138, Biography of Arthur Cayley 143, , |, , 7 Cosets and Lagrange’s Theorem 144, Properties of Cosets 144 | Lagrange’s Theorem and, Consequences 147 | An Application of Cosets to Permutation, Groups 151 | The Rotation Group of a Cube and a Soccer, Ball 153 | An Application of Cosets to the Rubik’s Cube 155, Exercises 156, Biography of Joseph Lagrange 161, , 8 External Direct Products 162, Definition and Examples 162 | Properties of External Direct, Products 163 | The Group of Units Modulo n as an External Direct, Product 166 | Applications 168, Exercises 174, Biography of Leonard Adleman 180, Supplementary Exercises for Chapters 5–8 181, , 9 Normal Subgroups and Factor Groups 185, Normal Subgroups 185 | Factor Groups 187 | Applications of, Factor Groups 193 | Internal Direct Products 195, Exercises 200, Biography of Évariste Galois 207, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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vi, , &RQWHQWV, , 10 Group Homomorphisms 208, Definition and Examples 208 | Properties of Homomorphisms, 210 | The First Isomorphism Theorem 214, Exercises 219, Biography of Camille Jordan 225, , 11 Fundamental Theorem of Finite, Abelian Groups 226, The Fundamental Theorem 226 | The Isomorphism Classes of, Abelian Groups 226 | Proof of the Fundamental Theorem 231, Exercises 234, Supplementary Exercises for Chapters 9–11 238, PART 3, , Rings 243, 12 Introduction to Rings 245, Motivation and Definition 245 | Examples of, Rings 246 | Properties of Rings 247 | Subrings 248, Exercises 250, Biography of I. N. Herstein 254, , 13 Integral Domains 255, Definition and Examples 255 | Fields 256, Ring 258, Exercises 261, Biography of Nathan Jacobson 266, , |, , Characteristic of a, , 14 Ideals and Factor Rings 267, Ideals 267 | Factor Rings 268 | Prime Ideals and Maximal, Ideals 272, Exercises 274, Biography of Richard Dedekind 279, Biography of Emmy Noether 280, Supplementary Exercises for Chapters 12–14 281, , 15 Ring Homomorphisms 285, Definition and Examples 285 | Properties of Ring, Homomorphisms 288 | The Field of Quotients 290, Exercises 292, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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vii, , &RQWHQWV, , 16 Polynomial Rings 298, Notation and Terminology 298 | The Division Algorithm and, Consequences 301, Exercises 305, Biography of Saunders Mac Lane 310, , 17 Factorization of Polynomials 311, Reducibility Tests 311 | Irreducibility Tests 314 | Unique, Factorization in Z[x] 319 | Weird Dice: An Application of Unique, Factorization 320, Exercises 322, Biography of Serge Lang 327, , 18 Divisibility in Integral Domains 328, Irreducibles, Primes 328 | Historical Discussion of Fermat’s Last, Theorem 331 | Unique Factorization Domains 334 | Euclidean, Domains 337, Exercises 341, Biography of Sophie Germain 345, Biography of Andrew Wiles 346, Supplementary Exercises for Chapters 15–18 347, PART 4, , Fields 349, 19 Vector Spaces 351, Definition and Examples 351 | Subspaces 352, Independence 353, Exercises 355, Biography of Emil Artin 358, Biography of Olga Taussky-Todd 359, , |, , Linear, , 20 Extension Fields 360, The Fundamental Theorem of Field Theory 360 | Splitting, Fields 362 | Zeros of an Irreducible Polynomial 368, Exercises 372, Biography of Leopold Kronecker 375, , 21 Algebraic Extensions 376, Characterization of Extensions 376 | Finite Extensions 378, Properties of Algebraic Extensions 382, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , |
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viii, , &RQWHQWV, , Exercises 384, Biography of Irving Kaplansky 387, , 22 Finite Fields 388, Classification of Finite Fields 388, Subfields of a Finite Field 393, Exercises 395, Biography of L. E. Dickson 398, , |, , Structure of Finite Fields 389, , 23 Geometric Constructions 399, Historical Discussion of Geometric Constructions 399, Constructible Numbers 400 | Angle-Trisectors and, Circle-Squarers 402, Exercises 402, Supplementary Exercises for Chapters 19–23 405, PART 5, , |, , Special Topics 407, 24 Sylow Theorems 409, Conjugacy Classes 409 | The Class Equation 410 |, The Probability That Two Elements Commute 411 | The Sylow, Theorems 412 | Applications of Sylow Theorems 417, Exercises 421, Biography of Ludwig Sylow 427, , 25 Finite Simple Groups 428, Historical Background 428 | Nonsimplicity Tests 433, The Simplicity of A5 437 | The Fields Medal 438 |, The Cole Prize 438, Exercises 439, Biography of Michael Aschbacher 442, Biography of Daniel Gorenstein 443, Biography of John Thompson 444, , |, , 26 Generators and Relations 445, Motivation 445 | Definitions and Notation 446 | Free, Group 447 | Generators and Relations 448 | Classification of, Groups of Order Up to 15 452 | Characterization of Dihedral, Groups 454 | Realizing the Dihedral Groups with Mirrors 455, Exercises 457, Biography of Marshall Hall, Jr. 460, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , |
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&RQWHQWV, , ix, , 27 Symmetry Groups 461, Isometries 461 | Classification of Finite Plane Symmetry, Groups 463 | Classification of Finite Groups of Rotations in R3 464, Exercises 466, , 28 Frieze Groups and Crystallographic Groups 469, The Frieze Groups 469 | The Crystallographic, Groups 475 | Identification of Plane Periodic Patterns 481, Exercises 487, Biography of M. C. Escher 492, Biography of George Pólya 493, Biography of John H. Conway 494, , 29 Symmetry and Counting 495, Motivation 495 | Burnside’s Theorem 496, Group Action 501, Exercises 502, Biography of William Burnside 505, , | Applications 498, , 30 Cayley Digraphs of Groups 506, Motivation 506 | The Cayley Digraph of a Group 506 |, Hamiltonian Circuits and Paths 510 | Some Applications 516, Exercises 519, Biography of William Rowan Hamilton 524, Biography of Paul Erdó´s 525, , 31 Introduction to Algebraic Coding Theory 526, Motivation 526 | Linear Codes 531 | Parity-Check Matrix, Decoding 536 | Coset Decoding 539 | Historical Note: The, Ubiquitous Reed–Solomon Codes 543, Exercises 545, Biography of Richard W. Hamming 550, Biography of Jessie MacWilliams 551, Biography of Vera Pless 552, , 32 An Introduction to Galois Theory 553, Fundamental Theorem of Galois Theory 553 | Solvability of, Polynomials by Radicals 560 | Insolvability of a Quintic 564, Exercises 565, Biography of Philip Hall 569, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , |
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x, , &RQWHQWV, , 33 Cyclotomic Extensions 570, Motivation 570 | Cyclotomic Polynomials 571, The Constructible Regular n-gons 575, Exercises 577, Biography of Carl Friedrich Gauss 579, Biography of Manjul Bhargava 580, Supplementary Exercises for Chapters 24–33 581, , |, , Selected Answers A1, Index of Mathematicians A45, Index of Terms A47, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Preface, 'HDU6LURU0DGDPZLOO\RXUHDGP\ERRNLWWRRNPH\HDUVWRZULWHZLOO\RX, WDNHDORRN", John Lennon and Paul McCartney, “Paperback Writer,” single, 1966*, , Although I wrote the first edition of this book more than 25 years ago,, my goals for it remain the same. I want students to receive a solid introduction to the traditional topics. I want readers to come away with the, view that abstract algebra is a contemporary subject—that its concepts, and methodologies are being used by working mathematicians, computer scientists, physicists, and chemists. I want students to see the, connections between abstract algebra and number theory and geometry. I want students to be able to do computations and to write proofs., I want students to enjoy reading the book. And I want to convey to the, reader my enthusiasm for this beautiful subject., Educational research has shown that an effective way of learning, mathematics is to interweave worked-out examples and practice problems. Thus, I have made examples and exercises the heart of the book., The examples elucidate the definitions, theorems, and proof techniques., The exercises facilitate understanding, provide insight, and develop the, ability of the students to do proofs. The exercises often foreshadow, definitions, concepts, and theorems to come. Many exercises focus on, special cases and ask the reader to generalize. Generalizing is a skill, that students should develop but rarely do. Even if an instructor chooses, not to spend class time on the applications in the book, I feel that having them there demonstrates to students the utility of the theory., Changes for the eighth edition include 200 new exercises, new examples, and a freshening of the quotations, historical notes, and biographies. These changes accentuate and enhance the hallmark features that, have made previous editions of the book a comprehensive, lively, and, engaging introduction to the subject:, r &YUFOTJWFDPWFSBHFPGHSPVQT SJOHT BOEGJFMET QMVTBWBSJFUZPG, nontraditional special topics, *Copyright © 1966 (Renewed) Stony/ATV Tunes LLC. All rights administered by, Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights, reserved. Used by permission., xi, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Page 19 : 3UHIDFH, , xiii, , r "OInstructor’s Solutions Manual with solutions to all the exercises, in the book and additional test questions and solutions, r "OPOMJOFMBCPSBUPSZNBOVBM XSJUUFOCZ+VMJBOOF3BJOCPMU XJUI, exercises designed to be done with the free computer algebra system, software GAP, r 0OMJOFJOTUSVDUPSBOTXFSLFZTUPUIFCPPLTDPNQVUFSFYFSDJTFTBOE, the exercises in the GAP lab manual, 4QFDJBMUIBOLTHPUPNZDPQZFEJUPSGPSUIJTFEJUJPO +FGG"OEFSTPO , and the accuracy checker, Roger Lipsett. I am grateful to each for their, careful attention to the manuscript. My appreciation also goes to Molly, Taylor, Shaylin Hogan, and Alex Gontar from Cengage Learning, as, well as Katie Costello and the Cengage production staff., The thoughtful input of the following people, who served as reviewers for the eighth edition, is also sincerely appreciated: Homer, Austin, Salisbury University; David Barth-Hart, Rochester Institute, PG5FDIOPMPHZ #SFU#FOFTI $PMMFHFPG4U#FOFEJDUBOE4U+PIOT, University; Daniel Daly, Southeast Missouri State University; Paul, Felt, University of Texas of the Permian Basin; Donald Hartig,, California Polytechnic State University, San Luis Obispo; Nancy, Ann Neudauer, Pacific University; Bingwu Wang, Eastern Michigan, University; Dana Williams, Dartmouth College; and Norbert Youmbi,, Saint Francis University., Over the years, many faculty and students have kindly sent me valuable comments and suggestions. They have helped to make each edition, better. I owe many thanks to my UMD colleague Robert McFarland for, giving me numerous exercises and comments that have been included, in this edition. Douglas Dunham, another UMD colleague, has generously provided the spectacular cover image for this edition. For an explanation of the mathematics underlying this image see www.d.umn, .edu/~jgallian/Dunhamimage. Please send any comments and suggestions you have to me at
[email protected], , Joseph A. Gallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PA R T, , 1, , Integers and, Equivalence Relations, , For online student resources, visit this textbook’s website at, www.CengageBrain.com, , 1, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 Preliminaries, , The whole of science is nothing more than a refinement, of everyday thinking., albert einstein, Physics and Reality, , Properties of Integers, Much of abstract algebra involves properties of integers and sets. In this, chapter we collect the properties we need for future reference., An important property of the integers, which we will often use, is the, so-called Well Ordering Principle. Since this property cannot be proved, from the usual properties of arithmetic, we will take it as an axiom., Well Ordering Principle, Every nonempty set of positive integers contains a smallest member., , The concept of divisibility plays a fundamental role in the theory of, numbers. We say a nonzero integer t is a divisor of an integer s if there, is an integer u such that s 5 tu. In this case, we write t | s (read “t, divides s”). When t is not a divisor of s, we write t B s. A prime is a, positive integer greater than 1 whose only positive divisors are 1 and, itself. We say an integer s is a multiple of an integer t if there is an integer u such that s 5 tu or, equivalently, if t is a divisor of s., As our first application of the Well Ordering Principle, we establish, a fundamental property of integers that we will use often., Theorem 0.1 Division Algorithm, Let a and b be integers with b . 0. Then there exist unique integers q, and r with the property that a 5 bq 1 r, where 0 # r , b., , PROOF We begin with the existence portion of the theorem. Consider, the set S 5 {a 2 bk | k is an integer and a 2 bk $ 0}. If 0 [ S, then b, 3, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4, , Integers and Equivalence Relations, , divides a and we may obtain the desired result with q 5 a/b and r 5 0., Now assume 0 n S. Since S is nonempty [if a . 0, a 2 b ? 0 [ S; if a ,, 0, a 2 b(2a) 5 a(1 2 2b) [ S; a 0 since 0 n S], we may apply the, Well Ordering Principle to conclude that S has a smallest member, say, r 5 a 2 bq. Then a 5 bq 1 r and r $ 0, so all that remains to be proved, is that r , b., If r $ b, then a 2 b(q 1 1) 5 a 2 bq 2 b 5 r 2 b $ 0, so that, a 2 b(q 1 1) [ S. But a 2 b(q 1 1) , a 2 bq, and a 2 bq is the, smallest member of S. So, r , b., To establish the uniqueness of q and r, let us suppose that there are, integers q, q9, r, and r9 such that, a 5 bq 1 r,, , 0 # r , b,, , and, , a 5 bq9 1 r9,, , 0 # r9 , b., , For convenience, we may also suppose that r9 $ r. Then bq 1 r 5, bq9 1 r9 and b(q 2 q9) 5 r9 2 r. So, b divides r9 2 r and 0 # r9 2 r #, r9 , b. It follows that r9 2 r 5 0, and therefore r9 5 r and q 5 q9., The integer q in the division algorithm is called the quotient upon dividing a by b; the integer r is called the remainder upon dividing a by b., EXAMPLE 1 For a 5 17 and b 5 5, the division algorithm gives, 17 5 5 ? 3 1 2; for a 5 223 and b 5 6, the division algorithm gives, 223 5 6(24) 1 1., Definitions Greatest Common Divisor, Relatively Prime Integers, , The greatest common divisor of two nonzero integers a and b is the, largest of all common divisors of a and b. We denote this integer by, gcd(a, b). When gcd(a, b) 5 1, we say a and b are relatively prime., , The following property of the greatest common divisor of two integers plays a critical role in abstract algebra. The proof provides an application of the division algorithm and our second application of the, Well Ordering Principle., Theorem 0.2 GCD Is a Linear Combination, For any nonzero integers a and b, there exist integers s and t such that, gcd(a, b) 5 as 1 bt. Moreover, gcd(a, b) is the smallest positive integer, of the form as 1 bt., , PROOF Consider the set S 5 {am 1 bn | m, n are integers and, am 1 bn . 0}. Since S is obviously nonempty (if some choice of m, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 5, , and n makes am 1 bn , 0, then replace m and n by 2m and 2n), the, Well Ordering Principle asserts that S has a smallest member, say,, d 5 as 1 bt. We claim that d 5 gcd(a, b). To verify this claim, use the, division algorithm to write a 5 dq 1 r, where 0 # r , d. If r . 0,, then r 5 a 2 dq 5 a 2 (as 1 bt)q 5 a 2 asq 2 btq 5 a(1 2 sq) 1, b(2tq) [ S, contradicting the fact that d is the smallest member of S., So, r 5 0 and d divides a. Analogously (or, better yet, by symmetry),, d divides b as well. This proves that d is a common divisor of a and b., Now suppose d9 is another common divisor of a and b and write a 5, d9h and b 5 d9k. Then d 5 as 1 bt 5 (d9h)s 1 (d9k)t 5 d9(hs 1 kt),, so that d9 is a divisor of d. Thus, among all common divisors of a and, b, d is the greatest., The special case of Theorem 0.2 when a and b are relatively prime is, so important in abstract algebra that we single it out as a corollary., Corollary, If a and b are relatively prime, then there exist integers s and t such, that as 1 bt 5 1., , EXAMPLE 2 gcd(4, 15) 5 1; gcd(4, 10) 5 2; gcd(22 ? 32 ? 5, 2 ? 33 ?, 72) 5 2 ? 32. Note that 4 and 15 are relatively prime, whereas 4 and 10 are, not. Also, 4 ? 4 1 15(21) 5 1 and 4(22) 1 10 ? 1 5 2., The next lemma is frequently used. It appeared in Euclid’s Elements., Euclid’s Lemma p | ab Implies p | a or p | b, If p is a prime that divides ab, then p divides a or p divides b., , PROOF Suppose p is a prime that divides ab but does not divide a. We, must show that p divides b. Since p does not divide a, there are, integers s and t such that 1 5 as 1 pt. Then b 5 abs 1 ptb, and since, p divides the right-hand side of this equation, p also divides b., Note that Euclid’s Lemma may fail when p is not a prime, since, 6 | (4 ? 3) but 6 B 4 and 6 B 3., Our next property shows that the primes are the building blocks for, all integers. We will often use this property without explicitly saying so., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6, , Integers and Equivalence Relations, , Theorem 0.3 Fundamental Theorem of Arithmetic, Every integer greater than 1 is a prime or a product of primes. This, product is unique, except for the order in which the factors appear., That is, if n 5 p1 p2 . . . pr and n 5 q1q2 . . . qs , where the p’s and q’s, are primes, then r 5 s and, after renumbering the q’s, we have pi 5 qi, for all i., , We will prove the existence portion of Theorem 0.3 later in this, chapter (Example 11). The uniqueness portion is a consequence of, Euclid’s Lemma (Exercise 31)., Another concept that frequently arises is that of the least common, multiple of two integers., Definition Least Common Multiple, , The least common multiple of two nonzero integers a and b is the, smallest positive integer that is a multiple of both a and b. We will, denote this integer by lcm(a, b)., , We leave it as an exercise (Exercise 10) to prove that every common, multiple of a and b is a multiple of lcm(a, b)., EXAMPLE 3 lcm(4, 6) 5 12; lcm(4, 8) 5 8; lcm(10, 12) 5 60;, lcm(6, 5) 5 30; lcm(22 ? 32 ? 5, 2 ? 33 ? 72) 5 22 ? 33 ? 5 ? 72., , Modular Arithmetic, Another application of the division algorithm that will be important to, us is modular arithmetic. Modular arithmetic is an abstraction of a, method of counting that you often use. For example, if it is now, September, what month will it be 25 months from now? Of course, the, answer is October, but the interesting fact is that you didn’t arrive at the, answer by starting with September and counting off 25 months., Instead, without even thinking about it, you simply observed that, 25 5 2 ? 12 1 1, and you added 1 month to September. Similarly, if it, is now Wednesday, you know that in 23 days it will be Friday. This, time, you arrived at your answer by noting that 23 5 7 ? 3 1 2, so you, added 2 days to Wednesday instead of counting off 23 days. If your, electricity is off for 26 hours, you must advance your clock 2 hours,, since 26 5 2 ? 12 1 2. Surprisingly, this simple idea has numerous, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 7, , important applications in mathematics and computer science. You will, see a few of them in this section. The following notation is convenient., When a 5 qn 1 r, where q is the quotient and r is the remainder, upon dividing a by n, we write a mod n 5 r. Thus,, 3 mod 2 5 1 since 3 5 1 ? 2 1 1,, 6 mod 2 5 0 since 6 5 3 ? 2 1 0,, 11 mod 3 5 2 since 11 5 3 ? 3 1 2,, 62 mod 85 5 62 since 62 5 0 ? 85 1 62,, 22 mod 15 5 13 since 22 5 (21)15 1 13., In general, if a and b are integers and n is a positive integer, then, a mod n 5 b mod n if and only if n divides a 2 b (Exercise 7)., In our applications, we will use addition and multiplication mod n., When you wish to compute ab mod n or (a 1 b) mod n, and a or b, is greater than n, it is easier to “mod first.” For example, to compute, (27 ? 36) mod 11, we note that 27 mod 11 5 5 and 36 mod 11 5 3, so, (27 ? 36) mod 11 5 (5 ? 3) mod 11 5 4. (See Exercise 9.), Modular arithmetic is often used in assigning an extra digit to identification numbers for the purpose of detecting forgery or errors. We, present two such applications., EXAMPLE 4 The United States Postal Service money order shown, in Figure 0.1 has an identification number consisting of 10 digits together, with an extra digit called a check. The check digit is the 10-digit number, modulo 9. Thus, the number 3953988164 has the check digit 2, since, , Figure 0.1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8, , Integers and Equivalence Relations, , 3953988164 mod 9 5 2.† If the number 39539881642 were incorrectly, entered into a computer (programmed to calculate the check digit) as,, say, 39559881642 (an error in the fourth position), the machine would, calculate the check digit as 4, whereas the entered check digit would be, 2. Thus, the error would be detected., EXAMPLE 5 Airline companies, the United Parcel Service, and, the rental-car companies Avis and National use the mod 7 values of, identification numbers to assign check digits. Thus, the identification, number 00121373147367 (see Figure 0.2) has the check digit 3 appended, , Figure 0.2, , Figure 0.3, †The value of N mod 9 is easy to compute with a calculator. If N 5 9q 1 r, where r is, the remainder upon dividing N by 9, then on a calculator screen N 4 9 appears as, q.rrrrr . . . , so the first decimal digit is the check digit. For example, 3953988164 4 9 5, 439332018.222, so 2 is the check digit. If N has too many digits for your calculator,, replace N by the sum of its digits and divide that number by 9. Thus, 3953988164, mod 9 5 56 mod 9 5 2. The value of 3953988164 mod 9 can also be computed by, searching Google for “3953988164 mod 9.”, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 9, , to it because 121373147367 mod 7 5 3. Similarly, the UPS pickup record number 768113999, shown in Figure 0.3, has the check digit 2, appended to it., The methods used by the Postal Service and the airline companies, do not detect all single-digit errors (see Exercises 41 and 45). However,, detection of all single-digit errors, as well as nearly all errors involving, the transposition of two adjacent digits, is easily achieved. One method, that does this is the one used to assign the so-called Universal Product, Code (UPC) to most retail items (see Figure 0.4). A UPC identification, number has 12 digits. The first six digits identify the manufacturer, the, next five identify the product, and the last is a check. (For many items,, the 12th digit is not printed, but it is always bar-coded.) In Figure 0.4,, the check digit is 8., , Figure 0.4, , To explain how the check digit is calculated, it is convenient to introduce the dot product notation for two k-tuples:, (a1, a2, . . . , ak) ? (w1, w2, . . . , wk) 5 a1w1 1 a2w2 1 ? ? ? 1 akwk., An item with the UPC identification number a1a2 ??? a12 satisfies the, condition, (a1, a2, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0., To verify that the number in Figure 0.4 satisfies this condition, we, calculate, (0 ? 3 1 2 ? 1 1 1 ? 3 1 0 ? 1 1 0 ? 3 1 0 ? 1 1 6 ? 3 1 5 ? 1, 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1) mod 10 5 90 mod 10 5 0., The fixed k-tuple used in the calculation of check digits is called the, weighting vector., Now suppose a single error is made in entering the number in, Figure 0.4 into a computer. Say, for instance, that 021000958978 is, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10, , Integers and Equivalence Relations, , entered (notice that the seventh digit is incorrect). Then the computer, calculates, 0?312?111?310?110?310?119?3, 1 5 ? 1 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1 5 99., Since 99 mod 10 ∞ 0, the entered number cannot be correct., In general, any single error will result in a sum that is not 0 modulo 10., The advantage of the UPC scheme is that it will detect nearly all, errors involving the transposition of two adjacent digits as well as all, errors involving one digit. For doubters, let us say that the identification number given in Figure 0.4 is entered as 021000658798. Notice, that the last two digits preceding the check digit have been transposed., But by calculating the dot product, we obtain 94 mod 10 ∞ 0, so we, have detected an error. In fact, the only undetected transposition, errors of adjacent digits a and b are those where |a 2 b| 5 5. To, verify this, we observe that a transposition error of the form, a1a2 ? ? ? aiai11 ? ? ? a12 → a1a2 ? ? ? ai11ai ? ? ? a12, is undetected if and only if, (a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0., That is, the error is undetected if and only if, (a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10, 5 (a1, a2, . . . , ai, ai11, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10., This equality simplifies to either, (3ai11 1 ai) mod 10 5 (3ai 1 ai11) mod 10, or, (ai11 1 3ai) mod 10 5 (ai 1 3ai11) mod 10,, depending on whether i is even or odd. Both cases reduce to 2(ai11 2 ai), mod 10 5 0. It follows that |ai11 2 ai| 5 5, if ai11 ∞ ai., In 2005, United States companies began to phase in the use of a 13th, digit to be in conformance with the 13-digit product identification numbers used in Europe. The weighting vector for 13-digit numbers is (1, 3,, 1, 3, . . . , 3, 1)., Identification numbers printed on bank checks (on the bottom left, between the two colons) consist of an eight-digit number a1a2 ? ? ? a8, and a check digit a9, so that, (a1, a2, . . . , a9) ? (7, 3, 9, 7, 3, 9, 7, 3, 9) mod 10 5 0., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 11, , As is the case for the UPC scheme, this method detects all singledigit errors and all errors involving the transposition of adjacent digits, a and b except when |a 2 b| 5 5. But it also detects most errors of the, form ? ? ? abc ? ? ? → ? ? ? cba ? ? ?, whereas the UPC method detects no, errors of this form., In Chapter 5, we will examine more sophisticated means of assigning check digits to numbers., What about error correction? Suppose you have a number such as, 73245018 and you would like to be sure that if even a single mistake, were made in entering this number into a computer, the computer, would nevertheless be able to determine the correct number. (Think of, it. You could make a mistake in dialing a telephone number but still get, the correct phone to ring!) This is possible using two check digits. One, of the check digits determines the magnitude of any single-digit error,, while the other check digit locates the position of the error. With these, two pieces of information, you can fix the error. To illustrate the idea, let, us say that we have the eight-digit identification number a1a2 ? ? ? a8. We, assign two check digits a9 and a10 so that, (a1 1 a2 1 ? ? ? 1 a9 1 a10) mod 11 5 0, and, (a1, a2, . . . , a9, a10) ? (1, 2, 3, . . . , 10) mod 11 5 0, are satisfied., Let’s do an example. Say our number before appending the two, check digits is 73245018. Then a9 and a10 are chosen to satisfy, (7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 1 a9 1 a10) mod 11 5 0, , (1), , and, (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1 0 ? 6, 1 1 ? 7 1 8 ? 8 1 a9 ? 9 1 a10 ? 10) mod 11 5 0., , (2), , Since 7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 5 30 and 30 mod 11 5 8,, Equation (1) reduces to, (8 1 a9 1 a10 ) mod 11 5 0., , (19), , Likewise, since (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1, 0 ? 6 1 1 ? 7 1 8 ? 8) mod 11 5 10, Equation (2) reduces to, (10 1 9a9 1 10a10) mod 11 5 0., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , (29)
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12, , Integers and Equivalence Relations, , Since we are using mod 11, we may rewrite Equation (29) as, (21 2 2a9 2 a10) mod 11 5 0, and add this to Equation (19) to obtain 7 2 a9 5 0. Thus a9 5 7. Now, substituting a9 5 7 into Equation (19) or Equation (29), we obtain, a10 5 7 as well. So, the number is encoded as 7324501877., Now let us suppose that this number is erroneously entered into, a computer programmed with our encoding scheme as 7824501877, (an error in position 2). Since the sum of the digits of the received, number mod 11 is 5, we know that some digit is 5 too large (assuming only one error has been made). But which one? Say the, error is in position i. Then the second dot product has the form a1 ? 1 1, a2 ? 2 1 ? ? ? 1 (ai 1 5)i 1 ai11 ? (i 1 1) 1 ? ? ? 1 a10 ? 10 5, (a1, a2, ? ? ?, a10) ? (1, 2, ? ? ? , 10) 1 5i. So, (7, 8, 2, 4, 5, 0, 1, 8, 7, 7) ?, (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) mod 11 5 5i mod 11. Since the left-hand, side mod 11 is 10, we see that i 5 2. Our conclusion: The digit in position 2 is 5 too large. We have successfully corrected the error., Modular arithmetic is often used to verify the validity of statements, about divisibility regarding all positive integers by checking only, finitely many cases., EXAMPLE 6 Consider the statement, “The sum of the cubes of any, three consecutive integers is divisible by 9.” This statement is equivalent to checking that the equation (n3 1 (n 1 1)3 1 (n 1 2)3) mod 9 5 0, is true for all integers n. Because of properties of modular arithmetic, to, prove this, all we need do is check the validity of the equation for n 5 0,, 1, …, 8., Modular arithmetic is occasionally used to show that certain equations have no rational number solutions., EXAMPLE 7 We use mod 3 arithmetic to show that there are no, integers a and b such that a2 2 6b 5 2. To see this, suppose that there, are such integers. Then, taking both sides modulo 3, there is an integer, solution to a2 mod 3 5 2. But trying a 5 0, 1, and 2 we obtain a contradiction., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 13, , Im, a + bi, , b, , r, , θ, , 0, , a, , Re, , Figure 0.5, , Complex Numbers, Recall that complex numbers are expressions of the form a 1 b 2 1,, where a and b are real numbers. The number 2 1 is defined to have, the property 2 12 5 21. It is customary to use i to denote 2 1., Then, i2 5 21. Complex numbers written in the form a 1 bi are said to, be in standard form. In some instances it is convenient to write a complex number a 1 bi in another form. To do this we represent a 1 bi as, the point (a,b) in a plane coordinatized by a horizontal axis called the, real axis and a vertical i axis called the imaginary axis. The distance, from the point a 1 bi to the origin is r 5 2a2 b2 and is often, denoted by |a 1 bi|. If we draw the line segment from the origin to, a 1 bi and denote the angle formed by the line segment and the positive, real axis by u, we can write a 1 bi as r(cos u 1 i sin u) (see Figure 0.5)., This form of a 1 bi is called the polar form. An advantage of the polar, form is demonstrated in parts 5 and 6 of Theorem 0.4., Theorem 0.4 Properties of Complex Numbers, 1. Closure under addition: (a 1 bi) 1 (c 1 di) 5 (a 1 c) 1 (b 1 d)i, 2. Closure under multiplication: (a 1 bi) (c 1 di) 5 (ac) 1 (ad)i 1, (bc)i 1 (bd)i2 5 (ac 2 bd) 1 (ad 1 bc)i, (a bi) (a bi) (c di), 5, 5, 3. Closure under division (c 1 di 0) :, (c di) (c di) (c di), (ac bd) (bc ad)i (ac bd) (bc ad), 5 2, 1 2, i, c2 d2, c d2, c d2, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14, , Integers and Equivalence Relations, , 4. Complex conjugation: (a 1 bi) (a 2 bi) 5 a2 1 b2, 5. Inverses: For every nonzero complex number a 1 bi there is a, complex number c 1 di such that (a 1 bi) (c 1 di) 5 1. (That is,, (a 1 bi)21 exists in C.), 6. Powers: For every complex number a 1 bi 5 r(cos u 1 i sin u) and, every positive integer n, we have (a 1 bi)n 5 3r1cos u i sin u24 n 5, r n (cos n u 1 i sin n u)., 7. Radicals: For every complex number a 1 bi 5 r(cos u 1 i sin u) and, 1, 1, every positive integer n, we have (a bi)n 5 3r1cos u i sin u24 n 5, 1, u, u, r n (cos n i sin n )., PROOF Parts 1 and 2 are definitions. Part 4 follows from part 2. Part 6, is proved in Example 10 in the next section of this chapter. Part 7 follows from Exercise 25 in this chapter., The next example illustrates properties of complex numbers., EXAMPLE 8 (3 1 5i) 1 (25 1 2i) 5 22 1 7i;, (3 1 5i)(25 1 2i) 5 225 1 (219)i 5 225 2 19i;, 3 5i, 2 7i, 29 31i, 29, 31, 3 5i, , 5, , , i;, 2 7i, 2 7i 2 7i, 53, 53, 53, (3 1 5i) (3 2 5i) 5 9 1 25 5 34;, 3, 5, , i., (3 1 5i)21 , 34, 34, 1, To find (3 1 5i)4 and (3 5i)4 we first note that if u 5 arctan 35 ,, 3, 5, then cos u 5, and sin u 5, . Thus, (3 1 5i)4 5 ((234(cos u 1, 234, 234, 1, i sin u))4 5 2344 (cos 4u 1 i sin 4u) and (3 5i)4 5 (234(cos u 1, 1, 1, u, u, i sin u))4 5 234 4 (cos 1 i sin )., 4, 4, , Mathematical Induction, There are two forms of proof by mathematical induction that we will, use. Both are equivalent to the Well Ordering Principle. The explicit, formulation of the method of mathematical induction came in the, 16th century. Francisco Maurolico (1494–1575), a teacher of Galileo,, used it in 1575 to prove that 1 1 3 1 5 1 ? ? ? 1 (2n 2 1) 5 n2, and, Blaise Pascal (1623–1662) used it when he presented what we now, call Pascal’s triangle for the coefficients of the binomial expansion., The term mathematical induction was coined by Augustus De Morgan., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 15, , Theorem 0.5 First Principle of Mathematical Induction, Let S be a set of integers containing a. Suppose S has the property that, whenever some integer n $ a belongs to S, then the integer n 1 1 also, belongs to S. Then, S contains every integer greater than or equal to a., , PROOF The proof is left as an exercise (Exercise 33)., So, to use induction to prove that a statement involving positive integers is true for every positive integer, we must first verify that the statement is true for the integer 1. We then assume the statement is true for, the integer n and use this assumption to prove that the statement is true, for the integer n 1 1., Our next example uses some facts about plane geometry. Recall that, given a straightedge and compass, we can construct a right angle., EXAMPLE 9 We use induction to prove that given a straightedge, a, compass, and a unit length, we can construct a line segment of length, 2n for every positive integer n. The case when n 5 1 is given. Now we, assume that we can construct a line segment of length 2n. Then use the, straightedge and compass to construct a right triangle with height 1 and, base 2n. The hypotenuse of the triangle has length 2n 1. So, by, induction, we can construct a line segment of length 2n for every positive integer n., EXAMPLE 10 DeMOIVRE’S THEOREM We use induction to prove, that for every positive integer n and every real number u, (cos u 1, i sin u)n 5 cos nu 1 i sin nu, where i is the complex number 2 1., Obviously, the statement is true for n 5 1. Now assume it is true for n., We must prove that (cos u 1 i sin u)n11 5 cos(n 1 1)u 1 i sin(n 1 1)u., Observe that, (cos u 1 i sin u)n11 5 (cos u 1 i sin u)n(cos u 1 i sin u), 5 (cos nu 1 i sin nu)(cos u 1 i sin u), 5 cos nu cos u 1 i(sin nu cos u, 1 sin u cos nu) 2 sin nu sin u., Now, using trigonometric identities for cos(a 1 b) and sin(a 1 b), we, see that this last term is cos(n 1 1)u 1 i sin(n 1 1)u. So, by induction,, the statement is true for all positive integers., In many instances, the assumption that a statement is true for an integer n does not readily lend itself to a proof that the statement is true, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16, , Integers and Equivalence Relations, , for the integer n 1 1. In such cases, the following equivalent form of, induction may be more convenient. Some authors call this formulation, the strong form of induction., Theorem 0.6 Second Principle of Mathematical Induction, Let S be a set of integers containing a. Suppose S has the property that, n belongs to S whenever every integer less than n and greater than or, equal to a belongs to S. Then, S contains every integer greater than or, equal to a., , PROOF The proof is left to the reader., To use this form of induction, we first show that the statement is true, for the integer a. We then assume that the statement is true for all integers that are greater than or equal to a and less than n, and use this assumption to prove that the statement is true for n., EXAMPLE 11 We will use the Second Principle of Mathematical, Induction with a 5 2 to prove the existence portion of the Fundamental, Theorem of Arithmetic. Let S be the set of integers greater than 1 that, are primes or products of primes. Clearly, 2 [ S. Now we assume that, for some integer n, S contains all integers k with 2 # k , n. We must, show that n [ S. If n is a prime, then n [ S by definition. If n is not a, prime, then n can be written in the form ab, where 1 , a , n and 1 , b, , n. Since we are assuming that both a and b belong to S, we know that, each of them is a prime or a product of primes. Thus, n is also a product, of primes. This completes the proof., Notice that it is more natural to prove the Fundamental Theorem of, Arithmetic with the Second Principle of Mathematical Induction than, with the First Principle. Knowing that a particular integer factors as a, product of primes does not tell you anything about factoring the next, larger integer. (Does knowing that 5280 is a product of primes help you, to factor 5281 as a product of primes?), The following problem appeared in the “Brain Boggler” section of, the January 1988 issue of the science magazine Discover.*, EXAMPLE 12 The Quakertown Poker Club plays with blue chips, worth $5.00 and red chips worth $8.00. What is the largest bet that, cannot be made?, *“Brain Boggler” by Maxwell Carver. Copyright © 1988 by Discover Magazine. Used, by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 17, , To gain insight into this problem, we try various combinations of, blue and red chips and obtain 5, 8, 10, 13, 15, 16, 18, 20, 21, 23, 24, 25,, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40. It appears that the, answer is 27. But how can we be sure? Well, we need only prove that, every integer greater than 27 can be written in the form a ? 5 1, b ? 8, where a and b are nonnegative integers. This will solve the problem, since a represents the number of blue chips and b the number of red, chips needed to make a bet of a ? 5 1 b ? 8. For the purpose of contrast,, we will give two proofs—one using the First Principle of Mathematical, Induction and one using the Second Principle., Let S be the set of all integers greater than or equal to 28 of the form, a ? 5 1 b ? 8, where a and b are nonnegative. Obviously, 28 [ S. Now, assume that some integer n [ S, say, n 5 a ? 5 1 b ? 8. We must show, that n 1 1 [ S. First, note that since n $ 28, we cannot have both a and, b less than 3. If a $ 3, then, n 1 1 5 (a ? 5 1 b ? 8) 1 (23 ? 5 1 2 ? 8), 5 (a 2 3) ? 5 1 (b 1 2) ? 8., (Regarding chips, this last equation says that we may increase a bet, from n to n 1 1 by removing three blue chips from the pot and adding, two red chips.) If b $ 3, then, n 1 1 5 (a ? 5 1 b ? 8) 1 (5 ? 5 2 3 ? 8), 5 (a 1 5) ? 5 1 (b 2 3) ? 8., (The bet can be increased by 1 by removing three red chips and adding, five blue chips.) This completes the proof., To prove the same statement by the Second Principle, we note that, each of the integers 28, 29, 30, 31, and 32 is in S. Now assume that, for some integer n . 32, S contains all integers k with 28 # k , n., We must show that n [ S. Since n 2 5 [ S, there are nonnegative, integers a and b such that n 2 5 5 a ? 5 1 b ? 8. But then, n 5 (a 1 1) ? 5 1 b ? 8. Thus n is in S., , Equivalence Relations, In mathematics, things that are considered different in one context may, be viewed as equivalent in another context. We have already seen one, such example. Indeed, the sums 2 1 1 and 4 1 4 are certainly different, in ordinary arithmetic, but are the same under modulo 5 arithmetic., Congruent triangles that are situated differently in the plane are not the, same, but they are often considered to be the same in plane geometry., In physics, vectors of the same magnitude and direction can produce, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18, , Integers and Equivalence Relations, , different effects—a 10-pound weight placed 2 feet from a fulcrum produces a different effect than a 10-pound weight placed 1 foot from a, fulcrum. But in linear algebra, vectors of the same magnitude and direction are considered to be the same. What is needed to make these, distinctions precise is an appropriate generalization of the notion of, equality; that is, we need a formal mechanism for specifying whether or, not two quantities are the same in a given setting. This mechanism is an, equivalence relation., Definition Equivalence Relation, , An equivalence relation on a set S is a set R of ordered pairs of, elements of S such that, 1. (a, a) [ R for all a [ S (reflexive property)., 2. (a, b) [ R implies (b, a) [ R (symmetric property)., 3. (a, b) [ R and (b, c) [ R imply (a, c) [ R (transitive property)., , When R is an equivalence relation on a set S, it is customary to write, aRb instead of (a, b) [ R. Also, since an equivalence relation is just a, generalization of equality, a suggestive symbol such as <, ;, or , is, usually used to denote the relation. Using this notation, the three conditions for an equivalence relation become a , a; a , b implies, b , a; and a , b and b , c imply a , c. If , is an equivalence relation, on a set S and a [ S, then the set [a] 5 {x [ S | x , a} is called the, equivalence class of S containing a., EXAMPLE 13 Let S be the set of all triangles in a plane. If a, b [ S,, define a , b if a and b are similar—that is, if a and b have corresponding angles that are the same. Then , is an equivalence relation on S., EXAMPLE 14 Let S be the set of all polynomials with real coefficients. If f, g [ S, define f , g if f 9 5 g9, where f 9 is the derivative of, f. Then , is an equivalence relation on S. Since two polynomials with, equal derivatives differ by a constant, we see that for any f in S, [ f ] 5, { f 1 c | c is real}., EXAMPLE 15 Let S be the set of integers and let n be a positive integer. If a, b [ S, define a ; b if a mod n 5 b mod n (that is, if a 2 b is, divisible by n). Then ; is an equivalence relation on S and [a] 5 {a 1, kn | k [ S}. Since this particular relation is important in abstract algebra, we will take the trouble to verify that it is indeed an equivalence, relation. Certainly, a 2 a is divisible by n, so that a ; a for all a in S., Next, assume that a ; b, say, a 2 b 5 rn. Then, b 2 a 5 (2r)n, and, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 19, , therefore b ; a. Finally, assume that a ; b and b ; c, say, a 2 b 5 rn, and b 2 c 5 sn. Then, we have a 2 c 5 (a 2 b) 1 (b 2 c) 5 rn 1 sn 5, (r 1 s)n, so that a ; c., EXAMPLE 16 Let ; be as in Example 15 and let n 5 7. Then we, have 16 ; 2; 9 ; 25; and 24 ; 3. Also, [1] 5 {. . . , 220, 213, 26, 1,, 8, 15, . . .} and [4] 5 {. . . , 217, 210, 23, 4, 11, 18, . . .}., EXAMPLE 17 Let S 5 {(a, b) | a, b are integers, b 2 0}. If, (a, b), (c, d ) [ S, define (a, b) < (c, d ) if ad 5 bc. Then < is an equivalence relation on S. [The motivation for this example comes from fractions. In fact, the pairs (a, b) and (c, d) are equivalent if the fractions a/b, and c/d are equal.], To verify that < is an equivalence relation on S, note that (a, b) < (a, b), requires that ab 5 ba, which is true. Next, we assume that (a, b) < (c, d),, so that ad 5 bc. We have (c, d) < (a, b) provided that cb 5 da, which is, true from commutativity of multiplication. Finally, we assume that (a, b), < (c, d ) and (c, d) < (e, f ) and prove that (a, b) < (e, f ). This amounts to, using ad 5 bc and cf 5 de to show that af 5 be. Multiplying both sides, of ad 5 bc by f and replacing cf by de, we obtain adf 5 bcf 5 bde. Since, d 2 0, we can cancel d from the first and last terms., Definition Partition, , A partition of a set S is a collection of nonempty disjoint subsets of S, whose union is S. Figure 0.6 illustrates a partition of a set into four, subsets., S, , Figure 0.6 Partition of S into four subsets., , EXAMPLE 18 The sets {0}, {1, 2, 3, . . .}, and {. . . , 23, 22, 21}, constitute a partition of the set of integers., EXAMPLE 19 The set of nonnegative integers and the set of nonpositive integers do not partition the integers, since both contain 0., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20, , Integers and Equivalence Relations, , The next theorem reveals that equivalence relations and partitions, are intimately intertwined., Theorem 0.7 Equivalence Classes Partition, The equivalence classes of an equivalence relation on a set S, constitute a partition of S. Conversely, for any partition P of S, there, is an equivalence relation on S whose equivalence classes are the, elements of P., , PROOF Let , be an equivalence relation on a set S. For any a [ S, the, reflexive property shows that a [ [a]. So, [a] is nonempty and the union, of all equivalence classes is S. Now, suppose that [a] and [b] are distinct, equivalence classes. We must show that [a] > [b] 5 0/. On the contrary,, assume c [ [a] > [b]. We will show that [a] # [b]. To this end, let x [ [a]., We then have c , a, c , b, and x , a. By the symmetric property, we, also have a , c. Thus, by transitivity, x , c, and by transitivity again,, x , b. This proves [a] # [b]. Analogously, [b] # [a]. Thus, [a] 5 [b],, in contradiction to our assumption that [a] and [b] are distinct equivalence classes., To prove the converse, let P be a collection of nonempty disjoint, subsets of S whose union is S. Define a , b if a and b belong to the, same subset in the collection. We leave it to the reader to show that , is, an equivalence relation on S (Exercise 61)., , Functions (Mappings), Although the concept of a function plays a central role in nearly every, branch of mathematics, the terminology and notation associated with, functions vary quite a bit. In this section, we establish ours., Definition Function (Mapping), , A function (or mapping) f from a set A to a set B is a rule that assigns, to each element a of A exactly one element b of B. The set A is called, the domain of f, and B is called the range of f. If f assigns b to a, then, b is called the image of a under f. The subset of B comprising all the, images of elements of A is called the image of A under f., , We use the shorthand f: A → B to mean that f is a mapping from, A to B. We will write f(a) 5 b or f: a → b to indicate that f carries, a to b., There are often different ways to denote the same element of a set. In, defining a function in such cases one must verify that the function, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 21, , values assigned to the elements depend not on the way the elements, are expressed but on only the elements themselves. For example, the, correspondence f from the rational numbers to the integers given by, f(a/b) 5 a 1 b does not define a function since 1/2 5 2/4 but f(1/2) ?, f(2/4). To verify that a correspondence is a function, you assume that, x1 5 x2 and prove that f(x1) 5 f (x2)., Definition Composition of Functions, , Let f: A → B and c: B → C. The composition cf is the mapping from, A to C defined by (cf)(a) 5 c(f(a)) for all a in A. The composition, function cf can be visualized as in Figure 0.7., , φ, , ψ, φ (a), , a, , ψφ, , ψ (φ (a)), , Figure 0.7 Composition of functions f and c., , In calculus courses, the composition of f with g is written ( f 8 g)(x) and, is defined by ( f 8 g)(x) 5 f (g(x)). When we compose functions, we omit, the “circle.”, EXAMPLE 20 Let f (x) 5 2x 1 3 and g(x) 5 x2 1 1. Then ( fg)(5) 5, f (g(5)) 5 f (26) 5 55; (g f )(5) 5 g ( f (5)) 5 g (13) 5 170. More generally,, ( fg)(x) 5 f (g(x)) 5 f (x2 1 1) 5 2(x2 1 1) 1 3 5 2x2 1 5 and (g f )(x) 5, g ( f (x)) 5 g (2x 1 3) 5 (2x 1 3)2 1 1 5 4x2 1 12x 1 9 1 1 5 4x2 1, 12x 1 10. Note that the function fg is not the same as the function g f., There are several kinds of functions that occur often enough to be, given names., Definition One-to-One Function, , A function f from a set A is called one-to-one if for every a1, a2 [ A,, f(a1) 5 f(a2) implies a1 5 a2., , The term one-to-one is suggestive, since the definition ensures that, one element of B can be the image of only one element of A. Alternatively,, f is one-to-one if a1 a2 implies f(a1) f(a2). That is, different elements of A map to different elements of B. See Figure 0.8., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22, , Integers and Equivalence Relations, , φ, , ψ, , a1, , φ (a1), , a1, , a2, , φ (a2), , a2, , ψ (a1) 5 ψ (a2), , ψ is not one-to-one, , φ is one-to-one, , Figure 0.8, Definition Function from A onto B, , A function f from a set A to a set B is said to be onto B if each element, of B is the image of at least one element of A. In symbols, f: A → B is, onto if for each b in B there is at least one a in A such that f(a) 5 b., See Figure 0.9., φ, , ψ, , φ is onto, , ψ is not onto, , Figure 0.9, , The next theorem summarizes the facts about functions we will need., Theorem 0.8 Properties of Functions, Given functions a: A → B, b: B → C, and g: C → D, then, 1. g(ba) 5 (gb)a (associativity)., 2. If a and b are one-to-one, then ba is one-to-one., 3. If a and b are onto, then ba is onto., 4. If a is one-to-one and onto, then there is a function a21 from B, onto A such that (a21a)(a) 5 a for all a in A and (aa21)(b) 5 b, for all b in B., , PROOF We prove only part 1. The remaining parts are left as exercises, (Exercise 57). Let a [ A. Then (g(ba))(a) 5 g((ba)(a)) 5 g(b(a(a)))., On the other hand, ((gb)a)(a) 5 (gb)(a(a)) 5 g(b(a(a))). So, g(ba) 5, (gb)a., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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23, , 0 | Preliminaries, , It is useful to note that if a is one-to-one and onto, the function a21, described in part 4 of Theorem 0.8 has the property that if a (s) 5 t,, then a21(t) 5 s. That is, the image of t under a21 is the unique element s, that maps to t under a. In effect, a21 “undoes” what a does., EXAMPLE 21 Let Z denote the set of integers, R the set of real, numbers, and N the set of nonnegative integers. The following table illustrates the properties of one-to-one and onto., Domain, Z, R, Z, Z, , Range, Z, R, N, Z, , Rule, x → x3, x → x3, x → |x|, x → x2, , One-to-One, Yes, Yes, No, No, , Onto, No, Yes, Yes, No, , To verify that x → x3 is one-to-one in the first two cases, notice that if, x3 5 y3, we may take the cube roots of both sides of the equation to obtain x 5 y. Clearly, the mapping from Z to Z given by x → x3 is not onto,, since 2 is the cube of no integer. However, x → x3 defines an onto function from R to R, since every real number is the cube of its cube root, 3, b → b). The remaining verifications are left to the reader., (that is, 2, , Exercises, I was interviewed in the Israeli Radio for five minutes and I said that more, than 2000 years ago, Euclid proved that there are infinitely many primes., Immediately the host interrupted me and asked: “Are there still infinitely, many primes?”, noga alon, , 1. For n 5 5, 8, 12, 20, and 25, find all positive integers less than n, and relatively prime to n., 2. Determine gcd(24 ? 32 ? 5 ? 72, 2 ? 33 ? 7 ? 11) and lcm(23 ? 32 ? 5,, 2 ? 33 ? 7 ? 11)., 3. Determine 51 mod 13, 342 mod 85, 62 mod 15, 10 mod 15, (82 ? 73), mod 7, (51 1 68) mod 7, (35 ? 24) mod 11, and (47 1 68) mod 11., 4. Find integers s and t such that 1 5 7 ? s 1 11 ? t. Show that s and t, are not unique., 5. Show that if a and b are positive integers, then ab 5 lcm(a, b) ?, gcd(a, b)., 6. Suppose a and b are integers that divide the integer c. If a and b are, relatively prime, show that ab divides c. Show, by example, that if, a and b are not relatively prime, then ab need not divide c., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24, , Integers and Equivalence Relations, , 7. If a and b are integers and n is a positive integer, prove that a mod n 5, b mod n if and only if n divides a 2 b., 8. Let d 5 gcd(a, b). If a 5 da9 and b 5 db9, show that gcd(a9, b9) 5 1., 9. Let n be a fixed positive integer greater than 1. If a mod n 5 a9 and, b mod n 5 b9, prove that (a 1 b) mod n 5 (a9 1 b9) mod n and (ab), mod n 5 (a9b9) mod n. (This exercise is referred to in Chapters 6,, 8, 10, and 15.), 10. Let a and b be positive integers and let d 5 gcd(a, b) and m 5, lcm(a, b). If t divides both a and b, prove that t divides d. If s is a, multiple of both a and b, prove that s is a multiple of m., 11. Let n and a be positive integers and let d 5 gcd(a, n). Show that the, equation ax mod n 5 1 has a solution if and only if d 5 1. (This, exercise is referred to in Chapter 2.), 12. Show that 5n 1 3 and 7n 1 4 are relatively prime for all n., 13. Suppose that m and n are relatively prime and r is any integer. Show, that there are integers x and y such that mx 1 ny 5 r., 14. Let p, q, and r be primes other than 3. Show that 3 divides p2 1, q2 1 r 2., 15. Prove that every prime greater than 3 can be written in the form, 6n 1 1 or 6n 1 5., 16. Determine 71000 mod 6 and 61001 mod 7., 17. Let a, b, s, and t be integers. If a mod st 5 b mod st, show that a, mod s 5 b mod s and a mod t 5 b mod t. What condition on s and, t is needed to make the converse true? (This exercise is referred to, in Chapter 8.), 18. Determine 8402 mod 5., 19. Show that gcd(a, bc) 5 1 if and only if gcd(a, b) 5 1 and, gcd(a, c) 5 1. (This exercise is referred to in Chapter 8.), 20. Let p1, p2, . . . , pn be primes. Show that p1 p2 ? ? ? pn 1 1 is divisible, by none of these primes., 21. Prove that there are infinitely many primes. (Hint: Use Exercise 20.), 22. Express (27 2 3i)21 in standard form., 5 2i, 23. Express, in standard form., 4 5i, 24. Express (cos 3608 1 i sin 3608)1/8 in standard form without trig, expressions. (Note that cos 3608 1 i sin 3608 5 1.), u, 25. Prove that for any positive integer n, (cos u 1 i sin u)1/n 5 cos n 1, u, i sin n ., 26. For every positive integer n, prove that 1 1 2 1 ? ? ? 1 n 5, n(n 1 1)/2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 25, , 27. For every positive integer n, prove that a set with exactly n elements, has exactly 2n subsets (counting the empty set and the entire set)., 28. Prove that 2n 32n 2 1 is always divisible by 17., 29. Prove that there is some positive integer n such that n, n 1 1,, n 1 2, ? ? ? , n 1 200 are all composite., 30. (Generalized Euclid’s Lemma) If p is a prime and p divides, a1a2 ? ? ? an, prove that p divides ai for some i., 31. Use the Generalized Euclid’s Lemma (see Exercise 30) to establish, the uniqueness portion of the Fundamental Theorem of Arithmetic., 32. What is the largest bet that cannot be made with chips worth $7.00, and $9.00? Verify that your answer is correct with both forms of, induction., 33. Prove that the First Principle of Mathematical Induction is a consequence of the Well Ordering Principle., 34. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . . In general, the Fibonacci numbers are defined by f1 5 1, f2 5 1, and for, n $ 3, fn 5 fn21 1 fn22. Prove that the nth Fibonacci number fn satisfies fn , 2n., 35. Prove by induction on n that for all positive integers n, n3 1, (n 1 1)3 1 (n 1 2)3 is a multiple of 9., 36. Suppose that there is a statement involving a positive integer, parameter n and you have an argument that shows that whenever, the statement is true for a particular n it is also true for n 1 2. What, remains to be done to prove the statement is true for every positive, integer? Describe a situation in which this strategy would be, applicable., 37. In the cut “As” from Songs in the Key of Life, Stevie Wonder mentions the equation 8 3 8 3 8 5 4. Find all integers n for which this, statement is true, modulo n., 38. Prove that for every integer n, n3 mod 6 5 n mod 6., 39. If it is 2:00 a.m. now, what time will it be 3736 hours from now?, 40. Determine the check digit for a money order with identification, number 7234541780., 41. Suppose that in one of the noncheck positions of a money order, number, the digit 0 is substituted for the digit 9 or vice versa. Prove, that this error will not be detected by the check digit. Prove that all, other errors involving a single position are detected., 42. Suppose that a money order identification number and check digit, of 21720421168 is erroneously copied as 27750421168. Will the, check digit detect the error?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26, , Integers and Equivalence Relations, , 43. A transposition error involving distinct adjacent digits is one of the, form . . . ab . . . → . . . ba . . . with a ≠ b. Prove that the money, order check-digit scheme will not detect such errors unless the, check digit itself is transposed., 44. Determine the check digit for the Avis rental car with identification, number 540047. (See Example 5.), 45. Show that a substitution of a digit ai9 for the digit ai (ai9 ≠ ai) in, a noncheck position of a UPS number is detected if and only, if |a i 2 a i9| ≠ 7., 46. Determine which transposition errors involving adjacent digits are, detected by the UPS check digit., 47. Use the UPC scheme to determine the check digit for the number, 07312400508., 48. Explain why the check digit for a money order for the number N is, the repeated decimal digit in the real number N 4 9., 49. The 10-digit International Standard Book Number (ISBN-10), a1a2a3a4a5a6a7a8 a9a10 has the property (a1, a2, . . . , a10) ? (10, 9, 8, 7,, 6, 5, 4, 3, 2, 1) mod 11 5 0. The digit a10 is the check digit. When, a10 is required to be 10 to make the dot product 0, the character X is, used as the check digit. Verify the check digit for the ISBN-10 assigned to this book., 50. Suppose that an ISBN-10 has a smudged entry where the question, mark appears in the number 0-716?-2841-9. Determine the missing, digit., 51. Suppose three consecutive digits abc of an ISBN-10 are scrambled as, bca. Which such errors will go undetected?, 52. The ISBN-10 0-669-03925-4 is the result of a transposition of two, adjacent digits not involving the first or last digit. Determine the, correct ISBN-10., 53. Suppose the weighting vector for ISBN-10s were changed to (1, 2,, 3, 4, 5, 6, 7, 8, 9, 10). Explain how this would affect the check digit., 54. Use the two-check-digit error-correction method described in this, chapter to append two check digits to the number 73445860., 55. Suppose that an eight-digit number has two check digits appended, using the error-correction method described in this chapter and it is, incorrectly transcribed as 4302511568. If exactly one digit is incorrect, determine the correct number., 56. The state of Utah appends a ninth digit a9 to an eight-digit driver’s, license number a1a2 . . . a8 so that (9a1 1 8a2 1 7a3 1 6a4 1 5a5 1, 4a6 1 3a7 1 2a8 1 a9) mod 10 5 0. If you know that the license, number 149105267 has exactly one digit incorrect, explain why the, error cannot be in position 2, 4, 6, or 8., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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0 | Preliminaries, , 27, , 57. Complete the proof of Theorem 0.8., 58. Let S be the set of real numbers. If a, b [ S, define a , b if a 2 b, is an integer. Show that , is an equivalence relation on S. Describe, the equivalence classes of S., 59. Let S be the set of integers. If a, b [ S, define aRb if ab $ 0. Is R an, equivalence relation on S?, 60. Let S be the set of integers. If a, b [ S, define aRb if a 1 b is even., Prove that R is an equivalence relation and determine the equivalence classes of S., 61. Complete the proof of Theorem 0.7 by showing that , is an equivalence relation on S., 62. Prove that 3, 5, and 7 are the only three consecutive odd integers, that are prime., 63. What is the last digit of 3100? What is the last digit of 2100?, 64. Prove that none of the integers 11, 111, 1111, 11111, . . . is a square, of an integer., 65. (Cancellation Property) Suppose a, b, and g are functions. If ag 5, bg and g is one-to-one and onto, prove that a 5 b., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Suggested Readings, Linda Deneen, “Secret Encryption with Public Keys,” The UMAP Journal, 8 (1987): 9–29., This well-written article describes several ways in which modular, arithmetic can be used to code secret messages. They range from a, simple scheme used by Julius Caesar to a highly sophisticated scheme, invented in 1978 and based on modular n arithmetic, where n has more, than 200 digits., J. A. Gallian, “Assigning Driver’s License Numbers,” Mathematics, Magazine 64 (1991): 13–22., This article describes various methods used by the states to assign driver’s license numbers. Several include check digits for error detection., This article can be downloaded at http://www.d.umn.edu/~jgallian/, license.pdf, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28, , Integers and Equivalence Relations, , J. A. Gallian, “The Mathematics of Identification Numbers,” The College, Mathematics Journal 22 (1991): 194–202., This article is a comprehensive survey of check-digit schemes that are, associated with identification numbers. This article can be downloaded, at http://www.d.umn.edu/~jgallian/ident.pdf, J. A. Gallian and S. Winters, “Modular Arithmetic in the Marketplace,”, The American Mathematical Monthly 95 (1988): 548–551., This article provides a more detailed analysis of the check-digit, schemes presented in this chapter. In particular, the error detection, rates for the various schemes are given. This article can be downloaded, at http://www.d.umn.edu/~jgallian/marketplace.pdf, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PA R T, , 2, , Groups, , For online student resources, visit this textbook’s website at, www.CengageBrain.com, , 29, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Introduction, 1 to Groups, Symmetry is a vast subject, significant in art and nature. Mathematics lies, at its root, and it would be hard to find a better one on which to, demonstrate the working of the mathematical intellect., hermann weyl, Symmetry, , Symmetries of a Square, Suppose we remove a square region from a plane, move it in some way,, then put the square back into the space it originally occupied. Our goal, in this chapter is to describe all possible ways in which this can be, done. More specifically, we want to describe the possible relationships, between the starting position of the square and its final position in, terms of motions. However, we are interested in the net effect of a motion, rather than in the motion itself. Thus, for example, we consider a, 908 rotation and a 4508 rotation as equal, since they have the same net, effect on every point. With this simplifying convention, it is an easy, matter to achieve our goal., To begin, we can think of the square region as being transparent, (glass, say), with the corners marked on one side with the colors blue,, white, pink, and green. This makes it easy to distinguish between motions that have different effects. With this marking scheme, we are now, in a position to describe, in simple fashion, all possible ways in which a, square object can be repositioned. See Figure 1.1. We now claim that, any motion—no matter how complicated—is equivalent to one of these, eight. To verify this claim, observe that the final position of the square, is completely determined by the location and orientation (that is, face, up or face down) of any particular corner. But, clearly, there are only, four locations and two orientations for a given corner, so there are, exactly eight distinct final positions for the corner., , 31, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32, , Groups, , P, , W, , G, , B, , P, , W, , R90 = Rotation of 90° (counterclockwise), G, , B, , P, , W, , R180 = Rotation of 180°, , V, , D, , W, , G, , B, , W, , B, , P, , G, , B, , G, , W, , P, , G, , P, , B, , W, , G, , B, , P, , W, , W, , P, , B, , G, , P, , G, , W, , B, , B, , W, , G, , P, , R90, , R180, , R270 = Rotation of 270°, , H, , P, R0, , R0 = Rotation of 0° (no change in position), , = Flip about a horizontal axis, , G, , B, , P, , W, , G, , B, , P, , W, , R270, , H, , = Flip about a vertical axis, , G, , B, , P, , W, , G, , B, , P, , W, , = Flip about the main diagonal, , D = Flip about the other diagonal, , G, , B, , P, , W, , G, , B, , V, , D, , D, , Figure 1.1, , Let’s investigate some consequences of the fact that every motion is, equal to one of the eight listed in Figure 1.1. Suppose a square is repositioned by a rotation of 908 followed by a flip about the horizontal axis, of symmetry., P, , W, , W, , B, , R90, G, , B, , P, , G, , W, , B, , H, P, , G, , Thus, we see that this pair of motions—taken together—is equal to, the single motion D. This observation suggests that we can compose, two motions to obtain a single motion. And indeed we can, since the, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33, , 1 | Introduction to Groups, , eight motions may be viewed as functions from the square region to, itself, and as such we can combine them using function composition., With this in mind, we write H R90 5 D because in lower level math, courses function composition f 8 g means “g followed by f.” The eight, motions R0, R90, R180, R270, H, V, D, and D9, together with the operation, composition, form a mathematical system called the dihedral group of, order 8 (the order of a group is the number of elements it contains). It is, denoted by D4. Rather than introduce the formal definition of a group, here, let’s look at some properties of groups by way of the example D4., To facilitate future computations, we construct an operation table or, Cayley table (so named in honor of the prolific English mathematician, Arthur Cayley, who first introduced them in 1854) for D 4 below. The, circled entry represents the fact that D 5 HR90. (In general, ab denotes, the entry at the intersection of the row with a at the left and the column, with b at the top.), , R0, R90, R180, R270, H, V, D, D9, , R0, , R90, , R180, , R270, , H, , V, , D, , R0, R90, R180, R270, H, V, D, D9, , R90, R180, R270, R0, D, D9, V, H, , R180, R270, R0, R90, V, H, D9, D, , R270, R0, R90, R180, D9, D, H, V, , H, D9, V, D, R0, R180, R270, R90, , V, D, H, D9, R180, R0, R90, R270, , D, H, D9, V, R90, R270, R0, R180, , D9, D9, V, D, H, R270, R90, R180, R0, , Notice how orderly this table looks! This is no accident. Perhaps the, most important feature of this table is that it has been completely filled, in without introducing any new motions. Of course, this is because, as, we have already pointed out, any sequence of motions turns out to be, the same as one of these eight. Algebraically, this says that if A and B, are in D4, then so is AB. This property is called closure, and it is one of, the requirements for a mathematical system to be a group. Next, notice, that if A is any element of D4, then AR0 5 R0 A 5 A. Thus, combining, any element A on either side with R0 yields A back again. An element, R0 with this property is called an identity, and every group must have, one. Moreover, we see that for each element A in D4, there is exactly, one element B in D4 such that AB 5 BA 5 R0. In this case, B is said to, be the inverse of A and vice versa. For example, R90 and R 270 are, inverses of each other, and H is its own inverse. The term inverse is a, descriptive one, for if A and B are inverses of each other, then B “undoes” whatever A “does,” in the sense that A and B taken together in either order produce R0, representing no change. Another striking feature, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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34, , Groups, , of the table is that every element of D4 appears exactly once in each, row and column. This feature is something that all groups must have,, and, indeed, it is quite useful to keep this fact in mind when constructing the table in the first place., Another property of D4 deserves special comment. Observe that, HD Z DH but R90R180 5 R180R90. Thus, in a group, ab may or may not, be the same as ba. If it happens that ab 5 ba for all choices of group, elements a and b, we say the group is commutative or—better yet—, Abelian (in honor of the great Norwegian mathematician Niels Abel)., Otherwise, we say the group is non-Abelian., Thus far, we have illustrated, by way of D4, three of the four conditions that define a group—namely, closure, existence of an identity,, and existence of inverses. The remaining condition required for a group, is associativity; that is, (ab)c 5 a(bc) for all a, b, c in the set. To be sure, that D4 is indeed a group, we should check this equation for each of the, 83 5 512 possible choices of a, b, and c in D4. In practice, however,, this is rarely done! Here, for example, we simply observe that the eight, motions are functions and the operation is function composition. Then,, since function composition is associative, we do not have to check the, equations., , The Dihedral Groups, The analysis carried out above for a square can similarly be done for, an equilateral triangle or regular pentagon or, indeed, any regular n-gon, (n $ 3). The corresponding group is denoted by Dn and is called the, dihedral group of order 2n., The dihedral groups arise frequently in art and nature. Many of the, decorative designs used on floor coverings, pottery, and buildings have, one of the dihedral groups as a group of symmetry. Corporation logos, are rich sources of dihedral symmetry [1]. Chrysler’s logo has D5 as a, symmetry group, and that of Mercedes-Benz has D3. The ubiquitous, five-pointed star has symmetry group D5. The phylum Echinodermata, contains many sea animals (such as starfish, sea cucumbers, feather, stars, and sand dollars) that exhibit patterns with D5 symmetry., Chemists classify molecules according to their symmetry. Moreover,, symmetry considerations are applied in orbital calculations, in determining energy levels of atoms and molecules, and in the study of molecular, vibrations. The symmetry group of a pyramidal molecule such as ammonia (NH3), depicted in Figure 1.2, is D3., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1 | Introduction to Groups, , 35, , N, , H, H, H, , Figure 1.2 A pyramidal molecule with symmetry group D3., , Tibor Zoltai, , Mineralogists determine the internal structures of crystals (that is,, rigid bodies in which the particles are arranged in three-dimensional, repeating patterns—table salt and table sugar are two examples) by, studying two-dimensional x-ray projections of the atomic makeup, of the crystals. The symmetry present in the projections reveals the, internal symmetry of the crystals themselves. Commonly occurring, symmetry patterns are D4 and D6 (see Figure 1.3). Interestingly, it is, mathematically impossible for a crystal to possess a Dn symmetry pattern with n 5 5 or n . 6., , Figure 1.3 X-ray diffraction photos revealing D4 symmetry patterns in crystals., , The dihedral group of order 2n is often called the group of symmetries of a regular n-gon. A plane symmetry of a figure F in a, plane is a function from the plane to itself that carries F onto F and, preserves distances; that is, for any points p and q in the plane, the, distance from the image of p to the image of q is the same as the, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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36, , Groups, , distance from p to q. (The term symmetry is from the Greek word, symmetros, meaning “of like measure.”) The symmetry group of a, plane figure is the set of all symmetries of the figure. Symmetries in, three dimensions are defined analogously. Obviously, a rotation of a, plane about a point in the plane is a symmetry of the plane, and a rotation about a line in three dimensions is a symmetry in three-dimensional, space. Similarly, any translation of a plane or of three-dimensional, space is a symmetry. A reflection across a line L is that function that, leaves every point of L fixed and takes any point q, not on L, to the point, q9 so that L is the perpendicular bisector of the line segment joining, q and q9 (see Figure 1.4). A reflection across a plane in three dimensions is defined analogously. Notice that the restriction of a 1808 rotation about a line L in three dimensions to a plane containing L is a, reflection across L in the plane. Thus, in the dihedral groups, the motions that we described as flips about axes of symmetry in three dimensions (for example, H, V, D, D9) are reflections across lines in two, dimensions. Just as a reflection across a line is a plane symmetry that, cannot be achieved by a physical motion of the plane in two dimensions, a reflection across a plane is a three-dimensional symmetry that, cannot be achieved by a physical motion of three-dimensional space. A, cup, for instance, has reflective symmetry across the plane bisecting, the cup, but this symmetry cannot be duplicated with a physical motion in three dimensions., q, q9, L, , Figure 1.4, , Many objects and figures have rotational symmetry but not reflective, symmetry. A symmetry group consisting of the rotational symmetries of, 08, 3608/n, 2(3608)/n, . . . , (n 2 1)3608/n, and no other symmetries, is, called a cyclic rotation group of order n and is denoted by 7R360/n8. Cyclic, rotation groups, along with dihedral groups, are favorites of artists, designers, and nature. Figure 1.5 illustrates with corporate logos the cyclic, rotation groups of orders 2, 3, 4, 5, 6, 8, 16, and 20., A study of symmetry in greater depth is given in Chapters 27 and 28., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1 | Introduction to Groups, , 37, , Figure 1.5 Logos with cyclic rotation symmetry groups., , Exercises, The only way to learn mathematics is to do mathematics., paul r. halmos, A Hilbert Space Problem Book, , 1. With pictures and words, describe each symmetry in D3 (the set of, symmetries of an equilateral triangle)., 2. Write out a complete Cayley table for D3. Is D3 Abelian?, 3. In D4, find all elements X such that, a. X3 5 V;, b. X3 5 R90;, c. X3 5 R0;, d. X2 5 R0;, e. X2 5 H., 4. Describe in pictures or words the elements of D5 (symmetries of a, regular pentagon)., 5. For n $ 3, describe the elements of Dn. (Hint: You will need to, consider two cases—n even and n odd.) How many elements, does Dn have?, 6. In Dn, explain geometrically why a reflection followed by a reflection must be a rotation., 7. In Dn, explain geometrically why a rotation followed by a rotation, must be a rotation., 8. In Dn, explain geometrically why a rotation and a reflection taken, together in either order must be a reflection., 9. Associate the number 1 with a rotation and the number 21 with a, reflection. Describe an analogy between multiplying these two, numbers and multiplying elements of Dn., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Groups, , 10. If r1, r2, and r3 represent rotations from Dn and f1, f2, and f3 represent, reflections from Dn, determine whether r1r2 f1r3 f2 f3r3 is a rotation, or a reflection., 11. Find elements A, B, and C in D4 such that AB 5 BC but A Z C., (Thus, “cross cancellation” is not valid.), 12. Explain what the following diagram proves about the group Dn., 1, , 1, , n, , F, 2, , n, , R360 / n, 2, 2, , 1, , 2, , n–1, 2, , R360/ n, n, , 1, , n, , F, 1, , 3, , 3, , 1, , 13. Describe the symmetries of a nonsquare rectangle. Construct the, corresponding Cayley table., 14. Describe the symmetries of a parallelogram that is neither a rectangle nor a rhombus. Describe the symmetries of a rhombus that is, not a rectangle., 15. Describe the symmetries of a noncircular ellipse. Do the same for, a hyperbola., 16. Consider an infinitely long strip of equally spaced H’s:, ???HHHH???, Describe the symmetries of this strip. Is the group of symmetries, of the strip Abelian?, 17. For each of the snowflakes in the figure, find the symmetry group, and locate the axes of reflective symmetry (disregard imperfections)., , Snow Crystals, by W. A. Bentley & W. J., Humphreys © Dover Publications, , 38, , Photographs of snowflakes from the Bentley and Humphreys atlas., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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1 | Introduction to Groups, , 39, , 18. Determine the symmetry group of the outer shell of the cross section of the human immunodeficiency virus (HIV) shown below., , 19. Does a fan blade have a cyclic symmetry group or a dihedral symmetry group?, 20. Bottle caps that are pried off typically have 22 ridges around the, rim. Find the symmetry group of such a cap., 21. What group theoretic property do uppercase letters F, G, J, L, P, Q, R, have that is not shared by the remaining uppercase letters in the, alphabet?, 22. What symmetry property does the word “zoonosis” have when, written in uppercase letters? (It means a disease of humans acquired, from animals.), 23. What symmetry property do the words “mow,” “sis,” and “swims”, have when written in uppercase letters?, 24. For each design below, determine the symmetry group (ignore, imperfections)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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40, , Groups, , Suggested Reading, Michael Field and Martin Golubitsky, Symmetry in Chaos, Oxford, University Press, 1992., This book has many beautiful symmetric designs that arise in, chaotic dynamic systems., , Suggested Website, http://britton.disted.camosun.bc.ca/jbsymteslk.htm, This spectacular website on symmetry and tessellations has numerous, activities and links to many other sites on related topics. It is a wonderful, website for K–12 teachers and students., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Niels Abel, He [Abel] has left mathematicians, something to keep them busy for five, hundred years., , Stock Montage, , charles hermite, , A 500-kroner bank note first issued, by Norway in 1948., , Niels Henrik Abel, one of the foremost, mathematicians of the 19th century, was, born in Norway on August 5, 1802. At the, age of 16, he began reading the classic mathematical works of Newton, Euler, Lagrange,, and Gauss. When Abel was 18 years old, his, father died, and the burden of supporting the, family fell upon him. He took in private pupils and did odd jobs, while continuing to do, mathematical research. At the age of 19,, Abel solved a problem that had vexed leading mathematicians for hundreds of years., He proved that, unlike the situation for equations of degree 4 or less, there is no finite, (closed) formula for the solution of the general fifth-degree equation., Although Abel died long before the advent of the subjects that now make up abstract algebra, his solution to the quintic, problem laid the groundwork for many of, these subjects. Just when his work was beginning to receive the attention it deserved,, Abel contracted tuberculosis. He died on, April 6, 1829, at the age of 26., , This stamp was issued in 1929, to commemorate the 100th, anniversary of Abel’s death., , In recognition of the fact that there is no, Nobel Prize for mathematics, in 2002 Norway, established the Abel Prize as the “Nobel Prize, in mathematics” in honor of its native son. At, approximately the $1,000,000 level, the Abel, Prize is now seen as an award equivalent to a, Nobel Prize., To find more information about Abel, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , 41, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 Groups, , A good stock of examples, as large as possible, is indispensable, for a thorough understanding of any concept, and when I want, to learn something new, I make it my first job to build one., paul r. halmos, , Definition and Examples of Groups, The term group was used by Galois around 1830 to describe sets of, one-to-one functions on finite sets that could be grouped together to, form a set closed under composition. As is the case with most fundamental concepts in mathematics, the modern definition of a group that, follows is the result of a long evolutionary process. Although this definition was given by both Heinrich Weber and Walther von Dyck in, 1882, it did not gain universal acceptance until the 20th century., Definition Binary Operation, , Let G be a set. A binary operation on G is a function that assigns each, ordered pair of elements of G an element of G., , A binary operation on a set G, then, is simply a method (or formula) by which the members of an ordered pair from G combine to, yield a new member of G. This condition is called closure. The most, familiar binary operations are ordinary addition, subtraction, and, multiplication of integers. Division of integers is not a binary operation on the integers because an integer divided by an integer need not, be an integer., The binary operations addition modulo n and multiplication modulo n on the set {0, 1, 2, . . . , n 2 1}, which we denote by Zn, play an, extremely important role in abstract algebra. In certain situations we, will want to combine the elements of Zn by addition modulo n only;, in other situations we will want to use both addition modulo n and, multiplication modulo n to combine the elements. It will be clear, , 42, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 43, , from the context whether we are using addition only or addition and, multiplication. For example, when multiplying matrices with entries, from Z n, we will need both addition modulo n and multiplication, modulo n., Definition Group, , Let G be a set together with a binary operation (usually called multiplication) that assigns to each ordered pair (a, b) of elements of G an element in G denoted by ab. We say G is a group under this operation if, the following three properties are satisfied., 1. Associativity. The operation is associative; that is, (ab)c 5 a(bc) for, all a, b, c in G., 2. Identity. There is an element e (called the identity) in G such that, ae 5 ea 5 a for all a in G., 3. Inverses. For each element a in G, there is an element b in G, (called an inverse of a) such that ab 5 ba 5 e., , In words, then, a group is a set together with an associative operation such that there is an identity, every element has an inverse, and any, pair of elements can be combined without going outside the set. Be, sure to verify closure when testing for a group (see Example 5). Notice, that if a is the inverse of b, then b is the inverse of a., If a group has the property that ab 5 ba for every pair of elements a, and b, we say the group is Abelian. A group is non-Abelian if there is, some pair of elements a and b for which ab 2 ba. When encountering, a particular group for the first time, one should determine whether or, not it is Abelian., Now that we have the formal definition of a group, our first job is, to build a good stock of examples. These examples will be used, throughout the text to illustrate the theorems. (The best way to grasp, the meat of a theorem is to see what it says in specific cases.) As we, progress, the reader is bound to have hunches and conjectures that, can be tested against the stock of examples. To develop a better understanding of the following examples, the reader should supply the, missing details., EXAMPLE 1 The set of integers Z (so denoted because the German, word for numbers is Zahlen), the set of rational numbers Q (for quotient), and the set of real numbers R are all groups under ordinary addition. In each case, the identity is 0 and the inverse of a is 2a., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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44, , Groups, , EXAMPLE 2 The set of integers under ordinary multiplication is not, a group. Since the number 1 is the identity, property 3 fails. For example, there is no integer b such that 5b 5 1., EXAMPLE 3 The subset {1, 21, i, 2i} of the complex numbers, is a group under complex multiplication. Note that 21 is its own inverse,, whereas the inverse of i is 2i, and vice versa., EXAMPLE 4 The set Q1 of positive rationals is a group under ordinary multiplication. The inverse of any a is 1/a 5 a21., EXAMPLE 5 The set S of positive irrational numbers together with 1, under multiplication satisfies the three properties given in the definition, of a group but is not a group. Indeed, 22 ? 22 5 2, so S is not closed, under multiplication., a b, EXAMPLE 6 A rectangular array of the form c, d is called a, c d, 2 3 2 matrix. The set of all 2 3 2 matrices with real entries is a group, under componentwise addition. That is,, c, , a1 b1, a b, a a2 b1 b2, d c 2 2d c 1, d, c1 d1, c2 d2, c1 c2 d1 d2, , The identity is c, , 0 0, a b, a b, d , and the inverse of c, d is c, d., 0 0, c d, c d, , EXAMPLE 7 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a group, under addition modulo n. For any j . 0 in Zn, the inverse of j is n 2 j., This group is usually referred to as the group of integers modulo n., As we have seen, the real numbers, the 2 3 2 matrices with real entries, and the integers modulo n are all groups under the appropriate, addition. But what about multiplication? In each case, the existence of, some elements that do not have inverses prevents the set from being a, group under the usual multiplication. However, we can form a group in, each case by simply throwing out the rascals. Examples 8, 9, and 11, illustrate this., EXAMPLE 8 The set R* of nonzero real numbers is a group under, ordinary multiplication. The identity is 1. The inverse of a is 1/a., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 45, , EXAMPLE 9† The determinant of the 2 3 2 matrix c, , a b, d is the, c d, number ad 2 bc. If A is a 2 3 2 matrix, det A denotes the determinant, of A. The set, GL(2, R) 5 e c, , a b, d ` a, b, c, d [ R, ad bc ? 0 f, c d, , of 2 3 2 matrices with real entries and nonzero determinants is a nonAbelian group under the operation, c, , a1 b1 a2 b2, a1a2 b1c2 a1b2 b1d2, dc, d c, d., c1 d1 c2 d2, c1a2 d1c2 c1b2 d1d2, , The first step in verifying that this set is a group is to show that the, product of two matrices with nonzero determinants also has a nonzero, determinant. This follows from the fact that for any pair of 2 3 2, matrices A and B, det (AB) 5 (det A)(det B)., Associativity can be verified by direct (but cumbersome) calcula1 0, a b, d ; the inverse of c, d is, tions. The identity is c, 0 1, c d, d, ≥, , ad, , b, bc ad, , c, ad, , bc, a, , bc ad, , ¥, , bc, , (explaining the requirement that ad 2 bc 2 0). This very important, non-Abelian group is called the general linear group of 2 3 2 matrices, over R., EXAMPLE 10 The set of all 2 3 2 matrices with real entries is not a, group under the operation defined in Example 9. Inverses do not exist, when the determinant is 0., Now that we have shown how to make subsets of the real numbers, and subsets of the set of 2 3 2 matrices into multiplicative groups, we, next consider the integers under multiplication modulo n., †For, , simplicity, we have restricted our matrix examples to the 2 3 2 case. However,, readers who have had linear algebra can readily generalize to n 3 n matrices., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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46, , Groups, , EXAMPLE 11 (L. EULER, 1761) By Exercise 11 in Chapter 0, an, integer a has a multiplicative inverse modulo n if and only if a and n are, relatively prime. So, for each n . 1, we define U(n) to be the set of all, positive integers less than n and relatively prime to n. Then U(n) is a, group under multiplication modulo n. (We leave it to the reader to check, that this set is closed under this operation.), For n 5 10, we have U(10) 5 {1, 3, 7, 9}. The Cayley table for, U(10) is, mod 10, , 1, , 3, , 7, , 9, , 1, 3, 7, 9, , 1, 3, 7, 9, , 3, 9, 1, 7, , 7, 1, 9, 3, , 9, 7, 3, 1, , (Recall that ab mod n is the unique integer r with the property a ? b 5, nq 1 r, where 0 # r , n and a ? b is ordinary multiplication.) In the, case that n is a prime, U(n) 5 {1, 2, . . . , n 2 1}., In his classic book Lehrbuch der Algebra, published in 1895, Heinrich, Weber gave an extensive treatment of the groups U(n) and described, them as the most important examples of finite Abelian groups., EXAMPLE 12 The set {0, 1, 2, 3} is not a group under multiplication modulo 4. Although 1 and 3 have inverses, the elements 0 and 2, do not., EXAMPLE 13 The set of integers under subtraction is not a group,, since the operation is not associative., With the examples given thus far as a guide, it is wise for the reader, to pause here and think of his or her own examples. Study actively!, Don’t just read along and be spoon-fed by the book., EXAMPLE 14 The complex numbers C 1 {a 1 bi | a, b [ R,, i2 5 21} are a group under the operation (a 1 bi) 1 (c 1 di) 5, (a 1 c) 1 (b 1 d)i. The inverse of a 1 bi is 2a 2bi. The nonzero, complex numbers C* are a group under the operation (a 1 bi), 1, 5, (c 1 di) 5 (ac 2 bd) 1 (ad 1 bc)i. The inverse of a 1 bi is, a bi, 1, 1, a bi, 1, 5 2, bi., 2a 2 2, a bi a bi a b, a b2, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 47, , EXAMPLE 15 For all integers n $ 1, the set of complex nth roots, of unity, e cos, , k # 360, k # 360, i sin, ` k 0, 1, 2, . . . , n 1 f, n, n, , (i.e., complex zeros of xn 2 1) is a group under multiplication. (See, DeMoivre’s Theorem—Example 10 in Chapter 0.) Compare this group, with the one in Example 3., Recall from Chapter 0 that the complex number cos u 1 i sin u can, be represented geometrically as the point (cos u, sin u) in a plane coordinatized by a real horizontal axis and a vertical imaginary axis, where, u is the angle formed by the line segment joining the origin and the, point (cos u, sin u) and the positive real axis. Thus, the six complex, zeros of x6 5 1 are located at points around the circle of radius 1, 60°, apart, as shown in Figure 2.1., Imaginary, – 1 + √3 i, 2, 2, , 1 + √3, 2, 2 i, , √3, 2, 60, , –1, , 1, , Real, , 1, 2, , – 1 – √3 i, 2, 2, , 1 – √3, 2, 2 i, , Figure 2.1, , EXAMPLE 16 The set Rn 5 {(a1, a2, . . . , an) U a1, a2, . . . , an [ R}, is a group under componentwise addition [i.e., (a1, a2, . . . , an) 1, (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn)]., EXAMPLE 17 For a fixed point (a, b) in R2, define Ta,b: R2 → R2, by (x, y) → (x 1 a, y 1 b). Then G 5 {Ta,b U a, b [ R} is a group, under function composition. Straightforward calculations show that, Ta,bTc,d 5 Ta1c,b1d. From this formula we may observe that G is closed,, T0,0 is the identity, the inverse of Ta,b is T2a,2b, and G is Abelian., Function composition is always associative. The elements of G are, called translations., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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48, , Groups, , EXAMPLE 18 The set of all 2 3 2 matrices with determinant 1 with entries from Q (rationals), R (reals), C (complex numbers), or Zp (p a prime), is a non-Abelian group under matrix multiplication. This group is called, the special linear group of 2 3 2 matrices over Q, R, C, or Zp, respectively., If the entries are from F, where F is any of the above, we denote this group, by SL(2, F). For the group SL(2, F), the formula given in Example 9 for, a b, d b, d simplifies to c, d . When the matrix, the inverse of c, c d, c, a, entries are from Zp, we use modulo p arithmetic to compute determinants, matrix products, and inverses. To illustrate the case SL(2, Z5),, 3 4, consider the element A 5 c, d . Then det A 5 (3 ? 4 2 4 ? 4) mod 5 5, 4 4, 4 4, 4 1, 24 mod 5 5 1, and the inverse of A is c, d c, d . Note, 4 3, 1 3, 3 4 4 1, 1 0, that c, dc, d c, d when the arithmetic is done modulo 5., 4 4 1 3, 0 1, Example 9 is a special case of the following general construction., EXAMPLE 19 Let F be any of Q, R, C, or Zp ( p a prime). The set, GL(2, F) of all 2 3 2 matrices with nonzero determinants and entries, from F is a non-Abelian group under matrix multiplication. As in, Example 18, when F is Zp, modulo p arithmetic is used to calculate, determinants, matrix products, and inverses. The formula given in, a b, d remains valid for elements from, Example 9 for the inverse of c, c d, GL(2, Zp), provided we interpret division by ad 2 bc as multiplication, by the inverse of (ad 2 bc) modulo p. For example, in GL(2, Z7),, 4 5, consider c, d. Then the determinant (ad 2 bc) mod 7 is (12 2 30), 6 3, mod 7 5 218 mod 7 5 3 and the inverse of 3 is 5 [since (3 ? 5), 4 5, 3?5 2?5, 1 3, mod 7 5 1]. So, the inverse of c, d is c, d c, d., 6 3, 1?5 4?5, 5 6, 4 5 1 3, 1 0, [The reader should check that c, d c, d c, d in GL(2, Z7)]., 6 3 5 6, 0 1, The group GL(n, F) is called the general linear group of n 3 n, matrices over F., EXAMPLE 20 The set {1, 2, . . . , n 2 1} is a group under multiplication modulo n if and only if n is prime., EXAMPLE 21 The set of all symmetries of the infinite ornamental, pattern in which arrowheads are spaced uniformly a unit apart along, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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49, , 2 | Groups, , a line is an Abelian group under composition. Let T denote a translation, to the right by one unit, T 21 a translation to the left by one unit, and H, a reflection across the horizontal line of the figure. Then, every member, of the group is of the form x1x2 ? ? ? xn, where each xi [, {T, T 21, H}. In this case, we say that T, T 21, and H generate the group., Table 2.1 summarizes many of the specific groups that we have, presented thus far., As the previous examples demonstrate, the notion of a group is a, very broad one indeed. The goal of the axiomatic approach is to find, properties general enough to permit many diverse examples having, these properties and specific enough to allow one to deduce many interesting consequences., The goal of abstract algebra is to discover truths about algebraic, systems (that is, sets with one or more binary operations) that are independent of the specific nature of the operations. All one knows, or needs to know is that these operations, whatever they may be, have, Table 2.1 Summary of Group Examples (F can be any of Q, R, C, or Zp; L is a reflection), Operation, , Z, Q1, , Addition, Multiplication, , 0, 1, , Zn, R*, , Addition mod n, Multiplication, , 0, 1, , k, m/n,, m, n . 0, k, x, , C*, , Multiplication, , 1, , a 1 bi, , GL(2, F), , Matrix, multiplication, , U(n), Rn, SL(2, F), , Dn, , Multiplication, mod n, Componentwise, addition, Matrix, multiplication, Composition, , Identity, , Form of, Element, , Group, , c, , 1 0, d, 0 1, , c, , a b, d, c d ,, , ad 2 bc 2 0, k,, gcd(k, n) 5 1, (0, 0, …, 0) (a1, a2, …, an), 1, , c, , 1 0, d, 0 1, R0, , c, , a b, d, c d ,, ad 2 bc 5 1, Ra, L, , Inverse, 2k, n/m, , Yes, Yes, , n2k, 1/x, , Yes, Yes, , 1, 1, a 2, bi, a2 b2, a b2, d, 2b, ad 2 bc ad 2 bc, ≥, ¥, 2c, a, ad 2 bc ad 2 bc, Solution to, kx mod n 5 1, (2a1, 2a2, …, 2an), c, , Abelian, , d, c, , b, d, a, , R360 2 a, L, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , Yes, , No, Yes, Yes, No, , No
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50, , Groups, , certain properties. We then seek to deduce consequences of these, properties. This is why this branch of mathematics is called abstract, algebra. It must be remembered, however, that when a specific group, is being discussed, a specific operation must be given (at least, implicitly)., , Elementary Properties of Groups, Now that we have seen many diverse examples of groups, we wish to, deduce some properties that they share. The definition itself raises, some fundamental questions. Every group has an identity. Could a, group have more than one? Every group element has an inverse. Could, an element have more than one? The examples suggest not. But examples can only suggest. One cannot prove that every group has a unique, identity by looking at examples, because each example inherently has, properties that may not be shared by all groups. We are forced to, restrict ourselves to the properties that all groups have; that is, we must, view groups as abstract entities rather than argue by example. The next, three theorems illustrate the abstract approach., Theorem 2.1 Uniqueness of the Identity, In a group G, there is only one identity element., , PROOF Suppose both e and e9 are identities of G. Then,, 1. ae 5 a for all a in G, and, 2. e9a 5 a for all a in G., The choices of a 5 e9 in (part 1) and a 5 e in (part 2) yield e9e 5 e9, and e9e 5 e. Thus, e and e9 are both equal to e9e and so are equal to, each other., Because of this theorem, we may unambiguously speak of “the identity” of a group and denote it by ‘e’ (because the German word for, identity is Einheit)., Theorem 2.2 Cancellation, In a group G, the right and left cancellation laws hold; that is,, ba 5 ca implies b 5 c, and ab 5 ac implies b 5 c., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 51, , PROOF Suppose ba 5 ca. Let a9 be an inverse of a. Then multiplying on the right by a9 yields (ba)a9 5 (ca)a9. Associativity yields, b(aa9) 5 c(aa9). Then be 5 ce and, therefore, b 5 c as desired. Similarly, one can prove that ab 5 ac implies b 5 c by multiplying by a9 on, the left., A consequence of the cancellation property is the fact that in a, Cayley table for a group, each group element occurs exactly once in, each row and column (see Exercise 31). Another consequence of the, cancellation property is the uniqueness of inverses., Theorem 2.3 Uniqueness of Inverses, For each element a in a group G, there is a unique element b in G, such that ab 5 ba 5 e., , PROOF Suppose b and c are both inverses of a. Then ab 5 e and, ac 5 e, so that ab 5 ac. Canceling the a on both sides gives b 5 c, as, desired., As was the case with the identity element, it is reasonable, in view, of Theorem 2.3, to speak of “the inverse” of an element g of a group;, in fact, we may unambiguously denote it by g21. This notation is suggested by that used for ordinary real numbers under multiplication., Similarly, when n is a positive integer, the associative law allows us to, use gn to denote the unambiguous product, gg ? ? ? g., n factors, , We define g0 5 e. When n is negative, we define gn 5 (g21)|n| [for example, g23 5 (g21)3]. Unlike for real numbers, in an abstract group we, do not permit noninteger exponents such as g1/2. With this notation, the, familiar laws of exponents hold for groups; that is, for all integers m and, n and any group element g, we have gmgn 5 gm1n and (gm)n 5 gmn., Although the way one manipulates the group expressions g mgn and, (gm)n coincides with the laws of exponents for real numbers, the laws, of exponents fail to hold for expressions involving two group elements., Thus, for groups in general, (ab)n Z anbn (see Exercise 23)., The important thing about the existence of a unique inverse for each, group element a is that for every element b in the group there is a unique, solution in the group of the equations ax 5 b and xa 5 b. Namely,, x 5 a21b in the first case and x 5 ba21 in the second case. In contrast,, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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52, , Groups, , in the set {0, 1, 2, 3, 4, 5}, the equation 2x 5 4 has the solutions x 5 2, and x 5 5 under the operation multiplication mod 6. However, this set, is not a group under multiplication mod 6., Also, one must be careful with this notation when dealing with a, specific group whose binary operation is addition and is denoted by, “1.” In this case, the definitions and group properties expressed in, multiplicative notation must be translated to additive notation. For, example, the inverse of g is written as 2g. Likewise, for example, g3, Table 2.2, Multiplicative Group, , a ? b or ab, e or 1, a21, an, ab21, , Multiplication, Identity or one, Multiplicative inverse of a, Power of a, Quotient, , Additive Group, , a1b, 0, 2a, na, a2b, , Addition, Zero, Additive inverse of a, Multiple of a, Difference, , means g 1 g 1 g and is usually written as 3g, whereas g23 means, (2g) 1 (2g) 1 (2g) and is written as 23g. When additive notation, is used, do not interpret “ng” as combining n and g under the group, operation; n may not even be an element of the group! Table 2.2 shows, the common notation and corresponding terminology for groups under, multiplication and groups under addition. As is the case for real numbers, we use a 2 b as an abbreviation for a 1 (2b)., Because of the associative property, we may unambiguously write, the expression abc, for this can be reasonably interpreted as only (ab)c, or a(bc), which are equal. In fact, by using induction and repeated application of the associative property, one can prove a general associative property that essentially means that parentheses can be inserted or, deleted at will without affecting the value of a product involving any, number of group elements. Thus,, a2(bcdb2) 5 a2b(cd )b2 5 (a2b)(cd )b2 5 a(abcdb)b,, and so on., Although groups do not have the property that (ab)n 5 a nb n, there is, a simple relationship between (ab)21 and a21 and b21., Theorem 2.4 Socks–Shoes Property, For group elements a and b, (ab)21 5 b21a21., , PROOF Since (ab)(ab)21 5 e and (ab)(b21a21) 5 a(bb21)a21 5, aea21 5 aa21 5 e, we have by Theorem 2.3 that (ab)21 5 b21a21., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 53, , Historical Note, We conclude this chapter with a bit of history concerning the noncommutativity of matrix multiplication. In 1925, quantum theory was, replete with annoying and puzzling ambiguities. It was Werner, Heisenberg who recognized the cause. He observed that the product of, the quantum-theoretical analogs of the classical Fourier series did not, necessarily commute. For all his boldness, this shook Heisenberg. As, he later recalled [2, p. 94]:, In my paper the fact that XY was not equal to YX was very disagreeable to me. I felt, this was the only point of difficulty in the whole scheme, otherwise I would be perfectly happy. But this difficulty had worried me and I was not able to solve it., , Heisenberg asked his teacher, Max Born, if his ideas were worth publishing. Born was fascinated and deeply impressed by Heisenberg’s new, approach. Born wrote [1, p. 217]:, After having sent off Heisenberg’s paper to the Zeitschrift für Physik for publication, I began to ponder over his symbolic multiplication, and was soon so involved, in it that I thought about it for the whole day and could hardly sleep at night. For I, felt there was something fundamental behind it, the consummation of our endeavors of many years. And one morning, about the 10 July 1925, I suddenly saw light:, Heisenberg’s symbolic multiplication was nothing but the matrix calculus, wellknown to me since my student days from Rosanes’ lectures in Breslau., , Born and his student, Pascual Jordan, reformulated Heisenberg’s ideas, in terms of matrices, but it was Heisenberg who was credited with the, formulation. In his autobiography, Born lamented [1, p. 219]:, Nowadays the textbooks speak without exception of Heisenberg’s matrices, Heisenberg’s commutation law, and Dirac’s field quantization., In fact, Heisenberg knew at that time very little of matrices and had to study, them., , Upon learning in 1933 that he was to receive the Nobel Prize, with Dirac and Schrödinger for this work, Heisenberg wrote to Born, [1, p. 220]:, If I have not written to you for such a long time, and have not thanked you for your, congratulations, it was partly because of my rather bad conscience with respect to, you. The fact that I am to receive the Nobel Prize alone, for work done in Göttingen, in collaboration—you, Jordan, and I—this fact depresses me and I hardly know, what to write to you. I am, of course, glad that our common efforts are now appreciated, and I enjoy the recollection of the beautiful time of collaboration. I also believe that all good physicists know how great was your and Jordan’s contribution to, the structure of quantum mechanics—and this remains unchanged by a wrong decision from outside. Yet I myself can do nothing but thank you again for all the fine, collaboration, and feel a little ashamed., , The story has a happy ending, however, because Born received the, Nobel Prize in 1954 for his fundamental work in quantum mechanics., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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54, , Groups, , Exercises, “For example” is not proof., jewish proverb, , 1. Which of the following binary operations are closed?, a. subtraction of positive integers, b. division of nonzero integers, c. function composition of polynomials with real coefficients, d. multiplication of 2 3 2 matrices with integer entries, 2. Which of the following binary operations are associative?, a. multiplication mod n, b. division of nonzero rationals, c. function composition of polynomials with real coefficients, d. multiplication of 2 3 2 matrices with integer entries, 3. Which of the following binary operations are commutative?, a. substraction of integers, b. division of nonzero real numbers, c. function composition of polynomials with real coefficients, d. multiplication of 2 3 2 matrices with real entries, 4. Which of the following sets are closed under the given operation?, a. {0, 4, 8, 12} addition mod 16, b. {0, 4, 8, 12} addition mod 15, c. {1, 4, 7, 13} multiplication mod 15, d. {1, 4, 5, 7} multiplication mod 9, 5. In each case, find the inverse of the element under the given, operation., a. 13 in Z20, b. 13 in U(14), c. n21 in U(n) (n . 2), d. 322i in C*, the group of nonzero complex numbers under multiplication, 6. In each case, perform the indicated operation., a. In C*, (7 1 5i)( 23 1 2i), 7 4, d, b. In GL(2, Z13), det c, 1 5, 6 3 1, c. In GL (2, R), c, d, 8 2, 6 3 1, d. In GL(2, Z13), c, d, 8 2, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 55, , 7. Give two reasons why the set of odd integers under addition is not, a group., 8. Referring to Example 13, verify the assertion that subtraction is not, associative., 9. Show that {1, 2, 3} under multiplication modulo 4 is not a group, but that {1, 2, 3, 4} under multiplication modulo 5 is a group., 10. Show that the group GL(2, R) of Example 9 is non-Abelian by exhibiting a pair of matrices A and B in GL(2, R) such that AB 2 BA., 2 6, d in GL(2, Z11)., 11. Find the inverse of the element c, 3 5, 12. Give an example of group elements a and b with the property that, a21ba 2 b., 13. Translate each of the following multiplicative expressions into its, additive counterpart. Assume that the operation is commutative., a. a2b3, b. a22(b21c)2, c. (ab2)23c2 5 e, 14. For group elements a, b, and c, express (ab)3 and (ab22 c)22 without parentheses., 15. Let G be a group and let H 5 {x21 | x [ G}. Show that G 5 H, as sets., 16. Show that the set {5, 15, 25, 35} is a group under multiplication, modulo 40. What is the identity element of this group? Can you see, any relationship between this group and U(8)?, 17. (From the GRE Practice Exam)* Let p and q be distinct primes., Suppose that H is a proper subset of the integers that is a group under addition that contains exactly three elements of the set {p, p 1 q,, pq, pq, qp}. Determine which of the following are the three elements, in H., a. pq, pq, qp, b. p 1 q, pq, pq, c. p, p 1 q, pq, d. p, pq, qp, e. p, pq, pq, 18. List the members of H 5 {x 2 | x [ D4} and K 5 {x [ D4 | x 2 5 e}., 19. Prove that the set of all 2 3 2 matrices with entries from R and, determinant 11 is a group under matrix multiplication., 20. For any integer n . 2, show that there are at least two elements in, U(n) that satisfy x2 5 1., 21. An abstract algebra teacher intended to give a typist a list of nine integers that form a group under multiplication modulo 91. Instead,, *GRE materials selected from the GRE Practice Exam, Question 9 by Educational Testing, Service. Reprinted by permission of Educational Testing Service, the copyright owner., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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56, , Groups, , 22., , 23., , 24., , 25., 26., 27., 28., 29., 30., 31., , 32., 33., , one of the nine integers was inadvertently left out, so that the list, appeared as 1, 9, 16, 22, 53, 74, 79, 81. Which integer was left out?, (This really happened!), Let G be a group with the property that for any x, y, z in the group,, xy 5 zx implies y 5 z. Prove that G is Abelian. (“Left-right cancellation” implies commutativity.), (Law of Exponents for Abelian Groups) Let a and b be elements of, an Abelian group and let n be any integer. Show that (ab)n 5 anbn., Is this also true for non-Abelian groups?, (Socks–Shoes Property) Draw an analogy between the statement, (ab)21 5 b21 a21 and the act of putting on and taking off your socks, and shoes. Find distinct nonidentity elements a and b from a, non-Abelian group such that (ab)21 5 a21 b21. Find an example, that shows that in a group, it is possible to have (ab)22 Z b22 a22., What would be an appropriate name for the group property, (abc)21 5 c21 b21 a21?, Prove that a group G is Abelian if and only if (ab)21 5 a21b21 for, all a and b in G., Prove that in a group, (a21)21 5 a for all a., For any elements a and b from a group and any integer n, prove, that (a21ba)n 5 a21bna., If a1, a2, . . . , an belong to a group, what is the inverse of a1a2 . . . an?, The integers 5 and 15 are among a collection of 12 integers that, form a group under multiplication modulo 56. List all 12., Give an example of a group with 105 elements. Give two examples, of groups with 44 elements., Prove that every group table is a Latin square†; that is, each element of the group appears exactly once in each row and each, column., Construct a Cayley table for U(12)., Suppose the table below is a group table. Fill in the blank entries., e, , a, , b, , c, , d, , e, a, b, c, , e, —, —, —, , —, b, c, d, , —, —, d, —, , —, —, e, a, , —, e, —, b, , d, , —, , —, , —, , —, , —, , †Latin squares are useful in designing statistical experiments. There is also a close connection between Latin squares and finite geometries., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 57, , 34. Prove that in a group, (ab)2 5 a2b2 if and only if ab 5 ba., 35. Let a, b, and c be elements of a group. Solve the equation axb 5 c, for x. Solve a21xa 5 c for x., 36. Let a and b belong to a group G. Find an x in G such that xabx21 5 ba., 37. Let G be a finite group. Show that the number of elements x of G, such that x3 5 e is odd. Show that the number of elements x of G, such that x2 2 e is even., 38. Give an example of a group with elements a, b, c, d, and x such, that axb 5 cxd but ab 2 cd. (Hence “middle cancellation” is not, valid in groups.), 39. Suppose that G is a group with the property that for every choice, of elements in G, axb 5 cxd implies ab 5 cd. Prove that G is, Abelian. (“Middle cancellation” implies commutativity.), 40. Find an element X in D4 such that R90VXH 5 D9., 41. Suppose F1 and F2 are distinct reflections in a dihedral group Dn., Prove that F1F2 2 R0., 42. Suppose F1 and F2 are distinct reflections in a dihedral group Dn, such that F1F2 5 F2F1. Prove that F1F2 5 R180., 43. Let R be any fixed rotation and F any fixed reflection in a dihedral, group. Prove that RkFRk 5 F., 44. Let R be any fixed rotation and F any fixed reflection in a dihedral, group. Prove that FRkF 5 R2k. Why does this imply that Dn is, non-Abelian?, 45. In the dihedral group Dn, let R 5 R360/n and let F be any reflection., Write each of the following products in the form Ri or RiF, where, 0 # i , n., a. In D4, FR22FR5, b. In D5, R23FR4FR22, c. In D6, FR5FR22F, 46. Prove that the set of all rational numbers of the form 3m6n, where m, and n are integers, is a group under multiplication., 47. Prove that if G is a group with the property that the square of every, element is the identity, then G is Abelian. (This exercise is referred, to in Chapter 26.), 48. Prove that the set of all 3 3 3 matrices with real entries of the form, 1 a b, £0 1 c §, 0 0 1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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58, , Groups, , is a group. (Multiplication is defined by, 1 a b, 1 a b, 1 a a b ac b, £ 0 1 c § £ 0 1 c § £ 0, 1, c c, §., 0 0 1, 0 0 1, 0, 0, 1, , 49., , 50., , 51., , 52., , 53., , 54., , This group, sometimes called the Heisenberg group after the Nobel, Prize–winning physicist Werner Heisenberg, is intimately related to, the Heisenberg Uncertainty Principle of quantum physics.), Prove the assertion made in Example 20 that the set {1, 2, . . . ,, n 2 1} is a group under multiplication modulo n if and only if n is, prime., In a finite group, show that the number of nonidentity elements, that satisfy the equation x5 5 e is a multiple of 5. If the stipulation, that the group be finite is omitted, what can you say about the, number of nonidentity elements that satisfy the equation x 5 5 e?, List the six elements of GL(2, Z2). Show that this group is nonAbelian by finding two elements that do not commute. (This exercise is referred to in Chapter 7.), a a, Let G e c, d 冟a [ R, a Z 0 f . Show that G is a group under, a a, matrix multiplication. Explain why each element of G has an inverse, even though the matrices have 0 determinants. (Compare with, Example 10.), Suppose that in the definition of a group G, the condition that there, exists an element e with the property ae 5 ea 5 a for all a in G is, replaced by ae 5 a for all a in G. Show that ea 5 a for all a in G., (Thus, a one-sided identity is a two-sided identity.), Suppose that in the definition of a group G, the condition that for, each element a in G there exists an element b in G with the property ab 5 ba 5 e is replaced by the condition ab 5 e. Show that, ba 5 e. (Thus, a one-sided inverse is a two-sided inverse.), , Computer Exercises, Software for the computer exercises in this chapter is available, at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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2 | Groups, , 59, , References, 1. Max Born, My Life: Recollections of a Nobel Laureate, New York:, Charles Scribner’s Sons, 1978., 2. J. Mehra and H. Rechenberg, The Historical Development of Quantum, Theory, Vol. 3, New York: Springer-Verlag, 1982., , Suggested Readings, Marcia Ascher, Ethnomathematics, Pacific Grove, CA: Brooks/Cole,, 1991., Chapter 3 of this book describes how the dihedral group of order 8 can, be used to encode the social structure of the kin system of family relationships among a tribe of native people of Australia., Arie Bialostocki, “An Application of Elementary Group Theory to Central, Solitaire,” The College Mathematics Journal, May 1998: 208–212., The author uses properties of groups to analyze the peg board game, central solitaire (which also goes by the name peg solitaire)., J. E. White, “Introduction to Group Theory for Chemists,” Journal of, Chemical Education 44 (1967): 128–135., Students interested in the physical sciences may find this article worthwhile. It begins with easy examples of groups and builds up to applications of group theory concepts and terminology to chemistry., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 Finite Groups;, Subgroups, In our own time, in the period 1960–1980, we have seen particle physics, emerge as the playground of group theory., FREEMAN DYSON, , Terminology and Notation, As we will soon discover, finite groups—that is, groups with finitely, many elements—have interesting arithmetic properties. To facilitate, the study of finite groups, it is convenient to introduce some terminology and notation., Definition Order of a Group, , The number of elements of a group (finite or infinite) is called its, order. We will use |G| to denote the order of G., , Thus, the group Z of integers under addition has infinite order,, whereas the group U(10) 5 {1, 3, 7, 9} under multiplication modulo, 10 has order 4., Definition Order of an Element, , The order of an element g in a group G is the smallest positive integer, n such that gn 5 e. (In additive notation, this would be ng 5 0.) If no, such integer exists, we say that g has infinite order. The order of an, element g is denoted by |g|., , So, to find the order of a group element g, you need only compute the, sequence of products g, g2, g3, . . . , until you reach the identity for the, first time. The exponent of this product (or coefficient if the operation is, addition) is the order of g. If the identity never appears in the sequence,, then g has infinite order., EXAMPLE 1 Consider U(15) 5 {1, 2, 4, 7, 8, 11, 13, 14} under, multiplication modulo 15. This group has order 8. To find the order of, 60, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 61, , the element 7, say, we compute the sequence 71 5 7, 72 5 4, 73 5 13,, 74 5 1, so |7| 5 4. To find the order of 11, we compute 111 5 11,, 112 5 1, so |11| 5 2. Similar computations show that |1| 5 1, |2| 5 4,, |4| 5 2, |8| 5 4, |13| 5 4, |14| 5 2. [Here is a trick that makes these, calculations easier. Rather than compute the sequence 131, 132, 133,, 134, we may observe that 13 5 22 mod 15, so that 132 5 (22)2 5 4,, 133 5 22 ? 4 5 28, 134 5 (22)(28) 5 1.]†, EXAMPLE 2 Consider Z10 under addition modulo 10. Since 1 ? 2 5 2,, 2 ? 2 5 4, 3 ? 2 5 6, 4 ? 2 5 8, 5 ? 2 5 0, we know that |2| 5 5. Similar, computations show that |0| 5 1, |7| 5 10, |5| 5 2, |6| 5 5. (Here 2 ? 2 is, an abbreviation for 2 1 2, 3 ? 2 is an abbreviation for 2 1 2 1 2, etc.), EXAMPLE 3 Consider Z under ordinary addition. Here every nonzero, element has infinite order, since the sequence a, 2a, 3a, . . . never includes, 0 when a 0., The perceptive reader may have noticed among our examples of, groups in Chapter 2 that some are subsets of others with the same, binary operation. The group SL(2, R) in Example 18, for instance, is a, subset of the group GL(2, R) in Example 9. Similarly, the group of, complex numbers {1, 21, i, 2i} under multiplication is a subset of the, group described in Example 15 for n equal to any multiple of 4. This, situation arises so often that we introduce a special term to describe it., Definition Subgroup, , If a subset H of a group G is itself a group under the operation of G, we, say that H is a subgroup of G., , We use the notation H # G to mean that H is a subgroup of G. If we, want to indicate that H is a subgroup of G but is not equal to G itself,, we write H , G. Such a subgroup is called a proper subgroup. The, subgroup {e} is called the trivial subgroup of G; a subgroup that is not, {e} is called a nontrivial subgroup of G., Notice that Zn under addition modulo n is not a subgroup of Z under, addition, since addition modulo n is not the operation of Z., , Subgroup Tests, When determining whether or not a subset H of a group G is a subgroup of G, one need not directly verify the group axioms. The next, † The website http://www.google.com provides a convenient way to do modular arithmetic. For example, to compute 134 mod 15, just type “13ˆ4 mod 15” in the search box., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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62, , Groups, , three results provide simple tests that suffice to show that a subset of a, group is a subgroup., Theorem 3.1 One-Step Subgroup Test, Let G be a group and H a nonempty subset of G. If ab21 is in H, whenever a and b are in H, then H is a subgroup of G. (In additive, notation, if a 2 b is in H whenever a and b are in H, then H is a, subgroup of G.), , PROOF Since the operation of H is the same as that of G, it is clear, that this operation is associative. Next, we show that e is in H. Since H, is nonempty, we may pick some x in H. Then, letting a 5 x and b 5 x, in the hypothesis, we have e 5 xx21 5 ab21 is in H. To verify that x21, is in H whenever x is in H, all we need to do is to choose a 5 e and, b 5 x in the statement of the theorem. Finally, the proof will be complete when we show that H is closed; that is, if x, y belong to H, we, must show that xy is in H also. Well, we have already shown that y21 is, in H whenever y is; so, letting a 5 x and b 5 y21, we have xy 5 x(y21)21 5, ab21 is in H., Although we have dubbed Theorem 3.1 the One-Step Subgroup Test,, there are actually four steps involved in applying the theorem. (After, you gain some experience, the first three steps will be routine.) Notice, the similarity between the last three steps listed below and the three, steps involved in the Second Principle of Mathematical Induction., 1. Identify the property P that distinguishes the elements of H; that is,, identify a defining condition., 2. Prove that the identity has property P. (This verifies that H is, nonempty.), 3. Assume that two elements a and b have property P., 4. Use the assumption that a and b have property P to show that, ab21 has property P., The procedure is illustrated in Examples 4 and 5., EXAMPLE 4 Let G be an Abelian group with identity e. Then H 5, {x [ G | x2 5 e} is a subgroup of G. Here, the defining property of H, is the condition x2 5 e. So, we first note that e2 5 e, so that H is nonempty. Now we assume that a and b belong to H. This means that a2 5 e, and b2 5 e. Finally, we must show that (ab21)2 5 e. Since G is, Abelian, (ab21)2 5 ab21ab21 5 a2(b21)2 5 a2(b2)21 5 ee21 5 e., Therefore, ab21 belongs to H and, by the One-Step Subgroup Test, H, is a subgroup of G., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 63, , In many instances, a subgroup will consist of all elements that have, a particular form. Then the property P is that the elements have that, particular form. This is illustrated in the following example., EXAMPLE 5 Let G be an Abelian group under multiplication with, identity e. Then H 5 {x2 | x [ G} is a subgroup of G. (In words, H is, the set of all “squares.”) Since e2 5 e, the identity has the correct form., Next, we write two elements of H in the correct form, say, a2 and b2. We, must show that a2(b2)21 also has the correct form; that is, a2(b2)21 is the, square of some element. Since G is Abelian, we may write a2(b2)21 as, (ab21)2, which is the correct form. Thus, H is a subgroup of G., Beginning students often prefer to use the next theorem instead of, Theorem 3.1., Theorem 3.2 Two-Step Subgroup Test, Let G be a group and let H be a nonempty subset of G. If ab is in H, whenever a and b are in H (H is closed under the operation), and a21, is in H whenever a is in H (H is closed under taking inverses), then H, is a subgroup of G., , PROOF By Theorem 3.1, it suffices to show that a, b [ H implies, ab21 [ H. So, we suppose that a, b [ H. Since H is closed under, taking inverses, we also have b21 [ H. Thus, ab21 [ H by closure under multiplication., When applying the Two-Step Subgroup Test, we proceed exactly as, in the case of the One-Step Subgroup Test, except we use the assumption that a and b have property P to prove that ab has property P and, that a21 has property P., EXAMPLE 6 Let G be an Abelian group. Then H 5 {x [ G | |x| is, finite} is a subgroup of G. Since e1 5 e, H Z u. To apply the Two-Step, Subgroup Test we assume that a and b belong to H and prove that ab, and a21 belong to H. Let |a| 5 m and |b| 5 n. Then, because G is, Abelian, we have (ab)mn 5 (am)n(bn)m 5 enem 5 e. Thus, ab has finite, order (this does not show that |ab| 5 mn). Moveover, (a21)m 5 (am)21 5, e21 5 e shows that a21 has finite order., We next illustrate how to use the Two-Step Subgroup Test by introducing an important technique for creating new subgroups of Abelian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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64, , Groups, , groups from existing ones. The method will be extended to some subgroups of non-Abelian groups in later chapters., EXAMPLE 7 Let G be an Abelian group and H and K be subgroups, of G. Then HK 5 {hk | h [ H, k [ K} is a subgroup of G. First note, that e 5 ee belongs to HK because e is in both H and K. Now suppose, that a and b are in HK. Then by definition of H there are elements h1,, h2 [ H and k1, k2 [ K such that a 5 h1k1 and b 5 h2k2. We must prove, that ab [ HK and a21 [ HK. Observe that because G is Abelian and H, and K are subgroups of G, we have ab 5 h1k1h2k2 5 (h1h2)(k1k2) [ HK., Likewise, a21 5 (h1k1)21 5 k121h121 5 h121k121 [ HK., How do you prove that a subset of a group is not a subgroup? Here, are three possible ways, any one of which guarantees that the subset is, not a subgroup:, 1. Show that the identity is not in the set., 2. Exhibit an element of the set whose inverse is not in the set., 3. Exhibit two elements of the set whose product is not in the set., EXAMPLE 8 Let G be the group of nonzero real numbers under, multiplication, H 5 {x [ G | x 5 1 or x is irrational} and K 5, {x [ G | x $ 1}. Then H is not a subgroup of G, since 22 [ H, but 22 ? 22 5 2 o H. Also, K is not a subgroup, since 2 [ K but, 221 o K., When dealing with finite groups, it is easier to use the following, subgroup test., Theorem 3.3 Finite Subgroup Test, Let H be a nonempty finite subset of a group G. If H is closed under, the operation of G, then H is a subgroup of G., , PROOF In view of Theorem 3.2, we need only prove that a21 [ H, whenever a [ H. If a 5 e, then a21 5 a and we are done. If a ∞ e,, consider the sequence a, a2, . . . . By closure, all of these elements, belong to H. Since H is finite, not all of these elements are distinct. Say, ai 5 aj and i . j. Then, ai2j 5 e; and since a ∞ e, i 2 j . 1. Thus,, aai2j21 5 ai2j 5 e and, therefore, ai2j21 5 a21. But i 2 j 2 1 $ 1, implies ai2j21 [ H and we are done., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 65, , Examples of Subgroups, The proofs of the next few theorems show how our subgroup tests, work. We first introduce an important notation. For any element a from, a group, we let kal denote the set {an | n [ Z}. In particular, observe, that the exponents of a include all negative integers as well as 0 and the, positive integers (a0 is defined to be the identity)., Theorem 3.4 kal Is a Subgroup, Let G be a group, and let a be any element of G. Then, kal is a subgroup of G., , PROOF Since a [ kal, kal is not empty. Let an, am [ kal. Then,, an(am)21 5 an2m [ kal; so, by Theorem 3.1, kal is a subgroup of G., The subgroup kal is called the cyclic subgroup of G generated by a. In, the case that G 5 kal, we say that G is cyclic and a is a generator of G., (A cyclic group may have many generators.) Notice that although the, list . . . , a22, a21, a0, a1, a2, . . . has infinitely many entries, the set, {an | n [ Z} might have only finitely many elements. Also note that,, since aia j 5 ai1j 5 a j1i 5 a j a i , every cyclic group is Abelian., EXAMPLE 9 In U(10), k3l 5 {3, 9, 7, 1} 5 U(10), for 31 5 3,, 32 5 9, 33 5 7, 34 5 1, 35 5 34 ? 3 5 1 ? 3, 36 5 34 ? 32 5 9, . . . ; 321 5 7, (since 3 ? 7 5 1), 322 5 9, 323 5 3, 324 5 1, 325 5 324 ? 321 5, 1 ? 7, 326 5 324 ? 322 5 1 ? 9 5 9, . . . ., EXAMPLE 10 In Z10, k2l 5 {2, 4, 6, 8, 0}. Remember, an means na, when the operation is addition., EXAMPLE 11 In Z, k21l 5 Z. Here each entry in the list . . . ,, 22(21), 21(21), 0(21), 1(21), 2(21), . . . represents a distinct group, element., EXAMPLE 12 In Dn, the dihedral group of order 2n, let R denote a, rotation of 360/n degrees. Then,, Rn 5 R360° 5 e,, , Rn11 5 R,, , Rn12 5 R2, . . . ., , Similarly, R21 5 Rn21, R22 5 Rn22, . . . , so that kRl 5 {e, R, . . . ,, Rn21}. We see, then, that the powers of R “cycle back” periodically, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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66, , Groups, , with period n. Visually, raising R to successive positive powers is the, same as moving counterclockwise around the following circle one, node at a time, whereas raising R to successive negative powers is the, same as moving around the circle clockwise one node at a time., Rn 5 e, Rn11 5 R, , Rn12 5 R2, , R21 5 Rn21, , R22 5 Rn22, , In Chapter 4 we will show that |kal| 5 |a|; that is, the order of the, subgroup generated by a is the order of a itself. (Actually, the definition, of |a| was chosen to ensure the validity of this equation.), For any element a of a group G, it is useful to think of kal as the, smallest subgroup of G containing a. This notion can be extended to, any collection S of elements from a group G by defining kSl as the, subgroup of G with the property that kSl contains S and if H is any, subgroup of G containing S, then H also contains kSl Thus, kSl is the, smallest subgroup of G that contains S. The set kSl is called the subgroup generated by S. We illustrate this concept in the next example., The verifications are left to the reader (Exercise 40)., EXAMPLE 13 In Z20, k8,14l 5 {0, 2, 4,…, 18} 5 k2l; in Z, k8, 13l 5, Z; in D4, kH, Vl 5 {H, H2, V, HV} 5 {R0, R180, H, V}; in D4, kR90, Vl 5, {R90, R902, R903, R904, V, R90V, R902V, R903V} 5 D4; in C*, the group of, nonzero complex numbers under multiplication, k1, il 5 {1, 21, i, –i} 5, kil; in C, the group of complex numbers under addition, k1, il 5 {a 1 bi, | a, b [ Z} (This group is called the “Gaussian integers”); in R, the, group of real numbers under addition, k2, p, 22l 5 {2a 1 bp 1 c22, | a, b, c [ Z}; in a group in which a, b, c, and d commute, ka, b, c, dl 5, {aqbrcsdt | q, r, s, t [ Z}., We next consider one of the most important subgroups., Definition Center of a Group, , The center, Z(G), of a group G is the subset of elements in G that, commute with every element of G. In symbols,, , Z(G) 5 {a [ G | ax 5 xa for all x in G}., [The notation Z(G) comes from the fact that the German word for, center is Zentrum. The term was coined by J. A. de Séguier in 1904.], , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 67, , Theorem 3.5 Center Is a Subgroup, The center of a group G is a subgroup of G., , PROOF For variety, we shall use Theorem 3.2 to prove this result., Clearly, e [ Z(G), so Z(G) is nonempty. Now, suppose a, b [ Z(G)., Then (ab)x 5 a(bx) 5 a(xb) 5 (ax)b 5 (xa)b 5 x(ab) for all x in G;, and, therefore, ab [ Z(G)., Next, assume that a [ Z(G). Then we have ax 5 xa for all x in G., What we want is a21x 5 xa21 for all x in G. Informally, all we need do, to obtain the second equation from the first one is simultaneously to, bring the a’s across the equals sign:, ax 5 xa, becomes xa21 5 a21x. (Be careful here; groups need not be commutative. The a on the left comes across as a21 on the left, and the a on the, right comes across as a21 on the right.) Formally, the desired equation, can be obtained from the original one by multiplying it on the left and, right by a21, like so:, a21(ax)a21 5 a21(xa)a21,, (a21a)xa21 5 a21x(aa21),, exa21 5 a21xe,, xa21 5 a21x., This shows that a21 [ Z(G) whenever a is., For practice, let’s determine the centers of the dihedral groups., EXAMPLE 14 For n $ 3,, Z1Dn 2 e, , 5 R0, R1806 when n is even,, 5 R5 06, when n is odd., , To verify this, first observe that since every rotation in Dn is a power, of R360/n, rotations commute with rotations. We now investigate when a, rotation commutes with a reflection. Let R be any rotation in Dn and let, F be any reflection in Dn. Observe that since RF is a reflection we have, RF 5 (RF )21 5 F21 R21 5 FR21. Thus, it follows that R and F commute, if and only if FR 5 RF 5 FR21. By cancellation, this holds if and only, if R 5 R21. But R 5 R21 only when R 5 R0 or R 5 R180, and R180 is in, Dn only when n is even. So, we have proved that Z(Dn) 5 {R0} when n, is odd and Z(Dn) 5 {R0, R180} when n is even., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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68, , Groups, , Although an element from a non-Abelian group does not necessarily, commute with every element of the group, there are always some, elements with which it will commute. For example, every element a, commutes with all powers of a. This observation prompts the next definition and theorem., Definition Centralizer of a in G, , Let a be a fixed element of a group G. The centralizer of a in G, C(a), is, the set of all elements in G that commute with a. In symbols, C(a) 5, {g [ G | ga 5 ag}., , EXAMPLE 15 In D4, we have the following centralizers:, C(R0) 5 D4 5 C(R180),, C(R90) 5 {R0, R90, R180, R270} 5 C(R270),, C(H) 5 {R0, H, R180, V} 5 C(V),, C(D) 5 {R0, D, R180, D9} 5 C(D9)., Notice that each of the centralizers in Example 15 is actually a subgroup of D4. The next theorem shows that this was not a coincidence., Theorem 3.6 C(a) Is a Subgroup, For each a in a group G, the centralizer of a is a subgroup of G., , PROOF A proof similar to that of Theorem 3.5 is left to the reader to, supply (Exercise 41)., Notice that for every element a of a group G, Z(G) # C(a). Also,, observe that G is Abelian if and only if C(a) 5 G for all a in G., , Exercises, The purpose of proof is to understand, not to verify., arnold ross, , 1. For each group in the following list, find the order of the group, and the order of each element in the group. What relation do you, see between the orders of the elements of a group and the order of, the group?, Z12,, , U(10),, , U(12),, , U(20),, , D4, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 69, , 2. Let Q be the group of rational numbers under addition and let Q*, be the group of nonzero rational numbers under multiplication., In Q, list the elements in k12l. In Q*, list the elements in k12l., 3. Let Q and Q* be as in Exercise 2. Find the order of each element in, Q and in Q*., 4. Prove that in any group, an element and its inverse have the same, order., 5. Without actually computing the orders, explain why the two elements in each of the following pairs of elements from Z 30 must, have the same order: {2, 28}, {8, 22}. Do the same for the following pairs of elements from U(15): {2, 8}, {7, 13}., 6. In the group Z12, find |a|, |b|, and |a 1 b| for each case., a. a 5 6, b 5 2, b. a 5 3, b 5 8, c. a 5 5, b 5 4, Do you see any relationship between |a|, |b|, and |a 1 b|?, 7. If a, b, and c are group elements and |a| 5 6, |b| 5 7, express, (a4c22b4)21 without using negative exponents., 8. What can you say about a subgroup of D3 that contains R240 and a, reflection F? What can you say about a subgroup of D3 that contains two reflections?, 9. What can you say about a subgroup of D4 that contains R270 and a, reflection? What can you say about a subgroup of D4 that contains, H and D? What can you say about a subgroup of D4 that contains H, and V?, 10. How many subgroups of order 4 does D4 have?, 11. Determine all elements of finite order in R*, the group of nonzero, real numbers under multiplication., 12. If a and b are group elements and ab Z ba, prove that aba Z e., 13. Suppose that H is a nonempty subset of a group G that is closed, under the group operation and has the property that if a is not in H, then a21 is not in H. Is H a subgroup?, 14. Let G be the group of polynomials under addition with coefficients, from Z10. Find the orders of f (x) 5 7x2 1 5x 1 4, g(x) 5 4x2 1 8x, 1 6, and f (x) 1 g(x) 5 x2 1 3x. If h(x) 5 anx n 1 an21x n21 1 …, 1 a0 belongs to G, determine |h(x)| given that gcd (a1, a2, …, an) 5 1;, gcd(a1, a2, …, an) 5 2; gcd(a1, a2, …, an) 5 5; and gcd(a1, a2, …,, an) 5 10., 15. If a is an element of a group G and |a| 5 7, show that a is the cube, of some element of G., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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70, , Groups, , 16. Suppose that H is a nonempty subset of a group G with the property that if a and b belong to H then a21b21 belongs to H. Prove or, disprove that this is enough to guarantee that H is a subgroup of G., 17. Prove that if an Abelian group has more than three elements of, order 2, then it has at least 7 elements of order 2. Find an example, that shows this is not true for non-Abelian groups., 18. Suppose that a is a group element and a6 5 e. What are the possibilities for |a|? Provide reasons for your answer., 19. If a is a group element and a has infinite order, prove that am ? an, when m ? n., 20. Let x belong to a group. If x2 2 e and x6 5 e, prove that x4 2 e and, x5 ∞ e. What can we say about the order of x?, 21. Show that if a is an element of a group G, then |a| # |G|., 22. Show that U(14) 5 k3l 5 k5l. [Hence, U(14) is cyclic.] Is, U(14) 5 k11l?, 23. Show that U(20) 2 kkl for any k in U(20). [Hence, U(20) is not, cyclic.], 24. Suppose n is an even positive integer and H is a subgroup of Zn., Prove that either every member of H is even or exactly half of the, members of H are even., 25. Prove that for every subgroup of Dn, either every member of the, subgroup is a rotation or exactly half of the members are rotations., 26. Prove that a group with two elements of order 2 that commute must, have a subgroup of order 4., 27. For every even integer n, show that Dn has a subgroup of order 4., 28. Suppose that H is a proper subgroup of Z under addition and H, contains 18, 30, and 40. Determine H., 29. Suppose that H is a proper subgroup of Z under addition and that H, contains 12, 30, and 54. What are the possibilities for H?, 30. Prove that the dihedral group of order 6 does not have a subgroup, of order 4., 31. For each divisor k . 1 of n, let Uk(n) 5 {x [ U(n) | x mod k 5 1}., [For example, U3(21) 5 {1, 4, 10, 13, 16, 19} and U7(21) 5 {1, 8}.], List the elements of U4(20), U5(20), U5(30), and U10(30). Prove that, Uk(n) is a subgroup of U(n). Let H 5 {x [ U(10) | x mod 3 5 1}. Is, H a subgroup of U(10)? (This exercise is referred to in Chapter 8.), 32. If H and K are subgroups of G, show that H > K is a subgroup of, G. (Can you see that the same proof shows that the intersection, of any number of subgroups of G, finite or infinite, is again a, subgroup of G?), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 71, , 33. Let G be a group. Show that Z(G) 5 >a[GC(a). [This means the, intersection of all subgroups of the form C(a).], 34. Let G be a group, and let a [ G. Prove that C(a) 5 C(a21)., 35. For any group element a and any integer k, show that C(a) # C(a k)., Use this fact to complete the following statement: “In a group, if x, commutes with a, then . . . .” Is the converse true?, 36. Complete the partial Cayley group table given below., , 1, 2, 3, 4, 5, 6, 7, 8, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 1, 2, 3, 4, 5, 6, 7, 8, , 2, 1, 4, 3, 6, 5, 8, 7, , 3, 4, 2, 1, 8, 7, 5, 6, , 4, 3, 1, 2, 7, 8, 6, 5, , 5, 6, 7, 8, 1, , 6, 5, 8, 7, , 7, 8, 6, 5, , 8, 7, 5, 6, , 1, 1, 1, , 37. Suppose G is the group defined by the following Cayley table., 1, 2, 3, 4, 5, 6, 7, 8, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 1, 2, 3, 4, 5, 6, 7, 8, , 2, 1, 4, 3, 6, 5, 8, 7, , 3, 8, 5, 2, 7, 4, 1, 6, , 4, 7, 6, 1, 8, 3, 2, 5, , 5, 6, 7, 8, 1, 2, 3, 4, , 6, 5, 8, 7, 2, 1, 4, 3, , 7, 4, 1, 6, 3, 8, 5, 2, , 8, 3, 2, 5, 4, 7, 6, 1, , a. Find the centralizer of each member of G., b. Find Z(G)., c. Find the order of each element of G. How are these orders arithmetically related to the order of the group?, 38. If a and b are distinct group elements, prove that either a2 2 b2 or, a 3 2 b 3., 39. Let S be a subset of a group and let H be the intersection of all subgroups of G that contain S., a. Prove that kSl 5 H., b. If S is nonempty, prove that kSl 5 {s1n1 s2n2 … smnm | m $ 1, si [ S,, ni [ Z}. (The si terms need not be distinct.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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72, , Groups, , 40. In the group Z, find, a. k8, 14l;, b. k8, 13l;, c. k6, 15l;, d. km, nl;, e. k12, 18, 45l., In each part, find an integer k such that the subgroup is kkl., 41. Prove Theorem 3.6., 42. If H is a subgroup of G, then by the centralizer C(H) of H we mean, the set {x [ G | xh 5 hx for all h [ H}. Prove that C(H) is a subgroup of G., 43. Must the centralizer of an element of a group be Abelian?, 44. Must the center of a group be Abelian?, 45. Let G be an Abelian group with identity e and let n be some fixed integer. Prove that the set of all elements of G that satisfy the equation, xn 5 e is a subgroup of G. Give an example of a group G in which the, set of all elements of G that satisfy the equation x2 5 e does not form, a subgroup of G. (This exercise is referred to in Chapter 11.), 46. Suppose a belongs to a group and |a| 5 5. Prove that C(a) 5 C(a3)., Find an element a from some group such that |a| 5 6 and C(a) ∞, C(a3)., 47. Let G be the set of all polynomials with coefficients from the set, {0, 1, 2, 3}. We can make G a group under addition by adding the, polynomials in the usual way, except that we use modulo 4 to combine the coefficients. With this group operation, determine the orders of the elements of G. Determine a necessary and sufficient, condition for an element of G to have order 2., 48. In each case, find elements a and b from a group such that |a| 5, |b| 5 2., a. |ab| 5 3, b. |ab| 5 4, c. |ab| 5 5, Can you see any relationship among |a|, |b|, and |ab|?, 49. Suppose a group contains elements a and b such that |a| 5 4,, |b| 5 2, and a3b 5 ba. Find |ab|., 50. Suppose a and b are group elements such that |a| 5 2, b ? e, and, aba 5 b 2. Determine |b|., 51. Let a be a group element of order n, and suppose that d is a positive divisor of n. Prove that |ad | 5 n/d., 0 1, 0, 1, d and B c, d from, 1 0, 1 1, SL(2, R). Find |A|, |B|, and |AB|. Does your answer surprise you?, , 52. Consider the elements A c, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 53. Consider the element A c, , 73, , 1 1, d in SL(2, R). What is the order of, 0 1, , A? If we view A c, , 1 1, d as a member of SL(2, Zp) (p is a prime),, 0 1, what is the order of A?, 54. For any positive integer n and any angle u, show that in the group, SL(2, R),, c, , n, , cos nu sin nu, cos u sin u, d., d c, sin nu, cos nu, sin u, cos u, , Use this formula to find the order of, cos 60 sin 60, cos 22 sin 22, d., c, d and c, sin 22 cos 22, sin 60 cos 60, (Geometrically, c, 55., 56., , 57., 58., 59., 60., 61., , 62., , 63., , cos u sin u, d represents a rotation of the plane, sin u, cos u, , u degrees.), Let G be the symmetry group of a circle. Show that G has elements, of every finite order as well as elements of infinite order., Let x belong to a group and |x| 5 6. Find |x2|, |x3|, |x4|, and |x5|. Let, y belong to a group and |y| 5 9. Find |y i| for i 5 2, 3, . . . , 8. Do, these examples suggest any relationship between the order of the, power of an element and the order of the element?, D4 has seven cyclic subgroups. List them., U(15) has six cyclic subgroups. List them., Prove that a group of even order must have an element of order 2., Suppose G is a group that has exactly eight elements of order 3., How many subgroups of order 3 does G have?, Let H be a subgroup of a finite group G. Suppose that g belongs to, G and n is the smallest positive integer such that gn [ H. Prove that, n divides |g|., Compute the orders of the following groups., a. U(3), U(4), U(12), b. U(5), U(7), U(35), c. U(4), U(5), U(20), d. U(3), U(5), U(15), On the basis of your answers, make a conjecture about the relationship among |U(r)|, |U(s)|, and |U(rs)|., Let R* be the group of nonzero real numbers under multiplication, and let H 5 {x [ R* | x2 is rational}. Prove that H is a subgroup of, R*. Can the exponent 2 be replaced by any positive integer and still, have H be a subgroup?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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74, , Groups, , 64. Compute |U(4)|, |U(10)|, and |U(40)|. Do these groups provide a, counterexample to your answer to Exercise 62? If so, revise your, conjecture., 65. Find a cyclic subgroup of order 4 in U(40)., 66. Find a noncyclic subgroup of order 4 in U(40)., 67. Let G 5 e c, , a b, d ` a, b, c, d [ Z f under addition. Let H 5, c d, , ec, , 68., 69., 70., , 71., , a b, d [ G | a bc d 0f . Prove that H is a subgroup of G., c d, What if 0 is replaced by 1?, Let H 5 {A [ GL(2, R) | det A is an integer power of 2}. Show that, H is a subgroup of GL(2, R)., Let H be a subgroup of R under addition. Let K 5 {2a | a [ H}., Prove that K is a subgroup of R* under multiplication., Let G be a group of functions from R to R*, where the operation, of G is multiplication of functions. Let H 5 { f [ G | f(2) 5 1}., Prove that H is a subgroup of G. Can 2 be replaced by any real, number?, a 0, d ` a and b are nonzero inteLet G 5 GL(2, R) and H e c, 0 b, gers f under the operation of matrix multiplication. Prove or, , disprove that H is a subgroup of GL(2, R)., 72. Let H 5 {a 1 bi | a, b [ R, ab $ 0}. Prove or disprove that H is a, subgroup of C under addition., 73. Let H 5 {a 1 bi | a, b [ R, a2 1 b2 5 1}. Prove or disprove that, H is a subgroup of C* under multiplication. Describe the elements, of H geometrically., 74. Let G be a finite Abelian group and let a and b belong to G. Prove, that the set Ka, bL 5 {ai b j | i, j [ Z} is a subgroup of G. What can, you say about |Ka, bL| in terms of |a| and |b|?, 75. Let H be a subgroup of a group G. Prove that the set HZ(G) 5, {hz | h [ H, z [ Z(G)} is a subgroup of G. This exercise is referred, to in this chapter., 76. Let G be a group and H a subgroup. For any element g of G, define, gH 5 {gh | h [ H}. If G is Abelian and g has order 2, show that the, set K 5 H < gH is a subgroup of G. Is your proof valid if we drop, the assumption that G is Abelian and let K = Z(G) < gZ(G)?, 77. Let a belong to a group and |a| 5 m. If n is relatively prime to m,, show that a can be written as the nth power of some element in the, group., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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3 | Finite Groups; Subgroups, , 75, , 78. Let F be a reflection in the dihedral group Dn and R a rotation in, Dn. Determine C(F) when n is odd. Determine C(F) when n is, even. Determine C(R)., 79. Let G 5 GL(2, R)., 1 1, d b., a. Find C a c, 1 0, 0 1, b. Find C a c, d b., 1 0, c. Find Z(G)., 80. Let G be a finite group with more than one element. Show that G, has an element of prime order., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Suggested Readings, Ruth Berger, “Hidden Group Structure,” Mathematics Magazine 78, (2005): 45–48., In this note, the author investigates groups obtained from U(n) by multiplying each element by some k in U(n). Such groups have identities, that are not obvious., J. Gallian and M. Reid, “Abelian Forcing Sets,” American Mathematical, Monthly 100 (1993): 580–582., A set S is called Abelian forcing if the only groups that satisfy (ab)n 5, anbn for all a and b in the group and all n in S are the Abelian ones., This paper characterizes the Abelian forcing sets. It can be downloaded, at http://www.d.umn.edu/~jgallian/forcing.pdf, Gina Kolata, “Perfect Shuffles and Their Relation to Math,” Science 216, (1982): 505–506., This is a delightful nontechnical article that discusses how group theory and computers were used to solve a difficult problem about shuffling a deck of cards. Serious work on the problem was begun by an, undergraduate student as part of a programming course., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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76, , Groups, , Suggested Software, Allen Hibbard and Kenneth Levasseur, Exploring Abstract Algebra with, Mathematica, New York: Springer-Verlag, 1999., This book, intended as a supplement for a course in abstract algebra,, consists of 14 group labs, 13 ring labs, and documentation for the, Abstract Algebra software on which the labs are based. The software uses, the Mathematica language, and only a basic familiarity with the program, is required. The software can be freely downloaded at http://www, .central.edu/eaam/ and can be used independently of the book., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 Cyclic Groups, The notion of a “group,” viewed only 30 years ago as the epitome of, sophistication, is today one of the mathematical concepts most widely, used in physics, chemistry, biochemistry, and mathematics itself., alexey sosinsky, 1991, , Properties of Cyclic Groups, Recall from Chapter 3 that a group G is called cyclic if there is an element a in G such that G 5 {an | n [ Z}. Such an element a is called a, generator of G. In view of the notation introduced in the preceding, chapter, we may indicate that G is a cyclic group generated by a by, writing G 5 kal., In this chapter, we examine cyclic groups in detail and determine, their important characteristics. We begin with a few examples., EXAMPLE 1 The set of integers Z under ordinary addition is cyclic., Both 1 and 21 are generators. (Recall that, when the operation is addition, 1n is interpreted as, 1 1 1 1 ??? 1 1, n terms, , when n is positive and as, (21) 1 (21) 1 ? ? ? 1 (21), |n| terms, , when n is negative.), EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a, cyclic group under addition modulo n. Again, 1 and 21 5 n 2 1 are, generators., , 77, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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78, , Groups, , Unlike Z, which has only two generators, Zn may have many generators (depending on which n we are given)., EXAMPLE 3 Z8 5 k1l 5 k3l 5 k5l 5 k7l. To verify, for instance, that, Z8 5 k3l, we note that k3l 5 {3, 3 1 3, 3 1 3 1 3, . . .} is the set {3, 6,, 1, 4, 7, 2, 5, 0} 5 Z8. Thus, 3 is a generator of Z8. On the other hand, 2, is not a generator, since k2l 5 {0, 2, 4, 6} 2 Z8., EXAMPLE 4 (See Example 11 in Chapter 2.), U(10) 5 {1, 3, 7, 9} 5 {30, 31, 33, 32} 5 k3l. Also, {1, 3, 7, 9} 5, {70, 73, 71, 72} 5 k7l. So both 3 and 7 are generators for U(10)., Quite often in mathematics, a “nonexample” is as helpful in understanding a concept as an example. With regard to cyclic groups, U(8), serves this purpose; that is, U(8) is not a cyclic group. How can we, verify this? Well, note that U(8) 5 {1, 3, 5, 7}. But, k1l 5 {1},, k3l 5 {3, 1},, k5l 5 {5, 1},, k7l 5 {7, 1},, so U(8) 2 kal for any a in U(8)., With these examples under our belts, we are now ready to tackle, cyclic groups in an abstract way and state their key properties., Theorem 4.1 Criterion for ai 5 a j, Let G be a group, and let a belong to G. If a has infinite order, then, ai 5 a j if and only if i 5 j. If a has finite order, say, n, then kal 5, {e, a, a2, . . . , an–1} and ai 5 a j if and only if n divides i – j., , PROOF If a has infinite order, there is no nonzero n such that an is the, identity. Since ai 5 a j implies ai2j 5 e, we must have i 2 j 5 0, and the, first statement of the theorem is proved., Now assume that |a| 5 n. We will prove that kal 5 {e, a, . . . , an21}., Certainly, the elements e, a, . . . , an21 are in kal., Now, suppose that ak is an arbitrary member of kal. By the division, algorithm, there exist integers q and r such that, k 5 qn 1 r, , with, , 0 # r , n., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 79, , Then ak 5 aqn1r 5 aqnar 5 (an)qar 5 ear 5 ar, so that ak [ {e, a,, a2, . . . , an21}. This proves that kal 5 {e, a, a2, . . . , an21}., Next, we assume that ai 5 a j and prove that n divides i 2 j. We, begin by observing that ai 5 a j implies ai2j 5 e. Again, by the division, algorithm, there are integers q and r such that, i 2 j 5 qn 1 r, , with, , 0 # r , n., , Then ai2j 5 aqn1r, and therefore e 5 ai2j 5 aqn1r 5 (an)qar 5 eqar 5, ear 5 ar. Since n is the least positive integer such that an is the identity,, we must have r 5 0, so that n divides i 2 j., Conversely, if i 2 j 5 nq, then a i2j 5 a nq 5 e q 5 e, so that, i, a 5 a j., Theorem 4.1 reveals the reason for the dual use of the notation and, terminology for the order of an element and the order of a group., Corollary 1 |a| 5 |kal|, For any group element a, |a| 5 |kal|., , One special case of Theorem 4.1 occurs so often that it deserves, singling out., Corollary 2 a k 5 e Implies That |a| Divides k, Let G be a group and let a be an element of order n in G. If a k 5 e,, then n divides k., , PROOF Since ak 5 e 5 a0, we know by Theorem 4.1 that n divides, k 2 0., Theorem 4.1 and its corollaries for the case |a| 5 6 are illustrated in, Figure 4.1., What is important about Theorem 4.1 in the finite case is that it says, that multiplication in kal is essentially done by addition modulo n. That, is, if (i 1 j) mod n 5 k, then aia j 5 ak. Thus, no matter what group G, is, or how the element a is chosen, multiplication in kal works the same, as addition in Zn whenever |a| 5 n. Similarly, if a has infinite order,, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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80, , Groups, , ... a – 6 = a 0 = a 6 ..., ... a – 5 = a = a 7..., , ... a –1 = a 5 = a 11..., , ... a – 4 = a 2 = a 8 ..., , ... a –2 = a 4 = a 10..., ... a – 3 = a 3 = a 9..., , Figure 4.1, , then multiplication in kal works the same as addition in Z, since aia j, 5 ai1j and no modular arithmetic is done., For these reasons, the cyclic groups Zn and Z serve as prototypes for, all cyclic groups, and algebraists say that there is essentially only one, cyclic group of each order. What is meant by this is that, although, there may be many different sets of the form {an | n [ Z}, there is, essentially only one way to operate on these sets. Algebraists do not, really care what the elements of a set are; they care only about the, algebraic properties of the set—that is, the ways in which the elements, of a set can be combined. We will return to this theme in the chapter, on isomorphisms (Chapter 6)., The next theorem provides a simple method for computing |ak|, knowing only |a|, and its first corollary provides a simple way to tell, when kail 5 ka j l., Theorem 4.2 kakl 5 kagcd(n,k)l and |ak| 5 n/gcd(n, k), Let a be an element of order n in a group and let k be a positive, integer. Then kakl 5 kagcd(n,k)l and |ak| 5 n/gcd(n, k)., , PROOF To simplify the notation, let d 5 gcd(n, k) and let k 5 dr., Since ak 5 (ad)r, we have by closure that kakl # kadl. By Theorem 0.2, (the gcd theorem), there are integers s and t such that d 5 ns 1 kt. So,, ad 5 ans1kt 5 ansakt 5 (an)s(ak)t 5 e(ak)t 5 (ak)t [ kakl. This, proves kadl # kakl. So, we have verified that kakl 5 kagcd(n,k)l., We prove the second part of the theorem by showing first that |ad| 5, n/d for any divisor d of n. Clearly, (ad)n/d 5 an 5 e, so that |ad| # n/d. On, the other hand, if i is a positive integer less than n/d, then (ad)i 2 e by definition of |a|. We now apply this fact with d 5 gcd(n, k) to obtain |ak| 5, |kakl| 5 |kagcd(n,k)l| 5 |agcd(n,k)| 5 n/gcd(n, k)., The advantage of Theorem 4.2 is that it allows us to replace one, generator of a cyclic subgroup with a more convenient one. For example,, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 81, , if |a| 5 30, we have ka26l 5 ka2l, ka23l 5 kal, ka22l 5 ka2l, ka21l 5 ka3l., From this we can easily see that |a23| 5 30 and |a22| 5 15. Moreover, if, one wants to list the elements of, say, ka21l, it is easier to list the elements, of ka3l instead. (Try it doing it both ways!)., Theorem 4.2 establishes an important relationship between the order, of an element in a finite cyclic group and the order of the group., Corollary 1 Orders of Elements in Finite Cyclic Groups, In a finite cyclic group, the order of an element divides the order, of the group., , Corollary 2 Criterion for kail ⫽ ka j l and |ai | ⫽ |a j |, Let |a| 5 n. Then kail 5 ka jl if and only if gcd(n, i) 5 gcd(n, j),, and |ai| 5 |a j| if and only if gcd(n, i) 5 gcd(n, j) ., , PROOF Theorem 4.2 shows that kail 5 kagcd(n,i)l and ka jl 5 kagcd(n,j)l,, so that the proof reduces to proving that kagcd(n,i)l 5 kagcd(n,j)l if and, only if gcd(n, i) 5 gcd(n, j). Certainly, gcd(n, i) 5 gcd(n, j) implies, that kagcd(n,i)l 5 kagcd(n, j)l. On the other hand, kagcd(n,i)l 5 kagcd(n,j)l, implies that |agcd(n,i)| 5 |agcd(n,j)|, so that by the second conclusion of, Theorem 4.2, we have n/gcd(n, i) 5 n/gcd(n, j), and therefore gcd(n, i) 5, gcd(n, j)., The second part of the corollary follows from the first part and, Corollary 1 of Theorem 4.1., The next two corollaries are important special cases of the preceding, corollary., Corollary 3 Generators of Finite Cyclic Groups, Let |a| 5 n. Then kal 5 ka jl if and only if gcd(n, j) 5 1, and, |a| 5 |ka jl| if and only if gcd(n, j) 5 1., , Corollary 4 Generators of Zn, An integer k in Zn is a generator of Zn if and only if gcd(n, k) 5 1., , The value of Corollary 3 is that once one generator of a cyclic group has, been found, all generators of the cyclic group can easily be determined., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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82, , Groups, , For example, consider the subgroup of all rotations in D6. Clearly, one, generator is R60. And, since |R60| 5 6, we see by Corollary 3 that the only, other generator is (R60)5 5 R300. Of course, we could have readily deduced this information without the aid of Corollary 3 by direct calculations. So, to illustrate the real power of Corollary 3, let us use it to find all, generators of the cyclic group U(50). First, note that direct computations, show that |U(50)| 5 20 and that 3 is one of its generators. Thus, in view of, Corollary 3, the complete list of generators for U(50) is, 3 mod 50 5 3,, 33 mod 50 5 27,, 37 mod 50 5 37,, 39 mod 50 5 33,, , 311 mod 50 5 47,, 313 mod 50 5 23,, 317 mod 50 5 13,, 319 mod 50 5 17., , Admittedly, we had to do some arithmetic here, but it certainly entailed, much less work than finding all the generators by simply determining, the order of each element of U(50) one by one., The reader should keep in mind that Theorem 4.2 and its corollaries, apply only to elements of finite order., , Classification of Subgroups, of Cyclic Groups, The next theorem tells us how many subgroups a finite cyclic group has, and how to find them., Theorem 4.3 Fundamental Theorem of Cyclic Groups, Every subgroup of a cyclic group is cyclic. Moreover, if |kal| 5 n,, then the order of any subgroup of kal is a divisor of n; and, for each, positive divisor k of n, the group kal has exactly one subgroup of, order k—namely, kan/k l., , Before we prove this theorem, let’s see what it means. Understanding what a theorem means is a prerequisite to understanding its proof., Suppose G 5 kal and G has order 30. The first and second parts of the, theorem say that if H is any subgroup of G, then H has the form ka30/kl for, some k that is a divisor of 30. The third part of the theorem says that G, has one subgroup of each of the orders 1, 2, 3, 5, 6, 10, 15, and 30—and, no others. The proof will also show how to find these subgroups., PROOF Let G 5 kal and suppose that H is a subgroup of G. We must, show that H is cyclic. If it consists of the identity alone, then clearly H is, cyclic. So we may assume that H 2 {e}. We now claim that H contains, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 83, , an element of the form at, where t is positive. Since G 5 kal, every, element of H has the form at; and when at belongs to H with t , 0, then, a2t belongs to H also and 2t is positive. Thus, our claim is verified. Now, let m be the least positive integer such that am [ H. By closure, kaml # H., We next claim that H 5 kaml. To prove this claim, it suffices to let b be an, arbitrary member of H and show that b is in kaml. Since b [ G 5 kal, we, have b 5 ak for some k. Now, apply the division algorithm to k and m to, obtain integers q and r such that k 5 mq 1 r where 0 # r , m. Then ak 5, amq1r 5 amqar, so that ar 5 a2mqak. Since ak 5 b [ H and a2mq 5, (am)2q is in H also, ar [ H. But, m is the least positive integer such that, am [ H, and 0 # r , m, so r must be 0. Therefore, b 5 ak 5 amq 5, (am)q [ kaml. This proves the assertion of the theorem that every subgroup of a cyclic group is cyclic., To prove the next portion of the theorem, suppose that |kal| 5 n and, H is any subgroup of kal. We have already shown that H 5 kaml, where, m is the least positive integer such that am [ H. Using e 5 b 5 an as in, the preceding paragraph, we have n 5 mq., Finally, let k be any positive divisor of n. We will show that kan/kl is, the one and only subgroup of kal of order k. From Theorem 4.2, we see, that kan/kl has order n/gcd(n, n/k) 5 n/(n/k) 5 k. Now let H be any, subgroup of kal of order k. We have already shown above that H 5 kaml,, where m is a divisor of n. Then m 5 gcd(n, m) and k 5 |am| 5 |agcd(n,m)| 5, n/gcd (n, m) 5 n/m. Thus, m 5 n/k and H 5 kan/kl., Returning for a moment to our discussion of the cyclic group kal,, where a has order 30, we may conclude from Theorem 4.3 that the subgroups of kal are precisely those of the form kaml, where m is a divisor, of 30. Moreover, if k is a divisor of 30, the subgroup of order k is, ka30/kl. So the list of subgroups of kal is:, kal 5 {e, a, a2, . . . , a29}, ka2l 5 {e, a2, a4, . . . , a28}, ka3l 5 {e, a3, a6, . . . , a27}, ka5l 5 {e, a5, a10, a15, a20, a25}, ka6l 5 {e, a6, a12, a18, a24}, ka10l 5 {e, a10, a20}, ka15l 5 {e, a15}, ka30l 5 {e}, , order 30,, order 15,, order 10,, order 6,, order 5,, order 3,, order 2,, order 1., , In general, if kal has order n and k divides n, then kan/kl is the unique, subgroup of order k., Taking the group in Theorem 4.3 to be Zn and a to be 1, we obtain, the following important special case., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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84, , Groups, , Corollary Subgroups of Zn, For each positive divisor k of n, the set kn/kl is the unique subgroup, of Zn of order k; moreover, these are the only subgroups of Zn., , EXAMPLE 5 The list of subgroups of Z30 is, k1l 5 {0, 1, 2, . . . , 29}, k2l 5 {0, 2, 4, . . . , 28}, k3l 5 {0, 3, 6, . . . , 27}, k5l 5 {0, 5, 10, 15, 20, 25}, k6l 5 {0, 6, 12, 18, 24}, k10l 5 {0, 10, 20}, k15l 5 {0, 15}, k30l 5 {0}, , order 30,, order 15,, order 10,, order 6,, order 5,, order 3,, order 2,, order 1., , Theorems 4.2 and 4.3 provide a simple way to find all the generators, of the subgroups of a finite cyclic group., EXAMPLE 6 To find the generators of the subgroup of order 9 in, Z36, we observe that 36/9 5 4 is one generator. To find the others, we, have from Corollary 3 of Theorem 4.2 that they are all elements of Z36, of the form 4j, where gcd(9, j) 5 1. Thus,, k4 ? 1l 5 k4 ? 2l 5 k4 ? 4l 5 k4 ? 5l 5 k4 ? 7l 5 k4 ? 8l., In the generic case, to find all the subgroups of kal of order 9 where, |a| 5 36, we have, k(a4)1l 5 k(a4)2l 5 k(a4)4l 5 k(a4)5l 5 k(a4)7l 5 k(a4)8l., In particular, note that once you have the generator an/d for the subgroup, of order d where d is a divisor of |a| 5 n, all the generators of kadl have, the form (ad) j where j [ U(d)., By combining Theorems 4.2 and 4.3, we can easily count the number of elements of each order in a finite cyclic group. For convenience,, we introduce an important number-theoretic function called the Euler, phi function. Let f(1) 5 1, and for any integer n . 1, let f(n) denote, the number of positive integers less than n and relatively prime to n., Notice that by definition of the group U(n), |U(n)| 5 f(n). The first 12, values of f(n) are given in Table 4.1., Table 4.1 Values of f(n), n, , 1, , 2 3 4, , 5 6, , 7, , 8, , 9 10, , 11 12, , f(n) 1, , 1 2 2, , 4 2, , 6, , 4, , 6, , 10, , 4, , 4, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 85, , Theorem 4.4 Number of Elements of Each Order in a Cyclic Group, If d is a positive divisor of n, the number of elements of order d in, a cyclic group of order n is f(d)., , PROOF By Theorem 4.3, the group has exactly one subgroup of, order d—call it kal. Then every element of order d also generates the, subgroup kal and, by Corollary 3 of Theorem 4.2, an element a k generates kal if and only if gcd(k, d ) 5 1. The number of such elements is, precisely f(d)., Notice that for a finite cyclic group of order n, the number of elements, of order d for any divisor d of n depends only on d. Thus, Z8, Z640, and, Z80000 each have f(8) 5 4 elements of order 8., Although there is no formula for the number of elements of each, order for arbitrary finite groups, we still can say something important, in this regard., Corollary Number of Elements of Order d in a Finite Group, In a finite group, the number of elements of order d is a multiple, of f(d)., , PROOF If a finite group has no elements of order d, the statement is, true, since f(d) divides 0. Now suppose that a [ G and |a| 5 d. By, Theorem 4.4, we know that kal has f(d) elements of order d. If all, elements of order d in G are in kal, we are done. So, suppose that there, is an element b in G of order d that is not in kal. Then, kbl also has f(d), elements of order d. This means that we have found 2f(d) elements of, order d in G provided that kal and kbl have no elements of order d in, common. If there is an element c of order d that belongs to both kal and, kbl, then we have kal 5 kcl 5 kbl, so that b [ kal, which is a contradiction. Continuing in this fashion, we see that the number of elements of, order d in a finite group is a multiple of f(d)., On its face, the value of Theorem 4.4 and its corollary seem limited, for large values of n, because it is tedious to determine the number of, positive integers less than or equal to n and relatively prime to n, by examining them one by one. However, the following properties of the, f function make computing f (n) simple: For any prime p, f (pn) 5, pn pn1 (see Exercise 85) and for relatively prime m and n, f(mn), 5 f (m) f (n). Thus, f (40) 5 f (8) f (5) 5 4?4 5 16; f (75) 5, f (52)f (3) 5 (25 2 5) ? 2 5 40., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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86, , Groups, , The relationships among the various subgroups of a group can be, illustrated with a subgroup lattice of the group. This is a diagram that includes all the subgroups of the group and connects a subgroup H at one, level to a subgroup K at a higher level with a sequence of line segments, if and only if H is a proper subgroup of K. Although there are many, ways to draw such a diagram, the connections between the subgroups, must be the same. Typically, one attempts to present the diagram in an, eye-pleasing fashion. The lattice diagram for Z30 is shown in Figure 4.2., Notice that k10l is a subgroup of both k2l and k5l, but k6l is not a subgroup of k10l., <1>, , <5>, , < 2>, <3>, <10>, , <15>, , <6>, , <0>, , Figure 4.2 Subgroup lattice of Z30., , The precision of Theorem 4.3 can be appreciated by comparing the, ease with which we are able to identify the subgroups of Z30 with that of, doing the same for, say, U(30) or D30. And these groups have relatively, simple structures among noncyclic groups., We will prove in Chapter 7 that a certain portion of Theorem 4.3, extends to arbitrary finite groups; namely, the order of a subgroup divides the order of the group itself. We will also see, however, that a finite, group need not have exactly one subgroup corresponding to each divisor, of the order of the group. For some divisors, there may be none at all,, whereas for other divisors, there may be many. Indeed, D4, the dihedral, group of order 8, has five subgroups of order 2 and three of order 4., One final remark about the importance of cyclic groups is appropriate. Although cyclic groups constitute a very narrow class of finite, groups, we will see in Chapter 11 that they play the role of building, blocks for all finite Abelian groups in much the same way that primes, are the building blocks for the integers and that chemical elements are, the building blocks for the chemical compounds., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 87, , Exercises, It is not unreasonable to use the hypothesis., arnold ross, , 1. Find all generators of Z6, Z8, and Z20., 2. Suppose that kal, kbl, and kcl are cyclic groups of orders 6, 8, and, 20, respectively. Find all generators of kal, kbl, and kcl., 3. List the elements of the subgroups k20l and k10l in Z30. Let a be a, group element of order 30. List the elements of the subgroups ka20l, and ka10l., 4. List the elements of the subgroups k3l and k15l in Z18. Let a be a, group element of order 18. List the elements of the subgroups ka3l, and ka15l., 5. List the elements of the subgroups k3l and k7l in U(20)., 6. What do Exercises 3, 4, and 5 have in common? Try to make a generalization that includes these three cases., 7. Find an example of a noncyclic group, all of whose proper subgroups are cyclic., 8. Let a be an element of a group and let |a| 5 15. Compute the orders of the following elements of G., a. a3, a6, a9, a12, b. a5, a10, c. a2, a4, a8, a14, 9. How many subgroups does Z20 have? List a generator for each of, these subgroups. Suppose that G 5 kal and |a| 5 20. How many, subgroups does G have? List a generator for each of these subgroups., 10. In Z24, list all generators for the subgroup of order 8. Let G 5 kal, and let |a| 5 24. List all generators for the subgroup of order 8., 11. Let G be a group and let a [ G. Prove that ka21l 5 kal., 12. In Z, find all generators of the subgroup k3l. If a has infinite order,, find all generators of the subgroup ka3l., 13. In Z24, find a generator for k21l > k10l. Suppose that |a| 5 24. Find, a generator for ka21l > ka10l. In general, what is a generator for the, subgroup kaml > kanl?, 14. Suppose that a cyclic group G has exactly three subgroups: G itself,, {e}, and a subgroup of order 7. What is |G|? What can you say if 7, is replaced with p where p is a prime?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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88, , Groups, , 15. Let G be an Abelian group and let H 5 {g [ G | |g| divides 12}., Prove that H is a subgroup of G. Is there anything special about 12, here? Would your proof be valid if 12 were replaced by some other, positive integer? State the general result., 16. Find a collection of distinct subgroups ka1l, ka2l, . . . , kanl of Z240, with the property that ka1l , ka2l , ? ? ? , kanl with n as large as, possible., 17. Complete the following statement: |a| 5 |a2| if and only if |a| . . . ., 18. If a cyclic group has an element of infinite order, how many elements of finite order does it have?, 19. List the cyclic subgroups of U(30)., 20. Suppose that G is an Abelian group of order 35 and every element, of G satisfies the equation x35 5 e. Prove that G is cyclic. Does, your argument work if 35 is replaced with 33?, 21. Let G be a group and let a be an element of G., a. If a12 5 e, what can we say about the order of a?, b. If am 5 e, what can we say about the order of a?, c. Suppose that |G| 5 24 and that G is cyclic. If a8 2 e and a12 2 e,, show that kal 5 G., 22. Prove that a group of order 3 must be cyclic., 23. Let Z denote the group of integers under addition. Is every subgroup of Z cyclic? Why? Describe all the subgroups of Z. Let a be, a group element with infinite order. Describe all subgroups of kal., 24. For any element a in any group G, prove that kal is a subgroup of, C(a) (the centralizer of a)., 25. If d is a positive integer, d 2 2, and d divides n, show that the number of elements of order d in Dn is f(d ). How many elements of, order 2 does Dn have?, 26. Find all generators of Z. Let a be a group element that has infinite, order. Find all generators of kal., 27. Prove that C*, the group of nonzero complex numbers under multiplication, has a cyclic subgroup of order n for every positive integer n., 28. Let a be a group element that has infinite order. Prove that kail 5, ka j l if and only if i 5 6j., 29. List all the elements of order 8 in Z8000000. How do you know your, list is complete? Let a be a group element such that |a| 8000000., List all elements of order 8 in kal. How do you know your list is, complete?, 30. Suppose a and b belong to a group, a has odd order, and aba21 5, b21. Show that b2 5 e., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 89, , 31. Let G be a finite group. Show that there exists a fixed positive integer, n such that an 5 e for all a in G. (Note that n is independent of a.), 32. Determine the subgroup lattice for Z12., 33. Determine the subgroup lattice for Zp2q, where p and q are distinct, primes., 34. Determine the subgroup lattice for Z8., 35. Determine the subgroup lattice for Zpn, where p is a prime and n is, some positive integer., 36. Prove that a finite group is the union of proper subgroups if and, only if the group is not cyclic., 37. Show that the group of positive rational numbers under multiplication is not cyclic., 38. Consider the set {4, 8, 12, 16}. Show that this set is a group under, multiplication modulo 20 by constructing its Cayley table. What, is the identity element? Is the group cyclic? If so, find all of its, generators., 39. Give an example of a group that has exactly 6 subgroups (including, the trivial subgroup and the group itself ). Generalize to exactly n, subgroups for any positive integer n., 40. Let m and n be elements of the group Z. Find a generator for the, group kml > knl., 41. Suppose that a and b are group elements that commute and have, orders m and n. If kal > kbl 5 {e}, prove that the group contains an, element whose order is the least common multiple of m and n., Show that this need not be true if a and b do not commute., 42. Suppose that a and b belong to a group G, a and b commute, and, |a| and |b| are finite. What are the possibilities for |ab|?, 43. Suppose that a and b belong to a group G, a and b commute, and |a|, and |b| are finite. Prove that G has an element of order lcm(|a|, |b|)., 44. Let F and F9 be distinct reflections in D21. What are the possibilities for |FF9|?, 45. Suppose that H is a subgroup of a group G and |H| 5 10. If a, belongs to G and a6 belongs to H, what are the possibilities for |a|?, 46. Which of the following numbers could be the exact number of, elements of order 21 in a group: 21600, 21602, 21604?, 47. If G is an infinite group, what can you say about the number of, elements of order 8 in the group? Generalize., 48. Suppose that K is a proper subgroup of D35 and K contains at least, two reflections. What are the possible orders of K? Explain your, reasoning., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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90, , Groups, , 49. For each positive integer n, prove that C*, the group of nonzero, complex numbers under multiplication, has exactly f(n) elements, of order n., 50. Prove or disprove that H 5 {n [ Z | n is divisible by both 8 and 10}, is a subgroup of Z., 51. Suppose that G is a finite group with the property that every nonidentity element has prime order (for example, D3 and D5). If Z(G), is not trivial, prove that every nonidentity element of G has the, same order., 52. Prove that an infinite group must have an infinite number of, subgroups., 53. Let p be a prime. If a group has more than p 2 1 elements of order p,, why can’t the group be cyclic?, 54. Suppose that G is a cyclic group and that 6 divides |G|. How many, elements of order 6 does G have? If 8 divides |G|, how many elements of order 8 does G have? If a is one element of order 8, list, the other elements of order 8., 55. List all the elements of Z40 that have order 10. Let |x| 5 40. List all, the elements of kxl that have order 10., 56. Reformulate the corollary of Theorem 4.4 to include the case when, the group has infinite order., 57. Determine the orders of the elements of D33 and how many there, are of each., 58. If G is a cyclic group and 15 divides the order of G, determine the, number of solutions in G of the equation x15 5 e. If 20 divides the, order of G, determine the number of solutions of x20 5 e., Generalize., 59. If G is an Abelian group and contains cyclic subgroups of orders 4, and 5, what other sizes of cyclic subgroups must G contain?, Generalize., 60. If G is an Abelian group and contains cyclic subgroups of orders 4, and 6, what other sizes of cyclic subgroups must G contain?, Generalize., 61. Prove that no group can have exactly two elements of order 2., 62. Given the fact that U(49) is cyclic and has 42 elements, deduce the, number of generators that U(49) has without actually finding any of, the generators., 63. Let a and b be elements of a group. If |a| 5 10 and |b| 5 21, show, that kal > kbl 5 {e}., 64. Let a and b belong to a group. If |a| and |b| are relatively prime,, show that kal > kbl 5 {e}., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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4 | Cyclic Groups, , 91, , 65. Let a and b belong to a group. If |a| 5 24 and |b| 5 10, what are, the possibilities for |kal > kbl|?, 66. Prove that U(2n) (n $ 3) is not cyclic., 67. Suppose that G is a group of order 16 and that, by direct computation, you know that G has at least nine elements x such that, x8 5 e. Can you conclude that G is not cyclic? What if G has at, least five elements x such that x4 5 e? Generalize., 68. Prove that Zn has an even number of generators if n . 2. What, does this tell you about f(n)?, 69. If |a5| 5 12, what are the possibilities for |a|? If |a4| 5 12, what, are the possibilities for |a|?, 70. Suppose that |x| 5 n. Find a necessary and sufficient condition on, r and s such that kx r l # kx s l., 71. Suppose a is a group element such that |a28| 10 and |a22| 20., Determine 冟a冟., 72. Let a be a group element such that |a| 48. For each part, find a, divisor k of 48 such that, a. ka21l 5 kakl;, b. ka14l 5 kakl;, c. ka18l 5 kakl., 73. Let p be a prime. Show that in a cyclic group of order pn 21, every, element is a pth power (that is, every element can be written in the, form ap for some a)., 1 n, 74. Prove that H e c, d ` n [ Z f is a cyclic subgroup of, 0 1, GL(2, R)., 75. Let a and b belong to a group. If |a| 5 12, |b| 5 22, and kal > kbl 2, {e}, prove that a6 5 b11., 76. (2008 GRE Practice Exam) If x is an element of a cyclic group of, order 15 and exactly two of x3, x5, and x9 are equal, determine |x13|., 77. Determine the number of cyclic subgroups of order 4 in Dn., 78. If n is odd, prove that Dn has no subgroup of order 4., 79. If n $ 4 and is even, show that Dn has exactly n/2 noncyclic, subgroups of order 4., 80. If n $ 4 and n is divisible by 2 but not by 4, prove that Dn has, exactly n/2 subgroups of order 4., 81. How many subgroups of order n does Dn have?, 82. Let G be the set of all polynomials of the form ax 2 1 bx 1 c with, coefficients from the set {0, 1, 2}. We can make G a group under, addition by adding the polynomials in the usual way, except that, we use modulo 3 to combine the coefficients. With this operation,, prove that G is a group of order 27 that is not cyclic., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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92, , Groups, , 83. Let a and b belong to some group. Suppose that |a| m, |b| n,, and m and n are relatively prime. If ak bk for some integer k,, prove that mn divides k., 84. For every integer n greater than 2, prove that the group U 1n2 12, is not cyclic., 85. Prove that for any prime p and positive integer n, f 1pn 2 5, pn pn1., 86. Give an example of an infinite group that has exactly two elements, of order 4., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Suggested Reading, Deborah L. Massari, “The Probability of Generating a Cyclic Group,”, Pi Mu Epsilon Journal 7 (1979): 3–6., In this easy-to-read paper, it is shown that the probability of a randomly chosen element from a cyclic group being a generator of the, group depends only on the set of prime divisors of the order of the, group, and not on the order itself. This article, written by an undergraduate student, received first prize in a Pi Mu Epsilon paper contest., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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James Joseph Sylvester, I really love my subject., j. j. sylvester, , James Joseph Sylvester was the most influential mathematician in America in the 19th, century. Sylvester was born on September 3,, 1814, in London and showed his mathematical genius early. At the age of 14, he studied, under De Morgan and won several prizes for, his mathematics, and at the unusually young, age of 25, he was elected a fellow of the, Royal Society., After receiving B.A. and M.A. degrees, from Trinity College in Dublin in 1841,, Sylvester began a professional life that was, to include academics, law, and actuarial careers. In 1876, at the age of 62, he was appointed to a prestigious position at the newly, founded Johns Hopkins University. During, his seven years at Johns Hopkins, Sylvester, pursued research in pure mathematics, with tremendous vigor and enthusiasm., He also founded the American Journal of, Mathematics, the first journal in America, devoted to mathematical research. Sylvester, returned to England in 1884 to a professorship at Oxford, a position he held until his, death on March 15, 1897., Sylvester’s major contributions to mathematics were in the theory of equations,, matrix theory, determinant theory, and invariant theory (which he founded with, Cayley). His writings and lectures—flowery, and eloquent, pervaded with poetic flights,, emotional expressions, bizarre utterances,, and paradoxes—reflected the personality of, this sensitive, excitable, and enthusiastic, , Stock Montage, , FPO, , man. We quote three of his students.† E. W., Davis commented on Sylvester’s teaching, methods., Sylvester’s methods! He had none. “Three lectures will be delivered on a New Universal, Algebra,” he would say; then, “The course, must be extended to twelve.” It did last all the, rest of that year. The following year the course, was to be Substitutions-Theorie, by Netto. We, all got the text. He lectured about three times,, following the text closely and stopping sharp, at the end of the hour. Then he began to think, about matrices again. “I must give one lecture, a week on those,” he said. He could not confine himself to the hour, nor to the one lecture, a week. Two weeks were passed, and Netto, was forgotten entirely and never mentioned, again. Statements like the following were not, infrequent in his lectures: “I haven’t proved, this, but I am as sure as I can be of anything, that it must be so. From this it will follow,, etc.” At the next lecture it turned out that what, he was so sure of was false. Never mind, he, kept on forever guessing and trying, and, presently a wonderful discovery followed,, then another and another. Afterward he would, go back and work it all over again, and surprise us with all sorts of side lights. He then, made another leap in the dark, more treasures, were discovered, and so on forever., , †F., , Cajori, Teaching and History of Mathematics in the United States, Washington: Government, Printing Office, 1890, 265–266., 93, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A short, broad man of tremendous vitality, . . ., Sylvester’s capacious head was ever lost in, the highest cloud-lands of pure mathematics., Often in the dead of night he would get his, favorite pupil, that he might communicate, the very last product of his creative thought., Everything he saw suggested to him something new in the higher algebra. This transmutation of everything into new mathematics, was a revelation to those who knew him, intimately. They began to do it themselves., , Sylvester had one remarkable peculiarity. He, seldom remembered theorems, propositions,, etc., but had always to deduce them when he, wished to use them. In this he was the very, antithesis of Cayley, who was thoroughly, conversant with everything that had been, done in every branch of mathematics., I remember once submitting to Sylvester, some investigations that I had been engaged, on, and he immediately denied my first statement, saying that such a proposition had never, been heard of, let alone proved. To his astonishment, I showed him a paper of his own in, which he had proved the proposition; in fact,, I believe the object of his paper had been the, very proof which was so strange to him., , Another characteristic of Sylvester, which, is very unusual among mathematicians, was, his apparent inability to remember mathematics! W. P. Durfee had the following to say., , For more information about Sylvester,, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , Sylvester’s enthusiasm for teaching and his, influence on his students are captured in the, following passage written by Sylvester’s first, student at Johns Hopkins, G. B. Halsted., , 94, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 1– 4, , 95, , Supplementary Exercises for Chapters 1–4, If you really want something in this life, you have to work for it. Now quiet,, they’re about to announce the lottery numbers!, homer simpson, , True/false questions for Chapters 1– 4 are available on the Web at:, http://www.d.umn.edu/~jgallian/ TF, 1. Let G be a group and let H be a subgroup of G. For any fixed x in, G, define xHx21 5 {xhx21 | h [ H}. Prove the following., a. xHx21 is a subgroup of G., b. If H is cyclic, then xHx21 is cyclic., c. If H is Abelian, then xHx21 is Abelian., The group xHx21 is called a conjugate of H. (Note that conjugation preserves structure.), 2. Let G be a group and let H be a subgroup of G. Define N(H) 5, {x [ G | xHx21 5 H}. Prove that N(H) (called the normalizer of, H) is a subgroup of G.†, 3. Let G be a group. For each a [ G, define cl(a) 5 {xax21 | x [ G}., Prove that these subsets of G partition G. [cl(a) is called the, conjugacy class of a.], 4. The group defined by the following table is called the group of, quaternions. Use the table to determine each of the following., a. The center, b. cl(a), c. cl(b), d. All cyclic subgroups, e, , a, , a2, , a3, , b, , ba, , ba2, , ba3, , e, a, a2, a3, b, ba, ba2, , e, a, a2, a3, b, ba, ba2, , a, a2, a3, e, ba, ba2, ba3, , a2, a3, e, a, ba2, ba3, b, , a3, e, a, a2, ba3, b, ba, , b, ba3, ba2, ba, a2, a, e, , ba, b, ba3, ba2, a3, a2, a, , ba2, ba, b, ba3, e, a3, a2, , ba3, ba2, ba, b, a, e, a3, , ba3, , ba3, , b, , ba, , ba2, , a3, , e, , a, , a2, , †This, , very important subgroup was first used by L. Sylow in 1872 to prove the existence of certain kinds of subgroups in a group. His work is discussed in Chapter 24., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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96, , Groups, , 5. (Conjugation preserves order.) Prove that, in any group, |xax21| 5, |a|. (This exercise is referred to in Chapter 24.), 6. Prove that, in any group, |ab| 5 |ba|., 7. If a and b are group elements, prove that |ab| 5 |a21b21|., 8. Prove that a group of order 4 cannot have a subgroup of order 3., 9. If a, b, and c are elements of a group, give an example to show that, it need not be the case that |abc| 5 |cba|., 10. Let a and b belong to a group G. Prove that there is an element x in, G such that xax 5 b if and only if ab 5 c2 for some element c in G., 11. Prove that if a is the only element of order 2 in a group, then a lies, in the center of the group., 12. Let G be the plane symmetry group of the infinite strip of equally, spaced H’s shown below., H, , H, , H, Axis 1, , 13., 14., 15., 16., 17., , 18., 19., , 20., , 21., , H, , H, , Axis 2, , Let x be the reflection about Axis 1 and let y be the reflection about, Axis 2. Calculate |x|, |y|, and |xy|. Must the product of elements, of finite order have finite order? (This exercise is referred to in, Chapter 27.), What are the orders of the elements of D15? How many elements, have each of these orders?, Prove that a group of order 4 is Abelian., Prove that a group of order 5 must be cyclic., Prove that an Abelian group of order 6 must be cyclic., Let G be an Abelian group and let n be a fixed positive integer. Let, Gn 5 {gn | g [ G}. Prove that Gn is a subgroup of G. Give an example showing that Gn need not be a subgroup of G when G is, non-Abelian. (This exercise is referred to in Chapter 11.), Let G 5 {a b226, where a and b are rational numbers not both, 0. Prove that G is a group under ordinary multiplication., (1969 Putnam Competition) Prove that no group is the union of, two proper subgroups. Does the statement remain true if “two” is, replaced by “three”?, Prove that the subset of elements of finite order in an Abelian, group forms a subgroup. (This subgroup is called the torsion subgroup.) Is the same thing true for non-Abelian groups?, Let p be a prime and let G be an Abelian group. Show that the set, of all elements whose orders are powers of p is a subgroup of G., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 1– 4, , 97, , 22. Suppose that a and b are group elements. If |b| 2 and bab a4,, determine the possibilities for |a|., 23. Suppose that a finite group is generated by two elements a and b, (that is, every element of the group can be expressed as some product of a’s and b’s). Given that a35 b2 5 e and ba2 5 ab, construct, the Cayley table for the group. We have already seen an example, of a group that satisfies these conditions. Name it., 24. If a is an element from a group and |a| n, prove that C(a) 5, C(ak) when k is relatively prime to n., 25. Let x and y belong to a group G. If xy [ Z(G), prove that xy 5 yx., 26. Suppose that H and K are nontrivial subgroups of Q under addition. Show that H > K is a nontrivial subgroup of Q. Is this true if, Q is replaced by R?, 27. Let H be a subgroup of G and let g be an element of G. Prove that, N(gHg21) 5 gN(H)g21. See Exercise 2 for the notation., 28. Let H be a subgroup of a group G and let |g| 5 n. If gm belongs to, H, and m and n are relatively prime, prove that g belongs to H., 29. Find a group that contains elements a and b such that |a| 5 2,, |b| 5 11, and |ab| 5 2., 30. Suppose that G is a group with exactly eight elements of order 10., How many cyclic subgroups of order 10 does G have?, 31. (1989 Putnam Competition) Let S be a nonempty set with an associative operation that is left and right cancellative (xy 5 xz implies, y 5 z, and yx 5 zx implies y 5 z). Assume that for every a in S the, set {an | n 5 1, 2, 3, . . .} is finite. Must S be a group?, 32. Let H1, H2, H3, . . . be a sequence of subgroups of a group with the, property that H1 # H2 # H3 . . . . Prove that the union of the sequence is a subgroup., 33. Let n be an integer greater than 1. Find a noncyclic subgroup of, U (4n) of order 4 that contains the element 2n 2 1., 34. Let G be an Abelian group and H 5 {x [ G | xn 5 e for some odd, integer n (n may vary with x)}. Prove that H is a subgroup of G. Is H, a subgroup if “odd” is replaced by “even”?, 35. Let H 5 {A [ GL(2, R) | det A is rational}. Prove or disprove that, H is a subgroup of GL(2, R). What if “rational” is replaced by “an, integer”?, 36. Suppose that G is a group that has exactly one nontrivial proper, subgroup. Prove that G is cyclic and |G| 5 p2, where p is prime., 37. Suppose that G is a group and G has exactly two nontrivial proper, subgroups. Prove that G is cyclic and |G| 5 pq, where p and q are, distinct primes, or that G is cyclic and |G| 5 p3, where p is prime., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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98, , Groups, , 38. If |a2| 5 |b2|, prove or disprove that |a| 5 |b|., 39. (1995 Putnam Competition) Let S be a set of real numbers that is, closed under multiplication. Let T and U be disjoint subsets of S, whose union is S. Given that the product of any three (not necessarily distinct) elements of T is in T and that the product of any, three elements of U is in U, show that at least one of the two subsets T and U is closed under multiplication., 40. If p is an odd prime, prove that there is no group that has exactly p, elements of order p., 41. Give an example of a group G with infinitely many distinct subgroups H1, H2, H3, . . . such that H1 , H2 , H3 . . . ., 42. Suppose a and b are group elements and b 2 e. If a21ba 5 b2 and, |a| 5 3, find |b|. What is |b|, if |a| 5 5? What can you say about |b|, in the case where |a| 5 k?, 43. Let a and b belong to a group G. Show that there is an element g in, G such that g21 abg 5 ba., 44. Suppose G is a group and x3y3 5 y3x3 for every x and y in G. Let, H 5 {x [ G | |x| is relatively prime to 3}. Prove that elements of H, commute with each other and that H is a subgroup of G. Is your, argument valid if 3 is replaced by an arbitrary positive integer n?, Explain why or why not., 45. Let G be a finite group and let S be a subset of G that contains, more than half of the elements of G. Show that every element of G, can be expressed in the form s1s2 where s1 and s2 belong to S., 46. Let G be a group and let f be a function from G to some set. Show, that H 5 {g [ G | f (xg) 5 f (x) for all x [ G} is a subgroup of G. In, the case that G is the group of real numbers under addition and, f (x) 5 sin x, describe H., 47. Let G be a cyclic group of order n and let H be the subgroup of, order d. Show that H 5 {x [ G | |x| divides d}., 48. Let a be an element of maximum order from a finite Abelian group, G. Prove that for any element b in G, |b| divides |a|. Show by, example that this need not be true for finite non-Abelian groups., 49. Define an operation * on the set of integers by a * b a b 1., Show that the set of integers under this operation is a cyclic group., 50. Let n be an integer greater than 1. Find a noncyclic subgroup of, U 14n2 of order 4 that contains the element 2n 1., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 Permutation Groups, , Wigner’s discovery about the electron permutation group was just the, beginning. He and others found many similar applications and nowadays, group theoretical methods—especially those involving characters and, representations—pervade all branches of quantum mechanics., george mackey, Proceedings of the, American Philosophical Society, , Definition and Notation, In this chapter, we study certain groups of functions, called permutation, groups, from a set A to itself. In the early and mid-19th century, groups, of permutations were the only groups investigated by mathematicians., It was not until around 1850 that the notion of an abstract group was, introduced by Cayley, and it took another quarter century before the, idea firmly took hold., Definitions Permutation of A, Permutation Group of A, , A permutation of a set A is a function from A to A that is both oneto-one and onto. A permutation group of a set A is a set of permutations of A that forms a group under function composition., , Although groups of permutations of any nonempty set A of objects, exist, we will focus on the case where A is finite. Furthermore, it is, customary, as well as convenient, to take A to be a set of the form, {1, 2, 3, . . . , n} for some positive integer n. Unlike in calculus, where, most functions are defined on infinite sets and are given by formulas,, in algebra, permutations of finite sets are usually given by an explicit, listing of each element of the domain and its corresponding functional, value. For example, we define a permutation a of the set {1, 2, 3, 4} by, specifying, a(1) 5 2,, , a(2) 5 3,, , a(3) 5 1,, , a(4) 5 4., , 99, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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100, , Groups, , A more convenient way to express this correspondence is to write a in, array form as, a c, , 1 2 3 4, d., 2 3 1 4, , Here a( j) is placed directly below j for each j. Similarly, the permutation b of the set {1, 2, 3, 4, 5, 6} given by, b(1) 5 5, b(2) 5 3, b(3) 5 1, b(4) 5 6,, , b(5) 5 2, b(6) 5 4, , is expressed in array form as, b c, , 1 2 3 4 5 6, d., 5 3 1 6 2 4, , Composition of permutations expressed in array notation is carried, out from right to left by going from top to bottom, then again from top, to bottom. For example, let, s c, , 1 2 3 4 5, d, 2 4 3 5 1, , and, g c, , 1 2 3 4 5, d;, 5 4 1 2 3, , then, gs 5 £, , 1 2 3 4 5, 5 4 1 2 3, , § £, , 1 2 3 4 5, , § 5c, , 2 4 3 5 1, , 1 2 3 4 5, d, 4 2 1 3 5, , On the right we have 4 under 1, since (gs)(1) 5 g(s(1)) 5 g(2) 5 4,, so gs sends 1 to 4. The remainder of the bottom row gs is obtained in, a similar fashion., We are now ready to give some examples of permutation groups., EXAMPLE 1 Symmetric Group S3 Let S3 denote the set of all, one-to-one functions from {1, 2, 3} to itself. Then S3, under function, composition, is a group with six elements. The six elements are, e c, , 1 2 3, d,, 1 2 3, , a c, , 1 2 3, d,, 2 3 1, , a2 c, , 1 2 3, d,, 3 1 2, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , b c, , 1 2 3, d,, 1 3 2, , Note that ba 5 c, , ab c, , 1 2 3, d,, 2 1 3, , a2b c, , 101, , 1 2 3, d., 3 2 1, , 1 2 3, d 5 a 2b 2 ab, so that S3 is non-Abelian., 3 2 1, , The relation ba 5 a 2b can be used to compute other products in S3, without resorting to the arrays. For example, ba2 5 (ba)a 5 (a 2b)a 5, a 2(ba) 5 a 2(a 2b) 5 a 4b 5 ab., Example 1 can be generalized as follows., EXAMPLE 2 Symmetric Group Sn Let A 5 {1, 2, . . . , n}. The set, of all permutations of A is called the symmetric group of degree n and is, denoted by Sn. Elements of Sn have the form, a c, , 1, 2, a(1) a(2), , p, p, , n, d., a(n), , It is easy to compute the order of Sn. There are n choices of a(1). Once, a(1) has been determined, there are n 2 1 possibilities for a(2) [since, a is one-to-one, we must have a(1) 2 a(2)]. After choosing a(2), there, are exactly n 2 2 possibilities for a(3). Continuing along in this fashion,, we see that Sn has n(n 2 1) ? ? ? 3 ? 2 ? 1 5 n! elements. We leave it to the, reader to prove that Sn is non-Abelian when n $ 3 (Exercise 45)., The symmetric groups are rich in subgroups. The group S4 has 30, subgroups, and S5 has well over 100 subgroups., EXAMPLE 3 Symmetries of a Square As a third example, we, associate each motion in D4 with the permutation of the locations of each, of the four corners of a square. For example, if we label the four corner, positions as in the figure below and keep these labels fixed for reference,, we may describe a 90° counterclockwise rotation by the permutation, 3, , 2, , 4, , 1, , r c, , 1 2 3 4, d,, 2 3 4 1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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102, , Groups, , whereas a reflection across a horizontal axis yields, f c, , 1 2 3 4, d., 2 1 4 3, , These two elements generate the entire group (that is, every element is, some combination of the r’s and f’s)., When D4 is represented in this way, we see that it is a subgroup, of S4., , Cycle Notation, There is another notation commonly used to specify permutations. It is, called cycle notation and was first introduced by the great French mathematician Cauchy in 1815. Cycle notation has theoretical advantages in, that certain important properties of the permutation can be readily determined when cycle notation is used., As an illustration of cycle notation, let us consider the permutation, a c, , 1 2 3 4 5 6, d., 2 1 4 6 5 3, , This assignment of values could be presented schematically as follows., 3, , 1, , 5, , α, α, , α, , α, 6, 2, , 4, , α, , α, , Although mathematically satisfactory, such diagrams are cumbersome. Instead, we leave out the arrows and simply write a 5 (1, 2), (3, 4, 6)(5). As a second example, consider, b c, , 1 2 3 4 5 6, d., 5 3 1 6 2 4, , In cycle notation, b can be written (2, 3, 1, 5)(6, 4) or (4, 6)(3, 1, 5, 2),, since both of these unambiguously specify the function b. An expression of the form (a1, a2, . . . , am) is called a cycle of length m or an, m-cycle., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 103, , A multiplication of cycles can be introduced by thinking of a cycle, as a permutation that fixes any symbol not appearing in the cycle., Thus, the cycle (4, 6) can be thought of as representing the, 1 2 3 4 5 6, permutation c, d . In this way, we can multiply cycles, 1 2 3 6 5 4, by thinking of them as permutations given in array form. Consider the, following example from S8. Let a 5 (13)(27)(456)(8) and b 5, (1237)(648)(5). (When the domain consists of single-digit integers, it is, common practice to omit the commas between the digits.) What, is the cycle form of ab? Of course, one could say that ab 5, (13)(27)(456)(8)(1237)(648)(5), but it is usually more desirable to express a permutation in a disjoint cycle form (that is, the various cycles, have no number in common). Well, keeping in mind that function composition is done from right to left and that each cycle that does not contain a symbol fixes the symbol, we observe that (5) fixes 1; (648) fixes 1;, (1237) sends 1 to 2; (8) fixes 2; (456) fixes 2; (27) sends 2 to 7; and (13), fixes 7. So the net effect of ab is to send 1 to 7. Thus, we begin, ab 5 (17 ? ? ?) ? ? ? . Now, repeating the entire process beginning with 7,, we have, cycle by cycle, right to left,, 7 → 7 → 7 → 1 → 1 → 1 → 1 → 3,, so that ab 5 (173 ? ? ?) ? ? ? . Ultimately, we have ab 5 (1732)(48)(56)., The important thing to bear in mind when multiplying cycles is to “keep, moving” from one cycle to the next from right to left. (Warning: Some authors, compose cycles from left to right. When reading another text, be sure to, determine which convention is being used.), To be sure you understand how to switch from one notation to the, other and how to multiply permutations, we will do one more example, of each., If array notations for a and b, respectively, are, c, , 1 2 3 4 5, d, 2 1 3 5 4, , and, , c, , 1 2 3 4 5, d,, 5 4 1 2 3, , then, in cycle notation, a 5 (12)(3)(45), b 5 (153)(24), and ab 5, (12)(3)(45)(153)(24)., To put ab in disjoint cycle form, observe that (24) fixes 1; (153), sends 1 to 5; (45) sends 5 to 4; and (3) and (12) both fix 4. So, ab sends, 1 to 4. Continuing in this way we obtain ab 5 (14)(253)., One can convert ab back to array form without converting each, cycle of ab into array form by simply observing that (14) means 1 goes, to 4 and 4 goes to 1; (253) means 2 → 5, 5 → 3, 3 → 2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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104, , Groups, , One final remark about cycle notation: Mathematicians prefer not to, write cycles that have only one entry. In this case, it is understood that any, missing element is mapped to itself. With this convention, the permutation, a above can be written as (12)(45). Similarly,, a c, , 1 2 3 4 5, d, 3 2 4 1 5, , can be written a 5 (134). Of course, the identity permutation consists, only of cycles with one entry, so we cannot omit all of these! In this, case, one usually writes just one cycle. For example,, e c, , 1 2 3 4 5, d, 1 2 3 4 5, , can be written as e 5 (5) or e 5 (1). Just remember that missing, elements are mapped to themselves., , Properties of Permutations, We are now ready to state several theorems about permutations and, cycles. The proof of the first theorem is implicit in our discussion of, writing permutations in cycle form., Theorem 5.1 Products of Disjoint Cycles, Every permutation of a finite set can be written as a cycle or as a, product of disjoint cycles., , PROOF Let a be a permutation on A 5 {1, 2, . . . , n}. To write a in, disjoint cycle form, we start by choosing any member of A, say a1, and let, a2 5 a(a1),, , a3 5 a(a(a1)) 5 a 2(a1),, , and so on, until we arrive at a1 5 a m(a1) for some m. We know that such, an m exists because the sequence a1, a(a1), a 2(a1), ? ? ? must be finite;, so there must eventually be a repetition, say a i(a1) 5 a j(a1) for some, i and j with i , j. Then a1 5 a m(a1), where m 5 j 2 i. We express this, relationship among a1, a2, . . . , am as, a 5 (a1, a2, . . . , am) ? ? ? ., The three dots at the end indicate the possibility that we may not have, exhausted the set A in this process. In such a case, we merely choose, any element b1 of A not appearing in the first cycle and proceed to, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 105, , create a new cycle as before. That is, we let b2 5 a(b1), b3 5 a2(b1), and, so on, until we reach b1 5 a k(b1) for some k. This new cycle will have, no elements in common with the previously constructed cycle. For, if, so, then a i(a1) 5 a j(b1) for some i and j. But then a i2j(a1) 5 b1, and, therefore b1 5 at for some t. This contradicts the way b1 was chosen., Continuing this process until we run out of elements of A, our permutation will appear as, a 5 (a1, a2, . . . , am)(b1, b2, . . . , bk) ? ? ? (c1, c2, . . . , cs)., In this way, we see that every permutation can be written as a product, of disjoint cycles., Theorem 5.2 Disjoint Cycles Commute, If the pair of cycles a 5 (a1, a2, . . . , am) and b 5 (b1,b2, . . . , bn ), have no entries in common, then ab 5 ba., , PROOF For definiteness, let us say that a and b are permutations of, the set, S 5 {a1, a2, . . . , am, b1, b2, . . . , bn, c1, c2, . . . , ck},, where the c’s are the members of S left fixed by both a and b (there, may not be any c’s). To prove that ab 5 ba, we must show that (ab)(x) 5, (ba)(x) for all x in S. If x is one of the a elements, say ai, then, (ab)(ai) 5 a(b(ai)) 5 a(ai) 5 ai11,, since b fixes all a elements. (We interpret ai11 as a1 if i 5 m.) For the, same reason,, (ba)(ai) 5 b(a(ai)) 5 b(ai11) 5 ai11., Hence, the functions of ab and ba agree on the a elements. A similar, argument shows that ab and ba agree on the b elements as well., Finally, suppose that x is a c element, say ci. Then, since both a and b, fix c elements, we have, (ab)(ci) 5 a(b(ci)) 5 a(ci) 5 ci, and, (ba)(ci) 5 b(a(ci)) 5 b(ci) 5 ci ., This completes the proof., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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106, , Groups, , In demonstrating how to multiply cycles, we showed that the product, (13)(27)(456)(8)(1237)(648)(5) can be written in disjoint cycle form as, (1732)(48)(56). Is economy in expression the only advantage to writing a permutation in disjoint cycle form? No. The next theorem shows, that the disjoint cycle form has the enormous advantage of allowing us, to “eyeball” the order of the permutation., Theorem 5.3 Order of a Permutation (Ruffini, 1799), The order of a permutation of a finite set written in disjoint cycle, form is the least common multiple of the lengths of the cycles., , PROOF First, observe that a cycle of length n has order n. (Verify this, yourself.) Next, suppose that a and b are disjoint cycles of lengths m, and n, and let k be the least common multiple of m and n. It follows from, Theorem 4.1 that both ak and bk are the identity permutation e and, since, a and b commute, (ab)k 5 a kb k is also the identity. Thus, we know by, Corollary 2 to Theorem 4.1 (ak 5 e implies that |a| divides k) that the, order of ab—let us call it t—must divide k. But then (ab)t 5 a tb t 5 e,, so that a t 5 b 2t. However, it is clear that if a and b have no common, symbol, the same is true for a t and b 2t, since raising a cycle to a power, does not introduce new symbols. But, if a t and b 2t are equal and have, no common symbol, they must both be the identity, because every symbol in a t is fixed by b 2t and vice versa (remember that a symbol not appearing in a permutation is fixed by the permutation). It follows, then,, that both m and n must divide t. This means that k, the least common, multiple of m and n, divides t also. This shows that k 5 t., Thus far, we have proved that the theorem is true in the cases, where the permutation is a single cycle or a product of two disjoint, cycles. The general case involving more than two cycles can be handled in an analogous way., Theorem 5.3 is an enomously powerful tool for calculating the orders of permutations and the number of permutations of a particular, order. We demonstrate this in the next two examples., EXAMPLE 4 To determine the orders of the 7! 5 5040 elements of, S7, we need only consider the possible disjoint cycle structures of the, elements of S7. For convenience, we denote an n-cycle by (n). Then,, arranging all possible disjoint cycle structures of elements of S7, according to longest cycle lengths left to right, we have, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 107, , (7), (6) (1), (5) (2), (5) (1) (1), (4) (3), (4) (2) (1), (4) (1) (1) (1), (3) (3) (1), (3) (2) (2), (3) (2) (1) (1), (3) (1) (1) (1) (1), (2) (2) (2) (1), (2) (2) (1) (1) (1), (2) (1) (1) (1) (1) (1), (1) (1) (1) (1) (1) (1) (1)., Now, from Theorem 5.3 we see that the orders of the elements of S7, are 7, 6, 10, 5, 12, 4, 3, 2, and 1. To do the same for the 10! 5 3628800, elements of S10 would be nearly as simple., EXAMPLE 5 We determine the number of elements of S7 of order 3., By Theorem 5.3, we need only count the number of permutations of, the forms (a1a2a3) and (a1a2a3) (a4a5a6). In the first case consider the, triple a1a2a3. Clearly there are 7 ? 6 ? 5 such triples. But this product, counts the permutation (a1a2a3) three times (for example, it counts 134,, 341, 413 as distinct triples whereas the cycles (134), (341), and (413), are the same group element). Thus, the number of permutations in S7 for, the form (a1a2a3) is (7 ? 6 ? 5)/3 5 70. For elements of S7 of the form, (a1a2a3) (a4a5a6) there are (7 ? 6 ? 5)/3 ways to create the first cycle and, (4 ? 3 ? 2)/3 to create the second cycle but the product of (7 ? 6 ? 5)/3 and, (4 ? 3 ? 2)/3) counts (a1a2a3) (a4a5a6) and (a4a5a6)(a3a2a1) as distinct when, they are equal group elements. Thus, the number of elements in S7 for the, form (a1a2a3) (a4a5a6) is (7 ? 6 ? 5)(4 ? 3 ? 2)/(3 ? 3 ? 2) 5 280. This gives, us 350 elements of order 3 in S7., As we will soon see, it is often greatly advantageous to write a permutation as a product of cycles of length 2—that is, as permutations of, the form (ab) where a 2 b. Many authors call these permutations transpositions, since the effect of (ab) is to interchange or transpose a and b., Example 6 and Theorem 5.4 show how this can always be done., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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108, , Groups, , EXAMPLE 6, (12345) 5 (15)(14)(13)(12), (1632)(457) 5 (12)(13)(16)(47)(45), Theorem 5.4 Product of 2-Cycles, Every permutation in Sn, n . 1, is a product of 2-cycles., , PROOF First, note that the identity can be expressed as (12)(12), and, so it is a product of 2-cycles. By Theorem 5.1, we know that every permutation can be written in the form, (a1a2 ? ? ? ak)(b1b2 ? ? ? bt) ? ? ? (c1c2 ? ? ? cs)., A direct computation shows that this is the same as, (a1ak)(a1ak21) ? ? ? (a1a2)(b1bt)(b1bt21) ? ? ? (b1b2), ? ? ? (c1cs)(c1cs21) ? ? ? (c1c2)., This completes the proof., The decomposition of a permutation into a product of 2-cycles given, in Example 6 and in the proof of Theorem 5.4 is not the only way a permutation can be written as a product of 2-cycles. Although the next, example shows that even the number of 2-cycles may vary from one, decomposition to another, we will prove in Theorem 5.5 (first proved by, Cauchy) that there is one aspect of a decomposition that never varies., EXAMPLE 7, (12345) 5 (54)(53)(52)(51), (12345) 5 (54)(52)(21)(25)(23)(13), We isolate a special case of Theorem 5.5 as a lemma., Lemma, If e 5 b1b2 ? ? ? br , where the b’s are 2-cycles, then r is even., , PROOF Clearly, r 2 1, since a 2-cycle is not the identity. If r 5 2, we, are done. So, we suppose that r . 2, and we proceed by induction., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 109, , Suppose that the rightmost 2-cycle is (ab). Then, since (ij) 5 ( ji), the, product br21br can be expressed in one of the following forms shown, on the right:, e 5 (ab)(ab),, (ab)(bc) 5 (ac)(ab),, (ac)(cb) 5 (bc)(ab),, (ab)(cd) 5 (cd)(ab)., If the first case occurs, we may delete br21 br from the original product, to obtain e 5 b1 b2 ? ? ? br22, and therefore, by the Second Principle of, Mathematical Induction, r 2 2 is even. In the other three cases, we, replace the form of br21 br on the right by its counterpart on the left to, obtain a new product of r 2-cycles that is still the identity, but where, the rightmost occurrence of the integer a is in the second-from-therightmost 2-cycle of the product instead of the rightmost 2-cycle. We now, repeat the procedure just described with br22 br21, and, as before, we, obtain a product of (r 2 2) 2-cycles equal to the identity or a new product, of r 2-cycles, where the rightmost occurrence of a is in the third 2-cycle, from the right. Continuing this process, we must obtain a product of, (r 2 2) 2-cycles equal to the identity, because otherwise we have a product equal to the identity in which the only occurrence of the integer a is in the, leftmost 2-cycle, and such a product does not fix a, whereas the identity, does. Hence, by the Second Principle of Mathematical Induction, r 2 2 is, even, and r is even as well., Theorem 5.5 Always Even or Always Odd, If a permutation a can be expressed as a product of an even (odd), number of 2-cycles, then every decomposition of a into a product of, 2-cycles must have an even (odd) number of 2-cycles. In symbols, if, , a 5 b1b2 ? ? ? br, , and, , a 5 g1g2 ? ? ? gs ,, , where the b’s and the g’s are 2-cycles, then r and s are both even or, both odd., , PROOF Observe that b1b2 ? ? ? br 5 g1g2 ? ? ? gs implies, e 5 g1g2 ? ? ? gsbr 21 ? ? ? b221b121, 5 g1g2 ? ? ? gsbr ? ? ? b2b1,, since a 2-cycle is its own inverse. Thus, the lemma on page 108 guarantees that s 1 r is even. It follows that r and s are both even or both, odd., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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110, , Groups, , Definition Even and Odd Permutations, , A permutation that can be expressed as a product of an even number, of 2-cycles is called an even permutation. A permutation that can, be expressed as a product of an odd number of 2-cycles is called an, odd permutation., , Theorems 5.4 and 5.5 together show that every permutation can be, unambiguously classified as either even or odd. The significance of, this observation is given in Theorem 5.6., Theorem 5.6 Even Permutations Form a Group, The set of even permutations in Sn forms a subgroup of Sn., , PROOF This proof is left to the reader (Exercise 17)., The subgroup of even permutations in Sn arises so often that we give, it a special name and notation., Definition Alternating Group of Degree n, , The group of even permutations of n symbols is denoted by An and is, called the alternating group of degree n., , The next result shows that exactly half of the elements of Sn(n . 1), are even permutations., Theorem 5.7, For n . 1, An has order n!/2., , PROOF For each odd permutation a, the permutation (12)a is even and,, by the cancellation property in groups, (12)a 2 (12)b when a 2 b. Thus,, there are at least as many even permutations as there are odd ones. On the, other hand, for each even permutation a, the permutation (12)a is odd and, (12)a 2 (12)b when a 2 b. Thus, there are at least as many odd permutations as there are even ones. It follows that there are equal numbers of, even and odd permutations. Since |Sn| 5 n!, we have |An| 5 n!/2., The names for the symmetric group and the alternating group of degree, n come from the study of polynomials over n variables. A symmetric, polynomial in the variables x1, x2, . . . , xn is one that is unchanged under, any transposition of two of the variables. An alternating polynomial is, one that changes signs under any transposition of two of the variables. For, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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111, , 5 | Permutation Groups, , example, the polynomial x1 x2 x3 is unchanged by any transposition of two, of the three variables, whereas the polynomial (x12x2)(x12x3)(x22x3), changes signs when any two of the variables are transposed. Since every, member of the symmetric group is the product of transpositions, the symmetric polynomials are those that are unchanged by members of the symmetric group. Likewise, since any member of the alternating group is the, product of an even number of transpositions, the alternating polynomials, are those that are unchanged by members of the alternating group., The alternating groups are among the most important examples of, groups. The groups A4 and A5 will arise on several occasions in later, chapters. In particular, A5 has great historical significance., A geometric interpretation of A4 is given in Example 8, and a multiplication table for A4 is given as Table 5.1., EXAMPLE 8 Rotations of a Tetrahedron, The 12 rotations of a regular tetrahedron can be conveniently described, with the elements of A4. The top row of Figure 5.1 illustrates the identity, and three 180° “edge” rotations about axes joining midpoints of two, edges. The second row consists of 120° “face” rotations about axes joining, a vertex to the center of the opposite face. The third row consists of 2120°, (or 240°) “face” rotations. Notice that the four rotations in the second row, can be obtained from those in the first row by left-multiplying the four in, the first row by the rotation (123), whereas those in the third row can be, obtained from those in the first row by left-multiplying the ones in the first, row by (132)., Table 5.1 The Alternating Group A4 of Even Permutations of {1, 2, 3, 4}, (In this table, the permutations of A4 are designated as a1, a2, . . . , a12 and an entry k inside, the table represents ak. For example, a3 a8 5 a6.), (1) 5 a1, (12)(34) 5 a2, (13)(24) 5 a3, (14)(23) 5 a4, (123) 5 a5, (243) 5 a6, (142) 5 a7, (134) 5 a8, (132) 5 a9, (143) 5 a10, (234) 5 a11, (124) 5 a12, , a1, , a2, , a3, , a4, , a5, , a6, , a7, , a8, , a9, , a10, , a11, , a12, , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, , 2, 1, 4, 3, 8, 7, 6, 5, 11, 12, 9, 10, , 3, 4, 1, 2, 6, 5, 8, 7, 12, 11, 10, 9, , 4, 3, 2, 1, 7, 8, 5, 6, 10, 9, 12, 11, , 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, , 6, 5, 8, 7, 12, 11, 10, 9, 3, 4, 1, 2, , 7, 8, 5, 6, 10, 9, 12, 11, 4, 3, 2, 1, , 8, 7, 6, 5, 11, 12, 9, 10, 2, 1, 4, 3, , 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, , 10, 9, 12, 11, 4, 3, 2, 1, 7, 8, 5, 6, , 11, 12, 9, 10, 2, 1, 4, 3, 8, 7, 6, 5, , 12, 11, 10, 9, 3, 4, 1, 2, 6, 5, 8, 7, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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112, , Groups, , 1, , 1, , D 4, , B, , 2, , C 4, , C C, , 2, , B 4, , D, A A, , 3, , 3, , 1, , 1, , D, , C, B, , 2, , 4, , A A, , BB, , 1, , 2, , A, , A 4, , D, , D, , C C, , 3, , 3, , 1, (143), , (124), , C 4, , 2 D, , AA, , B B, , 3, , 3, , C, , D D, , A, , 4, D, , 2, , 1, , 1, , B B, , B B, B 4, , C, D, , (234), , 2 C, , 1, (142), , 3, , 3, , (132), , 3, , A A, , D, , 4, D, , BB, , (243), , C, , A 4, , 2 C, , 1, , (134), , 2 A, , DD, , DD, , 3, , C, , (14)(23), C C, , B, , A, , (123), , 1, , (13)(24), B, , A A, , 2, , 1, , (12)(34), , (1), , 2, , B 4, , A, C C, , C, A, , 2, , 3, , B, , 4, , D D, 3, , Figure 5.1 Rotations of a regular tetrahedron., , Many molecules with chemical formulas of the form AB4, such as, methane (CH4) and carbon tetrachloride (CCl4), have A4 as their symmetry group. Figure 5.2 shows the form of one such molecule., Many games and puzzles can be analyzed using permutations., , Figure 5.2 A tetrahedral AB4 molecule., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 113, , EXAMPLE 9 (Loren Larson) A Sliding Disk Puzzle, Consider the puzzle shown below (the space in the middle is empty)., 1, 6, , 2, , 5, , 3, 4, , By sliding disks from one position to another along the lines indicated, without lifting or jumping them, can we obtain the following arrangement?, 3, 1, , 5, , 6, , 2, 4, , To answer this question, we view the positions as numbered in the, first figure above and consider two basic operations. Let r denote the, following operation: Move the disk in position 1 to the center position,, then move the disk in position 6 to position 1, the disk in position 5 to, position 6, the disk in position 4 to position 5, the disk in position 3 to, position 4, then the disk in the middle position to position 3. Let s, denote the operation: Move the disk in position 1 to the center position,, then move the disk in position 2 to position 1, then move the disk in position 3 to position 2, and finally move the disk in the center to position 3., In permutation notation, we have r 5 (13456) and s 5 (132). The, permutation for the arrangement we seek is (16523). Clearly, if we can, express (16523) as a string of r’s and s’s, we can achieve the desired, arangement. Rather than attempt to find an appropriate combination of, r’s and s’s by hand, it is easier to employ computer software that is designed for this kind of problem. One such software program is GAP (see, Suggested Software at the end of this chapter). With GAP, all we need to, do is use the following commands:, gap. G :5 SymmetricGroup(6);, gap. r :5 (1,3,4,5,6); s :5 (1, 3, 2);, gap. K :5 Subgroup(G,[r,s]);, gap. Factorization(K,(1,6,5,2,3));, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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114, , Groups, , The first three lines inform the computer that our group is the, subgroup of S6 generated by r 5 (13456) and s 5 (132). The fourth, line requests that (16523) be expressed in terms of r and s. If we include the command, gap. Size (K), we would find that the order of the subgroup generated by r and s is 360., Then, observing that r and s are even permutations and that |A6| 5 360,, we deduce that r and s can achieve any arrangement that corresponds to, an even permutation., Rubik's Cube, The Rubik’s Cube made from 48 cubes called “facets” is the quintessential example of a group theory puzzle. It was invented in 1974 by the, Hungarian Erró´ Rubik. By 2009 more than 350 million Rubik’s Cubes, had been sold. The current record time for solving it is under 7 seconds;, under 31 seconds blindfolded. Although it was proved in 1995 that, there was a starting configuration that required at least 20 moves to, solve, it was not until 2010 that it was determined that every cube could, be solved in at most 20 moves. This computer calculation utilized about, 35 CPU-years donated by Google to complete. In early discussions, about the minimum number of moves to solve the cube in the worst, possible case, someone called it “God’s number,” and the name stuck., A history of the quest to find God’s number is given at the website at, http://www.cube20.org/., The set of all configuration of the Rubik’s Cube form a group of, permutations of order 43,252,003,274,489,856,00. This order can be, computed using GAP by labeling the faces of the cube as shown here., , 9, 12, 14, , 10, left, 15, , 11, 13, 16, , 1, 4, 6, , 2, top, 7, , 3, 5, 8, , 17, 20, 22, , 18, front, 23, , 19, 21, 24, , 42, bottom, 47, , 43, 45, 48, , 41, 44, 46, , 25, 28, 30, , 26, right, 31, , 27, 29, 32, , 33, 36, 38, , 34, rear, 39, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , 35, 37, 40
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5 | Permutation Groups, , 115, , The group of permutations of the cube is generated by the following, rotations of the six layers., top 5 (1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19), left 5 (9,11,16,14)(10,13,15,12)(1,17,41,40)(4,20,44,37)(6,22,46,35), front 5 (17,19,24,22)(18,21,23,20)(6,25,43,16)(7,28,42,13)(8,30,41,11), right 5 (25,27,32,30)(26,29,31,28)(3,38,43,19)(5,36,45,21)(8,33,48,24), rear 5 (33,35,40,38)(34,37,39,36)(3,9,46,32)(2,12,47,29)(1,14,48,27), bottom 5 (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39), (16,24,32,40), , A Check-Digit Scheme Based on D5, In Chapter 0, we presented several schemes for appending a check digit, to an identification number. Among these schemes, only the International Standard Book Number method was capable of detecting all, single-digit errors and all transposition errors involving adjacent digits., However, recall that this success was achieved by introducing the alphabetical character X to handle the case where 10 was required to, make the dot product 0 modulo 11., In contrast, in 1969, J. Verhoeff [2] devised a method utilizing the, dihedral group of order 10 that detects all single-digit errors and all, transposition errors involving adjacent digits without the necessity of, avoiding certain numbers or introducing a new character. To describe, this method, consider the permutation s 5 (01589427)(36) and the dihedral group of order 10 as represented in Table 5.2. (Here we use 0, through 4 for the rotations, 5 through 9 for the reflections, and p for the, operation of D5.), Table 5.2 Multiplication for D5, *, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, , 1, 2, 3, 4, 0, 9, 5, 6, 7, 8, , 2, 3, 4, 0, 1, 8, 9, 5, 6, 7, , 3, 4, 0, 1, 2, 7, 8, 9, 5, 6, , 4, 0, 1, 2, 3, 6, 7, 8, 9, 5, , 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, , 6, 7, 8, 9, 5, 4, 0, 1, 2, 3, , 7, 8, 9, 5, 6, 3, 4, 0, 1, 2, , 8, 9, 5, 6, 7, 2, 3, 4, 0, 1, , 9, 5, 6, 7, 8, 1, 2, 3, 4, 0, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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116, , Groups, , Verhoeff’s idea was to view the digits 0 through 9 as the elements of, the group D5 and to replace ordinary addition with calculations done in, D5. In particular, to any string of digits a1a2 . . . an21, we append the, check digit an so that s(a1) p s 2 (a2 ) p ? ? ? p s n22(an22) p s n21(an21) p, s n (an ) 5 0. [Here s 2 (x) 5 s(s(x)), s 3 (x) 5 s(s 2 (x)), and so on.], Since s has the property that s i (a) 2 s i (b) if a 2 b, all single-digit errors are detected. Also, because, a p s(b) 2 b p s(a), , if a 2 b,, , (1), , as can be checked on a case-by-case basis (see Exercise 67), it follows, that all transposition errors involving adjacent digits are detected [since, Equation (1) implies that s i(a) p s i11(b) 2 si(b) p s i11(a) if a 2 b]., From 1990 until 2002, the German government used a minor modification of Verhoeff’s check-digit scheme to append a check digit to the, serial numbers on German banknotes. Table 5.3 gives the values of the, functions s, s 2 , . . . , s10 needed for the computations. [The functional, value s i ( j) appears in the row labeled with s i and the column labeled j.], Since the serial numbers on the banknotes use 10 letters of the alphabet in, addition to the 10 decimal digits, it is necessary to assign numerical values to the letters to compute the check digit. This assignment is shown in, Table 5.4., Table 5.3 Powers of s, , s, s2, s3, s4, s5, s6, s7, s8, s9, s10, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 1, 5, 8, 9, 4, 2, 7, 0, 1, 5, , 5, 8, 9, 4, 2, 7, 0, 1, 5, 8, , 7, 0, 1, 5, 8, 9, 4, 2, 7, 0, , 6, 3, 6, 3, 6, 3, 6, 3, 6, 3, , 2, 7, 0, 1, 5, 8, 9, 4, 2, 7, , 8, 9, 4, 2, 7, 0, 1, 5, 8, 9, , 3, 6, 3, 6, 3, 6, 3, 6, 3, 6, , 0, 1, 5, 8, 9, 4, 2, 7, 0, 1, , 9, 4, 2, 7, 0, 1, 5, 8, 9, 4, , 4, 2, 7, 0, 1, 5, 8, 9, 4, 2, , Table 5.4 Letter Values, A, , D, , G, , K, , L, , N, , S, , U, , Y, , Z, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 117, , To any string of digits a1a2 . . . a10 corresponding to a banknote serial, number, the check digit a11 is chosen such that s (a1) p s 2(a2) p ? ? ? p, s9(a9) p s10(a10) p a11 5 0 [instead of s(a1) p s 2 (a2) p ? ? ? p s10(a10) p, s11(a11) 5 0 as in the Verhoeff scheme]., To trace through a specific example, consider the banknote (featuring the mathematician Gauss) shown in Figure 5.3 with the number, AG8536827U7. To verify that 7 is the appropriate check digit, we observe that s(0) p s 2 (2) p s 3 (8) p s 4 (5) p s 5 (3) p s 6 (6) p s 7 (8) p, s 8 (2) p s 9 (7) p s 10(7) p 7 5 1 p 0 p 2 p 2 p 6 p 6 p 5 p 2 p 0 p 1 p, 7 5 0, as it should be. [To illustrate how to use the multiplication table, for D 5, we compute 1 p 0 p 2 p 2 5 (1 p 0) p 2 p 2 5 1 p 2 p 2 5, (1 p 2) p 2 5 3 p 2 5 0.], , Figure 5.3 German banknote with serial number AG8536827U and check digit 7., , One shortcoming of the German banknote scheme is that it does not, distinguish between a letter and its assigned numerical value. Thus, a, substitution of 7 for U (or vice versa) and the transposition of 7 and U, are not detected by the check digit. Moreover, the banknote scheme, does not detect all transpositions of adjacent characters involving the, check digit itself. For example, the transposition of D and 8 in positions 10 and 11 is not detected. Both of these defects can be avoided by, using the Verhoeff method with D18, the dihedral group of order 36, to, assign every letter and digit a distinct value together with an appropriate function s [1]. Using this method to append a check character, all, single-position errors and all transposition errors involving adjacent, digits will be detected., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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118, , Groups, , Exercises, When you feel how depressingly, slowly you climb,, it’s well to remember that, Things Take Time., piet hein, “t. t. t.,” Grooks (1966)†*, , 1. Let, a5 c, , 1 2 3 4 5 6, d, 2 1 3 5 4 6, , and b c, , 1 2 3 4 5 6, d., 6 1 2 4 3 5, , Compute each of the following., a. a21, b. ba, c. ab, 2. Let, 1 2 3 4 5 6 7 8, 1 2 3 4 5 6 7 8, d and b 5 c, d., a5 c, 2 3 4 5 1 7 8 6, 1 3 8 7 6 5 2 4, Write a, b, and ab as, a. products of disjoint cycles;, b. products of 2-cycles., 3. Write each of the following permutations as a product of disjoint, cycles., a. (1235)(413), b. (13256)(23)(46512), c. (12)(13)(23)(142), 4. Find the order of each of the following permutations., a. (14), b. (147), c. (14762), d. (a1a2 . . . ak), 5. What is the order of each of the following permutations?, a. (124)(357), b. (124)(3567), c. (124)(35), d. (124)(357869), e. (1235)(24567), f. (345)(245), †Hein, , is a Danish engineer and poet and is the inventor of the game Hex., *Piet Hein, “T.T.T.,” Grooks (1966) Copyright © Piet Hein Grooks. Reprinted with, kind permission from Piet Hein a/s, DK-5500 Middelfart, Denmark., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 119, , 6. What is the order of each of the following permutations?, 1 2 3 4 5 6, a. c, d, 2 1 5 4 6 3, b. c, , 7., 8., 9., 10., 11., , 12., , 13., 14., 15., , 16., 17., 18., , 19., 20., , 1 2 3 4 5 6 7, d, 7 6 1 2 3 4 5, What is the order of the product of a pair of disjoint cycles of, lengths 4 and 6?, Show that A8 contains an element of order 15., What are the possible orders for the elements of S6 and A6? What, about A7? (This exercise is referred to in Chapter 25.), What is the maximum order of any element in A10?, Determine whether the following permutations are even or odd., a. (135), b. (1356), c. (13567), d. (12)(134)(152), e. (1243)(3521), Show that a function from a finite set S to itself is one-to-one if and, only if it is onto. Is this true when S is infinite? (This exercise is referred to in Chapter 6.), Suppose that a is a mapping from a set S to itself and a(a(x)) 5 x, for all x in S. Prove that a is one-to-one and onto., Find eight elements in S6 that commute with (12)(34)(56). Do they, form a subgroup of S6?, Let n be a positive integer. If n is odd, is an n-cycle an odd or an, even permutation? If n is even, is an n-cycle an odd or an even permutation?, If a is even, prove that a21 is even. If a is odd, prove that a21 is odd., Prove Theorem 5.6., In Sn, let a be an r-cycle, b an s-cycle, and g a t-cycle. Complete, the following statements: ab is even if and only if r 1 s is . . . ;, abg is even if and only if r 1 s 1 t is . . . ., Let a and b belong to Sn. Prove that ab is even if and only if a, and b are both even or both odd., Associate an even permutation with the number 11 and an odd, permutation with the number 21. Draw an analogy between the, result of multiplying two permutations and the result of multiplying their corresponding numbers 11 or 21., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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120, , Groups, , 21. Let s be the permutation of the letters A through Z that takes each, letter to the one directly below it in the display following. Write s, in cycle form., A B C D E F G H I J K L M N O P Q R S T U V WX Y Z, HDB GJ E C M I L ONP F KR US AWQT V Z XY, 22. If a and b are distinct 2-cycles, what are the possibilities for |ab|?, 23. Show that if H is a subgroup of Sn, then either every member of H, is an even permutation or exactly half of the members are even., (This exercise is referred to in Chapter 25.), 24. Suppose that H is a subgroup of Sn of odd order. Prove that H is a, subgroup of An., 25. Give two reasons why the set of odd permutations in Sn is not a, subgroup., 26. Let a and b belong to S n . Prove that a 21 b 21 ab is an even, permutation., 27. Use Table 5.1 to compute the following., a. The centralizer of a3 5 (13)(24), b. The centralizer of a12 5 (124), 28. How many elements of order 5 are in S7?, 29. How many elements of order 4 does S6 have? How many elements, of order 2 does S6 have?, 30. Prove that (1234) is not the product of 3-cycles., 31. Let b [ S7 and suppose b 4 5 (2143567). Find b. What are the, possibilities for b if b [ S9?, 32. Let b 5 (123)(145). Write b 99 in disjoint cycle form., 33. Find three elements s in S 9 with the property that s 3 5, (157)(283)(469)., 34. What cycle is (a1a2 ? ? ? an)21?, 35. Let G be a group of permutations on a set X. Let a [ X and define, stab(a) 5 {a [ G | a(a) 5 a}. We call stab(a) the stabilizer of a in, G (since it consists of all members of G that leave a fixed). Prove, that stab(a) is a subgroup of G. (This subgroup was introduced by, Galois in 1832.) This exercise is referred to in Chapter 7., 36. Let b 5 (1,3,5,7,9,8,6)(2,4,10). What is the smallest positive integer n for which bn 5 b25?, 37. Let a 5 (1,3,5,7,9)(2,4,6)(8,10). If am is a 5-cycle, what can you, say about m?, 38. Let H 5 {b [ S5 | b(1) 5 1 and b(3) 5 3}. Prove that H is a subgroup of S5. How many elements are in H? Is your argument valid, when S5 is replaced by Sn for n $ 3? How many elements are in H, when S5 is replaced by An for n $ 4?, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 121, , 39. How many elements of order 5 are there in A6?, 40. In S4, find a cyclic subgroup of order 4 and a noncyclic subgroup, of order 4., 41. Suppose that b is a 10-cycle. For which integers i between 2 and, 10 is b i also a 10-cycle?, 42. In S3, find elements a and b such that |a| 5 2, |b| 5 2, and |ab| 5 3., 43. Find group elements a and b in S5 such that |a| 5 3, |b| 5 3, and, |ab| 5 5., 44. Represent the symmetry group of an equilateral triangle as a group, of permutations of its vertices (see Example 3)., 45. Prove that Sn is non-Abelian for all n $ 3., 46. Prove that An is non-Abelian for all n $ 4., 47. For n $ 3, let H 5 {b [ Sn | b(1) 5 1 or 2 and b(2) 5 1 or 2}., Prove that H is a subgroup of Sn. Determine |H|., 48. Show that in S7, the equation x2 5 (1234) has no solutions but the, equation x3 5 (1234) has at least two., 49. If (ab) and (cd) are distinct 2-cycles in Sn, prove that (ab) and (cd), commute if and only if they are disjoint., 50. Let a be a 2-cycle and b be a t-cycle in Sn. Prove that aba is a, t-cycle., 51. Use the previous exercise to prove that, if a and b belong to Sn and, b is the product of k-cycles of lengths n1, n2, . . . , nk, then aba21 is, the product of k-cycles of lengths n1, n2, . . . nk., 52. Let a and b belong to Sn. Prove that bab21 and a are both even or, both odd., 53. What is the smallest positive integer n such that Sn has an element, of order greater than 2n?, 54. Let n be an even positive integer. Prove that An has an element of, order greater than n if and only if n $ 8., 55. Let n be an odd positive integer. Prove that An has an element of, order greater than 2n if and only if n $ 13., 56. Let n be an even positive integer. Prove that An has an element of, order greater than 2n if and only if n $ 14., 57. Viewing the members of D4 as a group of permutations of a square, labeled 1, 2, 3, 4 as described in Example 3, which geometric symmetries correspond to even permutations?, 58. Viewing the members of D5 as a group of permutations of a regular, pentagon with consecutive vertices labeled 1, 2, 3, 4, 5, what geometric symmetry corresponds to the permutation (14253)? Which, symmetry corresponds to the permutation (25)(34)?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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122, , Groups, , 59. Let n be an odd integer greater than 1. Viewing Dn as a group of, permutations of a regular n-gon with consecutive vertices labeled, 1, 2, . . . , n, explain why the rotation subgroup of D n is a subgroup of An., 60. Let n be an integer greater than 1. Viewing Dn as a group of permutations of a regular n-gon with consecutive vertices labeled 1, 2, . . . ,, n, determine for which n all the permutations corresponding to reflections in Dn are even permutations. Hint: Consider the fours, cases for n mod 4., 61. Show that A5 has 24 elements of order 5, 20 elements of order 3, and, 15 elements of order 2. (This exercise is referred to in Chapter 25.), 62. Find a cyclic subgroup of A8 that has order 4., 63. Find a noncyclic subgroup of A8 that has order 4., 64. Compute the order of each member of A4. What arithmetic relationship do these orders have with the order of A4?, 65. Show that every element in A n for n $ 3 can be expressed as a, 3-cycle or a product of 3-cycle., 66. Show that for n $ 3, Z(Sn) 5 {e}., 67. Verify the statement made in the discussion of the Verhoeff check, digit scheme based on D5 that a * s(b) 2 b * s(a) for distinct a and, b. Use this to prove that s i(a) * s i11(b) 2 s i(b) * s i11(a) for all i., Prove that this implies that all transposition errors involving adjacent, digits are detected., 68. Use the Verhoeff check-digit scheme based on D5 to append a, check digit to 45723., 69. Prove that every element of Sn (n . 1) can be written as a product, of elements of the form (1k)., 70. (Indiana College Mathematics Competition) A card-shuffling machine always rearranges cards in the same way relative to the order, in which they were given to it. All of the hearts arranged in order, from ace to king were put into the machine, and then the shuffled, cards were put into the machine again to be shuffled. If the cards, emerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in what, order were the cards after the first shuffle?, 71. Show that a permutation with odd order must be an even permutation., 72. Let G be a group. Prove or disprove that H 5 {g2 | g [ G} is a subgroup of G. (Compare with Example 5 in Chapter 3.), 73. Let H 5 {a2 | a [ S4} and K 5 {a2 | a [ S5}. Prove H 5 A4 and, K 5 A5., 74. Let H 5 {a2 | a [ S6}. Prove H Z A6., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 123, , 75. Determine integers n for which H {a [ An | a2 e} is a subgroup of An., 76. Given that b and g are in S4 with bg 114322, gb 112432, and, b112 4, determine b and g., 77. Why does the fact that the orders of the elements of A4 are 1, 2, and, 3 imply that |Z(A4)| 5 1?, 78. Find five subgroups of S5 of order 24., 79. Find six subgroups of order 60 in S6., 80. For n . 1, let H be the set of all permutations in Sn that can be, expressed as a product of a multiple of four transpositions. Show, that H 5 An., 81. Shown below are four tire rotation patterns recommended by the, Dunlop Tire Company. Explain how these patterns can be represented as permutations in S4 and find the smallest subgroup of S4, that contains these four patterns. Is the subgroup Abelian?, , Rear Wheel Drive, Vehicles, , X Tires to, the Driven Axle, Front Wheel Drive, Vehicles, , FRONT, , FRONT, Modified, X, , Modified X, , 4 Wheel Drive, Vehicles, , Alternate Pattern, , FRONT, , FRONT, , X, , Normal, , 82. Label the four locations of tires on an automobile with the labels, 1, 2, 3, and 4, clockwise. Let a represent the operation of switching, the tires in positions 1 and 3 and switching the tires in positions, 2 and 4. Let b represent the operation of rotating the tires in positions 2, 3, and 4 clockwise and leaving the tire in position 1 as is., Let G be the group of all possible combinations of a and b. How, many elements are in G?, 83. What would be wrong with using the 2-cycle notation (11) instead, of the 1-cycle (1) to indicate that a cycle sends 1 to 1?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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124, , Groups, , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , References, 1. J. A. Gallian, “The Mathematics of Identification Numbers,” The College, Mathematics Journal 22 (1991): 194–202., 2. J. Verhoeff, Error Detecting Decimal Codes, Amsterdam: Mathematisch, Centrum, 1969., , Suggested Readings, Douglas E. Ensley, “Invariants Under Actions to Amaze Your Friends,” Mathematics Magazine, Dec. 1999: 383–387., This article explains some card tricks that are based on permutation, groups., Dmitry Fomin, “Getting It Together with ‘Polyominoes,’” Quantum,, Nov./Dec. 1991: 20–23., In this article, permutation groups are used to analyze various sorts of, checkerboard tiling problems., J. A. Gallian, “Error Detection Methods,” ACM Computing Surveys 28, (1996): 504–517., This article gives a comprehensive survey of error-detection methods that, use check digits. This article can be downloaded at http://www.d.umn, .edu/~jgallian/detection.pdf, I. N. Herstein and I. Kaplansky, Matters Mathematical, New York: Chelsea,, 1978., Chapter 3 of this book discusses several interesting applications of permutations to games., Douglas Hofstadter, “The Magic Cube’s Cubies Are Twiddled by Cubists and, Solved by Cubemeisters,” Scientific American 244 (1981): 20–39., This article, written by a Pulitzer Prize recipient, discusses the group theory involved in the solution of the Magic (Rubik’s) Cube. In particular,, permutation groups, subgroups, conjugates (elements of the form xyx21),, commutators (elements of the form xyx21y21), and the “always even or, always odd” theorem (Theorem 5.5) are prominently mentioned. At one, point, Hofstadter says, “It is this kind of marvelously concrete illustration, of an abstract notion of group theory that makes the Magic Cube one of, the most amazing things ever invented for teaching mathematical ideas.”, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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5 | Permutation Groups, , 125, , John O. Kiltinen, Oval Track & Other Permutation Puzzles & Just Enough, Group Theory to Solve Them, Washington, D.C.: Mathematical Association, of America, 2003., This book and the software that comes with it present the user with an array, of computerized puzzles, plus tools to vary them in thousands of ways. The, book provides the background needed to use the puzzle software to its fullest, potential, and also gives the reader a gentle, not-too-technical introduction to, the theory of permutation groups that is a prerequisite to a full understanding, of how to solve puzzles of this type. The website http://www-instruct.nmu, .edu/math_cs/kiltinen/web/mathpuzzles/ provides resources that expand, upon the book. It also has news about puzzle software—modules that add, functionality and fun to puzzles., Vladimir Dubrovsky, “Portrait of Three Puzzle Graces,” Quantum, Nov./Dec., 1991: 63–66., The author uses permutation groups to analyze solutions to the 15 puzzle,, Rubik’s Cube, and Rubik’s Clock., A. White and R. Wilson, “The Hunting Group,” Mathematical Gazette 79, (1995): 5–16., This article explains how permutation groups are used in bell ringing., S. Winters, “Error-Detecting Schemes Using Dihedral Groups,” UMAP, Journal 11, no. 4 (1990): 299–308., This article discusses error-detection schemes based on Dn for odd n., Schemes for both one and two check digits are analyzed., , Suggested Software, GAP is free for downloading. Versions are available for Unix, Windows,, and Macintosh at:, http://www.gap-system.org, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Augustin Cauchy, You see that little young, man? Well! He will supplant, all of us in so far as we are, mathematicians., , Stock Montage, , Spoken by Lagrange, to Laplace about the, 11-year-old Cauchy, , This stamp was issued by France, in Cauchy’s honor., , Augustin Louis Cauchy was born on, August 21, 1789, in Paris. By the time, he was 11, both Laplace and Lagrange had, recognized Cauchy’s extraordinary talent, for mathematics. In school he won prizes for, Greek, Latin, and the humanities. At the age, of 21, he was given a commission in, Napoleon’s army as a civil engineer. For the, next few years, Cauchy attended to his engineering duties while carrying out brilliant, mathematical research on the side., In 1815, at the age of 26, Cauchy was, made Professor of Mathematics at the École, Polytechnique and was recognized as the, leading mathematician in France. Cauchy, and his contemporary Gauss were among, the last mathematicians to know the whole, of mathematics as known at their time, and, both made important contributions to nearly, , every branch, both pure and applied, as well, as to physics and astronomy., Cauchy introduced a new level of rigor, into mathematical analysis. We owe our, contemporary notions of limit and continuity to him. He gave the first proof of the, Fundamental Theorem of Calculus. Cauchy, was the founder of complex function theory, and a pioneer in the theory of permutation, groups and determinants. His total written, output of mathematics fills 24 large volumes., He wrote more than 500 research papers, after the age of 50. Cauchy died at the age of, 67 on May 23, 1857., For more information about Cauchy,, visit:, http://www–groups.dcs, .st-and.ac.uk/~history/, , 126, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 Isomorphisms, , The basis for poetry and scientific discovery is the ability to comprehend, the unlike in the like and the like in the unlike., jacob bronowski, , Motivation, Suppose an American and a German are asked to count a handful of objects. The American says, “One, two, three, four, five, . . . ,” whereas the, German says, “Eins, zwei, drei, vier, fünf, . . . .” Are the two doing different things? No. They are both counting the objects, but they are using different terminology to do so. Similarly, when one person says, “Two plus, three is five” and another says, “Zwei und drei ist fünf,” the two are in, agreement on the concept they are describing, but they are using different, terminology to describe the concept. An analogous situation often occurs, with groups; the same group is described with different terminology. We, have seen two examples of this so far. In Chapter 1, we described the symmetries of a square in geometric terms (e.g., R90), whereas in Chapter 5 we, described the same group by way of permutations of the corners. In both, cases, the underlying group was the symmetries of a square. In Chapter 4,, we observed that when we have a cyclic group of order n generated by a,, the operation turns out to be essentially that of addition modulo n, since, aras 5 ak, where k 5 (r 1 s) mod n. For example, each of U(43) and U(49), is cyclic of order 42. So, each has the form kal, where aras 5 a (r 1 s)mod 42., , Definition and Examples, In this chapter, we give a formal method for determining whether two, groups defined in different terms are really the same. When this is the, case, we say that there is an isomorphism between the two groups. This, notion was first introduced by Galois about 180 years ago. The term, isomorphism is derived from the Greek words isos, meaning “same” or, “equal,” and morphe, meaning “form.” R. Allenby has colorfully, , 127, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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128, , Groups, , defined an algebraist as “a person who can’t tell the difference between, isomorphic systems.”, Definition Group Isomorphism, , An isomorphism f from a group G to a group G is a one-to-one mapping (or function) from G onto G that preserves the group operation., That is,, for all a, b in G., , f(ab) 5 f(a)f(b), , If there is an isomorphism from G onto G, we say that G and G are, isomorphic and write G < G., , This definition can be visualized as shown in Figure 6.1. The pairs, of dashed arrows represent the group operations., G, , φ, b, , φ, , a, , G, , φ (b), φ (a), , φ, φ (a)φ (b), , ab, , Figure 6.1, , It is implicit in the definition of isomorphism that isomorphic, groups have the same order. It is also implicit in the definition of, isomorphism that the operation on the left side of the equal sign is that, of G, whereas the operation on the right side is that of G. The four, cases involving ? and 1 are shown in Table 6.1., Table 6.1, , G Operation, , G Operation, , Operation Preservation, , ?, ?, 1, 1, , ?, 1, ?, 1, , f(a ? b) 5 f(a) ? f(b), f(a ? b) 5 f(a) 1 f(b), f(a 1 b) 5 f(a) ? f(b), f(a 1 b) 5 f(a) 1 f(b), , There are four separate steps involved in proving that a group G is, isomorphic to a group G., Step 1 “Mapping.” Define a candidate for the isomorphism; that is, define a function f from G to G., Step 2 “1–1.” Prove that f is one-to-one; that is, assume that f(a) 5, f(b) and prove that a 5 b., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 129, , Step 3 “Onto.” Prove that f is onto; that is, for any element g in G,, find an element g in G such that f(g) 5 g., Step 4 “O.P.” Prove that f is operation-preserving; that is, show that, f(ab) 5 f(a)f(b) for all a and b in G., None of these steps is unfamiliar to you. The only one that may appear, novel is the fourth one. It requires that one be able to obtain the same, result by combining two elements and then mapping, or by mapping, two elements and then combining them. Roughly speaking, this says, that the two processes—operating and mapping—can be done in either, order without affecting the result. This same concept arises in calculus, when we say, lim 1f 1x2 . g1x2 2 xSa, lim f 1x2 xSa, lim g1x2, xSa, , or, , 冮, , a, , b, , 1f g2 dx , , 冮, , a, , b, , b, , f dx , , 冮 g dx., a, , Before going any further, let’s consider some examples., EXAMPLE 1 Let G be the real numbers under addition and let G be, the positive real numbers under multiplication. Then G and G are isomorphic under the mapping f(x) 5 2x. Certainly, f is a function from, G to G. To prove that it is one-to-one, suppose that 2x 5 2y. Then log2 2x 5, log2 2y, and therefore x 5 y. For “onto,” we must find for any positive, real number y some real number x such that f(x) 5 y; that is, 2x 5 y., Well, solving for x gives log2 y. Finally,, f(x 1 y) 5 2x1y 5 2x ? 2y 5 f(x)f(y), for all x and y in G, so that f is operation-preserving as well., EXAMPLE 2 Any infinite cyclic group is isomorphic to Z. Indeed, if, a is a generator of the cyclic group, the mapping ak → k is an, isomorphism. Any finite cyclic group kal of order n is isomorphic, to Zn under the mapping ak → k mod n. That these correspondences are, functions and are one-to-one is the essence of Theorem 4.1. Obviously,, the mappings are onto. That the mappings are operation-preserving, follows from Exercise 9 in Chapter 0 in the finite case and from the, definitions in the infinite case., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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130, , Groups, , EXAMPLE 3 The mapping from R under addition to itself given by, f(x) 5 x3 is not an isomorphism. Although f is one-to-one and onto, it, is not operation-preserving, since it is not true that (x 1 y)3 5 x3 1 y3, for all x and y., EXAMPLE 4 U(10) < Z4 and U(5) < Z4. To verify this, one need, only observe that both U(10) and U(5) are cyclic of order 4. Then appeal to Example 2., EXAMPLE 5 U(10) ] U(12). This is a bit trickier to prove. First,, note that x2 5 1 for all x in U(12). Now, suppose that f is an isomorphism from U(10) onto U(12). Then,, f(9) 5 f(3 ? 3) 5 f(3)f(3) 5 1, and, f(1) 5 f(1 ? 1) 5 f(1)f(1) 5 1., Thus, f(9) 5 f(1), but 9 2 1, which contradicts the assumption that, f is one-to-one., EXAMPLE 6 There is no isomorphism from Q, the group of rational, numbers under addition, to Q*, the group of nonzero rational numbers, under multiplication. If f were such a mapping, there would be a rational number a such that f(a) 5 21. But then, 21 5 f(a) 5 f( 12a 1 12a) 5 f( 12a)f( 12a) 5 [f( 12a)]2., However, no rational number squared is 21., EXAMPLE 7 Let G 5 SL(2, R), the group of 2 3 2 real matrices, with determinant 1. Let M be any 2 3 2 real matrix with determinant 1., Then we can define a mapping from G to G itself by fM(A) 5 MAM21, for all A in G. To verify that fM is an isomorphism, we carry out the, four steps., Step 1 fM is a function from G to G. Here, we must show that fM(A) is, indeed an element of G whenever A is. This follows from properties of, determinants:, det (MAM21) 5 (det M)(det A)(det M)21 5 1 ? 1 ? 121 5 1., Thus, MAM21 is in G., Step 2 fM is one-to-one. Suppose that fM(A) 5 fM(B). Then MAM21 5, MBM21 and, by left and right cancellation, A 5 B., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 131, , Step 3 fM is onto. Let B belong to G. We must find a matrix A in G, such that fM(A) 5 B. How shall we do this? If such a matrix A is to exist, it must have the property that MAM21 5 B. But this tells us exactly, what A must be! For we can solve for A to obtain A 5 M21BM and, verify that fM(A) 5 MAM21 5 M(M21BM)M21 5 B., Step 4 fM is operation-preserving. Let A and B belong to G. Then,, fM(AB) 5 M(AB)M21 5 MA(M21M)BM21, 5 (MAM21)(MBM21) 5 fM(A)fM(B)., The mapping fM is called conjugation by M., , Cayley’s Theorem, Our first theorem is a classic result of Cayley. An important generalization of it will be given in Chapter 25., Theorem 6.1 Cayley’s Theorem (1854), Every group is isomorphic to a group of permutations., , PROOF To prove this, let G be any group. We must find a group G of, permutations that we believe is isomorphic to G. Since G is all we have, to work with, we will have to use it to construct G. For any g in G,, define a function Tg from G to G by, Tg(x) 5 gx, , for all x in G., , (In words, Tg is just multiplication by g on the left.) We leave it as an, exercise (Exercise 33) to prove that Tg is a permutation on the set of, elements of G. Now, let G 5 {Tg | g [ G}. Then, G is a group under, the operation of function composition. To verify this, we first observe, that for any g and h in G we have TgTh(x) 5 Tg(Th(x)) 5 Tg(hx) 5 g(hx) 5, (gh)x 5 Tgh(x), so that TgTh 5 Tgh. From this it follows that Te is the, identity and (Tg)21 5 Tg21 (see Exercise 9). Since function composition, is associative, we have verified all the conditions for G to be a group., The isomorphism f between G and G is now ready-made. For every, g in G, define f(g) 5 Tg. If Tg 5 Th, then Tg(e) 5 Th(e) or ge 5 he., Thus, g 5 h and f is one-to-one. By the way G was constructed, we, see that f is onto. The only condition that remains to be checked is that, f is operation-preserving. To this end, let a and b belong to G. Then, f(ab) 5 Tab 5 TaTb 5 f(a)f(b)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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132, , Groups, , The group G constructed previously is called the left regular representation of G., EXAMPLE 8 For concreteness, let us calculate the left regular representation U1122 for U(12) 5 {1, 5, 7, 11}. Writing the permutations of, U(12) in array form, we have (remember, Tx is just multiplication by x), T1 c, T7 c, , 1 5 7 11, d,, 1 5 7 11, , T5 c, , 1 5 7 11, d,, 5 1 11 7, , 1 5 7 11, d,, 7 11 1 5, , T11 c, , 1 5 7 11, d., 11 7 5 1, , It is instructive to compare the Cayley tables for U(12) and its left regular representation U1122., U(12), , 1, , 5, , 7, , 11, , U1122, , T1, , T5, , T7, , T11, , 1, 5, 7, 11, , 1, 5, 7, 11, , 5, 1, 11, 7, , 7, 11, 1, 5, , 11, 7, 5, 1, , T1, T5, T7, T11, , T1, T5, T7, T11, , T5, T1, T11, T7, , T7, T11, T1, T5, , T11, T7, T5, T1, , It should be abundantly clear from these tables that U(12) and U1122, are only notationally different., Cayley’s Theorem is important for two contrasting reasons. One is, that it allows us to represent an abstract group in a concrete way. A second is that it shows that the present-day set of axioms we have adopted, for a group is the correct abstraction of its much earlier predecessor—a, group of permutations. Indeed, Cayley’s Theorem tells us that abstract, groups are not different from permutation groups. Rather, it is the, viewpoint that is different. It is this difference of viewpoint that has, stimulated the tremendous progress in group theory and many other, branches of mathematics in the 20th century., It is sometimes very difficult to prove or disprove, whichever the, case may be, that two particular groups are isomorphic. For example, it, requires somewhat sophisticated techniques to prove the surprising fact, that the group of real numbers under addition is isomorphic to the, group of complex numbers under addition. Likewise, it is not easy to, prove the fact that the group of nonzero complex numbers under, multiplication is isomorphic to the group of complex numbers with absolute value of 1 under multiplication. In geometric terms, this says that,, as groups, the punctured plane and the unit circle are isomorphic [1]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 133, , Properties of Isomorphisms, Our next two theorems give a catalog of properties of isomorphisms, and isomorphic groups., Theorem 6.2 Properties of Isomorphisms Acting on Elements, Suppose that f is an isomorphism from a group G onto a group G., Then, 1. f carries the identity of G to the identity of G., 2. For every integer n and for every group element a in G, f(an) 5, [f(a)]n., 3. For any elements a and b in G, a and b commute if and only if, f(a) and f(b) commute., 4. G 5 kal if and only if G 5 kf(a)l., 5. |a| 5 |f(a)| for all a in G (isomorphisms preserve orders)., 6. For a fixed integer k and a fixed group element b in G, the, equation xk 5 b has the same number of solutions in G as does, the equation xk 5 f(b) in G., 7. If G is finite, then G and G have exactly the same number of, elements of every order., , PROOF We will restrict ourselves to proving only properties 1, 2, and 4,, but observe that property 5 follows from properties 1 and 2, property 6, follows from property 2, and property 7 follows from property 5. For, convenience, let us denote the identity in G by e and the identity in G, by e. Then, since e 5 ee, we have, f(e) 5 f(ee) 5 f(e)f(e)., Also, because f(e) [ G, we have f(e) 5 ef (e), as well. Thus, by cancellation, e 5 f(e). This proves property 1., For positive integers, property 2 follows from the definition of an, isomorphism and mathematical induction. If n is negative, then 2n is, positive, and we have from property 1 and the observation about the, positive integer case that e 5 f(e) 5 f(gng 2n) 5 f(g n)f(g 2n) 5, f(gn)(f(g))2n. Thus, multiplying both sides on the right by (f(g))n, we, have (f(g))n 5 f(gn). Property 1 takes care of the case n 5 0., To prove property 4, let G 5 kal and note that, by closure, kf(a)l #, G. Because f is onto, for any element b in G, there is an element ak in, G such that f(ak) 5 b. Thus, b 5 (f(a))k and so b [ kf(a)l. This, proves that G 5 kf(a)l., Now suppose that G 5 kf(a)l. Clearly, kal # G. For any element, b in G, we have f(b) [ kf(a)l. So, for some integer k we have, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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134, , Groups, , f(b) 5 (f(a))k 5 f(ak). Because f is one-to-one, b 5 ak. This proves, that kal 5 G., When the group operation is addition, property 2 of Theorem 6.2 is, f(na) 5 nf(a); property 4 says that an isomorphism between two, cyclic groups takes a generator to a generator., Property 6 is quite useful for showing that two groups are not isomorphic. Often b is picked to be the identity. For example, consider C*, and R*. Because the equation x4 5 1 has four solutions in C* but only, two in R*, no matter how one attempts to define an isomorphism from, C* to R*, property 6 cannot hold., Theorem 6.3 Properties of Isomorphisms Acting on Groups, Suppose that f is an isomorphism from a group G onto a group G., Then, f21 is an isomorphism from G onto G., G is Abelian if and only if G is Abelian., G is cyclic if and only if G is cyclic., If K is a subgroup of G, then f(K) 5 {f(k) | k [ K} is a, subgroup of G., 5. If K is a subgroup of G, then f21 (K) 5 {g [ G | f(g) [ K} is, a subgroup of G., 6. f(Z(G)) 5 Z(G)., , 1., 2., 3., 4., , PROOF Properties 1 and 4 are left as exercises (Exercises 31 and 32)., Properties 2 and 6 are a direct consequence of property 3 of Theorem 6.2., Property 3 follows from property 4 of Theorem 6.2 and property 1 of, Theorem 6.3. Property 5 follows from properties 1 and 4., Theorems 6.2 and 6.3 show that isomorphic groups have many properties in common. Actually, the definition is precisely formulated so, that isomorphic groups have all group theoretic properties in common., By this we mean that if two groups are isomorphic, then any property, that can be expressed in the language of group theory is true for one if, and only if it is true for the other. This is why algebraists speak of isomorphic groups as “equal” or “the same.” Admittedly, calling such, groups equivalent, rather than the same, might be more appropriate, but, we bow to long-standing tradition., , Automorphisms, Certain kinds of isomorphisms are referred to so often that they have, been given special names., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 135, , Definition Automorphism, , An isomorphism from a group G onto itself is called an automorphism, of G., , The isomorphism in Example 7 is an automorphism of SL(2, R)., Two more examples follow., EXAMPLE 9 The function f from C to C given by f(a 1 bi) 5, a 2 bi is an automorphism of the group of complex numbers under, addition. The restriction of f to C* is also an automorphism of the, group of nonzero complex numbers under multiplication. (See, Exercise 35.), EXAMPLE 10 Let R2 5 {(a, b) | a, b [ R}. Then f(a, b) 5 (b, a), is an automorphism of the group R2 under componentwise addition., Geometrically, f reflects each point in the plane across the line y 5 x., More generally, any reflection across a line passing through the, origin or any rotation of the plane about the origin is an automorphism of R2., The isomorphism in Example 7 is a particular instance of an automorphism that arises often enough to warrant a name and notation of, its own., Definition Inner Automorphism Induced by a, , Let G be a group, and let a [ G. The function fa defined by fa(x) 5, axa21 for all x in G is called the inner automorphism of G induced by a., , We leave it for the reader to show that fa is actually an automorphism of G. (Use Example 7 as a model.), EXAMPLE 11 The action of the inner automorphism of D4 induced, by R90 is given in the following table., x, R0, R90, R180, R270, H, V, D, D9, , fR90, →, →, →, →, →, →, →, →, , R 90 x R 90–1, R90R0R 90–1 5 R0, R90R90R 9021 5 R90, R90R180R 9021 5 R180, R90R270R 9021 5 R270, R90HR 9021 5 V, R90VR 9021 5 H, R90DR 9021 5 D9, , → R90D9R 9021 5 D, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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136, , Groups, , When G is a group, we use Aut(G) to denote the set of all automorphisms of G and Inn(G) to denote the set of all inner automorphisms of G. The reason these sets are noteworthy is demonstrated by, the next theorem., Theorem 6.4 Aut(G) and Inn(G) Are Groups†, The set of automorphisms of a group and the set of inner, automorphisms of a group are both groups under the operation of, function composition., , PROOF The proof of Theorem 6.4 is left as an exercise (Exercise 15)., , The determination of Inn(G) is routine. If G 5 {e, a, b, c. . . .}, then, Inn(G) 5 {fe, fa, fb, fc, . . .}. This latter list may have duplications,, however, since fa may be equal to fb even though a 2 b (see Exercise, 43). Thus, the only work involved in determining Inn(G) is deciding, which distinct elements give the distinct automorphisms. On the other, hand, the determination of Aut(G) is, in general, quite involved., EXAMPLE 12 Inn(D4), To determine Inn(D4), we first observe that the complete list of inner, automorphisms is fR0, fR90, fR180, fR270, fH, fV, fD, and fD9. Our job is, to determine the repetitions in this list. Since R180 [ Z(D4), we have, fR 180(x) 5 R180 xR180 21 5 x, so that fR 180 5 fR 0. Also, fR 270(x) 5, R270 xR27021 5 R90R180 xR18021R9021 5 R90 xR9021 5 fR90(x). Similarly,, since H 5 R180V and D9 5 R180D, we have fH 5 fV and fD 5 fD9., This proves that the previous list can be pared down to fR0, fR90, fH,, and fD. We leave it to the reader to show that these are distinct, (Exercise 13)., EXAMPLE 13 Aut(Z10), To compute Aut(Z10), we try to discover enough information about an, element a of Aut(Z10) to determine how a must be defined. Because Z10, is so simple, this is not difficult to do. To begin with, observe that once, we know a(1), we know a(k) for any k, because, , †The, , group Aut(G) was first studied by O. Hölder in 1893 and, independently, by E. H., Moore in 1894., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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137, , 6 | Isomorphisms, , a(k) 5 a(1 1 1 1 ? ? ? 1 1), k terms, 5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 ka(1)., k terms, So, we need only determine the choices for a(1) that make a an, automorphism of Z10. Since property 5 of Theorem 6.2 tells us that, |a(1)| 5 10, there are four candidates for a(1):, a(1) 5 1,, , a(1) 5 3,, , a(1) 5 7,, , a(1) 5 9., , To distinguish among the four possibilities, we refine our notation by, denoting the mapping that sends 1 to 1 by a 1, 1 to 3 by a 3, 1 to 7 by a 7,, and 1 to 9 by a 9. So the only possibilities for Aut(Z10) are a 1, a 3, a 7, and, a 9. But are all these automorphisms? Clearly, a1 is the identity. Let us, check a3. Since x mod 10 5 y mod 10 implies 3x mod 10 5 3y mod 10,, a3 is well defined. Moreover, because a3 112 3 is a generator of Z10, it, follows that a 3 is onto (and, by Exercise 12 in Chapter 5, it is also oneto-one). Finally, since a3(a 1 b) 5 3(a 1 b) 5 3a 1 3b 5 a3(a) 1 a3(b),, we see that a 3 is operation-preserving as well. Thus, a 3 [ Aut(Z10). The, same argument shows that a7 and a 9 are also automorphisms., This gives us the elements of Aut(Z10) but not the structure. For instance, what is a3a3? Well, (a3a3)(1) 5 a3(3) 5 3 ? 3 5 9 5 a9(1), so, a3a3 5 a 9. Similar calculations show that a33 5 a 7 and a34 5 a1, so, that |a3| 5 4. Thus, Aut(Z10 ) is cyclic. Actually, the following Cayley, tables reveal that Aut(Z10 ) is isomorphic to U(10)., U(10), , 1, , 3, , 7, , 9, , Aut(Z10), , a1, , a3, , a7, , a9, , 1, 3, 7, 9, , 1, 3, 7, 9, , 3, 9, 1, 7, , 7, 1, 9, 3, , 9, 7, 3, 1, , a1, a3, a7, a9, , a1, a3, a7, a9, , a3, a9, a1, a7, , a7, a1, a9, a3, , a9, a7, a3, a1, , With Example 13 as a guide, we are now ready to tackle the group, Aut(Zn). The result is particularly nice, since it relates the two kinds of, groups we have most frequently encountered thus far—the cyclic, groups Zn and the U-groups U(n)., Theorem 6.5 Aut(Zn) < U(n), For every positive integer n, Aut(Zn) is isomorphic to U(n)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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138, , Groups, , PROOF As in Example 13, any automorphism a is determined by the, value of a(1), and a(1) [ U(n). Now consider the correspondence, from Aut(Zn) to U(n) given by T: a → a(1). The fact that a(k) 5 ka(1), (see Example 13) implies that T is a one-to-one mapping. For if a and, b belong to Aut(Zn) and a(1) 5 b(1), then a(k) 5 ka(1) 5 kb(1) 5, b(k) for all k in Zn, and therefore a 5 b., To prove that T is onto, let r [ U(n) and consider the mapping a, from Zn to Zn defined by a(s) 5 sr (mod n) for all s in Zn. We leave it as, an exercise to verify that a is an automorphism of Zn (see Exercise 27)., Then, since T(a) 5 a(1) 5 r, T is onto U(n)., Finally, we establish the fact that T is operation-preserving. Let a,, b [ Aut(Zn). We then have, T(ab) 5 (ab)(1) 5 a(b(1)) 5 a(1 1 1 1 ? ? ? 1 1), b(1), , 5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 a(1)b(1), b(1), , 5 T(a)T(b)., This completes the proof., , Exercises, Being a mathematician is a bit like being a manic depressive: you spend, your life alternating between giddy elation and black despair., steven g. krantz, A Primer of Mathematical Writing, , 1. Find an isomorphism from the group of integers under addition to, the group of even integers under addition., 2. Find Aut(Z)., 3. Let R1 be the group of positive real numbers under multiplication., Show that the mapping f(x) 5 2x is an automorphism of R1., 4. Show that U(8) is not isomorphic to U(10)., 5. Show that U(8) is isomorphic to U(12)., 6. Prove that isomorphism is an equivalence relation. That is, for any, groups G, H, and K, G < G, G < H implies H < G, and G < H and, H < K implies G < K., 7. Prove that S4 is not isomorphic to D12., 8. Show that the mapping a → log10 a is an isomorphism from R+, under multiplication to R under addition., 9. In the notation of Theorem 6.1, prove that Te is the identity and, that (Tg)21 5 Tg ., 21, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 139, , 10. Let G be a group. Prove that the mapping a(g) 5 g21 for all g in G, is an automorphism if and only if G is Abelian., 11. If g and h are elements from a group, prove that fgfh 5 fgh., 12. Find two groups G and H such that G ] H, but Aut(G) < Aut(H)., 13. Prove the assertion in Example 12 that the inner automorphisms, fR , fR , fH, and fD of D4 are distinct., 0, 90, 14. Find Aut(Z6)., 15. If G is a group, prove that Aut(G) and Inn(G) are groups., 16. If a group G is isomorphic to H, prove that Aut(G) is isomorphic to, Aut(H)., 17. Suppose f belongs to Aut(Zn) and a is relatively prime to n., If f(a) 5 b, determine a formula for f(x)., 18. Let H be the subgroup of all rotations in Dn and let f be an automorphism of Dn. Prove that f(H) 5 H. (In words, an automorphism of Dn carries rotations to rotations.), 19. Let H 5 {b [ S5 | b(1) 5 1} and K 5 {b [ S5 | b(2) 5 2}. Prove, that H is isomorphic to K. Is the same true if S5 is replaced by Sn,, where n $ 3?, 20. Show that Z has infinitely many subgroups isomorphic to Z., 21. Let n be an even integer greater than 2 and let f be an automorphism of Dn. Determine f(R180)., 22. Let f be an automorphism of a group G. Prove that H 5 {x [ G |, f(x) 5 x} is a subgroup of G., 23. Give an example of a cyclic group of smallest order that contains a, subgroup isomorphic to Z12 and a subgroup isomorphic to Z20. No, need to prove anything, but explain your reasoning., 24. Suppose that f: Z20 S Z20 is an automorphism and f(5) 5 5. What, are the possibilities for f(x)?, 25. Identify a group G that has subgroups isomorphic to Zn for all positive integers n., 26. Prove that the mapping from U(16) to itself given by x → x3 is an, automorphism. What about x → x5 and x → x7? Generalize., 27. Let r [ U(n). Prove that the mapping a: Zn → Zn defined by a(s) 5, sr mod n for all s in Zn is an automorphism of Zn . (This exercise is, referred to in this chapter.), 1 a, 28. The group e c, d ` a [ Z f is isomorphic to what familiar, 0 1, group? What if Z is replaced by R?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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140, , Groups, , 29. If f and g are isomorphisms from the cyclic group HaI to some, group and f1a2 g1a2, prove that f g., 30. Suppose that f: Z50 S Z50 is an automorphism with f1112 13., Determine a formula for f1x2., 31. Prove property 1 of Theorem 6.3., 32. Prove property 4 of Theorem 6.3., 33. Referring to Theorem 6.1, prove that Tg is indeed a permutation on, the set G., 34. Prove or disprove that U(20) and U(24) are isomorphic., 35. Show that the mapping f(a 1 bi) 5 a 2 bi is an automorphism of, the group of complex numbers under addition. Show that f preserves complex multiplication as well—that is, f(xy) 5 f(x)f(y), for all x and y in C. (This exercise is referred to in Chapter 15.), 36. Let, G 5 {a 1 b22 | a, b are rational}, and, H5 ec, , a 2b, d ` a, b are rational f ., b a, Show that G and H are isomorphic under addition. Prove that G, and H are closed under multiplication. Does your isomorphism, preserve multiplication as well as addition? (G and H are examples, of rings—a topic we will take up in Part 3.), 37. Prove that Z under addition is not isomorphic to Q under addition., 38. Prove that the quaternion group (see Exercise 4, Supplementary Exercises for Chapters 1–4) is not isomorphic to the dihedral group D4., 39. Let C be the complex numbers and, M5 ec, , a, b, , b, d ` a, b [ R f ., a, , Prove that C and M are isomorphic under addition and that C* and, M*, the nonzero elements of M, are isomorphic under multiplication., 40. Let Rn 5 {(a1, a2, . . . , an) | ai [ R}. Show that the mapping f:, (a1, a2, . . . , an) → (2a1, 2a2, . . . , 2an) is an automorphism of, the group Rn under componentwise addition. This automorphism, is called inversion. Describe the action of f geometrically., 41. Consider the following statement: The order of a subgroup divides, the order of the group. Suppose you could prove this for finite, permutation groups. Would the statement then be true for all finite, groups? Explain., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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6 | Isomorphisms, , 141, , 42. Suppose that G is a finite Abelian group and G has no element of, order 2. Show that the mapping g → g2 is an automorphism of G., Show, by example, that there is an infinite Abelian group for which, the mapping g S g2 is one-to-one and operation-preserving but not, an automorphism., 43. Let G be a group and let g [ G. If z [ Z(G), show that the inner, automorphism induced by g is the same as the inner automorphism, induced by zg (that is, that the mappings fg and fzg are equal)., 44. Show that the mapping a → log10 a is an isomorphism from R+, under multiplication to R under addition., 45. Suppose that g and h induce the same inner automorphism of a, group G. Prove that h21g [ Z(G)., 46. Combine the results of Exercises 43 and 45 into a single “if and, only if” theorem., 47. If x and y are elements in Sn (n $ 3), prove that fx 5 fy implies, x 5 y. (Here, fx is the inner automorphism of Sn induced by x.), 48. Let f be an isomorphism from a group G to a group G and let a, belong to G. Prove that f(C(a)) 5 C(f(a))., 49. Suppose the f and g are isomorphisms of some group G to the, same group. Prove that H 5 {g [ G | f(g) 5 g(g)} is a subgroup, of G., 50. Suppose that b is an automorphism of a group G. Prove that H 5, {g [ G | b2 (g) 5 g} is a subgroup of G. Generalize., 51. Suppose that G is an Abelian group and f is an automorphism of, G. Prove that H 5 {x [ G | f(x) 5 x21} is a subgroup of G., 52. Given a group G, define a new group G* that has the same, elements as G with the operation * defines by a * b 5 ba for all, a and b in G*. Prove that the mapping from G to G* defined by, f(x) 5 x21 for all x in G is an isomorphism from G onto G*., 53. Let a belong to a group G and let |a| be finite. Let fa be the automorphism of G given by fa(x) 5 axa21. Show that |fa| divides |a|., Exhibit an element a from a group for which 1 , |fa| , |a|., 54. Let G 5 {0, 62, 64, 66, . . .} and H 5 {0, 63, 66, 69, . . .}., Show that G and H are isomorphic groups under addition. Does, your isomorphism preserve multiplication? Generalize to the case, when G kml and H knl, where m and n are integers., 55. Suppose that f is an automorphism of D4 such that f1R90 2 R270, and f1V2 V. Determine f1D2 and f1H2., 56. In Aut(Z9), let ai denote the automorphism that sends 1 to i where, gcd(i, 9) 5 1. Write a5 and a 8 as permutations of {0, 1, . . . , 8} in, disjoint cycle form. [For example, a2 5 (0)(124875)(36).], Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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142, , Groups, , 57. Write the permutation corresponding to R90 in the left regular representation of D4 in cycle form., 58. Show that every automorphism f of the rational numbers Q under, addition to itself has the form f(x) 5 xf(1)., 59. Prove that Q1, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup., 60. Prove that Q, the group of rational numbers under addition, is not, isomorphic to a proper subgroup of itself., 61. Prove that every automorphism of R*, the group of nonzero real, numbers under multiplication, maps positive numbers to positive, numbers and negative numbers to negative numbers., 62. Let G be a finite group. Show that in the disjoint cycle form of the, right regular representation Tg 1x2 xg of G, each cycle has, length 0 g 0 ., 63. Give a group theoretic proof that Q under addition is not isomorphic to R+ under multiplication., , Reference, 1. J. R. Clay, “The Punctured Plane Is Isomorphic to the Unit, Circle,” Journal of Number Theory 1 (1969): 500–501., , Computer Exercises, Software for the computer exercise in this chapter is available at the, website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Arthur Cayley, , The Granger Collection, New York, , Cayley is forging the weapons for future, generations of physicists., peter tait, , Arthur Cayley was born on August 16,, 1821, in England. His genius showed itself at, an early age. He published his first research, paper while an undergraduate of 20, and in, the next year he published eight papers., While still in his early 20s, he originated the, concept of n-dimensional geometry., After graduating from Trinity College,, Cambridge, Cayley stayed on for three years, as a tutor. At the age of 25, he began a 14year career as a lawyer. During this period,, he published approximately 200 mathematical papers, many of which are now classics., In 1863, Cayley accepted the newly established Sadlerian professorship of mathematics at Cambridge University. He spent, the rest of his life in that position. One of his, notable accomplishments was his role in the, successful effort to have women admitted to, Cambridge., , Among Cayley’s many innovations in, mathematics were the notions of an abstract, group and a group algebra, and the matrix, concept. He made major contributions to, geometry and linear algebra. Cayley and his, lifelong friend and collaborator J. J. Sylvester, were the founders of the theory of invariants,, which was later to play an important role in, the theory of relativity., Cayley’s collected works comprise 13, volumes, each about 600 pages in length., He died on January 26, 1895., To find more information about Cayley,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 143, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7, , Cosets and Lagrange’s, Theorem, It might be difficult, at this point, for students to see the extreme, importance of this result [Lagrange’s Theorem]. As we penetrate the subject, more deeply they will become more and more aware of its basic character., i. n. herstein, Topics in Algebra, , Properties of Cosets, In this chapter, we will prove the single most important theorem in finite, group theory—Lagrange’s Theorem. In his book on abstract algebra,, I. N. Herstein likened it to the ABC’s for finite groups. But first we introduce a new and powerful tool for analyzing a group—the notion of a, coset. This notion was invented by Galois in 1830, although the term, was coined by G. A. Miller in 1910., Definition Coset of H in G, , Let G be a group and let H be a nonempty subset of G. For any a [ G,, the set {ah | h [ H} is denoted by aH. Analogously, Ha 5 {ha | h [ H}, and aHa21 5 {aha21 | h [ H}. When H is a subgroup of G, the set aH is, called the left coset of H in G containing a, whereas Ha is called the right, coset of H in G containing a. In this case, the element a is called the coset, representative of aH (or Ha). We use |aH| to denote the number of elements in the set aH, and |Ha| to denote the number of elements in Ha., , EXAMPLE 1 Let G 5 S3 and H 5 {(1), (13)}. Then the left cosets of, H in G are, (1)H 5 H,, (12)H 5 {(12), (12)(13)} 5 {(12), (132)} 5 (132)H,, (13)H 5 {(13), (1)} 5 H,, (23)H 5 {(23), (23)(13)} 5 {(23), (123)} 5 (123)H., , 144, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 145, , EXAMPLE 2 Let _ 5 {R0, R180} in D4, the dihedral group of order 8., Then,, R0_ 5 _,, R90_ 5 {R90, R270} 5 R270_,, R180_ 5 {R180, R0} 5 _,, V_ 5 {V, H} 5 H_,, D_ 5 {D, D9} 5 D9_., EXAMPLE 3 Let H 5 {0, 3, 6} in Z9 under addition. In the case that, the group operation is addition, we use the notation a 1 H instead of, aH. Then the cosets of H in Z9 are, 0 1 H 5 {0, 3, 6} 5 3 1 H 5 6 1 H,, 1 1 H 5 {1, 4, 7} 5 4 1 H 5 7 1 H,, 2 1 H 5 {2, 5, 8} 5 5 1 H 5 8 1 H., The three preceding examples illustrate a few facts about cosets that, are worthy of our attention. First, cosets are usually not subgroups., Second, aH may be the same as bH, even though a is not the same as b., Third, since in Example 1 (12)H 5 {(12), (132)} whereas H(12) 5, {(12), (123)}, aH need not be the same as Ha., These examples and observations raise many questions. When does, aH 5 bH? Do aH and bH have any elements in common? When does, aH 5 Ha? Which cosets are subgroups? Why are cosets important? The, next lemma and theorem answer these questions. (Analogous results, hold for right cosets.), Lemma Properties of Cosets, Let H be a subgroup of G, and let a and b belong to G. Then,, 1., 2., 3., 4., 5., 6., 7., 8., 9., , a [ aH., aH 5 H if and only if a [ H., (ab)H 5 a(bH) and H(ab) 5 (Ha)b., aH 5 bH if and only if a [ bH., aH 5 bH or aH > bH 5 [., aH 5 bH if and only if a21b [ H., |aH| 5 |bH|., aH 5 Ha if and only if H 5 aHa21., aH is a subgroup of G if and only if a [ H., , PROOF, 1. a 5 ae [ aH., 2. To verify property 2, we first suppose that aH 5 H. Then a 5, ae [ aH 5 H. Next, we assume that a [ H and show that aH # H, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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146, , Groups, , 3., 4., 5., 6., 7., , 8., 9., , and H # aH. The first inclusion follows directly from the closure of, H. To show that H # aH, let h [ H. Then, since a [ H and h [ H, we, know that a21h [ H. Thus, h 5 eh 5 (aa21)h 5 a(a21h) [ aH., This follows directly from (ab)h 5 a(bh) and h(ab) 5 (ha)b., If aH 5 bH, then a 5 ae [ aH 5 bH. Conversely, if a [ bH we have, a 5 bh where h [ H, and therefore aH 5 (bh)H 5 b(hH) 5 bH., Property 5 follows directly from property 4, for if there is an element c in aH y bH, then cH 5 aH and cH 5 bH., Observe that aH 5 bH if and only if H 5 a21bH. The result now, follows from property 2., To prove that |aH| 5 |bH|, it suffices to define a one-to-one mapping from aH onto bH. Obviously, the correspondence ah → bh, maps aH onto bH. That it is one-to-one follows directly from the, cancellation property., Note that aH 5 Ha if and only if (aH)a21 5 (Ha)a21 5 H(aa–1) 5, H—that is, if and only if aHa21 5 H., If aH is a subgroup, then it contains the identity e. Thus, aH >, eH 2 [; and, by property 5, we have aH 5 eH 5 H. Thus, from, property 2, we have a [ H. Conversely, if a [ H, then, again by, property 2, aH 5 H., , Although most mathematical theorems are written in symbolic form,, one should also know what they say in words. In the preceding lemma,, property 1 says simply that the left coset of H containing a does contain a., Property 2 says that the H “absorbs” an element if and only if the element, belongs to H. Property 3 says that the left coset of H created by multiplying H on the left by ab is the same as the one created by multiplying H on, the left by b then multiplying the resulting coset bH on the left by a (and, analogously for multiplication on the right by ab). Property 4 shows that a, left coset of H is uniquely determined by any one of its elements. In particular, any element of a left coset can be used to represent the coset., Property 5 says—and this is very important—that two left cosets of H are, either identical or disjoint. Thus, a left coset of H is uniquely determined, by any one of its elements. In particular, any element of a left coset can be, used to represent the coset. Property 6 shows how we may transfer a question about equality of left cosets of H to a question about H itself and vice, versa. Property 7 says that all left cosets of H have the same size. Property, 8 is analogous to property 6 in that it shows how a question about the, equality of the left and right cosets of H containing a is equivalent to a, question about the equality of two subgroups of G. The last property of the, lemma says that H itself is the only coset of H that is a subgroup of G., Note that properties 1, 5, and 7 of the lemma guarantee that the, left cosets of a subgroup H of G partition G into blocks of equal size., Indeed, we may view the cosets of H as a partitioning of G into, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 147, , equivalence classes under the equivalence relation defined by a , b, if aH 5 bH (see Theorem 0.7)., In practice, the subgroup H is often chosen so that the cosets partition the group in some highly desirable fashion. For example, if G is, 3-space R3 and H is a plane through the origin, then the coset (a, b, c) 1, H (addition is done componentwise) is the plane passing through the, point (a, b, c) and parallel to H. Thus, the cosets of H constitute a partition of 3-space into planes parallel to H. If G 5 GL(2, R) and, H 5 SL(2, R), then for any matrix A in G, the coset AH is the set of all, 2 3 2 matrices with the same determinant as A. Thus,, c, , 2 0, d H, 0 1, , is the set of all 2 3 2 matrices of determinant 2, , and, c, , 1 2, d H is the set of all 2 3 2 matrices of determinant 23., 2 1, , Property 5 of the lemma is useful for actually finding the distinct, cosets of a subgroup. We illustrate this in the next example., EXAMPLE 4 To find the cosets of H 5 {1, 15} in G 5 U(32) 5, {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}, we begin with, H 5 {1, 15}. We can find a second coset by choosing any element not, in H, say 3, as a coset representative. This gives the coset 3H 5 {3, 13}., We find our next coset by choosing a representative not already appearing in the two previously chosen cosets, say 5. This gives us the coset 5H 5, {5, 11}. We continue to form cosets by picking elements from U(32), that have not yet appeared in the previous cosets as representatives of, the cosets until we have accounted for every element of U(32). We then, have the complete list of all distinct cosets of H., , Lagrange’s Theorem and Consequences, We are now ready to prove a theorem that has been around for more, than 200 years—longer than group theory itself! (This theorem was not, originally stated in group theoretic terms.) At this stage, it should come, as no surprise., Theorem 7.1 Lagrange’s Theorem†: |H| Divides |G|, If G is a finite group and H is a subgroup of G, then |H| divides |G|., Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|., †Lagrange, , stated his version of this theorem in 1770, but the first complete proof was, given by Pietro Abbati some 30 years later., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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148, , Groups, , PROOF Let a1H, a2H, . . . , ar H denote the distinct left cosets of H in, G. Then, for each a in G, we have aH 5 aiH for some i. Also, by property 1 of the lemma, a [ aH. Thus, each member of G belongs to one, of the cosets aiH. In symbols,, G 5 a1H < ? ? ? < ar H., Now, property 5 of the lemma shows that this union is disjoint, so that, |G| 5 |a1H| 1 |a2H| 1 ? ? ? 1 |ar H|., Finally, since |aiH| 5 |H| for each i, we have |G| 5 r|H|., We pause to emphasize that Lagrange’s Theorem is a subgroup candidate criterion; that is, it provides a list of candidates for the orders of, the subgroups of a group. Thus, a group of order 12 may have subgroups, of order 12, 6, 4, 3, 2, 1, but no others. Warning! The converse of Lagrange’s Theorem is false. For example, a group of order 12 need not, have a subgroup of order 6. We prove this in Example 5., A special name and notation have been adopted for the number of, left (or right) cosets of a subgroup in a group. The index of a subgroup, H in G is the number of distinct left cosets of H in G. This number, is denoted by |G:H|. As an immediate consequence of the proof of, Lagrange’s Theorem, we have the following useful formula for the, number of distinct left (or right) cosets of H in G., Corollary 1 |G:H| 5 |G|/|H|, If G is a finite group and H is a subgroup of G, then |G:H| 5 |G|/|H|., , Corollary 2 |a| Divides |G|, In a finite group, the order of each element of the group divides the, order of the group., , PROOF Recall that the order of an element is the order of the subgroup, generated by that element., Corollary 3 Groups of Prime Order Are Cyclic, A group of prime order is cyclic., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 149, , PROOF Suppose that G has prime order. Let a [ G and a 2 e. Then,, |kal| divides |G| and |kal| 2 1. Thus, |kal| 5 |G| and the corollary, follows., Corollary 4 a|G| 5 e, Let G be a finite group, and let a [ G. Then, a|G| 5 e., , PROOF By Corollary 2, |G| 5 |a|k for some positive integer k. Thus,, a|G| 5 a|a|k 5 ek 5 e., Corollary 5 Fermat’s Little Theorem, For every integer a and every prime p, a p mod p 5 a mod p., , PROOF By the division algorithm, a 5 pm 1 r, where 0 # r , p., Thus, a mod p 5 r, and it suffices to prove that r p mod p 5 r. If r 5 0,, the result is trivial, so we may assume that r [ U( p). [Recall that, U(p) 5 {1, 2, . . . , p 2 1} under multiplication modulo p.] Then, by the, preceding corollary, r p21 mod p 5 1 and, therefore, r p mod p 5 r., Fermat’s Little Theorem has been used in conjunction with computers to test for primality of certain numbers. One case concerned the, number p 5 2257 2 1. If p is prime, then we know from Fermat’s Little, Theorem that 10 p mod p 5 10 mod p and, therefore, 10 p11 mod p 5, 100 mod p. Using multiple precision and a simple loop, a computer, 257, was able to calculate 10 p11 mod p 5 10 2 mod p in a few seconds., The result was not 100, and so p is not prime., EXAMPLE 5 The Converse of Lagrange’s Theorem Is False.†, The group A4 of order 12 has no subgroups of order 6. To verify this,, recall that A4 has eight elements of order 3 (a5 through a12, in the notation, of Table 5.1) and suppose that H is a subgroup of order 6. Let a be any, element of order 3 in A4. If a is not in H, then A4 5 H c aH. But then, a2 is in H or a2 is in aH. If a2 is in H then so is (a2)2 5 a4 5 a, so this case, is ruled out. If a2 is in aH, then a2 5 ah for some h in H, but this also implies that a is in H. This argument shows that any subgroup of A4 of order, 6 must contain all eight elements of A4 of order 3, which is absurd., †The, , first counterexample to the converse of Lagrange’s Theorem was given by Paolo, Ruffini in 1799., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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150, , Groups, , Lagrange’s Theorem demonstrates that the finiteness of a group imposes severe restrictions on the possible orders of subgroups. The next, theorem also places powerful limits on the existence of certain subgroups in finite groups., Theorem 7.2 |HK| 5 |H||K|/|H y K|, For two finite subgroups H and K of a group, define the set, HK 5 {hk | h [ H, k [ K}. Then |HK| 5 |H||K|/|H y K|., PROOF Although the set HK has |H||K| products, not all of these, products need represent distinct group elements. That is, we may have, hk 5 h9k9 where h ? h9 and k ? k9. To determine |HK|, we must find, the extent to which this happens. For every t in H y K, the product hk, 5 (ht)(t21k), so each group element in HK is represented by at least, |H y K| products in HK. But hk 5 h9k9 implies t 5 h21h9 5 kk921 [ H, y K, so that h9 5 ht and k9 5 t21k. Thus, each element in HK is represented by exactly |H y K| products. So, |HK| 5 |H||K|/| H y K|., EXAMPLE 6 A group of order 75 can have at most one subgroup of, order 25. (It is shown in Chapter 24 that every group of order 75 has a, subgroup of order 25). To see that a group of order 75 cannot have two, subgroups of order 25, suppose H and K are two such subgroups. Since, |H y K| divides |H| 5 25 and |H y K| 5 1 or 5 results in |HK| 5, |H||K|/| H y K| 5 25 ? 25/| H y K| 5 625 or 125 elements, we have, that |H y K| 5 25 and therefore H 5 K., For any prime p . 2, we know that Z2p and Dp are nonisomorphic, groups of order 2p. This naturally raises the question of whether there, could be other possible groups of these orders. Remarkably, with just, the simple machinery available to us at this point, we can answer this, question., Theorem 7.3 Classification of Groups of Order 2p, Let G be a group of order 2p, where p is a prime greater than 2. Then, G is isomorphic to Z2p or Dp., , PROOF We assume that G does not have an element of order 2p and, show that G < Dp. We begin by first showing that G must have an, element of order p. By our assumption and Lagrange’s Theorem, any, nonidentity element of G must have order 2 or p. Thus, to verify our assertion, we may assume that every nonidentity element of G has order 2., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 151, , In this case, we have for all a and b in the group ab 5 (ab)21 5 b21a21 5 ba,, so that G is Abelian. Then, for any nonidentity elements a, b [ G with, a 2 b, the set {e, a, b, ab} is closed and therefore is a subgroup of G of, order 4. Since this contradicts Lagrange’s Theorem, we have proved, that G must have an element of order p; call it a., Now let b be any element not in kal. Then by Lagrange’s Theorem, and our assumption that G does not have an element of order 2p, we, have that |b| 5 2 or p. Because |kal y kbl| divides |kal| 5 p and kal ? kbl, we have that |kal y kbl| 5 1. But then |b| 5 2, for otherwise, by Theorem, 7.2 |kalkbl|5|kal||kbl| 5 p2 . 2p 5 |G|, which is impossible. So, any, element of G not in kal has order 2., Next consider ab. Since ab o kal, our argument above shows that, |ab| 5 2. Then ab 5 (ab)21 5 b21a21 5 ba21. Moreover, this relation, completely determines the multiplication table for G. [For example,, a3(ba4) 5 a2(ab)a4 5 a2(ba21)a4 5 a(ab)a3 5 a(ba21)a3 5 (ab)a2 5, (ba21)a2 5 ba.] Since the multiplication table for all noncyclic groups, of order 2p is uniquely determined by the relation ab 5 ba21, all, noncyclic groups of order 2p must be isomorphic to each other. But of, course, Dp, the dihedral group of order 2p, is one such group., As an immediate corollary, we have that the non-Abelian groups S3,, the symmetric group of degree 3, and GL(2, Z2), the group of 2 3 2, matrices with nonzero determinants with entries from Z2 (see Example, 19 and Exercise 51 in Chapter 2) are isomorphic to D3., , An Application of Cosets, to Permutation Groups, Lagrange’s Theorem and its corollaries dramatically demonstrate the, fruitfulness of the coset concept. We next consider an application of, cosets to permutation groups., Definition Stabilizer of a Point, , Let G be a group of permutations of a set S. For each i in S, let stabG(i) 5, {f [ G | f(i) 5 i}. We call stabG(i) the stabilizer of i in G., , The student should verify that stabG(i) is a subgroup of G. (See, Exercise 35 in Chapter 5.), Definition Orbit of a Point, , Let G be a group of permutations of a set S. For each s in S, let orbG(s) 5, {f(s) | f [ G}. The set orbG(s) is a subset of S called the orbit of s, under G. We use |orbG(s)| to denote the number of elements in orbG(s)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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152, , Groups, , Example 7 should clarify these two definitions., EXAMPLE 7 Let, G 5 {(1), (132)(465)(78), (132)(465), (123)(456),, (123)(456)(78), (78)}., Then,, orbG(1) 5 {1, 3, 2},, orbG(2) 5 {2, 1, 3},, orbG(4) 5 {4, 6, 5},, orbG(7) 5 {7, 8},, , stabG(1) 5 {(1), (78)},, stabG(2) 5 {(1), (78)},, stabG(4) 5 {(1), (78)},, stabG(7) 5 {(1), (132)(465), (123)(456)}., , EXAMPLE 8 We may view D4 as a group of permutations of a, square region. Figure 7.1(a) illustrates the orbit of the point p under D4,, and Figure 7.1(b) illustrates the orbit of the point q under D4. Observe, that stabD ( p) 5 {R0, D}, whereas stabD (q) 5 {R0}., 4, , 4, , p, , q, (b), , (a), , Figure 7.1, , The preceding two examples also illustrate the following theorem., Theorem 7.4 Orbit-Stabilizer Theorem, Let G be a finite group of permutations of a set S. Then, for, any i from S, |G| 5 |orbG (i)| |stabG(i)|., , PROOF By Lagrange’s Theorem, |G|/|stabG(i)| is the number of distinct left cosets of stabG(i) in G. Thus, it suffices to establish a oneto-one correspondence between the left cosets of stabG(i) and the, elements in the orbit of i. To do this, we define a correspondence T, by mapping the coset fstabG(i) to f(i) under T. To show that T is a welldefined function, we must show that astabG(i) 5 bstabG(i) implies a(i) 5, b(i). But astabG (i) 5 bstabG (i) implies a 21b [ stabG(i), so that, (a21b) (i) 5 i and, therefore, b(i) 5 a(i). Reversing the argument from, the last step to the first step shows that T is also one-to-one. We conclude, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 153, , the proof by showing that T is onto orbG(i). Let j [ orbG(i). Then a(i) 5 j, for some a [ G and clearly T(astabG(i)) 5 a(i) 5 j, so that T is onto., We leave as an exercise the proof of the important fact that the orbits, of the elements of a set S under a group partition S (Exercise 43)., , The Rotation Group of a Cube, and a Soccer Ball, It cannot be overemphasized that Theorem 7.4 and Lagrange’s Theorem, (Theorem 7.1) are counting theorems.† They enable us to determine the, numbers of elements in various sets. To see how Theorem 7.4 works, we, will determine the order of the rotation group of a cube and a soccer ball., That is, we wish to find the number of essentially different ways in, which we can take a cube or a soccer ball in a certain location in space,, physically rotate it, and then have it still occupy its original location., EXAMPLE 9 Let G be the rotation group of a cube. Label the six, faces of the cube 1 through 6. Since any rotation of the cube must carry, each face of the cube to exactly one other face of the cube and different, rotations induce different permutations of the faces, G can be viewed as, a group of permutations on the set {1, 2, 3, 4, 5, 6}. Clearly, there is, some rotation about a central horizontal or vertical axis that carries face, number 1 to any other face, so that |orbG(1)| 5 6. Next, we consider, stabG(1). Here, we are asking for all rotations of a cube that leave face, number 1 where it is. Surely, there are only four such motions—, rotations of 0°, 90°, 180°, and 270°—about the line perpendicular to, the face and passing through its center (see Figure 7.2). Thus, by, Theorem 7.4, |G| 5 |orbG(1)| |stabG(1)| 5 6 ? 4 5 24., , Figure 7.2 Axis of rotation of a cube., , Now that we know how many rotations a cube has, it is simple to determine the actual structure of the rotation group of a cube. Recall that, S4 is the symmetric group of degree 4., †“People, , who don’t count won’t count” (Anatole France)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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154, , Groups, , Theorem 7.5 The Rotation Group of a Cube, The group of rotations of a cube is isomorphic to S4., , PROOF Since the group of rotations of a cube has the same order as, S4, we need only prove that the group of rotations is isomorphic to a, subgroup of S4. To this end, observe that a cube has four diagonals and, that the rotation group induces a group of permutations on the four diagonals. But we must be careful not to assume that different rotations, correspond to different permutations. To see that this is so, all we need, do is show that all 24 permutations of the diagonals arise from rotations. Labeling the consecutive diagonals 1, 2, 3, and 4, it is obvious, that there is a 90° rotation that yields the permutation a 5 (1234); another 90° rotation about an axis perpendicular to our first axis yields, the permutation b 5 (1423). See Figure 7.3. So, the group of permutations induced by the rotations contains the eight-element subgroup, {e, a, a2, a3, b2, b2a, b2a2, b2a3} (see Exercise 63) and ab, which has, order 3. Clearly, then, the rotations yield all 24 permutations, since the, order of the rotation group must be divisible by both 8 and 3., EXAMPLE 10 A traditional soccer ball has 20 faces that are regular, hexagons and 12 faces that are regular pentagons. (The technical term, for this solid is truncated icosahedron.) To determine the number of rotational symmetries of a soccer ball using Theorem 7.4, we may choose, our set S to be the 20 hexagons or the 12 pentagons. Let us say that S is, the set of 12 pentagons. Since any pentagon can be carried to any other, , 2, , 1, , 2, , 1, , 4, , 3, , 4, , 3, , 3, , 4, , 3, , 4, 2, , 1, , 2, , 1, , β = (1423), , α = (1234), , Figure 7.3, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 155, , pentagon by some rotation, the orbit of any pentagon is S. Also, there, are five rotations that fix (stabilize) any particular pentagon. Thus, by, the Orbit-Stabilizer Theorem, there are 12 ? 5 5 60 rotational symmetries. (In case you are interested, the rotation group of a soccer ball is, isomorphic to A5.), , In 1985, chemists Robert Curl, Richard Smalley, and Harold Kroto, caused tremendous excitement in the scientific community when they, created a new form of carbon by using a laser beam to vaporize graphite., The structure of the new molecule was composed of 60 carbon atoms, arranged in the shape of a soccer ball! Because the shape of the new molecule reminded them of the dome structures built by the architect, R. Buckminster Fuller, Curl, Smalley, and Kroto named their discovery, “buckyballs.” Buckyballs are the roundest, most symmetric large molecules known. Group theory has been particularly useful in illuminating, the properties of buckyballs, since the absorption spectrum of a molecule, depends on its symmetries and chemists classify various molecular states, according to their symmetry properties. The buckyball discovery spurred, a revolution in carbon chemistry. In 1996, Curl, Smalley, and Kroto, received the Nobel Prize in chemistry for their discovery., , An Application of Cosets, to the Rubik’s Cube, Recall from Chapter 5 that in 2010 it was proved via a computer computation, which took 35 CPU-years to complete, that every Rubik’s, cube could be solved in at most 20 moves. To carry out this effort, the, research team of Morley Davidson, John Dethridge, Herbert Kociemba,, and Tomas Rokicki applied a program of Rokicki, which built on early, work of Kociemba, that checked the elements of the cosets of a subgroup H of order (8! · 8! · 4!)/2 5 19,508,428,800 to see if each cube in, a position corresponding to the elements in a coset could be solved, within 20 moves. In the rare cases where Rokicki’s program did not, work, an alternate method was employed. Using symmetry considerations, they were able to reduce the approximately 2 billion cosets of H, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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156, , Groups, , to about 56 million cosets for testing. Cosets played a role in this effort, because Rokicki’s program could handle the 19.51 billion elements in, the same coset in about 20 seconds., , Exercises, I don’t know, Marge. Trying is the first step towards failure., homer simpson, , 1. Let H 5 {(1), (12)(34), (13)(24), (14)(23)}. Find the left cosets of, H in A4 (see Table 5.1 on page 111)., 2. Let H be as in Exercise 1. How many left cosets of H in S4 are, there? (Determine this without listing them.), 3. Let H 5 {0, 63, 66, 69, . . .}. Find all the left cosets of H in Z., 4. Rewrite the condition a21b [ H given in property 5 of the lemma on, page 145 in additive notation. Assume that the group is Abelian., 5. Let H be as in Exercise 3. Use Exercise 4 to decide whether or not, the following cosets of H are the same., a. 11 1 H and 17 1 H, b. 21 1 H and 5 1 H, c. 7 1 H and 23 1 H, 6. Let n be a positive integer. Let H 5 {0, 6n, 62n, 63n, . . .}. Find, all left cosets of H in Z. How many are there?, 7. Find all of the left cosets of {1, 11} in U(30)., 8. Suppose that a has order 15. Find all of the left cosets of ka5l in kal., 9. Let |a| 5 30. How many left cosets of ka4l in kal are there? List them., 10. Give an example of a group G and subgroups H and K such that, HK 5 {h [ H, k [ K} is not a subgroup of G., 11. If H and K are subgroups of G and g belongs to G, show that, g(H y K) 5 gH y gK., 12. Let a and b be nonidentity elements of different orders in a group, G of order 155. Prove that the only subgroup of G that contains, a and b is G itself., 13. Let H be a subgroup of R*, the group of nonzero real numbers under multiplication. If R+ # H # R*, prove that H 5 R+ or H 5 R*., 14. Let C* be the group of nonzero complex numbers under multiplication and let H 5 {a + bi [ C* | a2 + b2 5 1}. Give a geometric description of the coset (3 + 4i)H. Give a geometric description of the, coset (c + di)H., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 157, , 15. Let G be a group of order 60. What are the possible orders for the, subgroups of G?, 16. Suppose that K is a proper subgroup of H and H is a proper subgroup of G. If |K| 5 42 and |G| 5 420, what are the possible orders, of H?, 17. Let G be a group with |G| 5 pq, where p and q are prime. Prove, that every proper subgroup of G is cyclic., 18. Recall that, for any integer n greater than 1, f(n) denotes the number of positive integers less than n and relatively prime to n. Prove, that if a is any integer relatively prime to n, then af(n) mod n 5 1., 19. Compute 515 mod 7 and 713 mod 11., 20. Use Corollary 2 of Lagrange’s Theorem (Theorem 7.1) to prove, that the order of U(n) is even when n . 2., 21. Suppose G is a finite group of order n and m is relatively prime to n., If g [ G and gm 5 e, prove that g 5 e., 22. Suppose H and K are subgroups of a group G. If |H| 5 12 and, |K| 5 35, find |H > K|. Generalize., 23. Suppose that H is a subgroup of S4 and that H contains (12) and, (234). Prove that H 5 S4., 24. Suppose that H and K are subgroups of G and there are elements, a and b in G such that aH 8 bK. Prove that H 8 K., 25. Suppose that G is an Abelian group with an odd number of elements., Show that the product of all of the elements of G is the identity., 26. Suppose that G is a group with more than one element and G has, no proper, nontrivial subgroups. Prove that |G| is prime. (Do not, assume at the outset that G is finite.), 27. Let |G| 5 15. If G has only one subgroup of order 3 and only one, of order 5, prove that G is cyclic. Generalize to |G| 5 pq, where p, and q are prime., 28. Let G be a group of order 25. Prove that G is cyclic or g5 5 e for, all g in G. Generalize to any group of order p2 where p is prime., Does your proof work for this generalization?, 29. Let |G| 5 33. What are the possible orders for the elements of G?, Show that G must have an element of order 3., 30. Let |G| 5 8. Show that G must have an element of order 2., 31. Can a group of order 55 have exactly 20 elements of order 11?, Give a reason for your answer., 32. Determine all finite subgroups of C*, the group of nonzero complex numbers under multiplication., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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158, , Groups, , 33. Let H and K be subgroups of a finite group G with H # K # G., Prove that |G:H| 5 |G:K| |K:H|., 34. Suppose that a group contains elements of orders 1 through 10., What is the minimum possible order of the group?, 35. Give an example of the dihedral group of smallest order that contains a subgroup isomorphic to Z12 and a subgroup isomorphic to, Z20. No need to prove anything, but explain your reasoning., 36. Show that in any group of order 100, either every element has order, that is a power of a prime or there is an element of order 10., 37. Suppose that a finite Abelian group G has at least three elements of, order 3. Prove that 9 divides |G|., 38. Prove that if G is a finite group, the index of Z(G) cannot be prime., 39. Find an example of a subgroup H of a group G and elements a and, b in G such that aH ? Hb and aH y Hb ? f. (Compare with property 5 of cosets.), 40. Prove that a group of order 63 must have an element of order 3., 41. Let G be a group of order 100 that has a subgroup H of order 25., Prove that every element of G of order 5 is in H., 42. Let G be a group of order n and k be any integer relatively prime to, n. Show that the mapping from G to G given by g S gk is one-toone. If G is also Abelian, show that the mapping given by, g S gk is an automorphism of G., 43. Let G be a group of permutations of a set S. Prove that the orbits of, the members of S constitute a partition of S. (This exercise is referred to in this chapter and in Chapter 29.), 44. Prove that every subgroup of Dn of odd order is cyclic., 45. Let G 5 {(1), (12)(34), (1234)(56), (13)(24), (1432)(56), (56)(13),, (14)(23), (24)(56)}., a. Find the stabilizer of 1 and the orbit of 1., b. Find the stabilizer of 3 and the orbit of 3., c. Find the stabilizer of 5 and the orbit of 5., 46. Prove that a group of order 12 must have an element of order 2., 47. Show that in a group G of odd order, the equation x2 5 a has a, unique solution for all a in G., 48. Let G be a group of order pqr, where p, q, and r are distinct primes., If H and K are subgroups of G with |H| 5 pq and |K| 5 qr, prove, that |H > K| 5 q., 49. Prove that a group that has more than one subgroup of order 5 must, have order at least 25., 50. Prove that A5 has a subgroup of order 12., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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7 | Cosets and Lagrange’s Theorem, , 159, , 51. Prove that A5 has no subgroup of order 30., 52. Prove that A5 has no subgroup of order 15 to 20., 53. Suppose that a is an element from a permutation group G and one, of its cycles in disjoint cycle form is (a1a2…ak). Show that {a1,, a2, …, ak} # orbG(ai) for 1 5 1, 2, …, k., 54. Let G be a group and suppose that H is a subgroup of G with the, property that for any a in G we have aH 5 Ha. (That is, every element of the form ah where h is some element of H can be written in, the form h1a for some h1 [ H.) If a has order 2, prove that the set, K 5 H c aH is a subgroup of G. Generalize to the case that |a| 5 k., 55. Prove that A5 is the only subgroup of S5 of order 60., 56. Why does the fact that A4 has no subgroup of order 6 imply that, |Z(A4)| 5 1?, 57. Let G 5 GL(2, R) and H 5 SL(2, R). Let A [ G and suppose that, det A 5 2. Prove that AH is the set of all 2 3 2 matrices in G that, have determinant 2., 58. Let G be the group of rotations of a plane about a point P in, the plane. Thinking of G as a group of permutations of the plane,, describe the orbit of a point Q in the plane. (This is the motivation, for the name “orbit.”), 59. Let G be the rotation group of a cube. Label the faces of the cube, 1 through 6, and let H be the subgroup of elements of G that carry, face 1 to itself. If s is a rotation that carries face 2 to face 1, give a, physical description of the coset Hs., 60. The group D4 acts as a group of permutations of the square regions, shown below. (The axes of symmetry are drawn for reference purposes.) For each square region, locate the points in the orbit of the, indicated point under D4. In each case, determine the stabilizer of, the indicated point., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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160, , Groups, , 61. Let G 5 GL(2, R), the group of 2 3 2 matrices over R with nonzero, determinant. Let H be the subgroup of matrices of determinant 61., If a, b [ G and aH 5 bH, what can be said about det (a) and, det (b)? Prove or disprove the converse. [Determinants have the, property that det (xy) 5 det (x)det (y).], 62. Calculate the orders of the following (refer to Figure 27.5 for, illustrations)., a. The group of rotations of a regular tetrahedron (a solid with four, congruent equilateral triangles as faces), b. The group of rotations of a regular octahedron (a solid with, eight congruent equilateral triangles as faces), c. The group of rotations of a regular dodecahedron (a solid with, 12 congruent regular pentagons as faces), d. The group of rotations of a regular icosahedron (a solid with, 20 congruent equilateral triangles as faces), 63. Prove that the eight-element set in the proof of Theorem 7.5 is a, group., 64. A soccer ball has 20 faces that are regular hexagons and 12 faces, that are regular pentagons. Use Theorem 7.4 to explain why a soccer ball cannot have a 60° rotational symmetry about a line through, the centers of two opposite hexagonal faces., 65. If G is a finite group with fewer than 100 elements and G has subgroups of orders 10 and 25, what is the order of G?, , Computer Exercises, A computer exercise for this chapter is available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Joseph Lagrange, Lagrange is the Lofty Pyramid of the, Mathematical Sciences., , Joseph Louis Lagrange was born in Italy of, French ancestry on January 25, 1736. He became captivated by mathematics at an early, age when he read an essay by Halley on, Newton’s calculus. At the age of 19, he became a professor of mathematics at the Royal, Artillery School in Turin. Lagrange made significant contributions to many branches of, mathematics and physics, among them the, theory of numbers, the theory of equations,, ordinary and partial differential equations,, the calculus of variations, analytic geometry,, fluid dynamics, and celestial mechanics. His, methods for solving third- and fourth-degree, polynomial equations by radicals laid the, groundwork for the group theoretic approach, to solving polynomials taken by Galois., Lagrange was a very careful writer with a, clear and elegant style., At the age of 40, Lagrange was appointed, head of the Berlin Academy, succeeding, Euler. In offering this appointment, Frederick, the Great proclaimed that the “greatest king, in Europe” ought to have the “greatest mathematician in Europe” at his court. In 1787,, Lagrange was invited to Paris by Louis XVI, and became a good friend of the king and his, wife, Marie Antoinette. In 1793, Lagrange, headed a commission, which included, Laplace and Lavoisier, to devise a new system, , The Granger Collection, New York, , napoleon bonaparte, , This stamp was issued by, France in Lagrange’s honor, in 1958., , of weights and measures. Out of this came, the metric system. Late in his life he was, made a count by Napoleon. Lagrange died on, April 10, 1813., To find more information about Lagrange,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 161, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8, , External Direct, Products, The universe is an enormous direct product of representations, of symmetry groups., steven weinberg†, , Definition and Examples, In this chapter, we show how to piece together groups to make larger, groups. In Chapter 9, we will show that we can often start with one, large group and decompose it into a product of smaller groups in much, the same way as a composite positive integer can be broken down into, a product of primes. These methods will later be used to give us a simple way to construct all finite Abelian groups., Definition External Direct Product, , Let G1, G2, . . . , Gn be a finite collection of groups. The external direct, product of G1, G2, . . . , Gn, written as G1 % G2 % ? ? ? % Gn, is the set of, all n-tuples for which the ith component is an element of Gi and the, operation is componentwise., In symbols,, , G1 % G2 % ? ? ? % Gn 5 {(g1, g2, . . . , gn) | gi [ Gi},, where (g1, g 2 , . . . , gn )(g 19, g 29, . . . , g n9) is defined to be (g1g 19,, g2g29, . . . , gngn9). It is understood that each product gigi9 is performed, with the operation of Gi. Note that in the case that each Gi is finite, we, have by properties of sets that |G1 % G2 % … % Gn | 5 |G1||G2| … |Gn|., We leave it to the reader to show that the external direct product of, groups is itself a group (Exercise 1)., This construction is not new to students who have had linear algebra or, physics. Indeed, R2 5 R % R and R3 5 R % R % R—the operation being, componentwise addition. Of course, there is also scalar multiplication, but, †Weinberg, , received the 1979 Nobel Prize in physics with Sheldon Glashow and Abdus, Salam for their construction of a single theory incorporating weak and electromagnetic, interactions., 162, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 163, , we ignore this for the time being, since we are interested only in the group, structure at this point., EXAMPLE 1, U(8) % U(10) 5 {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3),, (3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9),, (7, 1),(7, 3), (7, 7), (7, 9)}., The product (3, 7)(7, 9) 5 (5, 3), since the first components are combined by multiplication modulo 8, whereas the second components are, combined by multiplication modulo 10., EXAMPLE 2, Z2 % Z3 5 {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}., Clearly, this is an Abelian group of order 6. Is this group related to another Abelian group of order 6 that we know, namely, Z6? Consider the, subgroup of Z2 % Z3 generated by (1, 1). Since the operation in each component is addition, we have (1, 1) 5 (1, 1), 2(1, 1) 5 (0, 2), 3(1, 1) 5, (1, 0), 4(1, 1) 5 (0, 1), 5(1, 1) 5 (1, 2), and 6(1, 1) 5 (0, 0). Hence, Z2 % Z3 is cyclic. It follows that Z2 % Z3 is isomorphic to Z6., In Theorem 7.3 we classified the groups of order 2p where p is an, odd prime. Now that we have defined Z2 % Z2, it is easy to classify the, groups of order 4., EXAMPLE 3 Classification of Groups of Order 4, A group of order 4 is isomorphic to Z4 or Z2 % Z2. To verify this, let G 5, {e, a, b, ab}. If G is not cyclic, then it follows from Lagrange’s Theorem, that |a| 5 |b| 5 |ab| 5 2. Then the mapping e S (0, 0), a S (1, 0),, b S (0, 1), and ab S (1, 1) is an isomorphism from G onto Z2 % Z2., We see from Examples 2 and 3 that in some cases Zm % Zn is isomorphic to Zmn and in some cases it is not. Theorem 8.2 provides a simple, characterization for when the isomorphism holds., , Properties of External Direct Products, Our first theorem gives a simple method for computing the order of an, element in a direct product in terms of the orders of the component, pieces., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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164, , Groups, , Theorem 8.1 Order of an Element in a Direct Product, The order of an element in a direct product of a finite number of, finite groups is the least common multiple of the orders of the, components of the element. In symbols,, , |(g1, g2, . . . , gn)| 5 lcm(|g1|, |g2|, . . . , |gn|)., PROOF Denote the identity of Gi by ei. Let s 5 lcm(|g1|, |g2|, . . . , |gn|), and t 5|(g1, g2, . . . , gn)|. Because the fact that s is a multiple of each |gi|, implies that (g1, g2, . . . , gn)s 5 (gs1, gs2, . . . , gsn) 5 (e1, e2, . . . , en), we know, that t # s. On the other hand, from (g t1, g t2, . . . , g tn) 5 (g 1, g 2, . . . , g n)t 5, (e1, e2, . . . , en) we see that t is a common multiple of |g1|, |g2|, . . . , |gn|., Thus, s # t., The next two examples are applications of Theorem 8.1., EXAMPLE 4 We determine the number of elements of order 5 in, Z25 % Z5. By Theorem 8.1, we may count the number of elements, (a, b) in Z25 % Z5 with the property that 5 5 |(a, b)| 5 lcm(|a|, |b|)., Clearly this requires that either |a| 5 5 and |b| 5 1 or 5, or |b| 5 5 and, |a| 5 1 or 5. We consider two mutually exclusive cases., Case 1 |a| 5 5 and |b| 5 1 or 5. Here there are four choices for a, (namely, 5, 10, 15, and 20) and five choices for b. This gives 20 elements of order 5., Case 2 |a| 5 1 and |b| 5 5. This time there is one choice for a and four, choices for b, so we obtain four more elements of order 5., Thus, Z25 % Z5 has 24 elements of order 5., EXAMPLE 5 We determine the number of cyclic subgroups of order, 10 in Z100 % Z25. We begin by counting the number of elements (a, b) of, order 10., Case 1 |a| 5 10 and |b| 5 1 or 5. Since Z100 has a unique cyclic subgroup of order 10 and any cyclic group of order 10 has four generators, (Theorem 4.4), there are four choices for a. Similarly, there are five, choices for b. This gives 20 possibilities for (a, b)., Case 2 |a| 5 2 and |b| 5 5. Since any finite cyclic group of even order, has a unique subgroup of order 2 (Theorem 4.4), there is only one, choice for a. Obviously, there are four choices for b. So, this case yields, four more possibilities for (a, b)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 165, , Thus, Z100 % Z25 has 24 elements of order 10. Because each cyclic, subgroup of order 10 has four elements of order 10 and no two of the, cyclic subgroups can have an element of order 10 in common, there, must be 24/4 5 6 cyclic subgroups of order 10. (This method is analogous to determining the number of sheep in a flock by counting legs, and dividing by 4.), The direct product notation is convenient for specifying certain subgroups of a direct product., EXAMPLE 6 For each divisor r of m and s of n, the group Zm % Zn, has a subgroup isomorphic to Zr % Zs (see Exercise 19). To find a subgroup of, say, Z30 % Z12 isomorphic to Z6 % Z4, we observe that k5l is a, subgroup of Z30 of order 6 and k3l is a subgroup of Z12 of order 4, so, k5l % k3l is the desired subgroup., The next theorem and its first corollary characterize those direct, products of cyclic groups that are themselves cyclic., Theorem 8.2 Criterion for G % H to be Cyclic, Let G and H be finite cyclic groups. Then G % H is cyclic if and only, if |G| and |H| are relatively prime., , PROOF Let |G| 5 m and |H| 5 n, so that |G % H| 5 mn. To prove the, first half of the theorem, we assume G % H is cyclic and show that, m and n are relatively prime. Suppose that gcd(m, n) 5 d and (g, h) is a, generator of G % H. Since (g, h)mn/d 5 ((gm)n/d, (hn)m/d ) 5 (e, e), we, have mn 5 |(g, h)| # mn/d. Thus, d 5 1., To prove the other half of the theorem, let G 5 kgl and H 5 khl and suppose gcd(m, n) 5 1. Then, |(g, h)| 5 lcm(m, n) 5 mn 5 |G % H|, so that, (g, h) is a generator of G % H., As a consequence of Theorem 8.2 and an induction argument, we, obtain the following extension of Theorem 8.2., Corollary 1 Criterion for G1 % G2 % ? ? ? % Gn to Be Cyclic, An external direct product G1 % G2 % ? ? ? % Gn of a finite number, of finite cyclic groups is cyclic if and only if |Gi| and |Gj| are relatively, prime when i 2 j., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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166, , Groups, , Corollary 2 Criterion for Zn1n2 . . . nk < Zn1 % Zn2 % . . . % Znk, Let m 5 n1 n2 ? ? ? nk. Then Zm is isomorphic to Zn1 % Zn2 % ? ? ? % Znk, if and only if ni and nj are relatively prime when i 2 j., , By using the results above in an iterative fashion, one can express, the same group (up to isomorphism) in many different forms. For example, we have, Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5 < Z2 % Z30., Similarly,, Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5, < Z2 % Z3 % Z2 % Z5 < Z6 % Z10., Thus, Z2 % Z30 < Z6 % Z10. Note, however, that Z2 % Z30 ] Z60., , The Group of Units Modulo n as, an External Direct Product, The U-groups provide a convenient way to illustrate the preceding, ideas. We first introduce some notation. If k is a divisor of n, let, Uk(n) 5 {x [ U(n) | x mod k 5 1}., For example, U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92}. It can be readily, shown that Uk(n) is indeed a subgroup of U(n). (See Exercise 31 in, Chapter 3.), Theorem 8.3 U(n) as an External Direct Product, Suppose s and t are relatively prime. Then U(st) is isomorphic to the, external direct product of U(s) and U(t). In short,, U(st) < U(s) % U(t)., Moreover, Us(st) is isomorphic to U(t) and Ut(st) is isomorphic to U(s)., , PROOF An isomorphism from U(st) to U(s) % U(t) is x S (x mod s,, x mod t); an isomorphism from Us(st) to U(t) is x S x mod t; an isomorphism from Ut(st) to U(s) is x S x mod s. We leave the verification that, these mappings are operation-preserving, one-to-one, and onto to the, reader. (See Exercises 9, 17, and 19 in Chapter 0; see also [1].), As a consequence of Theorem 8.3, we have the following result., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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167, , 8 | External Direct Products, , Corollary, Let m 5 n1n2 ? ? ? nk , where gcd(ni , nj ) 5 1 for i 2 j. Then,, U(m) < U(n1) % U(n2) % ? ? ? % U(nk)., , To see how these results work, let’s apply them to U(105). We obtain, U(105) < U(7) % U(15),, U(105) < U(21) % U(5),, U(105) < U(3) % U(5) % U(7)., Moreover,, U(7) < U15(105) 5 {1, 16, 31, 46, 61, 76},, U(15) < U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92},, U(21) < U5(105) 5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101},, U(5) < U21(105) 5 {1, 22, 43, 64},, U(3) < U35(105) 5 {1, 71}., Among all groups, surely the cyclic groups Z n have the simplest, structures and, at the same time, are the easiest groups with which to, compute. Direct products of groups of the form Zn are only slightly, more complicated in structure and computability. Because of this, algebraists endeavor to describe a finite Abelian group as such a direct, product. Indeed, we shall soon see that every finite Abelian group can, be so represented. With this goal in mind, let us reexamine the, U-groups. Using the corollary to Theorem 8.3 and the facts (see, [2, p. 93]), first proved by Carl Gauss in 1801, that, U(2) < {0},, , U(4) < Z2,, , U(2n) < Z2 % Z2n22, , for n $ 3,, , and, U( pn) < Zpn2pn21, , for p an odd prime,, , we now can write any U-group as an external direct product of cyclic, groups. For example,, U(105) 5 U(3 ? 5 ? 7) < U(3) % U(5) % U(7), < Z2 % Z4 % Z6, and, U(720) 5 U(16 ? 9 ? 5) < U(16) % U(9) % U(5), < Z2 % Z4 % Z6 % Z4., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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168, , Groups, , What is the advantage of expressing a group in this form? Well, for one, thing, we immediately see that the orders of the elements U(720) can, only be 1, 2, 3, 4, 6, and 12. This follows from the observations that an, element from Z2 % Z4 % Z6 % Z4 has the form (a, b, c, d), where, |a| 5 1 or 2, |b| 5 1, 2, or 4, |c| 5 1, 2, 3, or 6, and |d| 5 1, 2, or 4, and, that |(a, b, c, d)| 5 lcm(|a|, |b|, |c|, |d|). For another thing, we can readily determine the number of elements of order 12, say, that U(720) has., Because U(720) is isomorphic to Z2 % Z4 % Z6 % Z4, it suffices to calculate the number of elements of order 12 in Z2 % Z4 % Z6 % Z4. But, this is easy. By Theorem 8.1, an element (a, b, c, d) has order 12 if and, only if lcm(|a|, |b|, |c|, |d|) 5 12. Since |a| 5 1 or 2, it does not matter, how a is chosen. So, how can we have lcm(|b|, |c|, |d|) 5 12? One way, is to have |b| 5 4, |c| 5 3 or 6, and d arbitrary. By Theorem 4.4, there, are two choices for b, four choices for c, and four choices for d. So, in, this case, we have 2 ? 4 ? 4 5 32 choices. The only other way to have, lcm(|b|, |c|, |d|) 5 12 is for |d| 5 4, |c| 5 3 or 6, and |b| 5 1 or 2 (we, exclude |b| 5 4, since this was already accounted for). This gives 2 ? 4 ?, 2 5 16 new choices. Finally, since a can be either of the two elements in, Z2, we have a total of 2(32 1 16) 5 96 elements of order 12., These calculations tell us more. Since Aut(Z720) is isomorphic to, U(720), we also know that there are 96 automorphisms of Z 720 of, order 12. Imagine trying to deduce this information directly from, U(720) or, worse yet, from Aut(Z720)! These results beautifully illustrate the advantage of being able to represent a finite Abelian group as, a direct product of cyclic groups. They also show the value of our theorems about Aut(Z n ) and U(n). After all, theorems are laborsaving devices. If you want to convince yourself of this, try to prove, directly from the definitions that Aut(Z720) has exactly 96 elements of, order 12., , Applications, We conclude this chapter with five applications of the material presented here—three to cryptography, the science of sending and deciphering secret messages, one to genetics, and one to electric circuits., Data Security, Because computers are built from two-state electronic components,, it is natural to represent information as strings of 0s and 1s called, binary strings. A binary string of length n can naturally be thought of, as an element of Z2 % Z2 % ? ? ? % Z2 (n copies) where the parentheses and the commas have been deleted. Thus the binary string, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 169, , 11000110 corresponds to the element (1, 1, 0, 0, 0, 1, 1, 0) in Z2 % Z2 %, Z2 % Z2 % Z2 % Z2 % Z2 % Z2. Similarly, two binary strings a1a2 ? ? ? an, and b1b2 ? ? ? bn are added componentwise modulo 2 just as their, corresponding elements in Z2 % Z2 % ? ? ? % Z2 are. For example,, 11000111 1 01110110 5 10110001, and, 10011100 1 10011100 5 00000000., The fact that the sum of two binary sequences a1a2 ? ? ? an 1 b1b2 ? ? ?, bn 5 00 ? ? ? 0 if and only if the sequences are identical is the basis for, a data security system used to protect Internet transactions., Suppose that you want to purchase a compact disc from http://www, .amazon.com. Need you be concerned that a hacker will intercept, your credit-card number during the transaction? As you might expect,, your credit-card number is sent to Amazon in a way that protects the, data. We explain one way to send credit-card numbers over the Web, securely. When you place an order with Amazon, the company sends, your computer a randomly generated string of 0’s and 1’s called a key., This key has the same length as the binary string corresponding to, your credit-card number and the two strings are added (think of this, process as “locking” the data). The resulting sum is then transmitted, to Amazon. Amazon in turn adds the same key to the received string,, which then produces the original string corresponding to your creditcard number (adding the key a second time “unlocks” the data)., To illustrate the idea, say you want to send an eight-digit binary string, such as s 5 10101100 to Amazon (actual credit-card numbers have, very long strings) and Amazon sends your computer the key, k 5 00111101. Your computer returns the string s 1 k 5 10101100 1, 00111101 5 10010001 to Amazon, and Amazon adds k to this string to, get 10010001 1 00111101 5 10101100, which is the string representing your credit-card number. If someone intercepts the number, s 1 k 5 10010001 during transmission it is no value without knowing k., The method is secure because the key sent by Amazon is randomly, generated and used only one time. You can tell when you are using an encryption scheme on a Web transaction by looking to see if the Web address begins with “https” rather than the customary “http.” You will also, see a small padlock in the status bar at the bottom of the browser window., Public Key Cryptography, Unlike auctions such as those on eBay, where each bid is known by, everyone, a silent auction is one in which each bid is secret. Suppose, that you wanted to use your Twitter account to run a silent auction., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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170, , Groups, , How could a scheme be devised so that users could post their bids in, such a way that the amounts are intelligible only to the account holder?, In the mid-1970s, Ronald Rivest, Adi Shamir, and Leonard Adleman, devised an ingenious method that permits each person who is to receive, a secret message to tell publicly how to scramble messages sent to him, or her. And even though the method used to scramble the message is, known publicly, only the person for whom it is intended will be able to, unscramble the message. The idea is based on the fact that there exist, efficient methods for finding very large prime numbers (say about, 100 digits long) and for multiplying large numbers, but no one knows, an efficient algorithm for factoring large integers (say about 200 digits, long). The person who is to receive the message chooses a pair of large, primes p and q and chooses an integer e (called the encryption exponent) with 1 , e , m, where m 5 lcm (p 2 1, q 2 1), such that e, is relatively prime to m (any such e will do). This person calculates, n 5 pq (n is called the key) and announces that a message M is to be, sent to him or her publicly as Me mod n. Although e, n, and Me are, available to everyone, only the person who knows how to factor n as pq, will be able to decipher the message., To present a simple example that nevertheless illustrates the principal features of the method, say we wish to send the messages “YES.”, We convert the message into a string of digits by replacing A by 01, B, by 02, . . . , Z by 26, and a blank by 00. So, the message YES becomes, 250519. To keep the numbers involved from becoming too unwieldy,, we send the message in blocks of four digits and fill in with blanks, when needed. Thus, the messages YES is represented by the two blocks, 2505 and 1900. The person to whom the message is to be sent has, picked two primes p and q, say p 5 37 and q 5 73, and a number e that, has no prime divisors in common with lcm (p 21, q 21) 5 72, say, e 5 5, and has published n 5 37 ? 73 5 2701 and e 5 5 in a public, forum. We will send the “scrambled” numbers (2505)5 mod 2701 and, (1900)5 mod 2701 rather than 2505 and 1900, and the receiver will unscramble them. We show the work involved for us and the receiver only, for the block 2505. We determine (2505)5 mod 2701 5 2415 by using a, modular arithmetic calculator such as the one at http://users.wpi, .edu/~martin/mod.html.†, †Provided, , that the numbers are not too large, the Google search engine at http://www, .google.com will do modular arithmetic. For example, entering 2505^2 mod 2701 in, the search box yields 602. Be careful, however: Entering 2505^5 mod 2701 does not, return a value, because 25055 is too large. Instead, we can use Google to compute, smaller powers such as 2505^2 mod 2701 and 2505^3 mod 2701 (which yields 852), and then enter (852 3 602) mod 2701., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 171, , Thus, the number 2415 is sent to the receiver. Now the receiver must, take this number and convert it back to 2505. To do so, the receiver, takes the two factors of 2701, p 5 37 and q 5 73, and calculates the, least common multiple of p 2 1 5 36 and q 2 1 5 72, which is 72., (This is where the knowledge of p and q is necessary.) Next, the receiver must find e21 5 d (called the decryption exponent) in U(72)—, that is, solve the equation 5 ? d 5 1 mod 72. This number is 29. See, http://www.d.umn.edu/~jgallian/msproject06/chap8.html, #chap8ex5 or use a Google search box to compute 5k for each divisor k, of |U(72)| 5 f(9) ? f(8) 5 24 starting with 2 until we reach 5k mod 72, 5 1. Doing so, we obtain 56 mod 72 5 1, which implies that 55 mod 72, 5 29 is 521 in U(72)., Then the receiver takes the number received, 2415, and calculates, (2415)29 mod 2701 5 2505, the encoded number. Thus, the receiver correctly determines the code for “YE.” On the other hand, without knowing how pq factors, one cannot find the modulus (in our case, 72) that is, needed to determine the decryption exponent d., The procedure just described is called the RSA public key encryption, scheme in honor of the three people (Rivest, Shamir, and Adleman) who, discovered the method. It is widely used in conjunction with web servers, and browsers, e-mail programs, remote login sessions, and electronic financial transactions. The algorithm is summarized below., Receiver, 1. Pick very large primes p and q and compute n 5 pq., 2. Compute the least common multiple of p – 1 and q – 1; let us call, it m., 3. Pick e relatively prime to m., 4. Find d such that ed mod m 5 1., 5. Publicly announce n and e., Sender, 1. Convert the message to a string of digits., 2. Break up the message into uniform blocks of digits; call them M1,, M2,…, Mk., 3. Check to see that the greatest common divisor of each Mi and n is 1., If not, n can be factored and our code is broken. (In practice, the, primes p and q are so large that they exceed all Mi, so this step may, be omitted.), 4. Calculate and send Ri 5 Mie mod n., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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172, , Groups, , Receiver, 1. For each received message Ri, calculate Rid mod n., 2. Convert the string of digits back to a string of characters., Why does this method work? Well, we know that U(n) < U(p) %, U(q) < Zp21 % Zq21. Thus, an element of the form xm in U(n) corresponds under an isomorphism to one of the form (mx1, mx2) in Zp21 %, Zq21. Since m is the least common multiple of p 2 1 and q 2 1, we may, write m 5 s(p 2 1) and m 5 t(q 2 1) for some integers s and t. Then, (mx1, mx2) 5 (s(p 2 1)x1, t(q 2 1)x2) 5 (0, 0) in Zp21 % Zq21, and it, follows that xm 5 1 for all x in U(n). So, because each message Mi is an, element of U(n) and e was chosen so that ed 5 1 1 km for some k, we, have, modulo n,, Rid 5 (Mie)d 5 Mied 5 Mi 11km 5 Mi(Mim)k 5 Mi1k 5 Mi., In 2002, Ronald Rivest, Adi Shamir, and Leonard Adleman received, the Association for Computing Machinery A. M. Turing Award, which, is considered the “Nobel Prize of computing,” for their contribution to, public key cryptography., An RSA calculator that does all the calculations is provided at http://, www.d.umn.edu/~jgallian/msproject06/chap8.html#chap8ex5. A list, of primes can be found by searching the Web for “list of primes.”, Digital Signatures, With so many financial transactions now taking place electronically, the, problem of authenticity is paramount. How is a stockbroker to know that, an electronic message she receives that tells her to sell one stock and buy, another actually came from her client? The technique used in public key, cryptography allows for digital signatures as well. Let us say that person, A wants to send a secret message to person B in such a way that only B, can decode the message and B will know that only A could have sent it., Abstractly, let EA and DA denote the algorithms that A uses for encryption and decryption, respectively, and let EB and DB denote the algorithms that B uses for encryption and decryption, respectively. Here, we assume that EA and EB are available to the public, whereas DA is, known only to A and DB is known only to B, and that DBEB and EADA, applied to any message leaves the message unchanged. Then A sends, a message M to B as EB (DA(M)) and B decodes the received message, by applying the function EADB to it to obtain, (EADB) (EB (DA(M)) 5 EA(DBEB )(DA(M)) 5 EA(DA(M)) 5 M., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 173, , Notice that only A can execute the first step (i.e., create DA(M)) and, only B can implement the last step (i.e., apply EADB to the received, message)., Transactions using digital signatures became legally binding in the, United States in October 2000., Genetics†, The genetic code can be conveniently modeled using elements of Z4 %, Z4 % ? ? ? % Z4, where we omit the parentheses and the commas and, just use strings of 0’s, 1’s, 2’s, and 3’s and add componentwise modulo, 4. A DNA molecule is composed of two long strands in the form of a, double helix. Each strand is made up of strings of the four nitrogen, bases adenine (A), thymine (T), guanine (G), and cytosine (C). Each, base on one strand binds to a complementary base on the other strand., Adenine always is bound to thymine, and guanine always is bound to, cytosine. To model this process, we identify A with 0, T with 2, G with 1,, and C with 3. Thus, the DNA segment ACGTAACAGGA and its complement segment TGCATTGTCCT are denoted by 03120030110 and, 21302212332. Noting that in Z4, 0 1 2 5 2, 2 1 2 5 0, 1 1 2 5 3, and, 3 1 2 5 1, we see that adding 2 to elements of Z4 interchanges 0 and 2, and 1 and 3. So, for any DNA segment a1a2 ? ? ? an represented by elements of Z4 % Z4 % ? ? ? % Z4, we see that its complementary segment, is represented by a1a2 ? ? ? an 1 22 ? ? ? 2., Electric Circuits, Many homes have light fixtures that are operated by a pair of switches., They are wired so that when either switch is thrown, the light changes, its status (from on to off or vice versa). Suppose the wiring is done so, that the light is on when both switches are in the up position. We can, conveniently think of the states of the two switches as being matched, with the elements of Z2 % Z2, with the two switches in the up position, corresponding to (0, 0) and the two switches in the down position corresponding to (1, 1). Each time a switch is thrown, we add 1 to the, corresponding component in the group Z2 % Z2. We then see that the, lights are on when the switches correspond to the elements of the subgroup k(1, 1)l and are off when the switches correspond to the elements, in the coset (1, 0) 1 k(1, 1)l. A similar analysis applies in the case of, three switches, with the subgroup {(0, 0, 0), (1, 1, 0), (0, 1, 1), (1, 0, 1)}, corresponding to the lights-on situation., †This, , discussion is adapted from [3]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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174, , Groups, , Exercises, What’s the most difficult aspect of your life as a mathematician, Diane, Maclagan, an assistant professor at Rutgers, was asked. “Trying to prove, theorems,” she said. And the most fun? “Trying to prove theorems.”, , 1. Prove that the external direct product of any finite number of, groups is a group. (This exercise is referred to in this chapter.), 2. Show that Z2 % Z2 % Z2 has seven subgroups of order 2., 3. Let G be a group with identity eG and let H be a group with identity eH. Prove that G is isomorphic to G % {eH} and that H is isomorphic to {eG} % H., 4. Show that G % H is Abelian if and only if G and H are Abelian., State the general case., 5. Prove or disprove that Z % Z is a cyclic group., 6. Prove, by comparing orders of elements, that Z 8 % Z2 is not isomorphic to Z4 % Z4., 7. Prove that G1 % G2 is isomorphic to G2 % G1. State the general, case., 8. Is Z3 % Z9 isomorphic to Z27? Why?, 9. Is Z3 % Z5 isomorphic to Z15? Why?, 10. How many elements of order 9 does Z3 % Z9 have? (Do not do this, exercise by brute force.), 11. How many elements of order 4 does Z4 % Z4 have? (Do not do this, by examining each element.) Explain why Z4 % Z4 has the same, number of elements of order 4 as does Z8000000 % Z400000. Generalize to the case Zm % Zn., 12. Give examples of four groups of order 12, no two of which are, isomorphic. Give reasons why no two are isomorphic., 13. For each integer n . 1, give examples of two nonisomorphic, groups of order n2., 14. The dihedral group Dn of order 2n (n $ 3) has a subgroup of n rotations and a subgroup of order 2. Explain why Dn cannot be isomorphic to the external direct product of two such groups., 15. Prove that the group of complex numbers under addition is isomorphic to R % R., 16. Suppose that G1 < G2 and H1 < H2. Prove that G1 % H1 < G2 %, H2. State the general case., 17. If G % H is cyclic, prove that G and H are cyclic. State the general, case., 18. In Z40 % Z30, find two subgroups of order 12., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 175, , 19. If r is a divisor of m and s is a divisor of n, find a subgroup of Zm %, Zn that is isomorphic to Zr % Zs., 20. Find a subgroup of Z12 % Z18 that is isomorphic to Z9 % Z4., 21. Let G and H be finite groups and (g, h) [ G % H. State a necessary, and sufficient condition for k(g, h)l 5 kgl % khl., 22. Determine the number of elements of order 15 and the number of, cyclic subgroups of order 15 in Z30 % Z20., 23. What is the order of any nonidentity element of Z3 % Z3 % Z3?, Generalize., 24. Let m . 2 be an even integer and let n . 2 be an odd integer. Find, a formula for the number of elements of order 2 in Dm % Dn., 25. Let M be the group of all real 2 3 2 matrices under addition. Let, N 5 R % R % R % R under componentwise addition. Prove that, M and N are isomorphic. What is the corresponding theorem for, the group of m 3 n matrices under addition?, 26. The group S3 % Z2 is isomorphic to one of the following groups:, Z12, Z6 % Z2, A4, D6. Determine which one by elimination., 27. Let G be a group, and let H 5 {(g, g) | g [ G}. Show that H is a, subgroup of G % G. (This subgroup is called the diagonal of, G % G.) When G is the set of real numbers under addition,, describe G % G and H geometrically., 28. Find a subgroup of Z4 % Z2 that is not of the form H % K, where H, is a subgroup of Z4 and K is a subgroup of Z2., 29. Find all subgroups of order 3 in Z9 % Z3., 30. Find all subgroups of order 4 in Z4 % Z4., 31. What is the largest order of any element in Z30 % Z20?, 32. What is the order of the largest cyclic subgroup of Z6 % Z10 % Z15?, What is the order of the largest cyclic subgroup of Zn % Zn % …, 1, 2, % Zn ?, k, 33. Find three cyclic subgroups of maximum possible order in Z6 %, Z10 % Z15 of the form kal % kbl % kcl, where a [ Z6, b [ Z10, and, c [ Z15., 34. How many elements of order 2 are in Z2000000 % Z4000000? Generalize., 35. Find a subgroup of Z800 % Z200 that is isomorphic to Z2 % Z4., 36. Find a subgroup of Z12 % Z4 % Z15 that has order 9., 37. Prove that R* % R* is not isomorphic to C*. (Compare this with, Exercise 15.), 38. Let, 1 a b, H • £ 0 1 0 t † a, b [ Z3 ¶ ., 0 0 1, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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176, , Groups, , 39., 40., 41., 42., 43., 44., 45., 46., 47., 48., , 49., 50., 51., , 52., 53., 54., 55., 56., 57., 58., 59., 60., , (See Exercise 48 in Chapter 2 for the definition of multiplication.), Show that H is an Abelian group of order 9. Is H isomorphic to Z9, or to Z3 % Z3?, Let G 5 {3m6n | m, n [ Z} under multiplication. Prove that G is isomorphic to Z % Z. Does your proof remain valid if G 5 {3m9n | m, n [ Z}?, Let (a1, a2, . . . , an) [ G1 % G2 % ? ? ? % Gn. Give a necessary and, sufficient condition for |(a1, a2, . . . , an)| 5 `., Prove that D3 % D4 ] D12% Z2., Determine the number of cyclic subgroups of order 15 in Z90 % Z36., Provide a generator for each of the subgroups of order 15., List the elements in the groups U5(35) and U7(35)., Prove or disprove that U(40) % Z6 is isomorphic to U(72) % Z4., Prove or disprove that C* has a subgroup isomorphic to Z2 % Z2., Let G be a group isomorphic to Zn1 % Zn2 % . . . % Znk. Let x be the, product of all elements in G. Describe all possibilities for x., If a group has exactly 24 elements of order 6, how many cyclic, subgroups of order 6 does it have?, For any Abelian group G and any positive integer n, let Gn 5 {gn |, g [ G} (see Exercise 17, Supplementary Exercises for Chapters, 1– 4). If H and K are Abelian, show that (H % K)n 5 Hn % Kn., Express Aut(U(25)) in the form Zm % Zn., Determine Aut(Z2 % Z2)., Suppose that n1, n2, . . . , nk are positive even integers. How many, elements of order 2 does Zn1 % Zn2 % . . . % Zn have ? How many are, k, there if we drop the requirement that n1, n2, . . . , nk must be even?, Is Z10 % Z12 % Z6 ^ Z60 % Z6 % Z2?, Is Z10 % Z12 % Z6 ^ Z15 % Z4 % Z12?, Find an isomorphism from Z12 to Z4 % Z3., How many isomorphisms are there from Z12 to Z4 % Z3?, Suppose that f is an isomorphism from Z3 % Z5 to Z15 and, f(2, 3) 5 2. Find the element in Z3 % Z5 that maps to 1., If f is an isomorphism from Z4 % Z3 to Z12, what is f(2, 0)? What, are the possibilities for f(1, 0)? Give reasons for your answer., Prove that Z5 % Z5 has exactly six subgroups of order 5., Let (a, b) belong to Zm % Zn. Prove that |(a, b)| divides lcm(m, n)., Let G 5 {ax2 1 bx 1 c | a, b, c [ Z3}. Add elements of G as you, would polynomials with integer coefficients, except use modulo 3, addition. Prove that G is isomorphic to Z3 % Z3 % Z3. Generalize., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 177, , 61. Determine all cyclic groups that have exactly two generators., 62. Explain a way that a string of length n of the four nitrogen bases A,, T, G, and C could be modeled with the external direct product of n, copies of Z2 % Z2., 63. Let p be a prime. Prove that Zp % Zp has exactly p 1 1 subgroups, of order p., 64. Give an example of an infinite non-Abelian group that has exactly, six elements of finite order., 65. Give an example to show that there exists a group with elements a, and b such that |a| 5 `, |b| 5 `, and |ab| 5 2., 66. Express U(165) as an external direct product of cyclic groups of, the form Zn., 67. Express U(165) as an external direct product of U-groups in four, different ways., 68. Without doing any calculations in Aut(Z20), determine how many, elements of Aut(Z20) have order 4. How many have order 2?, 69. Without doing any calculations in Aut(Z720), determine how many, elements of Aut(Z720) have order 6., 70. Without doing any calculations in U(27), decide how many subgroups U(27) has., 71. What is the largest order of any element in U(900)?, 72. Let p and q be odd primes and let m and n be positive integers., Explain why U( pm) % U(qn) is not cyclic., 73. Use the results presented in this chapter to prove that U(55) is, isomorphic to U(75)., 74. Use the results presented in this chapter to prove that U(144) is, isomorphic to U(140)., 75. For every n . 2, prove that U(n)2 5 {x2 | x [ U(n)} is a proper, subgroup of U(n)., 76. Show that U(55)3 5 {x3 | x [ U(55)} is U(55)., 77. Find an integer n such that U(n) contains a subgroup isomorphic to, Z5 % Z5., 78. Find a subgroup of order 6 in U(700)., 79. Show that there is a U-group containing a subgroup isomorphic, to Z3 % Z3., 80. Find an integer n such that U(n) is isomorphic to Z2 % Z4 % Z9., 81. What is the smallest positive integer k such that xk 5 e for all x in, U(7 ? 17)? Generalize to U(pq) where p and q are distinct primes., 82. If k divides m and m divides n, how are Um(n) and Uk(n) related?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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178, , Groups, , 83. Let p1, p2,…, pk be distinct odd primes and n1, n2,…, nk be, positive integers. Determine the number of elements of order 2 in, U(p1n1 p2n2… pknk). How many are there in U(2np1n1 p2n2… pknk) where, n is at least 3?, 84. Show that no U-group has order 14., 85. Show that there is a U-group containing a subgroup isomorphic, to Z14., 86. Show that no U-group is isomorphic to Z4 % Z4., 87. Show that there is a U-group containing a subgroup isomorphic to, Z4 % Z4., 88. Using the RSA scheme with p 5 37, q 5 73, and e 5 5, what number would be sent for the message “RM”?, 89. Assuming that a message has been sent via the RSA scheme with, p 5 37, q 5 73, and e 5 5, decode the received message “34.”, , Computer Exercises, Computer exercises in this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , References, 1. J. A. Gallian and D. Rusin, “Factoring Groups of Integers Modulo n,”, Mathematics Magazine 53 (1980): 33–36., 2. D. Shanks, Solved and Unsolved Problems in Number Theory, 2nd ed.,, New York: Chelsea, 1978., 3. S. Washburn, T. Marlowe, and C. Ryan, Discrete Mathematics, Reading,, MA: Addison-Wesley, 1999., , Suggested Readings, Y. Cheng, “Decompositions of U-Groups,” Mathematics Magazine 62, (1989): 271–273., This article explores the decomposition of U(st), where s and t are, relatively prime, in greater detail than we have provided., David J. Devries, “The Group of Units in Zm,” Mathematics Magazine 62, (1989): 340–342., This article provides a simple proof that U(n) is not cyclic when n is, not of the form 1, 2, 4, pk, or 2pk, where p is an odd prime., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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8 | External Direct Products, , 179, , David R. Guichard, “When Is U(n) Cyclic? An Algebraic Approach,”, Mathematics Magazine 72 (1999): 139–142., The author provides a group theoretic proof of the fact that U(n) is, cyclic if and only if n is 1, 2, 4, pk, or 2pk, where p is an odd prime., Markku Niemenmaa, “A Check Digit System for Hexadecimal Numbers,”, Applicable Algebra in Engineering, Communication, and Computing 22, (2011):109–112., This article provides a new check-digit system for hexadecimal numbers that is based on the use of a suitable automorphism of the group, Z2 % Z2 % Z2 % Z2. It is able to detect all single errors, adjacent transpositions, twin errors, jump transpositions, and jump twin errors., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Leonard Adleman, “For their ingenious contribution for making, public-key cryptography useful in practice.”, , Leonard Aldeman, , Citation for the ACM A. M. Turing Award, , Leonard Adleman was born on December, 31, 1945 in San Francisco, California. He received a B.A. degree in mathematics in 1968, and a Ph.D. degree in computer science in, 1976 from the University of California,, Berkeley. He spent 1976–1980 as professor, of mathematics at the Massachusetts Institute, of Technology where he met Ronald Rivest, and Adi Shamir. Rivest and Shamir were attempting to devise a secure public key cryptosystem and asked Adleman if he could, break their codes. Eventually, they invented, what is now known as the RSA code that was, simple to implement yet secure., In 1983, Adleman, Shamir, and Rivest, formed the RSA Data Security company to, license their algorithm. Their algorithm has, become the primary cryptosystem used for, security on the World Wide Web. They sold, their company for $200 million in 1996., In the early 1990s, Adleman became interested in trying to find out a way to use, DNA as a computer. His pioneering work on, this problem lead to the field now called, “DNA computing.”, , Among his many honors are: the, Association for Computing Machinery, A. M. Turing Award, the Kanallakis Award, for Theory and Practice, and election to the, National Academy of Engineering, the, American Academy of Arts and Sciences,, and the National Academy of Sciences., Adleman’s current position is the Henry, Salvatori Distinguished Chair in Computer, Science and Professor of Computer Science, and Biological Sciences at the University of, Southern California, where he has been, since 1980., For more information on Adleman, visit:, http://www.wikipedia.com, and, http://www.nytimes.com/1994/12/13/, science/scientist-at-work-leonardadleman-hitting-the-high-spots-ofcomputer-theory.html?, pagewanted=all&src=pm, , 180, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 5–8, My mind rebels at stagnation. Give me problems, give me work, give me, the most obstruse cryptogram, or the most intricate analysis, and I am in, my own proper atmosphere., sherlock holmes, The Sign of Four, , True/false questions for Chapters 5–8 are available on the Web at:, www.d.umn.edu/~jgallian/TF, 1. A subgroup N of a group G is called a characteristic subgroup if, f(N) 5 N for all automorphisms f of G. (The term characteristic, was first applied by G. Frobenius in 1895.) Prove that every subgroup of a cyclic group is characteristic., 2. Prove that the center of a group is characteristic., 3. The commutator subgroup G9 of a group G is the subgroup generated by the set {x21y21xy | x, y [ G}. (That is, every element of G9, has the form a1i1a2i2 ? ? ? a kik, where each aj has the form x21y21xy,, each ij 5 61, and k is any positive integer.) Prove that G9 is a characteristic subgroup of G. (This subgroup was first introduced by, G. A. Miller in 1898.), 4. Prove that the property of being a characteristic subgroup is transitive. That is, if N is a characteristic subgroup of K and K is a characteristic subgroup of G, then N is a characteristic subgroup of G., 5. Let G 5 Z3 % Z3 % Z3 and let H be the subgroup of SL(3, Z3), consisting of, 1, H • £0, 0, , a, 1, 0, , b, c t † a, b, c, 1, , [ Z3 ¶ ., , (See Exercise 48 in Chapter 2 for the definition of multiplication.), Determine the number of elements of each order in G and H. Are G, and H isomorphic? (This exercise shows that two groups with the, same number of elements of each order need not be isomorphic.), 6. Let H and K be subgroups of a group G and let HK 5 {hk | h [ H,, k [ K} and KH 5 {kh | k [ K, h [ H}. Prove that HK is a group if, and only if HK 5 KH., 7. Let G be a finite Abelian group in which every nonidentity element, has order 2. If |G| . 2, prove that the product of all the elements in, G is the identity., 181, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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182, , Groups, , 8. Prove that S4 is not isomorphic to D4 % Z3., 9. Let G be a group. For any element g of G, define gZ(G) 5 {gh | h, [ Z(G)}. If a is an element of G of order 4, prove that H 5 Z(G) x, aZ(G) x a2Z(G) x a3Z(G) is a subgroup of G. Generalize to the, case that |a| 5 k., 10. The exponent of a group is the smallest positive integer n such that, xn 5 e for all x in the group. Prove that every finite group has an exponent that divides the order of the group., 11. Determine all U-groups of exponent 2., 12. Suppose that H and K are subgroups of a group and that |H| and |K|, are relatively prime. Show that H y K 5 {e}., 13. Let R1 denote the multiplicative group of positive real numbers and, let T 5 {a 1 bi [ C* | a2 1 b2 5 1} be the multiplicative group of, complex numbers on the unit circle. Show that every element of C*, can be uniquely expressed in the form rz, where r [ R1 and z [ T., 14. Prove that Q* under multiplication is not isomorphic to R* under, multiplication., 15. Prove that Q under addition is not isomorphic to R under addition., 16. Prove that R under addition is not isomorphic to R* under, multiplication., 17. Show that Q1 (the set of positive rational numbers) under multiplication is not isomorphic to Q under addition., 18. Suppose that G 5 {e, x, x2, y, yx, yx2} is a non-Abelian group with, |x| 5 3 and |y| 5 2. Show that xy 5 yx2., 19. Let p be an odd prime. Show that 1 is the only solution of x p22 5 1, in U(p)., 20. Let G be an Abelian group under addition. Let n be a fixed positive, integer and let H 5 {(g, ng) | g [ G}. Show that H is a subgroup of, G % G. When G is the set of real numbers under addition, describe, H geometrically., 21. Find a subgroup of Z12 % Z20 that is isomorphic to Z4 % Z5., 22. Suppose that G 5 G1 % G2 % ? ? ? % Gn. Prove that Z(G) 5, Z(G1) % Z(G2) % ? ? ? % Z(Gn)., 23. Exhibit four nonisomorphic groups of order 18., 24. What is the order of the largest cyclic subgroup in Aut(Z720)? (Hint:, It is not necessary to consider automorphisms of Z720.), 25. Let G be the group of all permutations of the positive integers. Let, H be the subset of elements of G that can be expressed as a product, of a finite number of cycles. Prove that H is a subgroup of G., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 5– 8, , 183, , 26. Let G be a group and let g [ G. Show that Z(G)kgl is a subgroup of G., 27. Show that D11 % Z3 ] D3 % Z11. (This exercise is referred to in, Chapter 24.), 28. Show that D33 ] D11 % Z3. (This exercise is referred to in Chapter 24.), 29. Show that D33 ] D3 % Z11. (This exercise is referred to in Chapter 24.), 30. Exhibit four nonisomorphic groups of order 66. (This exercise is, referred to in Chapter 24.), 31. Prove that |Inn(G)| 5 1 if and only if G is Abelian., 32. Prove that x100 5 1 for all x in U(1000)., 33. Find a subgroup of order 6 in U(450)., 34. List four elements of Z20 % Z5 % Z60 that form a noncyclic, subgroup., 35. In S10, let b 5 (13)(17)(265)(289). Find an element in S10 that commutes with b but is not a power of b., 36. Prove or disprove that Z4 % Z15 < Z6 % Z10., 37. Prove or disprove that D12 < Z3 % D4., 38. Describe a three-dimensional solid whose symmetry group is isomorphic to D5., 39. Let G 5 U(15) % Z10 % S5. Find the order of (2, 3, (123)(15)). Find, the inverse of (2, 3, (123)(15))., 40. Let G 5 Z % Z10 and let H 5 {g [ G | |g| 5 ` or |g| 5 1}. Prove, or disprove that H is a subgroup of G., 41. Find a subgroup H of Zp2 % Zp2 such that (Zp2 % Zp2)/H is isomorphic to Zp % Zp., 42. Find three subgroups H1, H2, and H3 of Zp2 % Zp2 such that (Zp2 %, Zp2)/Hi is isomorphic to Zp2 for i 5 1, 2, 3., 43. Find an element of order 10 in A9., 44. In the left regular representation for D4, write TR90 and TH in matrix, form and in cycle form., 45. How many elements of order 6 are in S7?, 46. Prove that S3 % S4 is not isomorphic to a subgroup of S6., 47. Find a permutation b such that b2 5 (13579)(268)., 48. In R % R under componentwise addition, let H 5 {(x, 3x) | x [ R}., (Note that H is the subgroup of all points on the line y 5 3x.) Show, that (2, 5) 1 H is a straight line passing through the point (2, 5) and, parallel to the line y 5 3x., 49. In R % R, suppose that H is the subgroup of all points lying on a, line through the origin. Show that any left coset of H is a line parallel to H., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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184, , Groups, , 50. Let G be a group of permutations on the set {1, 2, . . . , n}. Recall, that stabG(1) 5 {a [ G | a(1) 5 1}. If g sends 1 to k, prove that, g stabG(1) 5 {b [ G | b(1) 5 k}., 51. Let H be a subgroup of G and let a, b [ G. Show that aH 5 bH if, and only if Ha21 5 Hb21., 52. Suppose that G is a finite Abelian group that does not contain a, subgroup isomorphic to Zp % Zp for any prime p. Prove that G is, cyclic., 53. Let p be a prime. Determine the number of elements of order p in, Zp2 % Zp2., 54. Show that Zp2 % Zp2 has exactly one subgroup isomorphic to Zp % Zp., 55. Let p be a prime. Determine the number of subgroups of Zp2 % Zp2, that are isomorphic to Zp2., 56. Find a group of order 32 ? 52 ? 72 ? 28 that contains a subgroup isomorphic to A8., 57. Let p and q be distinct odd primes. Let n 5 lcm(p 2 1, q 2 1)., Prove that xn 5 1 for all x [ U( pq)., 58. Give a simple characterization of all positive integers n for which, Zn 艐 H % Zn/H for every subgroup H of Zn., 59. Prove that the permutations (12) and (123 . . . n) generate Sn. (That, is, every member of Sn can be expressed as some combination of, these elements.), 60. Suppose that n is even and s is an (n 2 1)-cycle in Sn. Show that s, does not commute with any element of order 2., 61. Suppose that n is odd and s is an n-cycle in Sn. Prove that s does, not commute with any element of order 2., 62. Let H 5 {a [ Sn | a maps the set {1, 2} to itself}. Prove that, C ((12)) 5 H., 63. Let m be a positive integer. For any n-cycle s, show that sm is the, product of gcd(m, n) disjoint cycles, each of length n/gcd(m, n)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Normal Subgroups, 9, and Factor Groups, It is tribute to the genius of Galois that he recognized that those subgroups, for which the left and right cosets coincide are distinguished ones. Very, often in mathematics the crucial problem is to recognize and to discover, what are the relevant concepts; once this is accomplished the job may be, more than half done., i. n. herstein, Topics in Algebra, , Normal Subgroups, As we saw in Chapter 7, if G is a group and H is a subgroup of G, it is not, always true that aH 5 Ha for all a in G. There are certain situations where, this does hold, however, and these cases turn out to be of critical importance in the theory of groups. It was Galois, about 180 years ago, who first, recognized that such subgroups were worthy of special attention., Definition Normal Subgroup, , A subgroup H of a group G is called a normal subgroup of G if aH 5, Ha for all a in G. We denote this by H v G., , You should think of a normal subgroup in this way: You can switch, the order of a product of an element a from the group and an element h, from the normal subgroup H, but you must “fudge” a bit on the element, from the normal subgroup H by using some h9 from H rather than h., That is, there is an element h9 in H such that ah 5 h9a. Likewise, there, is some h0 in H such that ha 5 ah0. (It is possible that h9 5 h or h0 5 h,, but we may not assume this.), There are several equivalent formulations of the definition of normality. We have chosen the one that is the easiest to use in applications., However, to verify that a subgroup is normal, it is usually better to use, Theorem 9.1, which is a weaker version of property 8 of the lemma in, Chapter 7. It allows us to substitute a condition about two subgroups of, G for a condition about two cosets of G., , 185, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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186, , Groups, , Theorem 9.1 Normal Subgroup Test, A subgroup H of G is normal in G if and only if xHx21 # H, for all x in G., , PROOF If H is normal in G, then for any x [ G and h [ H there is an h9, in H such that xh 5 h9x. Thus, xhx21 5 h9, and therefore xHx21 # H., Conversely, if xHx21 # H for all x, then, letting x 5 a, we have, aHa21 # H or aH # Ha. On the other hand, letting x 5 a21, we have, a21H(a21)21 5 a21Ha # H or Ha # aH., EXAMPLE 1 Every subgroup of an Abelian group is normal. (In this, case, ah 5 ha for a in the group and h in the subgroup.), EXAMPLE 2 The center Z(G) of a group is always normal. [Again,, ah 5 ha for any a [ G and any h [ Z(G).], EXAMPLE 3 The alternating group An of even permutations is a normal subgroup of Sn. [Note, for example, that for (12) [ Sn and (123) [, An, we have (12)(123) 2 (123)(12) but (12)(123) 5 (132)(12) and, (132) [ An.], EXAMPLE 4 Every subgroup of Dn consisting solely of rotations is, normal in Dn. (For any rotation R and any reflection F, we have FR 5, R21F and any two rotations commute.), The next example illustrates a way to use a normal subgroup to create new subgroups from existing ones., EXAMPLE 5 Let H be a normal subgroup of a group G and K be any, subgroup of G. Then HK 5 {hk | h [ H, k [ K} is a subgroup of G., To verify this, note that e 5 ee is in HK. Then for any a 5 h1k1 and, b 5 h2k2, where h1, h2 are in H and k1, k2 are in K, there is an element h9, in H such that ab21 5 h1k1k221h221 5 h1(k1k221)h221 5 (h1h9)(k1k221)., So, ab21 is in HK., Be careful not to assume that for any subgroups H and K of a group, G, the set HK is a subgroup of G. See Exercise 57., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 187, , Combining Examples 4 and 5, we form a non-Abelian subgroup of, D8 of order 8., EXAMPLE 6 In D8, let H 5 {R0, R90, R180, R270} and K 5 {R0, F},, where F is any reflection. Then HK 5 {R0, R90, R180, R270, R0F, R90F,, R180F, R270F} is a subgroup of D8., EXAMPLE 7 If a group G has a unique subgroup H of some finite, order, then H is normal in G. To see that this is so, observe that for any, g [ G, gHg21 is a subgroup of G and |gHg21| 5 |H|., EXAMPLE 8 The group SL(2, R) of 2 3 2 matrices with determinant, 1 is a normal subgroup of GL(2, R), the group of 2 3 2 matrices with, nonzero determinant. To verify this, we use the Normal Subgroup Test, given in Theorem 9.1. Let x [ GL(2, R) 5 G, h [ SL(2, R) 5 H, and, note that det xhx21 5 (det x)(det h)(det x)21 5 (det x)(det x)21 5 1. So,, xhx21 [ H, and, therefore, xHx21 # H., EXAMPLE 9 Referring to the group table for A4 given in Table 5.1 on, page 111, we may observe that H 5 {a1, a2, a3, a4} is a normal, subgroup of A4 , whereas K 5 {a1, a5, a9} is not a normal subgroup, of A4. To see that H is normal, simply note that for any b in A4, bHb21 is, a subgroup of order 4 and H is the only subgroup of A4 of order 4, (see Table 5.1). Thus, bHb21 5 H. In contrast, a2a5a 21, 2 5 a7, so that, a2Ka 221 s K., , Factor Groups, We have yet to explain why normal subgroups are of special significance., The reason is simple. When the subgroup H of G is normal, then the set, of left (or right) cosets of H in G is itself a group—called the factor group, of G by H (or the quotient group of G by H). Quite often, one can obtain, information about a group by studying one of its factor groups. This, method will be illustrated in the next section of this chapter., Theorem 9.2 Factor Groups (O. Hölder, 1889), Let G be a group and let H be a normal subgroup of G. The set, G/H 5 {aH | a [ G} is a group under the operation (aH)(bH) 5 abH.†, , †The, , notation G/H was first used by C. Jordan., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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188, , Groups, , PROOF Our first task is to show that the operation is well-defined; that, is, we must show that the correspondence defined above from G/H 3, G/H into G/H is actually a function. To do this, we assume that for, some elements a, a9, b, and b9 from G, we have aH 5 a9H and bH 5, b9H, and verify that aHbH 5 a9Hb9H. That is, verify that abH 5 a9b9H., (This shows that the definition of multiplication depends on only the, cosets and not on the coset representatives.) From aH 5 a9H and bH 5, b9H , we have a9 5 ah1 and b9 5 bh2 for some h1, h2 in H, and therefore, a9b9H 5 ah1bh2H 5 ah1bH 5 ah1Hb 5 aHb 5 abH. Here we have made, multiple use of associativity, property 2 of the lemma in Chapter 7, and, the fact that H v G. The rest is easy: eH 5 H is the identity; a21H is the, inverse of aH; and (aHbH)cH 5 (ab)HcH 5 (ab)cH 5 a(bc)H 5, aH(bc)H 5 aH(bHcH). This proves that G/H is a group., Although it is merely a curiosity, we point out that the converse of, Theorem 9.2 is also true; that is, if the correspondence aHbH 5 abH, defines a group operation on the set of left cosets of H in G, then H is, normal in G., The next few examples illustrate the factor group concept., EXAMPLE 10 Let 4Z 5 {0, 64, 68, . . .}. To construct Z/4Z, we, first must determine the left cosets of 4Z in Z. Consider the following, four cosets:, 0 1 4Z 5 4Z 5 {0, 64, 68, . . .},, 1 1 4Z 5 {1, 5, 9, . . . ; 23, 27, 211, . . .},, 2 1 4Z 5 {2, 6, 10, . . . ; 22, 26, 210, . . .},, 3 1 4Z 5 {3, 7, 11, . . . ; 21, 25, 29, . . .}., We claim that there are no others. For if k [ Z, then k 5 4q 1 r, where, 0 # r , 4; and, therefore, k 1 4Z 5 r 1 4q 1 4Z 5 r 1 4Z. Now that, we know the elements of the factor group, our next job is to determine, the structure of Z/4Z. Its Cayley table is, , 0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z, , 0 1 4Z, , 1 1 4Z, , 2 1 4Z, , 3 1 4Z, , 0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z, , 1 1 4Z, 2 1 4Z, 3 1 4Z, 0 1 4Z, , 2 1 4Z, 3 1 4Z, 0 1 4Z, 1 1 4Z, , 3 1 4Z, 0 1 4Z, 1 1 4Z, 2 1 4Z, , Clearly, then, Z/4Z L Z4. More generally, if for any n . 0 we let nZ 5, {0, 6n, 62n, 63n, . . .}, then Z/nZ is isomorphic to Zn., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 189, , EXAMPLE 11 Let G 5 Z18 and let H 5 k6l 5 {0, 6, 12}. Then G/H 5, {0 1 H, 1 1 H, 2 1 H, 3 1 H, 4 1 H, 5 1 H}. To illustrate how the, group elements are combined, consider (5 1 H) 1 (4 1 H). This, should be one of the six elements listed in the set G/H. Well, (5 1 H) 1, (4 1 H) 5 5 1 4 1 H 5 9 1 H 5 3 1 6 1 H 5 3 1 H, since H absorbs all multiples of 6., A few words of caution about notation are warranted here. When H, is a normal subgroup of G, the expression |aH| has two possible interpretations. One could be thinking of aH as a set of elements and |aH|, as the size of the set; or, as is more often the case, one could be thinking of aH as a group element of the factor group G/H and |aH| as the, order of the element aH in G/H. In Example 11, for instance, the set, 3 1 H has size 3, since 3 1 H 5 {3, 9, 15}. But the group element, 3 1 H has order 2, since (3 1 H) 1 (3 1 H) 5 6 1 H 5 0 1 H. As is, usually the case when one notation has more than one meaning, the appropriate interpretation will be clear from the context., EXAMPLE 12 Let _ 5 {R0, R180}, and consider the factor group of, the dihedral group D4 (see the back inside cover for the multiplication, table for D4), D4/_ 5 {_, R90_, H_, D_}., The multiplication table for D4/_ is given in Table 9.1. (Notice that, even though R90H 5 D9, we have used D_ in Table 9.1 for R90_H_, because D9_ 5 D_.), Table 9.1, , , R90, H, D, , , , R90, , H, , D, , _, R90_, H_, D_, , R90_, _, D_, H_, , H_, D_, _, R90_, , D_, H_, R90_, _, , D 4/_ provides a good opportunity to demonstrate how a factor, group of G is related to G itself. Suppose we arrange the heading of the, Cayley table for D4 in such a way that elements from the same coset of, _ are in adjacent columns (Table 9.2). Then, the multiplication table, for D4 can be blocked off into boxes that are cosets of _, and the substitution that replaces a box containing the element x with the coset x_, yields the Cayley table for D4/_., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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191, , 9 | Normal Subgroups and Factor Groups, , Table 9.3, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 1, 2, 3, 4, , 1, 2, 3, 4, , 2, 1, 4, 3, , 3, 4, 1, 2, , 4, 3, 2, 1, , 5, 6, 7, 8, , 6, 5, 8, 7, , 7, 8, 5, 6, , 8, 7, 6, 5, , 9, 10, 11, 12, , 10, 9, 12, 11, , 11, 12, 9, 10, , 12, 11, 10, 9, , 5, 6, 7, 8, , 5, 6, 7, 8, , 8, 7, 6, 5, , 6, 5, 8, 7, , 7, 8, 5, 6, , 9, 10, 11, 12, , 12, 11, 10, 9, , 10, 9, 12, 11, , 11, 12, 9, 10, , 1, 2, 3, 4, , 4, 3, 2, 1, , 2, 1, 4, 3, , 3, 4, 1, 2, , 9, 10, 11, 12, , 9, 10, 11, 12, , 11, 12, 9, 10, , 12, 11, 10, 9, , 10, 9, 12, 11, , 1, 2, 3, 4, , 3, 4, 1, 2, , 4, 3, 2, 1, , 2, 1, 4, 3, , 5, 6, 7, 8, , 7, 8, 5, 6, , 8, 7, 6, 5, , 6, 5, 8, 7, , Table 9.4, , 1H, 5H, 9H, , 1H, , 5H, , 9H, , 1H, 5H, 9H, , 5H, 9H, 1H, , 9H, 1H, 5H, , This procedure can be illustrated more vividly with colors. Let’s say, we had printed the elements of H in green, the elements of 5H in red,, and the elements of 9H in blue. Then, in Table 9.3, each box would, consist of elements of a uniform color. We could then think of, the factor group as consisting of the three colors that define a group, table isomorphic to G/H., , Green, Red, Blue, , Green, , Red, , Blue, , Green, Red, Blue, , Red, Blue, Green, , Blue, Green, Red, , It is instructive to see what happens if we attempt the same procedure with a group G and a subgroup H that is not normal in G—that is,, if we arrange the headings of the Cayley table so that the elements, from the same coset of H are in adjacent columns and attempt to block, off the table into boxes that are also cosets of H to produce a Cayley, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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192, , Groups, , table for the set of cosets. Say, for instance, we were to take G to be A4, and H 5 {1, 5, 9}. The cosets of H would be H, 2H 5 {2, 6, 10},, 3H 5 {3, 7, 11}, and 4H 5 {4, 8, 12}. Then the first three rows of the, rearranged Cayley table for A4 would be the following., , 1, 5, 9, , 1, , 5, , 9, , 2, , 6, , 10, , 3, , 7, , 11, , 4, , 8, , 12, , 1, 5, 9, , 5, 9, 1, , 9, 1, 5, , 2, 8, 11, , 6, 12, 3, , 10, 4, 7, , 3, 6, 12, , 7, 10, 4, , 11, 2, 8, , 4, 7, 10, , 8, 11, 2, , 12, 3, 6, , But already we are in trouble, for blocking these off into 3 3 3 boxes, yields boxes that contain elements of different cosets. Hence, it is impossible to represent an entire box by a single element of the box in the, same way we could for boxes made from the cosets of a normal subgroup. Had we printed the rearranged table in four colors with all, members of the same coset having the same color, we would see multicolored boxes rather than the uniformly colored boxes produced by a, normal subgroup., In Chapter 11, we will prove that every finite Abelian group is, isomorphic to a direct product of cyclic groups. In particular, an, Abelian group of order 8 is isomorphic to one of Z8, Z4 % Z2, or Z2 %, Z2 % Z2. In the next two examples, we examine Abelian factor groups, of order 8 and determine the isomorphism type of each., EXAMPLE 14 Let G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,, 23, 25, 27, 29, 31} and H 5 U16(32) 5 {1, 17}. Then G/H is an Abelian, group of order 16/2 5 8. Which of the three Abelian groups of order 8, is it—Z8, Z4 % Z2, or Z2 % Z2 % Z2? To answer this question, we need, only determine the elements of G/H and their orders. Observe that the, eight cosets, 1H 5 {1, 17}, 3H 5 {3, 19},, 5H 5 {5, 21},, 9H 5 {9, 25}, 11H 5 {11, 27}, 13H 5 {13, 29},, , 7H 5 {7, 23},, 15H 5 {15, 31}, , are all distinct, so that they form the factor group G/H. Clearly,, (3H)2 5 9H 2 H, and so 3H has order at least 4. Thus, G/H is not, Z2 % Z2 % Z2. On the other hand, direct computations show that both, 7H and 9H have order 2, so that G/H cannot be Z8 either, since a cyclic, group of even order has exactly one element of order 2 (Theorem 4.4)., This proves that U(32)/U16(32) L Z4 % Z2, which (not so incidentally!), is isomorphic to U(16)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 193, , EXAMPLE 15 Let G 5 U(32) and K 5 {1, 15}. Then |G/K| 5 8,, and we ask, which of the three Abelian groups of order 8 is G/K? Since, (3K)4 5 81K 5 17K 2 K, |3K| 5 8. Thus, G/K L Z8., It is crucial to understand that when we factor out by a normal subgroup H, what we are essentially doing is defining every element in H, to be the identity. Thus, in Example 12, we are making R180_ 5 _ the, identity. Likewise, R270_ 5 R90R180_ 5 R90_. Similarly, in Example 10,, we are declaring any multiple of 4 to be 0 in the factor group Z/4Z. This, is why 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z, and so on. In Example 14, we, have 3H 5 19H, since 19 5 3 ? 17 in U(32) and going to the factor, group makes 17 the identity. Algebraists often refer to the process of, creating the factor group G/H as “killing” H., , Applications of Factor Groups, Why are factor groups important? Well, when G is finite and H 2 {e},, G/H is smaller than G, and its structure is usually less complicated than, that of G. At the same time, G/H simulates G in many ways. In fact, we, may think of a factor group of G as a less complicated approximation, of G (similar to using the rational number 3.14 for the irrational, number p). What makes factor groups important is that one can often, deduce properties of G by examining the less complicated group G/H, instead. We illustrate this by giving another proof that A4 has no subgroup of order 6., EXAMPLE 16 A4 has no subgroup of order 6., The group A4 of even permutations on the set {1, 2, 3, 4} has no subgroup H of order 6. To see this, suppose that A4 does have a subgroup H, of order 6. By Exercise 9 in this chapter, we know that H v A4. Thus,, the factor group A4/H exists and has order 2. Since the order of an, element divides the order of the group, we have for all a [ A4 that, a2H 5 (aH)2 5 H. Thus, a2 [ H for all a in A4. Referring to the main, diagonal of the group table for A4 given in Table 5.1 on page 111, however, we observe that A4 has nine different elements of the form a2, all, of which must belong to H, a subgroup of order 6. This is clearly, impossible, so a subgroup of order 6 cannot exist in A4.†, The next three theorems illustrate how knowledge of a factor group, of G reveals information about G itself., †“How, , often have I said to you that when you have eliminated the impossible, whatever, remains, however improbable, must be the truth.” Sherlock Holmes, The Sign of Four, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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194, , Groups, , Theorem 9.3 G/Z Theorem, Let G be a group and let Z(G) be the center of G. If G/Z(G) is cyclic,, then G is Abelian., , PROOF Since G is Abelian is equivalent to Z(G) 5 G, it suffices to, show that the only element of G/Z(G) is the identity coset Z(G). To this end,, let G/Z(G) 5 kgZ(G)l and let a [ G. Then there exists an integer i such that, aZ(G) 5 (gZ(G)) i 5 g i Z(G). Thus, a 5 g i z for some z in Z(G). Since both, g i and z belong to C(g), so does a. Because a is an arbitrary element of G, this means that every element of G commutes with g so g [ Z(G). Thus,, gZ(G) 5 Z(G) is the only element of G/Z(G)., A few remarks about Theorem 9.3 are in order. First, our proof shows, that a better result is possible: If G/H is cyclic, where H is a subgroup of, Z(G), then G is Abelian. Second, in practice, it is the contrapositive of, the theorem that is most often used—that is, if G is non-Abelian, then, G/Z(G) is not cyclic. For example, it follows immediately from this, statement and Lagrange’s Theorem that a non-Abelian group of order, pq, where p and q are primes, must have a trivial center. Third, if G/Z(G), is cyclic, it must be trivial., Theorem 9.4 G/Z(G) < Inn(G), For any group G, G/Z(G) is isomorphic to Inn(G)., , PROOF Consider the correspondence from G/Z(G) to Inn(G) given by, T : gZ(G) → fg [where, recall, fg(x) 5 gxg21 for all x in G]. First, we, show that T is well defined. To do this, we assume that, gZ1G2 hZ1G2 and verify that fg fh. (This shows that the image, of a coset of Z1G2 depends only on the coset itself and not on the element representing the coset.) From gZ1G2 hZ1G2, we have that, h1g belongs to Z(G). Then, for all x in G, h 21gx 5 xh 21g. Thus,, gxg21 5 hxh21 for all x in G, and, therefore, fg 5 fh. Reversing this, argument shows that T is one-to-one, as well. Clearly, T is onto., That T is operation-preserving follows directly from the fact that, fgfh 5 fgh for all g and h in G., As an application of Theorems 9.3 and 9.4, we may easily determine, Inn(D6) without looking at Inn(D6)!, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 195, , EXAMPLE 17 We know from Example 14 in Chapter 3 that, |Z(D6)| 5 2. Thus, |D6 /Z (D6)| 5 6. So, by our classification of groups, of order 6 (Theorem 7.3), we know that Inn(D6) is isomorphic to D3, or Z6. Now, if Inn(D6) were cyclic, then, by Theorem 9.4, D6 /Z(D6), would be also. But then, Theorem 9.3 would tell us that D6 is Abelian., So, Inn(D6) is isomorphic to D3., The next theorem demonstrates one of the most powerful proof techniques available in the theory of finite groups—the combined use of, factor groups and induction., Theorem 9.5 Cauchy’s Theorem for Abelian Groups, Let G be a finite Abelian group and let p be a prime that divides the, order of G. Then G has an element of order p., , PROOF Clearly, this statement is true for the case in which G has, order 2. We prove the theorem by using the Second Principle of Mathematical Induction on |G|. That is, we assume that the statement is true, for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has, elements of prime order, for if |x| 5 m and m 5 qn, where q is prime,, then |x n| 5 q. So let x be an element of G of some prime order q, say. If, q 5 p, we are finished; so assume that q 2 p. Since every subgroup of, an Abelian group is normal, we may construct the factor group G 5, G/kxl. Then G is Abelian and p divides |G|, since |G| 5 |G|/q. By, induction, then, G has an element—call it ykxl—of order p., Then, (ykxl) p = y pkxl = kxl and therefore y p [ kxl. If y p = e, we are, done. If not, then y p has order q and yq has order p., , Internal Direct Products, As we have seen, the external direct product provides a method of putting groups together to get a larger group in such a way that we can, determine many properties of the larger group from the properties of, the component pieces. For example: If G 5 H % K, then |G| 5 |H||K|;, every element of G has the form (h, k) where h [ H and k [ K; if |h|, and |k| are finite, then |(h, k)| 5 lcm(|h|, |k|); if H and K are Abelian,, then G is Abelian; if H and K are cyclic and |H| and |K| are relatively, prime, then H % K is cyclic. It would be quite useful to be able to reverse, this process—that is, to be able to start with a large group G and break, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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196, , Groups, , it down into a product of subgroups in such a way that we could glean, many properties of G from properties of the component pieces. It is occasionally possible to do this., Definition Internal Direct Product of H and K, , We say that G is the internal direct product of H and K and write, G 5 H 3 K if H and K are normal subgroups of G and, G 5 HK, , H > K 5 {e}., , and, , The wording of the phrase “internal direct product” is easy to justify., We want to call G the internal direct product of H and K if H and K are, subgroups of G, and if G is naturally isomorphic to the external direct, product of H and K. One forms the internal direct product by starting, with a group G and then proceeding to find two subgroups H and K, within G such that G is isomorphic to the external direct product of H, and K. (The definition ensures that this is the case—see Theorem 9.6.), On the other hand, one forms an external direct product by starting with, any two groups H and K, related or not, and proceeding to produce the, larger group H % K. The difference between the two products is that the, internal direct product can be formed within G itself, using subgroups, of G and the operation of G, whereas the external direct product can be, formed with totally unrelated groups by creating a new set and a new, operation. (See Figures 9.1 and 9.2.), G, , H, , K, e, , Figure 9.1 For the internal direct product,, H and K must be subgroups of the same group., , H, , K, , Figure 9.2 For the external, direct product, H and K can, be any groups., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 197, , Perhaps the following analogy with integers will be useful in clarifying the distinction between the two products of groups discussed, in the preceding paragraph. Just as we may take any (finite) collection of integers and form their product, we may also take any collection of groups and form their external direct product. Conversely,, just as we may start with a particular integer and express it as a product of certain of its divisors, we may be able to start with a particular, group and factor it as an internal direct product of certain of its subgroups., EXAMPLE 18 In D6, the dihedral group of order 12, let F denote, some reflection and let Rk denote a rotation of k degrees. Then,, D6 5 {R0, R120, R240, F, R120F, R240F} 3 {R0, R180}., Students should be cautioned about the necessity of having all conditions of the definition of internal direct product satisfied to ensure, that HK L H % K. For example, if we take, G 5 S3,, , H 5 k(123)l,, , and, , K 5 k(12)l,, , then G 5 HK, and H > K 5 {(1)}. But G is not isomorphic to H % K,, since, by Theorem 8.2, H % K is cyclic, whereas S3 is not. Note that K, is not normal., A group G can also be the internal direct product of a collection of, subgroups., Definition Internal Direct Product H1 3 H2 3 ? ? ? 3 Hn, , Let H1, H2, . . . , Hn be a finite collection of normal subgroups of G. We, say that G is the internal direct product of H1, H2, . . . , Hn and write, G 5 H1 3 H2 3 ? ? ? 3 Hn, if, 1. G 5 H1H2 ? ? ? Hn 5 {h1h2 ? ? ? hn | hi [ Hi},, 2. (H1H2 ? ? ? Hi) > Hi11 5 {e} for i 5 1, 2, . . . , n 2 1., , This definition is somewhat more complicated than the one given for, two subgroups. The student may wonder about the motivation for it—, that is, why should we want the subgroups to be normal and why is it, desirable for each subgroup to be disjoint from the product of all previous ones? The reason is quite simple. We want the internal direct product to be isomorphic to the external direct product. As the next theorem, shows, the conditions in the definition of internal direct product were, chosen to ensure that the two products are isomorphic., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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198, , Groups, , Theorem 9.6 H1 3 H2 3 ? ? ? 3 Hn L H1 % H2 % ? ? ? % Hn, If a group G is the internal direct product of a finite number of, subgroups H1, H2, . . . , Hn, then G is isomorphic to the external, direct product of H1, H2, . . . , Hn., , PROOF We first show that the normality of the H’s together with the, second condition of the definition guarantees that h’s from different, Hi’s commute. For if hi [ Hi and hj [ Hj with i 2 j, then, (hihj hi21)hj21 [ Hj hj 21 5 Hj, and, hi (hj hi21hj21) [ hiHi 5 Hi., Thus, hihj hi21hj 21 [ Hi > Hj 5 {e} (see Exercise 5), and, therefore,, hi hj 5 hjhi. We next claim that each member of G can be expressed, uniquely in the form h1h2 ? ? ? hn, where hi [ Hi. That there is at least one, such representation is the content of condition 1 of the definition. To, prove uniqueness, suppose that g 5 h1h2 ? ? ? hn and g 5 h19 h29 ? ? ? hn9,, where hi and hi9 belong to Hi for i 5 1, . . . , n. Then, using the fact that, the h’s from different Hi’s commute, we can solve the equation, h1h2 ? ? ? hn 5 h19 h29 ? ? ? hn9, , (1), , for hn9 h n21 to obtain, hn9 hn21 5 (h91)21h1(h29)21h2 ? ? ? (h9n21)21hn21., But then, h n9 hn21 [ H1H2 ? ? ? Hn21 > Hn 5 {e},, so that hn9 hn 21 5 e and, therefore, hn9 5 hn. At this point, we can cancel, hn and hn9 from opposite sides of the equal sign in Equation (1) and repeat, 9 . Continuing in this fashthe preceding argument to obtain hn21 5 hn21, ion, we eventually have hi 5 hi9 for i 5 1, . . . , n. With our claim established, we may now define a function f from G to H1 % H2 % ? ? ? % Hn, by f(h1h2 ? ? ? hn) 5 (h1, h2, . . . , hn). We leave to the reader the easy verification that f is an isomorphism., The next theorem provides an important application of Theorem 9.6., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 199, , Theorem 9.7 Classification of Groups of Order p2, Every group of order p 2, where p is a prime, is isomorphic to Zp2 or, Zp % Zp., , PROOF Let G be a group of order p2, where p is a prime. If G has an, element of order p2, then G is isomorphic to Zp2. So, by Corollary 2 of, Lagrange’s Theorem, we may assume that every nonidentity element of, G has order p. First we show that for any element a, the subgroup kal is, normal in G. If this is not the case, then there is an element b in G such, that bab1 is not in kal. Then kal and kbab1l are distinct subgroups of, order p. Since kal > kbab1l is a subgroup of both kal and kbab1l,, we have that kal > kbab1l {e}. From this it follows that the distinct, left cosets of kbab1l are kbab1l, akbab1l, a2kbab1l, . . . ,, ap1kbab1l. Since b1 must lie in one of these cosets, we may write, b1 in the form b1 ai 1bab1 2 j aiba jb1 for some i and j. Canceling the b1 terms, we obtain e aibaj and therefore b aij [ kal., This contradiction verifies our assertion that every subgroup of the form, kal is normal in G. To complete the proof, let x be any nonidentity element in G and y be any element of G not in kxl. Then, by comparing orders and using Theorem 9.6, we see that G kxl kyl ⬇ Zp % Zp., As an immediate corollary of Theorem 9.7, we have the following, important fact., Corollary, If G is a group of order p 2, where p is a prime, then G is Abelian., , We mention in passing that if G 5 H1 % H2 % ? ? ? % Hn, then G can, be expressed as the internal direct product of subgroups isomorphic to, H1, H2, . . . , Hn. For example, if G 5 H1 % H2, then G 5 H1 3 H2,, where H1 5 H1 % {e} and H2 5 {e} % H2., The topic of direct products is one in which notation and terminology vary widely. Many authors use H 3 K to denote both the internal, direct product and the external direct product of H and K, making no, notational distinction between the two products. A few authors define, only the external direct product. Many people reserve the notation, H % K for the situation where H and K are Abelian groups under addition and call it the direct sum of H and K. In fact, we will adopt this terminology in the section on rings (Part 3), since rings are always Abelian, groups under addition., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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200, , Groups, , The U-groups provide a convenient way to illustrate the preceding, ideas and to clarify the distinction between internal and external direct, products. It follows directly from Theorem 8.3, its corollary, and, Theorem 9.6 that if m 5 n1n2 ? ? ? nk, where gcd(ni, nj) 5 1 for i 2 j, then, U(m) 5 Um/n1(m) 3 Um/n2(m) 3 ? ? ? 3 Um/nk(m), L U(n1) % U(n2) % ? ? ? % U(nk)., Let us return to the examples given following Theorem 8.3., U(105) 5 U(15 ? 7) 5 U15(105) 3 U7(105), 5 {1, 16, 31, 46, 61, 76} 3 {1, 8, 22, 29, 43, 64, 71, 92}, L U(7) % U(15),, U(105) 5 U(5 ? 21) 5 U5(105) 3 U21(105), 5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101}, 3 {1, 22, 43, 64} L U(21) % U(5),, U(105) 5 U(3 ? 5 ? 7) 5 U35(105) 3 U21(105) 3 U15(105), 5 {1, 71} 3 {1, 22, 43, 64} 3 {1, 16, 31, 46, 61, 76}, L U(3) % U(5) % U(7)., , Exercises, The heart of mathematics is its problems., Paul Halmos, , 1. Let H 5 {(1), (12)}. Is H normal in S3?, 2. Prove that An is normal in Sn., 3. In D4, let K 5 {R0, R90, R180, R270}. Write HR90 in the form xH,, where x [ K. Write DR270 in the form xD, where x [ K. Write R90V, in the form Vx, where x [ K., 4. Write (12)(13)(14) in the form a(12), where a [ A4. Write (1234), (12)(23), in the form a(1234), where a [ A4., 5. Show that if G is the internal direct product of H1, H2, . . . , Hn and, i 2 j with 1 # i # n, 1 # j # n, then Hi > Hj 5 {e}. (This exercise, is referred to in this chapter.), a b, d ` a, b, d [ R, ad ? 0 f . Is H a normal sub6. Let H e c, 0 d, group of GL(2, R)?, 7. Let G 5 GL(2, R) and let K be a subgroup of R*. Prove that H 5, {A [ G | det A [ K} is a normal subgroup of G., 8. Viewing k3l and k12l as subgroups of Z, prove that k3l/k12l is isomorphic to Z4. Similarly, prove that k8l/k48l is isomorphic to Z6., Generalize to arbitrary integers k and n., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 201, , 9. Prove that if H has index 2 in G, then H is normal in G. (This exercise is referred to in Chapters 24 and 25 and this chapter.), 10. Let H 5 {(1), (12)(34)} in A4., a. Show that H is not normal in A4., b. Referring to the multiplication table for A4 in Table 5.1 on page, 111, show that, although a6H 5 a7H and a9H 5 a11H, it is not, true that a6a9H 5 a7a11H. Explain why this proves that the left, cosets of H do not form a group under coset multiplication., 11. Let G 5 Z4 % U(4), H 5 k(2, 3)l, and K 5 k(2, 1)l. Show that G/H, is not isomorphic to G/K. (This shows that H L K does not imply, that G/H L G/K.), 12. Prove that a factor group of a cyclic group is cyclic., 13. Prove that a factor group of an Abelian group is Abelian., 14. What is the order of the element 14 1 k8l in the factor group, Z24/k8l?, 15. What is the order of the element 4U5(105) in the factor group, U(105)/U5(105)?, 16. Recall that Z(D6) 5 {R0, R180}. What is the order of the element, R60Z(D6) in the factor group D6/Z(D6)?, 17. Let G 5 Z/k20l and H 5 k4l/k20l. List the elements of H and G/H., 18. What is the order of the factor group Z60/k15l?, 19. What is the order of the factor group (Z10 % U(10))/k(2, 9)l?, 20. Construct the Cayley table for U(20)/U5(20)., 21. Prove that an Abelian group of order 33 is cyclic., 22. Determine the order of (Z % Z)/k(2, 2)l. Is the group cyclic?, 23. Determine the order of (Z % Z)/k(4, 2)l. Is the group cyclic?, 24. The group (Z4 % Z12)/k(2, 2)l is isomorphic to one of Z8, Z4 % Z2, or, Z2 % Z2 % Z2. Determine which one by elimination., 25. Let G 5 U(32) and H 5 {1, 31}. The group G/H is isomorphic to, one of Z8, Z4 % Z2, or Z2 % Z2 % Z2. Determine which one by, elimination., 26. Let G be the group of quaternions given by the table in Exercise 4, of the Supplementary Exercises for Chapters 1–4, and let H be the, subgroup {e, a2}. Is G/H isomorphic to Z4 or Z2 % Z2?, 27. Let G 5 U(16), H 5 {1, 15}, and K 5 {1, 9}. Are H and K isomorphic? Are G/H and G/K isomorphic?, 28. Let G 5 Z4 % Z4, H 5 {(0, 0), (2, 0), (0, 2), (2, 2)}, and K 5 k(1, 2)l., Is G/H isomorphic to Z4 or Z2 % Z2? Is G/K isomorphic to Z4 or, Z2 % Z2?, 29. Prove that A4 % Z3 has no subgroup of order 18., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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202, , Groups, , 30. Express U(165) as an internal direct product of proper subgroups, in four different ways., 31. Let R* denote the group of all nonzero real numbers under multiplication. Let R1 denote the group of positive real numbers under, multiplication. Prove that R* is the internal direct product of R1, and the subgroup {1, 21}., 32. Prove that D4 cannot be expressed as an internal direct product of, two proper subgroups., 33. Let H and K be subgroups of a group G. If G 5 HK and g 5 hk,, where h [ H and k [ K, is there any relationship among |g|, |h|,, and |k|? What if G 5 H 3 K?, 34. In Z, let H 5 k5l and K 5 k7l. Prove that Z 5 HK. Does Z 5 H 3 K?, 35. Let G 5 {3a 6 b10 c | a, b, c [ Z} under multiplication and H 5, {3a 6 b12c | a, b, c [ Z} under multiplication. Prove that G 5 k3l 3, k6l 3 k10l, whereas H 2 k3l 3 k6l 3 k12l., 36. Determine all subgroups of R* (nonzero reals under multiplication) of index 2., 37. Let G be a finite group and let H be a normal subgroup of G. Prove, that the order of the element gH in G/H must divide the order, of g in G., 38. Let H be a normal subgroup of G and let a belong to G. If the element aH has order 3 in the group G/H and |H| 510, what are the, possibilities for the order of a?, 39. If H is a normal subgroup of a group G, prove that C(H), the centralizer of H in G, is a normal subgroup of G., 40. Let f be an isomorphism from a group G onto a group G. Prove, that if H is a normal subgroup of G, then f(H) is a normal subgroup of G., 41. Show that Q, the group of rational numbers under addition, has no, proper subgroup of finite index., 42. An element is called a square if it can be expressed in the form b2, for some b. Suppose that G is an Abelian group and H is a subgroup of G. If every element of H is a square and every element of, G/H is a square, prove that every element of G is a square. Does, your proof remain valid when “square” is replaced by “nth power,”, where n is any integer?, 43. Show, by example, that in a factor group G/H it can happen that, aH 5 bH but |a| 2 |b|., 44. Observe from the table for A4 given in Table 5.1 on page 111 that, the subgroup given in Example 9 of this chapter is the only subgroup of A4 of order 4. Why does this imply that this subgroup, must be normal in A4? Generalize this to arbitrary finite groups., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 203, , 45. Let p be a prime. Show that if H is a subgroup of a group of order, 2p that is not normal, then H has order 2., 46. Show that D13 is isomorphic to Inn(D13)., 47. Suppose that N is a normal subgroup of a finite group G and H is a, subgroup of G. If 0 G/N 0 is prime, prove that H is contained in N or, that NH G., 48. If G is a group and |G: Z(G)| 5 4, prove that G/Z(G) < Z2 % Z2., 49. Suppose that G is a non-Abelian group of order p3, where p is a, prime, and Z(G) 2 {e}. Prove that |Z(G)| 5 p., 50. If |G| 5 pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| 5 1 or pq., 51. Let N be a normal subgroup of G and let H be a subgroup of G. If, N is a subgroup of H, prove that H/N is a normal subgroup of G/N, if and only if H is a normal subgroup of G., 52. Let G be an Abelian group and let H be the subgroup consisting of, all elements of G that have finite order. (See Exercise 20 in the, Supplementary Exercises for Chapters 1– 4.) Prove that every nonidentity element in G/H has infinite order., 53. Determine all subgroups of R* that have finite index., 54. Let G 5 {61, 6i, 6j, 6k}, where i 2 5 j 2 5 k 2 5 21, 2i 5 (21)i,, 12 5 (21)2 5 1, ij 5 2ji 5 k, jk 5 2kj 5 i, and ki 5 2ik 5 j., a. Construct the Cayley table for G., b. Show that H 5 {1, 21} v G., c. Construct the Cayley table for G/H. Is G/H isomorphic to Z4 or, Z2 % Z2?, (The rules involving i, j, and k can be remembered by using the circle below., i, , k, , j, , Going clockwise, the product of two consecutive elements is the third, one. The same is true for going counterclockwise, except that we obtain the negative of the third element.) This is the group of quaternions that was given in another form in Exercise 4 in the Supplementary, Exercises for Chapters 1–4. It was invented by William Hamilton, in 1843. The quaternions are used to describe rotations in threedimensional space, and they are used in physics. The quaternions can, be used to extend the complex numbers in a natural way., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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204, , Groups, , 55. In D4, let K 5 {R0, D} and let L 5 {R0, D, D9, R180}. Show that K v, L v D4, but that K is not normal in D4. (Normality is not transitive., Compare Exercise 4, Supplementary Exercises for Chapters 5–8.), 56. Show that the intersection of two normal subgroups of G is a normal subgroup of G. Generalize., 57. Give an example of subgroups H and K of a group G such that HK, is not a subgroup of G., 58. If N and M are normal subgroups of G, prove that NM is also a, normal subgroup of G., 59. Let N be a normal subgroup of a group G. If N is cyclic, prove that, every subgroup of N is also normal in G. (This exercise is referred, to in Chapter 24.), 60. Without looking at inner automorphisms of Dn, determine the number of such automorphisms., 61. Let H be a normal subgroup of a finite group G and let x [ G. If, gcd(|x|, |G/H|) 5 1, show that x [ H. (This exercise is referred to, in Chapter 25.), 62. Let G be a group and let G9 be the subgroup of G generated by the, set S 5 {x21y21xy | x, y [ G}. (See Exercise 3, Supplementary, Exercises for Chapters 5–8, for a more complete description of G9.), a. Prove that G9 is normal in G., b. Prove that G/G9 is Abelian., c. If G/N is Abelian, prove that G9 # N., d. Prove that if H is a subgroup of G and G9 # H, then H is normal, in G., 63. If N is a normal subgroup of G and |G/N| 5 m, show that xm [ N, for all x in G., 64. Suppose that a group G has a subgroup of order n. Prove that the, intersection of all subgroups of G of order n is a normal subgroup, of G., 65. If G is non-Abelian, show that Aut(G) is not cyclic., 66. Let |G| 5 p nm, where p is prime and gcd( p, m) 5 1. Suppose that, H is a normal subgroup of G of order pn. If K is a subgroup of G of, order pk, show that K # H., 67. Suppose that H is a normal subgroup of a finite group G. If G/H, has an element of order n, show that G has an element of order n., Show, by example, that the assumption that G is finite is necessary., 68. Recall that a subgroup N of a group G is called characteristic if, f(N) 5 N for all automorphisms f of G. If N is a characteristic, subgroup of G, show that N is a normal subgroup of G., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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9 | Normal Subgroups and Factor Groups, , 205, , 69. In D4, let _ 5 {R0, H}. Form an operation table for the cosets _,, D_, V_, and D9_. Is the result a group table? Does your answer, contradict Theorem 9.2?, 70. Prove that A4 is the only subgroup of S4 of order 12., 71. If |G| 5 30 and |Z(G)| 5 5, what is the structure of G/Z(G)?, 72. If H is a normal subgroup of G and |H| 5 2, prove that H is contained in the center of G., 73. Prove that A5 cannot have a normal subgroup of order 2., 74. Let G be a finite group and let H be an odd-order subgroup of G of, index 2. Show that the product of all the elements of G (taken in, any order) cannot belong to H., 75. Let G be a group and p a prime. Suppose that H 5 {g p Z g [ G} is, a subgroup of G. Show that H is normal and that every nonidentity, element of G/H has order p., 76. Suppose that H is a normal subgroup of G. If |H| 5 4 and gH has, order 3 in G/H, find a subgroup of order 12 in G., 77. Let G be a group and H an odd-order subgroup of G of index 2., Show that H contains every element of G of odd order., 78. A proper subgroup H of a group G is called maximal if there is no, subgroup K such that H ( K ( G (that is, there is no subgroup K, properly contained between H and G). Show that Z(G) is never a, maximal subgroup of a group G., 79. Let G be a group of order 100 that has exactly one subgroup of, order 5. Prove that it has a subgroup of order 10., , Suggested Readings, Michael Brennan and Des MacHale, “Variations on a Theme: A4 Definitely Has No Subgroup of Order Six!,” Mathematics Magazine 73, (2000): 36–40., The authors offer 11 proofs that A4 has no subgroup of order 6. These, proofs provide a review of many of the ideas covered thus far in this, text., J. A. Gallian, R. S. Johnson, and S. Peng, “On Quotient Structures of, Z n,” Pi Mu Epsilon Journal 9 (1993): 524–526., The authors determine the structure of the group (Z % Z)/k(a, b)l and, related groups., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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206, , Groups, , Tony Rothman, “Genius and Biographers: The Fictionalization of Évariste, Galois,” The American Mathematical Monthly 89 (1982): 84–106., The author argues that many popular accounts of Galois’s life have, been greatly embroidered., Paul F. Zweifel, “Generalized Diatonic and Pentatonic Scales: A Grouptheoretic Approach,” Perspectives of New Music 34 (1996): 140–161., The author discusses how group theoretic notions such as subgroups,, cosets, factor groups, and isomorphisms of Z12 and Z20 relate to musical, scales, tuning, temperament, and structure., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Évariste Galois, , The Granger Collection, New York, , Galois at seventeen was making discoveries of epochal significance in the theory of, equations, discoveries whose consequences are not yet exhausted after more, than a century., e. t. bell, Men of Mathematics, , This French stamp was issued as part of, the 1984 “Celebrity Series” in support of, the Red Cross Fund., , Évariste Galois (pronounced gal-WAH), was born on October 25, 1811, near Paris., Although he had mastered the works of, Legendre and Lagrange at age 15, Galois, twice failed his entrance examination to the, École Polytechnique. He did not know, some basic mathematics, and he did mathematics almost entirely in his head, to the, annoyance of the examiner., At 18, Galois wrote his important research, on the theory of equations and submitted it to, the French Academy of Sciences for publication. The paper was given to Cauchy for refereeing. Cauchy, impressed by the paper,, agreed to present it to the academy, but he, never did. At the age of 19, Galois entered a, , paper of the highest quality in the competition for the Grand Prize in Mathematics,, given by the French Academy of Sciences., The paper was given to Fourier, who died, shortly thereafter. Galois’s paper was never, seen again., Galois spent most of the last year and a, half of his life in prison for revolutionary political offenses. While in prison, he attempted, suicide and prophesied that he would die in a, duel. On May 30, 1832, Galois was shot in a, duel; he died the next day at the age of 20., Among the many concepts introduced by, Galois are normal subgroups, isomorphisms,, simple groups, finite fields, and Galois theory., His work provided a method for disposing, of several famous constructability problems,, such as trisecting an arbitrary angle and doubling a cube. Galois’s entire collected works, fill only 60 pages., To find more information about Galois,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 207, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Group, 10, Homomorphisms, When it comes to laws, there is absolutely no doubt that symmetry and, group theory are extremely useful concepts. Without the introduction of, symmetry and the language of groups into particle physics the description, of the elementary particles and their interactions would have been an, intricate nightmare. Groups truly flesh out order and identify patterns like, no other mathematical machinery., mario livio, The Equation That Couldn't be Solved, , Definition and Examples, In this chapter, we consider one of the most fundamental ideas of, algebra—homomorphisms. The term homomorphism comes from the, Greek words homo, “like,” and morphe, “form.” We will see that a homomorphism is a natural generalization of an isomorphism and that, there is an intimate connection between factor groups of a group and, homomorphisms of a group. The concept of group homomorphisms, was introduced by Camille Jordan in 1870, in his influential book Traité, des substitutions., Definition Group Homomorphism, , A homomorphism f from a group G to a group G is a mapping, from G into G that preserves the group operation; that is, f(ab) 5, f(a)f(b) for all a, b in G., , Before giving examples and stating numerous properties of, homomorphisms, it is convenient to introduce an important subgroup, that is intimately related to the image of a homomorphism. (See, property 4 of Theorem 10.1.), Definition Kernel of a Homomorphism, , The kernel of a homomorphism f from a group G to a group with, identity e is the set {x [ G | f(x) 5 e}. The kernel of f is denoted by, Ker f., , 208, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 209, , EXAMPLE 1 Any isomorphism is a homomorphism that is also onto, and one-to-one. The kernel of an isomorphism is the trivial subgroup., EXAMPLE 2 Let R* be the group of nonzero real numbers under, multiplication. Then the determinant mapping A → det A is a, homomorphism from GL(2, R) to R*. The kernel of the determinant, mapping is SL(2, R)., EXAMPLE 3 The mapping f from R* to R*, defined by f(x) 5 |x|,, is a homomorphism with Ker f 5 {1, 21}., EXAMPLE 4 Let R[x] denote the group of all polynomials with real, coefficients under addition. For any f in R[x], let f 9 denote the derivative of f. Then the mapping f S f 9 is a homomorphism from R[x] to itself. The kernel of the derivative mapping is the set of all constant, polynomials., EXAMPLE 5 The mapping f from Z to Zn, defined by f(m) 5 m, mod n, is a homomorphism (see Exercise 9 in Chapter 0). The kernel of, this mapping is knl., EXAMPLE 6 The mapping f(x) 5 x2 from R*, the nonzero real, numbers under multiplication, to itself is a homomorphism, since, f(ab) 5 (ab)2 5 a2b2 5 f(a)f(b) for all a and b in R*. (See Exercise 5.), The kernel is {1, –1}., EXAMPLE 7 The mapping f(x) 5 x2 from R, the real numbers, under addition, to itself is not a homomorphism, since f(a 1 b) 5, (a 1 b)2 5 a2 1 2ab 1 b2, whereas f(a) 1 f(b) 5 a2 1 b2., When defining a homomorphism from a group in which there are, several ways to represent the elements, caution must be exercised to, ensure that the correspondence is a function. (The term well-defined is, often used in this context.) For example, since 3(x 1 y) 5 3x 1 3y in, Z6, one might believe that the correspondence x 1 k3l S 3x from Z/k3l to, Z6 is a homomorphism. But it is not a function, since 0 1 k3l 5 3 1, k3l in Z/k3l but 3 ? 0 2 3 ? 3 in Z6., For students who have had linear algebra, we remark that every, linear transformation is a group homomorphism and the null-space is, the same as the kernel. An invertible linear transformation is a group, isomorphism., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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210, , Groups, , Properties of Homomorphisms, Theorem 10.1 Properties of Elements Under Homomorphisms, Let f be a homomorphism from a group G to a group G and let g be, an element of G. Then, 1. f carries the identity of G to the identity of G., 2. f(gn) 5 (f(g))n for all n in Z., 3. If |g| is finite, then |f(g)| divides |g|., 4. Ker f is a subgroup of G., 5. f(a) 5 f(b) if and only if aKer f 5 bKer f., 6. If f(g) 5 g9, then f21(g9) 5 {x [ G | f(x) 5 g9} 5 gKer f., , PROOF The proofs of properties 1 and 2 are identical to the proofs of, properties 1 and 2 of isomorphisms in Theorem 6.2. To prove property 3,, notice that properties 1 and 2 together with gn 5 e imply that e 5, f(e) 5 f(gn) 5 (f(g))n. So, by Corollary 2 to Theorem 4.1, we have, |f(g)| divides n., By property 1 we know that Ker f is not empty. So, to prove property 4, we assume that a, b [ Ker f and show that ab21 [ Ker f., Since f(a) 5 e and f(b) 5 e, we have f(ab21) 5 f(a)f(b 21) 5, f(a)(f(b))21 5 ee21 5 e. So, ab21 [ Ker f., To prove property 5, first assume that f(a) 5 f(b). Then, e 5 (f(b)) 21f(a) 5 f(b21)f(a) 5 f(b 21a), so that b 21a[ Ker f., It now follows from property 6 of the lemma in Chapter 7 that, bKer f 5 aKer f. Reversing this argument completes the proof., To prove property 6, we must show that f21(g9) # gKer f and that, gKer f # f21(g9). For the first inclusion, let x [ f21(g9), so that, f(x) 5 g9. Then f(g) 5 f(x) and by property 5 we have gKer f 5, xKer f and therefore x [ gKer f. This completes the proof that, f21(g9) # gKer f. To prove that gKer f # f21(g9), suppose that k [, Ker f. Then f(gk) 5 f(g)f(k) 5 g9e 5 g9. Thus, by definition, gk [, f21(g9)., Since homomorphisms preserve the group operation, it should not be, a surprise that they preserve many group properties., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 211, , Theorem 10.2 Properties of Subgroups Under Homomorphisms, Let f be a homomorphism from a group G to a group G and let H be, a subgroup of G. Then, 1. f(H) 5 {f(h) | h [ H} is a subgroup of G., 2. If H is cyclic, then f(H) is cyclic., 3. If H is Abelian, then f(H) is Abelian., 4. If H is normal in G, then f(H) is normal in f(G)., 5. If |Ker f| 5 n, then f is an n-to-1 mapping from G onto f(G)., 6. If |H| 5 n, then |f(H)| divides n., 7. If K is a subgroup of G, then f21(K) 5 {k [ G | f(k) [ K}, is a subgroup of G., 8. If K is a normal subgroup of G, then f21(K) 5 {k [ G |, f(k) [ K} is a normal subgroup of G., 9. If f is onto and Ker f 5 {e}, then f is an isomorphism, from G to G., , PROOF First note that the proofs of properties 1, 2, and 3 are identical to the proofs of properties 4, 3, and 2, respectively, of Theorem, 6.3, since those proofs use only the fact that an isomorphism is an, operation-preserving mapping., To prove property 4, let f(h) [ f(H) and f(g) [ f(G). Then, f(g)f(h)f(g)21 5 f(ghg21) [ f(H), since H is normal in G., Property 5 follows directly from property 6 of Theorem 10.1 and the, fact that all cosets of Ker f 5 f21(e) have the same number of elements., To prove property 6, let fH denote the restriction of f to the, elements of H. Then fH is a homomorphism from H onto f(H)., Suppose |Ker fH| 5 t. Then, by property 5, fH is a t-to-1 mapping. So,, |f(H)|t 5 |H|., To prove property 7, we use the One-Step Subgroup Test. Clearly,, e [ f21(K), so that f21(K) is not empty. Let k1, k2 [ f21(K). Then,, by the definition of f21(K), we know that f(k1), f(k 2) [ K. Thus,, f(k2)21 [ K as well and f(k 1k 221) 5 f(k1)f(k2)21 [ K. So, by the, definition of f21(K), we have k 1k 221 [ f21(K)., To prove property 8, we use the normality test given in Theorem 9.1., Note that every element in xf21(K)x21 has the form xkx21, where f(k) [, K. Thus, since K is normal in G, f(xkx21) 5 f(x)f(k)(f(x))21 [ K,, and, therefore, xkx21 [ f21(K)., Finally, property 9 follows directly from property 5., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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212, , Groups, , A few remarks about Theorems 10.1 and 10.2 are in order. Students, should remember the various properties of these theorems in words. For, example, properties 2 and 3 of Theorem 10.2 say that the homomorphic, image of a cyclic group is cyclic and the homomorphic image of an, Abelian group is Abelian. Property 4 of Theorem 10.2 says that the homomorphic image of a normal subgroup of G is normal in the image of, G. Property 5 of Theorem 10.2 says that if f is a homomorphism from, G to G, then every element of G that gets “hit” by f gets hit the same, number of times as does the identity. The set f21(g9) defined in property 6 of Theorem 10.1 is called the inverse image of g9 (or the pullback, of g9). Note that the inverse image of an element is a coset of the kernel, and that every element in that coset has the same image. Similarly, the, set f21(K) defined in property 7 of Theorem 10.2 is called the inverse, image of K (or the pullback of K)., Property 6 of Theorem 10.1 is reminiscent of something from linear, algebra and differential equations. Recall that if x is a particular solution to a system of linear equations and S is the entire solution set of the, corresponding homogeneous system of linear equations, then x 1 S is, the entire solution set of the nonhomogeneous system. In reality, this, statement is just a special case of property 6. Properties 1 and 6 of, Theorem 10.1 and property 5 of Theorem 10.2 are pictorially represented in Figure 10.1., The special case of property 8 of Theorem 10.2, where K 5 {e}, is, of such importance that we single it out., Corollary Kernels Are Normal, Let f be a group homomorphism from G to G. Then Ker f is a normal subgroup of G., , The next two examples illustrate several properties of Theorems 10.1, and 10.2., EXAMPLE 8 Consider the mapping f from C* to C* given by, f(x) 5 x4. Since (xy)4 5 x4y4, f is a homomorphism. Clearly,, Ker f 5 {x | x4 5 1} 5 {1, 21, i, 2i}. So, by property 5 of Theorem, 10.2, we know that f is a 4-to-1 mapping. Now let’s find all elements, 4, 2 ) 5 2. Then, by property 6 of, that map to, say, 2. Certainly, f(2, 4, 2 Ker f 5, Theorem 10.1, the set of all elements that map to 2 is 2, 4, 4, 4, 4, {22 , 222 ,22 i, 222 i}., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 213, , G, Ker φ = φ 21(e), , gKer φ = φ 21(g9), , e = g1, g2,..., gn, , g, gg2,..., ggn, , φ, , φ, G, , φ (G), e, , φ (g) = g9, , Figure 10.1, , Finally, we verify a specific instance of property 3 of Theorem 10.1, and of properties 2 and 6 of Theorem 10.2. Let H 5 kcos 30° 1 i sin, 30°l. It follows from DeMoivre’s Theorem (Example 10 in Chapter 0), that |H| 5 12, f(H) 5 kcos 120° 1 i sin 120°l, and |f(H)| 5 3., EXAMPLE 9 Define f: Z12 → Z12 by f(x) 5 3x. To verify that f is a, homomorphism, we observe that in Z12, 3(a 1 b) 5 3a 1 3b (since the, group operation is addition modulo 12). Direct calculations show that, Ker f 5 {0, 4, 8}. Thus, we know from property 5 of Theorem 10.2 that, f is a 3-to-1 mapping. Since f(2) 5 6, we have by property 6 of, Theorem 10.1 that f21(6) 5 2 1 Ker f 5 {2, 6, 10}. Notice also that k2l, is cyclic and f(k2l) 5 {0, 6} is cyclic. Moreover, |2| 5 6 and |f(2)| 5, |6| 5 2, so |f(2)| divides |2| in agreement with property 3 of Theorem, 10.1. Letting K5 {0, 6}, we see that the subgroup f21(K) 5 {0, 2, 4, 6,, 8, 10}. This verifies property 7 of Theorem 10.2 in this particular case., The next example illustrates how one can easily determine all homomorphisms from a cyclic group to a cyclic group., EXAMPLE 10 We determine all homomorphisms from Z12 to Z30., By property 2 of Theorem 10.1, such a homomorphism is completely, specified by the image of 1. That is, if 1 maps to a, then x maps to xa., Lagrange’s Theorem and property 3 of Theorem 10.1 require that |a| divide both 12 and 30. So, |a| 5 1, 2, 3, or 6. Thus, a 5 0, 15, 10, 20,, 5, or 25. This gives us a list of candidates for the homomorphisms. That, each of these six possibilities yields an operation-preserving, welldefined function can now be verified by direct calculations. [Note that, gcd(12, 30) 5 6. This is not a coincidence!], , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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214, , Groups, , ), 3, , 3, , ), , (1, , ), 2, ), (1, , (1, , ), 2, 3, , ), , (1, , 3, (2, ), 2, , ), , (1, ), , 1, , 1, , (2, , 3, , ), , (1, , (2, , 3, , 3, , ), , (1, , (1, , 2, , ), , 3, , 3, , ), , ), , (1, , (1, , 2, , (1, 3, 2, ), , (1, , 2, , ), , (2, , (1, ), 3, 2, , ), 3, 2, , ), 3, (1, ), 3, , ), 2, 3, (1, , 3, ), (1, , ), (1, , ), 3, 2, (1, , ), (1, , ), 3, 2, (1, , (1, 2, 3, ), , (1, , (2, , (1, , 3, , 3, , ), , 2, , ), , (1, ), (1, , ), 2, 3, (1, , ), (1, , 2, , ), 2, (1, , ), 3, , ), 2, 3, (1, , ), 3, 2, (1, , ), (1, ), (1, , 3, , ), , (2, , 2, (1, , ), 2, 3, (1, , ), 3, 2, (1, , (1, , ), , (1, , 2, , (1, , ), , 3, , (1, , ), , 2, , ), , (1, , 3, , ), , (2, , 3, , ), , (2, , 3, , ), , EXAMPLE 11 The mapping from Sn to Z 2 that takes an even permutation to 0 and an odd permutation to 1 is a homomorphism. Figure 10.2, illustrates the telescoping nature of the mapping., , O, O, , O, , 1, , 1, , O, , Figure 10.2 Homomorphism from S3 to Z2., , The First Isomorphism Theorem, In Chapter 9, we showed that for a group G and a normal subgroup H,, we could arrange the Cayley table of G into boxes that represented the, cosets of H in G, and that these boxes then became a Cayley table for, G/H. The next theorem shows that for any homomorphism f of G and, the normal subgroup Ker f, the same process produces a Cayley table, isomorphic to the homomorphic image of G. Thus, homomorphisms,, like factor groups, cause a systematic collapse of a group to a simpler, but closely related group. This can be likened to viewing a group, through the reverse end of a telescope—the general features of the, group are present, but the apparent size is diminished. The important, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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215, , 10 | Group Homomorphisms, , relationship between homomorphisms and factor groups given below is, often called the Fundamental Theorem of Group Homomorphisms., Theorem 10.3 First Isomorphism Theorem (Jordan, 1870), Let f be a group homomorphism from G to G. Then the mapping, from G/Ker f to f(G), given by gKer f → f(g), is an isomorphism., In symbols, G/Ker f L f(G)., , PROOF Let us use c to denote the correspondence gKer f S f(g)., That c is well-defined (that is, the correspondence is independent of, the particular coset representative chosen) and one-to-one follows, directly from property 5 of Theorem 10.1. To show that c is operationpreserving, observe that c(xKer f yKer f) 5 c(xyKer f) 5 f(xy) 5, f(x) f(y) 5 c(xKer f)c(yKer f)., The next corollary follows directly from Theorem 10.3, property 1 of, Theorem 10.2, and Lagrange’s Theorem., Corollary, If f is a homomorphism from a finite group G to G, then |f(G)|, divides |G| and |G|., , EXAMPLE 12 To illustrate Theorem 10.3 and its proof, consider the, homomorphism f from D4 to itself given by the following., R180, , R0, R0, , R90, , R270, , H, , H, , V, , D, , R180, , D9, V, , Then Ker f 5 {R0, R180}, and the mapping c in Theorem 10.3 is, R0Ker f S R0, R90Ker f S H, HKer f S R180, DKer f S V. It is, straightforward to verify that the mapping c is an isomorphism., Mathematicians often give a pictorial representation of Theorem, 10.3, as follows:, φ, , G, γ, , φ (G), ψ, , G/Ker φ, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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216, , Groups, , where g: G S G/Ker f is defined as g(g) 5 gKer f. The mapping g, is called the natural mapping from G to G/Ker f. Our proof of, Theorem 10.3 shows that cg 5 f. In this case, one says that the preceding diagram is commutative., As a consequence of Theorem 10.3, we see that all homomorphic images of G can be determined using G. We may simply consider the various, factor groups of G. For example, we know that the homomorphic image of, an Abelian group is Abelian because the factor group of an Abelian group, is Abelian. We know that the number of homomorphic images of a cyclic, group G of order n is the number of divisors of n, since there is exactly one, subgroup of G (and therefore one factor group of G) for each divisor of n., (Be careful: The number of homomorphisms of a cyclic group of order n, need not be the same as the number of divisors of n, since different homomorphisms can have the same image.), An appreciation for Theorem 10.3 can be gained by looking at a few, examples., EXAMPLE 13 Z/8n9 艐 Zn, Consider the mapping from Z to Zn defined in Example 5. Clearly, its, kernel is knl. So, by Theorem 10.3, Z/knl L Zn., EXAMPLE 14 Wrapping Function, Recall the wrapping function W from trigonometry. The real number, line is wrapped around a unit circle in the plane centered at (0, 0) with, the number 0 on the number line at the point (1, 0), the positive reals, in the counterclockwise direction and the negative reals in the, clockwise direction (see Figure 10.3). The function W assigns to each, real number a the point a radians from (1, 0) on the circle. This mapping is a homomorphism from the group R under addition onto the, circle group (the group of complex numbers of magnitude 1 under, multiplication). Indeed, it follows from elementary facts of trigonometry that W(x) 5 cos x 1 i sin x and W(x 1 y) 5 W(x)W(y). Since W is, periodic of period 2p, Ker W 5 k2pl. So, from the First Isomorphism, Theorem, we see that R/k2pl is isomorphic to the circle group., W(2), , W(3), , W(1), , (1, 0) W(0), (0, 0), , W(21), , Figure 10.3, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 217, , Our next example is a theorem that is used repeatedly in Chapters 24, and 25., EXAMPLE 15 N/C Theorem, Let H be a subgroup of a group G. Recall that the normalizer of H in, G is N(H) 5 {x [ G | xHx21 5 H} and the centralizer of H in G is, C(H) 5 {x [ G | xhx21 5 h for all h in H}. Consider the mapping from, N(H) to Aut(H) given by g S fg, where fg is the inner automorphism of, H induced by g [that is, fg(h) 5 ghg21 for all h in H]. This mapping is a, homomorphism with kernel C(H). So, by Theorem 10.3, N(H)/C(H) is, isomorphic to a subgroup of Aut(H)., As an application of the N/C Theorem, we will show that every group, of order 35 is cyclic., EXAMPLE 16 Let G be a group of order 35. By Lagrange’s, Theorem, every nonidentity element of G has order 5, 7, or 35. If, some element has order 35, G is cyclic. So we may assume that all, nonidentity elements have order 5 or 7. However, not all such, elements can have order 5, since elements of order 5 come 4 at a time, (if |x| 5 5, then |x2| 5 |x3| 5 |x4| 5 5) and 4 does not divide 34., Similarly, since 6 does not divide 34, not all nonidentity elements can, have order 7. So, G has elements of order 7 and order 5. Since G has, an element of order 7, it has a subgroup of order 7. Let us call it H. In, fact, H is the only subgroup of G of order 7, for if K is another subgroup of G of order 7, we have by Theorem 7.2 that |HK| 5 |H||K|/|, H > K| 5 7 ? 7/1 5 49. But, of course, this is impossible in a group of, order 35. Since for every a in G, aHa21 is also a subgroup of G of, order 7 (see Exercise 1 of the Supplementary Exercises for Chapters, 1–4), we must have aHa21 5 H. So, N(H) 5 G. Since H has prime, order, it is cyclic and therefore Abelian. In particular, C(H) contains, H. So, 7 divides |C(H)| and |C(H)| divides 35. It follows, then, that, C(H) 5 G or C(H) 5 H. If C(H) 5 G, then we may obtain an element, x of order 35 by letting x 5 hk, where h is a nonidentity element of H, and k has order 5. On the other hand, if C(H) 5 H, then |C(H)| 5 7 and, |N(H)/C(H)| 5 35/7 5 5. However, 5 does not divide |Aut(H)| 5, |Aut(Z 7)| 5 6. This contradiction shows that G is cyclic., The corollary of Theorem 10.2 says that the kernel of every homomorphism of a group is a normal subgroup of the group. We conclude, this chapter by verifying that the converse of this statement is also true., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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218, , Groups, , Theorem 10.4 Normal Subgroups Are Kernels, Every normal subgroup of a group G is the kernel of a homomorphism of G. In particular, a normal subgroup N is the kernel, of the mapping g S gN from G to G/N., , PROOF Define g: G S G/N by g(g) 5 gN. (This mapping is called the, natural homomorphism from G to G/N.) Then, g(xy) 5 (xy)N 5 xNyN 5, g(x)g(y). Moreover, g [ Ker g if and only if gN 5 g(g) 5 N, which is, true if and only if g [ N (see property 2 of the lemma in Chapter 7)., Examples 13, 14, and 15 illustrate the utility of the First Isomorphism, Theorem. But what about homomorphisms in general? Why would one, care to study a homomorphism of a group? The answer is that, just as, was the case with factor groups of a group, homomorphic images of a, group tell us some of the properties of the original group. One measure, of the likeness of a group and its homomorphic image is the size of the, kernel. If the kernel of the homomorphism of group G is the identity,, then the image of G tells us everything (group theoretically) about G (the, two being isomorphic). On the other hand, if the kernel of the homomorphism is G itself, then the image tells us nothing about G. Between these, two extremes, some information about G is preserved and some is lost., The utility of a particular homomorphism lies in its ability to preserve, the group properties we want, while losing some inessential ones. In this, way, we have replaced G by a group less complicated (and therefore easier to study) than G; but, in the process, we have saved enough information to answer questions that we have about G itself. For example, if G is, a group of order 60 and G has a homomorphic image of order 12 that is, cyclic, then we know from properties 5, 7, and 8 of Theorem 10.2 that G, has normal subgroups of orders 5, 10, 15, 20, 30, and 60. To illustrate, further, suppose we are asked to find an infinite group that is the union, of three proper subgroups. Instead of attempting to do this directly, we, first make the problem easier by finding a finite group that is the union, of three proper subgroups. Observing that Z 2 % Z 2 is the union of H1 5, k1, 0l, H2 5 k0, 1l, and H3 5 k1, 1l, we have found our finite group. Now, all we need do is think of an infinite group that has Z 2 % Z 2 as a homomorphic image and pull back H1, H2, and H3, and our original problem is, solved. Clearly, the mapping from Z 2 % Z 2 % Z onto Z 2 % Z 2 given by, f(a, b, c) 5 (a, b) is such a mapping, and therefore Z 2 % Z 2 % Z is the, union of f21(H1) 5 {(a, 0, c,) | a [ Z 2, c [ Z}, f21(H2) 5 {(0, b, c) | b, [ Z2, c [ Z}, and f21(H3) 5 {(a, a, c) | a [ Z2, c [ Z}., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 219, , Although an isomorphism is a special case of a homomorphism, the, two concepts have entirely different roles. Whereas isomorphisms, allow us to look at a group in an alternative way, homomorphisms act as, investigative tools. The following analogy between homomorphisms, and photography may be instructive.† A photograph of a person cannot, tell us the person’s exact height, weight, or age. Nevertheless, we may, be able to decide from a photograph whether the person is tall or short,, heavy or thin, old or young, male or female. In the same way, a homomorphic image of a group gives us some information about the group., In certain branches of group theory, and especially in physics and, chemistry, one often wants to know all homomorphic images of a group, that are matrix groups over the complex numbers (these are called group, representations). Here, we may carry our analogy with photography one, step further by saying that this is like wanting photographs of a person, from many different angles (front view, profile, head-to-toe view, closeup, etc.), as well as x-rays! Just as this composite information from the, photographs reveals much about the person, several homomorphic images of a group reveal much about the group., , Exercises, The greater the difficulty, the more glory in surmounting it. Skillful pilots, gain their reputation from storms and tempests., epicurus, , 1., 2., 3., 4., 5., , Prove that the mapping given in Example 2 is a homomorphism., Prove that the mapping given in Example 3 is a homomorphism., Prove that the mapping given in Example 4 is a homomorphism., Prove that the mapping given in Example 11 is a homomorphism., Let R* be the group of nonzero real numbers under multiplication,, and let r be a positive integer. Show that the mapping that takes x to, xr is a homomorphism from R* to R* and determine the kernel., Which values of r yield an isomorphism?, 6. Let G be the group of all polynomials with real coefficients under addition. For each f in G, let ∫f denote the antiderivative of f that passes, through the point (0, 0). Show that the mapping f S ∫f from G to G is, a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if ∫f denotes the antiderivative of f that passes, through (0, 1)?, , †“All, , perception of truth is the detection of an analogy.” Henry David Thoreau, Journal., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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220, , Groups, , 7. If f is a homomorphism from G to H and s is a homomorphism, from H to K, show that sf is a homomorphism from G to K. How, are Ker f and Ker sf related? If f and s are onto and G is finite,, describe [Ker sf:Ker f] in terms of |H| and |K|., 8. Let G be a group of permutations. For each s in G, define, 1 if s is an even permutation,, sgn1s2 e, 1 if s is an odd permutation., Prove that sgn is a homomorphism from G to the multiplicative, group {11, 21}. What is the kernel? Why does this homomorphism allow you to conclude that An is a normal subgroup of Sn of, index 2? Why does this prove Exercise 23 of Chapter 5?, 9. Prove that the mapping from G % H to G given by (g, h) S g is a, homomorphism. What is the kernel? This mapping is called the, projection of G % H onto G., 10. Let G be a subgroup of some dihedral group. For each x in G, define, 1 if x is a rotation,, f1x2 e, 1 if x is a reflection., , 11., 12., 13., 14., 15., 16., 17., 18., 19., 20., 21., , Prove that f is a homomorphism from G to the multiplicative, group {1, 1}. What is the kernel? Why does this prove Exercise, 25 of Chapter 3?, Prove that (Z % Z)/(k(a, 0)l 3 k(0, b)l) is isomorphic to Za % Zb., Suppose that k is a divisor of n. Prove that Zn/kkl L Zk., Prove that (A % B)/(A % {e}) L B., Explain why the correspondence x → 3x from Z12 to Z10 is not a, homomorphism., Suppose that f is a homomorphism from Z30 to Z30 and Ker f 5, {0, 10, 20}. If f(23) 5 9, determine all elements that map to 9., Prove that there is no homomorphism from Z8 % Z2 onto Z4 % Z4., Prove that there is no homomorphism from Z16 % Z2 onto Z4 % Z4., Can there be a homomorphism from Z4 % Z4 onto Z8? Can there be, a homomorphism from Z16 onto Z2 % Z2? Explain your answers., Suppose that there is a homomorphism f from Z17 to some group, and that f is not one-to-one. Determine f., How many homomorphisms are there from Z20 onto Z8? How many, are there to Z8?, If f is a homomorphism from Z30 onto a group of order 5, determine the kernel of f., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 221, , 22. Suppose that f is a homomorphism from a finite group G onto G, and that G has an element of order 8. Prove that G has an element, of order 8. Generalize., 23. Suppose that f is a homomorphism from Z36 to a group of order 24., a. Determine the possible homomorphic images., b. For each image in part a, determine the corresponding kernel of f., 24. Suppose that f: Z50 S Z15 is a group homomorphism with f(7) 5 6., a. Determine f(x)., b. Determine the image of f., c. Determine the kernel of f., d. Determine f21(3). That is, determine the set of all elements, that map to 3., 25. How many homomorphisms are there from Z20 onto Z10? How, many are there to Z10?, 26. Determine all homomorphisms from Z4 to Z2 % Z2., 27. Determine all homomorphisms from Zn to itself., 28. Suppose that f is a homomorphism from S4 onto Z2. Determine, Ker f. Determine all homomorphisms from S4 to Z2., 29. Suppose that there is a homomorphism from a finite group G onto, Z10. Prove that G has normal subgroups of indexes 2 and 5., 30. Suppose that f is a homomorphism from a group G onto Z6 % Z2, and that the kernel of f has order 5. Explain why G must have normal subgroups of orders 5, 10, 15, 20, 30, and 60., 31. Suppose that f is a homomorphism from U(30) to U(30) and, that Ker f 5 {1, 11}. If f(7) 5 7, find all elements of U(30) that, map to 7., 32. Find a homomorphism f from U(30) to U(30) with kernel {1, 11}, and f(7) 5 7., 33. Suppose that f is a homomorphism from U(40) to U(40) and that, Ker f 5 {1, 9, 17, 33}. If f(11) 5 11, find all elements of U(40), that map to 11., 34. Find a homomorphism f from U(40) to U(40) with kernel {1, 9,, 17, 33} and f(11) 5 11., 35. Prove that the mapping f: Z % Z S Z given by (a, b) S a 2 b is a, homomorphism. What is the kernel of f? Describe the set f21(3), (that is, all elements that map to 3)., 36. Suppose that there is a homomorphism f from Z % Z to a group G, such that f((3, 2)) 5 a and f((2, 1)) 5 b. Determine f((4, 4)) in, terms of a and b. Assume that the operation of G is addition., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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222, , Groups, , 37. Let H 5 {z [ C* | |z| 5 1}. Prove that C*/H is isomorphic to R1,, the group of positive real numbers under multiplication., 38. Let a be a homomorphism from G1 to H1 and b be a homomorphism from G2 to H2. Determine the kernel of the homomorphism, g from G1 % G2 to H1 % H2 defined by g(g1, g2) 5 (a(g1), b(g2))., 39. Prove that the mapping x S x6 from C* to C* is a homomorphism., What is the kernel?, 40. For each pair of positive integers m and n, we can define a homomorphism from Z to Zm % Zn by x S (x mod m, x mod n). What is, the kernel when (m, n) 5 (3, 4)? What is the kernel when (m, n) 5, (6, 4)? Generalize., 41. (Second Isomorphism Theorem) If K is a subgroup of G and N is, a normal subgroup of G, prove that K/(K > N) is isomorphic, to KN/N., 42. (Third Isomorphism Theorem) If M and N are normal subgroups of, G and N # M, prove that (G/N)/(M/N) L G/M., 43. Let f(d) denote the Euler phi function of d (see page 85). Show, that the number of homomorphisms from Zn to Zk is Sf(d), where, the sum runs over all common divisors d of n and k. [It follows, from number theory that this sum is actually gcd(n, k).], 44. Let k be a divisor of n. Consider the homomorphism from U(n) to, U(k) given by x S x mod k. What is the relationship between this, homomorphism and the subgroup Uk(n) of U(n)?, 45. Determine all homomorphic images of D4 (up to isomorphism)., 46. Let N be a normal subgroup of a finite group G. Use the theorems, of this chapter to prove that the order of the group element gN in, G/N divides the order of g., 47. Suppose that G is a finite group and that Z10 is a homomorphic, image of G. What can we say about |G|? Generalize., 48. Suppose that Z10 and Z15 are both homomorphic images of a finite, group G. What can be said about |G|? Generalize., 49. Suppose that for each prime p, Zp is the homomorphic image of a, group G. What can we say about |G|? Give an example of such a, group., 50. (For students who have had linear algebra.) Suppose that x is a, particular solution to a system of linear equations and that S is the, entire solution set of the corresponding homogeneous system of, linear equations. Explain why property 6 of Theorem 10.1 guarantees that x 1 S is the entire solution set of the nonhomogeneous, system. In particular, describe the relevant groups and the homomorphism between them., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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10 | Group Homomorphisms, , 223, , 51. Let N be a normal subgroup of a group G. Use property 7 of, Theorem 10.2 to prove that every subgroup of G/N has the form, H/N, where H is a subgroup of G. (This exercise is referred to in, Chapter 24.), 52. Show that a homomorphism defined on a cyclic group is completely determined by its action on a generator of the group., 53. Use the First Isomorphism Theorem to prove Theorem 9.4., 54. Let a and b be group homomorphisms from G to G and let H 5, {g [ G | a(g) 5 b(g)}. Prove or disprove that H is a subgroup of G., 55. Let Z[x] be the group of polynomials in x with integer coefficients, under addition. Prove that the mapping from Z[x] into Z given by, f(x) S f(3) is a homomorphism. Give a geometric description of, the kernel of this homomorphism. Generalize., 56. Prove that the mapping from R under addition to GL(2, R) that, takes x to, cos x, sin x, c, d, sin x, cos x, 57., 58., 59., , 60., , 61., 62., 63., 64., , 65., , is a group homomorphism. What is the kernel of the homomorphism?, Suppose there is a homomorphism f from G onto Z2 % Z2. Prove, that G is the union of three proper normal subgroups., If H and K are normal subgroups of G and H > K 5 {e}, prove that, G is isomorphic to a subgroup of G/H % G/K., Suppose that H and K are distinct subgroups of G of index 2. Prove, that H > K is a normal subgroup of G of index 4 and that G/(H > K), is not cyclic., Suppose that the number of homomorphisms from G to H is n., How many homomorphisms are there from G to H % H % ? ? ? % H, (s terms)? When H is Abelian, how many homomorphisms are there, from G % G % ? ? ? % G (s terms) to H?, Prove that every group of order 77 is cyclic., Determine all homomorphisms from Z onto S3. Determine all, homomorphisms from Z to S3., Let G be an Abelian group. Determine all homomorphisms from, S3 to G., If f is an isomorphism from a group G under addition to a group G, under addition, prove that for any integer n, the mapping from G to, G defined by g(x) 5 nf(x) is a homomorphism from G to G., Prove that the mapping from C* to C* given by f(z) 5 z2 is a homomorphism and that C*/ {1, 21} is isomorphic to C*., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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224, , Groups, , 66. Let p be a prime. Determine the number of homomorphisms from, Zp % Zp into Zp., 67. Suppose G is an Abelian group under addition with the property, that for every positive integer n, the set nG 5{ng | g [ G} 5 G., Show that every proper subgroup of G is properly contained in a, proper subgroup of G. Name two familiar groups that satisfy the, hypothesis., , Computer Exercise, A computer exercise for this chapter is available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Camille Jordan, , j. dieudonné, Dictionary of, Scientific Biography, , Camille Jordan was born into a well-to-do, family on January 5, 1838, in Lyons, France., Like his father, he graduated from the École, Polytechnique and became an engineer., Nearly all of his 120 research papers in, mathematics were written before his retirement from engineering in 1885. From 1873, until 1912, Jordan taught simultaneously at, the École Polytechnique and at the College, of France., In the great French tradition, Jordan was, a universal mathematician who published in, nearly every branch of mathematics. Among, the concepts named after him are the Jordan, canonical form in matrix theory, the Jordan, curve theorem from topology, and the, Jordan–Hölder Theorem from group theory., , The Granger Collection, New York, , Although these contributions [to, analysis and topology] would have been, enough to rank Jordan very high among, his mathematical contemporaries, it is, chiefly as an algebraist that he reached, celebrity when he was barely thirty; and, during the next forty years he was, universally regarded as the undisputed, master of group theory., , His classic book Traité des substitutions,, published in 1870, was the first to be devoted solely to group theory and its applications to other branches of mathematics., Another book that had great influence, and set a new standard for rigor was his, Cours d’analyse. This book gave the first, clear definitions of the notions of volume, and multiple integral. Nearly 100 years, after this book appeared, the distinguished, mathematician and mathematical historian, B. L. van der Waerden wrote, “For me, every, single chapter of the Cours d’analyse is a, pleasure to read.” Jordan died in Paris on, January 22, 1922., To find more information about Jordan,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 225, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Fundamental, 11 Theorem of Finite, Abelian Groups, By a small sample we may judge of the whole piece., miguel de cervantes, Don Quixote, , The Fundamental Theorem, In this chapter, we present a theorem that describes to an algebraist’s, eye (that is, up to isomorphism) all finite Abelian groups in a standardized way. Before giving the proof, which is long and difficult, we, discuss some consequences of the theorem and its proof. The first proof, of the theorem was given by Leopold Kronecker in 1858., Theorem 11.1 Fundamental Theorem of Finite Abelian Groups, Every finite Abelian group is a direct product of cyclic groups of, prime-power order. Moreover, the number of terms in the product, and the orders of the cyclic groups are uniquely determined by the, group., , Since a cyclic group of order n is isomorphic to Zn, Theorem 11.1, shows that every finite Abelian group G is isomorphic to a group of, the form, Zp1n1 % Zp2n2 % ? ? ? % Z pknk,, where the p i’s are not necessarily distinct primes and the prime, powers p1n1, p2n2, . . . , pknk are uniquely determined by G. Writing a, group in this form is called determining the isomorphism class of G., , The Isomorphism Classes, of Abelian Groups, The Fundamental Theorem is extremely powerful. As an application,, we can use it as an algorithm for constructing all Abelian groups of any, order. Let’s look at groups whose orders have the form pk, where p is, 226, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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11 | Fundamental Theorem of Finite Abelian Groups, , 227, , prime and k # 4. In general, there is one group of order pk for each set, of positive integers whose sum is k (such a set is called a partition of k);, that is, if k can be written as, k 5 n 1 1 n2 1 ? ? ? 1 nt ,, where each ni is a positive integer, then, Zp n1 % Zp n2 % ? ? ? % Zp nt, is an Abelian group of order pk., , Partitions of k, 1, 2, , Possible direct, products for G, Zp, Z p2, , p3, , 111, 3, , Z p % Zp, Z p3, , p4, , 211, 11111, 4, , Zp2 % Zp, Zp % Zp % Zp, Z p4, , 311, 212, , Zp3 % Zp, Zp2 % Zp2, , 21111, 1111111, , Zp2 % Zp % Zp, Zp % Zp % Zp % Zp, , Order of G, p, p2, , Furthermore, the uniqueness portion of the Fundamental Theorem, guarantees that distinct partitions of k yield distinct isomorphism, classes. Thus, for example, Z9 % Z3 is not isomorphic to Z3 % Z3 % Z3., A reliable mnemonic for comparing external direct products is the cancellation property: If A is finite, then, A%BLA%C, , if and only if, , BLC, , (see [1])., , Thus, Z4 % Z4 is not isomorphic to Z4 % Z2 % Z2, because Z4 is not, isomorphic to Z2 % Z2., To appreciate fully the potency of the Fundamental Theorem, contrast, the ease with which the Abelian groups of order pk, k # 4, were, determined with the corresponding problem for non-Abelian groups., Even a description of the two non-Abelian groups of order 8 is a challenge (see Theorem 26.4), and a description of the nine non-Abelian, groups of order 16 is well beyond the scope of this text., Now that we know how to construct all the Abelian groups of primepower order, we move to the problem of constructing all Abelian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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228, , Groups, , groups of a certain order n, where n has two or more distinct prime, divisors. We begin by writing n in prime-power decomposition form, n 5 p1n1p2n2 ? ? ? pk n k. Next, we individually form all Abelian groups of, order p1n 1, then p2n 2, and so on, as described earlier. Finally, we form all, possible external direct products of these groups. For example, let n 5, 1176 5 23 ? 3 ? 72. Then, the complete list of the distinct isomorphism, classes of Abelian groups of order 1176 is, Z8 % Z3 % Z49,, Z4 % Z2 % Z3 % Z49,, Z2 % Z2 % Z2 % Z3 % Z49,, Z8 % Z3 % Z7 % Z7,, Z4 % Z2 % Z3 % Z7 % Z7,, Z2 % Z2 % Z2 % Z3 % Z7 % Z7., If we are given any particular Abelian group G of order 1176, the, question we want to answer about G is: Which of the preceding six isomorphism classes represents the structure of G? We can answer this, question by comparing the orders of the elements of G with the orders of, the elements in the six direct products, since it can be shown that two finite Abelian groups are isomorphic if and only if they have the same, number of elements of each order. For instance, we could determine, whether G has any elements of order 8. If so, then G must be isomorphic, to the first or fourth group above, since these are the only ones with elements of order 8. To narrow G down to a single choice, we now need, only check whether or not G has an element of order 49, since the first, product above has such an element, whereas the fourth one does not., What if we have some specific Abelian group G of order p1n 1p2 n 2, ? ? ? pk n k, where the pi’s are distinct primes? How can G be expressed as, an internal direct product of cyclic groups of prime-power order? For, simplicity, let us say that the group has 2n elements. First, we must, compute the orders of the elements. After this is done, pick an element, of maximum order 2r, call it a1. Then ka1l is one of the factors in the, desired internal direct product. If G 2 ka1l, choose an element a2 of, maximum order 2s such that s # n 2 r and none of a2, a2 2, a2 4, . . . ,, s21, a22 is in ka1l. Then ka2l is a second direct factor. If n 2 r 1 s, select, an element a3 of maximum order 2t such that t # n 2 r 2 s and none of, t21, a3, a32, a34, . . . , a32 is in ka1l 3 ka2l 5 {a1i a 2 j | 0 # i , 2r, 0 #, j , 2s}. Then ka3l is another direct factor. We continue in this fashion, until our direct product has the same order as G., A formal presentation of this algorithm for any Abelian group G of, prime-power order pn is as follows., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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229, , 11 | Fundamental Theorem of Finite Abelian Groups, , Greedy Algorithm for an Abelian Group of Order pn, 1. Compute the orders of the elements of the group G., 2. Select an element a1 of maximum order and define G1 5 ka1l., Set i 5 1., 3. If |G| 5 |Gi|, stop. Otherwise, replace i by i 1 1., 4. Select an element ai of maximum order p k such that p k #, 2, k21, |G|/|Gi21| and none of ai , ai p, ai p , . . . , ai p is in Gi21, and define, Gi 5 Gi21 3 kail., 5. Return to step 3., In the general case where |G| 5 p1 n 1 p2 n 2 ? ? ? pk n k, we simply use the, algorithm to build up a direct product of order p1n 1, then another of, order p2 n 2, and so on. The direct product of all of these pieces is the, desired factorization of G. The following example is small enough that, we can compute the appropriate internal and external direct products, by hand., EXAMPLE 1 Let G 5 {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51,, 53, 57, 64} under multiplication modulo 65. Since G has order 16, we, know it is isomorphic to one of, Z16,, Z8 % Z2,, Z4 % Z4,, Z4 % Z2 % Z2,, Z2 % Z2 % Z2 % Z2., To decide which one, we dirty our hands to calculate the orders of the, elements of G., Element 1, , 8, , 12, , 14, , 18, , 21, , 27, , 31, , 34 38, , 44, , 47 51 53, , 57 64, , Order, , 4, , 4, , 2, , 4, , 4, , 4, , 4, , 4, , 4, , 4, , 4, , 1, , 4, , 2, , 4, , 2, , From the table of orders, we can instantly rule out all but Z4 % Z4 and, Z4 % Z2 % Z2 as possibilities. Finally, we observe that since this latter, group has a subgroup isomorphic to Z2 % Z2 % Z2, it has more than, three elements of order 2, and therefore we must have G L Z4 % Z4., Expressing G as an internal direct product is even easier. Pick an element of maximum order, say the element 8. Then k8l is a factor in the, product. Next, choose a second element, say a, so that a has order 4 and, a and a2 are not in k8l 5 {1, 8, 64, 57}. Since 12 has this property, we, have G 5 k8l 3 k12l., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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230, , Groups, , Example 1 illustrates how quickly and easily one can write an Abelian, group as a direct product given the orders of the elements of the group., But calculating all those orders is certainly not an appealing prospect!, The good news is that, in practice, a combination of theory and calculation of the orders of a few elements will usually suffice., EXAMPLE 2 Let G 5 {1, 8, 17, 19, 26, 28, 37, 44, 46, 53, 62,, 64, 71, 73, 82, 89, 91, 98, 107, 109, 116, 118, 127, 134} under multiplication modulo 135. Since G has order 24, it is isomorphic to, one of, Z8 % Z3 L Z24,, Z4 % Z2 % Z3 L Z12 % Z2,, Z2 % Z2 % Z2 % Z3 L Z6 % Z2 % Z2., Consider the element 8. Direct calculations show that 86 5 109 and 812 5 1., (Be sure to mod as you go. For example, 83 mod 135 5 512 mod, 135 5 107, so compute 84 as 8 ? 107 rather than 8 ? 512.) But now we, know G. Why? Clearly, |8| 5 12 rules out the third group in the list. At, the same time, |109| 5 2 5 |134| (remember, 134 5 21 mod 135) implies that G is not Z24 (see Theorem 4.4). Thus, G L Z12 % Z2, and G 5, k8l 3 k134l., Rather than express an Abelian group as a direct product of cyclic, groups of prime-power orders, it is often more convenient to combine, the cyclic factors of relatively prime order, as we did in Example 2, to, obtain a direct product of the form Zn1 % Zn2 % ? ? ? % Znk , where ni divides ni21. For example, Z4 % Z4 % Z2 % Z9 % Z3 % Z5 would be written, as Z180 % Z12 % Z2 (see Exercise 11). The algorithm above is easily, adapted to accomplish this by replacing step 4 by 49: Select an element, ai of maximum order m such that m # |G|/|Gi21| and none of ai, ai 2, . . . ,, ai m21 is in Gi21, and define Gi 5 Gi21 3 kail., As a consequence of the Fundamental Theorem of Finite Abelian, Groups, we have the following corollary, which shows that the converse, of Lagrange’s Theorem is true for finite Abelian groups., Corollary Existence of Subgroups of Abelian Groups, If m divides the order of a finite Abelian group G, then G has a, subgroup of order m., , It is instructive to verify this corollary for a specific case. Let us say, that G is an Abelian group of order 72 and we wish to produce a subgroup, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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231, , 11 | Fundamental Theorem of Finite Abelian Groups, , of order 12. According to the Fundamental Theorem, G is isomorphic to, one of the following six groups:, Z8 % Z9,, Z4 % Z2 % Z9,, Z2 % Z2 % Z2 % Z9,, , Z8 % Z3 % Z3,, Z4 % Z2 % Z3 % Z3,, Z2 % Z2 % Z2 % Z3 % Z3., , Obviously, Z8 % Z9 L Z72 and Z4 % Z2 % Z3 % Z3 L Z12 % Z6 both, have a subgroup of order 12. To construct a subgroup of order 12 in Z4, % Z2 % Z9, we simply piece together all of Z4 and the subgroup of order, 3 in Z9; that is, {(a, 0, b) | a [ Z4, b [ {0, 3, 6}}. A subgroup of order, 12 in Z8 % Z3 % Z3 is given by {(a, b, 0) | a [ {0, 2, 4, 6}, b [ Z3}. An, analogous procedure applies to the remaining cases and indeed to any, finite Abelian group., , Proof of the Fundamental Theorem, Because of the length and complexity of the proof of the Fundamental, Theorem of Finite Abelian Groups, we will break it up into a series of, lemmas., Lemma 1, Let G be a finite Abelian group of order pnm, where p is a prime that, n, does not divide m. Then G 5 H 3 K, where H 5 {x [ G | x p 5 e}, and K 5 {x [ G | x m 5 e}. Moreover, |H| 5 pn., , PROOF It is an easy exercise to prove that H and K are subgroups of G, (see Exercise 45 in Chapter 3). Because G is Abelian, to prove that G 5, H 3 K we need only prove that G 5 HK and H > K 5 {e}. Since we, have gcd(m, pn) 5 1, there are integers s and t such that 1 5 sm 1 tpn., n, n, For any x in G, we have x 5 x1 5 x sm1tp 5 x smx tp and, by Corollary 4, n, of Lagrange’s Theorem (Theorem 7.1), x sm [ H and x tp [ K. Thus,, G 5 HK. Now suppose that some x [ H > K. Then x pn 5 e 5 x m and,, by Corollary 2 of Theorem 4.1, |x| divides both pn and m. Since p does, not divide m, we have | x| 5 1 and, therefore, x 5 e., To prove the second assertion of the lemma, note that pnm 5, |HK| 5 |H||K|/|H > K| 5 |H||K| (Theorem 7.2). It follows from, Theorem 9.5 and Corollary 2 to Theorem 4.1 that p does not divide |K|, and therefore |H| 5 pn., Given an Abelian group G with |G| 5 p1 n 1 p2 n 2 ? ? ? pk n k, where the, n, p’s are distinct primes, we let G(pi) denote the set {x [ G | x pi 5 e}., i, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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232, , Groups, , It then follows immediately from Lemma 1 and induction that G 5, G( p1) 3 G(p2) 3 ? ? ? 3 G(pk) and |G(pi)| 5 pi n i. Hence, we turn our, attention to groups of prime-power order., Lemma 2, Let G be an Abelian group of prime-power order and let a be an, element of maximum order in G. Then G can be written in the form, kal 3 K., , PROOF We denote |G| by pn and induct on n. If n 5 1, then G 5, kal 3 kel. Now assume that the statement is true for all Abelian, groups of order p k, where k , n. Among all the elements of G, choose, a of maximum order pm. Then x pm 5 e for all x in G. We may assume, that G 2 kal, for otherwise there is nothing to prove. Now, among all, the elements of G, choose b of smallest order such that b o kal. We, claim that kal > kbl 5 {e}. Since |b p| 5 |b|/p, we know that b p [ kal, by the manner in which b was chosen. Say b p 5 ai. Notice that e 5, m21, b pm 5 (b p) pm21 5 (ai) p , so |ai| # p m21. Thus, ai is not a generator of, kal and, therefore, by Corollary 3 to Theorem 4.2, gcd(p m, i) 2 1., This proves that p divides i, so that we can write i 5 pj. Then b p 5, ai 5 a pj. Consider the element c 5 a2jb. Certainly, c is not in kal, for, if it were, b would be, too. Also, c p 5 a2jpb p 5 a2ib p 5 b2pb p 5 e., Thus, we have found an element c of order p such that c o kal. Since, b was chosen to have smallest order such that b o kal, we conclude, that b also has order p. It now follows that kal > kbl 5 {e}, because, any nonidentity element of the intersection would generate kbl and, thus contradict b o kal., Now consider the factor group G 5 G/kbl. To simplify the notation,, m21, we let x denote the coset xkbl in G. If |a| , |a| 5 pm, then ap 5 e. This, m21, m21, m21, means that (akbl) p 5 a p kbl 5 kbl, so that ap [ kal > kbl 5 {e},, m, contradicting the fact that |a| 5 p . Thus, |a| 5 |a| 5 pm, and therefore, a is an element of maximum order in G. By induction, we know that G, can be written in the form kal 3 K for some subgroup K of G. Let K be, the pullback of K under the natural homomorphism from G to G (that, is, K 5 {x [ G | x [ K}). We claim that kal > K 5 {e}. For if x [ kal, > K, then x [ kal > K 5 {e} 5 kbl and x [ kal > kbl 5 {e}. It now, follows from an order argument (see Exercise 35) that G 5 kalK, and, therefore G 5 kal 3 K., Lemma 2 and induction on the order of the group now give the, following., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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11 | Fundamental Theorem of Finite Abelian Groups, , 233, , Lemma 3, A finite Abelian group of prime-power order is an internal direct, product of cyclic groups., , Let us pause to determine where we are in our effort to prove the, Fundamental Theorem of Finite Abelian Groups. The remark following, Lemma 1 shows that G 5 G( p1) 3 G( p2) 3 ? ? ? 3 G( pn), where each, G( pi) is a group of prime-power order, and Lemma 3 shows that each of, these factors is an internal direct product of cyclic groups. Thus, we have, proved that G is an internal direct product of cyclic groups of primepower order. All that remains to be proved is the uniqueness of the factors., Certainly the groups G(pi) are uniquely determined by G, since they, comprise the elements of G whose orders are powers of pi. So we must, prove that there is only one way (up to isomorphism and rearrangement, of factors) to write each G( pi) as an internal direct product of cyclic, groups., Lemma 4, Suppose that G is a finite Abelian group of prime-power order. If, G 5 H1 3 H2 3 ? ? ? 3 Hm and G 5 K1 3 K2 3 ? ? ? 3 Kn, where the, H’s and K’s are nontrivial cyclic subgroups with |H1| $ |H2| $ ? ? ? $, |Hm| and |K1| $ |K2| $ ? ? ? $ |Kn|, then m 5 n and |Hi| 5 |Ki|, for all i., , PROOF We proceed by induction on |G|. Clearly, the case where |G| 5, p is true. Now suppose that the statement is true for all Abelian groups, of order less than |G|. For any Abelian group L, the set L p 5 {x p | x [ L}, is a subgroup of L (see Exercise 17 in the Supplementary Exercises for, Chapters 1– 4) and, by Theorem 9.5, is a proper subgroup if p, divides |L|. It follows that G p 5 H1 p 3 H2 p 3 ? ? ? 3 Hm9 p, and G p 5, K1 p 3 K2 p 3 ? ? ? 3 Kn9 p, where m9 is the largest integer i such that, |Hi| . p, and n9 is the largest integer j such that |Kj| . p. (This ensures, that our two direct products for G p do not have trivial factors.) Since |G p|, , |G|, we have, by induction, m9 5 n9 and |Hi p| 5 |Ki p| for i 5 1, . . . ,, m9. Since |Hi| 5 p|Hi p|, this proves that |Hi| 5 |Ki| for all i 5 1, . . . , m9., All that remains to be proved is that the number of Hi of order p equals, the number of Ki of order p; that is, we must prove that m 2 m9 5 n 2 n9, (since n9 5 m9). This follows directly from the facts that |H1||H2| ? ? ?, |Hm9|pm2m9 5 |G| 5 |K1||K2| ? ? ? |Kn9|pn2n9, |Hi| 5 |Ki|, and m9 5 n9., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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234, , Groups, , Exercises, You know it ain’t easy, you know how hard it can be., john lennon and paul mccartney,, “The Ballad of John and Yoko”*, , 1. What is the smallest positive integer n such that there are two nonisomorphic groups of order n? Name the two groups., 2. What is the smallest positive integer n such that there are three, nonisomorphic Abelian groups of order n? Name the three groups., 3. What is the smallest positive integer n such that there are exactly, four nonisomorphic Abelian groups of order n? Name the four, groups., 4. Calculate the number of elements of order 2 in each of Z16, Z8 % Z2,, Z4 % Z4, and Z4 % Z2 % Z2. Do the same for the elements of order 4., 5. Prove that any Abelian group of order 45 has an element of order 15., Does every Abelian group of order 45 have an element of order 9?, 6. Show that there are two Abelian groups of order 108 that have, exactly one subgroup of order 3., 7. Show that there are two Abelian groups of order 108 that have, exactly four subgroups of order 3., 8. Show that there are two Abelian groups of order 108 that have, exactly 13 subgroups of order 3., 9. Suppose that G is an Abelian group of order 120 and that G has, exactly three elements of order 2. Determine the isomorphism class, of G., 10. Find all Abelian groups (up to isomorphism) of order 360., 11. Prove that every finite Abelian group can be expressed as the, (external) direct product of cyclic groups of orders n1, n2, . . . , nt,, where ni11 divides ni for i 5 1, 2, . . . , t 2 1. (This exercise is referred to in this chapter.), 12. Suppose that the order of some finite Abelian group is divisible by, 10. Prove that the group has a cyclic subgroup of order 10., 13. Show, by example, that if the order of a finite Abelian group is divisible by 4, the group need not have a cyclic subgroup of order 4., 14. On the basis of Exercises 12 and 13, draw a general conclusion, about the existence of cyclic subgroups of a finite Abelian group., *Copyright © 1969 (Renewed) Stony/ATV Tunes LLC. All rights administered by, Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights, reserved. Used by permission., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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11 | Fundamental Theorem of Finite Abelian Groups, , 235, , 15. How many Abelian groups (up to isomorphism) are there, a. of order 6?, b. of order 15?, c. of order 42?, d. of order pq, where p and q are distinct primes?, e. of order pqr, where p, q, and r are distinct primes?, f. Generalize parts d and e., 16. How does the number (up to isomorphism) of Abelian groups of, order n compare with the number (up to isomorphism) of Abelian, groups of order m where, a. n 5 32 and m 5 52?, b. n 5 24 and m 5 54?, c. n 5 pr and m 5 qr, where p and q are prime?, d. n 5 pr and m 5 prq, where p and q are distinct primes?, e. n 5 pr and m 5 prq2, where p and q are distinct primes?, 17. Up to isomorphism, how many additive Abelian groups of order 16, have the property that x 1 x 1 x 1 x 5 0 for all x in the group?, 18. Let p1, p2, p , pn be distinct primes. Up to isomorphism, how many, Abelian groups are there of order p41 p42 . . . pn4?, 19. The symmetry group of a nonsquare rectangle is an Abelian group, of order 4. Is it isomorphic to Z4 or Z2 % Z2?, 20. Verify the corollary to the Fundamental Theorem of Finite, Abelian Groups in the case that the group has order 1080 and the, divisor is 180., 21. The set {1, 9, 16, 22, 29, 53, 74, 79, 81} is a group under multiplication modulo 91. Determine the isomorphism class of this group., 22. Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic., 23. Characterize those integers n such that the only Abelian groups of, order n are cyclic., 24. Characterize those integers n such that any Abelian group of order, n belongs to one of exactly four isomorphism classes., 25. Refer to Example 1 in this chapter and explain why it is unnecessary to compute the orders of the last five elements listed to determine the isomorphism class of G., 26. Let G 5 {1, 7, 17, 23, 49, 55, 65, 71} under multiplication modulo, 96. Express G as an external and an internal direct product of cyclic, groups., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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236, , Groups, , 27. Let G 5 {1, 7, 43, 49, 51, 57, 93, 99, 101, 107, 143, 149, 151, 157,, 193, 199} under multiplication modulo 200. Express G as an external and an internal direct product of cyclic groups., 28. The set G 5 {1, 4, 11, 14, 16, 19, 26, 29, 31, 34, 41, 44} is a group, under multiplication modulo 45. Write G as an external and an internal direct product of cyclic groups of prime-power order., 29. Suppose that G is an Abelian group of order 9. What is the maximum number of elements (excluding the identity) of which one, needs to compute the order to determine the isomorphism class of, G? What if G has order 18? What about 16?, 30. Suppose that G is an Abelian group of order 16, and in computing, the orders of its elements, you come across an element of order 8, and two elements of order 2. Explain why no further computations, are needed to determine the isomorphism class of G., 31. Let G be an Abelian group of order 16. Suppose that there are elements a and b in G such that |a| 5 |b| 5 4 and a2 2 b2. Determine, the isomorphism class of G., 32. Prove that an Abelian group of order 2n (n $ 1) must have an odd, number of elements of order 2., 33. Without using Lagrange’s Theorem, show that an Abelian group of, odd order cannot have an element of even order., 34. Let G be the group of all n 3 n diagonal matrices with 61 diagonal entries. What is the isomorphism class of G?, 35. Prove the assertion made in the proof of Lemma 2 that G 5 kalK., 36. Suppose that G is a finite Abelian group. Prove that G has order pn,, where p is prime, if and only if the order of every element of G is a, power of p., 37. Dirichlet’s Theorem says that, for every pair of relatively prime, integers a and b, there are infinitely many primes of the form at 1 b., Use Dirichlet’s Theorem to prove that every finite Abelian group is, isomorphic to a subgroup of a U-group., 38. Determine the isomorphism class of Aut(Z2 % Z3 % Z5)., 39. Give an example to show that Lemma 2 is false if G is non-Abelian., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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11 | Fundamental Theorem of Finite Abelian Groups, , 237, , Reference, 1. R. Hirshon, “On Cancellation in Groups,” American Mathematical, Monthly 76 (1969): 1037–1039., , Suggested Readings, J. A. Gallian, “Computers in Group Theory,” Mathematics Magazine, 49 (1976): 69–73., This paper discusses several computer-related projects in group theory, done by undergraduate students., J. Kane, “Distribution of Orders of Abelian Groups,” Mathematics Magazine, 49 (1976): 132–135., In this article, the author determines the percentages of integers k between, 1 and n, for sufficiently large n, that have exactly one isomorphism class of, Abelian groups of order k, exactly two isomorphism classes of Abelian, groups of order k, and so on, up to 13 isomorphism classes., G. Mackiw, “Computing in Abstract Algebra,” The College Mathematics, Journal 27 (1996): 136–142., This article explains how one can use computer software to implement the, algorithm given in this chapter for expressing an Abelian group as an internal direct product., , Suggested Website, To find more information about the development of group theory, visit:, , http://www-groups.dcs.st-and.ac.uk/~history/, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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238, , Groups, , Supplementary Exercises for Chapters 9–11, Every prospector drills many a dry hole, pulls out his rig, and moves on., john l. hess, , True/false questions for Chapters 9–11 are available on the Web at:, http://www.d.umn.edu/~jgallian/TF, 1. Suppose that H is a subgroup of G and that each left coset of H in, G is some right coset of H in G. Prove that H is normal in G., 2. Use a factor group-induction argument to prove that a finite, Abelian group of order n has a subgroup of order m for every positive divisor m of n., 3. Let diag(G) 5 {(g, g) | g [ G}. Prove that diag(G) v G % G if, and only if G is Abelian. When G is finite, what is the index of, diag(G) in G % G?, 4. Let H be any group of rotations in Dn. Prove that H is normal in Dn., 5. Prove that Inn(G) v Aut(G)., 6. Let H be a subgroup of G. Prove that H is a normal subgroup if and, only if, for all a and b in G, ab [ H implies ba [ H., 7. The factor group GL(2, R)/SL(2, R) is isomorphic to some very, familiar group. What is the group?, 8. Let k be a divisor of n. The factor group (Z/knl)/(kkl/knl) is isomorphic to some very familiar group. What is the group?, 9. Let, 1, H • £0, 0, , a, 1, 0, , b, c t † a, b, c [ Q ¶, 1, , under matrix multiplication., a. Find Z1H2., b. Prove that Z(H) is isomorphic to Q under addition., c. Prove that H/Z(H) is isomorphic to Q % Q., d. Are your proofs for parts a and b valid when Q is replaced by, R? Are they valid when Q is replaced by Zp, where p is prime?, 10. Prove that D4/Z(D4) is isomorphic to Z2 % Z2., 11. Prove that Q/Z under addition is an infinite group in which every, element has finite order., 12. Show that the intersection of any collection of normal subgroups of, a group is a normal subgroup., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 9– 11, , 239, , 13. Let n . 1 be a fixed integer and let G be a group. If the set H 5, {x [ G | |x| 5 n} together with the identity forms a subgroup of, G, prove that it is a normal subgroup of G. In the case where such, a subgroup exists, what can be said about n? Give an example of, a non-Abelian group that has such a subgroup. Give an example, of a group G and a prime n for which the set H together with the, identity is not a subgroup., 14. Show that Q/Z has a unique subgroup of order n for each positive, integer n., 15. If H and K are normal Abelian subgroups of a group and H > K 5, {e}, prove that HK is Abelian., 16. Let G be a group of odd order. Prove that the mapping x S x2 from, G to itself is one-to-one., 17. Suppose that G is a group of permutations on some set. If |G| 5 60, and orbG(5) 5 {1, 5}, prove that stabG(5) is normal in G., 18. Suppose that G 5 H 3 K and that N is a normal subgroup of H., Prove that N is normal in G., 19. Show that there is no homomorphism from Z8 % Z2 % Z2 onto, Z4 % Z4., 20. Show that there is no homomorphism from A4 onto a group of, order 2, 4, or 6, but that there is a homomorphism from A4 onto a, group of order 3., 21. Let H be a normal subgroup of S4 of order 4. Prove that S4/H is isomorphic to S3., 22. Suppose that f is a homomorphism of U(36), Ker f 5 {1, 13, 25},, and f(5) 5 17. Determine all elements that map to 17., 23. Let n 5 2m, where m is odd. How many elements of order 2, does Dn/Z(Dn) have? How many elements are in the subgroup, kR360/nl/Z(Dn)? How do these numbers compare with the number, of elements of order 2 in Dm?, 24. Suppose that H is a normal subgroup of a group G of odd order and, that |H| 5 5. Show that H # Z(G)., 25. Let G be an Abelian group and let n be a positive integer. Let Gn 5, {g | gn 5 e} and Gn 5 {gn | g [ G}. Prove that G/Gn is isomorphic, to Gn., 26. Let R1 denote the multiplicative group of positive reals and let T 5, {a 1 bi [ C | a2 1 b2 5 1} be the multiplicative group of complex, numbers of norm 1. Show that C* is the internal direct product of R1, and T., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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240, , Groups, , 27. Let G be a finite group and let p be a prime. If p2 . |G|, show that, any subgroup of order p is normal in G., 28. Let G 5 Z % Z and H 5 {(x, y) | x and y are even integers}. Show, that H is a subgroup of G. Determine the order of G/H. To which, familiar group is G/H isomorphic?, 29. Let n be a positive integer. Prove that every element of order n in, Q/Z is contained in k1/n 1 Zl., 30. (1997 Putnam Competition) Let G be a group and let f: G S G be, a function such that, f(g1)f(g2)f(g3) 5 f(h1)f(h2)f(h3), , 31., 32., 33., 34., 35., 36., , 37., , 38., , whenever g1g2g3 5 e 5 h1h2h3. Prove that there exists an element a, in G such that c(x) 5 af(x) is a homomorphism., Prove that every homomorphism from Z % Z into Z has the form, (x, y) S ax 1 by, where a and b are integers., Prove that every homomorphism from Z % Z into Z % Z has the, form (x, y) S (ax 1 by, cx 1 dy), where a, b, c, and d are integers., Prove that Q/Z is not isomorphic to a proper subgroup of itself., Prove that for each positive integer n, the group Q/Z has exactly, f(n) elements of order n (f is the Euler phi function)., Show that any group with more than two elements has an automorphism other than the identity mapping., A proper subgroup H of a group G is called maximal if there is no, subgroup K such that H , K , G. Prove that Q under addition has, no maximal subgroups., Let G be the group of quaternions as given in Exercise 4 of the, Supplementary Exercises for Chapters 1–4 and let H ka2l., Determine whether G/H is isomorphic to Z4 or Z2 % Z2. Is G/H isomorphic to a subgroup of G?, Write the dihedral group D8 as {R0, R45, R90, R135, R180, R225, R270,, R315, F1, F2, F3, F4, F5, F6, F7, F8} and let N 5 {R0, R90, R180, R270}., Prove that N is normal in D8. Given that F1N 5 {F1, F4, F3, F2},, determine whether D8/N is cyclic., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 9– 11, , 39. Let G be the group e c, , 1, 0, , 241, , a, d ` where a, b [ R, b Z 0 f and, b, , 1 x, d ` where x [ R f . Show that H is a subgroup of G. Is, 0 1, H a normal subgroup of G? Justify your answer., 40. Find a subgroup H of Zp2 % Zp2 such that (Zp2 % Zp2)/H is isomorphic, to Zp % Zp., 41. Recall that H is a characteristic subgroup of K if f1H2 H for, every automorphism f of K. Prove that if H is a characteristic subgroup of K, and K is a normal subgroup of G, then H is a normal, subgroup of G., H ec, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PA R T, , 3, , Rings, , For online student resources, visit this textbook’s website at, www.CengageBrain.com, , 243, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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12, , Introduction, to Rings, Example is the school of mankind, and they will learn at no other., edmund burke, On a Regicide Peace, , Motivation and Definition, Many sets are naturally endowed with two binary operations: addition, and multiplication. Examples that quickly come to mind are the integers, the integers modulo n, the real numbers, matrices, and polynomials. When considering these sets as groups, we simply used addition and, ignored multiplication. In many instances, however, one wishes to take, into account both addition and multiplication. One abstract concept that, does this is the concept of a ring.† This notion was originated in the, mid-19th century by Richard Dedekind, although its first formal abstract, definition was not given until Abraham Fraenkel presented it in 1914., Definition Ring, , A ring R is a set with two binary operations, addition (denoted by, a 1 b) and multiplication (denoted by ab), such that for all a, b, c in R:, 1. a 1 b 5 b 1 a., 2. (a 1 b) 1 c 5 a 1 (b 1 c)., 3. There is an additive identity 0. That is, there is an element 0 in R, such that a 1 0 5 a for all a in R., 4. There is an element 2a in R such that a 1 (2a) 5 0., 5. a(bc) 5 (ab)c., 6. a(b 1 c) 5 ab 1 ac and (b 1 c) a 5 ba 1 ca., , So, a ring is an Abelian group under addition, also having an associative multiplication that is left and right distributive over addition., Note that multiplication need not be commutative. When it is, we say, that the ring is commutative. Also, a ring need not have an identity, †The, , term ring was first applied in 1897 by the German mathematician David Hilbert, (1862–1943)., 245, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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246, , Rings, , under multiplication. A unity (or identity) in a ring is a nonzero element, that is an identity under multiplication. A nonzero element of a commutative ring with unity need not have a multiplicative inverse. When it, does, we say that it is a unit of the ring. Thus, a is a unit if a21 exists., The following terminology and notation are convenient. If a and b, belong to a commutative ring R and a is nonzero, we say that a divides, b (or that a is a factor of b) and write a | b, if there exists an element c, in R such that b 5 ac. If a does not divide b, we write a B b., Recall that if a is an element from a group under the operation of, addition and n is a positive integer, na means a 1 a 1 ? ? ? 1 a, where, there are n summands. When dealing with rings, this notation can cause, confusion, since we also use juxtaposition for the ring multiplication., When there is the potential for confusion, we will use n ? a to mean, a 1 a 1 ? ? ? 1 a (n summands)., For an abstraction to be worthy of study, it must have many diverse, concrete realizations. The following list of examples shows that the, ring concept is pervasive., , Examples of Rings, EXAMPLE 1 The set Z of integers under ordinary addition and, multiplication is a commutative ring with unity 1. The units of Z are, 1 and 21., EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} under addition and, multiplication modulo n is a commutative ring with unity 1. The set of, units is U(n)., EXAMPLE 3 The set Z[x] of all polynomials in the variable x with, integer coefficients under ordinary addition and multiplication is a, commutative ring with unity f(x) 5 1., EXAMPLE 4 The set M2(Z) of 2 3 2 matrices with integer entries, is a noncommutative ring with unity c, , 1 0, d., 0 1, , EXAMPLE 5 The set 2Z of even integers under ordinary addition, and multiplication is a commutative ring without unity., EXAMPLE 6 The set of all continuous real-valued functions of a, real variable whose graphs pass through the point (1, 0) is a commutative ring without unity under the operations of pointwise addition and, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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12 | Introduction to Rings, , 247, , multiplication [that is, the operations ( f 1 g)(a) 5 f(a) 1 g(a) and, ( fg)(a) 5 f(a)g(a)]., EXAMPLE 7 Let R1, R2, . . . , Rn be rings. We can use these to construct a new ring as follows. Let, R1 % R2 % ? ? ? % Rn 5 {(a1, a2, . . . , an) | ai [ Ri}, and perform componentwise addition and multiplication; that is, define, (a1, a 2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn), and, (a1, a2, . . . , an)(b1, b2, . . . , bn) 5 (a1b1, a2b2, . . . , anbn)., This ring is called the direct sum of R1, R2, . . . , Rn., , Properties of Rings, Our first theorem shows how the operations of addition and multiplication intertwine. We use b 2 c to denote b 1 (2c)., Theorem 12.1 Rules of Multiplication, Let a, b, and c belong to a ring R. Then, 1., 2., 3., 4., , a0 5 0a 5 0., a(2b) 5 (2a)b 5 2(ab)., (2a)(2b) 5 ab.†, a(b 2 c) 5 ab 2 ac and, , (b 2 c)a 5 ba 2 ca., , Furthermore, if R has a unity element 1, then, 5. (21)a 5 2a., 6. (21)(21) 5 1., , PROOF We will prove rules 1 and 2 and leave the rest as easy exercises, (see Exercise 11). To prove statements such as those in Theorem 12.1, we, need only “play off ” the distributive property against the fact that R is a, group under addition with additive identity 0. Consider rule 1. Clearly,, 0 1 a0 5 a0 5 a(0 1 0) 5 a0 1 a0., So, by cancellation, 0 5 a0. Similarly, 0a 5 0., †“Minus, , times minus equals plus., The reason for this we need not discuss.”, W. H. Auden, A Certain World: A Commonplace Book, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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248, , Rings, , To prove rule 2, we observe that a(2b) 1 ab 5 a(2b 1 b) 5, a0 5 0. So, adding 2(ab) to both sides yields a(2b) 5 2(ab). The remainder of rule 2 is done analogously., Recall that in the case of groups, the identity and inverses are unique., The same is true for rings, provided that these elements exist. The proofs, are identical to the ones given for groups and therefore are omitted., Theorem 12.2 Uniqueness of the Unity and Inverses, If a ring has a unity, it is unique. If a ring element has a multiplicative inverse, it is unique., , Many students have the mistaken tendency to treat a ring as if it were, a group under multiplication. It is not. The two most common errors are, the assumptions that ring elements have multiplicative inverses—they, need not—and that a ring has a multiplicative identity—it need not. For, example, if a, b, and c belong to a ring, a 2 0 and ab 5 ac, we cannot, conclude that b 5 c. Similarly, if a2 5 a, we cannot conclude that a 5 0, or 1 (as is the case with real numbers). In the first place, the ring need, not have multiplicative cancellation, and in the second place, the ring, need not have a multiplicative identity. There is an important class of, rings that contains Z and Z[x] wherein multiplicative identities exist and, for which multiplicative cancellation holds. This class is taken up in the, next chapter., , Subrings, In our study of groups, subgroups played a crucial role. Subrings, the, analogous structures in ring theory, play a much less prominent role than, their counterparts in group theory. Nevertheless, subrings are important., Definition Subring, , A subset S of a ring R is a subring of R if S is itself a ring with the, operations of R., , Just as was the case for subgroups, there is a simple test for subrings., Theorem 12.3 Subring Test, A nonempty subset S of a ring R is a subring if S is closed under, subtraction and multiplication—that is, if a 2 b and ab are in S, whenever a and b are in S., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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12 | Introduction to Rings, , 249, , PROOF Since addition in R is commutative and S is closed under subtraction, we know by the One-Step Subgroup Test (Theorem 3.1) that S, is an Abelian group under addition. Also, since multiplication in R is, associative as well as distributive over addition, the same is true for, multiplication in S. Thus, the only condition remaining to be checked, is that multiplication is a binary operation on S. But this is exactly what, closure means., We leave it to the student to confirm that each of the following examples is a subring., EXAMPLE 8 {0} and R are subrings of any ring R. {0} is called the, trivial subring of R., EXAMPLE 9 {0, 2, 4} is a subring of the ring Z6, the integers modulo 6. Note that although 1 is the unity in Z6, 4 is the unity in, {0, 2, 4}., EXAMPLE 10 For each positive integer n, the set, nZ 5 {0, 6n, 62n, 63n, . . .}, is a subring of the integers Z., EXAMPLE 11 The set of Gaussian integers, Z[i] 5 {a 1 bi | a, b [ Z}, is a subring of the complex numbers C., EXAMPLE 12 Let R be the ring of all real-valued functions of a single real variable under pointwise addition and multiplication. The subset S of R of functions whose graphs pass through the origin forms a, subring of R., EXAMPLE 13 The set, ec, , a 0, d ` a, b [ Z f, 0 b, , of diagonal matrices is a subring of the ring of all 2 3 2 matrices, over Z., We can picture the relationship between a ring and its various subrings by way of a subring lattice diagram. In such a diagram, any ring, is a subring of all the rings that it is connected to by one or more upward lines. Figure 12.1 shows the relationships among some of the, rings we have already discussed., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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250, , Rings, , C, R, Z[ i] = {a 1 bi | a, b [ Z}, , Q(√2) = {a 1 b√2 | a, b [ Q}, Q, Z, , 5Z, , 10Z, , 2Z, , 4Z, , 8Z, , 7Z, , 3Z, , 6Z, , 12Z, , 9Z, , 18Z, , Figure 12.1 Partial subring lattice diagram of C., , In the next several chapters, we will see that many of the fundamental concepts of group theory can be naturally extended to rings. In particular, we will introduce ring homomorphisms and factor rings., , Exercises, There is no substitute for hard work., thomas alva edison, Life, , 1. Give an example of a finite noncommutative ring. Give an example, of an infinite noncommutative ring that does not have a unity., 2. The ring {0, 2, 4, 6, 8} under addition and multiplication modulo, 10 has a unity. Find it., 3. Give an example of a subset of a ring that is a subgroup under, addition but not a subring., 4. Show, by example, that for fixed nonzero elements a and b in a, ring, the equation ax 5 b can have more than one solution. How, does this compare with groups?, 5. Prove Theorem 12.2., 6. Find an integer n that shows that the rings Zn need not have the following properties that the ring of integers has., a. a2 5 a implies a 5 0 or a 5 1., b. ab 5 0 implies a 5 0 or b 5 0., c. ab 5 ac and a 2 0 imply b 5 c., Is the n you found prime?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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12 | Introduction to Rings, , 251, , 7. Show that the three properties listed in Exercise 6 are valid for Zp,, where p is prime., 8. Show that a ring is commutative if it has the property that ab 5 ca, implies b 5 c when a 2 0., 9. Prove that the intersection of any collection of subrings of a ring R, is a subring of R., 10. Verify that Examples 8 through 13 in this chapter are as stated., 11. Prove rules 3 through 6 of Theorem 12.1., 12. Let a, b, and c be elements of a commutative ring, and suppose that, a is a unit. Prove that b divides c if and only if ab divides c., 13. Describe all the subrings of the ring of integers., 14. Let a and b belong to a ring R and let m be an integer. Prove that, m ? (ab) 5 (m ? a)b 5 a(m ? b)., 15. Show that if m and n are integers and a and b are elements from a, ring, then (m ? a)(n ? b) 5 (mn) ? (ab). (This exercise is referred to, in Chapters 13 and 15.), 16. Show that if n is an integer and a is an element from a ring, then, n ? (2a) 5 2(n ? a)., 17. Show that a ring that is cyclic under addition is commutative., 18. Let a belong to a ring R. Let S 5 {x [ R | ax 5 0}. Show that S is, a subring of R., 19. Let R be a ring. The center of R is the set {x [ R | ax 5 xa for all, a in R}. Prove that the center of a ring is a subring., 20. Describe the elements of M2(Z) (see Example 4) that have multiplicative inverses., 21. Suppose that R1, R2, . . . , Rn are rings that contain nonzero elements. Show that R1 % R2 % ? ? ? % Rn has a unity if and only if, each Ri has a unity., 22. Let R be a commutative ring with unity and let U(R) denote the set, of units of R. Prove that U(R) is a group under the multiplication of, R. (This group is called the group of units of R.), 23. Determine U(Z[i]) (see Example 11)., 24. If R1, R2, . . . , Rn are commutative rings with unity, show that, U(R1 % R2 % ? ? ? % Rn) 5 U(R1) % U(R2) % ? ? ? % U(Rn)., 25. Determine U(Z[x]). (This exercise is referred to in Chapter 17.), 26. Determine U(R[x])., 27. Show that a unit of a ring divides every element of the ring., 28. In Z6, show that 4 | 2; in Z8, show that 3 | 7; in Z15, show that 9 | 12., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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252, , Rings, , 29. Suppose that a and b belong to a commutative ring R with unity. If, a is a unit of R and b2 5 0, show that a 1 b is a unit of R., 30. Suppose that there is an integer n . 1 such that xn 5 x for all elements, x of some ring. If m is a positive integer and am 5 0 for some a, show, that a 5 0., 31. Give an example of ring elements a and b with the properties that, ab 5 0 but ba 2 0., 32. Let n be an integer greater than 1. In a ring in which xn 5 x for all x,, show that ab 5 0 implies ba 5 0., 33. Suppose that R is a ring such that x3 5 x for all x in R. Prove that, 6x 5 0 for all x in R., 34. Suppose that a belongs to a ring and a4 5 a2. Prove that a2n 5 a2, for all n $ 1., 35. Find an integer n . 1 such that an 5 a for all a in Z6. Do the same, for Z10. Show that no such n exists for Zm when m is divisible by the, square of some prime., 36. Let m and n be positive integers and let k be the least common multiple of m and n. Show that mZ > nZ 5 kZ., 37. Explain why every subgroup of Zn under addition is also a subring, of Zn., 38. Is Z6 a subring of Z12?, 39. Suppose that R is a ring with unity 1 and a is an element of R such, that a2 5 1. Let S 5 {ara | r [ R}. Prove that S is a subring of R., Does S contain 1?, 40. Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5, a, ab, ec, d ` a, b [ Z f . Prove or disprove that R is a subring, ab, b, of M2(Z)., 41. Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5, a, ab, ec, d ` a, b [ Z f . Prove or disprove that R is a subring, ab, b, of M2(Z)., a a, 42. Let R 5 e c, d ` a, b [ Z f . Prove or disprove that R is a subring, b b, of M2(Z)., 43. Let R 5 Z % Z % Z and S 5 {(a, b, c) [ R | a 1 b 5 c}. Prove or, disprove that S is a subring of R., 44. Suppose that there is a positive even integer n such that an 5 a for, all elements a of some ring. Show that 2a 5 a for all a in the ring., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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12 | Introduction to Rings, , 253, , 45. Let R be a ring with unity 1. Show that S 5 {n ? 1 | n [ Z} is a, subring of R., 46. Show that 2Z < 3Z is not a subring of Z., 47. Determine the smallest subring of Q that contains 1/2. (That is,, find the subring S with the property that S contains 1/2 and, if T is, any subring containing 1/2, then T contains S.), 48. Determine the smallest subring of Q that contains 2/3., 49. Let R be a ring. Prove that a2 2 b2 5 (a 1 b)(a 2 b) for all a, b in, R if and only if R is commutative., 50. Suppose that R is a ring and that a2 5 a for all a in R. Show that R, is commutative. [A ring in which a2 5 a for all a is called a, Boolean ring, in honor of the English mathematician George Boole, (1815–1864).], 51. Give an example of a Boolean ring with four elements. Give an example of an infinite Boolean ring., 52. If a, b, and c are elements of a ring, does the equation ax 1 b 5 c, always have a solution x? If it does, must the solution be unique?, Answer the same questions given that a is a unit., , Computer Exercises, Software for the computer exercises in this chapter is available at the, website:, http://www.d.umn.edu/~jgallian, , Suggested Reading, D. B. Erickson, “Orders for Finite Noncommutative Rings,” American, Mathematical Monthly 73 (1966): 376–377., In this elementary paper, it is shown that there exists a noncommutative ring, of order m . 1 if and only if m is divisible by the square of a prime., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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I. N. Herstein, , St. Johns University, Collegeville, MN, , A whole generation of textbooks and an entire, generation of mathematicians, myself, included, have been profoundly influenced, by that text [Herstein’s Topics in Algebra]., georgia benkart, , I. N. Herstein was born on March 28, 1923,, in Poland. His family moved to Canada, when he was seven. He grew up in a poor and, tough environment, on which he commented, that in his neighborhood you became either a, gangster or a college professor. During his, school years he played football, hockey, golf,, tennis, and pool. During this time he worked, as a steeplejack and as a barber at a fair., Herstein received a B.S. degree from the, University of Manitoba, an M.A. from the, University of Toronto, and, in 1948, a Ph.D., degree from Indiana University under the supervision of Max Zorn. Before permanently, settling at the University of Chicago in 1962,, he held positions at the University of Kansas,, the Ohio State University, the University of, Pennsylvania, and Cornell University., Herstein wrote more than 100 research, papers and a dozen books. Although his, principal interest was noncommutative ring, , theory, he also wrote papers on finite, groups, linear algebra, and mathematical, economics. His textbook Topics in Algebra,, first published in 1964, dominated the field, for 20 years and has become a classic., Herstein had great influence through his, teaching and his collaboration with colleagues. He had 30 Ph.D. students, and, traveled and lectured widely. His nonmathematical interests included languages and, art. He spoke Italian, Hebrew, Polish, and, Portuguese. Herstein died on February 9,, 1988, after a long battle with cancer., To find more information about Herstein,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 254, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 Integral Domains, Don’t just read it! Ask your own questions, look for your own examples,, discover your own proofs. Is the hypothesis necessary? Is the converse, true? What happens in the classical special case? Where does the proof, use the hypothesis?, paul halmos, , Definition and Examples, To a certain degree, the notion of a ring was invented in an attempt to, put the algebraic properties of the integers into an abstract setting. A, ring is not the appropriate abstraction of the integers, however, for too, much is lost in the process. Besides the two obvious properties of commutativity and existence of a unity, there is one other essential feature, of the integers that rings in general do not enjoy—the cancellation, property. In this chapter, we introduce integral domains—a particular, class of rings that have all three of these properties. Integral domains, play a prominent role in number theory and algebraic geometry., Definition Zero-Divisors, , A zero-divisor is a nonzero element a of a commutative ring R such, that there is a nonzero element b [ R with ab 5 0., Definition Integral Domain, , An integral domain is a commutative ring with unity and no, zero-divisors., , Thus, in an integral domain, a product is 0 only when one of the, factors is 0; that is, ab 5 0 only when a 5 0 or b 5 0. The following, examples show that many familiar rings are integral domains and some, familiar rings are not. For each example, the student should verify the, assertion made., EXAMPLE 1 The ring of integers is an integral domain., , 255, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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256, , Rings, , EXAMPLE 2 The ring of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z}, is an integral domain., EXAMPLE 3 The ring Z[x] of polynomials with integer coefficients, is an integral domain., EXAMPLE 4 The ring Z[22] 5 {a 1 b22 | a, b [ Z} is an integral, domain., EXAMPLE 5 The ring Zp of integers modulo a prime p is an integral, domain., EXAMPLE 6 The ring Zn of integers modulo n is not an integral domain when n is not prime., EXAMPLE 7 The ring M2(Z) of 2 3 2 matrices over the integers is, not an integral domain., EXAMPLE 8 Z % Z is not an integral domain., What makes integral domains particularly appealing is that they have, an important multiplicative group theoretic property, in spite of the fact, that the nonzero elements need not form a group under multiplication., This property is cancellation., Theorem 13.1 Cancellation, Let a, b, and c belong to an integral domain. If a 2 0 and ab 5 ac,, then b 5 c., , PROOF From ab 5 ac, we have a(b 2 c) 5 0. Since a 2 0, we must, have b 2 c 5 0., Many authors prefer to define integral domains by the cancellation, property—that is, as commutative rings with unity in which the cancellation property holds. This definition is equivalent to ours., , Fields, In many applications, a particular kind of integral domain called a field, is necessary., Definition Field, , A field is a commutative ring with unity in which every nonzero, element is a unit., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 | Integral Domains, , 257, , To verify that every field is an integral domain, observe that if a and, b belong to a field with a 2 0 and ab 5 0, we can multiply both sides, of the last expression by a21 to obtain b 5 0., It is often helpful to think of ab21 as a divided by b. With this in, mind, a field can be thought of as simply an algebraic system that, is closed under addition, subtraction, multiplication, and division, (except by 0). We have had numerous examples of fields: the complex, numbers, the real numbers, the rational numbers. The abstract theory of, fields was initiated by Heinrich Weber in 1893. Groups, rings, and, fields are the three main branches of abstract algebra. Theorem 13.2, says that, in the finite case, fields and integral domains are the same., Theorem 13.2 Finite Integral Domains Are Fields, A finite integral domain is a field., , PROOF Let D be a finite integral domain with unity 1. Let a be any, nonzero element of D. We must show that a is a unit. If a 5 1, a is its, own inverse, so we may assume that a 2 1. Now consider the following, sequence of elements of D: a, a2, a3, . . . . Since D is finite, there must, be two positive integers i and j such that i . j and ai 5 a j. Then, by cancellation, ai2j 5 1. Since a 2 1, we know that i 2 j . 1, and we have, shown that ai2j21 is the inverse of a., Corollary Zp Is a Field, For every prime p, Zp, the ring of integers modulo p is a field., , PROOF According to Theorem 13.2, we need only prove that Zp has, no zero-divisors. So, suppose that a, b [ Zp and ab 5 0. Then ab 5 pk, for some integer k. But then, by Euclid’s Lemma (see Chapter 0), p, divides a or p divides b. Thus, in Zp, a 5 0 or b 5 0., Putting the preceding corollary together with Example 6, we see that, Zn is a field if and only if n is prime. In Chapter 22, we will describe, how all finite fields can be constructed. For now, we give one example, of a finite field that is not of the form Zp., EXAMPLE 9 Field with Nine Elements, Let, Z3[i] 5 {a 1 bi | a, b [ Z3}, 5 {0, 1, 2, i, 1 1 i, 2 1 i, 2i, 1 1 2i, 2 1 2i},, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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258, , Rings, , where i2 5 21. This is the ring of Gaussian integers modulo 3. Elements are added and multiplied as in the complex numbers, except that, the coefficients are reduced modulo 3. In particular, 21 5 2. Table 13.1, is the multiplication table for the nonzero elements of Z3[i]., Table 13.1 Multiplication Table for Z3[i]*, , 1, 2, i, 11i, 21i, 2i, 1 1 2i, 2 1 2i, , 1, , 2, , i, , 11i, , 21i, , 2i, , 1 1 2i, , 2 1 2i, , 1, 2, i, 11i, 21i, 2i, 1 1 2i, 2 1 2i, , 2, 1, 2i, 2 1 2i, 1 1 2i, i, 21i, 11i, , i, 2i, 2, 21i, 2 1 2i, 1, 11i, 1 1 2i, , 11i, 2 1 2i, 21i, 2i, 1, 1 1 2i, 2, i, , 21i, 1 1 2i, 2 1 2i, 1, i, 11i, 2i, 2, , 2i, i, 1, 1 1 2i, 11i, 2, 2 1 2i, 21i, , 1 1 2i, 21i, 11i, 2, 2i, 2 1 2i, i, 1, , 2 1 2i, 11i, 1 1 2i, i, 2, 21i, 1, 2i, , EXAMPLE 10 Let Q[22] 5 {a 1 b22 | a, b [ Q}. It is easy to see, that Q[22] is a ring. Viewed as an element of R, the multiplicative inverse of any nonzero element of the form a 1 b22 is simply 1/(a 1, b22). To verify that Q[22] is a field, we must show that 1/(a 1 b22), can be written in the form c 1 d22. In high school algebra, this process, is called “rationalizing the denominator.” Specifically,, 1, 1, a b22, a, b, , 2, 22., 2 2, a 2b, a 2b2, a b2 2, a b2 2 a b2 2, (Note that a 1 b22 2 0 guarantees that a 2 b22 2 0.), , Characteristic of a Ring, Note that for any element x in Z3[i], we have 3x 5 x 1 x 1 x 5 0, since, addition is done modulo 3. Similarly, in the subring {0, 3, 6, 9} of Z12,, we have 4x 5 x 1 x 1 x 1 x 5 0 for all x. This observation motivates, the following definition., Definition Characteristic of a Ring, , The characteristic of a ring R is the least positive integer n such that, nx 5 0 for all x in R. If no such integer exists, we say that R has characteristic 0. The characteristic of R is denoted by char R., , Thus, the ring of integers has characteristic 0, and Zn has characteristic n. An infinite ring can have a nonzero characteristic. Indeed, the, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 | Integral Domains, , 259, , ring Z2[x] of all polynomials with coefficients in Z2 has characteristic 2., (Addition and multiplication are done as for polynomials with ordinary, integer coefficients except that the coefficients are reduced modulo 2.), When a ring has a unity, the task of determining the characteristic is, simplified by Theorem 13.3., Theorem 13.3 Characteristic of a Ring with Unity, Let R be a ring with unity 1. If 1 has infinite order under addition,, then the characteristic of R is 0. If 1 has order n under addition,, then the characteristic of R is n., , PROOF If 1 has infinite order, then there is no positive integer n such, that n ? 1 5 0, so R has characteristic 0. Now suppose that 1 has additive order n. Then n ? 1 5 0, and n is the least positive integer with this, property. So, for any x in R, we have, n ? x 5 x 1 x 1 ? ? ? 1 x (n summands), 5 1x 1 1x 1 ? ? ? 1 1x (n summands), 5 (1 1 1 1 ? ? ? 1 1)x (n summands), 5 (n ? 1)x 5 0x 5 0., Thus, R has characteristic n., In the case of an integral domain, the possibilities for the characteristic are severely limited., Theorem 13.4 Characteristic of an Integral Domain, The characteristic of an integral domain is 0 or prime., , PROOF By Theorem 13.3, it suffices to show that if the additive order, of 1 is finite, it must be prime. Suppose that 1 has order n and that n 5 st,, where 1 # s, t # n. Then, by Exercise 15 in Chapter 12,, 0 5 n ? 1 5 (st) ? 1 5 (s ? 1)(t ? 1)., So, s ? 1 5 0 or t ? 1 5 0. Since n is the least positive integer with the, property that n ? 1 5 0, we must have s 5 n or t 5 n. Thus, n is, prime., We conclude this chapter with a brief discussion of polynomials, with coefficients from a ring—a topic we will consider in detail in, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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260, , Rings, , later chapters. The existence of zero-divisors in a ring causes unusual, results when one is finding zeros of polynomials with coefficients in, the ring. Consider, for example, the equation x2 2 4x 1 3 5 0. In the, integers, we could find all solutions by factoring, x2 2 4x 1 3 5 (x 2 3)(x 2 1) 5 0, and setting each factor equal to 0. But notice that when we say we can, find all solutions in this manner, we are using the fact that the only way, for a product to equal 0 is for one of the factors to be 0—that is, we are, using the fact that Z is an integral domain. In Z12, there are many pairs of, nonzero elements whose products are 0: 2 ? 6 5 0, 3 ? 4 5 0, 4 ? 6 5 0,, 6 ? 8 5 0, and so on. So, how do we find all solutions of x2 2 4x 1 3 5 0, in Z12? The easiest way is simply to try every element! Upon doing so,, we find four solutions: x 5 1, x 5 3, x 5 7, and x 5 9. Observe that we, can find all solutions of x2 2 4x 1 3 5 0 over Z11 or Z13, say, by setting, the two factors x 2 3 and x 2 1 equal to 0. Of course, the reason this, works for these rings is that they are integral domains. Perhaps this will, convince you that integral domains are particularly advantageous rings., Table 13.2 gives a summary of some of the rings we have introduced and, their properties., Table 13.2 Summary of Rings and Their Properties, Ring, Z, Zn, n composite, Zp, p prime, Z[x], nZ, n . 1, M2(Z), M2(2Z), Z[i], Z3[i], Z[22], Q[22], Z%Z, , Integral, Commutative Domain Field Characteristic, , Form of Element, , Unity, , k, k, k, anxn 1 ? ? ? 1, a1x 1 a0, nk, a b, c, d, c d, 2a 2b, c, d, 2c 2d, a 1 bi, a 1 bi; a, b [ Z3, a 1 b22; a, b [ Z, a 1 b22; a, b [ Q, (a, b), , 1, 1, 1, f(x) 5 1, , Yes, Yes, Yes, Yes, , Yes, No, Yes, Yes, , No, No, Yes, No, , 0, n, p, 0, , None, 1 0, c, d, 0 1, , Yes, , No, , No, , 0, , No, , No, , No, , 0, , None, , No, , No, , No, , 0, , 1, 1, 1, 1, (1, 1), , Yes, Yes, Yes, Yes, Yes, , Yes, Yes, Yes, Yes, No, , No, Yes, No, Yes, No, , 0, 3, 0, 0, 0, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 | Integral Domains, , 261, , Exercises, It looked absolutely impossible. But it so happens that you go on worrying, away at a problem in science and it seems to get tired, and lies down and, lets you catch it., william lawrence bragg†, , 1. Verify that Examples 1 through 8 are as claimed., 2. Which of Examples 1 through 5 are fields?, 3. Show that a commutative ring with the cancellation property, (under multiplication) has no zero-divisors., 4. List all zero-divisors in Z20. Can you see a relationship between the, zero-divisors of Z20 and the units of Z20?, 5. Show that every nonzero element of Zn is a unit or a zero-divisor., 6. Find a nonzero element in a ring that is neither a zero-divisor nor a, unit., 7. Let R be a finite commutative ring with unity. Prove that every, nonzero element of R is either a zero-divisor or a unit. What happens if we drop the “finite” condition on R?, 8. Let a 2 0 belong to a commutative ring. Prove that a is a zerodivisor if and only if a2b 5 0 for some b Z 0., 9. Find elements a, b, and c in the ring Z % Z % Z such that ab, ac,, and bc are zero-divisors but abc is not a zero-divisor., 10. Describe all zero-divisors and units of Z % Q % Z., 11. Let d be an integer. Prove that Z[2d] 5 {a 1 b2d | a, b [ Z} is, an integral domain. (This exercise is referred to in Chapter 18.), 12. In Z7, give a reasonable interpretation for the expressions 1/2,, 22/3, 23, and 21/6., 13. Give an example of a commutative ring without zero-divisors that, is not an integral domain., 14. Find two elements a and b in a ring such that both a and b are zerodivisors, a 1 b 2 0, and a 1 b is not a zero-divisor., 15. Let a belong to a ring R with unity and suppose that an 5 0 for, some positive integer n. (Such an element is called nilpotent.), Prove that 1 2 a has a multiplicative inverse in R. [Hint: Consider, (1 2 a)(1 1 a 1 a2 1 ? ? ? 1 an21).], †Bragg,, , at age 24, won the Nobel Prize for the invention of x-ray crystallography. He, remains the youngest person ever to receive the Nobel Prize., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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262, , Rings, , 16. Show that the nilpotent elements of a commutative ring form a, subring., 17. Show that 0 is the only nilpotent element in an integral domain., 18. A ring element a is called an idempotent if a2 5 a. Prove that the, only idempotents in an integral domain are 0 and 1., 19. Let a and b be idempotents in a commutative ring. Show that each, of the following is also an idempotent: ab, a ab, a b ab,, a b 2ab., 20. Show that Zn has a nonzero nilpotent element if and only if n is divisible by the square of some prime., 21. Let R be the ring of real-valued continuous functions on [21, 1]., Show that R has zero-divisors., 22. Prove that if a is a ring idempotent, then an a for all positive integers n., 23. Determine all ring elements that are both nilpotent elements and, idempotents., 24. Find a zero-divisor in Z5[i] 5 {a 1 bi | a, b [ Z5}., 25. Find an idempotent in Z5[i] 5 {a 1 bi | a, b [ Z5}., 26. Find all units, zero-divisors, idempotents, and nilpotent elements, in Z3 % Z6., 27. Determine all elements of a ring that are both units and idempotents., 28. Let R be the set of all real-valued functions defined for all real, numbers under function addition and multiplication., a. Determine all zero-divisors of R., b. Determine all nilpotent elements of R., c. Show that every nonzero element is a zero-divisor or a unit., 29. (Subfield Test) Let F be a field and let K be a subset of F with at, least two elements. Prove that K is a subfield of F if, for any, a, b (b 2 0) in K, a 2 b and ab21 belong to K., 30. Let d be a positive integer. Prove that Q[ 2d] 5 {a 1 b 2d |, a, b [ Q} is a field., 31. Let R be a ring with unity 1. If the product of any pair of nonzero, elements of R is nonzero, prove that ab 5 1 implies ba 5 1., 32. Let R 5 {0, 2, 4, 6, 8} under addition and multiplication modulo, 10. Prove that R is a field., 33. Formulate the appropriate definition of a subdomain (that is, a, “sub” integral domain). Let D be an integral domain with unity 1., Show that P 5 {n ? 1 | n [ Z} (that is, all integral multiples of 1), is a subdomain of D. Show that P is contained in every subdomain, of D. What can we say about the order of P?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 | Integral Domains, , 263, , 34. Prove that there is no integral domain with exactly six elements. Can, your argument be adapted to show that there is no integral domain, with exactly four elements? What about 15 elements? Use these observations to guess a general result about the number of elements in, a finite integral domain., 35. Let F be a field of order 2n. Prove that char F 5 2., 36. Determine all elements of an integral domain that are their own, inverses under multiplication., 37. Characterize those integral domains for which 1 is the only element that is its own multiplicative inverse., 38. Determine all integers n 7 1 for which 1n 12! is a zero-divisor, in Zn., 39. Suppose that a and b belong to an integral domain., a. If a5 5 b5 and a3 5 b3, prove that a 5 b., b. If am 5 bm and an 5 bn, where m and n are positive integers that, are relatively prime, prove that a 5 b., 40. Find an example of an integral domain and distinct positive integers m and n such that am 5 bm and an 5 bn, but a 2 b., 41. If a is an idempotent in a commutative ring, show that 1 2 a is also, an idempotent., 42. Construct a multiplication table for Z2[i], the ring of Gaussian integers modulo 2. Is this ring a field? Is it an integral domain?, 43. The nonzero elements of Z3[i] form an Abelian group of order 8 under multiplication. Is it isomorphic to Z8, Z4 % Z2, or Z2 % Z2 % Z2?, 44. Show that Z 7[23] 5 {a 1 b23 | a, b [ Z 7} is a field. For any, positive integer k and any prime p, determine a necessary and sufficient condition for Zp[2k] 5 {a 1 b2k | a, b [ Zp} to be a field., 45. Show that a finite commutative ring with no zero-divisors and at, least two elements has a unity., 46. Suppose that a and b belong to a commutative ring and ab is a, zero-divisor. Show that either a or b is a zero-divisor., 47. Suppose that R is a commutative ring without zero-divisors. Show, that all the nonzero elements of R have the same additive order., 48. Suppose that R is a commutative ring without zero-divisors. Show, that the characteristic of R is 0 or prime., 49. Let x and y belong to a commutative ring R with prime characteristic p., a. Show that (x 1 y) p 5 x p 1 y p., n, n, n, b. Show that, for all positive integers n, (x 1 y) p 5 x p 1 y p ., c. Find elements x and y in a ring of characteristic 4 such that, (x 1 y)4 2 x4 1 y4. (This exercise is referred to in Chapter 20.), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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264, , Rings, , 50. Let R be a commutative ring with unity 1 and prime characteristic., If a [ R is nilpotent, prove that there is a positive integer k such that, (1 1 a)k 5 1., 51. Show that any finite field has order pn, where p is a prime. Hint: Use, facts about finite Abelian groups. (This exercise is referred to in, Chapter 22.), 52. Give an example of an infinite integral domain that has characteristic 3., 53. Let R be a ring and let M2(R) be the ring of 2 3 2 matrices with entries, from R. Explain why these two rings have the same characteristic., 54. Let R be a ring with m elements. Show that the characteristic of R, divides m., 55. Explain why a finite ring must have a nonzero characteristic., 56. Find all solutions of x2 2 x 1 2 5 0 over Z3[i]. (See Example 9.), 57. Consider the equation x2 2 5x 1 6 5 0., a. How many solutions does this equation have in Z7?, b. Find all solutions of this equation in Z8., c. Find all solutions of this equation in Z12., d. Find all solutions of this equation in Z14., 58. Find the characteristic of Z4 % 4Z., 59. Suppose that R is an integral domain in which 20 ? 1 5 0 and, 12 ? 1 5 0. (Recall that n ? 1 means the sum 1 1 1 1 ? ? ? 1 1 with, n terms.) What is the characteristic of R?, 60. In a commutative ring of characteristic 2, prove that the idempotents form a subring., 61. Describe the smallest subfield of the field of real numbers that contains 22. (That is, describe the subfield K with the property that K, contains 22 and if F is any subfield containing 22, then F contains K.), 62. Let F be a finite field with n elements. Prove that xn21 5 1 for all, nonzero x in F., 63. Let F be a field of prime characteristic p. Prove that K 5 {x [ F |, x p 5 x} is a subfield of F., 64. Suppose that a and b belong to a field of order 8 and that a2 1 ab 1, b2 5 0. Prove that a 5 0 and b 5 0. Do the same when the field has, order 2n with n odd., 65. Let F be a field of characteristic 2 with more than two elements., Show that (x 1 y)3 2 x3 1 y3 for some x and y in F., 66. Suppose that F is a field with characteristic not 2, and that the nonzero elements of F form a cyclic group under multiplication. Prove, that F is finite., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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13 | Integral Domains, , 265, , 67. Suppose that D is an integral domain and that f is a nonconstant, function from D to the nonnegative integers such that f(xy) 5, f(x)f(y). If x is a unit in D, show that f(x) 5 1., 68. Let F be a field of order 32. Show that the only subfields of F are, F itself and {0, 1}., 69. Suppose that F is a field with 27 elements. Show that for every, element a [ F, 5a a., 70. Let, R ec, , a b, d ` a, b [ Z7 f, b a, , with the usual matrix addition and multiplication and mod 7 addition and multiplication of the entries. Prove that R is a commutative, ring. How many elements are in R? Is R a field? What happens, when Z7 is replaced by Z5?, , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Suggested Readings, Eric Berg, “A Family of Fields,” Pi Mu Epsilon 9 (1990): 154–155., In this article, the author uses properties of logarithms and exponents, to define recursively an infinite family of fields starting with the real, numbers., N. A. Khan, “The Characteristic of a Ring,” American Mathematical Monthly, 70 (1963): 736–738., Here it is shown that a ring has nonzero characteristic n if and only, if n is the maximum of the orders of the elements of R., K. Robin McLean, “Groups in Modular Arithmetic,” The Mathematical, Gazette 62 (1978): 94–104., This article explores the interplay between various groups of integers under multiplication modulo n and the ring Zn. It shows how to construct, groups of integers in which the identity is not obvious; for example, 1977, is the identity of the group {1977, 5931} under multiplication modulo, 7908., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Nathan Jacobson, , American Mathematical Society, , Few mathematicians have been as productive over such a long career or have had as, much influence on the profession as has, Professor Jacobson., Citation for the Steele Prize, for Lifetime Achievement, , Nathan Jacobson was born on September 8,, 1910, in Warsaw, Poland. After arriving in, the United States in 1917, Jacobson grew up, in Alabama, Mississippi, and Georgia, where, his father owned small clothing stores. He, received a B.A. degree from the University of, Alabama in 1930 and a Ph.D. from Princeton, in 1934. After brief periods as a professor at, Bryn Mawr, the University of Chicago, the, University of North Carolina, and Johns, Hopkins, Jacobson accepted a position at, Yale, where he remained until his retirement, in 1981., Jacobson’s principal contributions to algebra were in the areas of rings, Lie algebras,, and Jordan algebras. In particular, he developed structure theories for these systems. He, was the author of nine books and numerous, articles, and he had 33 Ph.D. students., , Jacobson held visiting positions in, France, India, Italy, Israel, China, Australia,, and Switzerland. Among his many honors, were the presidency of the American, Mathematical Society, memberships in the, National Academy of Sciences and the, American Academy of Arts and Sciences, a, Guggenheim Fellowship, and an honorary, degree from the University of Chicago., Jacobson died on December 5, 1999, at the, age of 89., To find more information about Jacobson,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 266, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 Ideals and Factor Rings, , The secret of science is to ask the right questions, and it is the choice of, problem more than anything else that marks the man of genius in the, scientific world., sir henry tizard in c. p. snow,, A postscript to Science and Government, , Ideals, Normal subgroups play a special role in group theory—they permit us, to construct factor groups. In this chapter, we introduce the analogous, concepts for rings—ideals and factor rings., Definition Ideal, , A subring A of a ring R is called a (two-sided) ideal of R if for, every r [ R and every a [ A both ra and ar are in A., , So, a subring A of a ring R is an ideal of R if A “absorbs” elements, from R—that is, if rA 5 {ra | a [ A} # A and Ar 5 {ar | a [ A} # A, for all r [ R., An ideal A of R is called a proper ideal of R if A is a proper subset, of R. In practice, one identifies ideals with the following test, which is, an immediate consequence of the definition of ideal and the subring, test given in Theorem 12.3., Theorem 14.1 Ideal Test, A nonempty subset A of a ring R is an ideal of R if, 1. a 2 b [ A whenever a, b [ A., 2. ra and ar are in A whenever a [ A and r [ R., , 267, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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268, , Rings, , EXAMPLE 1 For any ring R, {0} and R are ideals of R. The ideal {0}, is called the trivial ideal., EXAMPLE 2 For any positive integer n, the set nZ 5 {0, 6n,, 62n, . . .} is an ideal of Z., EXAMPLE 3 Let R be a commutative ring with unity and let a [ R., The set kal 5 {ra | r [ R} is an ideal of R called the principal ideal, generated by a. (Notice that kal is also the notation we used for, the cyclic subgroup generated by a. However, the intended meaning, will always be clear from the context.) The assumption that R is commutative is necessary in this example (see Exercise 31 in the Supplementary Exercises for Chapters 12–14)., EXAMPLE 4 Let R[x] denote the set of all polynomials with real coefficients and let A denote the subset of all polynomials with constant, term 0. Then A is an ideal of R[x] and A 5 kxl., EXAMPLE 5 Let R be a commutative ring with unity and let a1,, a2, . . . , an belong to R. Then I 5 ka1, a2, . . . , anl 5 {r1a1 1 r2a2 1, ? ? ? 1 rnan | ri [ R} is an ideal of R called the ideal generated by a1,, a2, . . . , an. The verification that I is an ideal is left as an easy exercise, (Exercise 3)., EXAMPLE 6 Let Z[x] denote the ring of all polynomials with integer coefficients and let I be the subset of Z[x] of all polynomials with, even constant terms. Then I is an ideal of Z[x] and I 5 kx, 2l (see, Exercise 37)., EXAMPLE 7 Let R be the ring of all real-valued functions of a real, variable. The subset S of all differentiable functions is a subring of R, but not an ideal of R., , Factor Rings, Let R be a ring and let A be an ideal of R. Since R is a group under addition and A is a normal subgroup of R, we may form the factor group, R/A 5 {r 1 A | r [ R}. The natural question at this point is: How may, we form a ring of this group of cosets? The addition is already taken care, of, and, by analogy with groups of cosets, we define the product of two, cosets of s 1 A and t 1 A as st 1 A. The next theorem shows that this definition works as long as A is an ideal, and not just a subring, of R., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 | Ideals and Factor Rings, , 269, , Theorem 14.2 Existence of Factor Rings, Let R be a ring and let A be a subring of R. The set of cosets {r 1 A |, r [ R} is a ring under the operations (s 1 A) 1 (t 1 A) 5 s 1 t 1 A, and (s 1 A)(t 1 A) 5 st 1 A if and only if A is an ideal of R., , PROOF We know that the set of cosets forms a group under addition., Once we know that multiplication is indeed a binary operation on the, cosets, it is trivial to check that the multiplication is associative and, that multiplication is distributive over addition. Hence, the proof boils, down to showing that multiplication is well-defined if and only if A is, an ideal of R. To do this, let us suppose that A is an ideal and let s 1 A 5, s9 1 A and t 1 A 5 t9 1 A. Then we must show that st 1 A 5 s9t9 1 A., Well, by definition, s 5 s9 1 a and t 5 t9 1 b, where a and b belong, to A. Then, st 5 (s9 1 a)(t9 1 b) 5 s9t9 1 at9 1 s9b 1 ab,, and so, st 1 A 5 s9t9 1 at9 1 s9b 1 ab 1 A 5 s9t9 1 A,, since A absorbs at9 1 s9b 1 ab. Thus, multiplication is well-defined, when A is an ideal., On the other hand, suppose that A is a subring of R that is not an, ideal of R. Then there exist elements a [ A and r [ R such that ar o A, or ra o A. For convenience, say ar o A. Consider the elements a 1 A 5, 0 1 A and r 1 A. Clearly, (a 1 A)(r 1 A) 5 ar 1 A but (0 1 A) ?, (r 1 A) 5 0 ? r 1 A 5 A. Since ar 1 A 2 A, the multiplication is not, well-defined and the set of cosets is not a ring., Let’s look at a few factor rings., EXAMPLE 8 Z/4Z 5 {0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z}. To see how, to add and multiply, consider 2 1 4Z and 3 1 4Z., (2 1 4Z) 1 (3 1 4Z) 5 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z,, (2 1 4Z)(3 1 4Z) 5 6 1 4Z 5 2 1 4 1 4Z 5 2 1 4Z., One can readily see that the two operations are essentially modulo 4, arithmetic., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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270, , Rings, , EXAMPLE 9 2Z/6Z 5 {0 1 6Z, 2 1 6Z, 4 1 6Z}. Here the operations are essentially modulo 6 arithmetic. For example, (4 1 6Z) 1, (4 1 6Z) 5 2 1 6Z and (4 1 6Z)(4 1 6Z) 5 4 1 6Z., Here is a noncommutative example of an ideal and factor ring., EXAMPLE 10 Let R 5 e c, , a1 a2, d ` ai [ Z f and let I be the, a3 a4, subset of R consisting of matrices with even entries. It is easy to, show that I is indeed an ideal of R (Exercise 21). Consider the factor, ring R/I. The interesting question about this ring is: What is its size?, r1 r2, d I 0 ri [{0, 1}f ., We claim R/I has 16 elements; in fact, R/I 5 e c, r3 r4, An example illustrates the typical situation. Which of the 16 elements, 7 8, 7, 8, 1 0, d I ? We l l , o b s e r ve t h a t c, d I c, d , is c, 5 3, 5 3, 1 1, 6 8, 1 0, c, d I c, d I, since an ideal absorbs its own elements., 4 4, 1 1, The general case is left to the reader (Exercise 23)., EXAMPLE 11 Consider the factor ring of the Gaussian integers, R 5 Z[i]/k2 2 il. What does this ring look like? Of course, the elements, of R have the form a 1 bi 1 k2 2 il, where a and b are integers, but the, important question is: What do the distinct cosets look like? The fact, that 2 2 i 1 k2 2 il 5 0 1 k2 2 il means that when dealing with coset, representatives, we may treat 2 2 i as equivalent to 0, so that 2 5 i. For, example, the coset 3 1 4i 1 k2 2 il 5 3 1 8 1 k2 2 il 5 11 1 k2 2 il., Similarly, all the elements of R can be written in the form a 1 k2 2 il,, where a is an integer. But we can further reduce the set of distinct coset, representatives by observing that when dealing with coset representatives, 2 5 i implies (by squaring both sides) that 4 5 21 or 5 5 0., Thus, the coset 3 1 4i 1 k2 2 il 5 11 1 k2 2 il 5 1 1 5 1 5 1 k2 2 il 5, 1 1 k2 2 il. In this way, we can show that every element of R is equal to, one of the following cosets: 0 1 k2 2 il, 1 1 k2 2 il, 2 1 k2 2 il, 3 1, k2 2 il, 4 1 k2 2 il. Is any further reduction possible? To demonstrate, that there is not, we will show that these five cosets are distinct. It suffices to show that 1 1 k2 2 il has additive order 5. Since 5(1 1 k2 2 il) 5, 5 1 k2 2 il 5 0 1 k2 2 il, 1 1 k2 2 il has order 1 or 5. If the order is, actually 1, then 1 1 k2 2 il 5 0 1 k2 2 il, so 1 [ k2 2 il. Thus, 1 5, (2 2 i) (a 1 bi) 5 2a 1 b 1 (2a 1 2b)i for some integers a and b. But, this equation implies that 1 5 2a 1 b and 0 5 2a 1 2b, and solving these, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 | Ideals and Factor Rings, , 271, , simultaneously yields b 5 1/5, which is a contradiction. It should be, clear that the ring R is essentially the same as the field Z5., EXAMPLE 12 Let R[x] denote the ring of polynomials with real coefficients and let kx2 1 1l denote the principal ideal generated by, x2 1 1; that is,, kx2 1 1l 5 {f(x)(x2 1 1) | f(x) [ R[x]}., Then, R[x]/kx2 1 1l 5 {g(x) 1 kx2 1 1l | g(x) [ R[x]}, 5 {ax 1 b 1 kx2 1 1l | a, b [ R}., To see this last equality, note that if g(x) is any member of R[x], then, we may write g(x) in the form q(x)(x2 1 1) 1 r(x), where q(x) is the, quotient and r(x) is the remainder upon dividing g(x) by x2 1 1. In, particular, r(x) 5 0 or the degree of r(x) is less than 2, so that r(x) 5, ax 1 b for some a and b in R. Thus,, g(x) 1 kx2 1 1l 5 q(x)(x2 1 1) 1 r(x) 1 kx2 1 1l, 5 r(x) 1 kx2 1 1l,, since the ideal kx2 1 1l absorbs the term q(x)(x2 1 1)., How is multiplication done? Since, x2 1 1 1 kx2 1 1l 5 0 1 kx2 1 1l,, one should think of x2 1 1 as 0 or, equivalently, as x2 5 21. So, for, example,, (x 1 3 1 kx2 1 1l) ? (2x 1 5 1 kx2 1 1l), 5 2x2 1 11x 1 15 1 kx2 1 1l 5 11x 1 13 1 kx2 1 1l., In view of the fact that the elements of this ring have the form ax 1, b 1 kx2 1 1l, where x2 1 kx2 1 1l 5 21 1 kx2 1 1l, it is perhaps not, surprising that this ring turns out to be algebraically the same ring as, the ring of complex numbers. This observation was first made by Cauchy in 1847., Examples 11 and 12 illustrate one of the most important applications of factor rings—the construction of rings with highly desirable, properties. In particular, we shall show how one may use factor rings to, construct integral domains and fields., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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272, , Rings, , Prime Ideals and Maximal Ideals, Definition Prime Ideal, Maximal Ideal, , A prime ideal A of a commutative ring R is a proper ideal of R such, that a, b [ R and ab [ A imply a [ A or b [ A. A maximal ideal of a, commutative ring R is a proper ideal of R such that, whenever B is an, ideal of R and A # B # R, then B 5 A or B 5 R., , So, the only ideal that properly contains a maximal ideal is the entire ring. The motivation for the definition of a prime ideal comes from, the integers., EXAMPLE 13 Let n be an integer greater than 1. Then, in the ring of, integers, the ideal nZ is prime if and only if n is prime (Exercise 9)., ({0} is also a prime ideal of Z.), EXAMPLE 14 The lattice of ideals of Z36 (Figure 14.1) shows that, only k2l and k3l are maximal ideals., EXAMPLE 15 The ideal kx2 1 1l is maximal in R[x]. To see this,, assume that A is an ideal of R[x] that properly contains kx2 1 1l. We will, prove that A 5 R[x] by showing that A contains some nonzero real, number c. [This is the constant polynomial h(x) 5 c for all x.] Then 1 5, (1/c)c [ A and therefore, by Exercise 15, A 5 R[x]. To this end, let, f(x) [ A, but f(x) o kx2 1 1l. Then, f(x) 5 q(x)(x2 1 1) 1 r(x),, where r(x) 2 0 and the degree of r(x) is less than 2. It follows that, r(x) 5 ax 1 b, where a and b are not both 0, and, ax 1 b 5 r(x) 5 f(x) 2 q(x)(x2 1 1) [ A., Z 36, , <2>, , <4>, , <3>, , <6>, , <9>, , <18>, , <12 >, , <0>, , Figure 14.1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 | Ideals and Factor Rings, , 273, , Thus,, a2x2 2 b2 5 (ax 1 b)(ax 2 b) [ A, , and, , a2(x2 1 1) [ A., , So,, 0 2 a2 1 b2 5 (a2x2 1 a2) 2 (a2x2 2 b2) [ A., EXAMPLE 16 The ideal kx2 1 1l is not prime in Z2[x], since it contains (x 1 1)2 5 x2 1 2x 1 1 5 x2 1 1 but does not contain x 1 1., The next two theorems are useful for determining whether a particular ideal is prime or maximal., Theorem 14.3 R/A Is an Integral Domain If and Only If A Is Prime, Let R be a commutative ring with unity and let A be an ideal of R., Then R/A is an integral domain if and only if A is prime., , PROOF Suppose that R/A is an integral domain and ab [ A. Then, (a 1 A)(b 1 A) 5 ab 1 A 5 A, the zero element of the ring R/A. So,, either a 1 A 5 A or b 1 A 5 A; that is, either a [ A or b [ A. Hence,, A is prime., To prove the other half of the theorem, we first observe that R/A is a, commutative ring with unity for any proper ideal A. Thus, our task is, simply to show that when A is prime, R/A has no zero-divisors. So, suppose that A is prime and (a 1 A)(b 1 A) 5 0 1 A 5 A. Then ab [ A, and, therefore, a [ A or b [ A. Thus, one of a 1 A or b 1 A is the zero, coset in R/A., For maximal ideals, we can do even better., Theorem 14.4 R/A Is a Field If and Only If A Is Maximal, Let R be a commutative ring with unity and let A be an ideal of R., Then R/A is a field if and only if A is maximal., , PROOF Suppose that R/A is a field and B is an ideal of R that properly, contains A. Let b [ B but b o A. Then b 1 A is a nonzero element, of R/A and, therefore, there exists an element c 1 A such that, (b 1 A) ? (c 1 A) 5 1 1 A, the multiplicative identity of R/A. Since, b [ B, we have bc [ B. Because, 1 1 A 5 (b 1 A)(c 1 A) 5 bc 1 A,, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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274, , Rings, , we have 1 2 bc [ A , B. So, 1 5 (1 2 bc) 1 bc [ B. By Exercise 15,, B 5 R. This proves that A is maximal., Now suppose that A is maximal and let b [ R but b o A. It suffices, to show that b 1 A has a multiplicative inverse. (All other properties, for a field follow trivially.) Consider B 5 {br 1 a | r [ R, a [ A}. This, is an ideal of R that properly contains A (Exercise 25). Since A is maximal, we must have B 5 R. Thus, 1 [ B, say, 1 5 bc 1 a9, where a9 [ A., Then, 1 1 A 5 bc 1 a9 1 A 5 bc 1 A 5 (b 1 A)(c 1 A)., When a commutative ring has a unity, it follows from Theorems, 14.3 and 14.4 that a maximal ideal is a prime ideal. The next example, shows that a prime ideal need not be maximal., EXAMPLE 17 The ideal kxl is a prime ideal in Z[x] but not a maximal ideal in Z[x]. To verify this, we begin with the observation that, kxl 5 {f(x) [ Z[x] | f(0) 5 0} (see Exercise 29). Thus, if g(x)h(x) [ kxl,, then g(0)h(0) 5 0. And since g(0) and h(0) are integers, we have g(0) 5 0, or h(0) 5 0., To see that kxl is not maximal, we simply note that kxl , kx, 2l ,, Z[x] (see Exercise 37)., , Exercises, Problems worthy of attack, prove their worth by hitting back., piet hein, “Problems,” Grooks*, , 1. Verify that the set defined in Example 3 is an ideal., 2. Verify that the set A in Example 4 is an ideal and that A 5 kxl., 3. Verify that the set I in Example 5 is an ideal and that if J is any, ideal of R that contains a1, a2, . . . , an, then I # J. (Hence, ka1,, a2, . . . , anl is the smallest ideal of R that contains a1, a2, . . . , an .), 4. Find a subring of Z % Z that is not an ideal of Z % Z., 5. Let S 5 {a 1 bi | a, b [ Z, b is even}. Show that S is a subring of, Z[i], but not an ideal of Z[i]., 6. Find all maximal ideals in, b. Z10., c. Z12., d. Zn., a. Z8., 7. Let a belong to a commutative ring R. Show that aR 5 {ar | r [ R} is, an ideal of R. If R is the ring of even integers, list the elements of 4R., *Piet Hein, “Problems,” Grooks (1966) Copyright © Piet Hein Grooks. Reprinted with kind, permission from Piet Hein a/s, DK-5500 Middelfart, Denmark., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 | Ideals and Factor Rings, , 275, , 8. Prove that the intersection of any set of ideals of a ring is an ideal., 9. If n is an integer greater than 1, show that knl 5 nZ is a prime ideal, of Z if and only if n is prime. (This exercise is referred to in this, chapter.), 10. If A and B are ideals of a ring, show that the sum of A and B, A 1 B 5, {a 1 b | a [ A, b [ B}, is an ideal., 11. In the ring of integers, find a positive integer a such that, a. kal 5 k2l 1 k3l., b. kal 5 k6l 1 k8l., c. kal 5 kml 1 knl., 12. If A and B are ideals of a ring, show that the product of A and B,, AB 5 {a1b1 1 a2b2 1 ? ? ? 1 anbn | ai [ A, bi [ B, n a positive, integer}, is an ideal., 13. Find a positive integer a such that, a. kal 5 k3lk4l., b. kal 5 k6lk8l., c. kal 5 kmlknl., 14. Let A and B be ideals of a ring. Prove that AB # A > B., 15. If A is an ideal of a ring R and 1 belongs to A, prove that A 5 R., (This exercise is referred to in this chapter.), 16. If A and B are ideals of a commutative ring R with unity and A 1 B 5 R,, show that A > B 5 AB., 17. If an ideal I of a ring R contains a unit, show that I 5 R., 18. Suppose that in the ring Z, the ideal k35l is a proper ideal of J and J, is a proper ideal of I. What are the possibilities for J? What are the, possibilities for I?, 19. Give an example of a ring that has exactly two maximal ideals., 20. Suppose that R is a commutative ring and |R| 5 30. If I is an ideal, of R and |R| 5 10, prove that I is a maximal ideal., 21. Let R and I be as described in Example 10. Prove that I is an ideal, of R., 22. Let I 5 k2l. Prove that I[x] is not a maximal ideal of Z[x] even, though I is a maximal ideal of Z., 23. Verify the claim made in Example 10 about the size of R/I., 24. Give an example of a commutative ring that has a maximal ideal, that is not a prime ideal., 25. Show that the set B in the latter half of the proof of Theorem 14.4, is an ideal of R. (This exercise is referred to in this chapter.), 26. If R is a commutative ring with unity and A is a proper ideal of R,, show that R/A is a commutative ring with unity., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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276, , Rings, , 27. Prove that the only ideals of a field F are {0} and F itself., 28. Show that R[x]/kx2 1 1l is a field., 29. In Z[x], the ring of polynomials with integer coefficients, let I 5, {f (x) [ Z[x] | f (0) 5 0}. Prove that I 5 kxl. (This exercise is referred to in this chapter and in Chapter 15.), 30. Show that A 5 {(3x, y) | x, y [ Z} is a maximal ideal of Z % Z., Generalize. What happens if 3x is replaced by 4x? Generalize., 31. Let R be the ring of continuous functions from R to R. Show that, A 5 {f [ R | f (0) 5 0} is a maximal ideal of R., 32. Let R 5 Z8 % Z30. Find all maximal ideals of R, and for each maximal ideal I, identify the size of the field R/I., 33. How many elements are in Z[i]/k3 1 il? Give reasons for your, answer., 34. In Z[x], the ring of polynomials with integer coefficients, let I 5, { f (x) [ Z[x] | f (0) 5 0}. Prove that I is not a maximal ideal., 35. In Z % Z, let I 5 {(a, 0) | a [ Z}. Show that I is a prime ideal but, not a maximal ideal., 36. Let R be a ring and let I be an ideal of R. Prove that the factor ring, R/I is commutative if and only if rs 2 sr [ I for all r and s in R., 37. In Z[x], let I 5 { f(x) [ Z[x] | f (0) is an even integer}. Prove that, I 5 kx, 2l. Is I a prime ideal of Z[x]? Is I a maximal ideal? How, many elements does Z[x]/I have? (This exercise is referred to in, this chapter.), 38. Prove that I 5 k2 1 2il is not a prime ideal of Z[i]. How many, elements are in Z[i]/I? What is the characteristic of Z[i]/I?, 39. In Z5[x], let I 5 kx2 1 x 1 2l. Find the multiplicative inverse of 2x 1, 3 1 I in Z5[x]/I., 40. Let R be a ring and let p be a fixed prime. Show that Ip 5 {r [ R |, additive order of r is a power of p} is an ideal of R., 41. An integral domain D is called a principal ideal domain if every, ideal of D has the form kal 5 {ad | d [ D} for some a in D. Show, that Z is a principal ideal domain. (This exercise is referred to in, Chapter 18.), 42. Let R 5 e c, , a b, r s, d ` a, b, d [ Z f and S e c, d ` r, s, t [ Z, s, 0 d, 0 t, , is even f . If S is an ideal of R, what can you say about r and t?, 43. If R and S are principal ideal domains, prove that R % S is a principal ideal ring. (See Exercise 41 for the definition.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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14 | Ideals and Factor Rings, , 277, , 44. Let a and b belong to a commutative ring R. Prove that {x [ R |, ax [ bR} is an ideal., 45. Let R be a commutative ring and let A be any subset of R. Show, that the annihilator of A, Ann(A) 5 {r [ R | ra 5 0 for all a in A},, is an ideal., 46. Let R be a commutative ring and let A be any ideal of R. Show that, the nil radical of A, N(A) 5 {r [ R | r n [ A for some positive integer n (n depends on r)}, is an ideal of R. [N(k0l) is called the nil, radical of R.], 47. Let R 5 Z27. Find, a. N(k0l)., b. N(k3l)., c. N(k9l)., 48. Let R 5 Z36. Find, a. N(k0l)., b. N(k4l)., c. N(k6l)., 49. Let R be a commutative ring. Show that R/N(k0l) has no nonzero, nilpotent elements., 50. Let A be an ideal of a commutative ring. Prove that N(N(A)) 5 N(A)., 51. Let Z2[x] be the ring of all polynomials with coefficients in Z2 (that, is, coefficients are 0 or 1, and addition and multiplication of coefficients are done modulo 2). Show that Z2[x]/kx2 1 x 1 1l is a field., 52. List the elements of the field given in Exercise 51, and make an addition and multiplication table for the field., 53. Show that Z 3[x]/kx2 1 x 1 1l is not a field., 54. Let R be a commutative ring without unity, and let a [ R. Describe, the smallest ideal I of R that contains a (that is, if J is any ideal that, contains a, then I # J)., 55. Let R be the ring of continuous functions from R to R. Let A 5, { f [ R | f (0) is an even integer}. Show that A is a subring of R,, but not an ideal of R., 56. Show that Z[i]/k1 2 il is a field. How many elements does this, field have?, 57. If R is a principal ideal domain and I is an ideal of R, prove that, every ideal of R/I is principal (see Exercise 41)., 58. How many elements are in Z5[i]/k1 1 il?, 59. Let R be a commutative ring with unity that has the property that, a2 5 a for all a in R. Let I be a prime ideal in R. Show that |R/I| 5 2., 60. Let R be a commutative ring with unity, and let I be a proper ideal, with the property that every element of R that is not in I is a unit of R., Prove that I is the unique maximal ideal of R., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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278, , Rings, , 61. Let I0 5 { f(x) [ Z[x] | f(0) 5 0}. For any positive integer n, show, that there exists a sequence of strictly increasing ideals such that, I0 , I1 , I2 , ? ? ? , In , Z[x]., 62. Let R 5 {(a1, a2, a3, . . .)}, where each ai [ Z. Let I 5 {(a1, a2,, a3, . . . )}, where only a finite number of terms are nonzero. Prove, that I is not a principal ideal of R., 63. Let R be a commutative ring with unity and let a, b [ R. Show that, ka, bl, the smallest ideal of R containing a and b, is I 5 {ra 1 sb |, r, s [ R}. That is, show that I contains a and b and that any ideal, that contains a and b also contains I., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Richard Dedekind, Richard Dedekind was not only, a mathematician, but one of the wholly, great in the history of mathematics, now, and in the past, the last hero of a great, epoch, the last pupil of Gauss, for four, decades himself a classic, from whose, works not only we, but our teachers and, the teachers of our teachers, have drawn., edmund landau,, Commemorative Address, to the Royal Society of Göttingen, , This stamp was issued by East Germany, in 1981 to commemorate the 150th, anniversary of Dedekind’s birth. Notice, that it features the representation of an, ideal as the product of powers of prime, ideals., , Richard Dedekind was born on October 6,, 1831, in Brunswick, Germany, the birthplace of Gauss. Dedekind was the youngest, of four children of a law professor. His early, interests were in chemistry and physics, but, he obtained a doctor’s degree in mathematics at the age of 21 under Gauss at the, University of Göttingen. Dedekind continued his studies at Göttingen for a few years,, and in 1854 he began to lecture there., , Dedekind spent the years 1858–1862 as a, professor in Zürich. Then he accepted a position at an institute in Brunswick where he, had once been a student. Although this, school was less than university level,, Dedekind remained there for the next, 50 years. He died in Brunswick in 1916., During his career, Dedekind made numerous fundamental contributions to mathematics. His treatment of irrational numbers,, “Dedekind cuts,” put analysis on a firm,, logical foundation. His work on unique, factorization led to the modern theory of, algebraic numbers. He was a pioneer in the, theory of rings and fields. The notion of, ideals as well as the term itself are attributed, to Dedekind. Mathematics historian Morris, Kline has called him “the effective founder, of abstract algebra.”, To fi n d m o r e i n f o r m a t i o n a b o u t, Dedekind, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 279, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Emmy Noether, . . . she discovered methods which have, proved of enormous importance in the, development of the present-day younger, generation of mathematicians., , Emmy Noether was born on March 23,, 1882, in Germany. When she entered the, University of Erlangen, she was one of, only two women among the 1000 students., Noether completed her doctorate in 1907., In 1916, Noether went to Göttingen and,, under the influence of David Hilbert and, Felix Klein, became interested in general, relativity. While there, she made a major, contribution to physics with her theorem, that whenever there is a symmetry in nature,, there is also a conservation law, and vice, versa. Hilbert tried unsuccessfully to obtain, a faculty appointment at Göttingen for, Noether, saying, “I do not see that the sex of, the candidate is an argument against her admission as Privatdozent. After all, we are a, university and not a bathing establishment.”, It was not until she was 38 that Noether’s, true genius revealed itself. Over the next, , The Granger Collection, New York, , albert einstein, The New York Time, , 13 years, she used an axiomatic method to, develop a general theory of ideals and noncommutative algebras. With this abstract, theory, Noether was able to weld together, many important concepts. Her approach was, even more important than the individual, results. Hermann Weyl said of Noether,, “She originated above all a new and epochmaking style of thinking in algebra.”, With the rise of Hitler in 1933, Noether,, a Jew, fled to the United States and took a, position at Bryn Mawr College. She died, suddenly on April 14, 1935, following an, operation., To find more information about Noether,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 280, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 12– 14, , 281, , Supplementary Exercises for Chapters 12–14, If at first you do succeed—try to hide your astonishment., harry f. banks, , True/false questions for Chapters 12–14 are available on the Web at:, http://www.d.umn.edu/~jgallian/TF, 1. Find all idempotents in Z10, Z20, and Z30. (Recall that a is an idempotent if a2 5 a.), 2. If m and n are relatively prime integers greater than 1, prove that, Zmn has at least two idempotents besides 0 and 1., 3. Suppose that R is a ring in which a2 5 0 implies a 5 0. Show that, R has no nonzero nilpotent elements. (Recall that b is nilpotent if, bn 5 0 for some positive integer n.), 4. Let R be a commutative ring with more than one element. Prove, that if for every nonzero element a of R we have aR 5 R, then R is, a field., 5. Let A, B, and C be ideals of a ring R. If AB # C and C is a prime, ideal of R, show that A # C or B # C. (Compare this with Euclid’s, Lemma in Chapter 0.), 6. Show, by example, that the intersection of two prime ideals need, not be a prime ideal., 7. Let R denote the ring of real numbers. Determine all ideals of R % R., What happens if R is replaced by any field F?, 8. Determine all factor rings of Z., 9. Suppose that n is a square-free positive integer (that is, n is not, divisible by the square of any prime). Prove that Zn has no nonzero, nilpotent elements., 10. Let R be a commutative ring with unity. Suppose that a is a unit, and b is nilpotent. Show that a 1 b is a unit. (Hint: See Exercise 29, in Chapter 12.), 11. Let A, B, and C be subrings of a ring R. If A # B < C, show that, A # B or A # C., 12. For any element a in a ring R, define kal to be the smallest ideal of, R that contains a. If R is a commutative ring with unity, show that, kal 5 aR 5 {ar | r [ R}. Show, by example, that if R is commutative but does not have a unity, then kal and aR may be different., 13. Let R be a ring with unity. Show that kal 5 {s1at1 1 s2at2 1 ? ? ? 1, snatn | si, ti [ R and n is a positive integer}., 14. Show that Zn[x] has characteristic n., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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282, , Rings, , 15. Let A and B be ideals of a ring R. If A > B 5 {0}, show that ab 5 0, when a [ A and b [ B., 16. Show that the direct sum of two integral domains is not an integral, domain., 17. Consider the ring R 5 {0, 2, 4, 6, 8, 10} under addition and multiplication modulo 12. What is the characteristic of R?, 18. What is the characteristic of Zm % Zn? Generalize., 19. Let R be a commutative ring with unity. Suppose that the only ideals of R are {0} and R. Show that R is a field., 20. Suppose that I is an ideal of J and that J is an ideal of R. Prove that, if I has a unity, then I is an ideal of R. (Be careful not to assume that, the unity of I is the unity of R. It need not be—see Exercise 2 in, Chapter 12.), 21. Show that in the ring Z[x]/K2x 1 1L, the element x 1 K2x 1 1L, is a unit., 22. Let a [ Z. Show that KaL is not a maximal ideal in Z[x]., 23. Recall that an idempotent b in a ring is an element with the property, that b2 5 b. Find a nontrivial idempotent (that is, not 0 and not 1), in Q[x]/kx4 1 x2l., 24. In a principal ideal domain, show that every nontrivial prime ideal, is a maximal ideal., 25. Find an example of a commutative ring R with unity such that a,, b [ R, a 2 b, an 5 bn, and am 5 bm, where n and m are positive integers that are relatively prime. (Compare with Exercise 39, part b, in, Chapter 13.), 3, 26. Let Q(2, 2 ) denote the smallest subfield of R that contains Q and, 3, 3, 3, ., [That, is, Q(2, 22, 2 ) is the subfield with the property that Q(2, 2), 3, 3, contains Q and22 and if F is any subfield containing Q and22 ,, 3, 3, then F contains Q(2, 2 ).] Describe the elements of Q(2, 2 )., 27. Let R be an integral domain with nonzero characteristic. If A is a, proper ideal of R, show that R/A has the same characteristic as R., 28. Let F be a field of order pn. Determine the group isomorphism, class of F under the operation addition., 29. If R is a finite commutative ring with unity, prove that every prime, ideal of R is a maximal ideal of R., 30. Let R be a noncommutative ring and let C(R) be the center of R, (see Exercise 19 in Chapter 12). Prove that the additive group of, R/C(R) is not cyclic., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 12– 14, , 283, , 31. Let, R ec, , a b, d ` a, b, c, d [ Z2 f, c d, , with ordinary matrix addition and multiplication modulo 2. Show that, A ec, , 32., 33., 34., , 35., 36., 37., , 38., 39., 40., 41., 42., , 43., 44., 45., , 1 0, dr ` r [ Rf, 0 0, , is not an ideal of R. (Hence, in Exercise 7 in Chapter 14, the commutativity assumption is necessary.), If R is an integral domain and A is a proper ideal of R, must R/A be, an integral domain?, Let A 5 {a 1 bi | a, b [ Z, a mod 2 5 b mod 2}. How many elements does Z[i]/A have? Show that A is a maximal ideal of Z[i]., Suppose that R is a commutative ring with unity such that for each, a in R there is a positive integer n greater than 1 (n depends on a), such that an 5 a. Prove that every prime ideal of R is a maximal, ideal of R., State a “finite subfield test”; that is, state conditions that guarantee, that a finite subset of a field is a subfield., Let F be a finite field with more than two elements. Prove that the, sum of all of the elements of F is 0., Show that if there are nonzero elements a and b in Zn such that a2 1, b2 5 0, then the ring Zn[i] 5 {x 1 yi | x, y [ Zn} has zero-divisors., Use this fact to find a zero-divisor in Z13[i]., Suppose that R is a ring with no zero-divisors and that R contains a, nonzero element b such that b2 5 b. Show that b is a unity for R., Find the characteristic of Z[i]/k2 1 il., Show that the characteristic of Z[i]/ka 1 bil divides a2 1 b2., Show that 4x2 1 6x 1 3 is a unit in Z8[x]., For any commutative ring R, R[x, y] is the ring of polynomials in x, and y with coefficients in R (that is, R[x, y] consists of all finite sums, of terms of the form axiyj, where a [ R and i and j are nonnegative, integers). (For example, x4 2 3x2y 2 y3 [ Z[x, y].) Prove that kx, yl, is a prime ideal in Z[x, y] but not a maximal ideal in Z[x, y]., Prove that kx, yl is a maximal ideal in Z5[x, y]., Prove that k2, x, yl is a maximal ideal in Z[x, y]., Let R and S be rings. Prove that (a, b) is nilpotent in R % S if and, only if both a and b are nilpotent., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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284, , Rings, , 46. Let R and S be commutative rings. Prove that (a, b) is a zero-divisor, in R % S if and only if a or b is a zero-divisor or exactly one of a or, b is 0., 47. Determine all idempotents in Zp k, where p is a prime., 48. Let R be a commutative ring with unity 1. Show that a is an idempotent if and only if there exists an element b in R such that ab 5 0, and a 1 b 5 1., 49. Let Zn[22] 5 {a 1 b22 | a, b [ Zn}. Define addition and multiplication as in Z[22], except that modulo n arithmetic is used to, combine the coefficients. Show that Z3[22] is a field but Z7[22], is not., 50. Let p be a prime. Prove that every zero-divisor in Zp n is a nilpotent, element., 51. If x is a nilpotent element in a commutative ring R, prove that rx is, nilpotent for all r in R., 52. List the distinct elements in the ring Z[x]/K3, x2 1 1L. Show that this, ring is a field., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 Ring Homomorphisms, If there is one central idea which is common to, all aspects of modern algebra it is the notion of homomorphism., i. n. herstein, Topics in Algebra, , Definition and Examples, In our work with groups, we saw that one way to discover information, about a group is to examine its interaction with other groups by way of, homomorphisms. It should not be surprising to learn that this concept, extends to rings with equally profitable results., Just as a group homomorphism preserves the group operation, a ring, homomorphism preserves the ring operations., Definitions Ring Homomorphism, Ring Isomorphism, , A ring homomorphism f from a ring R to a ring S is a mapping from, R to S that preserves the two ring operations; that is, for all a, b in R,, , f(a 1 b) 5 f(a) 1 f(b), , and, , f(ab) 5 f(a)f(b)., , A ring homomorphism that is both one-to-one and onto is called a, ring isomorphism., , As is the case for groups, in the preceding definition the operations, on the left of the equal signs are those of R, whereas the operations on, the right of the equal signs are those of S., Again as with group theory, the roles of isomorphisms and homomorphisms are entirely distinct. An isomorphism is used to show that two, rings are algebraically identical; a homomorphism is used to simplify a, ring while retaining certain of its features., A schematic representation of a ring homomorphism is given in, Figure 15.1. The dashed arrows indicate the results of performing the, ring operations., The following examples illustrate ring homomorphisms. The reader, should supply the missing details., 285, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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286, , Rings, , φ, , R, , S, , φ (b), , b, , φ, , a, , φ, , a.b, a1b, , φ (a), , φ, , φ (a) . φ (b), φ (a) 1 φ (b), , Figure 15.1, , EXAMPLE 1 For any positive integer n, the mapping k S k mod n is, a ring homomorphism from Z onto Zn (see Exercise 9 in Chapter 0)., This mapping is called the natural homomorphism from Z to Zn., EXAMPLE 2 The mapping a 1 bi S a 2 bi is a ring isomorphism, from the complex numbers onto the complex numbers (see Exercise 35, in Chapter 6)., EXAMPLE 3 Let R[x] denote the ring of all polynomials with real, coefficients. The mapping f (x) S f(1) is a ring homomorphism from, R[x] onto R., EXAMPLE 4 The correspondence f: x S 5x from Z4 to Z10, is a ring homomorphism. Although showing that f(x 1 y) 5, f(x) 1 f(y) appears to be accomplished by the simple statement that, 5(x 1 y) 5 5x 1 5y, we must bear in mind that the addition on the left is, done modulo 4, whereas the addition on the right and the multiplication, on both sides are done modulo 10. An analogous difficulty arises in showing that f preserves multiplication. So, to verify that f preserves both operations, we write x 1 y 5 4q1 1 r1 and xy 5 4q2 1 r2, where 0 # r1 , 4, and 0 # r2 , 4. Then f(x 1 y) 5 f(r1) 5 5r1 5 5(x 1 y 2 4q1) 5 5x 1, 5y 2 20q1 5 5x 1 5y 5 f(x) 1 f(y) in Z10. Similarly, using the fact that, 5 ? 5 5 5 in Z10, we have f(xy) 5 f(r2) 5 5r2 5 5(xy 2 4q2) 5 5xy 2, 20q2 5 (5 ? 5)xy 5 5x5y 5 f(x)f(y) in Z10., EXAMPLE 5 We determine all ring homomorphisms from Z12 to Z30., By Example 10 in Chapter 10, the only group homomorphisms from Z12, to Z30 are x S ax, where a 5 0, 15, 10, 20, 5, or 25. But, since 1 ? 1 5 1, in Z12, we must have a ? a 5 a in Z30. This requirement rules out 20 and 5, as possibilities for a. Finally, simple calculations show that each of the, remaining four choices does yield a ring homomorphism., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 | Ring Homomorphisms, , 287, , EXAMPLE 6 Let R be a commutative ring of characteristic 2. Then, the mapping a S a2 is a ring homomorphism from R to R., EXAMPLE 7 Although 2Z, the group of even integers under addition, is group-isomorphic to the group Z under addition, the ring 2Z is, not ring-isomorphic to the ring Z. (Quick! What does Z have that 2Z, doesn’t?), Our next two examples are applications to number theory of the natural homomorphism given in Example 1., EXAMPLE 8 Test for Divisibility by 9, An integer n with decimal representation akak21 ? ? ? a0 is divisible by 9, if and only if ak 1 ak21 1 ? ? ? 1 a0 is divisible by 9. To verify this, observe that n 5 ak10k 1 ak2110k21 1 ? ? ? 1 a0. Then, letting a denote, the natural homomorphism from Z to Z9 [in particular, a(10) 5 1], we, note that n is divisible by 9 if and only if, 0 5 a(n) 5 a(ak)(a(10))k 1 a(ak21)(a(10))k21 1 ? ? ? 1 a(a0), 5 a(ak) 1 a(ak21) 1 ? ? ? 1 a(a0), 5 a(ak 1 ak21 1 ? ? ? 1 a0)., But a(ak 1 ak21 1 ? ? ? 1 a0) 5 0 is equivalent to ak 1 ak21 1 ? ? ? 1, a0 being divisible by 9., EXAMPLE 9 Theorem of Gersonides, Among the most important unsolved problems in number theory is the, so-called “abc conjecture.” This conjecture is a natural generalization, of a theorem first proved in the fourteenth century by the Rabbi, Gersonides. Gersonides proved that the only pairs of positive integers, that are powers of 2 and powers of 3 which differ by 1 are 1, 2; 2, 3; 3,, 4; and 8, 9. That is, these four pairs are the only solutions to the equations 2m 5 3n 6 1. To verify that this is so for 2m 5 3n 1 1, observe that, for all n we have 3n mod 8 5 3 or 1. Thus, 3n 1 1 mod 8 5 4 or 2. On, the other hand, for m . 2, we have 2m mod 8 5 0. To handle the case, where 2m 5 3n 2 1, we first note that for all n, 3n mod 16 5 3, 9, 11, or, 1, depending on the value of n mod 4. Thus, (3n 2 1) mod 16 5 2, 8, 10,, or 0. Since 2m mod 16 5 0 for m $ 4, we have ruled out the cases where, n mod 4 5 1, 2, or 3. Because 34k mod 5 5 (34)k mod 5 5 1k mod 5 5, 1, we know that (34k 2 1) mod 5 5 0. But the only values for 2m mod 5, are 2, 4, 3, and 1. This contradiction completes the proof., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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288, , Rings, , Properties of Ring Homomorphisms, Theorem 15.1 Properties of Ring Homomorphisms, Let f be a ring homomorphism from a ring R to a ring S. Let A be a, subring of R and let B be an ideal of S., 1. For any r [ R and any positive integer n, f(nr) 5 nf(r) and, f(r n) 5 (f(r))n., 2. f(A) 5 {f(a) | a [ A} is a subring of S., 3. If A is an ideal and f is onto S, then f(A) is an ideal., 4. f21(B) 5 {r [ R | f(r) [ B} is an ideal of R., 5. If R is commutative, then f(R) is commutative., 6. If R has a unity 1, S 2 {0}, and f is onto, then f(1) is the unity, of S., 7. f is an isomorphism if and only if f is onto and Ker f 5, {r [ R | f(r) 5 0} 5 {0}., 8. If f is an isomorphism from R onto S, then f21 is an, isomorphism from S onto R., , PROOF The proofs of these properties are similar to those given in, Theorems 10.1 and 10.2 and are left as exercises (Exercise 1)., The student should learn the various properties of Theorem 15.1, in words in addition to the symbols. Property 2 says that the homomorphic image of a subring is a subring. Property 4 says that the pullback, of an ideal is an ideal, and so on., The next three theorems parallel results we had for groups. The, proofs are nearly identical to their group theory counterparts and are, left as exercises (Exercises 2, 3, and 4)., Theorem 15.2 Kernels Are Ideals, Let f be a ring homomorphism from a ring R to a ring S. Then Ker f, 5 {r [ R | f(r) 5 0} is an ideal of R., , Theorem 15.3 First Isomorphism Theorem for Rings, Let f be a ring homomorphism from R to S. Then the mapping from, R/Ker f to f(R), given by r 1 Ker f S f(r), is an isomorphism. In, symbols, R/Ker f < f(R)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 | Ring Homomorphisms, , 289, , Theorem 15.4 Ideals Are Kernels, Every ideal of a ring R is the kernel of a ring homomorphism of R., In particular, an ideal A is the kernel of the mapping r S r 1 A, from R to R/A., , The homomorphism from R to R/A given in Theorem 15.4 is called, the natural homomorphism from R to R/A. Theorem 15.3 is often referred to as the Fundamental Theorem of Ring Homomorphisms., In Example 17 in Chapter 14 we gave a direct proof that kxl is a, prime ideal of Z[x] but not a maximal ideal. In the following example, we illustrate a better way to do this kind of problem., EXAMPLE 10 Since the mapping f from Z[x] onto Z given by, f( f(x)) 5 f(0) is a ring homomorphism with Ker f 5 kxl (see Exercise 29, in Chapter 14), we have, by Theorem 15.3, Z[x]/kxl < Z. And because, Z is an integral domain but not a field, we know by Theorems 14.3 and, 14.4 that the ideal kxl is prime but not maximal in Z[x]., Theorem 15.5 Homomorphism from Z to a Ring with Unity, Let R be a ring with unity 1. The mapping f: Z S R given by n S n ? 1, is a ring homomorphism., , PROOF Since the multiplicative group property am+n 5 aman translates to, (m 1 n)a 5 ma 1 na when the operation is addition, we have f(m 1 n) 5, (m 1 n) ? 1 5 m ? 1 1 n ? 1. So, f preserves addition., That f also preserves multiplication follows from Exercise 15 in, Chapter 12, which says that (m ? a)(n ? b) 5 (mn) ? (ab) for all integers, m and n. Thus, f(mn) 5 (mn) ? 1 5 (mn) ? ((1)(1)) 5 (m ? 1)(n ? 1) 5, f(m)f(n). So, f preserves multiplication as well., Corollary 1 A Ring with Unity Contains Zn or Z, If R is a ring with unity and the characteristic of R is n . 0, then, R contains a subring isomorphic to Zn. If the characteristic of R is 0,, then R contains a subring isomorphic to Z., , PROOF Let 1 be the unity of R and let S 5 {k ? 1 | k [ Z}. Theorem 15.5, shows that the mapping f from Z to S given by f(k) 5 k ? 1 is a homomorphism, and by the First Isomorphism Theorem for rings, we have, Z/Ker f < S. But, clearly, Ker f 5 knl, where n is the additive order of 1, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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290, , Rings, , and, by Theorem 13.3, n is also the characteristic of R. So, when R, has characteristic n, S < Z/knl < Zn. When R has characteristic 0, S <, Z/k0l < Z., Corollary 2 Zm Is a Homomorphic Image of Z, For any positive integer m, the mapping of f: Z S Zm given by x S, x mod m is a ring homomorphism., , PROOF This follows directly from the statement of Theorem 15.5,, since in the ring Zm, the integer x mod m is x ? 1. (For example, in Z3, if, x 5 5, we have 5 ? 1 5 1 1 1 1 1 1 1 1 1 5 2.), Corollary 3 A Field Contains Zp or Q (Steinitz, 1910), If F is a field of characteristic p, then F contains a subfield, isomorphic to Zp. If F is a field of characteristic 0, then F contains, a subfield isomorphic to the rational numbers., , PROOF By Corollary 1, F contains a subring isomorphic to Zp if F has, characteristic p, and F has a subring S isomorphic to Z if F has characteristic 0. In the latter case, let, T 5 {ab21 | a, b [ S, b 2 0}., Then T is isomorphic to the rationals (Exercise 63)., Since the intersection of all subfields of a field is itself a subfield, (Exercise 11), every field has a smallest subfield (that is, a subfield, that is contained in every subfield). This subfield is called the prime, subfield of the field. It follows from Corollary 3 that the prime, subfield of a field of characteristic p is isomorphic to Zp, whereas the, prime subfield of a field of characteristic 0 is isomorphic to Q. (See, Exercise 67.), , The Field of Quotients, Although the integral domain Z is not a field, it is at least contained in a, field—the field of rational numbers. And notice that the field of rational, numbers is nothing more than quotients of integers. Can we mimic the, construction of the rationals from the integers for other integral domains? Yes. The field constructed in Theorem 15.6 is called the field of, quotients of D. Throughout the proof of Theorem 15.6, you should keep, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 | Ring Homomorphisms, , 291, , in mind that we are using the construction of the rationals from the integers as a model for our construction of the field of quotients of D., Theorem 15.6 Field of Quotients, Let D be an integral domain. Then there exists a field F (called the, field of quotients of D) that contains a subring isomorphic to D., , PROOF Let S 5 {(a, b) | a, b [ D, b 2 0}. We define an equivalence, relation on S by (a, b) ; (c, d ) if ad 5 bc (compare with Example 17, in Chapter 0). Now, let F be the set of equivalence classes of S under, the relation ; and denote the equivalence class that contains (x, y) by, x/y. We define addition and multiplication on F by, a/b 1 c/d 5 (ad 1 bc)/(bd), , and, , a/b ? c/d 5 (ac)/(bd)., , (Notice that here we need the fact that D is an integral domain to ensure, that multiplication is closed; that is, bd 2 0 whenever b 2 0 and d 2 0.), Since there are many representations of any particular element of F, ( just as in the rationals, we have 1/2 5 3/6 5 4/8), we must show that, these two operations are well-defined. To do this, suppose that a/b 5 a9/b9, and c/d 5 c9/d9, so that ab9 5 a9b and cd9 5 c9d. It then follows that, (ad 1 bc)b9d9 5 adb9d9 1 bcb9d9 5 (ab9)dd9 1 (cd9)bb9, 5 (a9b)dd9 1 (c9d)bb9 5 a9d9bd 1 b9c9bd, 5 (a9d9 1 b9c9)bd., Thus, by definition, we have, (ad 1 bc)/(bd) 5 (a9d9 1 b9c9)/(b9d9),, and, therefore, addition is well-defined. We leave the verification that, multiplication is well-defined as an exercise (Exercise 55). That F is a, field is straightforward. Let 1 denote the unity of D. Then 0/1 is the, additive identity of F. The additive inverse of a/b is 2a/b; the multiplicative inverse of a nonzero element a/b is b/a. The remaining field, properties can be checked easily., Finally, the mapping f: D S F given by x S x/1 is a ring isomorphism from D to f(D) (see Exercise 7)., EXAMPLE 11 Let D 5 Z[x]. Then the field of quotients of D is {f(x)/, g(x) | f(x), g(x) [ D, where g(x) is not the zero polynomial}., When F is a field, the field of quotients of F[x] is traditionally denoted by F(x)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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292, , Rings, , EXAMPLE 12 Let p be a prime. Then Zp(x) 5 {f(x)/g(x) | f(x), g(x), [ Zp[x], g(x) 2 0} is an infinite field of characteristic p., , Exercises, We can work it out., john lennon and paul mccartney,, “We Can Work It Out,” single*, , 1., 2., 3., 4., 5., 6., 7., 8., 9., , 10., 11., 12., 13., , Prove Theorem 15.1., Prove Theorem 15.2., Prove Theorem 15.3., Prove Theorem 15.4., Show that the correspondence x S 5x from Z5 to Z10 does not preserve addition., Show that the correspondence x S 3x from Z4 to Z12 does not preserve multiplication., Show that the mapping f: D S F in the proof of Theorem 15.6 is a, ring homomorphism., Prove that every ring homomorphism f from Zn to itself has the, form f(x) 5 ax, where a2 5 a., Suppose that f is a ring homomorphism from Zm to Zn. Prove that, if f(1) 5 a, then a2 5 a. Give an example to show that the converse, is false., a. Is the ring 2Z isomorphic to the ring 3Z?, b. Is the ring 2Z isomorphic to the ring 4Z?, Prove that the intersection of any collection of subfields of a field F, is a subfield of F. (This exercise is referred to in this chapter.), Let Z3[i] 5 {a 1 bi | a, b [ Z3} (see Example 9 in Chapter 13). Show, that the field Z3[i] is ring-isomorphic to the field Z3[x]/kx2 1 1l., Let, Sec, , a b, d ` a, b [ Rf ., b a, Show that f: C S S given by, a b, f(a 1 bi) 5 c, d, b a, is a ring isomorphism., *Copyright © 1965 (Renewed) Stony/ATV Tunes LLC. All rights administered by, Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights, reserved. Used by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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293, , 15 | Ring Homomorphisms, , 14. Let Z[22] 5 {a 1 b22 | a, b [ Z} and, H ec, , a 2b, d ` a, b [ Z f ., b a, , Show that Z[22] and H are isomorphic as rings., 15. Consider the mapping from M2(Z ) into Z given by c, , a, c, , b, d S a., d, , Prove or disprove that this is a ring homomorphism., 16. Let R 5 e c, , a, 0, , b, d ` a, b, c [ Z f . Prove or disprove that the mapc, , ping c, , 17., , 18., 19., 20., 21., 22., 23., 24., 25., 26., , 27., 28., 29., 30., 31., , a b, d S a is a ring homomorphism., 0 c, Is the mapping from Z5 to Z30 given by x S 6x a ring homomorphism? Note that the image of the unity is the unity of the image, but not the unity of Z30., Is the mapping from Z10 to Z10 given by x S 2x a ring homomorphism?, Describe the kernel of the homomorphism given in Example 3., Recall that a ring element a is called an idempotent if a2 5 a. Prove, that a ring homomorphism carries an idempotent to an idempotent., Determine all ring homomorphisms from Z6 to Z6. Determine all, ring homomorphisms from Z20 to Z30., Determine all ring isomorphisms from Zn to itself., Determine all ring homomorphisms from Z to Z., Suppose f is a ring homomorphism from Z % Z into Z % Z. What, are the possibilities for f((1, 0))?, Determine all ring homomorphisms from Z % Z into Z % Z., In Z, let A 5 k2l and B 5 k8l. Show that the group A/B is isomorphic to the group Z4 but that the ring A/B is not ring-isomorphic to, the ring Z4., Let R be a ring with unity and let f be a ring homomorphism from R, onto S where S has more than one element. Prove that S has a unity., Show that (Z % Z)/(kal % kbl) is ring-isomorphic to Za % Zb., Determine all ring homomorphisms from Z % Z to Z., Prove that the sum of the squares of three consecutive integers cannot be a square., Let m be a positive integer and let n be an integer obtained from m, by rearranging the digits of m in some way. (For example, 72345 is, a rearrangement of 35274.) Show that m 2 n is divisible by 9., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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294, , Rings, , 32. (Test for Divisibility by 11) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 11 if and only, if a0 2 a1 1 a2 2 ? ? ? (21)kak is divisible by 11., 33. Show that the number 7,176,825,942,116,027,211 is divisible by 9, but not divisible by 11., 34. Show that the number 9,897,654,527,609,805 is divisible by 99., 35. (Test for Divisibility by 3) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 3 if and only, if ak 1 ak21 1 ? ? ? 1 a1 1 a0 is divisible by 3., 36. (Test for Divisibility by 4) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 4 if and only, if a1a0 is divisible by 4., 37. Show that no integer of the form 111,111,111, . . . ,111 is prime., 38. Consider an integer n of the form a,111,111,111,111,111,111,, 111,111,12b. Find values for a and b such that n is divisible by 99., 39. Suppose n is a positive integer written in the form n 5 ak3k 1, ak213k21 1 ? ? ? 1 a13 1 a0, where each of the ai’s is 0, 1, or 2 (the, base 3 representative of n). Show that n is even if and only if ak 1, ak21 1 ? ? ? 1 a1 1 a0 is even., 40. Find an analog of the condition given in the previous exercise for, characterizing divisibility by 4., 41. In your head, determine (2 ? 1075 1 2)100 mod 3 and (10100 1 1)99, mod 3., 42. Determine all ring homomorphisms from Q to Q., 43. Let R and S be commutative rings with unity. If f is a homomorphism from R onto S and the characteristic of R is nonzero, prove, that the characteristic of S divides the characteristic of R., 44. Let R be a commutative ring of prime characteristic p. Show that, the Frobenius map x S xp is a ring homomorphism from R to R., 45. Is there a ring homomorphism from the reals to some ring whose, kernel is the integers?, 46. Show that a homomorphism from a field onto a ring with more, than one element must be an isomorphism., 47. Suppose that R and S are commutative rings with unities. Let f be a, ring homomorphism from R onto S and let A be an ideal of S., a. If A is prime in S, show that f21(A) 5 {x [ R | f(x) [ A} is, prime in R., b. If A is maximal in S, show that f21(A) is maximal in R., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 | Ring Homomorphisms, , 295, , 48. A principal ideal ring is a ring with the property that every ideal, has the form kal. Show that the homomorphic image of a principal, ideal ring is a principal ideal ring., 49. Let R and S be rings., a. Show that the mapping from R % S onto R given by (a, b) S a, is a ring homomorphism., b. Show that the mapping from R to R % S given by a S (a, 0) is a, one-to-one ring homomorphism., c. Show that R % S is ring-isomorphic to S % R., 50. Show that if m and n are distinct positive integers, then mZ is not, ring-isomorphic to nZ., 51. Prove or disprove that the field of real numbers is ring-isomorphic, to the field of complex numbers., 52. Show that the only ring automorphism of the real numbers is the, identity mapping., 53. Determine all ring homomorphisms from R to R., 54. Suppose that n divides m and that a is an idempotent of Zn (that is,, a2 5 a). Show that the mapping x S ax is a ring homomorphism, from Zm to Zn. Show that the same correspondence need not yield a, ring homomorphism if n does not divide m., 55. Show that the operation of multiplication defined in the proof of, Theorem 15.6 is well-defined., 56. Let Q[22] 5 {a 1 b22 | a, b [ Q} and Q[25] 5 {a 1 b25 |, a, b [ Q}. Show that these two rings are not ring-isomorphic., 57. Let Z[i] 5 {a 1 bi | a, b [ Z}. Show that the field of quotients of, Z[i] is ring-isomorphic to Q[i] 5 {r 1 si | r, s [ Q}. (This exercise, is referred to in Chapter 18.), 58. Let F be a field. Show that the field of quotients of F is ringisomorphic to F., 59. Let D be an integral domain and let F be the field of quotients of D., Show that if E is any field that contains D, then E contains a, subfield that is ring-isomorphic to F. (Thus, the field of quotients, of an integral domain D is the smallest field containing D.), 60. Explain why a commutative ring with unity that is not an integral domain cannot be contained in a field. (Compare with Theorem 15.6.), 61. Show that the relation ; defined in the proof of Theorem 15.6 is an, equivalence relation., 62. Give an example of a ring without unity that is contained in a field., 63. Prove that the set T in the proof of Corollary 3 to Theorem 15.5 is, ring-isomorphic to the field of rational numbers., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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296, , Rings, , 64. Suppose that f: R S S is a ring homomorphism and that the, image of f is not {0}. If R has a unity and S is an integral domain,, show that f carries the unity of R to the unity of S. Give an example to show that the preceding statement need not be true if S, is not an integral domain., 65. Let f(x) [ R[x]. If a 1 bi is a complex zero of f(x) (here i 5 21),, show that a 2 bi is a zero of f(x). (This exercise is referred to in, Chapter 32.), 66. Let R 5 e c, , a, b, , b, d ` a, b [ Z f , and let f be the mapping that, a, , a b, d to a 2 b., b a, a. Show that f is a homomorphism., b. Determine the kernel of f., c. Show that R/Ker f is isomorphic to Z., d. Is Ker f a prime ideal?, e. Is Ker f a maximal ideal?, Show that the prime subfield of a field of characteristic p is ringisomorphic to Zp and that the prime subfield of a field of characteristic 0 is ring-isomorphic to Q. (This exercise is referred to in, this chapter.), Let n be a positive integer. Show that there is a ring isomorphism, from Z2 to a subring of Z2n if and only if n is odd., Show that Zmn is ring-isomorphic to Zm % Zn when m and n are relatively prime., Prove that every integer with decimal representation of the form, abcabc (for example, 916916) is divisible by 11., takes c, , 67., , 68., 69., 70., , Suggested Readings, J. A. Gallian and J. Van Buskirk, “The Number of Homomorphisms from, Zm into Zn,” American Mathematical Monthly 91 (1984): 196–197., In this article, formulas are given for the number of group homomorphisms from Zm into Zn and the number of ring homomorphisms from, Zm into Zn., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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15 | Ring Homomorphisms, , 297, , Lillian Kinkade and Joyce Wagner, “When Polynomial Rings Are, Principal Ideal Rings,” Journal of Undergraduate Mathematics 23, (1991): 59–62., In this article written by undergraduates, it is shown that R[x] is a, principal ideal ring if and only if R < R1 % R2 % ? ? ? % Rn, where, each Ri is a field., Mohammad Saleh and Hasan Yousef, “The Number of Ring Homomorphisms from Zm1 % ? ? ? % Zmr into Zk1 % ? ? ? % Zk s ,” American Mathematical Monthly 105 (1998): 259–260., This article gives a formula for the number described in the title., , Suggested Website, http://www.d.umn.edu/~jgallian/puzzle, This site has a math puzzle that is based on the ideas presented in this, chapter. The user selects an integer and then proceeds through a series of, steps to produce a new integer. Finally, another integer is created by using, all but one of the digits of the previous integer in any order. The software, then determines the digit not used., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 Polynomial Rings, , Wit lies in recognizing the resemblance among things which differ and the, difference between things which are alike., madame de staël, , Notation and Terminology, One of the mathematical concepts that students are most familiar with, and most comfortable with is that of a polynomial. In high school,, students study polynomials with integer coefficients, rational coefficients, real coefficients, and perhaps even complex coefficients. In earlier chapters of this book, we introduced something that was probably, new—polynomials with coefficients from Zn. Notice that all of these, sets of polynomials are rings, and, in each case, the set of coefficients is, also a ring. In this chapter, we abstract all of these examples into one., Definition Ring of Polynomials over R, , Let R be a commutative ring. The set of formal symbols, R[x] 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | ai [ R,, n is a nonnegative integer}, is called the ring of polynomials over R in the indeterminate x., Two elements, anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0, and, bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0, of R[x] are considered equal if and only if ai 5 bi for all nonnegative, integers i. (Define ai 5 0 when i . n and bi 5 0 when i . m.), , 298, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 | Polynomial Rings, , 299, , In this definition, the symbols x, x 2, . . . , x n do not represent, “unknown” elements or variables from the ring R. Rather, their purpose, is to serve as convenient placeholders that separate the ring elements, an, an21, . . . , a0. We could have avoided the x’s by defining a polynomial as an infinite sequence a 0, a1, a2, . . . , an, 0, 0, 0, . . . , but our, method takes advantage of the student’s experience in manipulating, polynomials where x does represent a variable. The disadvantage of our, method is that one must be careful not to confuse a polynomial with the, function determined by a polynomial. For example, in Z3[x], the polynomials f (x) 5 x and g(x) 5 x3 determine the same function from Z3 to, Z3, since f(a) 5 g(a) for all a in Z3.† But f(x) and g(x) are different elements of Z3[x]. Also, in the ring Zn[x], be careful to reduce only the, coefficients and not the exponents modulo n. For example, in Z3[x],, 5x 5 2x, but x5 2 x2., To make R[x] into a ring, we define addition and multiplication in, the usual way., Definition Addition and Multiplication in R[x], , Let R be a commutative ring and let, f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0, and, g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0, belong to R[x]. Then, f(x) 1 g(x) 5 (as 1 bs)xs 1 (as21 1 bs21)xs21, 1 ? ? ? 1 (a1 1 b1)x 1 a0 1 b0,, where s is the maximum of m and n, ai 5 0 for i . n, and bi 5 0 for, i . m. Also,, f(x)g(x) 5 cm1nxm1n 1 cm1n21xm1n21 1 ? ? ? 1 c1x 1 c0,, where, ck 5 akb0 1 ak21b1 1 ? ? ? 1 a1bk21 1 a0bk, for k 5 0, . . . , m 1 n., , Although the definition of multiplication might appear complicated,, it is just a formalization of the familiar process of using the distributive, †In, , general, given f(x) in R[x] and a in R, f(a) means substitute a for x in the formula, for f(x). This substitution is a homomorphism from R[x] to R., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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300, , Rings, , property and collecting like terms. So, just multiply polynomials over a, commutative ring R in the same way that polynomials are always multiplied. Here is an example., Consider f(x) 5 2x3 1 x2 1 2x 1 2 and g(x) 5 2x2 1 2x 1 1 in Z3[x]., Then, in our preceding notation, a5 5 0, a4 5 0, a3 5 2, a2 5 1, a1 5 2,, a0 5 2, and b5 5 0, b4 5 0, b3 5 0, b2 5 2, b1 5 2, b0 5 1. Now, using, the definitions and remembering that addition and multiplication of the, coefficients are done modulo 3, we have, f(x) 1 g(x) 5 (2 1 0)x3 1 (1 1 2)x2 1 (2 1 2)x 1 (2 1 1), 5 2x3 1 0x2 1 1x 1 0, 5 2x3 1 x, and, f(x) ? g(x) 5 (0 ? 1 1 0 ? 2 1 2 ? 2 1 1 ? 0 1 2 ? 0 1 2 ? 0)x5, 1 (0 ? 1 1 2 ? 2 1 1 ? 2 1 2 ? 0 1 2 ? 0)x4, 1 (2 ? 1 1 1 ? 2 1 2 ? 2 1 2 ? 0)x3, 1 (1 ? 1 1 2 ? 2 1 2 ? 2)x2 1 (2 ? 1 1 2 ? 2)x 1 2 ? 1, 5 x5 1 0x4 1 2x3 1 0x2 1 0x 1 2, 5 x5 1 2x3 1 2., Our definitions for addition and multiplication of polynomials were, formulated so that they are commutative and associative, and so that, multiplication is distributive over addition. We leave the verification, that R[x] is a ring to the reader., It is time to introduce some terminology for polynomials. If, f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0,, where an 2 0, we say that f(x) has degree n; the term an is called the, leading coefficient of f(x), and if the leading coefficient is the multiplicative identity element of R, we say that f(x) is a monic polynomial., The polynomial f(x) 5 0 has no degree. Polynomials of the form, f(x) 5 a0 are called constant. We often write deg f(x) 5 n to indicate, that f(x) has degree n. As with polynomials with real coefficients, we, may insert or delete terms of the form 0xk; 1xk is the same as xk; and, 1(2ak)xk is the same as 2akxk., Very often properties of R carry over to R[x]. Our first theorem is a, case in point., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 | Polynomial Rings, , 301, , Theorem 16.1 D an Integral Domain Implies D[x] an Integral Domain, If D is an integral domain, then D [x] is an integral domain., , PROOF Since we already know that D[x] is a ring, all we need, to show is that D[x] is commutative with a unity and has no zero-divisors., Clearly, D[x] is commutative whenever D is. If 1 is the unity element of, D, it is obvious that f(x) 5 1 is the unity element of D[x]. Finally, suppose that, f (x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0, and, g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b0,, where an 2 0 and bm 2 0. Then, by definition, f(x)g(x) has leading coefficient anbm and, since D is an integral domain, anbm 2 0., , The Division Algorithm, and Consequences, One of the properties of integers that we have used repeatedly is the, division algorithm: If a and b are integers and b 2 0, then there exist, unique integers q and r such that a 5 bq 1 r, where 0 # r , |b|. The, next theorem is the analogous statement for polynomials over a field., Theorem 16.2 Division Algorithm for F[x], Let F be a field and let f(x), g(x) [ F[x] with g(x) 2 0. Then, there exist unique polynomials q(x) and r(x) in F[x] such that f(x) 5, g(x)q(x) 1 r(x) and either r(x) 5 0 or deg r(x) , deg g(x)., , PROOF We begin by showing the existence of q(x) and r(x). If, f(x) 5 0 or deg f(x) , deg g(x), we simply set q(x) 5 0 and r(x) 5 f(x)., So, we may assume that n 5 deg f(x) $ deg g(x) 5 m and let f(x) 5, anxn 1 ? ? ? 1 a0 and g(x) 5 bmxm 1 ? ? ? 1 b0. The idea behind this, proof is to begin just as if you were going to “long divide” g(x) into, f(x), then use the Second Principle of Mathematical Induction on, deg f(x) to finish up. Thus, resorting to long division, we let f1(x) 5, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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302, , Rings, , f(x) 2 anbm21xn2mg(x).† Then, f1(x) 5 0 or deg f1(x) , deg f(x); so, by, our induction hypothesis, there exist q1(x) and r1(x) in F[x] such that, f1(x) 5 g(x)q1(x) 1 r1(x), where r1(x) 5 0 or deg r1(x) , deg g(x)., [Technically, we should get the induction started by proving the case, in which deg f(x) 5 0, but this is trivial.] Thus,, f(x) 5 anbm 21xn2mg(x) 1 f1(x), 5 anbm 21xn2mg(x) 1 q1(x)g(x) 1 r1(x), 5 [anbm 21xn2m 1 q1(x)]g(x) 1 r1(x)., So, the polynomials q(x) 5 anbm 21xn2m 1 q1(x) and r(x) 5 r1(x) have, the desired properties., To prove uniqueness, suppose that f(x) 5 g(x)q(x) 1 r(x) and f(x) 5, g(x) q(x) 1 r(x), where r(x) 5 0 or deg r(x) , deg g(x) and r(x) 5 0, or deg r(x) , deg g(x). Then, subtracting these two equations, we obtain, 0 5 g(x)[q(x) 2 q(x)] 1 [r(x) 2 r(x)], or, r(x) 2 r(x) 5 g(x)[q(x) 2 q(x)]., Thus, r(x) 2 r(x) is 0, or the degree of r(x) 2 r(x) is at least that of, g(x). Since the latter is clearly impossible, we have r(x) 5 r(x) and, q(x) 5 q(x) as well., The polynomials q(x) and r(x) in the division algorithm are called, the quotient and remainder in the division of f(x) by g(x). When the, ring of coefficients of a polynomial ring is a field, we can use the long, division process to determine the quotient and remainder., , †For, , example,, (3>2)x2, , 2x2 1 2 q3x4, , 1x11, 3x4 1 3x2, 2 3x2 1 x 1 1, , So,, 23x2 1 x 1 1 5 3x4 1 x 1 1 2 (3/2)x2(2x2 1 2), In general,, , anbm21xn2m, ., ., ., bm x 1, qan xn 1 . . ., a xn 1 . . ., m, , n, , f1 (x), , So,, f1(x) 5 (anxn 1 ? ? ?) 2 anbm 21xn2m(bmxm 1 ? ? ?), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 | Polynomial Rings, , 303, , EXAMPLE 1 To find the quotient and remainder upon dividing, f(x) 5 3x4 1 x3 1 2x2 1 1 by g(x) 5 x2 1 4x 1 2, where f(x) and g(x), belong to Z5[x], we may proceed by long division, provided we keep in, mind that addition and multiplication are done modulo 5. Thus,, 3x2 1 4x, x2 1 4x 1 2 q3x4 1 x3 1 2x2, 11, 4, 3, 2, 3x 1 2x 1 x, 4x3 1 x2, 11, 4x3 1 x2 1 3x, 2x 1 1, So, 3x2 1 4x is the quotient and 2x 1 1 is the remainder. Therefore,, 3x4 1 x3 1 2x2 1 1 5 (x2 1 4x 1 2)(3x2 1 4x) 1 2x 1 1., Let D be an integral domain. If f(x) and g(x) [ D[x], we say that g(x), divides f(x) in D[x] [and write g(x) | f(x)] if there exists an h(x) [ D[x], such that f(x) 5 g(x)h(x). In this case, we also call g(x) a factor of f(x)., An element a is a zero (or a root) of a polynomial f(x) if f(a) 5 0., [Recall that f(a) means substitute a for x in the expression for f(x).], When F is a field, a [ F, and f(x) [ F[x], we say that a is a zero of, multiplicity k (k $ 1) if (x 2 a)k is a factor of f(x) but (x 2 a)k11 is not, a factor of f(x). With these definitions, we may now give several important corollaries of the division algorithm. No doubt you have seen these, for the special case where F is the field of real numbers., Corollary 1 Remainder Theorem, Let F be a field, a [ F, and f(x) [ F[x]. Then f(a) is the remainder in, the division of f(x) by x 2 a., , PROOF The proof of Corollary 1 is left as an exercise (Exercise 5)., Corollary 2 Factor Theorem, Let F be a field, a [ F, and f(x) [ F[x]. Then a is a zero of f(x) if, and only if x 2 a is a factor of f(x)., , PROOF The proof of Corollary 2 is left as an exercise (Exercise 9)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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304, , Rings, , Corollary 3 Polynomials of Degree n Have at Most n Zeros, A polynomial of degree n over a field has at most n zeros, counting, multiplicity., , PROOF We proceed by induction on n. Clearly, a polynomial of, degree 0 over a field has no zeros. Now suppose that f(x) is a polynomial of degree n over a field and a is a zero of f(x) of multiplicity k., Then, f(x) 5 (x 2 a)kq(x) and q(a) 2 0; and, since n 5 deg f(x) 5 deg, (x 2 a)k q(x) 5 k 1 deg q(x), we have k # n (see Exercise 19). If f(x), has no zeros other than a, we are done. On the other hand, if b 2 a and, b is a zero of f(x), then 0 5 f(b) 5 (b 2 a)kq(b), so that b is also a zero, of q(x) with the same multiplicity as it has for f(x) (see Exercise 21)., By the Second Principle of Mathematical Induction, we know that, q(x) has at most deg q(x) 5 n 2 k zeros, counting multiplicity. Thus,, f(x) has at most k 1 n 2 k 5 n zeros, counting multiplicity., We remark that Corollary 3 is not true for arbitrary polynomial rings., For example, the polynomial x2 1 3x 1 2 has four zeros in Z6. (See, Exercise 3.) Lagrange was the first to prove Corollary 3 for polynomials in Zp[x]., EXAMPLE 2 The Complex Zeros of xn 2 1, We find all complex zeros of x n 2 1. Let v 5 cos(360°/n) 1, i sin(360°/n). It follows from DeMoivre’s Theorem (see Example 10, in Chapter 0) that vn 5 1 and vk 2 1 for 1 # k , n. Thus, each of 1,, v, v2, . . . , vn21 is a zero of xn 2 1 and, by Corollary 3, there are no, others., The complex number v in Example 2 is called a primitive nth root of, unity., We conclude this chapter with an important theoretical application, of the division algorithm, but first an important definition., Definition Principal Ideal Domain (PID), , A principal ideal domain is an integral domain R in which every ideal, has the form kal 5 {ra | r [ R} for some a in R., , Theorem 16.3 F[x] Is a PID, Let F be a field. Then F[x] is a principal ideal domain., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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305, , 16 | Polynomial Rings, , PROOF By Theorem 16.1, we know that F[x] is an integral domain., Now, let I be an ideal in F[x]. If I 5 {0}, then I 5 k0l. If I 2 {0}, then, among all the elements of I, let g(x) be one of minimum degree. We will, show that I 5 kg(x)l. Since g(x) [ I, we have kg(x)l # I. Now, let f(x) [ I. Then, by the division algorithm, we may write f(x) 5, g(x)q(x) 1 r(x), where r(x) 5 0 or deg r(x) , deg g(x). Since r(x) 5 f(x) 2, g(x)q(x) [ I, the minimality of deg g(x) implies that the latter condition, cannot hold. So, r(x) 5 0 and, therefore, f(x) [ kg(x)l. This shows that, I # kg(x)l., The proof of Theorem 16.3 also establishes the following., Theorem 16.4 Criterion for I 5 kg(x)l, Let F be a field, I a nonzero ideal in F [x], and g(x) an element of, F[x]. Then, I 5 8 g(x)9 if and only if g(x) is a nonzero polynomial of, minimum degree in I., , As an application of the First Isomorphism Theorem for Rings, (Theorem 15.3) and Theorem 16.4, we verify the remark we made in, Example 12 in Chapter 14 that the ring R[x]/kx2 1 1l is isomorphic to, the ring of complex numbers., EXAMPLE 3 Consider the homomorphism f from R[x] onto C given, by f(x) S f(i) (that is, evaluate a polynomial in R[x] at i). Then, x2 1 1 [ Ker f and is clearly a polynomial of minimum degree in Ker f., Thus, Ker f 5 kx2 1 1l and R[x]/kx2 1 1l is isomorphic to C., , Exercises, If I feel unhappy, I do mathematics to become happy. If I am happy, I do, mathematics to keep happy., paul turán, , 1. Let f(x) 5 4x3 1 2x2 1 x 1 3 and g(x) 5 3x4 1 3x3 1 3x2 1 x 1 4,, where f(x), g(x) [ Z5[x]. Compute f(x) 1 g(x) and f(x) ? g(x)., 2. In Z3[x], show that the distinct polynomials x4 1 x and x2 1 x, determine the same function from Z3 to Z3., 3. Show that x2 1 3x 1 2 has four zeros in Z6. (This exercise is, referred to in this chapter.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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306, , Rings, , 4. If R is a commutative ring, show that the characteristic of R[x] is, the same as the characteristic of R., 5. Prove Corollary 1 of Theorem 16.2., 6. List all the polynomials of degree 2 in Z2[x]. Which of these are, equal as functions from Z2 to Z2?, 7. Find two distinct cubic polynomials over Z2 that determine the, same function from Z2 to Z2., 8. For any positive integer n, how many polynomials are there of, degree n over Z2? How many distinct polynomial functions from Z2, to Z2 are there?, 9. Prove Corollary 2 of Theorem 16.2., 10. Let R be a commutative ring. Show that R[x] has a subring isomorphic to R., 11. If f: R S S is a ring homomorphism, define f: R[x] S S[x] by, (anxn 1 ? ? ? 1 a0) S f(an)xn 1 ? ? ? 1 f(a0). Show that f is a ring, homomorphism. (This exercise is referred to in Chapter 33.), 12. If the rings R and S are isomorphic, show that R[x] and S[x] are, isomorphic., 13. Let f(x) 5 5x4 1 3x3 1 1 and g(x) 5 3x2 1 2x 1 1 in Z7[x]., Determine the quotient and remainder upon dividing f(x) by g(x)., 14. Let f(x) and g(x) be cubic polynomials with integer coefficients, such that f (a) 5 g(a) for four integer values of a. Prove that f (x) 5, g(x). Generalize., 15. Show that the polynomial 2x 1 1 in Z4[x] has a multiplicative inverse in Z4[x]., 16. Are there any nonconstant polynomials in Z[x] that have multiplicative inverses? Explain your answer., 17. Let p be a prime. Are there any nonconstant polynomials in Zp[x], that have multiplicative inverses? Explain your answer., 18. Show that Corollary 3 of Theorem 16.2 is false for any commutative ring that has a zero divisor., 19. (Degree Rule) Let D be an integral domain and f(x), g(x) [ D[x]., Prove that deg ( f(x) ? g(x)) 5 deg f(x) 1 deg g(x). Show, by example, that for commutative ring R it is possible that deg f(x)g(x) ,, deg f(x) 1 deg g(x), where f(x) and g(x) are nonzero elements in, R[x]. (This exercise is referred to in this chapter, Chapter 17, and, Chapter 18.), 20. Prove that the ideal kxl in Q[x] is maximal., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 | Polynomial Rings, , 307, , 21. Let f(x) belong to F[x], where F is a field. Let a be a zero of f(x) of, multiplicity n, and write f(x) 5 (x 2 a)nq(x). If b Z a is a zero of, q(x), show that b has the same multiplicity as a zero of q(x) as it, does for f(x). (This exercise is referred to in this chapter.), 22. Prove that for any positive integer n, a field F can have at most a, finite number of elements of multiplicative order at most n., 23. Let F be an infinite field and let f(x) [ F[x]. If f(a) 5 0 for infinitely many elements a of F, show that f(x) 5 0., 24. Let F be an infinite field and let f(x), g(x) [ F[x]. If f(a) 5 g(a) for, infinitely many elements a of F, show that f(x) 5 g(x)., 25. Let F be a field and let p(x) [ F[x]. If f(x), g(x) [ F[x] and, deg f(x) , deg p(x) and deg g(x) , deg p(x), show that f(x) 1, kp(x)l 5 g(x) 1 kp(x)l implies f(x) 5 g(x). (This exercise is, referred to in Chapter 20.), 26. Prove that Z[x] is not a principal ideal domain. (Compare this with, Theorem 16.3.), 27. Find a polynomial with integer coefficients that has 1/2 and 21/3, as zeros., 28. Let f(x) [ R[x]. Suppose that f(a) 5 0 but f9(a) 2 0, where f9(x) is, the derivative of f(x). Show that a is a zero of f(x) of multiplicity 1., 29. Show that Corollary 2 of Theorem 16.2 is true over any commutative ring with unity., 30. Show that Corollary 3 of Theorem 16.2 is true for polynomials, over integral domains., 31. Let F be a field and let, I 5 {anxn 1 an21xn21 1 ? ? ? 1 a0 | an, an21, . . . , a0 [ F and, an 1 an21 1 ? ? ? 1 a0 5 0}., Show that I is an ideal of F[x] and find a generator for I., 32. Let F be a field and let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ F[x]., Prove that x 2 1 is a factor of f(x) if and only if an 1 an21 1 ? ? ? 1, a0 5 0., 33. Let m be a fixed positive integer. For any integer a, let a denote, a mod m. Show that the mapping of f: Z[x] S Zm[x] given by, f(anxn 1 an21xn21 1 ? ? ? 1 a0) 5 anxn 1 an21xn21 1 ? ? ? 1 a0, is a ring homomorphism. (This exercise is referred to in Chapters, 17 and 33.), 34. Find infinitely many polynomials f(x) in Z3[x] such that f(a) 5 0 for, all a in Z3., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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308, , Rings, , 35. For every prime p, show that, xp21 2 1 5 (x 2 1)(x 2 2) ? ? ? [x 2 ( p 2 1)], 36., , 37., 38., 39., 40., 41., 42., 43., 44., 45., , 46., 47., 48., , 49., , 50., 51., , in Zp[x]., Let f be the ring homomorphism from Z[x] to Z given by f(f(x)) 5, f(1). Find a polynomial g(x) in Z[x] such that Ker f 5 kg(x)l. Is, there more than one possibility for g(x)? To what familiar ring is, Z[x]/Ker f isomorphic? Do this exercise with Z replaced by Q., Give an example of a field that properly contains the field of complex numbers C., (Wilson’s Theorem) For every integer n . 1, prove that (n 2 1)!, mod n 5 n 2 1 if and only if n is prime., For every prime p, show that ( p 2 2)! mod p 5 1., Find the remainder upon dividing 98! by 101., Prove that (50!)2 mod 101 5 21 mod 101., If I is an ideal of a ring R, prove that I[x] is an ideal of R[x]., Give an example of a commutative ring R with unity and a, maximal ideal I of R such that I[x] is not a maximal ideal of R[x]., Let R be a commutative ring with unity. If I is a prime ideal of R,, prove that I[x] is a prime ideal of R[x]., Let F be a field, and let f(x) and g(x) belong to F[x]. If there is no, polynomial of positive degree in F[x] that divides both f(x) and g(x), [in this case, f(x) and g(x) are said to be relatively prime], prove that, there exist polynomials h(x) and k(x) in F[x] with the property that, f(x)h(x) 1 g(x)k(x) 5 1. (This exercise is referred to in Chapter 20.), Prove that Q[x]/kx2 2 2l is ring-isomorphic to Q[22] 5 {a 1, b22 | a, b [ Q}., Let f(x) [ R[x]. If f(a) 5 0 and f 9(a) 5 0 [f 9(a) is the derivative of, f(x) at a], show that (x 2 a)2 divides f(x)., Let F be a field and let I 5 {f(x) [ F[x] | f (a) 5 0 for all a in F}., Prove that I is an ideal in F[x]. Prove that I is infinite when F is finite and I 5 {0} when F is infinite. When F is finite, find a monic, polynomial g(x) such that I 5 kg(x)l., Let g(x) and h(x) belong to Z[x] and let h(x) be monic. If h(x) divides g(x) in Q[x], show that h(x) divides g(x) in Z[x]. (This exercise is referred to in Chapter 33.), Let R be a ring and x be an indeterminate. Prove that the rings R[x], and R[x2] are ring-isomorphic., Let f (x) be a nonconstant element of Z[x]. Prove that f (x) takes on, infinitely many values in Z., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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16 | Polynomial Rings, , 309, , 52. Let f (x) be a nonconstant element in Z[x]. Prove that k f (x)l is not, maximal in Z[x]., 53. Let Zn[x] be the ring of polynomials in x with coefficients from Zn, and ordinary addition and multiplication. If n can be written in the, form t2m, show that tmx 1 1 is a unit in Zn[x]., 54. Let f (x) belong to Zp[x]. Prove that if f (b) 5 0, then f (bp) 5 0., 55. Suppose f (x) is a polynomial with odd integer coefficients and even, degree. Prove that f (x) has no rational zeros., 56. For any field F, recall that F(x) denotes the field of quotients of the, ring F[x]. Prove that there is no element in F(x) whose square is x., 57. Let F be a field. Show that there exist a, b [ F with the property, that x2 1 x 1 1 divides x43 1 ax 1 b., 58. Let f(x) 5 am x m 1 am21x m21 1 ? ? ? 1 a0 and g(x) 5 bn 1 bn21x n21 1, ? ? ? 1 b0 belong to Q[x] and suppose that f (x)g(x) belongs to Z[x]., Prove that aibj is an integer for every i and j., 59. Let f (x) belong to Z[x]. If a mod m 5 b mod m, prove that f (a), mod m 5 f(b) mod m. Prove that if both f(0) and f(1) are odd, then, f has no zero in Z., 60. Find the remainder when x51 is divided by x 1 4 in Z7[x]., 61. Show that 1 is the only solution of x25 2 1 5 0 in Z37., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Saunders Mac Lane, , Osa Mac Lane, , The 1986 Steele Prize for cumulative, influence is awarded to Saunders Mac Lane, for his many contributions to algebra and, algebraic topology, and in particular for his, pioneering work in homological and, categorical algebra., Citation for the Steele Prize, , Saunders Mac Lane ranks among the most, influential mathematicians in the 20th century. He was born on August 4, 1909, in, Norwich, Connecticut. In 1933, at the height, of the Depression, he was newly married; despite having degrees from Yale, the University, of Chicago, and the University of Göttingen,, he had no prospects for a position at a college, or university. After applying for employment, as a master at a private preparatory school for, boys, Mac Lane received a two-year instructorship at Harvard in 1934. He then spent a, year at Cornell and a year at the University, of Chicago before returning to Harvard, in 1938. In 1947, he went back to Chicago, permanently., Much of Mac Lane’s work focuses on the, interconnections among algebra, topology,, , and geometry. His book Survey of Modern, Algebra, coauthored with Garrett Birkhoff,, influenced generations of mathematicians, and is now a classic. Mac Lane served as, president of the Mathematical Association of, America and the American Mathematical, Society. He was elected to the National, Academy of Sciences, received the National, Medal of Science and the American, Mathematical Society’s Steele Prize for, Lifetime Achievement, and supervised 41, Ph.D. theses. Mac Lane died April 14, 2005,, at age 95., To find more information about Mac, Lane, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 310, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Factorization, 17 of Polynomials, The value of a principle is the number of things it will explain., ralph waldo emerson, , Reducibility Tests, In high school, students spend much time factoring polynomials and, finding their zeros. In this chapter, we consider the same problems in a, more abstract setting., To discuss factorization of polynomials, we must first introduce the, polynomial analog of a prime integer., Definition Irreducible Polynomial, Reducible Polynomial, , Let D be an integral domain. A polynomial f(x) from D[x] that is, neither the zero polynomial nor a unit in D[x] is said to be irreducible, over D if, whenever f(x) is expressed as a product f(x) 5 g(x)h(x), with, g(x) and h(x) from D[x], then g(x) or h(x) is a unit in D[x]. A nonzero,, nonunit element of D[x] that is not irreducible over D is called, reducible over D., , In the case that an integral domain is a field F, it is equivalent and more, convenient to define a nonconstant f(x) [ F[x] to be irreducible if f(x) cannot be expressed as a product of two polynomials of lower degree., EXAMPLE 1 The polynomial f(x) 5 2x2 1 4 is irreducible over Q, but reducible over Z, since 2x2 1 4 5 2(x2 1 2) and neither 2 nor x2 1 2, is a unit in Z[x]., EXAMPLE 2 The polynomial f(x) 5 2x2 1 4 is irreducible over R, but reducible over C., EXAMPLE 3 The polynomial x2 2 2 is irreducible over Q but reducible over R., , 311, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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312, , Rings, , EXAMPLE 4 The polynomial x2 1 1 is irreducible over Z3 but reducible over Z5., In general, it is a difficult problem to decide whether or not a particular polynomial is reducible over an integral domain, but there are special cases when it is easy. Our first theorem is a case in point. It applies, to the four preceding examples., Theorem 17.1 Reducibility Test for Degrees 2 and 3, Let F be a field. If f(x) [ F[x] and deg f(x) is 2 or 3, then f(x) is, reducible over F if and only if f(x) has a zero in F., , PROOF Suppose that f(x) 5 g(x)h(x), where both g(x) and h(x) belong, to F[x] and have degrees less than that of f(x). Since deg f(x) 5 deg g(x) 1, deg h(x) (Exercise 19 in Chapter 16) and deg f(x) is 2 or 3, at least one, of g(x) and h(x) has degree 1. Say g(x) 5 ax 1 b. Then, clearly, 2a21b, is a zero of g(x) and therefore a zero of f(x) as well., Conversely, suppose that f(a) 5 0, where a [ F. Then, by the Factor, Theorem, we know that x 2 a is a factor of f(x) and, therefore, f(x) is, reducible over F., Theorem 17.1 is particularly easy to use when the field is Zp, because, in this case we can check for reducibility of f(x) by simply testing, to see if f(a) 5 0 for a 5 0, 1, . . . , p 2 1. For example, since 2 is a, zero of x2 1 1 over Z5 , x2 1 1 is reducible over Z5 . On the other hand,, because neither 0, 1, nor 2 is a zero of x2 1 1 over Z3, x2 1 1 is irreducible over Z3., Note that polynomials of degree larger than 3 may be reducible over, a field even though they do not have zeros in the field. For example, in, Q[x], the polynomial x4 1 2x2 1 1 is equal to (x2 1 1)2, but has no, zeros in Q., Our next three tests deal with polynomials with integer coefficients., To simplify the proof of the first of these, we introduce some terminology and isolate a portion of the argument in the form of a lemma., Definition Content of a Polynomial, Primitive Polynomial, , The content of a nonzero polynomial anxn 1 an21xn21 1 ? ? ? 1 a0,, where the a’s are integers, is the greatest common divisor of the, integers an, an21, . . . , a0. A primitive polynomial is an element of Z[x], with content 1., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 313, , Gauss’s Lemma, The product of two primitive polynomials is primitive., , PROOF Let f(x) and g(x) be primitive polynomials, and suppose that, f(x)g(x) is not primitive. Let p be a prime divisor of the content of, f(x)g(x), and let f (x), g(x), and f(x)g(x) be the polynomials obtained, from f(x), g(x), and f(x)g(x) by reducing the coefficients modulo p., Then, f (x) and g(x) belong to the integral domain Zp[x] and f (x)g(x) 5, f(x)g(x) 5 0, the zero element of Zp[x] (see Exercise 33 in Chapter 16)., Thus, f (x) 5 0 or g(x) 5 0. This means that either p divides every coefficient of f(x) or p divides every coefficient of g(x). Hence, either f(x), is not primitive or g(x) is not primitive. This contradiction completes, the proof., Remember that the question of reducibility depends on which ring, of coefficients one permits. Thus, x2 2 2 is irreducible over Z but, reducible over Q[22]. In Chapter 20, we will prove that every polynomial of degree greater than 1 with coefficients from an integral, domain is reducible over some field. Theorem 17.2 shows that in the, case of polynomials irreducible over Z, this field must be larger than, the field of rational numbers., Theorem 17.2 Reducibility over Q Implies Reducibility over Z, Let f(x) [ Z[x]. If f(x) is reducible over Q, then it is reducible over Z., , PROOF Suppose that f (x) 5 g(x)h(x), where g(x) and h(x) [ Q[x]., Clearly, we may assume that f (x) is primitive because we can divide, both f(x) and g(x) by the content of f (x). Let a be the least common, multiple of the denominators of the coefficients of g(x), and b the least, common multiple of the denominators of the coefficients of h(x). Then, abf(x) 5 ag(x) ? bh(x), where ag(x) and bh(x) [ Z[x]. Let c1 be the content of ag(x) and let c2 be the content of bh(x). Then ag(x) 5 c1g1(x) and, bh(x) 5 c2h1(x), where both g1(x) and h1(x) are primitive, and abf(x) 5, c1c2g1(x)h1(x). Since f(x) is primitive, the content of abf(x) is ab. Also,, since the product of two primitive polynomials is primitive, it follows, that the content of c1c2g1(x)h1(x) is c1c2. Thus, ab 5 c1c2 and f(x) 5, g1(x)h1(x), where g1(x) and h1(x) [ Z[x] and deg g1(x) 5 deg g(x) and, deg h1(x) 5 deg h(x)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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314, , Rings, , EXAMPLE 5 We illustrate the proof of Theorem 17.2 by tracing, through it for the polynomial f (x) 5 6x2 1 x 2 2 5 (3x 2 3/2)(2x 1, 4/3) 5 g(x)h(x). In this case we have a 5 2, b 5 3, c1 5 3, c2 5 2, g1(x) 5, 2x 2 1, and h1(x) 5 3x 1 2, so that 2 ? 3(6x2 1 x 2 2) 5 3 ? 2(2x 2, 1)(3x 1 2) or 6x2 1 x 2 2 5 (2x 2 1)(3x 1 2)., , Irreducibility Tests, Theorem 17.1 reduces the question of irreducibility of a polynomial of, degree 2 or 3 to one of finding a zero. The next theorem often allows us, to simplify the problem even further., Theorem 17.3 Mod p Irreducibility Test, Let p be a prime and suppose that f(x) [ Z[x] with deg f(x) $ 1., Let f (x) be the polynomial in Zp[x] obtained from f(x) by reducing, all the coefficients of f(x) modulo p. If f (x) is irreducible over Zp and, deg f (x) 5 deg f(x), then f(x) is irreducible over Q., , PROOF It follows from the proof of Theorem 17.2 that if f(x) is reducible over Q, then f(x) 5 g(x)h(x) with g(x), h(x) [ Z[x], and both, g(x) and h(x) have degree less than that of f(x). Let f (x), g(x), and h(x), be the polynomials obtained from f(x), g(x), and h(x) by reducing all, the coefficients modulo p. Since deg f(x) 5 deg f (x), we have deg, g(x) # deg g(x) , deg f (x) and deg h(x) # deg h(x) , deg f (x). But,, f (x) 5 g(x)h(x), and this contradicts our assumption that f (x) is irreducible over Zp., EXAMPLE 6 Let f(x) 5 21x3 2 3x2 1 2x 1 9. Then, over Z2, we, have f (x) 5 x3 1 x2 1 1 and, since f 102 5 1 and f 112 5 1, we see that, f (x) is irreducible over Z2. Thus, f (x) is irreducible over Q. Notice that,, over Z3, f (x) 5 2x is irreducible, but we may not apply Theorem 17.3 to, conclude that f(x) is irreducible over Q., Be careful not to use the converse of Theorem 17.3. If f(x) [ Z[x], and f (x) is reducible over Zp for some p, f(x) may still be irreducible, over Q. For example, consider f(x) 5 21x3 2 3x2 1 2x 1 8. Then, over, Z2, f (x) 5 x3 1 x2 5 x2(x 1 1). But over Z5, f (x) has no zeros and, therefore is irreducible over Z5. So, f(x) is irreducible over Q. Note that, this example shows that the Mod p Irreducibility Test may fail for, some p and work for others. To conclude that a particular f(x) in Z[x] is, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 315, , irreducible over Q, all we need to do is find a single p for which the corresponding polynomial f 1x2 in Zp is irreducible. However, this is not always possible, since f(x) 5 x4 1 1 is irreducible over Q but reducible, over Zp for every prime p. (See Exercise 29.), The Mod p Irreducibility Test can also be helpful in checking for, irreducibility of polynomials of degree greater than 3 and polynomials, with rational coefficients., EXAMPLE 7 Let f(x) 5 (3/7)x4 2 (2/7)x2 1 (9/35)x 1 3/5. We will, show that f(x) is irreducible over Q. First, let h(x) 5 35f(x) 5 15x4 2, 10x2 1 9x 1 21. Then f(x) is irreducible over Q if h(x) is irreducible, over Z. Next, applying the Mod 2 Irreducibility Test to h(x), we get, h1x2 5 x4 1 x 1 1. Clearly, h1x2 has no zeros in Z2. Furthermore, h1x2, has no quadratic factor in Z2[x] either. [For if so, the factor would have, to be either x2 1 x 1 1 or x2 1 1. Long division shows that x2 1 x 1 1, is not a factor, and x2 1 1 cannot be a factor because it has a zero,, whereas h1x2 does not.] Thus, h1x2 is irreducible over Z2[x]. This guarantees that h(x) is irreducible over Q., EXAMPLE 8 Let f(x) 5 x 5 1 2x 1 4. Obviously, neither Theorem, 17.1 nor the Mod 2 Irreducibility Test helps here. Let’s try mod 3., Substitution of 0, 1, and 2 into f (x) does not yield 0, so there are no linear, factors. But f (x) may have a quadratic factor. If so, we may assume it has, the form x2 1 ax 1 b (see Exercise 5). This gives nine possibilities to, check. We can immediately rule out each of the nine that has a zero over, Z3, since f (x) does not have one. This leaves only x2 1 1, x2 1 x 1 2, and, x2 1 2x 1 2 to check. These are eliminated by long division. So, since, f (x) is irreducible over Z3, f(x) is irreducible over Q. (Why is it unnecessary to check for cubic or fourth-degree factors?), Another important irreducibility test is the following one, credited to, Ferdinand Eisenstein (1823–1852), a student of Gauss. The corollary, was first proved by Gauss by a different method., Theorem 17.4 Eisenstein’s Criterion (1850), Let, f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]., If there is a prime p such that p B an, p | an21, . . . , p | a0 and p2 B a0,, then f(x) is irreducible over Q., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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316, , Rings, , PROOF If f (x) is reducible over Q, we know by Theorem 17.2 that, there exist elements g(x) and h(x) in Z[x] such that f(x) 5 g(x)h(x),, 1 # deg g(x), and 1 # deg h(x) , n. Say g(x) 5 br xr 1 ? ? ? 1 b0 and, h(x) 5 cs xs 1 ? ? ? 1 c0. Then, since p | a0, p2 B a0, and a0 5 b0c0, it follows that p divides one of b0 and c0 but not the other. Let us say p | b0, and p B c0. Also, since p B an 5 brcs, we know that p B br. So, there is a, least integer t such that p B bt. Now, consider at 5 btc0 1 bt21c1 1 ? ? ?, 1 b0ct. By assumption, p divides at and, by choice of t, every summand, on the right after the first one is divisible by p. Clearly, this forces p to, divide btc0 as well. This is impossible, however, since p is prime and p, divides neither bt nor c0., Corollary Irreducibility of pth Cyclotomic Polynomial, For any prime p, the pth cyclotomic polynomial, xp 1, 5 xp21 1 xp22 1 ? ? ? 1 x 1 1, x1, is irreducible over Q., Fp(x) 5, , PROOF Let, f(x)£ p(x 1), , (x 1)p 1, p, p, p, xp1a b xp2 a b xp3 . . . a b., 1, 2, 1, (x 1) 1, , Then, since every coefficient except that of xp21 is divisible by p and, the constant term is not divisible by p2, by Eisenstein’s Criterion, f(x) is, irreducible over Q. So, if Fp(x) 5 g(x)h(x) were a nontrivial factorization of Fp(x) over Q, then f(x) 5 Fp(x 1 1) 5 g(x 1 1) ? h(x 1 1), would be a nontrivial factorization of f(x) over Q. Since this is impossible, we conclude that Fp(x) is irreducible over Q., EXAMPLE 9 The polynomial 3x5 1 15x4 2 20x3 1 10x 1 20 is, irreducible over Q because 5 B 3 and 25 B 20 but 5 does divide 15,, 220, 10, and 20., The principal reason for our interest in irreducible polynomials, stems from the fact that there is an intimate connection among them,, maximal ideals, and fields. This connection is revealed in the next theorem and its first corollary., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 317, , Theorem 17.5 kp(x)l Is Maximal If and Only If p(x) Is Irreducible, Let F be a field and let p(x) [ F[x]. Then kp(x)l is a maximal ideal, in F[x] if and only if p(x) is irreducible over F., , PROOF Suppose first that kp(x)l is a maximal ideal in F[x]. Clearly,, p(x) is neither the zero polynomial nor a unit in F[x], because neither, {0} nor F[x] is a maximal ideal in F[x]. If p(x) 5 g(x)h(x) is a factorization of p(x) over F, then k p(x)l # kg(x)l # F[x]. Thus, k p(x)l 5 kg(x)l, or F[x] 5 kg(x)l. In the first case, we must have deg p(x) 5 deg g(x). In, the second case, it follows that deg g(x) 5 0 and, consequently, deg h(x) 5, deg p(x). Thus, p(x) cannot be written as a product of two polynomials, in F[x] of lower degree., Now, suppose that p(x) is irreducible over F. Let I be any ideal of, F[x] such that k p(x)l # I # F[x]. Because F[x] is a principal ideal domain, we know that I 5 kg(x)l for some g(x) in F[x]. So, p(x) [ kg(x)l, and, therefore, p(x) 5 g(x)h(x), where h(x) [ F[x]. Since p(x) is irreducible over F, it follows that either g(x) is a constant or h(x) is a constant. In the first case, we have I 5 F[x]; in the second case, we have, k p(x)l 5 kg(x)l 5 I. So, k p(x)l is maximal in F[x]., Corollary 1 F[x]/k p(x)l Is a Field, Let F be a field and p(x) be an irreducible polynomial over F. Then, F[x]/k p(x)l is a field., , PROOF This follows directly from Theorems 17.5 and 14.4., The next corollary is a polynomial analog of Euclid’s Lemma for, primes (see Chapter 0)., Corollary 2 p(x) | a(x)b(x) Implies p(x) | a(x) or p(x) | b(x), Let F be a field and let p(x), a(x), b(x) [ F[x]. If p(x) is irreducible, over F and p(x) | a(x)b(x), then p(x) | a(x) or p(x) | b(x)., , PROOF Since p(x) is irreducible, F[x]/kp(x)l is a field and, therefore, an, integral domain. From Theorem 14.3, we know that kp(x)l is a prime, ideal, and since p(x) divides a(x)b(x), we have a(x)b(x) [ kp(x)l. Thus,, a(x) [ k p(x)l or b(x) [ k p(x)l. This means that p(x) | a(x) or p(x) | b(x)., The next two examples put the theory to work., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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318, , Rings, , EXAMPLE 10 We construct a field with eight elements. By Theorem, 17.1 and Corollary 1 of Theorem 17.5, it suffices to find a cubic polynomial over Z2 that has no zero in Z2. By inspection, x3 1x 1 1 fills the, bill. Thus, Z2[x]/kx3 1 x 1 1l 5 {ax2 1 bx 1 c 1 kx3 1 x 1 1l | a, b, c, [ Z2} is a field with eight elements. For practice, let us do a few calculations in this field. Since the sum of two polynomials of the form ax2 1, bx 1 c is another one of the same form, addition is easy. For example,, (x2 1 x 1 1 1 kx3 1 x 1 1l) 1 (x2 1 1 1 kx3 1 x 1 1l), 5 x 1 kx3 1 x 1 1l., On the other hand, multiplication of two coset representatives need not, yield one of the original eight coset representatives:, (x2 1 x 1 1 1 kx3 1 x 1 1l) ? (x2 1 1 1 kx3 1 x 1 1l), 5 x4 1 x3 1 x 1 1 1 kx3 1 x 1 1l 5 x4 1 kx3 1 x 1 1l, (since the ideal absorbs the last three terms). How do we express this in, the form ax2 1 bx 1 c 1 kx3 1 x 1 1l? One way is to long divide x4 by, x3 1 x 1 1 to obtain the remainder of x2 1 x (just as one reduces, 12 1 k5l to 2 1 k5l by dividing 12 by 5 to obtain the remainder 2)., Another way is to observe that x3 1 x 1 1 1 kx3 1 x 1 1l 5 0 1, kx3 1 x 1 1l implies x3 1 kx3 1 x 1 1l 5 x 1 1 1 kx3 1 x 1 1l. Thus,, we may multiply both sides by x to obtain, x4 1 kx3 1 x 1 1l 5 x2 1 x 1 kx3 1 x 1 1l., Similarly,, (x2 1 x 1 kx3 1 x 1 1l) ? (x 1 kx3 1 x 1 1l), 5 x3 1 x2 1 kx3 1 x 1 1l, 5 x2 1 x 1 1 1 kx3 1 x 1 1l., A partial multiplication table for this field is given in Table 17.1. To, simplify the notation, we indicate a coset by its representative only., Table 17.1 A Partial Multiplication Table for Example 10, , 1, x, x11, x2, x2 1 1, , 1, , x, , x11, , x2, , x2 1 1, , x2 1 x, , x2 1 x 1 1, , 1, x, x11, x2, x2 1 1, , x, x2, x2 1 x, x11, 1, , x11, x2 1 x, x2 1 1, x2 1 x 1 1, x2, , x2, x11, x2 1 x 1 1, x2 1 x, x, , x2 1 1, 1, x2, x, x2 1 x 1 1, , x2 1 x, x2 1 x 1 1, 1, x2 1 1, x11, , x2 1 x 1 1, x2 1 1, x, 1, x2 1 x, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 319, , (Complete the table yourself. Keep in mind that x3 can be replaced by, x 1 1 and x4 by x2 1 x.), EXAMPLE 11 Since x2 1 1 has no zero in Z3, it is irreducible over, Z3. Thus, Z3[x]/kx2 1 1l is a field. Analogous to Example 12 in Chapter 14,, Z3[x]/kx2 1 1l 5 {ax 1 b 1 kx2 1 1l | a, b [ Z3}. Thus, this field has, nine elements. A multiplication table for this field can be obtained from, Table 13.1 by replacing i by x. (Why does this work?), , Unique Factorization in Z[x], As a further application of the ideas presented in this chapter, we next, prove that Z[x] has an important factorization property. In Chapter 18,, we will study this property in greater depth. The first proof of Theorem, 17.6 was given by Gauss. In reading this theorem and its proof, keep in, mind that the units in Z[x] are precisely f(x) 5 1 and f(x) 5 21 (see, Exercise 25 in Chapter 12), the irreducible polynomials of degree 0, over Z are precisely those of the form f(x) 5 p and f(x) 5 2p where p is, a prime, and every nonconstant polynomial from Z[x] that is irreducible, over Z is primitive (see Exercise 3)., Theorem 17.6 Unique Factorization in Z[x], Every polynomial in Z[x] that is not the zero polynomial or a unit, in Z[x] can be written in the form b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x),, where the bi’s are irreducible polynomials of degree 0 and the pi(x)’s, are irreducible polynomials of positive degree. Furthermore, if, b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ? qn(x),, where the bi’s and ci’s are irreducible polynomials of degree 0 and the, pi(x)’s and qi(x)’s are irreducible polynomials of positive degree, then, s 5 t, m 5 n, and, after renumbering the c’s and q(x)’s, we have bi 5, 6ci for i 5 1, . . . , s and pi(x) 5 6qi(x) for i 5 1, . . . , m., , PROOF Let f(x) be a nonzero, nonunit polynomial from Z[x]. If, deg f(x) 5 0, then f(x) is constant and the result follows from the, Fundamental Theorem of Arithmetic. If deg f(x) . 0, let b denote the, content of f(x), and let b1b2 ? ? ? bs be the factorization of b as a product, of primes. Then, f(x) 5 b1b2 ? ? ? bs f1(x), where f1(x) belongs to Z[x], is, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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320, , Rings, , primitive and deg f1(x) 5 deg f(x). Thus, to prove the existence portion, of the theorem, it suffices to show that a primitive polynomial f(x) of, positive degree can be written as a product of irreducible polynomials, of positive degree. We proceed by induction on deg f(x). If deg f(x) 5 1,, then f(x) is already irreducible and we are done. Now suppose that, every primitive polynomial of degree less than deg f(x) can be written, as a product of irreducibles of positive degree. If f(x) is irreducible,, there is nothing to prove. Otherwise, f(x) 5 g(x)h(x), where both g(x), and h(x) are primitive and have degree less than that of f(x). Thus, by induction, both g(x) and h(x) can be written as a product of irreducibles of, positive degree. Clearly, then, f(x) is also such a product., To prove the uniqueness portion of the theorem, suppose that, f(x) 5 b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ?, qn(x), where the bi’s and ci’s are irreducible polynomials of degree 0, and the pi(x)’s and qi(x)’s are irreducible polynomials of positive degree., Let b 5 b1b2 ? ? ? bs and c 5 c1c2 ? ? ? ct. Since the p(x)’s and q(x)’s are, primitive, it follows from Gauss’s Lemma that p1(x)p2(x) ? ? ? pm(x) and, q1(x)q2(x) ? ? ? qn(x) are primitive. Hence, both b and c must equal plus, or minus the content of f(x) and, therefore, are equal in absolute value., It then follows from the Fundamental Theorem of Arithmetic that s 5 t, and, after renumbering, bi 5 6ci for i 5 1, 2, . . . , s. Thus, by canceling the constant terms in the two factorizations for f(x), we have, p1(x)p2(x) ? ? ? pm(x) 5 6q1(x) q2(x) ? ? ? qn(x). Now, viewing the p(x)’s, and q(x)’s as elements of Q[x] and noting that p1(x) divides q1(x) ? ? ?, qn(x), it follows from Corollary 2 of Theorem 17.5 and induction (see, Exercise 28) that p1(x) | qi(x) for some i. By renumbering, we may assume i 5 1. Then, since q1(x) is irreducible, we have q1(x) 5 (r/s)p1(x),, where r, s [ Z. However, because both q1(x) and p1(x) are primitive, we, must have r/s 5 61. So, q1(x) 5 6p1(x). Also, after canceling, we have, p2(x) ? ? ? pm(x) 5 6q2(x) ? ? ? qn(x). Now, we may repeat the argument, above with p2(x) in place of p1(x). If m , n, after m such steps we, would have 1 on the left and a nonconstant polynomial on the right., Clearly, this is impossible. On the other hand, if m . n, after n steps we, would have 61 on the right and a nonconstant polynomial on the left—, another impossibility. So, m 5 n and pi(x) 5 6qi(x) after suitable, renumbering of the q(x)’s., , Weird Dice: An Application, of Unique Factorization, EXAMPLE 12 Consider an ordinary pair of dice whose faces are, labeled 1 through 6. The probability of rolling a sum of 2 is 1/36, the, probability of rolling a sum of 3 is 2/36, and so on. In a 1978 issue of, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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321, , 17 | Factorization of Polynomials, , Scientific American [1], Martin Gardner remarked that if one were to, label the six faces of one cube with the integers 1, 2, 2, 3, 3, 4 and the six, faces of another cube with the integers 1, 3, 4, 5, 6, 8, then the probability of obtaining any particular sum with these dice (called Sicherman, dice) would be the same as the probability of rolling that sum with ordinary dice (that is, 1/36 for a 2, 2/36 for a 3, and so on). See Figure 17.1., In this example, we show how the Sicherman labels can be derived, and, that they are the only possible such labels besides 1 through 6. To do so,, we utilize the fact that Z[x] has the unique factorization property., , 2, , 3, , 4, , 5, , 6, , 7, , 2, , 3, , 3, , 4, , 4, , 5, , 3, , 4, , 5, , 6, , 7, , 8, , 4, , 5, , 5, , 6, , 6, , 7, , 4, , 5, , 6, , 7, , 8, , 9, , 5, , 6, , 6, , 7, , 7, , 8, , 5, , 6, , 7, , 8, , 9, , 10, , 6, , 7, , 7, , 8, , 8, , 9, , 6, , 7, , 8, , 9, , 10, , 11, , 7, , 8, , 8, , 9, , 9, , 10, , 7, , 8, , 9, , 10, , 11, , 12, , 9, , 10, , 10, , 11, , 11, , 12, , Figure 17.1, , To begin, let us ask ourselves how we may obtain a sum of 6, say, with, an ordinary pair of dice. Well, there are five possibilities for the two faces:, (5, 1), (4, 2), (3, 3), (2, 4), and (1, 5). Next we consider the product of the, two polynomials created by using the ordinary dice labels as exponents:, (x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x)., Observe that we pick up the term x6 in this product in precisely the following ways: x5 ? x1, x4 ? x2 , x3 ? x3, x2 ? x4, x1 ? x5. Notice the correspondence between pairs of labels whose sums are 6 and pairs of terms, whose products are x6. This correspondence is one-to-one, and it is valid, for all sums and all dice—including the Sicherman dice and any other, dice that yield the desired probabilities. So, let a1, a2, a3, a4, a5, a6 and, b1, b2, b3, b4, b5, b6 be any two lists of positive integer labels for the faces, of a pair of cubes with the property that the probability of rolling any, particular sum with these dice (let us call them weird dice) is the same as, the probability of rolling that sum with ordinary dice labeled 1 through, 6. Using our observation about products of polynomials, this means that, (x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x), 5 (xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6) ?, (1), (xb1 1 xb2 1 xb3 1 xb4 1 xb5 1 xb6)., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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322, , Rings, , Now all we have to do is solve this equation for the a’s and b’s. Here is, where unique factorization in Z[x] comes in. The polynomial x6 1 x5 1, x4 1 x3 1 x2 1 x factors uniquely into irreducibles as, x(x 1 1)(x2 1 x 1 1)(x2 2 x 1 1), so that the left-hand side of Equation (1) has the irreducible factorization, x2(x 1 1)2(x2 1 x 1 1)2(x2 2 x 1 1)2., So, by Theorem 17.6, this means that these factors are the only possible, irreducible factors of P(x) 5 xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6. Thus,, P(x) has the form, xq(x 1 1)r(x2 1 x 1 1)t(x2 2 x 1 1)u,, where 0 # q, r, t, u # 2., To restrict further the possibilities for these four parameters, we evaluate P(1) in two ways. P(1) 5 1 a 1 1 1 a 2 1 ? ? ? 1 1 a 6 5 6 and, P(1) 5 1q2r3t1u. Clearly, this means that r 5 1 and t 5 1. What about q?, Evaluating P(0) in two ways shows that q 2 0. On the other hand, if, q 5 2, the smallest possible sum one could roll with the corresponding, labels for dice would be 3. Since this violates our assumption, we have, now reduced our list of possibilities for q, r, t, and u to q 5 1, r 5 1,, t 5 1, and u 5 0, 1, 2. Let’s consider each of these possibilities in turn., When u 5 0, P(x) 5 x4 1 x3 1 x3 1 x2 1 x2 1 x, so the die labels, are 4, 3, 3, 2, 2, 1—a Sicherman die., When u 5 1, P(x) 5 x6 1 x5 1 x4 1 x3 1 x2 1 x, so the die labels, are 6, 5, 4, 3, 2, 1—an ordinary die., When u 5 2, P(x) 5 x8 1 x6 1 x5 1 x4 1 x3 1 x, so the die labels, are 8, 6, 5, 4, 3, 1—the other Sicherman die., This proves that the Sicherman dice do give the same probabilities, as ordinary dice and that they are the only other pair of dice that have, this property., , Exercises, No matter how good you are at something, there’s always about a million, people better than you., homer simpson, , 1. Verify the assertion made in Example 2., 2. Suppose that D is an integral domain and F is a field containing D., If f(x) [ D[x] and f(x) is irreducible over F but reducible over D,, what can you say about the factorization of f(x) over D?, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 323, , 3. Show that a nonconstant polynomial from Z[x] that is irreducible, over Z is primitive. (This exercise is referred to in this chapter.), 4. Suppose that f(x) 5 xn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]. If r is rational and x 2 r divides f(x), show that r is an integer., 5. Let F be a field and let a be a nonzero element of F., a. If af(x) is irreducible over F, prove that f(x) is irreducible over F., b. If f(ax) is irreducible over F, prove that f(x) is irreducible over F., c. If f(x 1 a) is irreducible over F, prove that f(x) is irreducible, over F., d. Use part c to prove that 8x3 2 6x 1 1 is irreducible over Q., (This exercise is referred to in this chapter.), 6. Let F be a field and f(x) [ F[x]. Show that, as far as deciding upon, the irreducibility of f(x) over F is concerned, we may assume that, f(x) is monic. (This assumption is useful when one uses a computer, to check for irreducibility.), 7. Explain how the Mod p Irreducibility Test (Theorem 17.3) can be, used to test members of Q[x] for irreducibility., 8. Suppose that f(x) [ Zp[x] and f(x) is irreducible over Zp, where p is, a prime. If deg f(x) 5 n, prove that Zp[x]/k f(x)l is a field with pn, elements., 9. Construct a field of order 25., 10. Construct a field of order 27., 11. Show that x3 1 x2 1 x 1 1 is reducible over Q. Does this fact contradict the corollary to Theorem 17.4?, 12. Determine which of the polynomials below is (are) irreducible, over Q., a. x5 1 9x4 1 12x2 1 6, b. x4 1 x 1 1, c. x4 1 3x2 1 3, d. x5 1 5x2 1 1, e. (5/2)x5 1 (9/2)x4 1 15x3 1 (3/7)x2 1 6x 1 3/14, 13. Show that x4 1 1 is irreducible over Q but reducible over R. (This, exercise is referred to in this chapter.), 14. Show that x2 1 x 1 4 is irreducible over Z11., 15. Let f(x) 5 x3 1 6 [ Z7[x]. Write f(x) as a product of irreducible, polynomials over Z7., 16. Let f(x) 5 x3 1 x2 1 x 1 1 [ Z2[x]. Write f(x) as a product of irreducible polynomials over Z2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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324, , Rings, , 17. Let p be a prime., a. Show that the number of reducible polynomials over Zp of the, form x2 1 ax 1 b is p(p 1 1)/2., b. Determine the number of reducible quadratic polynomials over Zp., 18. Let p be a prime., a. Determine the number of irreducible polynomials over Zp of the, form x2 1 ax 1 b., b. Determine the number of irreducible quadratic polynomials, over Zp., 19. Show that for every prime p there exists a field of order p2., 20. Prove that, for every positive integer n, there are infinitely many, polynomials of degree n in Z[x] that are irreducible over Q., 21. Show that the field given in Example 11 in this chapter is isomorphic to the field given in Example 9 in Chapter 13., 22. Let f(x) [ Zp[x]. Prove that if f(x) has no factor of the form x2 1, ax 1 b, then it has no quadratic factor over Zp., 23. Find all monic irreducible polynomials of degree 2 over Z3., 24. Given that p is not the zero of a nonzero polynomial with rational, coefficients, prove that p 2 cannot be written in the form ap 1 b,, where a and b are rational., 25. Find all the zeros and their multiplicities of x5 1 4x4 1 4x3 2 x2 2, 4x 1 1 over Z5., 26. Find all zeros of f(x) 5 3x2 1 x 1 4 over Z7 by substitution. Find, all zeros of f(x) by using the quadratic formula (2b 6 2b2 4ac) ?, (2a)21 (all calculations are done in Z7). Do your answers agree?, Should they? Find all zeros of g(x) 5 2x2 1 x 1 3 over Z5 by substitution. Try the quadratic formula on g(x). Do your answers, agree? State necessary and sufficient conditions for the quadratic, formula to yield the zeros of a quadratic from Zp[x], where p is a, prime greater than 2., 27. (Rational Root Theorem) Let, f (x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x], and an 2 0. Prove that if r and s are relatively prime integers and, f(r/s) 5 0, then r | a0 and s | an., 28. Let F be a field and let p(x), a1(x), a2(x), . . . , ak(x) [ F[x], where, p(x) is irreducible over F. If p(x) | a1(x)a2(x) ? ? ? ak(x), show that, p(x) divides some ai(x). (This exercise is referred to in the proof of, Theorem 17.6.), 29. Show that x4 1 1 is reducible over Zp for every prime p. (This exercise is referred to in this chapter.), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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17 | Factorization of Polynomials, , 325, , 30. If p is a prime, prove that xp21 2 xp22 1 xp23 2 ? ? ? 2 x 1 1 is, irreducible over Q., 31. Let F be a field and let p(x) be irreducible over F. If E is a field, that contains F and there is an element a in E such that p(a) 5 0,, show that the mapping f: F[x] S E given by f(x) S f(a) is a ring, homomorphism with kernel kp(x)l. (This exercise is referred to in, Chapter 20.), 32. Prove that the ideal kx2 1 1l is prime in Z[x] but not maximal in Z[x]., 33. Let F be a field and let p(x) be irreducible over F. Show that {a 1, k p(x)l | a [ F} is a subfield of F[x]/kp(x)l isomorphic to F. (This, exercise is referred to in Chapter 20.), 34. Let F be a field and let f(x) be a polynomial in F[x] that is reducible, over F. Prove that kf(x)l is not a prime ideal in F[x]., 35. Example 1 in this chapter shows the converse of Theorem 17.2 is, not true. That is, a polynomial f(x) in Z[x] can be reducible over Z, but irreducible over Q. State a condition on f(x) that makes the converse true., 36. Suppose there is a real number r with the property that r 1 1/r is, an odd integer. Prove that r is irrational., 37. In the game of Monopoly, would the probabilities of landing on, various properties be different if the game were played with, Sicherman dice instead of ordinary dice? Why?, 38. Carry out the analysis given in Example 12 for a pair of tetrahedrons instead of a pair of cubes. (Define ordinary tetrahedral dice, as the ones labeled 1 through 4.), 39. Suppose in Example 12 that we begin with n (n . 2) ordinary dice, each labeled 1 through 6, instead of just two. Show that the only, possible labels that produce the same probabilities as n ordinary, dice are the labels 1 through 6 and the Sicherman labels., 40. Show that one two-sided die labeled with 1 and 4 and another 18sided die labeled with 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8, yield the same probabilities as an ordinary pair of cubes labeled, 1 through 6. Carry out an analysis similar to that given in Example, 12 to derive these labels., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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326, , Rings, , Reference, 1. Martin Gardner, “Mathematical Games,” Scientific American 238/2, (1978): 19–32., , Suggested Readings, Duane Broline, “Renumbering the Faces of Dice,” Mathematics Magazine, 52 (1979): 312–315., In this article, the author extends the analysis we carried out in, Example 12 to dice in the shape of Platonic solids., J. A. Gallian and D. J. Rusin, “Cyclotomic Polynomials and Nonstandard, Dice,” Discrete Mathematics 27 (1979): 245–259., Here Example 12 is generalized to the case of n dice each with m, labels for all n and m greater than 1., Randall Swift and Brian Fowler, “Relabeling Dice,” The College Mathematics Journal 30 (1999): 204–208., The authors use the method presented in this chapter to derive positive, integer labels for a pair of dice that are not six-sided but give the, same probabilities for the sum of the faces as a pair of cubes labeled, 1 through 6., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Serge Lang, , Bogdan Oporowski, , Lang’s Algebra changed the way graduate, algebra is taught . . . . It has affected all, subsequent graduate-level algebra books., Citation for the Steele Prize, , Serge Lang was a prolific mathematician,, inspiring teacher, and political activist. He, was born near Paris on May 19, 1927. His, family moved to Los Angeles when he was a, teenager. Lang received a B.A. in physics, from Caltech in 1946 and a Ph.D. in mathematics from Princeton in 1951 under Emil, Artin (see the biography in Chapter 19). His, first permanent position was at Columbia, University in 1955, but in 1971 Lang resigned his position at Columbia as a protest, against Columbia’s handling of Vietnam antiwar protesters. He joined Yale University in, 1972 and remained there until his retirement., Lang made significant contributions to, number theory, algebraic geometry, differential geometry, and analysis. He wrote more, than 120 research articles and 60 books., His most famous and influential book was, his graduate-level Algebra. Lang was a, , prize-winning teacher known for his extraordinary devotion to students. Lang often, got into heated discussions about mathematics, the arts, and politics. In one incident, he, threatened to hit a fellow mathematician, with a bronze bust for not conceding it was, self-evident that the Beatles were greater, musicians than Beethoven., Among Lang’s honors were the Steele, Prize for Mathematical Exposition from the, American Mathematical Society, the Cole, Prize in Algebra (see Chapter 25), and election to the National Academy of Sciences., Lang died on September 25, 2005, at the, age of 78., For more information about Lang, visit:, http://wikipedia.org/wiki/, Serge_Lang, , 327, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Divisibility in, 18 Integral Domains, Give me a fruitful error anytime, full of seeds, bursting with its own, corrections. You can keep your sterile truth for yourself., vilfredo pareto, , Irreducibles, Primes, In the preceding two chapters, we focused on factoring polynomials, over the integers or a field. Several of those results—unique factorization in Z[x] and the division algorithm for F[x], for instance—are natural counterparts to theorems about the integers. In this chapter and the, next, we examine factoring in a more abstract setting., Definition Associates, Irreducibles, Primes, , Elements a and b of an integral domain D are called associates if, a 5 ub, where u is a unit of D. A nonzero element a of an integral, domain D is called an irreducible if a is not a unit and, whenever b,, c [ D with a 5 bc, then b or c is a unit. A nonzero element a of an, integral domain D is called a prime if a is not a unit and a | bc implies, a | b or a | c., , Roughly speaking, an irreducible is an element that can be factored, only in a trivial way. Notice that an element a is a prime if and only if, kal is a prime ideal., Relating the definitions above to the integers may seem a bit confusing, since in Chapter 0 we defined a positive integer to be a prime if it, satisfies our definition of an irreducible, and we proved that a prime integer satisfies the definition of a prime in an integral domain (Euclid’s, Lemma). The source of the confusion is that in the case of the integers,, the concepts of irreducibles and primes are equivalent, but in general, as, we will soon see, they are not., The distinction between primes and irreducibles is best illustrated by, integral domains of the form Z[2d] 5 {a 1 b2d | a, b [ Z}, where d is, not 1 and is not divisible by the square of a prime. (These rings are of, fundamental importance in number theory.) To analyze these rings, we, need a convenient method of determining their units, irreducibles, and, 328, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 329, , primes. To do this, we define a function N, called the norm, from Z[2d], into the nonnegative integers by N(a 1 b2d) 5 |a2 2 db2|. We leave it, to the reader (Exercise 1) to verify the following four properties: N(x) 5 0, if and only if x 5 0; N(xy) 5 N(x)N(y) for all x and y; x is a unit if and, only if N(x) 5 1; and, if N(x) is prime, then x is irreducible in Z[2d]., EXAMPLE 1 We exhibit an irreducible in Z[23] that is not prime., Here, N(a 1 b23) 5 a2 1 3b2. Consider 1 1 23. Suppose that we, can factor this as xy, where neither x nor y is a unit. Then N(xy) 5, N(x)N(y) 5 N(1 1 23) 5 4, and it follows that N(x) 5 2. But there are, no integers a and b that satisfy a2 1 3b2 5 2. Thus, x or y is a unit and, 1 1 23 is an irreducible. To verify that it is not prime, we observe that, (1 1 23)(1 2 23) 5 4 5 2 ? 2, so that 1 1 23 divides 2 ? 2. On the, other hand, for integers a and b to exist so that 2 5 (1 1 23)(a 1, b23) 5 (a 2 3b) 1 (a 1 b)23, we must have a 2 3b 5 2 and a 1, b 5 0, which is impossible., Showing that an element of a ring of the form Z[2d] is irreducible is, more difficult when d . 1. The next example illustrates one method of, doing this. The example also shows that the converse of the fourth, property above for the norm is not true. That is, it shows that x may be, irreducible even if N(x) is not prime., EXAMPLE 2 The element 7 is irreducible in the ring Z[25]. To verify, this assertion, suppose that 7 5 xy, where neither x nor y is a unit. Then, 49 5 N(7) 5 N(x)N(y), and since x is not a unit, we cannot have N(x) 5, 1. This leaves only the case N(x) 5 7. Let x 5 a 1 b25. Then there are, integers a and b satisfying |a2 2 5b2| 5 7. This means that a2 2 5b2 5, 67. Viewing this equation modulo 5 and trying all possible cases for a, reveals that the only solution is a 5 0. But this means that a is divisible, by 5, and this implies that |a2 2 5b2| 5 7 is divisible by 5, which is, false., Example 1 raises the question of whether or not there is an integral, domain containing a prime that is not an irreducible. The answer: no., Theorem 18.1 Prime Implies Irreducible, In an integral domain, every prime is an irreducible., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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330, , Rings, , PROOF Suppose that a is a prime in an integral domain and a 5 bc., We must show that b or c is a unit. By the definition of prime, we know, that a | b or a | c. Say at 5 b. Then 1b 5 b 5 at 5 (bc)t 5 b(ct) and,, by cancellation, 1 5 ct. Thus, c is a unit., Recall that a principal ideal domain is an integral domain in which, every ideal has the form kal. The next theorem reveals a circumstance, in which primes and irreducibles are equivalent., Theorem 18.2 PID Implies Irreducible Equals Prime, In a principal ideal domain, an element is an irreducible if and only, if it is a prime., , PROOF Theorem 18.1 shows that primes are irreducibles. To prove the, converse, let a be an irreducible element of a principal ideal domain D, and suppose that a | bc. We must show that a | b or a | c. Consider the, ideal I 5 {ax 1 by | x, y [ D} and let kdl 5 I. Since a [ I, we can write, a 5 dr, and because a is irreducible, d is a unit or r is a unit. If d is a, unit, then I 5 D and we may write 1 5 ax 1 by. Then c 5 acx 1 bcy,, and since a divides both terms on the right, a also divides c., On the other hand, if r is a unit, then kal 5 kdl 5 I, and, because b [ I,, there is an element t in D such that at 5 b. Thus, a divides b., It is an easy consequence of the respective division algorithms for Z, and F[x], where F is a field, that Z and F[x] are principal ideal domains, (see Exercise 41 in Chapter 14 and Theorem 16.3). Our next example, shows, however, that one of the most familiar rings is not a principal, ideal domain., EXAMPLE 3 We show that Z[x] is not a principal ideal domain., Consider the ideal I 5 k2, xl. We claim that I is not of the form kh(x)l. If, this were so, there would be f(x) and g(x) in Z[x] such that 2 5 h(x)f(x), and x 5 h(x)g(x), since both 2 and x belong to I. By the degree rule, (Exercise 19 in Chapter 16), 0 5 deg 2 5 deg h(x) 1 deg f(x), so that, h(x) is a constant polynomial. To determine which constant, we observe, that 2 5 h(1)f(1). Thus, h(1) 5 61 or 62. Since 1 is not in I, we must, have h(x) 5 62. But then x 5 62g(x), which is nonsense., We have previously proved that the integral domains Z and Z[x] have, important factorization properties: Every integer greater than 1 can be, uniquely factored as a product of irreducibles (that is, primes), and, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 331, , every nonzero, nonunit polynomial can be uniquely factored as a product of irreducible polynomials. It is natural to ask whether all integral, domains have this property. The question of unique factorization in integral domains first arose with the efforts to solve a famous problem in, number theory that goes by the name Fermat’s Last Theorem., , Historical Discussion, of Fermat’s Last Theorem, There are infinitely many nonzero integers x, y, z that satisfy the equation x2 1 y2 5 z2. But what about the equation x3 1 y3 5 z3 or, more, generally, xn 1 yn 5 zn, where n is an integer greater than 2 and x, y, z, are nonzero integers? Well, no one has ever found a single solution of, this equation, and for more than three centuries many have tried to, prove that there is none. The tremendous effort put forth by the likes of, Euler, Legendre, Abel, Gauss, Dirichlet, Cauchy, Kummer, Kronecker,, and Hilbert to prove that there are no solutions to this equation has, greatly influenced the development of ring theory., About a thousand years ago, Arab mathematicians gave an incorrect, proof that there were no solutions when n 5 3. The problem lay dormant until 1637, when the French mathematician Pierre de Fermat, (1601–1665) wrote in the margin of a book, “. . . it is impossible to, separate a cube into two cubes, a fourth power into two fourth powers,, or, generally, any power above the second into two powers of the same, degree: I have discovered a truly marvelous demonstration [of this general theorem] which this margin is too narrow to contain.”, Because Fermat gave no proof, many mathematicians tried to prove, the result. The case where n 5 3 was done by Euler in 1770, although, his proof was incomplete. The case where n 5 4 is elementary and was, done by Fermat himself. The case where n 5 5 was done in 1825 by, Dirichlet, who had just turned 20, and by Legendre, who was past 70., Since the validity of the case for a particular integer implies the validity for all multiples of that integer, the next case of interest was n 5 7., This case resisted the efforts of the best mathematicians until it was, done by Gabriel Lamé in 1839. In 1847, Lamé stirred excitement by, announcing that he had completely solved the problem. His approach, was to factor the expression xp 1 yp, where p is an odd prime, into, (x 1 y)(x 1 ay) ? ? ? (x 1 a p21y),, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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332, , Rings, , where a is the complex number cos(2p/p) 1 i sin(2p/p). Thus, his, factorization took place in the ring Z[a] 5 {a0 1 a1a 1 ? ? ? 1, ap21a p21 | ai [ Z}. But Lamé made the mistake of assuming that, in, such a ring, factorization into the product of irreducibles is unique. In, fact, three years earlier, Ernst Eduard Kummer had proved that this is, not always the case. Undaunted by the failure of unique factorization,, Kummer began developing a theory to “save” factorization by creating a new type of number. Within a few weeks of Lamé’s announcement, Kummer had shown that Fermat’s Last Theorem is true for all, primes of a special type. This proved that the theorem was true for all, exponents less than 100, prime or not, except for 37, 59, 67, and 74., Kummer’s work has led to the theory of ideals as we know it today., Over the centuries, many proposed proofs have not held up under, scrutiny. The famous number theorist Edmund Landau received so many, of these that he had a form printed with “On page ____, lines ____ to, ____, you will find there is a mistake.” Martin Gardner, “Mathematical, Games” columnist of Scientific American, had postcards printed to, decline requests from readers asking him to examine their proofs., Recent discoveries tying Fermat’s Last Theorem closely to modern, mathematical theories gave hope that these theories might eventually, lead to a proof. In March 1988, newspapers and scientific publications, worldwide carried news of a proof by Yoichi Miyaoka (see Figure 18.1)., Within weeks, however, Miyaoka’s proof was shown to be invalid. In, June 1993, excitement spread through the mathematics community, with the announcement that Andrew Wiles of Princeton University had, proved Fermat’s Last Theorem (see Figure 18.2). The Princeton mathematics department chairperson was quoted as saying, “When we, heard it, people started walking on air.” But once again a proof did not, hold up under scrutiny. This story does have a happy ending. The mathematical community has agreed on the validity of the revised proof, given by Wiles and Richard Taylor in September of 1994., In view of the fact that so many eminent mathematicians were unable to prove Fermat’s Last Theorem, despite the availability of the, vastly powerful theories, it seems highly improbable that Fermat had a, correct proof. Most likely, he made the error that his successors made, of assuming that the properties of integers, such as unique factorization, carry over to integral domains in general., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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333, , Times Magazine, Science News #133. LA Times, , 18 | Divisibility in Integral Domains, , Figure 18.1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Rings, , AP Photo/Charles Rex Arbogast, , 334, , Figure 18.2, , Andrew Wiles, , Unique Factorization Domains, We now have the necessary terminology to formalize the idea of, unique factorization., Definition Unique Factorization Domain (UFD), , An integral domain D is a unique factorization domain if, 1. every nonzero element of D that is not a unit can be written as a, product of irreducibles of D; and, 2. the factorization into irreducibles is unique up to associates and, the order in which the factors appear., , Another way to formulate part 2 of this definition is the following:, If p1 n 1 p 2 n 2 ? ? ? pr n r and q1 m 1 q 2m 2 ? ? ? q sms are two factorizations of some, element as a product of irreducibles, where no two of the pi’s are associates and no two of the q j’s are associates, then r 5 s, each p i is an, associate of one and only one q j , and ni = mj ., Of course, the Fundamental Theorem of Arithmetic tells us that the, ring of integers is a unique factorization domain, and Theorem 17.6, says that Z[x] is a unique factorization domain. In fact, as we shall soon, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 335, , see, most of the integral domains we have encountered are unique factorization domains., Before proving our next theorem, we need the ascending chain condition for ideals., Lemma Ascending Chain Condition for a PID, In a principal ideal domain, any strictly increasing chain of ideals, I1 , I2 , ? ? ? must be finite in length., , PROOF Let I1 , I2 , ? ? ? be a chain of strictly increasing ideals in, an integral domain D, and let I be the union of all the ideals in this chain., We leave it as an exercise (Exercise 3) to verify that I is an ideal of D., Then, since D is a principal ideal domain, there is an element a in D, such that I 5 kal. Because a [ I and I 5 <Ik, a belongs to some member of the chain, say a [ In. Clearly, then, for any member Ii of the, chain, we have Ii # I 5 kal # In, so that In must be the last member of, the chain., Theorem 18.3 PID Implies UFD, Every principal ideal domain is a unique factorization domain., , PROOF Let D be a principal ideal domain and let a0 be any nonzero, nonunit in D. We will show that a0 is a product of irreducibles (the, product might consist of only one factor). We begin by showing that, a0 has at least one irreducible factor. If a0 is irreducible, we are done., Thus, we may assume that a0 5 b1a1, where neither b1 nor a1 is a unit, and a1 is nonzero. If a1 is not irreducible, then we can write a1 5 b2a2,, where neither b2 nor a2 is a unit and a2 is nonzero. Continuing in this, fashion, we obtain a sequence b1, b2, . . . of elements that are not units, in D and a sequence a0, a1, a2, . . . of nonzero elements of D with an 5, bn11 an11 for each n. Hence, ka0l , ka1l , ? ? ? is a strictly increasing, chain of ideals (see Exercise 5), which, by the preceding lemma, must, be finite, say, ka0l , ka1l , ? ? ? , karl. In particular, ar is an irreducible factor of a0. This argument shows that every nonzero nonunit, in D has at least one irreducible factor., Now write a0 5 p1c1, where p1 is irreducible and c1 is not a unit. If c1, is not irreducible, then we can write c1 5 p2c2, where p2 is irreducible, and c2 is not a unit. Continuing in this fashion, we obtain, as before, a, strictly increasing sequence ka0l , kc1l , kc2l , ? ? ? , which must end, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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336, , Rings, , in a finite number of steps. Let us say that the sequence ends with kcsl., Then cs is irreducible and a0 5 p1p2 ? ? ? pscs, where each pi is also irreducible. This completes the proof that every nonzero nonunit of a principal ideal domain is a product of irreducibles., It remains to be shown that the factorization is unique up to associates and the order in which the factors appear. To do this, suppose that, some element a of D can be written, a 5 p1p2 ? ? ? pr 5 q1q2 ? ? ? qs,, where the p’s and q’s are irreducible and repetition is permitted. We use, induction on r. If r 5 1, then a is irreducible and, clearly, s 5 1 and, p1 5 q1. So we may assume that any element that can be expressed as a, product of fewer than r irreducible factors can be so expressed in only, one way (up to order and associates). Since D is a principal ideal, domain, by Theorem 18.2, each irreducible pi in the product p1p2 ? ? ?, pr is prime. Then because p1 divides q1q2 ? ? ? qs, p1 must divide some qi, (see Exercise 33), say p1 | q1. Then, q1 5 up1, where u is a unit of D., Since, up1p2 ? ? ? pr 5 uq1q2 ? ? ? qs 5 q1(uq2) ? ? ? qs, and, up1 5 q1,, we have, by cancellation,, p2 ? ? ? pr 5 (uq2) ? ? ? qs., The induction hypothesis now tells us that these two factorizations are, identical up to associates and the order in which the factors appear., Hence, the same is true about the two factorizations of a., In the existence portion of the proof of Theorem 18.3, the only, way we used the fact that the integral domain D is a principal ideal, domain was to say that D has the property that there is no infinite,, strictly increasing chain of ideals in D. An integral domain with this, property is called a Noetherian domain, in honor of Emmy Noether,, who inaugurated the use of chain conditions in algebra. Noetherian, domains are of the utmost importance in algebraic geometry. One, reason for this is that, for many important rings R, the polynomial, ring R[x] is a Noetherian domain but not a principal ideal domain., One such example is Z[x]. In particular, Z[x] shows that a UFD need, not be a PID (see Example 3)., As an immediate corollary of Theorem 18.3, we have the following fact., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 337, , Corollary F[x] Is a UFD, Let F be a field. Then F[x] is a unique factorization domain., , PROOF By Theorem 16.3, F[x] is a principal ideal domain. So, F[x] is, a unique factorization domain, as well., As an application of the preceding corollary, we give an elegant, proof, due to Richard Singer, of Eisenstein’s Criterion (Theorem 17.4)., EXAMPLE 4 Let, f (x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x],, and suppose that p is prime such that, p B an, p | an21, . . . , p | a0, , and, , p2 B a0., , We will prove that f(x) is irreducible over Q. If f(x) is reducible over Q,, we know by Theorem 17.2 that there exist elements g(x) and h(x) in Z[x], such that f(x) 5 g(x)h(x), 1 # deg g(x) , n, and 1 # degh (x) , n. Let, f (x), g(x), and h(x) be the polynomials in Zp[x] obtained from f (x),, g(x), and h(x) by reducing all coefficients modulo p. Then, since p divides all the coefficients of f(x) except an, we have anxn 5 f (x) 5 g(x)?, h(x). Since Zp is a field, Zp[x] is a unique factorization domain. Thus,, x | g(x) and x | h(x). So, g(0) 5 h(0) 5 0 and, therefore, p | g(0) and p | h(0)., But then p2 | g(0)h(0) 5 f(0) 5 a0, which is a contradiction., , Euclidean Domains, Another important kind of integral domain is a Euclidean domain., Definition Euclidean Domain (ED), , An integral domain D is called a Euclidean domain if there is a, function d (called the measure) from the nonzero elements of D to, the nonnegative integers such that, 1. d(a) # d(ab) for all nonzero a, b in D; and, 2. if a, b [ D, b 2 0, then there exist elements q and r in D such, that a 5 bq 1 r, where r 5 0 or d(r) , d(b)., , EXAMPLE 5 The ring Z is a Euclidean domain with d(a) 5 |a| (the, absolute value of a)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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338, , Rings, , EXAMPLE 6 Let F be a field. Then F[x] is a Euclidean domain with, d( f(x)) 5 deg f(x) (see Theorem 16.2)., Examples 5 and 6 illustrate just one of many similarities between the, rings Z and F[x]. Additional similarities are summarized in Table 18.1., Table 18.1 Similarities Between Z and F[x], Z, , F[x], , Euclidean domain:, d(a) 5 |a|, Units:, a is a unit if and only if |a| 5 1, Division algorithm:, For a, b [ Z, b 2 0, there exist q, r [ Z, such that a 5 bq 1 r, 0 # r , |b|, , ↔, , PID:, Every nonzero ideal I 5 kal, where, a 2 0 and |a| is minimum, Prime:, No nontrivial factors, UFD:, Every element is a “unique” product of, primes, , ↔, , ↔, , ↔, ↔, , Euclidean domain:, d( f(x)) 5 deg f(x), Units:, f(x) is a unit if and only if deg f(x) 5 0, Division algorithm:, For f(x), g(x) [ F[x], g(x) 2 0, there, exist q(x), r(x) [ F[x] such that f(x), 5 g(x)q(x) 1 r(x), 0 # deg r(x) ,, deg g(x) or r(x) 5 0, PID:, Every nonzero ideal I 5 kf(x)l, where, deg f(x) is minimum, Irreducible:, No nontrivial factors, UFD:, Every element is a “unique” product of, irreducibles, , EXAMPLE 7 The ring of Gaussian integers, Z[i] 5 {a 1 bi | a, b [ Z}, is a Euclidean domain with d(a 1 bi) 5 a2 1 b2. Unlike the previous, two examples, in this example the function d does not obviously satisfy the necessary conditions. That d(x) # d(xy) for x, y [ Z[i] follows, directly from the fact that d(xy) 5 d(x)d(y) (Exercise 7). To verify that, condition 2 holds, observe that if x, y [ Z[i] and y 2 0, then xy21 [, Q[i], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say, xy21 5 s 1 ti, where s, t [ Q. Now let m be the integer nearest s, and, let n be the integer nearest t. (These integers may not be uniquely, determined, but that does not matter.) Thus, |m 2 s| # 1/2 and |n 2 t|, # 1/2. Then, xy21 5 s 1 ti 5 (m 2 m 1 s) 1 (n 2 n 1 t)i, 5 (m 1 ni) 1 [(s 2 m) 1 (t 2 n)i]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 339, , So,, x 5 (m 1 ni)y 1 [(s 2 m) 1 (t 2 n)i]y., We claim that the division condition of the definition of a Euclidean, domain is satisfied with q 5 m 1 ni and, r 5 [(s 2 m) 1 (t 2 n)i]y., Clearly, q belongs to Z[i], and since r 5 x 2 qy, so does r. Finally,, d(r) 5 d([(s 2 m) 1 (t 2 n)i])d(y), 5 [(s 2 m)2 1 (t 2 n)2]d(y), 1, 1, # a b d(y) , d(y)., 4, 4, Theorem 18.4 ED Implies PID, Every Euclidean domain is a principal ideal domain., , PROOF Let D be a Euclidean domain and I a nonzero ideal of D. Among, all the nonzero elements of I, let a be such that d(a) is a minimum. Then, I 5 kal. For, if b [ I, there are elements q and r such that b 5 aq 1 r,, where r 5 0 or d(r) , d(a). But r 5 b 2 aq [ I, so d(r) cannot be less, than d(a). Thus, r 5 0 and b [ kal. Finally, the zero ideal is k0l., Although it is not easy to verify, we remark that there are principal, ideal domains that are not Euclidean domains. The first such example, was given by T. Motzkin in 1949. A more accessible account of, Motzkin’s result can be found in [2]., As an immediate consequence of Theorems 18.3 and 18.4, we have, the following important result., Corollary ED Implies UFD, Every Euclidean domain is a unique factorization domain., , We may summarize our theorems and remarks as follows:, ED ⇒ PID ⇒ UFD;, UFD ⇒, / PID ⇒, / ED., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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340, , Rings, , (You can remember these implications by listing the types alphabetically.), In Chapter 17, we proved that Z[x] is a unique factorization domain., Since Z is a unique factorization domain, the next theorem is a broad, generalization of this fact. The proof is similar to that of the special, case, and we therefore omit it., Theorem 18.5 D a UFD Implies D[x] a UFD, If D is a unique factorization domain, then D[x] is a unique, factorization domain., , We conclude this chapter with an example of an integral domain that, is not a unique factorization domain., EXAMPLE 8 The ring Z[25] 5 {a 1 b25 | a, b [ Z} is an integral domain but not a unique factorization domain. It is straightforward, that Z[25] is an integral domain (see Exercise 11 in Chapter 13). To, verify that unique factorization does not hold, we mimic the method, used in Example 1 with N(a 1 b25) 5 a2 1 5b2. Since N(xy) 5 N(x), N(y) and N(x) 5 1 if and only if x is a unit (see Exercise 1), it follows, that the only units of Z[25] are 61., Now consider the following factorizations:, 46 5 2 ? 23,, 46 5 (1 1 325)(1 2 325)., We claim that each of these four factors is irreducible over Z[25]., Suppose that, say, 2 5 xy, where x, y [ Z[25] and neither is a unit., Then 4 5 N(2) 5 N(x)N(y) and, therefore, N(x) 5 N(y) 5 2, which is, impossible. Likewise, if 23 5 xy were a nontrivial factorization, then, N(x) 5 23. Thus, there would be integers a and b such that a2 1 5b2 5, 23. Clearly, no such integers exist. The same argument applies to 1 6, 325., In light of Examples 7 and 8, one can’t help but wonder for which d , 0, is Z[2d] a unique factorization domain. The answer is only when d 5 21, or 22 (see [1], p. 297). The case where d 5 21 was first proved, naturally, enough, by Gauss., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 341, , Exercises, I tell them that if they will occupy themselves with the study of mathematics they will find in it the best remedy against lust of the flesh., thomas mann, The Magic Mountain, , 1. For the ring Z[2d] 5 {a 1 b2d | a, b [ Z}, where d 2 1 and d is, not divisible by the square of a prime, prove that the norm N(a 1, b2d) 5 |a2 2 db2| satisfies the four assertions made preceding, Example 1. (This exercise is referred to in this chapter.), 2. In an integral domain, show that a and b are associates if and only, if kal 5 kbl., 3. Show that the union of a chain I1 , I2 , ? ? ? of ideals of a ring R is, an ideal of R. (This exercise is referred to in this chapter.), 4. In an integral domain, show that the product of an irreducible and a, unit is an irreducible., 5. Suppose that a and b belong to an integral domain, b 2 0, and a is, not a unit. Show that kabl is a proper subset of kbl. (This exercise is, referred to in this chapter.), 6. Let D be an integral domain. Define a , b if a and b are associates., Show that this defines an equivalence relation on D., 7. In the notation of Example 7, show that d(xy) 5 d(x)d(y)., 8. Let D be a Euclidean domain with measure d. Prove that u is a unit, in D if and only if d(u) 5 d(1)., 9. Let D be a Euclidean domain with measure d. Show that if a and b, are associates in D, then d(a) 5 d(b)., 10. Let D be a principal ideal domain and let p [ D. Prove that kpl is a, maximal ideal in D if and only if p is irreducible., 11. Trace through the argument given in Example 7 to find q and r in, Z[i] such that 3 2 4i 5 (2 1 5i)q 1 r and d(r) , d(2 1 5i)., 12. Let D be a principal ideal domain. Show that every proper ideal of, D is contained in a maximal ideal of D., 13. In Z[25], show that 21 does not factor uniquely as a product of, irreducibles., 14. Show that 1 2 i is an irreducible in Z[i]., 15. Show that Z[26] is not a unique factorization domain. (Hint:, Factor 10 in two ways.) Why does this show that Z[26] is not a, principal ideal domain?, 16. Give an example of a unique factorization domain with a subdomain that does not have a unique factorization., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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342, , Rings, , 17. In Z[i], show that 3 is irreducible but 2 and 5 are not., 18. Prove that 7 is irreducible in Z[26], even though N(7) is not prime., 19. Prove that if p is a prime in Z that can be written in the form a2 1 b2,, then a 1 bi is irreducible in Z[i]. Find three primes that have this, property and the corresponding irreducibles., 20. Prove that Z[23] is not a principal ideal domain., 21. In Z[25], prove that 1 1 325 is irreducible but not prime., 22. In Z[25], prove that both 2 and 1 1 25 are irreducible but not, prime., 23. Prove that Z[25] is not a unique factorization domain., 24. Let F be a field. Show that in F[x] a prime ideal is a maximal ideal., 25. Let d be an integer less than 21 that is not divisible by the square, of a prime. Prove that the only units of Z[2d] are 11 and 21., 26. In Z[22] 5 {a 1 b22 | a, b [ Z}, show that every element of the, form (3 1 222)n is a unit, where n is a positive integer., 27. If a and b belong to Z[2d], where d is not divisible by the square, of a prime and ab is a unit, prove that a and b are units., 28. For a commutative ring with unity we may define associates, irreducibles, and primes exactly as we did for integral domains. With, these definitions, show that both 2 and 3 are prime in Z12 but 2 is, irreducible and 3 is not., 29. Let n be a positive integer and p a prime that divides n. Prove that p, is prime in Zn. (See Exercise 28)., 30. Let p be a prime divisor of a positive integer n. Prove that p is irreducible in Zn if and only if p2 divides n. (See Exercise 28)., 31. Prove or disprove that if D is a principal ideal domain, then D[x] is, a principal ideal domain., 32. Determine the units in Z[i]., 33. Let p be a prime in an integral domain. If p | a1a2 ? ? ? an, prove that, p divides some ai. (This exercise is referred to in this chapter.), 34. Show that 3x2 1 4x 1 3 [ Z5[x] factors as (3x 1 2)(x 1 4) and, (4x 1 1)(2x 1 3). Explain why this does not contradict the corollary of Theorem 18.3., 35. Let D be a principal ideal domain and p an irreducible element of D., Prove that D/kpl is a field., 36. Show that an integral domain with the property that every strictly, decreasing chain of ideals I1 . I2 . ? ? ? must be finite in length is, a field., 37. An ideal A of a commutative ring R with unity is said to be finitely, generated if there exist elements a1, a2, . . . , an of A such that, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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18 | Divisibility in Integral Domains, , 38., 39., 40., 41., 42., , 43., 44., , 45., 46., , 343, , A 5 ka1, a2, . . . , anl. An integral domain R is said to satisfy the, ascending chain condition if every strictly increasing chain of ideals I1 , I2 , ? ? ? must be finite in length. Show that an integral, domain R satisfies the ascending chain condition if and only if, every ideal of R is finitely generated., Prove or disprove that a subdomain of a Euclidean domain is a, Euclidean domain., Show that for any nontrivial ideal I of Z[i], Z[i]/I is finite., Find the inverse of 1 1 22 in Z[22]. What is the multiplicative, order of 1 1 22?, In Z[27], show that N(6 1 227) 5 N(1 1 327) but 6 1 2, 27 and 1 1 327 are not associates., Let R 5 Z % Z % ? ? ? (the collection of all sequences of integers, under componentwise addition and multiplication). Show that R, has ideals I1, I2, I3, . . . with the property that I1 , I2 , I3 , ? ? ?., (Thus R does not have the ascending chain condition.), Prove that in a unique factorization domain, an element is irreducible if and only if it is prime., Let F be a field and let R be the integral domain in F 3x4 generated by, x2 and x3. (That is, R is contained in every integral domain in F 3x4 that, contains x2 and x3.) Show that R is not a unique factorization domain., Prove that for every field F, there are infinitely many irreducible, elements in F 3x4 ., Find a mistake in the statement shown in Figure 18.2., , Computer Exercise, Software for a computer exercise is available at the website:, http://www.d.umn.edu/~jgallian, , References, 1. H. M. Stark, An Introduction to Number Theory, Chicago: Markham,, 1970., 2. J. C. Wilson, “A Principal Ideal Ring That Is Not a Euclidean Ring,”, Mathematics Magazine 46 (1973): 34–38., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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344, , Rings, , Suggested Readings, Oscar Campoli, “A Principal Ideal Domain That Is Not a Euclidean, Domain,” The American Mathematical Monthly 95 (1988): 868–871., The author shows that {a 1 bu | a, b [ Z, u 5 (1 1 219)/2} is a, PID that is not an ED., Gina Kolata, “At Last, Shout of ‘Eureka!’ in Age-Old Math Mystery,” The, New York Times, June 24, 1993., This front-page article reports on Andrew Wiles’s announced proof of, Fermat’s Last Theorem., C. Krauthhammer, “The Joy of Math, or Fermat’s Revenge,” Time, April 18,, 1988: 92., The demise of Miyaoka’s proof of Fermat’s Last Theorem is charmingly lamented., Sahib Singh, “Non-Euclidean Domains: An Example,” Mathematics Magazine 49 (1976): 243., This article gives a short proof that Z[2n] 5 {a 1 b2n | a, b [ Z}, is an integral domain that is not Euclidean when n . 2 and 2n mod, 4 5 2 or 2n mod 4 5 3., Simon Singh and Kenneth Ribet, “Fermat’s Last Stand,” Scientific American 277 (1997): 68–73., This article gives an accessible description of Andrew Wiles’s proof of, Fermat’s Last Theorem., , Suggested Video, The Proof, Nova, http://www.pbs.org/wgbh/nova/proof, This documentary film shown on PBS’s Nova program in 1997 chronicles, the seven-year effort of Andrew Wiles to prove Fermat’s Last Theorem. It, can be viewed in five segments at http://www.youtube.com., , Suggested Websites, http://en.wikipedia.org/wiki/Fermat's_Last_Theorem, This website provides a concise history of the efforts to prove Fermat’s Last, Theorem. It includes photographs, references, and links., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Sophie Germain, , Stock Montage, , One of the very few women to overcome, the prejudice and discrimination that, tended to exclude women from the pursuit, of higher mathematics in her time was, Sophie Germain., , Sophie Germain was born in Paris on April 1,, 1776. She educated herself by reading the, works of Newton and Euler in Latin and the, lecture notes of Lagrange. In 1804, Germain, wrote to Gauss about her work in number, theory but used the pseudonym Monsieur, LeBlanc because she feared that Gauss would, not take seriously the efforts of a woman., Gauss gave Germain’s results high praise and, a few years later, upon learning her true identity, wrote to her:, But how to describe to you my admiration, and astonishment at seeing my esteemed correspondent Mr. LeBlanc metamorphose himself into this illustrious personage who gives, such a brilliant example of what I would find, it difficult to believe. A taste for the abstract, sciences in general and above all the mysteries of numbers is excessively rare: it is not a, subject which strikes everyone; the enchanting charms of this sublime science reveal, , themselves only to those who have the, courage to go deeply into it. But when a, person of the sex which, according to our, customs and prejudices, must encounter, infinitely more difficulties than men to, familiarize herself with these thorny researches, succeeds nevertheless in surmounting these obstacles and penetrating the most, obscure parts of them, then without doubt she, must have the noblest courage, quite extraordinary talents, and a superior genius.*, , Germain is best known for her work on, Fermat’s Last Theorem. She died on June, 27, 1831, in Paris., For more information about Germain,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history, , *Quote from Math’s Hidden Woman, Nova Online, http://www.pbs.org/wgbh/nova/proof/germain, .html (accessed Nov 5, 2008)., 345, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Andrew Wiles, For spectacular contributions to number, theory and related fields, for major, advances on fundamental conjectures,, and for settling Fermat’s Last Theorem., Princeton University, , Citation for the Wolf Prize, , Postage stamp issued by the, Czech Republic in honor of, Fermat’s Last Theorem., , In 1993, Andrew Wiles of Princeton electrified the mathematics community by announcing that he had proved Fermat’s Last Theorem, after seven years of effort. His proof, which, ran 200 pages, relied heavily on ring theory, and group theory. Because of Wiles’s solid, reputation and because his approach was, based on deep results that had already shed, much light on the problem, many experts in, the field believed that Wiles had succeeded, where so many others had failed. Wiles’s, achievement was reported in newspapers and, magazines around the world. The New York, Times ran a front-page story on it, and one TV, network announced it on the evening news., Wiles even made People magazine’s list of the, 25 most intriguing people of 1993! In San, Francisco a group of mathematicians rented a, 1200-seat movie theater and sold tickets for, $5.00 each for public lectures on the proof., , Scalpers received as much as $25.00 a ticket, for the sold-out event., The bubble soon burst when experts, had an opportunity to scrutinize Wiles’s, manuscript. By December, Wiles released a, statement saying he was working to resolve, a gap in the proof. In September of 1994, a, paper by Wiles and Richard Taylor, a former, student of his, circumvented the gap in the, original proof. Since then, many experts, have checked the proof and have found no, errors. One mathematician was quoted as, saying, “The exuberance is back.” In 1997,, Wiles’s proof was the subject of a PBS Nova, program., Wiles was born in 1953 in Cambridge,, England. He obtained his bachelor’s degree at, Oxford and his doctoral degree at Cambridge, University in 1980. He was a professor at Harvard before moving to Princeton in 1982. In, 2011, he became a research professor at, Oxford. He has received many prestigious, awards., To find more information about Wiles,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 346, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 15– 18, , 347, , Supplementary Exercises for Chapters 15–18, The intelligence is proved not by ease of learning, but by understanding, what we learn., joseph whitney, , True/false questions for Chapters 15–18 are available on the Web at:, http://www.d.umn.edu/~jgallian/TF, 1. Suppose that F is a field and there is a ring homomorphism from Z, onto F. Show that F is isomorphic to Zp for some prime p., 2. Let Q[22] 5 {r 1 s22 | r, s [ Q}. Determine all ring automorphisms of Q[22]., 3. (Second Isomorphism Theorem for Rings) Let A be a subring, of R and let B be an ideal of R. Show that A > B is an ideal of A, and that A/(A > B) is isomorphic to (A 1 B)/B. (Recall that A 1 B 5, {a 1 b | a [ A, b [ B}.), 4. (Third Isomorphism Theorem for Rings) Let A and B be ideals of a, ring R with B # A. Show that A/B is an ideal of R/B and, (R/B)/(A/B) is isomorphic to R/A., 5. Let f(x) and g(x) be irreducible polynomials over a field F. If f(x), and g(x) are not associates, prove that F[x]/k f(x)g(x)l is isomorphic, to F[x]/k f(x)l % F[x]/kg(x)l., 6. (Chinese Remainder Theorem for Rings) If R is a commutative, ring and I and J are two proper ideals with I 1 J 5 R, prove that, R/(I > J) is isomorphic to R/I % R/J. Explain why Exercise 5 is a, special case of this theorem., 7. Prove that the set of all polynomials whose coefficients are all even, is a prime ideal in Z[x]., 8. Let R 5 Z[25] and let I 5 {a 1 b25 | a, b [ Z, a 2 b is, even}. Show that I is a maximal ideal of R., 9. Let R be a ring with unity and let a be a unit in R. Show that the mapping from R into itself given by x S axa21 is a ring automorphism., 10. Let a 1 b25 belong to Z[25] with b 2 0. Show that 2 does, not belong to ka 1 b25l., 11. Show that Z[i]/k2 1 il is a field. How many elements does it have?, 12. Is the homomorphic image of a principal ideal domain a principal, ideal domain?, 13. For any f(x) [ Zp[x], show that f(xp) 5 ( f(x)) p., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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348, , Rings, , 14. Let p be a prime. Show that there is exactly one ring homomorphism from Zm to Zpk if pk does not divide m, and exactly two ring, homomorphisms from Zm to Zpk if pk does divide m., 15. Recall that a is an idempotent if a2 5 a. Show that if 1 1 k is an, idempotent in Zn, then n 2 k is an idempotent in Zn., 16. Show that Zn (where n . 1) always has an even number of idempotents. (The number is 2d, where d is the number of distinct prime, divisors of n.), 17. Show that the equation x2 1 y2 5 2003 has no solutions in the, integers., 18. Prove that if both k and k 1 1 are idempotents in Zn and k 2 0, then, n 5 2k., 19. Prove that x4 1 15x3 1 7 is irreducible over Q., 20. For any integers m and n, prove that the polynomial x3 1 (5m 1 1)x 1, 5n 1 1 is irreducible over Z., 21. Prove that k22l is a maximal ideal in Z[22]. How many elements, are in the ring Z[22]/k22l?, 22. Prove that Z[22] and Z[22] are unique factorization domains., (Hint: Mimic Example 7 in Chapter 18.), 23. Is k3l a maximal ideal in Z[i]?, 24. Express both 13 and 5 1 i as products of irreducibles from Z[i]., 25. Let R 5 {a/b | a, b [ Z, 3 B b}. Prove that R is an integral domain., Find its field of quotients., 26. Give an example of a ring that contains a subring isomorphic to Z, and a subring isomorphic to Z3., 27. Show that Z[i]/k3l is not ring-isomorphic to Z3 % Z3., a 0, 28. For any n . 1, prove that R 5 e c, d ` a, b [ Zn f is ring0 b, isomorphic to Zn % Zn., 29. Suppose that R is a commutative ring and I is an ideal of R. Prove, that R[x]/I[x] is isomorphic to (R/I)[x]., 30. Find an ideal I of Z8[x] such that the factor ring Z8[x]/I is a field., 31. Find an ideal I of Z8[x] such that the factor ring Z8[x]/I is an integral, domain but not a field., 32. Find an ideal I of Z[x] such that Z[x]/I is ring-isomorphic to Z3., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PA R T, , 4, , Fields, , For online student resources, visit this textbook’s website at, www.CengageBrain.com, , 349, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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19 Vector Spaces, Still round the corner there may wait, A new road or a secret gate., j. r. r. tolkien, The Fellowship of the Ring, , Definition and Examples, Abstract algebra has three basic components: groups, rings, and fields., Thus far we have covered groups and rings in some detail, and we have, touched on the notion of a field. To explore fields more deeply, we need, some rudiments of vector space theory that are covered in a linear algebra course. In this chapter, we provide a concise review of this material., Definition Vector Space, , A set V is said to be a vector space over a field F if V is an Abelian, group under addition (denoted by 1) and, if for each a [ F and, v [ V, there is an element av in V such that the following conditions, hold for all a, b in F and all u, v in V., 1., 2., 3., 4., , a(v 1 u) 5 av 1 au, (a 1 b)v 5 av 1 bv, a(bv) 5 (ab)v, 1v 5 v, , The members of a vector space are called vectors. The members of, the field are called scalars. The operation that combines a scalar a and, a vector v to form the vector av is called scalar multiplication. In general, we will denote vectors by letters from the end of the alphabet,, such as u, v, w, and scalars by letters from the beginning of the alphabet, such as a, b, c., , 351, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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352, , Fields, , EXAMPLE 1 The set Rn 5 {(a1, a2, . . . , an) | ai [ R} is a vector, space over R. Here the operations are the obvious ones:, (a1, a2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn), and, b(a1, a2, . . . , an) 5 (ba1, ba2, . . . , ban)., EXAMPLE 2 The set M2(Q) of 2 3 2 matrices with entries from Q is, a vector space over Q. The operations are, c, , a 1 a2, b1 b2, a1 b1 a2 b2, d c, d c, d, a3 a4, b3 b4, a3 b3 a4 b4, , and, bc, , a1 a2, ba1 ba2, d., d c, ba3 ba4, a3 a4, , EXAMPLE 3 The set Zp[x] of polynomials with coefficients from Zp, is a vector space over Zp, where p is a prime., EXAMPLE 4 The set of complex numbers C 5 {a 1 bi | a, b [ R}, is a vector space over R. The vector addition and scalar multiplication, are the usual addition and multiplication of complex numbers., The next example is a generalization of Example 4. Although it appears rather trivial, it is of the utmost importance in the theory of fields., EXAMPLE 5 Let E be a field and let F be a subfield of E. Then E is a, vector space over F. The vector addition and scalar multiplication are, the operations of E., , Subspaces, Of course, there is a natural analog of subgroup and subring., Definition Subspace, , Let V be a vector space over a field F and let U be a subset of V. We, say that U is a subspace of V if U is also a vector space over F under, the operations of V., , EXAMPLE 6 The set {a2x2 1 a1x 1 a0 | a0, a1, a2 [ R} is a subspace of the vector space of all polynomials with real coefficients, over R., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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19 | Vector Spaces, , 353, , EXAMPLE 7 Let V be a vector space over F and let v1, v2, . . . , vn be, (not necessarily distinct) elements of V. Then the subset, kv1, v2, . . . , vnl 5 {a1v1 1 a2v2 1 ? ? ? 1 anvn | a1, a2, . . . , an [ F}, is called the subspace of V spanned by v 1, v2, . . . , vn. Any sum of, the form a1v1 1 a2v2 1 ? ? ? 1 anvn is called a linear combination of, v1, v2, . . . , vn. If kv1, v2, . . . , vnl 5 V, we say that {v1, v2, . . . , vn}, spans V., , Linear Independence, The next definition is the heart of the theory., Definition Linearly Dependent, Linearly Independent, , A set S of vectors is said to be linearly dependent over the field F if, there are vectors v1, v2, . . . , vn from S and elements a1, a2, . . . , an from, F, not all zero, such that a1v1 1 a2v2 1 ? ? ? 1 anvn 5 0. A set of vectors, that is not linearly dependent over F is called linearly independent, over F., , In other words, a set of vectors is linearly dependent over F if there, is a nontrivial linear combination of them over F equal to 0., EXAMPLE 8 In R3 the vectors (1, 0, 0), (1, 0, 1), and (1, 1, 1) are linearly independent over R. To verify this, assume that there are real, numbers a, b, and c such that a(1, 0, 0) 1 b(1, 0, 1) 1 c(1, 1, 1) 5, (0, 0, 0). Then (a 1 b 1 c, c, b 1 c) 5 (0, 0, 0). From this we see that, a 5 b 5 c 5 0., Certain kinds of linearly independent sets play a crucial role in the, theory of vector spaces., Definition Basis, , Let V be a vector space over F. A subset B of V is called a basis for V, if B is linearly independent over F and every element of V is a linear, combination of elements of B., , The motivation for this definition is twofold. First, if B is a basis for, a vector space V, then every member of V is a unique linear combination of the elements of B (see Exercise 19). Second, with every vector, space spanned by finitely many vectors, we can use the notion of basis, to associate a unique integer that tells us much about the vector space., (In fact, this integer and the field completely determine the vector space, up to isomorphism—see Exercise 31.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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354, , Fields, , a, ab, d ` a, b [ R f, ab, b, is a vector space over R (see Exercise 17). We claim that the set, 1 1, 0 1, B ec, d, c, d f is a basis for V over R. To prove that the set, 1 0, 1 1, EXAMPLE 9 The set V 5 e c, , B is linearly independent, suppose that there are real numbers a and b, such that, 1 1, 0 1, 0 0, a c, d b c, d c, d., 1 0, 1 1, 0 0, This gives c, , a, ab, 0, d c, ab, b, 0, , 0, d , so that a 5 b 5 0. On the other, 0, , hand, since every member of V has the form, c, , a, ab, 1 1, 0 1, d ac, d bc, d,, ab, b, 1 0, 1 1, , we see that B spans V., We now come to the main result of this chapter., Theorem 19.1 Invariance of Basis Size, If {u1, u2, . . . , um} and {w1, w2, . . . , wn} are both bases of a vector, space V over a field F, then m 5 n., , PROOF Suppose that m 2 n. To be specific, let us say that m , n., Consider the set {w1, u1, u2 , . . . , um}. Since the u’s span V, we, know that w1 is a linear combination of the u’s, say, w1 5 a1u1 1, a2u2 1 ? ? ? 1 amum, where the a’s belong to F. Clearly, not all the a’s are, 0. For convenience, say a1 2 0. Then {w1, u2, . . . , um} spans V (see, Exercise 21). Next, consider the set {w1, w2, u2, . . . , um}. This time, w2, is a linear combination of w1, u2, . . . , um, say, w2 b1w1 b2u2 . . ., 1 bmum, where the b’s belong to F. Then at least one of b2, . . . , bm is, nonzero, for otherwise the w’s are not linearly independent. Let us say, b2 2 0. Then w1, w2, u3, . . . , um span V. Continuing in this fashion, we, see that {w1, w2, . . . , wm} spans V. But then wm11 is a linear combination of w1, w2, . . . , wm and, therefore, the set {w1, . . . , wn} is not, linearly independent. This contradiction finishes the proof., Theorem 19.1 shows that any two finite bases for a vector space have, the same size. Of course, not all vector spaces have finite bases., However, there is no vector space that has a finite basis and an infinite, basis (see Exercise 25)., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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19 | Vector Spaces, , 355, , Definition Dimension, , A vector space that has a basis consisting of n elements is said, to have dimension n. For completeness, the trivial vector space {0} is, said to be spanned by the empty set and to have dimension 0., , Although it requires a bit of set theory that is beyond the scope of, this text, it can be shown that every vector space has a basis. A vector, space that has a finite basis is called finite dimensional; otherwise, it is, called infinite dimensional., , Exercises, Somebody who thinks logically is a nice contrast to the real world., the law of thumb, , 1. Verify that each of the sets in Examples 1– 4 satisfies the axioms, for a vector space. Find a basis for each of the vector spaces in, Examples 1–4., 2. (Subspace Test) Prove that a nonempty subset U of a vector space, V over a field F is a subspace of V if, for every u and u9 in U and, every a in F, u 1 u9 [ U and au [ U. (In words, a nonempty set U, is a subspace of V if it is closed under the two operations of V.), 3. Verify that the set in Example 6 is a subspace. Find a basis for this, subspace. Is {x2 1 x 1 1, x 1 5, 3} a basis?, 4. Verify that the set kv1, v2, . . . , vnl defined in Example 7 is a subspace., 5. Determine whether or not the set {(2, 21, 0), (1, 2, 5), (7, 21, 5)} is, linearly independent over R., 6. Determine whether or not the set, ec, , 7., 8., 9., 10., , 2 1, 0 1, 1 1, d, c, d, c, df, 1 0, 1 2, 1 1, , is linearly independent over Z5., If {u, v, w} is a linearly independent subset of a vector space, show, that {u, u 1 v, u 1 v 1 w} is also linearly independent., If {v1, v2, . . . , vn} is a linearly dependent set of vectors, prove that, one of these vectors is a linear combination of the other., (Every spanning collection contains a basis.) If {v1, v2, . . . , vn} spans, a vector space V, prove that some subset of the v’s is a basis for V., (Every independent set is contained in a basis.) Let V be a finitedimensional vector space and let {v 1, v2, . . . , vn} be a linearly, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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356, , Fields, , independent subset of V. Show that there are vectors w1, w2, . . . , wm, such that {v1, v2, . . . , vn, w1, . . . , wm} is a basis for V., 11. If V is a vector space over F of dimension 5 and U and W are subspaces of V of dimension 3, prove that U > W 2 {0}. Generalize., 12. Show that the solution set to a system of equations of the form, a11x1 1 ? ? ? 1 a1nxn 5 0, a21x1 1 ? ? ? 1 a2nxn 5 0, ?, ?, ?, ?, ?, ?, ?, ?, ?, am1x1 1 ? ? ? 1 amnxn 5 0,, where the a’s are real, is a subspace of Rn., 13. Let V be the set of all polynomials over Q of degree 2 together, with the zero polynomial. Is V a vector space over Q?, 14. Let V 5 R3 and W 5 {(a, b, c) [ V | a2 1 b2 5 c2}. Is W a subspace of V? If so, what is its dimension?, 15. Let V 5 R3 and W 5 {(a, b, c) [ V | a 1 b 5 c}. Is W a subspace, of V? If so, what is its dimension?, 16. Let V 5 e c, 17., 18., 19., , 20., 21., 22., 23., 24., , a b, d ` a, b, c [ Q f . Prove that V is a vector space, b c, over Q, and find a basis for V over Q., Verify that the set V in Example 9 is a vector space over R., Let P 5 {(a, b, c) | a, b, c [ R, a 5 2b 1 3c}. Prove that P is a subspace of R3. Find a basis for P. Give a geometric description of P., Let B be a subset of a vector space V. Show that B is a basis for V if, and only if every member of V is a unique linear combination of, the elements of B. (This exercise is referred to in this chapter and, in Chapter 20.), If U is a proper subspace of a finite-dimensional vector space V,, show that the dimension of U is less than the dimension of V., Referring to the proof of Theorem 19.1, prove that {w1, u2, . . . , um}, spans V., If V is a vector space of dimension n over the field Zp, how many, elements are in V?, Let S 5 {(a, b, c, d) | a, b, c, d [ R, a 5 c, d 5 a 1 b}. Find a, basis for S., Let U and W be subspaces of a vector space V. Show that U > W is, a subspace of V and that U 1 W 5 {u 1 w | u [ U, w [ W} is a, subspace of V., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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19 | Vector Spaces, , 357, , 25. If a vector space has one basis that contains infinitely many elements, prove that every basis contains infinitely many elements., (This exercise is referred to in this chapter.), 26. Let u 5 (2, 3, 1), v 5 (1, 3, 0), and w 5 (2, 23, 3). Since (1/2)u 2, (2/3)v 2 (1/6)w 5 (0, 0, 0), can we conclude that the set {u, v, w}, is linearly dependent over Z7?, 27. Define the vector space analog of group homomorphism and ring, homomorphism. Such a mapping is called a linear transformation., Define the vector space analog of group isomorphism and ring isomorphism., 28. Let T be a linear transformation from V to W. Prove that the image, of V under T is a subspace of W., 29. Let T be a linear transformation of a vector space V. Prove that, {v [ V | T(v) 5 0}, the kernel of T, is a subspace of V., 30. Let T be a linear transformation of V onto W. If {v1, v2, . . . , vn}, spans V, show that {T(v1), T(v2), . . . , T(vn)} spans W., 31. If V is a vector space over F of dimension n, prove that V is isomorphic as a vector space to Fn 5 {(a1, a2, . . . , an) | ai [ F}. (This, exercise is referred to in this chapter.), 32. Let V be a vector space over an infinite field. Prove that V is not the, union of finitely many proper subspaces of V., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Emil Artin, , Michael Artin, , For Artin, to be a mathematician meant to, participate in a great common effort, to, continue work begun thousands of years, ago, to shed new light on old discoveries,, to seek new ways to prepare the developments of the future. Whatever standards, we use, he was a great mathematician., richard brauer,, Bulletin of the American, Mathematical Society, , Emil Artin was one of the leading mathematicians of the 20th century and a major, contributor to linear algebra and abstract algebra. Artin was born on March 3, 1898, in, Vienna, Austria, and grew up in what was, recently known as Czechoslovakia. He received a Ph.D. in 1921 from the University, of Leipzig. From 1923 until he emigrated, to America in 1937, he was a professor at, the University of Hamburg. After one year, at Notre Dame, Artin went to Indiana, University. In 1946 he moved to Princeton,, where he stayed until 1958. The last four, years of his career were spent where it, began, at Hamburg., Artin’s mathematics is both deep and, broad. He made contributions to number theory, group theory, ring theory, field theory,, Galois theory, geometric algebra, algebraic, , topology, and the theory of braids—a field, he invented. Artin received the American, Mathematical Society’s Cole Prize in number, theory, and he solved one of the 23 famous, problems posed by the eminent mathematician David Hilbert in 1900., Artin was an outstanding teacher of, mathematics at all levels, from freshman, calculus to seminars for colleagues. Many, of his Ph.D. students as well as his son, Michael have become leading mathematicians. Through his research, teaching, and, books, Artin exerted great influence among, his contemporaries. He died of a heart attack, at the age of 64, in 1962., For more information about Artin, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 358, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Olga Taussky-Todd, “Olga Taussky-Todd was a distinguished, and prolific mathematician who wrote, about 300 papers.”, , MaryAnn McLoughlin, , edith luchins and mary ann mcloughlin,, Notices of the American, Mathematical Society, 1996, , Olga Taussky-Todd was born on August 30,, 1906, in Olmütz in the Austro-Hungarian, Empire. Taussky-Todd received her doctoral, degree in 1930 from the University of Vienna., In the early 1930s she was hired as an assistant at the University of Göttingen to edit, books on the work of David Hilbert. She also, edited lecture notes of Emil Artin and assisted Richard Courant. She spent 1934 and, 1935 at Bryn Mawr and the next two years, at Girton College in Cambridge, England., In 1937, she taught at the University of, London. In 1947, she moved to the United, States and took a job at the National Bureau, of Standards’ National Applied Mathematics, Laboratory. In 1957, she became the first, woman to teach at the California Institute of, Technology as well as the first woman to, receive tenure and a full professorship in, mathematics, physics, or astronomy there., Thirteen Caltech Ph.D. students wrote their, Ph.D. theses under her direction., , In addition to her influential contributions to linear algebra, Taussky-Todd did, important work in number theory., Taussky-Todd received many honors and, awards. She was elected a Fellow of the, American Association for the Advancement, of Science and vice president of the American, Mathematical Society. In 1990, Caltech established an instructorship named in her honor., Taussky-Todd died on October 7, 1995, at the, age of 89., For more information about TausskyTodd, visit:, http://www-groups.dcs, .st-and.ac.uk/~history, http://www.agnesscott, .edu/lriddle/women/women.htm, , 359, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 Extension Fields, In many respects this [Kronecker’s Theorem] is the, fundamental theorem of algebra., richard a. dean, Elements of Abstract Algebra, , The Fundamental Theorem, of Field Theory, In our work on rings, we came across a number of fields, both finite, and infinite. Indeed, we saw that Z3[x]/kx2 1 1l is a field of order 9,, whereas R[x]/kx2 1 1l is a field isomorphic to the complex numbers., In the next three chapters, we take up, in a systematic way, the subject, of fields., Definition Extension Field, , A field E is an extension field of a field F if F # E and the operations, of F are those of E restricted to F., , Cauchy’s observation in 1847 that R[x]/kx2 1 1l is a field that contains a zero of x2 1 1 prepared the way for the following sweeping, generalization of that fact., Theorem 20.1 Fundamental Theorem of Field Theory, (Kronecker’s Theorem, 1887), Let F be a field and let f(x) be a nonconstant polynomial in F[x]., Then there is an extension field E of F in which f (x) has a zero., , PROOF Since F[x] is a unique factorization domain, f (x) has an irreducible factor, say, p(x). Clearly, it suffices to construct an extension, field E of F in which p(x) has a zero. Our candidate for E is F[x]/kp(x)l., We already know that this is a field from Corollary 1 of Theorem 17.5., Also, since the mapping of f: F S E given by f(a) 5 a 1 kp(x)l is, one-to-one and preserves both operations, E has a subfield isomorphic, 360, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 | Extension Fields, , 361, , to F. We may think of E as containing F if we simply identify the coset, a 1 kp(x)l with its unique coset representative a that belongs to F [that, is, think of a 1 kp(x)l as just a and vice versa; see Exercise 33 in, Chapter 17]., Finally, to show that p(x) has a zero in E, write, p(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0., Then, in E, x 1 kp(x)l is a zero of p(x), because, p(x 1 kp(x)l) 5 an(x 1 kp(x)l)n 1 an21(x 1 kp(x)l)n21 1 ? ? ? 1 a0, 5 an(xn 1 kp(x)l) 1 an21(xn21 1 kp(x)l) 1 ? ? ? 1 a0, 5 anxn 1 an21xn21 1 ? ? ? 1 a0 1 kp(x)l, 5 p(x) 1 kp(x)l 5 0 1 kp(x)l., EXAMPLE 1 Let f(x) 5 x2 1 1 [ Q[x]. Then, viewing f (x) as an, element of E[x] 5 (Q[x]/kx2 1 1l)[x], we have, f (x 1 kx2 1 1l) 5 (x 1 kx2 1 1l)2 1 1, 5 x2 1 kx2 1 1l 1 1, 5 x2 1 1 1 kx2 1 1l, 5 0 1 kx2 1 1l., Of course, the polynomial x2 1 1 has the complex number 21 as a, zero, but the point we wish to emphasize here is that we have constructed a field that contains the rational numbers and a zero for the, polynomial x2 1 1 by using only the rational numbers. No knowledge, of complex numbers is necessary. Our method utilizes only the field we, are given., EXAMPLE 2 Let f (x) 5 x5 1 2x2 1 2x 1 2 [ Z3[x]. Then, the irreducible factorization of f (x) over Z3 is (x2 1 1)(x3 1 2x 1 2). So, to find, an extension E of Z3 in which f (x) has a zero, we may take E 5 Z3[x]/, kx2 1 1l, a field with nine elements, or E 5 Z3[x]/kx3 1 2x 1 2l, a field, with 27 elements., Since every integral domain is contained in its field of quotients, (Theorem 15.6), we see that every nonconstant polynomial with coefficients from an integral domain always has a zero in some field containing the ring of coefficients. The next example shows that this is not, true for commutative rings in general., EXAMPLE 3 Let f(x) 5 2x 1 1 [ Z4[x]. Then f(x) has no zero in any, ring containing Z4 as a subring, because if b were a zero in such a ring,, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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362, , Fields, , then 0 5 2b 1 1, and therefore 0 5 2(2b 1 1) 5 2(2b) 1 2 5, (2 ? 2)b 1 2 5 0 ? b 1 2 5 2. But 0 2 2 in Z4., , Splitting Fields, To motivate the next definition and theorem, let’s return to Example 1 for, a moment. For notational convenience, in Q[x]/kx 2 1 1l, let a 5, x 1 kx2 1 1l. Then, since a and 2a are both zeros of x2 1 1 in (Q[x]/, kx2 1 1l)[x], it should be the case that x2 1 1 5 (x 2 a)(x 1 a). Let’s, check this out. First note that, (x 2 a)(x 1 a) 5 x2 2 a2 5 x2 2 (x2 1 kx2 1 1l)., At the same time,, x2 1 kx2 1 1l 5 21 1 kx2 1 1l, and we have agreed to identify 21 and 21 1 kx2 1 1l, so, (x 2 a)(x 1 a) 5 x2 2 (21) 5 x2 1 1., This shows that x2 1 1 can be written as a product of linear factors in, some extension of Q. That was easy and you might argue coincidental., The polynomial given in Example 2 presents a greater challenge. Is, there an extension of Z3 in which that polynomial factors as a product, of linear factors? Yes, there is. But first some notation and a definition., Let F be a field and let a1, a2, . . . , an be elements of some extension E of F. We use F(a1,a2, . . . , an) to denote the smallest subfield, of E that contains F and the set {a1, a2, . . . , an}. We leave it as an, exercise (Exercise 35) to show that F(a1, a2, . . . , an) is the intersection of all subfields of E that contain F and the set {a1, a2, . . . , an}., Definition Splitting Field, , Let E be an extension field of F and let f(x) [ F[x] with degree at least, 1. We say that f(x) splits in E if there are elements a [ F and a1, a2, . . .,, an [ E such that, f(x) 5 a(x 2 a1)(x 2 a2) . . . (x 2 an)., We call E a splitting field for f(x) over F if, E 5 F(a1, a2, . . ., an)., , Note that a splitting field of a polynomial over a field depends not, only on the polynomial but on the field as well. Indeed, a splitting field, of f (x) over F is just a smallest extension field of F in which f(x) splits., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 | Extension Fields, , 363, , The next example illustrates how a splitting field of a polynomial f (x), over field F depends on F., EXAMPLE 4 Consider the polynomial f(x) 5 x2 1 1 [ Q[x]. Since, , x2 1 1 5 (x 1 21)(x 2 21), we see that f(x) splits in C, but a splitting, , field over Q is Q(i) 5 {r 1 si | r, s [ Q}. A splitting field for x2 1 1 over, R is C. Likewise, x2 2 2 [ Q[x] splits in R, but a splitting field over Q is, Q(22) 5 {r 1 s 22 | r, s [ Q}., , There is a useful analogy between the definition of a splitting field and, the definition of an irreducible polynomial. Just as it makes no sense to, say “f(x) is irreducible,” it makes no sense to say “E is a splitting field for, f(x).” In each case, the underlying field must be specified; that is, one must, say “f(x) is irreducible over F” and “E is a splitting field for f(x) over F.”, Our notation in Example 4 appears to be inconsistent with the notation that we used in earlier chapters. For example, we denoted the set, {a 1 b22 | a, b [ Z} by Z[22] and the set {a 1 b22 | a, b [ Q} by, Q(22). The difference is that Z[22] is merely a ring, whereas Q(22) is, a field. In general, parentheses are used when one wishes to indicate, that the set is a field, although no harm would be done by using, say,, Q[22] to denote {a 1 b22 | a, b [ Q} if we were concerned with its, ring properties only. Using parentheses rather than brackets simply, conveys a bit more information about the set., Theorem 20.2 Existence of Splitting Fields, Let F be a field and let f(x) be a nonconstant element of F[x]. Then, there exists a splitting field E for f (x) over F., , PROOF We proceed by induction on deg f (x). If deg f (x) 5 1, then, f (x) is linear. Now suppose that the statement is true for all fields and, all polynomials of degree less than that of f (x). By Theorem 20.1,, there is an extension E of F in which f(x) has a zero, say, a1. Then we, may write f (x) 5 (x 2 a1)g(x), where g(x) [ E[x]. Since deg g(x) ,, deg f (x), by induction, there is a field K that contains E and all the, zeros of g(x), say, a2, . . . , an. Clearly, then, a splitting field for f (x), over F is F(a1, a2, . . . , an)., EXAMPLE 5 Consider, f(x) 5 x4 2 x2 2 2 5 (x2 2 2)(x2 1 1), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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364, , Fields, , over Q. Obviously, the zeros of f (x) in C are 622 and 6i. So a splitting field for f(x) over Q is, Q(22, i) 5 Q(22)(i) 5 {a 1 bi | a, b [ Q(22)}, 5 {(a 1 b 22) 1 (c 1 d 22)i | a, b, c, d [ Q}., EXAMPLE 6 Consider f (x) 5 x2 1 x 1 2 over Z3. Then Z3(i) 5, {a 1 bi | a, b [ Z3} (see Example 9 in Chapter 13) is a splitting field, for f(x) over Z3 because, f (x) 5 [x 2 (1 1 i)][x 2 (1 2 i)]., At the same time, we know by the proof of Kronecker’s Theorem that, the element x 1 kx2 1 x 1 2l of, F 5 Z3[x]/kx2 1 x 1 2l, is a zero of f (x). Since f(x) has degree 2, it follows from the Factor, Theorem (Corollary 2 of Theorem 16.2) that the other zero of f (x) must, also be in F. Thus, f (x) splits in F, and because F is a two-dimensional, vector space over Z3, we know that F is also a splitting field of f (x) over, Z3. But how do we factor f(x) in F? Factoring f(x) in F is confusing because we are using the symbol x in two distinct ways: It is used as a, placeholder to write the polynomial f(x), and it is used to create the coset, representatives of the elements of F. This confusion can be avoided by, simply identifying the coset 1 1 kx2 1 x 1 2l with the element 1 in Z3 and, denoting the coset x 1 kx2 1 x 1 2l by b. With this identification, the field, Z3[x]/kx2 1 x 1 2l can be represented as {0, 1, 2, b, 2b, b 1 1, 2b 1 1,, b 1 2, 2b 1 2}. These elements are added and multiplied just as polynomials are, except that we use the observation that x2 1 x 1 2 1 kx2 1, x 1 2l 5 0 implies that b2 1 b 1 2 5 0, so that b2 5 2b 2 2 5 2b 1 1., For example, (2b 1 1)(b 1 2) 5 2b2 1 5b 1 2 5 2(2b 1 1) 1 5b 1, 2 5 9b 1 4 5 1. To obtain the factorization of f(x) in F, we simply long, divide, as follows:, x 1 (b 1 1), x 2 b qx 2 1 x 1 2, x 2 2 bx, (b 1 1)x 1 2, (b 1 1)x 2 (b 1 1)b, (b 1 1)b 1 2 5 b2 1 b 1 2 5 0., So, x2 1 x 1 2 5 (x 2 b)(x 1 b 1 1). Thus, we have found two splitting fields for x2 1 x 1 2 over Z3, one of the form F(a) and one of the, form F[x]/kp(x)l [where F 5 Z3 and p(x) 5 x2 1 x 1 2]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 | Extension Fields, , 365, , The next theorem shows how the fields F(a) and F[x]/kp(x)l are, related in the case where p(x) is irreducible over F and a is a zero of, p(x) in some extension of F., Theorem 20.3 F(a) < F[x]/kp(x)l, Let F be a field and let p(x) [ F[x] be irreducible over F. If a is a, zero of p(x) in some extension E of F, then F(a) is isomorphic to, F[x]/kp(x)l. Furthermore, if deg p(x) 5 n, then every member of F(a), can be uniquely expressed in the form, cn21an21 1 cn22an22 1 ? ? ? 1 c1a 1 c0,, where c0 , c1, . . . , cn21 [ F., , PROOF Consider the function f from F[x] to F(a) given by f( f (x)) 5, f(a). Clearly, f is a ring homomorphism. We claim that Ker f 5 kp(x)l., (This is Exercise 31 in Chapter 17.) Since p(a) 5 0, we have kp(x)l #, Ker f. On the other hand, we know by Theorem 17.5 that kp(x)l is a, maximal ideal in F[x]. So, because Ker f 2 F[x] [it does not contain, the constant polynomial f (x) 5 1], we have Ker f 5 kp(x)l. At this, point it follows from the First Isomorphism Theorem for Rings and, Corollary 1 of Theorem 17.5 that f(F[x]) is a subfield of F(a). Noting, that f(F[x]) contains both F and a and recalling that F(a) is the smallest such field, we have F[x]/kp(x)l < f(F[x]) 5 F(a)., The final assertion of the theorem follows from the fact that every, element of F[x]/kp(x)l can be expressed uniquely in the form, cn21x n21 1 ? ? ? 1 c0 1 kp(x)l,, where c0, . . . , cn21 [ F (see Exercise 25 in Chapter 16), and the, natural isomorphism from F[x]/kp(x)l to F(a) carries c kxk 1 kp(x)l, to ckak., As an immediate corollary of Theorem 20.3, we have the following, attractive result., Corollary F(a) < F(b), Let F be a field and let p(x) [ F[x] be irreducible over F. If a is a, zero of p(x) in some extension E of F and b is a zero of p(x) in some, extension E9 of F, then the fields F(a) and F(b) are isomorphic., , PROOF From Theorem 20.3, we have, F(a) < F[x]/kp(x)l < F(b)., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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366, , Fields, , Recall that a basis for an n-dimensional vector space over a field F, is a set of n vectors v1, v2, . . . , vn with the property that every member, of the vector space can be expressed uniquely in the form a1v1 1, a2v2 1 ? ? ? 1 anvn, where the a’s belong to F (Exercise 19 in Chapter 19)., So, in the language of vector spaces, the latter portion of Theorem 20.3, says that if a is a zero of an irreducible polynomial over F of degree n, then, the set {1, a, . . . , an21} is a basis for F(a) over F., Theorem 20.3 often provides a convenient way of describing the, elements of a field., EXAMPLE 7 Consider the irreducible polynomial f(x) 5 x6 2 2, 6, over Q. Since 12 is a zero of f(x), we know from Theorem 20.3 that the, 6, 1/6, 2/6, set {1, 2 , 2 , 23/6, 24/6, 25/6} is a basis for Q(12) over Q. Thus,, 6, Q(12) 5 {a0 1 a121/6 1 a222/6 1 a323/6 1 a424/6 1 a525/6 | ai [ Q}., , This field is isomorphic to Q[x]/kx6 2 2l., In 1882, Ferdinand von Lindemann (1852–1939) proved that p is, not the zero of any polynomial in Q[x]. Because of this important result, Theorem 20.3 does not apply to Q(p) (see Exercise 11). Fields of, the form F(a) where a is in some extension field of F but not the zero, of an element of F(x) are discussed in the next chapter., In Example 6, we produced two splitting fields for the polynomial, 2, x 1 x 1 2 over Z3. Likewise, it is an easy exercise to show that both, Q[x]/kx2 1 1l and Q(i) 5 {r 1 si | r, s [ Q} are splitting fields of the, polynomial x2 1 1 over Q. But are these different-looking splitting fields, algebraically different? Not really. We conclude our discussion of splitting fields by proving that splitting fields are unique up to isomorphism., To make it easier to apply induction, we will prove a more general result., We begin by observing first that any ring isomorphism f from F, to F9 has a natural extension from F[x] to F9[x] given by cnx n 1, cn21xn21 1 ? ? ? 1 c1x 1 c0 S f(cn)x n 1 f(cn21)x n21 1 ? ? ? 1, f(c1)x 1 f(c0). Since this mapping agrees with f on F, it is convenient and natural to use f to denote this mapping as well., Lemma, Let F be a field, let p(x) [ F[x] be irreducible over F, and let a be a, zero of p(x) in some extension of F. If f is a field isomorphism from, F to F9 and b is a zero of f(p(x)) in some extension of F9, then there, is an isomorphism from F(a) to F9(b) that agrees with f on F and, carries a to b., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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367, , 20 | Extension Fields, , PROOF First observe that since p(x) is irreducible over F, f(p(x)) is, irreducible over F9. It is straightforward to check that the mapping, from F[x]/kp(x)l to F9[x]/kf(p(x))l given by, f (x) 1 kp(x)l S f(f (x)) 1 kf(p(x))l, is a field isomorphism. By a slight abuse of notation, we denote this, mapping by f also. (If you object, put a bar over the f.) From the proof, of Theorem 20.3, we know that there is an isomorphism a from F(a) to, F[x]/kp(x)l that is the identity on F and carries a to x 1 kp(x)l. Similarly, there is an isomorphism b from F9[x]/kf(p(x))l to F9(b) that is the, identity on F9 and carries x 1 kf(p(x))l to b. Thus, bfa is the desired, mapping from F1a2 to F1b2. See Figure 20.1., F'[x]/ φ ( p (x)), , F[x]/ p(x), , F(a), α, , φ, , F, , φ, , F'(b), β, , F', , Figure 20.1, , Theorem 20.4 Extending f: F S F9, Let f be an isomorphism from a field F to a field F9 and let, f(x) [ F[x]. If E is a splitting field for f (x) over F and E9 is a, splitting field for f(f(x)) over F9, then there is an isomorphism, from E to E9 that agrees with f on F., , PROOF We use induction on deg f (x). If deg f (x) 5 1, then E 5 F, and E9 5 F9, so that f itself is the desired mapping. If deg f (x) . 1,, let p(x) be an irreducible factor of f (x), let a be a zero of p(x) in E, and, let b be a zero of f(p(x)) in E9. By the preceding lemma, there is an, isomorphism a from F(a) to F9(b) that agrees with f on F and carries, a to b. Now write f(x) 5 (x 2 a)g(x), where g(x) [ F(a)[x]. Then E is a, splitting field for g(x) over F(a) and E9 is a splitting field for a(g(x)), over F9(b). Since deg g(x) , deg f (x), there is an isomorphism from E, to E9 that agrees with a on F(a) and therefore with f on F., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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368, , Fields, , Corollary Splitting Fields Are Unique, Let F be a field and let f(x) [ F[x]. Then any two splitting fields, of f(x) over F are isomorphic., , PROOF Suppose that E and E9 are splitting fields of f(x) over F. The, result follows immediately from Theorem 20.4 by letting f be the, identity from F to F., In light of the corollary above, we may refer to “the” splitting field, of a polynomial over F without ambiguity., 6, 6, Even though x6 2 2 has a zero in Q(12), it does not split in Q(12)., The splitting field is easy to obtain, however., EXAMPLE 8 The Splitting Field of xn 2 a over Q, Let a be a positive rational number and let v be a primitive nth root of, unity (see Example 2 in Chapter 16). Then each of, a1/n, va1/n, v2a1/n, . . . , vn21a1/n, n, , is a zero of xn 2 a in Q( 2a, v)., , Zeros of an Irreducible Polynomial, Now that we know that every nonconstant polynomial over a field, splits in some extension, we ask whether irreducible polynomials must, split in some special way. Yes, they do. To discover how, we borrow, something whose origins are in calculus., Definition Derivative, , Let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 belong to F[x]. The, derivative of f(x), denoted by f9(x), is the polynomial nanx n21 1, (n 2 1)an21x n22 1 ? ? ? 1 a1 in F[x]., , Notice that our definition does not involve the notion of a limit. The, standard rules for handling sums and products of functions in calculus, carry over to arbitrary fields as well., Lemma Properties of the Derivative, Let f(x) and g(x) [ F[x] and let a [ F. Then, 1. (f(x) 1 g(x))9 5 f 9(x) 1 g9(x)., 2. (af(x))9 5 af 9(x)., 3. (f(x)g(x))9 5 f(x)g9(x) 1 g(x)f 9(x)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 | Extension Fields, , 369, , PROOF Properties 1 and 2 follow from straightforward applications, of the definition. Using property 1 and induction on deg f(x), property 3, reduces to the special case in which f (x) 5 anx n. This also follows directly from the definition., Before addressing the question of the nature of the zeros of an irreducible polynomial, we establish a general result concerning zeros of, multiplicity greater than 1. Such zeros are called multiple zeros., Theorem 20.5 Criterion for Multiple Zeros, A polynomial f(x) over a field F has a multiple zero in some, extension E if and only if f (x) and f9(x) have a common factor of, positive degree in F[x]., , PROOF If a is a multiple zero of f (x) in some extension E, then, there is a g(x) in E[x] such that f (x) 5 (x 2 a)2g(x). Since f 9(x) 5, (x 2 a)2g9(x) 1 2(x 2 a)g(x), we see that f 9(a) 5 0. Thus, x 2 a is a factor of both f(x) and f 9(x) in the extension E of F. Now if f(x) and f 9(x), have no common divisor of positive degree in F[x], there are polynomials, h(x) and k(x) in F[x] such that f(x)h(x) 1 f 9(x)k(x) 5 1 (see Exercise 45, in Chapter 16). Viewing f (x)h(x) 1 f 9(x)k(x) as an element of E[x], we, see also that x 2 a is a factor of 1. Since this is nonsense, f (x) and f 9(x), must have a common divisor of positive degree in F[x]., Conversely, suppose that f (x) and f 9(x) have a common factor of positive degree. Let a be a zero of the common factor. Then a is a zero of f (x), and f 9(x). Since a is a zero of f (x), there is a polynomial q(x) such that, f (x) 5 (x 2 a)q(x). Then f 9(x) 5 (x 2 a)q9(x) 1 q(x) and 0 5 f 9(a) 5, q(a). Thus, x 2 a is a factor of q(x) and a is a multiple zero of f (x)., Theorem 20.6 Zeros of an Irreducible, Let f(x) be an irreducible polynomial over a field F. If F has, characteristic 0, then f(x) has no multiple zeros. If F has characteristic p 2 0, then f(x) has a multiple zero only if it is of the, form f(x) 5 g(x p) for some g(x) in F[x]., , PROOF If f (x) has a multiple zero, then, by Theorem 20.5, f (x) and, f 9(x) have a common divisor of positive degree in F[x]. Since the only, divisor of positive degree of f(x) in F[x] is f(x) itself (up to associates),, we see that f (x) divides f 9(x). Because a polynomial over a field cannot, divide a polynomial of smaller degree, we must have f 9(x) 5 0., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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370, , Fields, , Now what does it mean to say that f 9(x) 5 0? If we write f(x) 5 anxn 1, an21x n21 1 ? ? ? 1 a1x 1 a0, then f9(x) 5 nanx n21 1 (n 2 1)an21x n22 1, ? ? ? 1 a1. Thus, f 9(x) 5 0 only when kak 5 0 for k 5 1, . . . , n., So, when char F 5 0, we have f(x) 5 a0, which is not an irreducible, polynomial. This contradicts the hypothesis that f(x) is irreducible over, F. Thus, f(x) has no multiple zeros., When char F 5 p 2 0, we have ak 5 0 when p does not divide k. Thus,, the only powers of x that appear in the sum anx n 1 ? ? ? 1 a1x 1 a0 are, those of the form x pj 5 (x p) j. It follows that f (x) 5 g(x p) for some, g(x) [ F[x]. [For example, if f(x) 5 x4p 1 3x2p 1 x p 1 1, then g(x) 5, x4 1 3x2 1 x 1 1.], Theorem 20.6 shows that an irreducible polynomial over a field of, characteristic 0 cannot have multiple zeros. The desire to extend this result to a larger class of fields motivates the following definition., Definition Perfect Field, , A field F is called perfect if F has characteristic 0 or if F has, characteristic p and Fp 5 {ap | a [ F} 5 F., , The most important family of perfect fields of characteristic p is the, finite fields., Theorem 20.7 Finite Fields Are Perfect, Every finite field is perfect., , PROOF Let F be a finite field of characteristic p. Consider the mapping f from F to F defined by f(x) 5 xp for all x [ F. We claim that, f is a field automorphism. Obviously, f(ab) 5 (ab) p 5 a pb p 5, p, f(a)f(b). Moreover, f(a 1 b) 5 (a 1 b) p 5 a p 1 a b a p21b 1, 1, p p22 2, p, a ba b 1???1a, b ab p21 1 b p 5 a p 1 b p, since each, 2, p1, p, a b is divisible by p. Finally, since xp 2 0 when x 2 0, Ker f 5 {0}., i, Thus, f is one-to-one and, since F is finite, f is onto. This proves that, Fp 5 F., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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371, , 20 | Extension Fields, , Theorem 20.8 Criterion for No Multiple Zeros, If f(x) is an irreducible polynomial over a perfect field F, then f(x) has, no multiple zeros., , PROOF The case where F has characteristic 0 has been done. So, let us assume that f (x) [ F[x] is irreducible over a perfect field F of, characteristic p and that f (x) has multiple zeros. From Theorem 20.6, we know that f (x) 5 g(x p) for some g(x) [ F[x], say, g(x) 5 anx n 1, an21x n21 1 ? ? ? 1 a1x 1 a0. Since Fp 5 F, each ai in F can be written, in the form b p for some bi in F. So, using Exercise 49a in Chapter 13,, i, we have, p, , f(x) 5 g(x p) 5 bn px pn 1 bn21 px p(n21) 1 ? ? ? 1 b1 x p 1 b0, , p, , 5 (bn x n 1 bn21x n21 1 ? ? ? 1 b1x 1 b0)p 5 (h(x))p,, where h(x) [ F[x]. But then f(x) is not irreducible., The next theorem shows that when an irreducible polynomial does, have multiple zeros, there is something striking about the multiplicities., Theorem 20.9 Zeros of an Irreducible over a Splitting Field, Let f(x) be an irreducible polynomial over a field F and let E be a, splitting field of f(x) over F. Then all the zeros of f (x) in E have the, same multiplicity., , PROOF Let a and b be distinct zeros of f(x) in E. If a has multiplicity m,, then in E[x] we may write f(x) 5 (x 2 a)mg(x). It follows from the, lemma preceding Theorem 20.4 and from Theorem 20.4 that there is a, field isomorphism f from E to itself that carries a to b and acts as the, identity on F. Thus,, f (x) 5 f( f (x)) 5 (x 2 b)mf(g(x)),, and we see that the multiplicity of b is greater than or equal to the multiplicity of a. By interchanging the roles of a and b, we observe that the, multiplicity of a is greater than or equal to the multiplicity of b. So, we, have proved that a and b have the same multiplicity., As an immediate corollary of Theorem 20.9 we have the following, appealing result., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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372, , Fields, , Corollary Factorization of an Irreducible over a Splitting Field, Let f(x) be an irreducible polynomial over a field F and let E be a, splitting field of f(x). Then f(x) has the form, a(x 2 a1)n(x 2 a2)n ? ? ? (x 2 at)n,, where a1, a2, . . . , at are distinct elements of E and a [ F., , We conclude this chapter by giving an example of an irreducible, polynomial over a field that does have a multiple zero. In particular,, notice that the field we use is not perfect., EXAMPLE 9 Let F 5 Z2(t) be the field of quotients of the ring Z2[t], of polynomials in the indeterminate t with coefficients from Z2. (We, must introduce a letter other than x, since the members of F are going to, be our coefficients for the elements in F[x].) Consider f (x) 5 x2 2 t [, F[x]. To see that f (x) is irreducible over F, it suffices to show that it has, no zeros in F. Well, suppose that h(t)/k(t) is a zero of f (x). Then (h(t)/, k(t))2 5 t, and therefore (h(t))2 5 t(k(t))2. Since h(t), k(t) [ Z2[t], we, then have h(t2) 5 tk(t2) (see Exercise 49 in Chapter 13). But deg h(t2) is, even, whereas deg tk(t2) is odd. So, f(x) is irreducible over F., Finally, since t is a constant in F[x] and the characteristic of F is 2, we, have f9(x) 5 0, so that f9(x) and f(x) have f(x) as a common factor. So, by, Theorem 20.5, f(x) has a multiple zero in some extension of F. (Indeed, it, has a single zero of multiplicity 2 in K 5 F[x]/kx2 2 tl.), , Exercises, I have yet to see any problem, however complicated, which, when you, looked at it in the right way, did not become still more complicated., paul anderson, New Scientist, 3, 1. Describe the elements of Q(1, 5)., 2. Show that Q(22, 23) 5 Q(22 1 23)., 3. Find the splitting field of x3 2 1 over Q. Express your answer in, the form Q(a)., 4. Find the splitting field of x4 1 1 over Q., 5. Find the splitting field of, , x4 1 x2 1 1 5 (x2 1 x 1 1)(x2 2 x 1 1), over Q., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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20 | Extension Fields, , 373, , 6. Let a, b [ R with b 2 0. Show that R(a 1 bi) 5 C., 7. Find a polynomial p(x) in Q[x] such that Q(21 25) is ringisomorphic to Q[x]/kp(x)l., 8. Let F 5 Z2 and let f (x) 5 x3 1 x 1 1 [ F[x]. Suppose that a is a, zero of f(x) in some extension of F. How many elements does F(a), have? Express each member of F(a) in terms of a. Write out a, complete multiplication table for F(a)., 9. Let F(a) be the field described in Exercise 8. Express each of a5,, a22, and a100 in the form c2a2 1 c1a 1 c0., 10. Let F(a) be the field described in Exercise 8. Show that a2 and a2 1 a, are zeros of x3 1 x 1 1., 11. Describe the elements in Q(p)., 12. Let F 5 Q(p3). Find a basis for F(p) over F., 13. Write x7 2 x as a product of linear factors over Z3. Do the same for, x10 2 x., 3, 14. Find all ring automorphisms of Q( 15)., 15. Let F be a field of characteristic p and let f(x) 5 x p 2 a [ F[x]., Show that f(x) is irreducible over F or f(x) splits in F., 16. Suppose that b is a zero of f(x) 5 x 4 1 x 1 1 in some extension, field E of Z2. Write f (x) as a product of linear factors in E[x]., 17. Find a, b, c in Q such that, 3, 3, 3, 3, (1 1 14)/(2 2 12) 5 a 1 b 12 1 c 14., , Note that such a, b, c exist, since, 3, 3, 3, 3, 3, (1 1 14)/(2 2 12) [ Q(12) 5 {a 1 b12 1 c14 | a, b, c [ Q}., , 18. Express (3 1 422)21 in the form a 1 b22, where a, b [ Q., 19. Show that Q(4 2 i) 5 Q(1 1 i), where i 5 21., 20. Let F be a field, and let a and b belong to F with a 2 0. If c, belongs to some extension of F, prove that F(c) 5 F(ac 1 b)., (F “absorbs” its own elements.), 21. Let f (x) [ F[x] and let a [ F. Show that f (x) and f (x 1 a) have the, same splitting field over F., 22. Recall that two polynomials f (x) and g(x) from F[x] are said to be, relatively prime if there is no polynomial of positive degree in F[x], that divides both f (x) and g(x). Show that if f (x) and g(x) are relatively prime in F[x], they are relatively prime in K[x], where K is, any extension of F., 23. Determine all of the subfields of Q(22)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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374, , Fields, , 24. Let E be an extension of F and let a and b belong to E. Prove that, F(a, b) 5 F(a)(b) 5 F(b)(a)., 25. Write x3 1 2x 1 1 as a product of linear polynomials over some, extension field of Z3., 26. Express x8 2 x as a product of irreducibles over Z2., 27. Prove or disprove that Q(23) and Q(23) are ring-isomorphic., 28. For any prime p, find a field of characteristic p that is not perfect., 29. If b is a zero of x2 1 x 1 2 over Z5, find the other zero., 30. Show that x4 1 x 1 1 over Z2 does not have any multiple zeros in, any extension field of Z2., 31. Show that x 21 1 2x 8 1 1 does not have multiple zeros in any, extension of Z3., 32. Show that x21 1 2x9 1 1 has multiple zeros in some extension of Z3., 33. Let F be a field of characteristic p 2 0. Show that the polynomial, n, f(x) 5 x p 2 x over F has distinct zeros., 34. Find the splitting field for f (x) 5 (x2 1 x 1 2)(x2 1 2x 1 2) over, Z3[x]. Write f(x) as a product of linear factors., 35. Let F be a field and E an extension field of F that contains a1,, a2, . . ., an. Prove that F(a1, a2, . . ., an) is the intersection of all, subfields of E that contain F and the set {a1, a2, . . ., an}. (This exercise is referred to in this chapter.), 36. Suppose that a is algebraic over a field F. Show that a and, 1 1 a21 have the same degree over F., 37. Suppose that f1x2 is a fifth-degree polynomial that is irreducible, over Z2. Prove that every nonidentity element is a generator of the, cyclic group 1Z23x4/kf1x2l2 *., 38. Show that Q(27, i) is the splitting field for x4 2 6x2 27., 39. Let p be a prime, F 5 Zp(t) (the field of quotients of the ring Zp[x]),, and f(x) 5 xp 2 t. Prove that f(x) is irreducible over F and has a, multiple zero in K 5 F[x]/kxp 2 tl., 40. Let f(x) be an irreducible polynomial over a field F. Prove that the, number of distinct zeros of f(x) in a splitting field divides deg f(x)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Leopold Kronecker, He [Kronecker] wove together the three, strands of his greatest interests—the, theory of numbers, the theory of equations, and elliptic functions—into, one beautiful pattern., e. t. bell, , Leopold Kronecker was born on December, 7, 1823, in Liegnitz, Prussia. As a schoolboy,, he received special instruction from the great, algebraist Kummer. Kronecker entered the, University of Berlin in 1841 and completed, his Ph.D. dissertation in 1845 on the units in a, certain ring., Kronecker devoted the years 1845–1853 to, business affairs, relegating mathematics to a, hobby. Thereafter, being well-off financially,, he spent most of his time doing research in algebra and number theory. Kronecker was one, of the early advocates of the abstract approach, to algebra. He innovatively applied rings and, fields in his investigations of algebraic numbers, established the Fundamental Theorem of, Finite Abelian Groups, and was the first mathematician to master Galois’s theory of fields., , Kronecker advocated constructive methods for all proofs and definitions. He believed, that all mathematics should be based on relationships among integers. He went so far as, to say to Lindemann, who proved that p is, transcendental, that irrational numbers do, not exist. His most famous remark on the, matter was “God made the integers, all the, rest is the work of man.” Henri Poincaré, once remarked that Kronecker was able to, produce fine work in number theory and, algebra only by temporarily forgetting his, own philosophy., Kronecker died on December 29, 1891,, at the age of 68., For more information about Kronecker,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 375, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 Algebraic Extensions, , Banach once told me, “Good mathematicians see analogies between, theorems or theories, the very best ones see analogies between analogies.”, s. m. ulam, Adventures of a Mathematician, , Characterization of Extensions, In Chapter 20, we saw that every element in the field Q(22) has the, particularly simple form a 1 b22, where a and b are rational. On the, other hand, the elements of Q(p) have the more complicated form, (anpn 1 an21pn21 1 ? ? ? 1 a0)/(bmpm 1 bm21pm21 1 ? ? ? 1 b0),, where the a’s and b’s are rational. The fields of the first type have a, great deal of algebraic structure. This structure is the subject of this, chapter., Definition Types of Extensions, , Let E be an extension field of a field F and let a [ E. We call a, al gebraic over F if a is the zero of some nonzero polynomial in F[x]. If, a is not algebraic over F, it is called transcendental over F. An extension E of F is called an algebraic extension of F if every element of E is, algebraic over F. If E is not an algebraic extension of F, it is called a, transcendental extension of F. An extension of F of the form F(a) is, called a simple extension of F., , Leonhard Euler used the term transcendental for numbers that are, not algebraic because “they transcended the power of algebraic methods.” Although Euler made this distinction in 1744, it wasn’t until 1844, that the existence of transcendental numbers over Q was proved by, Joseph Liouville. Charles Hermite proved that e is transcendental over, Q in 1873, and Lindemann showed that p is transcendental over Q in, 1882. To this day, it is not known whether p 1 e is transcendental over Q., With a precise definition of “almost all,” it can be shown that almost all, real numbers are transcendental over Q., , 376, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 | Algebraic Extensions, , 377, , Theorem 21.1 shows why we make the distinction between elements, that are algebraic over a field and elements that are transcendental over, a field. Recall that F(x) is the field of quotients of F[x]; that is,, F(x) 5 {f(x)/g(x) | f(x), g(x) [ F[x], g(x) 2 0}., Theorem 21.1 Characterization of Extensions, Let E be an extension field of the field F and let a [ E. If a is, transcendental over F, then F(a) < F(x). If a is algebraic over F, then, F(a) < F[x]/kp(x)l, where p(x) is a polynomial in F[x] of minimum, degree such that p(a) 5 0. Moreover, p(x) is irreducible over F., , PROOF Consider the homomorphism f: F[x] S F(a) given by, f (x) S f (a). If a is transcendental over F, then Ker f 5 {0}, and so, we may extend f to an isomorphism f: F(x) S F(a) by defining, f( f(x)/g(x)) 5 f(a)/g(a)., If a is algebraic over F, then Ker f 2 {0}; and, by Theorem 16.4,, there is a polynomial p(x) in F[x] such that Ker f 5 kp(x)l and p(x) has, minimum degree among all nonzero elements of Ker f. Thus, p(a) 5 0, and, since p(x) is a polynomial of minimum degree with this property,, it is irreducible over F., The proof of Theorem 21.1 can readily be adapted to yield the next, two results also. The details are left to the reader (see Exercise 1)., Theorem 21.2 Uniqueness Property, If a is algebraic over a field F, then there is a unique monic irreducible polynomial p(x) in F[x] such that p(a) 5 0., , The polynomial with the property specified in Theorem 21.2 is, called the minimal polynomial for a over F., Theorem 21.3 Divisibility Property, Let a be algebraic over F, and let p(x) be the minimal polynomial for, a over F. If f(x) [ F[x] and f(a) 5 0, then p(x) divides f(x) in F[x]., , If E is an extension field of F, we may view E as a vector space over F, (that is, the elements of E are the vectors and the elements of F are the, scalars). We are then able to use such notions as dimension and basis in, our discussion., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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378, , Fields, , Finite Extensions, Definition Degree of an Extension, , Let E be an extension field of a field F. We say that E has degree n, over F and write [E:F] 5 n if E has dimension n as a vector space, over F. If [E:F] is finite, E is called a finite extension of F; otherwise,, we say that E is an infinite extension of F., , Figure 21.1 illustrates a convenient method of depicting the degree, of a field extension over a field., Q(√2), , 3, , 6, , Q(√2), , Q(√2), , 2, , 3, , 6, , n, , [Q(√2):Q] = 2, , [Q(√2):Q] = 3, , [Q(√2):Q] = 6, , [E:F] = n, , Q, , Q, , E, , Q, , 3, , 6, , F, , Figure 21.1, , EXAMPLE 1 The field of complex numbers has degree 2 over the, reals, since {1, i} is a basis. The field of complex numbers is an infinite, extension of the rationals., EXAMPLE 2 If a is algebraic over F and its minimal polynomial, over F has degree n, then, by Theorem 20.3, we know that {1, a, . . . ,, an21} is a basis for F(a) over F; and, therefore, [F(a):F] 5 n. In this, case, we say that a has degree n over F., Theorem 21.4 Finite Implies Algebraic, If E is a finite extension of F, then E is an algebraic extension of F., , PROOF Suppose that [E:F] 5 n and a [ E. Then the set {1, a, . . . , an}, is linearly dependent over F; that is, there are elements c0, c1, . . . , cn in F,, not all zero, such that, cnan 1 cn21an21 1 ? ? ? 1 c1a 1 c0 5 0., Clearly, then, a is a zero of the nonzero polynomial, f (x) 5 cnx n 1 cn21x n21 1 ? ? ? 1 c1x 1 c0., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 | Algebraic Extensions, , 379, , The converse of Theorem 21.4 is not true, for otherwise, the degrees of the elements of every algebraic extension of E over F would, 3, 4, be bounded. But Q(22, 22, 22, . . .) is an algebraic extension of Q, that contains elements of every degree over Q (see Exercise 3)., The next theorem is the field theory counterpart of Lagrange’s Theorem for finite groups. Like all counting theorems, it has far-reaching, consequences., Theorem 21.5 [K:F] 5 [K:E][E:F], Let K be a finite extension field of the field E and let E be a finite, extension field of the field F. Then K is a finite extension field, of F and [K:F] 5 [K:E][E:F]., , PROOF Let X 5 {x1, x2, . . . , xn} be a basis for K over E, and let, Y 5 {y1, y2, . . . , ym} be a basis for E over F. It suffices to prove that, YX 5 {yj xi | 1 # j # m, 1 # i # n}, is a basis for K over F. To do this, let a [ K. Then there are elements, b1, b2, . . . , bn [ E such that, a 5 b1x1 1 b2x2 1 ? ? ? 1 bnxn, and, for each i 5 1, . . . , n, there are elements ci1, ci2, . . . , cim [ F, such that, bi 5 ci1y1 1 ci2 y2 1 ? ? ? 1 cimym., Thus,, a g bi xi g a g cij yj b xi g cij 1yj xi 2., n, , n, , i1, , m, , i, j, , i1 j1, , This proves that YX spans K over F., Now suppose there are elements cij in F such that, 0 g cij (yj xi) g 1 g (cijyj)2xi., i,j, , i, , j, , Then, since each g cij yj [ E and X is a basis for K over E, we have, j, , g cij yj 0, j, , for each i. But each cij [ F and Y is a basis for E over F, so each cij 5 0., This proves that the set YX is linearly independent over F., Using the fact that for any field extension L of a field J, [L:J] 5 n if, and only if L is isomorphic to Jn as vector spaces (see Exercise 29), we, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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380, , Fields, K, n, nm, , E, m, F, , [K:F] = [K:E ][E:F ], , Figure 21.2, , may give a concise conceptual proof of Theorem 21.5, as follows. Let, [K:E] 5 n and [E:F] 5 m. Then K < En and E < Fm, so that K < En <, (Fm)n < Fmn. Thus, [K:F] 5 mn., The content of Theorem 21.5 can be pictured as in Figure 21.2. Examples 3, 4, and 5 show how Theorem 21.5 is often utilized., EXAMPLE 3 Since {1, 23} is a basis for Q(23, 25) over Q(25), (see Exercise 7) and {1, 25} is a basis for Q(25) over Q, the proof of, Theorem 21.5 shows that {1, 23, 25, 215} is a basis for Q(23, 25) over, Q. (See Figure 21.3.), 3, 4, 3, 4, EXAMPLE 4 Consider Q(2, 2, 2, 3). Then [Q(2, 2, 2, 3):Q] 5 12. For,, 3, 4, 3, 4, 3, 3, 3, clearly, [Q(22, 23):Q] 5 [Q(22, 23):Q(22)][Q(22):Q] and [Q(22,, 4, 3, 4, 4, 4, 3, 23):Q] 5 [Q(22, 23):Q(23)][Q(23):Q] show that both 3 5 [Q(22):, 4, 3, 4, 3, 4, Q] and 4 5 [Q(23):Q] divide [Q(22, 23):Q]. Thus, [Q(22, 23):Q], 3, 4, 3, 4, $ 12. On the other hand, [Q(22, 23):Q(22)] is at most 4, since 23 is a, 3, 3, 4, 3, 4, zero of x4 2 3 [ Q(22)[x]. Therefore, [Q(22, 23):Q] 5 [Q(22, 23):, 3, 3, Q(22)][Q(22):Q] # 4 ? 3 5 12. (See Figure 21.4.), , Q(√2, √3), 3, , Q(√3,√5 ), 2, , 2, , Q(√3), 2, , 4, Q(√5), , 4, 2, Q, , Figure 21.3, , 4, , 3, , 3, , Q(√2), , 4, , Q(√3), , 12, 3, , 4, Q, , Figure 21.4, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 | Algebraic Extensions, , 381, , Theorem 21.5 can sometimes be used to show that a field does not, contain a particular element., EXAMPLE 5 Recall from Example 7 in Chapter 17 that h(x) 5 15x4 2, 10x2 1 9x 1 21 is irreducible over Q. Let b be a zero of h(x) in some extension of Q. Then, even though we don’t know what b is, we can still, 3, 3, prove that 22 is not an element of Q(b). For, if so, then Q , Q(22) #, 3, 3, Q(b) and 4 5 [Q(b):Q] 5 [Q(b):Q(22)][Q(22):Q] implies that 3, divides 4. Notice that this argument cannot be used to show that 22 is, not contained in Q(b)., EXAMPLE 6 Consider Q(23, 25). We claim that Q(23, 25) 5, Q(23 1 25). The inclusion Q(23 1 25) # Q(23, 25) is clear., Now note that since, 1, 23 25, 1, ., 5 (23 2 25),, 2, 23 25, 23 25, we know that 23 2 25 belongs to Q (23 1 25). It follows that, [(23 1 25) 1 (23 2 25)]/2 5 23 and [(23 1 25 ) 2 (23 2 25)]/2, 5 25 both belong to Q(23 1 25), and therefore Q(23, 25) #, Q(23 1 25)., (23 1 25)21 5, , EXAMPLE 7 It follows from Example 6 and Theorem 20.3 that the, minimal polynomial for 23 1 25 over Q has degree 4. How can we, find this polynomial? We begin with x 5 23 1 25. Then x2 5 3 1, 2215 1 5. From this we obtain x2 2 8 5 2215 and, by squaring both, sides, x4 2 16x 1 64 5 60. Thus, 23 1 25 is a zero of x4 2 16x 1 4., We know that this is the minimal polynomial of 23 1 25 over Q since, it is monic and has degree 4., Example 6 shows that an extension obtained by adjoining two elements to a field can sometimes be obtained by adjoining a single, element to the field. Our next theorem shows that, under certain conditions, this can always be done., Theorem 21.6 Primitive Element Theorem (Steinitz, 1910), If F is a field of characteristic 0, and a and b are algebraic over F,, then there is an element c in F(a, b) such that F(a, b) 5 F(c)., , PROOF Let p(x) and q(x) be the minimal polynomials over F for a and, b, respectively. In some extension K of F, let a1, a2, . . . , am and b1, b2,, . . . , bn be the distinct zeros of p(x) and q(x), respectively, where a 5 a1, and b 5 b 1. Among the infinitely many elements of F, choose an, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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382, , Fields, , element d not equal to (ai 2 a)/(b 2 bj) for all i $ 1 and all j . 1. In, particular, ai 2 a 1 d(b 2 bj) for j . 1., We shall show that c 5 a 1 db has the property that F(a, b) 5 F(c)., Certainly, F(c) # F(a, b). To verify that F(a, b) # F(c), it suffices to prove, that b [ F(c), for then b, c, and d belong to F(c) and a 5 c 2 bd. Consider the polynomials q(x) and r(x) 5 p(c 2 dx) [that is, r(x) is obtained, by substituting c 2 dx for x in p(x)] over F(c). Since both q(b) 5 0 and, r (b) 5 p(c 2 db) 5 p(a) 5 0, both q(x) and r (x) are divisible by the, minimal polynomial s(x) for b over F(c) (see Theorem 21.3). Because s(x), [ F(c)[x], we may complete the proof by proving that s(x) 5 x 2 b. Since, s(x) is a common divisor of q(x) and r(x), the only possible zeros of s(x) in, K are the zeros of q(x) that are also zeros of r(x). But r(bj) 5 p(c 2 dbj) 5, p(a 1db 2 dbj) 5 p(a 1 d(b 2 bj)) and d was chosen such that a 1, d(b 2 bj) 2 ai for j . 1. It follows that b is the only zero of s(x) in K[x], and, therefore, s(x) 5 (x 2 b)u. Since s(x) is irreducible and F has characteristic 0, Theorem 20.6 guarantees that u 5 1., In the terminology introduced earlier, it follows from Theorem 21.6, and induction that any finite extension of a field of characteristic 0 is a, simple extension. An element a with the property that E 5 F(a) is, called a primitive element of E., , Properties of Algebraic Extensions, Theorem 21.7 Algebraic over Algebraic Is Algebraic, If K is an algebraic extension of E and E is an algebraic extension, of F, then K is an algebraic extension of F., , PROOF Let a [ K. It suffices to show that a belongs to some finite, extension of F. Since a is algebraic over E, we know that a is the zero, of some irreducible polynomial in E[x], say, p(x) 5 bnx n 1 ? ? ? 1 b0., Now we construct a tower of extension fields of F, as follows:, F0 5 F(b0),, F1 5 F0(b1), . . . , Fn 5 Fn21(bn)., In particular,, Fn 5 F(b0, b1, . . . , bn),, so that p(x) [ Fn[x]. Thus, [Fn(a):Fn] 5 n; and, because each bi is algebraic over F, we know that each [Fi11:Fi] is finite. So,, [Fn(a):F ] 5 [Fn(a):Fn][Fn:Fn21] ? ? ? [F1:F0][F0:F ], is finite. (See Figure 21.5.), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 | Algebraic Extensions, , 383, , K, E, , Fn(a), Fn, Fn 21, F1, F0, F, , Figure 21.5, , Corollary Subfield of Algebraic Elements, Let E be an extension field of the field F. Then the set of all elements, of E that are algebraic over F is a subfield of E., , PROOF Suppose that a, b [ E are algebraic over F and b 2 0. To, show that a 1 b, a 2 b, ab, and a/b are algebraic over F, it suffices to, show that [F(a, b):F] is finite, since each of these four elements belongs to F(a, b). But note that, [F(a, b):F] 5 [F(a, b):F(b)][F(b):F]., Also, since a is algebraic over F, it is certainly algebraic over F(b)., Thus, both [F(a, b):F(b)] and [F(b):F] are finite., For any extension E of a field F, the subfield of E of the elements, that are algebraic over F is called the algebraic closure of F in E., One might wonder if there is such a thing as a maximal algebraic, extension of a field F—that is, whether there is an algebraic extension E, of F that has no proper algebraic extensions. For such an E to exist, it is, necessary that every polynomial in E[x] splits in E. Otherwise, it follows, from Kronecker’s Theorem that E would have a proper algebraic extension. This condition is also sufficient. If every member of E[x] splits in E,, and K is an algebraic extension of E, then every member of K is a zero of, some element of E[x]. But the zeros of elements of E[x] are in E. A field, that has no proper algebraic extension is called algebraically closed. In, 1910, Ernst Steinitz proved that every field F has a unique (up to isomorphism) algebraic extension that is algebraically closed. This field is, called the algebraic closure of F. A proof of this result requires a sophisticated set theory background., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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384, , Fields, , In 1799, Gauss, at the age of 22, proved that C is algebraically, closed. This fact was considered so important at the time that it was, called the Fundamental Theorem of Algebra. Over a 50-year period,, Gauss found three additional proofs of the Fundamental Theorem., Today more than 100 proofs exist. In view of the ascendancy of abstract, algebra in the 20th century, a more appropriate phrase for Gauss’s result, would be the Fundamental Theorem of Classical Algebra., , Exercises, It matters not what goal you seek, Its secret here reposes:, You’ve got to dig from week to week, To get Results or Roses., edgar guest, , 1. Prove Theorem 21.2 and Theorem 21.3., 2. Let E be the algebraic closure of F. Show that every polynomial in, F[x] splits in E., 3, 4, 3. Prove that Q(22, 22, 22, . . .) is an algebraic extension of Q but, not a finite extension of Q. (This exercise is referred to in this, chapter.), 4. Let E be an algebraic extension of F. If every polynomial in F[x], splits in E, show that E is algebraically closed., 5. Suppose that F is a field and every irreducible polynomial in F[x], is linear. Show that F is algebraically closed., 6. Suppose that f (x) and g(x) are irreducible over F and that deg f (x), and deg g(x) are relatively prime. If a is a zero of f (x) in some extension of F, show that g(x) is irreducible over F(a)., 7. Let a and b belong to Q with b 2 0. Show that Q(2a) 5 Q(2b) if, and only if there exists some c [ Q such that a 5 bc2., 8. Find the degree and a basis for Q(23 1 25) over Q(215). Find, 3, 4, the degree and a basis for Q(22, 22, 22) over Q., 9. Suppose that E is an extension of F of prime degree. Show that, for, every a in E, F(a) 5 F or F(a) 5 E., 10. Let a be a complex number that is algebraic over2 n Q. Show that, 2a is algebraic over Q. Why does this prove that 2a is algebraic, over Q?, 11. Suppose that E is an extension of F and a, b [ E. If a is algebraic, over F of degree m, and b is algebraic over F of degree n, where m, and n are relatively prime, show that [F(a, b):F ] 5 mn., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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21 | Algebraic Extensions, , 385, , 12. Find an example of a field F and elements a and b from some, extension field such that F(a, b) 2 F(a), F(a, b) 2 F(b), and, [F(a, b):F] , [F(a):F ][F(b):F ]., 13. Let K be a field extension of F and let a [ K. Show that, [F(a):F(a3)] # 3. Find examples to illustrate that [F(a):F(a3)] can, be 1, 2, or 3., 14. Find the minimal polynomial for 23 1 22 over Q., 15. Let K be an extension of F. Suppose that E1 and E2 are contained, in K and are extensions of F. If [E1:F ] and [E2:F ] are both prime,, show that E1 5 E2 or E1 > E2 5 F., 3, 3, 16. Find the minimal polynomial for 22 1 24 over Q., 17. Let E be a finite extension of R. Use the fact that C is algebraically, closed to prove that E 5 C or E 5 R., 18. Suppose that [E:Q] 5 2. Show that there is an integer d such that, E 5 Q(2d) where d is not divisible by the square of any prime., 19. Suppose that p(x) [ F[x] and E is a finite extension of F. If p(x) is, irreducible over F, and deg p(x) and [E:F ] are relatively prime,, show that p(x) is irreducible over E., 20. Let E be an extension field of F. Show that [E:F ] is finite if and only, if E 5 F(a1, a2, . . . , an), where a1, a2, . . . , an are algebraic over F., 21. If a and b are real numbers and a and b are transcendental over Q,, show that either ab or a 1 b is also transcendental over Q., 22. Let f (x) be a nonconstant element of F[x]. If a belongs to some, extension of F and f (a) is algebraic over F, prove that a is algebraic over F., 23. Let f (x) 5 ax2 1 bx 1 c [ Q[x]. Find a primitive element for the, splitting field for f (x) over Q., 24. Find the splitting field for x4 2 x2 2 2 over Z3., 25. Let f (x) [ F[x]. If deg f (x) 5 2 and a is a zero of f (x) in some, extension of F, prove that F(a) is the splitting field for f (x) over F., 26. Let a be a complex zero of x 2 1 x 1 1 over Q. Prove that, Q(2a) 5 Q(a)., 27. If F is a field and the multiplicative group of nonzero elements of, F is cyclic, prove that F is finite., 28. Let a be a complex number that is algebraic over Q and let r be a, rational number. Show that ar is algebraic over Q., 29. Prove that, if K is an extension field of F, then [K:F ] 5 n if and, only if K is isomorphic to Fn as vector spaces. (See Exercise 27 in, Chapter 19 for the appropriate definition. This exercise is referred, to in this chapter.), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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386, , Fields, , 30. Let a be a positive real number and let n be an integer greater than 1., Prove or disprove that [Q(a1/n):Q] 5 n., 31. Let a and b belong to some extension field of F and let b be algebraic over F. Prove that [F(a, b):F(a)] # [F(a, b):F]., 32. Let f (x) and g(x) be irreducible polynomials over a field F and let, a and b belong to some extension E of F. If a is a zero of f (x) and, b is a zero of g(x), show that f (x) is irreducible over F(b) if and, only if g(x) is irreducible over F(a)., 33. Let b be a zero of f(x) 5 x5 1 2x 1 4 (see Example 8 in Chapter 17)., 3, 4, Show that none of 22, 22, 22 belongs to Q(b)., 3, 6, 34. Prove that Q(22, 22) 5 Q(22)., 35. Let a and b be rational numbers. Show that Q( 2a, 2b) 5, Q(2a 1 2b)., 36. Let F, K, and L be fields with F 8 K 8 L. If L is a finite extension of F, and [L:F] 5 [L:K], prove that F K., 37. Let F be a field and K a splitting field for some nonconstant polynomial over F. Show that K is a finite extension of F., 38. Prove that C is not the splitting field of any polynomial in Q[x]., 39. Prove that 22 is not an element of Q1p2., 2p, 2p, 2p, 40. Let a cos 2p, 7 i sin 7 and b cos 5 i sin 5 . Prove that b, is not in Q1a2., 41. Suppose that a is algebraic over a field F. Show that a and 1 a1, have the same degree over F., 42. Suppose K is an extension of F of degree n. Prove that K can be, written in the form F(x1, x2, ), xn) for some x1, x2, ), xn in K., , Suggested Readings, R. L. Roth, “On Extensions of Q by Square Roots,” American Mathematical, Monthly 78 (1971): 392–393., In this paper, it is proved that if p1, p2, . . . , pn are distinct primes,, then [Q(2p1,2p2, . . . , 2pn):Q] 5 2n., Paul B. Yale, “Automorphisms of the Complex Numbers,” Mathematics, Magazine 39 (1966): 135–141., This award-winning expository paper is devoted to various results on, automorphisms of the complex numbers., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Irving Kaplansky, He got to the top of the heap, by being a first-rate doer and, expositor of algebra., , Irving Kaplansky was born on March 22,, 1917, in Toronto, Canada, a few years after, his parents emigrated from Poland. Although his parents thought he would pursue, a career in music, Kaplansky knew early on, that mathematics was what he wanted to do., As an undergraduate at the University of, Toronto, Kaplansky was a member of the, winning team in the first William Lowell, Putnam Competition, a mathematical contest for United States and Canadian college, students. Kaplansky received a B.A. degree, from Toronto in 1938 and an M.A. in 1939. In, 1939, he entered Harvard University to earn, his doctorate as the first recipient of a Putnam, Fellowship. After receiving his Ph.D. from, Harvard in 1941, Kaplansky stayed on as, Benjamin Peirce Instructor until 1944. After, one year at Columbia University, he went to, the University of Chicago, where he remained, until his retirement in 1984. He then became, the director of the Mathematical Sciences, Research Institute at the University of California, Berkeley., , American Mathematical Society, , paul r. halmos, I Have a, Photographic Memory, , Kaplansky’s interests were broad, including areas such as ring theory, group theory,, field theory, Galois theory, ergodic theory,, algebras, metric spaces, number theory, statistics, and probability., Among the many honors Kaplansky, received are election to both the National, Academy of Sciences and the American, Academy of Arts and Sciences, election to, the presidency of the American Mathematical Society, and the 1989 Steele Prize for, cumulative influence from the American, Mathematical Society. The Steele Prize citation says, in part, “. . . he has made striking, changes in mathematics and has inspired, generations of younger mathematicians.”, Kaplansky died on June 25, 2006, at the age, of 89., For more information about Kaplansky,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 387, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 Finite Fields, , This theory [of finite fields] is of considerable interest in its own right and it, provides a particularly beautiful example of how the general theory of the, preceding chapters fits together to provide a rather detailed description of, all finite fields., richard a. dean, Elements of Abstract Algebra, , Classification of Finite Fields, In this, our final chapter on field theory, we take up one of the most, beautiful and important areas of abstract algebra—finite fields. Finite, fields were first introduced by Galois in 1830 in his proof of the unsolvability of the general quintic equation. When Cayley invented matrices a, few decades later, it was natural to investigate groups of matrices over, finite fields. To this day, matrix groups over finite fields are among the, most important classes of groups. In the past 50 years, there have been, important applications of finite fields in computer science, coding theory, information theory, and cryptography. But, besides the many uses of, finite fields in pure and applied mathematics, there is yet another good, reason for studying them. They are just plain fun!, The most striking fact about finite fields is the restricted nature of, their order and structure. We have already seen that every finite field, has prime-power order (Exercise 51 in Chapter 13). A converse of sorts, is also true., Theorem 22.1 Classification of Finite Fields, For each prime p and each positive integer n, there is, up to, isomorphism, a unique finite field of order p n., , PROOF Consider the splitting field E of f(x) 5 xpn 2 x over Zp. We, will show that |E| 5 pn. Since f(x) splits in E, we know that f(x) has exactly pn zeros in E, counting multiplicity. Moreover, by Theorem 20.5,, every zero of f(x) has multiplicity 1. Thus, f(x) has pn distinct zeros in E., 388, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 | Finite Fields, , 389, , On the other hand, the set of zeros of f(x) in E is closed under addition,, subtraction, multiplication, and division by nonzero elements (see, Exercise 37), so that the set of zeros of f(x) is itself an extension field of, Zp in which f(x) splits. Thus, the set of zeros of f(x) is E and, therefore,, |E| 5 pn., , To show that there is a unique field for each prime-power, suppose, that K is any field of order pn. Then K has a subfield isomorphic to Zp, (generated by 1), and, because the nonzero elements of K form a multiplicative group of order pn 2 1, every element of K is a zero of f(x) 5, n, xp 2 x (see Exercise 27). So, K must be a splitting field for f (x) over Zp., By the corollary to Theorem 20.4, there is only one such field up to, isomorphism., The existence portion of Theorem 22.1 appeared in the works of, Galois and Gauss in the first third of the 19th century. Rigorous proofs, were given by Dedekind in 1857 and by Jordan in 1870 in his classic, book on group theory. The uniqueness portion of the theorem was, proved by E. H. Moore in an 1893 paper concerning finite groups. The, mathematics historian E. T. Bell once said that this paper by Moore, marked the beginning of abstract algebra in America., Because there is only one field for each prime-power pn, we may unambiguously denote it by GF( pn), in honor of Galois, and call it the, Galois field of order pn., , Structure of Finite Fields, The next theorem tells us the additive and multiplicative group structure of a field of order pn., Theorem 22.2 Structure of Finite Fields, As a group under addition, GF(pn) is isomorphic to, Zp % Zp % ? ? ? % Zp., n factors, As a group under multiplication, the set of nonzero elements of, GF( pn) is isomorphic to Zpn21 (and is, therefore, cyclic)., , PROOF Since GF( p n) has characteristic p (Theorem 13.3), every, nonzero element of GF( pn) has additive order p. Then by the Fundamental Theorem of Finite Abelian Groups, GF( pn) under addition is, isomorphic to a direct product of n copies of Zp., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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390, , Fields, , To see that the multiplicative group GF( pn)* of nonzero elements of, GF(pn) is cyclic, we first note that by the Fundamental Theorem of, Finite Abelian Groups (Theorem 11.1), GF(pn)* is isomorphic to a direct, product of the form Zn1 % Zn2 % ? ? ? % Znm. If the orders of these components are pairwise relatively prime, then it follows from Corollary 1 of, Theorem 8.2 that GF(pn)* is cyclic. Hence we may assume that there is, an integer d . 1 that divides the orders of two of the components. From, the Fundamental Theorem of Cyclic Groups (Theorem 4.3) we know that, each of these components has a subgroup of order d. This means that, GF(pn)* has two distinct subgroups of order d, call them H and K. But, then every element of H and K is a zero of xd 2 1, which contradicts the, fact that a polynomial of degree d over a field can have at most d zeros, (Corollary 3 of Theorem 16.2)., Some students misinterpret Theorem 22.2 to mean that Z p % Zp, % ? ? ? % Zp is a field of order pn. Since any element of Zp % Zp % ? ? ? %, Zp that has at least one coordinate equal to 0 cannot have an inverse, it, is not a field., Since Zp % Zp % ? ? ? % Zp is a vector space over Zp with {(1, 0,, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1)} as a basis, we have the, following useful and aesthetically appealing formula., Corollary 1, [GF(p n):GF(p)] 5 n, , Corollary 2 GF(pn) Contains an Element of Degree n, Let a be a generator of the group of nonzero elements of GF( pn), under multiplication. Then a is algebraic over GF( p) of degree n., , PROOF Observe that [GF( p)(a):GF( p)] 5 [GF( pn):GF( p)] 5 n., EXAMPLE 1 Let’s examine the field GF(16) in detail. Since x4 1, x 1 1 is irreducible over Z2, we know that, GF(16) < {ax3 1 bx2 1 cx 1 d 1 kx4 1 x 1 1l | a, b, c, d [ Z2}., Thus, we may think of GF(16) as the set, F 5 {ax3 1 bx2 1 cx 1 d | a, b, c, d [ Z2},, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 | Finite Fields, , 391, , where addition is done as in Z2[x], but multiplication is done modulo, x4 1 x 1 1. For example,, (x3 1 x2 1 x 1 1)(x3 1 x) 5 x3 1 x2,, since the remainder upon dividing, (x3 1 x2 1 x 1 1)(x3 1 x) 5 x6 1 x5 1 x2 1 x, by x4 1 x 1 1 in Z2[x] is x3 1 x2. An easier way to perform the same, calculation is to observe that in this context x4 1 x 1 1 is 0, so, x4 5 2x 2 1 5 x 1 1,, x5 5 x2 1 x,, x6 5 x3 1 x2., Thus,, x6 1 x5 1 x2 1 x 5 (x3 1 x2) 1 (x2 1 x) 1 x2 1 x 5 x3 1 x2., Another way to simplify the multiplication process is to make use of, the fact that the nonzero elements of GF(16) form a cyclic group of, order 15. To take advantage of this, we must first find a generator of this, group. Since any element F* must have a multiplicative order that divides 15, all we need to do is find an element a in F* such that a3 2 1, and a5 2 1. Obviously, x has these properties. So, we may think of, GF(16) as the set {0, 1, x, x2, . . . , x14}, where x15 5 1. This makes multiplication in F trivial, but, unfortunately, it makes addition more difficult. For example, x10 ? x7 5 x17 5 x2, but what is x10 1 x7? So, we face, a dilemma. If we write the elements of F* in the additive form ax3 1, bx2 1 cx 1 d, then addition is easy and multiplication is hard. On the, other hand, if we write the elements of F* in the multiplicative form xi,, then multiplication is easy and addition is hard. Can we have the best of, both? Yes, we can. All we need to do is use the relation x4 5 x 1 1 to, make a two-way conversion table, as in Table 22.1., So, we see from Table 22.1 that, x10 1 x7 5 (x2 1 x 1 1) 1 (x3 1 x 1 1), 5 x3 1 x2 5 x6, and, (x3 1 x2 1 1)(x3 1 x2 1 x 1 1) 5 x13 ? x12, 5 x25 5 x10 5 x2 1 x 1 1., Don’t be misled by the preceding example into believing that the, element x is always a generator for the cyclic multiplicative group, of nonzero elements. It is not. (See Exercise 19.) Although any two, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 | Finite Fields, , 393, , So, yes, it is. Next we try a3., f(a3) 5 (a3)3 1 (a3)2 1 1 5 a9 1 a6 1 1, 5 a2 1 (a2 1 a) 1 1 5 a 1 1 2 0., Now a4., f(a4) 5 (a4)3 1 (a4)2 1 1 5 a12 1 a8 1 1, 5 a5 1 a 1 1 5 (a 1 1) 1 a 1 1 5 0., So, a4 is our remaining zero. Thus, f(x) 5 (x 2 a)(x 2 a2)(x 2 a4) 5, (x 1 a)(x 1 a2)(x 1 a4), since char F 5 2., We may check this factorization by expanding the product and using, a conversion table to obtain f(x) 5 x3 1 x2 1 1., , Subfields of a Finite Field, Theorem 22.1 gives us a complete description of all finite fields. The, following theorem gives us a complete description of all the subfields, of a finite field. Notice the close analogy between this theorem and, Theorem 4.3, which describes all the subgroups of a finite cyclic group., Theorem 22.3 Subfields of a Finite Field, For each divisor m of n, GF( pn) has a unique subfield of order pm., Moreover, these are the only subfields of GF( pn)., , PROOF To show the existence portion of the theorem, suppose that, m divides n. Then, since, pn 2 1 5 ( pm 2 1)( pn2m 1 pn22m 1 ? ? ? 1 pm 1 1),, we see that pm 2 1 divides pn 2 1. For simplicity, write pn 2 1 5, m, (pm 2 1)t. Let K 5 {x [ GF(pn) | x p 5 x}. We leave it as an easy exercise for the reader to show that K is a subfield of GF(pn) (Exercise 25)., m, Since the polynomial x p 2 x has at most pm zeros in GF(pn), we have, m, |K| # pm. Let kal 5 GF(pn)*. Then |at| 5 pm 2 1, and since (a t) p 21 5 1,, t, n, m, it follows that a [ K. So, K is a subfield of GF(p ) of order p ., The uniqueness portion of the theorem follows from the observation, that if GF(pn) had two distinct subfields of order pm, then the polynom, mial x p 2 x would have more than pm zeros in GF(pn). This contradicts Corollary 3 of Theorem 16.2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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394, , Fields, , Finally, suppose that F is a subfield of GF( pn). Then F is isomorphic to, GF( pm) for some m and, by Theorem 21.5,, n 5 [GF( pn):GF( p)], 5 [GF( pn):GF( pm)][GF( pm):GF( p)], 5 [GF( pn):GF( pm)]m., Thus, m divides n., Theorems 22.2 and 22.3, together with Theorem 4.3, make the task, of finding the subfields of a finite field a simple exercise in arithmetic., EXAMPLE 3 Let F be the field of order 16 given in Example 1. Then, there are exactly three subfields of F, and their orders are 2, 4, and 16., Obviously, the subfield of order 2 is {0, 1} and the subfield of order 16, is F itself. To find the subfield of order 4, we merely observe that the, three nonzero elements of this subfield must be the cyclic subgroup of, F* 5 kxl of order 3. So the subfield of order 4 is, {0, 1, x5, x10} 5 {0, 1, x2 1 x, x2 1 x 1 1}., EXAMPLE 4 If F is a field of order 36 5 729 and a is a generator of, F*, then the subfields of F are, 1. GF(3) 5 {0} < ka364l 5 {0, 1, 2},, 2. GF(9) 5 {0} < ka91l,, 3. GF(27) 5 {0} < ka28l,, 4. GF(729) 5 {0} < kal., EXAMPLE 5 The subfield lattice of GF(224) is the following., GF(224 ), GF(212 ), 8, , GF(2 ), GF(26 ), GF(24 ), GF(2 3 ), 2, , GF(2 ), , GF(2), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 | Finite Fields, , 395, , Exercises, No pressure, no diamonds., mary case, , 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., , 12., 13., 14., 15., , 16., 17., 18., , Find [GF(729):GF(9)] and [GF(64):GF(8)]., If m divides n, show that [GF( pn):GF( pm)] 5 n/m., Draw the lattice of subfields of GF(64)., Let a be a zero of x3 1 x2 1 1 in some extension field of Z2. Find, the multiplicative inverse of a 1 1 in Z2[a]., Let a be a zero of f (x) 5 x2 1 2x 1 2 in some extension field of Z3., Find the other zero of f (x) in Z3[a]., Let a be a zero of f (x) 5 x3 1 x 1 1 in some extension field of Z2., Find the other zeros of f (x) in Z2[a]., Let K be a finite extension field of a finite field F. Show that there, is an element a in K such that K 5 F(a)., How many elements of the cyclic group GF(81)* are generators?, Let f (x) be a cubic irreducible over Z2. Prove that the splitting field, of f(x) over Z2 has order 8., Prove that the rings Z3[x]/kx2 1 x 1 2l and Z3[x]/kx2 1 2x 1 2l are, isomorphic., Show that the Frobenius mapping f: GF( pn) S GF( pn), given by, a S ap, is a ring automorphism of order n (that is, fn is the identity, mapping). (This exercise is referred to in Chapter 32.), Determine the possible finite fields whose largest proper subfield, is GF(25)., Prove that the degree of any irreducible factor of x 8 2 x over Z2 is, 1 or 3., Find the smallest field that has exactly 6 subfields., Find the smallest field of characteristic 2 that contains an element, whose multiplicative order is 5 and the smallest field of characteristic 3 that contains an element whose multiplicative order is 5., Verify that the factorization for f (x) 5 x3 1 x2 1 1 over Z2 given in, Example 2 is correct by expanding., Show that x is a generator of the cyclic group (Z3[x]/kx3 1 2x 1 1l)*., Suppose that f (x) is a fifth-degree polynomial that is irreducible, over Z2. Prove that x is a generator of the cyclic group (Z2[x]/, k f (x)l)*., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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396, , Fields, , 19. Show that x is not a generator of the cyclic group (Z3[x]/kx3 1, 2x 1 2l)*. Find one such generator., 20. If f (x) is a cubic irreducible polynomial over Z3, prove that either x, or 2x is a generator for the cyclic group (Z3[x]/kf(x)l)*., 21. Prove the uniqueness portion of Theorem 22.3 using a group, theoretic argument., 22. Suppose that a and b belong to GF(81)*, with |a| 5 5 and |b| 5 16., Show that ab is a generator of GF(81)*., 23. Construct a field of order 9 and carry out the analysis as in Example 1, including the conversion table., 24. Show that any finite subgroup of the multiplicative group of a field, is cyclic., 25. Show that the set K in the proof of Theorem 22.3 is a subfield., n, 26. If g(x) is irreducible over GF( p) and g(x) divides x p 2 x, prove, that deg g(x) divides n., 27. Use a purely group theoretic argument to show that if F is a field, n, of order pn, then every element of F* is a zero of x p 2 x. (This exercise is referred to in the proof of Theorem 22.1.), 28. Draw the subfield lattices of GF(318) and of GF(230)., 29. How does the subfield lattice of GF(230) compare with the subfield, lattice of GF(330)?, 30. If p(x) is a polynomial in Zp[x] with no multiple zeros, show that, p(x) divides x pn 2 x for some n., 31. Suppose that p is a prime and p 2 2. Let a be a nonsquare in, GF( p)—that is, a does not have the form b2 for any b in GF( p)., Show that a is a nonsquare in GF( pn) if n is odd and that a is a, square in GF( pn) if n is even., 32. Let f (x) be a cubic irreducible over Zp, where p is a prime. Prove, that the splitting field of f (x) over Zp has order p3 or p6., 33. Show that every element of GF( pn) can be written in the form ap, for some unique a in GF( pn)., 34. Suppose that F is a field of order 1024 and F* 5 kal. List the elements of each subfield of F., 35. Suppose that F is a field of order 125 and F* 5 kal. Show that, a62 5 21., 36. Show that no finite field is algebraically closed., n, 37. Let E be the splitting field of f(x) 5 x p 2 x over Zp. Show that the, set of zeros of f(x) in E is closed under addition, subtraction, multiplication, and division (by nonzero elements). (This exercise is, referred to in the proof of Theorem 22.1.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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22 | Finite Fields, , 397, , 38. Suppose that L and K are subfields of GF( pn). If L has ps elements, and K has pt elements, how many elements does L > K have?, 39. Give an example to show that the mapping a S ap need not be an, automorphism for arbitrary fields of prime characteristic p., 40. In the field GF(pn), show that for every positive divisor d of n,, n, xp x has an irreducible factor over GF(p) of degree d., 41. Let a be a primitive element for the field GF(pn), where p is an odd, prime and n is a positive integer. Find the smallest positive integer, k such that ak 5 p 2 1., 42. Let a be a primitive element for the field GF(5n), where n is a positive integer. Find the smallest positive integer k such that ak 5 2., 43. Let p be a prime such that p mod 4 5 1. How many elements of, order 4 are in GF(pn)*?, 44. Let p be a prime such that p mod 4 5 3. How many elements of, order 4 are in GF(pn)*?, , Computer Exercises, Software for the computer exercises in this chapter is available at the, website:, http://www.d.umn.edu/~jgallian, , Suggested Reading, Judy L. Smith and J. A. Gallian, “Factoring Finite Factor Rings,”, Mathematics Magazine 58 (1985): 93–95., This paper gives an algorithm for finding the group of units of the ring, F[x]/kg(x)ml., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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L .E. Dickson, , American Mathematical Society, , One of the books [written by L. E. Dickson], is his major, three-volume History of the, Theory of Numbers which would be a life’s, work by itself for a more ordinary man., a. a. albert,, Bulletin of the American, Mathematical Society, , Leonard Eugene Dickson was born in, Independence, Iowa, on January 22, 1874., In 1896, he received the first Ph.D. to be, awarded in mathematics at the University of, Chicago. After spending a few years at the, University of California and the University, of Texas, he was appointed to the faculty at, Chicago and remained there until his retirement in 1939., Dickson was one of the most prolific, mathematicians of the 20th century, writing, 267 research papers and 18 books. His principal interests were matrix groups, finite, fields, algebra, and number theory., Dickson had a disdainful attitude toward, applicable mathematics; he would often say,, “Thank God that number theory is unsullied, by any applications.” He also had a sense of, , humor. Dickson would often mention his, honeymoon: “It was a great success,” he, said, “except that I only got two research, papers written.”, Dickson received many honors in his, career. He was the first to be awarded the, prize from the American Association for the, Advancement of Science for the most notable, contribution to the advancement of science,, and the first to receive the Cole Prize in algebra from the American Mathematical Society., The University of Chicago has research instructorships named after him. Dickson died, on January 17, 1954., For more information about Dickson, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 398, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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23 Geometric, Constructions, At the age of eleven, I began Euclid. . . . This was one of the great events, of my life, as dazzling as first love., bertrand russell, , Historical Discussion, of Geometric Constructions, The ancient Greeks were fond of geometric constructions. They were, especially interested in constructions that could be achieved using only a, straightedge without markings and a compass. They knew, for example,, that any angle can be bisected, and they knew how to construct an equilateral triangle, a square, a regular pentagon, and a regular hexagon. But, they did not know how to trisect every angle or how to construct a regular seven-sided polygon (heptagon). Another problem that they attempted was the duplication of the cube—that is, given any cube, they, tried to construct a new cube having twice the volume of the given one, using only an unmarked straightedge and a compass. Legend has it that, the ancient Athenians were told by the oracle at Delos that a plague, would end if they constructed a new altar to Apollo in the shape of a cube, with double the volume of the old altar, which was also a cube. Besides, “doubling the cube,” the Greeks also attempted to “square the circle”—to, construct a square with area equal to that of a given circle. They knew, how to solve all these problems using other means, such as a compass, and a straightedge with two marks, or an unmarked straightedge and a, spiral, but they could not achieve any of the constructions with a compass, and an unmarked straightedge alone. These problems vexed mathematicians for over 2000 years. The resolution of these perplexities was made, possible when they were transferred from questions of geometry to questions of algebra in the 19th century., The first of the famous problems of antiquity to be solved was that of, the construction of regular polygons. It had been known since Euclid that, regular polygons with a number of sides of the form 2k, 2k ? 3, 2k ? 5, and, 2k ? 3 ? 5 could be constructed, and it was believed that no others were, 399, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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400, , Fields, , possible. In 1796, while still a teenager, Gauss proved that the, 17-sided regular polygon is constructible. In 1801, Gauss asserted that a, regular polygon of n sides is constructible if and only if n hass the form, 2kp1 p2 ? ? ? pi, where the p’s are distinct primes of the form 22 1 1. We, provide a proof of this statement in Theorem 33.5., Thus, regular polygons with 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, and 20, sides are possible to construct, whereas those with 7, 9, 11, 13, 14,, 18, and 19 sides are not. How these constructions can be effected is another matter. One person spent 10 years trying to determine a way to, construct the 65,537-sided polygon., Gauss’s result on the constructibility of regular n-gons eliminated, another of the famous unsolved problems, because the ability to trisect, a 60° angle enables one to construct a regular 9-gon. Thus, there is no, method for trisecting a 60° angle with an unmarked straightedge and a, compass. In 1837, Wantzel proved that it was not possible to double, the cube. The problem of the squaring of a circle resisted all attempts, until 1882, when Ferdinand Lindemann proved that p is transcendental, since, as we will show, all constructible numbers are algebraic., , Constructible Numbers, With the field theory we now have, it is an easy matter to solve the following, problem: Given an unmarked straightedge, a compass, and a unit length, what, other lengths can be constructed? To begin, we call a real number a constructible if, by means of an unmarked straightedge, a compass, and a line segment, of length 1, we can construct a line segment of length |a| in a finite number of, steps. It follows from plane geometry that if a and b (b 2 0) are constructible, numbers, then so are a 1 b, a 2 b, a ? b, and a/b. (See the exercises for, hints.) Thus, the set of constructible numbers contains Q and is a subfield of, the real numbers. What we desire is an algebraic characterization of this field., To derive such a characterization, let F be any subfield of the reals. Call, the subset {(x, y) [ R2 | x, y [ F} of the real plane the plane of F, call, any line joining two points in the plane of F a line in F, and call any circle, whose center is in the plane of F and whose radius is in F a circle in F. Then a, line in F has an equation of the form, ax 1 by 1 c 5 0,, , where a, b, c [ F,, , and a circle in F has an equation of the form, x2 1 y2 1 ax 1 by 1 c 5 0,, , where a, b, c [ F., , In particular, note that to find the point of intersection of a pair of lines, in F or the points of intersection of a line in F and a circle in F, one, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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23 | Geometric Constructions, , 401, , need only solve a linear or quadratic equation in F. We now come to, the crucial question. Starting with points in the plane of some field F,, which points in the real plane can be obtained with an unmarked, straightedge and a compass? Well, there are only three ways to construct points, starting with points in the plane of F., 1. Intersect two lines in F., 2. Intersect a circle in F and a line in F., 3. Intersect two circles in F., In case 1, we do not obtain any new points, because two lines in F intersect in a point in the plane of F. In case 2, the point of intersection is, the solution to either a linear equation in F or a quadratic equation in, F. So, the point lies in the plane of F or in the plane of F(2a), where, a [ F and a is positive. In case 3, no new points are obtained, because,, if the two circles are given by x2 1 y 2 1 ax 1 by 1 c 5 0 and, x2 1 y2 1 a9x 1 b9y 1 c9 5 0, then we have (a 2 a9)x 1 (b 2 b9)y 1, (c 2 c9) 5 0, which is a line in F. So, the points of intersection are in F., It follows, then, that the only points in the real plane that can be, constructed from the plane of a field F are those whose coordinates, lie in fields of the form F(2a), where a [ F and a is positive. Of, course, we can start over with F 1 5 F( 2a) and construct points, whose coordinates lie in fields of the form F2 5 F1(2b), where b [, F1 and b is positive. Continuing in this fashion, we see that a real, number c is constructible if and only if there is a series of fields Q 5, F1 # F2 # ? ? ? # Fn # R such that Fi11 5 Fi (2ai), where ai [ Fi, and c [ Fn. Since [Fi11:Fi] 5 1 or 2, we see by Theorem 21.5 that if, c is constructible, then [Q(c):Q] 5 2k for some nonnegative integer k., We now dispatch the problems that plagued the Greeks. Consider doubling the cube of volume 1. The enlarged cube would have an edge of, 3, 3, length 22. But [Q(22):Q] 5 3, so such a cube cannot be constructed., Next consider the possibility of trisecting a 60° angle. If it were possible to trisect an angle of 60°, then cos 20° would be constructible. (See, Figure 23.1.) In particular, [Q(cos 20°):Q] 5 2k for some k. Now, using, the trigonometric identity cos 3u 5 4 cos3 u 2 3 cos u, with u 5 20°, we, , (cos 20°, sin 20°), , (0, 0), , (1, 0), , Figure 23.1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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402, , Fields, , see that 1/2 5 4 cos3 20° 2 3 cos 20°, so that cos 20° is a zero of, 8x3 2 6x 2 1. But, since 8x3 2 6x 2 1 is irreducible over Q (see, Exercise 13), we must also have [Q(cos 20°):Q] 5 3. This contradiction, shows that trisecting a 60° angle is impossible., The remaining problems are relegated to the reader as Exercises 14,, 15, and 17., , Angle-Trisectors and Circle-Squarers, Down through the centuries, hundreds of people have claimed to have, achieved one or more of the impossible constructions. In 1775, the Paris, Academy, so overwhelmed with these claims, passed a resolution to no, longer examine these claims or claims of machines purported to exhibit, perpetual motion. Although it has been more than 100 years since the last, of the constructions was shown to be impossible, there continues to be a, steady parade of people who claim to have done one or more of them., Most of these people have heard that this is impossible but have refused, to believe it. One person insisted that he could trisect any angle with a, straightedge alone [2, p. 158]. Another found his trisection in 1973 after, 12,000 hours of work [2, p. 80]. One got his from God [2, p. 73]. In, 1971, a person with a Ph.D. in mathematics asserted that he had a valid, trisection method [2, p. 127]. Many people have claimed the hat trick:, trisecting the angle, doubling the cube, and squaring the circle. Two men, who did this in 1961 succeeded in having their accomplishment noted in, the Congressional Record [2, p. 110]. Occasionally, newspapers and, magazines have run stories about “doing the impossible,” often giving, the impression that the construction may be valid. Many angle-trisectors, and circle-squarers have had their work published at their own expense, and distributed to colleges and universities. One had his printed in four, languages! There are two delightful books written by mathematicians, about their encounters with these people. The books are full of wit,, charm, and humor ([1] and [2])., , Exercises, Only prove to me that it is impossible, and I will set about it this very, evening., Spoken by a member of the audience after De Morgan gave a, lecture on the impossibility of squaring the circle., , 1. If a and b are constructible numbers and a $ b . 0, give a geometric proof that a 1 b and a 2 b are constructible., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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23 | Geometric Constructions, , 403, , 2. If a and b are constructible, give a geometric proof that ab is constructible. (Hint: Consider the following figure. Notice that all segments in the figure can be made with an unmarked straightedge and a, compass.), , a, , 1, b, , 3. Prove that if c is a constructible number, then so is 2|c|. (Hint:, Consider the following semicircle with diameter 1 1 |c|.) (This exercise is referred to in Chapter 33.), α, d, , α, , ⏐c⏐, , 1, , 4. If a and b (b 2 0) are constructible numbers, give a geometric proof, that a/b is constructible. (Hint: Consider the following figure.), b, , 0, , 1, a, , 5. Prove that sin u is constructible if and only if cos u is constructible., 6. Prove that an angle u is constructible if and only if sin u is constructible., 7. Prove that cos 2u is constructible if and only if cos u is constructible., 8. Prove that 30° is a constructible angle., 9. Prove that a 45° angle can be trisected with an unmarked straightedge and a compass., 10. Prove that a 40° angle is not constructible., 11. Show that the point of intersection of two lines in the plane of a, field F lies in the plane of F., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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404, , Fields, , 12. Show that the points of intersection of a circle in the plane of a field, F and a line in the plane of F are points in the plane of F or in the, plane of F(2a), where a [ F and a is positive. Give an example, of a circle and a line in the plane of Q whose points of intersection, are not in the plane of Q., 13. Prove that 8x3 2 6x 2 1 is irreducible over Q., 14. Use the fact that 8 cos3(2p/7) 1 4 cos2(2p/7) 2 4 cos(2p/7) 2 1 5 0, to prove that a regular seven-sided polygon is not constructible with, an unmarked straightedge and a compass., 15. Show that a regular 9-gon cannot be constructed with an unmarked, straightedge and a compass., 16. Show that if a regular n-gon is constructible, then so is a regular, 2n-gon., 17. (Squaring the Circle) Show that it is impossible to construct, with, an unmarked straightedge and a compass, a square whose area, equals that of a circle of radius 1. You may use the fact that p is, transcendental over Q., 18. Use the fact that 4 cos2(2p/5) 1 2 cos(2p/5) 2 1 5 0 to prove that, a regular pentagon is constructible., 19. Can the cube be “tripled”?, 20. Can the cube be “quadrupled”?, 21. Can the circle be “cubed”?, 22. If a, b, and c are constructible, show that the real roots of ax2 1, bx 1 c are constructible., , References, 1. Augustus De Morgan, A Budget of Paradoxes, Dover Publications, 1954,, books.google.com., 2. Underwood Dudley, A Budget of Trisections, New York:, Springer-Verlag, 1987., , Suggested Website, http://en.wikipedia.org/wiki/Squaring_the_circle, This website provides an excellent account of efforts to square the, circle, and links for articles about trisecting the angle and doubling, the cube., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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405, , Supplementary Exercises for Chapters 19–23, , Supplementary Exercises for Chapters 19–23, Difficulties strengthen the mind, as labor does the body., seneca, , True/false questions for Chapters 19–23 are available on the Web at:, http://www.d.umn.edu/~jgallian/TF, x50, , 1. Show that, 2 1 has no multiple zeros in any extension of Z3., 2. Suppose that p(x) is a quadratic polynomial with rational coefficients and is irreducible over Q. Show that p(x) has two zeros in, Q[x]/kp(x)l., 3. Let F be a finite field of order q and let a be a nonzero element in, F. If n divides q 2 1, prove that the equation x n 5 a has either no, solutions in F or n distinct solutions in F., 4. Without using the Primitive Element Theorem, prove that if [K:F], is prime, then K has a primitive element., 5. Let a be a zero of x2 1 x 1 1. Express (5a2 1 2)/a in the form c 1, ba, where c and b are rational., 4, 6. Describe the elements of the extension Q(22) over the field Q(22)., 7. If [F(a):F] 5 5, find [F(a3):F]. Does your argument apply equally, well if a3 is replaced with a2 or a4?, 8. If p(x) [ F[x] and deg p(x) 5 n, show that the splitting field for, p(x) over F has degree at most n!., 9. Let a be a nonzero algebraic element over F of degree n. Show that, a21 is also algebraic over F of degree n., 10. Prove that p2 2 1 is algebraic over Q(p3)., 11. If ab is algebraic over F and b 2 0, prove that a is algebraic over F(b)., 12. Let E be an algebraic extension of a field F. If R is a ring and E $, R $ F, show that R must be a field., 13. If a is transcendental over F, show that every element of F(a) that, is not in F is transcendental over F., 14. What is the order of the splitting field of x5 1 x4 1 1 5 (x2 1 x 1 1) ?, (x3 1 x 1 1) over Z2?, 15. Show that a finite extension of a finite field is a simple extension., 16. Let R be an integral domain that contains a field F as a subring. If, R is finite dimensional when viewed as a vector space over F, prove, that R is a field., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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406, , Fields, , 17. Show that it is impossible to find a basis for the vector space of, n 3 n (n . 1) matrices such that each pair of elements in the basis, commutes under multiplication., 18. Let Pn 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | each ai is a real, number}. Is it possible to have a basis for Pn such that every element of the basis has x as a factor?, 19. Find a basis for the vector space { f [ P3 | f (0) 5 0}. (See Exercise 18, for notation.), 20. Given that f is a polynomial of degree n in Pn, show that { f, f 9,, f 0, . . . , f (n)} is a basis for Pn. ( f (k) denotes the kth derivative of f.), 21. Suppose that K is an extension field of a field F of characteristic, n, p 2 0. Let L 5 {a [ K | a p [ F for some nonnegative integer n}., Prove that L is a subfield of K that contains F., 22. In which fields does x n 2 x have a multiple zero?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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PA R T, , 5, , Special, Topics, , For online student resources, visit this textbook’s website at, www.CengageBrain.com, , 407, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 Sylow Theorems, , Generally these three results are implied by the expression “Sylow’s, Theorem.” All of them are of fundamental importance. In fact, if the, theorems of group theory were arranged in order of their importance, Sylow’s Theorem might reasonably occupy the second place—coming next, to Lagrange’s Theorem in such an arrangement., g. a. miller, Theory and Application, of Finite Groups, , Conjugacy Classes, In this chapter, we derive several important arithmetic relationships, between a group and certain of its subgroups. Recall from Chapter 7, that Lagrange’s Theorem was proved by showing that cosets of a subgroup partition the group. Another fruitful method of partitioning the, elements of a group is by way of conjugacy classes., Definition Conjugacy Class of a, , Let a and b be elements of a group G. We say that a and b are, conjugate in G (and call b a conjugate of a) if xax21 5 b for some x, in G. The conjugacy class of a is the set cl(a) 5 {xax21 | x [ G}., , We leave it to the reader (Exercise 1) to prove that conjugacy is an, equivalence relation on G, and that the conjugacy class of a is the equivalence class of a under conjugacy. Thus, we may partition any group into, disjoint conjugacy classes. Let’s look at one example. In D4 we have, cl(H) 5 {R0HR021, R90 HR9021, R180 HR18021, R270HR27021,, HHH21, VHV21, DHD21, D9HD921} 5 {H, V}., Similarly, one may verify that, cl(R0) 5 {R0},, cl(R90) 5 {R90, R270} 5 cl(R270),, cl(R180) 5 {R180},, cl(V) 5 {V, H} 5 cl(H),, cl(D) 5 {D, D9} 5 cl(D9)., 409, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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410, , Special Topics, , Theorem 24.1 gives an arithmetic relationship between the size of, the conjugacy class of a and the size of C(a), the centralizer of a., Theorem 24.1 Number of Conjugates of a, Let G be a finite group and let a be an element of G. Then,, |cl(a)| 5 |G:C(a)|., , PROOF Consider the function T that sends the coset xC(a) to the, conjugate xax21 of a. A routine calculation shows that T is well-defined, is, one-to-one, and maps the set of left cosets onto the conjugacy class of a., Thus, the number of conjugates of a is the index of the centralizer of a., Corollary 1 |cl(a)| Divides |G|, In a finite group, |cl(a)| divides |G|., , The Class Equation, Since the conjugacy classes partition a group, the following important, counting principle is a corollary to Theorem 24.1., Corollary 2 Class Equation, For any finite group G,, , |G| 5 S|G:C(a)|,, where the sum runs over one element a from each conjugacy class of G., , In finite group theory, counting principles such as this corollary are, powerful tools.† Theorem 24.2 is the single most important fact about, finite groups of prime-power order (a group of order pn, where p is a, prime, is called a p-group)., Theorem 24.2 p-Groups Have Nontrivial Centers, Let G be a nontrivial finite group whose order is a power of a prime p., Then Z(G) has more than one element., , †“Never, , underestimate a theorem that counts something.” John Fraleigh, A First Course, in Abstract Algebra., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 411, , PROOF First observe that cl(a) 5 {a} if and only if a [ Z(G) (see, Exercise 4). Thus, by culling out these elements, we may write the, class equation in the form, |G| 5 |Z(G)| 1 S|G:C(a)|,, where the sum runs over representatives of all conjugacy classes with, more than one element (this set may be empty). But |G:C(a)| 5, |G|/|C(a)|, so each term in S|G:C(a)| has the form pk with k $ 1. Hence,, |G| 2 S|G:C(a)| 5 |Z(G)|,, where each term on the left is divisible by p. It follows, then, that p also, divides |Z(G)|, and hence |Z(G)| 2 1., Corollary Groups of Order p2 Are Abelian, If |G| 5 p2, where p is prime, then G is Abelian., , PROOF By Theorem 24.2 and Lagrange’s Theorem, |Z(G)| 5 p or p2., If |Z(G)| 5 p2, then G 5 Z(G) and G is Abelian. If |Z(G)| 5 p, then, |G/Z(G)| 5 p, so that G/Z(G) is cyclic. But, then, by Theorem 9.3, G is, Abelian., , The Probability That Two, Elements Commute, Before proceeding to the main goal of this chapter, we pause for an interesting application of Theorem 24.1 and the class equation. (Our discussion is based on [1] and [2].) Suppose we select two elements, at random (with replacement) from a finite group. What is the probability that these two elements commute? Well, suppose that G is a finite group of order n. Then the probability Pr(G) that two elements, selected at random from G commute is |K|/n2, where K 5 {(x, y) [, G % G | xy 5 yx}. Now notice that for each x [ G we have (x, y) [ K, if and only if y [ C(x). Thus,, 0 K 0 a 0C1x2 0 ., x[G, , Also, it follows from Theorem 24.1 that if x and y are in the same, conjugacy class, then |C(x)| 5 |C(y)| (see Exercise 71). If, for example, cl(a) 5 {a1, a2, . . . , at}, then, |C(a1)| 1 |C(a2)| 1 ? ? ? 1 |C(at)| 5 t|C(a)|, 5 |G:C(a)||C(a)| 5 |G| 5 n., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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412, , Special Topics, , So, by choosing one representative from each conjugacy class, say, x1,, x2, . . . , xm, we have, m, , 0 K 0 a 0 C1x2 0 a 0 G:C1xi 2 0 0 C1xi 2 0 m # n., x[G, , i1, , Thus, the answer to our question is mn/n2 5 m/n, where m is the number of conjugacy classes in G and n is the number of elements of G., Obviously, when G is non-Abelian, Pr(G) is less than 1. But how much, less than 1? Clearly, the more conjugacy classes there are, the larger Pr(G), is. Consequently, Pr(G) is large when the sizes of the conjugacy classes, are small. Noting that |cl(a)| 5 1 if and only if a [ Z(G), we obtain the, maximum number of conjugacy classes when |Z(G)| is as large as possible and all other conjugacy classes have exactly two elements in each., Since G is non-Abelian, it follows from Theorem 9.3 that |G/Z(G)| $ 4, and, therefore, |Z(G)| # |G|/4. Thus, in the extreme case, we would have, |Z(G)| 5 |G|/4, and the remaining (3/4)|G| elements would be distributed, in conjugacy classes with two elements each. So, in a non-Abelian group,, the number of conjugacy classes is no more than |G|/4 1 (1/2)(3/4)|G|,, and Pr(G) is less than or equal to 5/8. The dihedral group D4 is an example of a group that has probability equal to 5/8., , The Sylow Theorems, Now to the Sylow theorems. Recall that the converse of Lagrange’s, Theorem is false; that is, if G is a group of order m and n divides m,, G need not have a subgroup of order n. Our next theorem is a partial, converse of Lagrange’s Theorem. It, as well as Theorem 24.2, was first, proved by the Norwegian mathematician Ludwig Sylow (1832–1918)., Sylow’s Theorem and Lagrange’s Theorem are the two most important, results in finite group theory.† The first gives a sufficient condition for, the existence of subgroups, and the second gives a necessary condition., Theorem 24.3 Existence of Subgroups of Prime-Power Order, (Sylow’s First Theorem, 1872), Let G be a finite group and let p be a prime. If p k divides |G|, then G, has at least one subgroup of order p k., , †My candidate for the third most important result is the Fundamental Theorem of, Finite Abelian Groups., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 413, , PROOF We proceed by induction on |G|. If |G| 5 1, Theorem 24.3 is, trivially true. Now assume that the statement is true for all groups of, order less than |G|. If G has a proper subgroup H such that pk divides, |H|, then, by our inductive assumption, H has a subgroup of order pk, and we are done. Thus, we may henceforth assume that p k does not, divide the order of any proper subgroup of G. Next, consider the class, equation for G in the form, |G| 5 |Z(G)| 1 S|G:C(a)|,, where we sum over a representative of each conjugacy class cl(a), where, a o Z(G). Since pk divides |G| 5 |G:C(a)||C(a)| and pk does not divide, |C(a)|, we know that p must divide |G:C(a)| for all a o Z(G). It then follows from the class equation that p divides |Z(G)|. The Fundamental, Theorem of Finite Abelian Groups (Theorem 11.1), or Theorem 9.5, then, guarantees that Z(G) contains an element of order p, say x. Since x is in, the center of G, kxl is a normal subgroup of G, and we may form the factor group G/kxl. Now observe that pk21 divides |G/kxl|. Thus, by the, induction hypothesis, G/kxl has a subgroup of order pk21 and, by Exercise 51 in Chapter 10, this subgroup has the form H/kxl, where H is a, subgroup of G. Finally, note that |H/kxl| 5 pk21 and |kxl| 5 p imply that, |H| 5 pk. Thus, we have produced a subgroup of order pk, which contradicts our assumption that no such subgroup exists. Therefore, we, must have originally had a subgroup of order pk, and we can apply the, induction hypothesis to that subgroup., Let’s be sure we understand exactly what Sylow’s First Theorem, means. Say we have a group G of order 23 ? 32 ? 54 ? 7. Then Sylow’s, First Theorem says that G must have at least one subgroup of each, of the following orders: 2, 4, 8, 3, 9, 5, 25, 125, 625, and 7. On the, other hand, Sylow’s First Theorem tells us nothing about the possible, existence of subgroups of order 6, 10, 15, 30, or any other divisor of, |G| that has two or more distinct prime factors. Because certain subgroups guaranteed by Sylow’s First Theorem play a central role in the, theory of finite groups, they are given a special name., Definition Sylow p-Subgroup, , Let G be a finite group and let p be a prime. If pk divides |G| and pk11, does not divide |G|, then any subgroup of G of order pk is called a, Sylow p-subgroup of G.†, , †Note, , that it follows from Sylow’s First Theorem and the definition that the trivial subgroup {e} is a Sylow p-subgroup of G if and only if p does not divide |G|., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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414, , Special Topics, , So, returning to our group G of order 23 ? 32 ? 54 ? 7, we call any subgroup of order 8 a Sylow 2-subgroup of G, any subgroup of order 625 a Sylow 5-subgroup of G, and so on. Notice that a Sylow, p-subgroup of G is a subgroup whose order is the largest power of, p consistent with Lagrange’s Theorem., Since any subgroup of order p is cyclic, we have the following generalization of Theorem 9.5, first proved by Cauchy in 1845. His proof, ran nine pages!, Corollary Cauchy’s Theorem, Let G be a finite group and let p be a prime that divides the order, of G. Then G has an element of order p., , Sylow’s First Theorem is so fundamental to finite group theory that, many different proofs of it have been published over the years [our proof, is essentially the one given by Georg Frobenius (1849–1917) in 1895]., Likewise, there are scores of generalizations of Sylow’s Theorem., Observe that the corollary to the Fundamental Theorem of Finite, Abelian Groups and Sylow’s First Theorem show that the converse of, Lagrange’s Theorem is true for all finite Abelian groups and all finite, groups of prime-power order., There are two more Sylow theorems that are extremely valuable, tools in finite group theory. But first we introduce a new term., Definition Conjugate Subgroups, , Let H and K be subgroups of a group G. We say that H and K are, conjugate in G if there is an element g in G such that H 5 gKg21., , Recall from Chapter 7 that if G is a finite group of permutations on a, set S and i [ S, then orbG(i) 5 {f(i) | f [ G} and |orbG(i)| divides |G|., Theorem 24.4 Sylow’s Second Theorem, If H is a subgroup of a finite group G and |H| is a power of a prime p,, then H is contained in some Sylow p-subgroup of G., , PROOF Let K be a Sylow p-subgroup of G and let C 5 {K1, K2, . . . , Kn}, with K 5 K1 be the set of all conjugates of K in G. Since conjugation is an, automorphism, each element of C is a Sylow p-subgroup of G. Let SC, denote the group of all permutations of C. For each g [ G, define, fg: C S C by fg(Ki) 5 gKig21. It is easy to show that each fg [ SC., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 415, , Now define a mapping T: G S SC by T(g) 5 fg. Since fgh(Ki) 5, (gh)K i (gh) 21 5 g(hK i h 21 )g 21 5 gf h (K i )g 21 5 f g (f h (K i )) 5, (f gf h)(Ki), we have fgh 5 fgfh, and therefore T is a homomorphism, from G to SC., Next consider T(H), the image of H under T. Since |H| is a power, of p, so is |T(H)| (see property 6 of Theorem 10.2). Thus, by the OrbitStabilizer Theorem (Theorem 7.3), for each i, |orbT(H)(Ki)| divides, |T(H)|, so that |orbT(H)(Ki)| is a power of p. Now we ask: Under what, condition does |orbT(H)(Ki)| 5 1? Well, |orbT(H)(Ki)| 5 1 means that, fg(Ki) 5 gKig21 5 Ki for all g [ H; that is, |orbT(H)(Ki)| 5 1 if and, only if H # N(Ki). But the only elements of N(Ki) that have orders that, are powers of p are those of Ki (see Exercise 13). Thus, |orbT(H)(Ki)| 5 1, if and only if H # Ki., So, to complete the proof, all we need to do is show that for some i,, |orbT(H)(Ki)| 5 1. Analogous to Theorem 24.1, we have |C| 5 |G:N(K)|, (see Exercise 5). And since |G:K| 5 |G:N(K)||N(K):K| is not divisible, by p, neither is |C|. Because the orbits partition C, |C| is the sum of, powers of p. If no orbit has size 1, then p divides each summand and,, therefore, p divides |C|, which is a contradiction. Thus, there is an orbit, of size 1, and the proof is complete., Theorem 24.5 Sylow’s Third Theorem, Let p be a prime and let G be a group of order pkm, where p does not, divide m. Then the number n of Sylow p-subgroups of G is equal to, 1 modulo p and divides m. Furthermore, any two Sylow p-subgroups, of G are conjugate., , PROOF Let K be any Sylow p-subgroup of G and let C 5 {K 1 ,, K 2, . . . , Kn}, with K 5 K1, be the set of all conjugates of K in G. We, first prove that n mod p 5 1., Let S C and T be as in the proof of Theorem 24.4. This time, we consider T(K), the image of K under T. As before, we have, |orbT(K)(Ki)| is a power of p for each i and |orbT(K)(Ki)| 5 1 if and only, if K # Ki. Thus, |orbT(K)(K1)| 5 1 and |orbT(K)(Ki)| is a power of p, greater than 1 for all i 2 1. Since the orbits partition C, it follows that,, modulo p, n 5 |C| 5 1., Next we show that every Sylow p-subgroup of G belongs to C. To do, this, suppose that H is a Sylow p-subgroup of G that is not in C. Let, SC and T be as in the proof of Theorem 24.4, and this time consider, T(H). As in the previous paragraph, |C| is the sum of the orbits’ sizes, under the action of T(H). However, no orbit has size 1, since H is not, in C. Thus, |C| is a sum of terms each divisible by p, so that, modulo p,, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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416, , Special Topics, , n 5 |C| 5 0. This contradiction proves that H belongs to C, and that, n is the number of Sylow p-subgroups of G., Finally, that n divides m follows directly from the fact that n 5, |G:N(K)| (see Exercise 5) and n is relatively prime to p., It is convenient to let np denote the number of Sylow p-subgroups of, a group. Observe that the first portion of Sylow’s Third Theorem is a, counting principle.† As an important consequence of Sylow’s Third, Theorem, we have the following corollary., Corollary A Unique Sylow p-Subgroup Is Normal, A Sylow p-subgroup of a finite group G is a normal subgroup of G if, and only if it is the only Sylow p-subgroup of G., , We illustrate Sylow’s Third Theorem with two examples., EXAMPLE 1 Consider the Sylow 2-subgroups of S3. They are, {(1), (12)}, {(1), (23)}, and {(1), (13)}. According to Sylow’s Third, Theorem, we should be able to obtain the latter two of these from the, first by conjugation. Indeed,, (13){(1), (12)}(13)21 5 {(1), (23)},, (23){(1), (12)}(23)21 5 {(1), (13)}., EXAMPLE 2 Consider the Sylow 3-subgroups of A4. They are {a1, a5,, a9}, {a1, a6, a11}, {a1, a7, a12}, and {a1, a8, a10}. (See Table 5.1.) Then,, a2{a1, a5, a9}a221 5 {a1, a7, a12},, a3{a1, a5, a9}a321 5 {a1, a8, a10},, a4{a1, a5, a9}a421 5 {a1, a6, a11}., Thus, the number of Sylow 3-subgroups is 1 modulo 3, and the four, Sylow 3-subgroups are conjugate., Figure 24.1 shows the subgroup lattices for S3 and A4. We have connected the Sylow p-groups with dashed circles to indicate that they belong to one orbit under conjugation. Notice that the three subgroups of, order 2 in A4 are contained in a Sylow 2-group, as required by Sylow’s, Second Theorem. As it happens, these three subgroups also belong to, one orbit under conjugation, but this is not a consequence of Sylow’s, Third Theorem., † “Whenever, , you can, count.” Sir Francis Galton (1822–1911), The World of, , Mathematics., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 417, , S3, , A3 = <(123)>, , <(12)>, , <(13)>, <(23)>, , <(1)>, A4, <α8 >, , {α 1, α 2, α 3, α 4}, < α5 >, < α 2>, , < α 4>, , < α 7>, < α6 >, , < α3 >, , < α 1>, , Figure 24.1 Lattices of subgroups for S3 and A4., , In contrast to the two preceding examples, observe that the, dihedral group of order 12 has seven subgroups of order 2, but that, conjugating {R0, R180} does not yield any of the other six. (Why?), , Applications of Sylow Theorems, A few numerical examples will make the Sylow theorems come to life., EXAMPLE 3 Say G is a group of order 40. What do the Sylow theorems tell us about G? A great deal! Since 1 is the only divisor of 40 that, is congruent to 1 modulo 5, we know that G has exactly one subgroup of, order 5, and therefore it is normal. Similarly, G has either one or five, subgroups of order 8. If there is only one subgroup of order 8, it is normal. If there are five subgroups of order 8, none is normal and all, five can be obtained by starting with any particular one, say H, and, computing xHx21 for various x’s. Finally, if we let K denote the normal subgroup of order 5 and let H denote any subgroup of order 8, then, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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418, , Special Topics, , G 5 HK. (See Example 5 in Chapter 9.) If H happens to be normal, we, can say even more: G 5 H 3 K., EXAMPLE 4 Consider a group of order 30. By Sylow’s Third, Theorem, it must have either one or six subgroups of order 5 and one or, 10 subgroups of order 3. However, G cannot have both six subgroups of, order 5 and 10 subgroups of order 3 (for then G would have more than, 30 elements). Thus, the subgroup of order 3 is unique or the subgroup, of order 5 is unique (or both are unique) and therefore is normal in G. It, follows, then, that the product of a subgroup of order 3 and one of order, 5 is a group of order 15 that is both cyclic (Exercise 33) and normal, (Exercise 9 in Chapter 9) in G. [This, in turn, implies that both the subgroup of order 3 and the subgroup of order 5 are normal in G (Exercise, 59 in Chapter 9).] So, if we let y be a generator of the cyclic subgroup, of order 15 and let x be an element of order 2 (the existence of which is, guaranteed by Cauchy’s Theorem), we see that, G 5 {x iy j | 0 # i # 1, 0 # j # 14}., EXAMPLE 5 We show that any group G of order 72 must have a, proper, nontrivial normal subgroup. Our arguments are a preview of, those in Chapter 25. By Sylow’s Third Theorem, the number of Sylow, 3-subgroups of G is equal to 1 mod 3 and divides 8. Thus, the number, is 1 or 4. If there is only one, then it is normal by the corollary of, Sylow’s Third Theorem. Otherwise, let H and H9 be two distinct Sylow, 3-subgroups. By Theorem 7.2, we have that |H H9| 5 |H||H9|/|H y H9| 5, 81/|H y H9|. Since |G| 5 72 and |H y H9| is a subgroup of H and, H9, we know that |H y H9| 5 3. By the corollary to Theorem 24.2,, N(H y H9) contains both H and H9. Thus, |N(H y H9)| divides 72, is, divisible by 9, and has at least |H H9|’ 5 27 elements. This leaves only, 36 or 72 for |N(H y H9)|. In the first case, we have from Exercise 9 of, Chapter 9 that N(H y H9) is normal in G. In the second case, we have, by definition that H y H9 is normal in G., Note that in these examples we were able to deduce all of this information from knowing only the order of the group—so many conclusions, from one assumption! This is the beauty of finite group theory., In Chapter 7 we saw that the only group (up to isomorphism) of, prime order p is Zp. As a further illustration of the power of the Sylow, theorems, we next give a sufficient condition that guarantees that a, group of order pq, where p and q are primes, must be Zpq., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 419, , Theorem 24.6 Cyclic Groups of Order pq, If G is a group of order pq, where p and q are primes, p , q,, and p does not divide q 2 1, then G is cyclic. In particular, G is, isomorphic to Zpq., , PROOF Let H be a Sylow p-subgroup of G and let K be a Sylow, q-subgroup of G. Sylow’s Third Theorem states that the number of Sylow, p-subgroups of G is of the form 1 1 kp and divides pq. So 1 1 kp 5 1,, p, q, or pq. From this and the fact that p B q 2 1, it follows that k 5 0, and, therefore H is the only Sylow p-subgroup of G., Similarly, there is only one Sylow q-subgroup of G. Thus, by the, corollary to Theorem 24.5, H and K are normal subgroups of G. Let, H 5 kxl and K 5 kyl. To show that G is cyclic, it suffices to show that x, and y commute, for then |xy| 5 |x||y| 5 pq. But observe that, since H, and K are normal, we have, xyx21y21 5 (xyx21)y21 [ Ky21 5 K, and, xyx21y21 5 x(yx21y21) [ xH 5 H., Thus, xyx21y21 [ K > H 5 {e}, and hence xy 5 yx., Theorem 24.6 demonstrates the power of the Sylow theorems in, classifying the finite groups whose orders have small numbers of prime, factors. Similar results exist for groups of orders p2q, p2q2, p3, and p4,, where p and q are prime., For your amusement, Figure 24.2 lists the number of nonisomorphic, groups with order at most 100. Note in particular the large number of, groups of order 64. Also observe that, generally speaking, it is not the size, of the group that gives rise to a large number of groups of that size but the, number of prime factors involved. In all, there are 1047 nonisomorphic, groups with 100 or fewer elements. Contrast this with the fact that there are, 49,487,365,422 groups of order 1024 5 210. The number of groups of any, order less than 2048 is given at http://oeis.org/A000001/b000001.txt., As a final application of the Sylow theorems, you might enjoy seeing, a determination of the groups of order 99, 66, and 255. In fact, our arguments serve as a good review of much of our work in group theory., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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420, , Special Topics, , Order, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9 10 11 12 13 14 15, , Number, , 1, , 1, , 1, , 2, , 1, , 2, , 1, , 5, , 2, , Order, , 21 22 23, 2, , Number, Order, Number, , 1, , 2, , 2, , 5, , 4, , 44 45 46 47 48, , 1, , 1 52, , 6, , 61 62 63, 1, , Number, , 24 25 26 27 28, 15, , 41 42 43, 1, , Order, , 2, , 2, , 4, , 2, , 2, , 64 65 66 67 68, , 4 267, , 1, , 4, , 1, , 5, , 2, , 1, , 5, , 1, , 2, , 1, , 29 30 31 32 33 34 35, 1, , 4, , 1 51, , 1, , 2, , 1, , 49 50 51 52 53 54 55, 2, , 5, , 1, , 5, , 1 15, , 2, , 69 70 71 72 73 74 75, 1, , 4, , 1 50, , 1, , 2, , 3, , Order, , 81 82 83, , 84 85 86 87 88 89 90 91 92 93 94 95, , Number, , 15, , 15, , 2, , 1, , 1, , 2, , 1 12, , 1 10, , 1, , 4, , 2, , 2, , 16 17 18 19, 14, , 1, , 5, , 20, , 1, , 5, , 36 37 38 39, , 40, , 14, , 2, , 14, , 56 57 58 59, , 60, , 1, , 2, , 1, , 13, , 76 77 78 79, , 80, , 13, , 4, , 2, , 1, , 2, , 6, , 1, , 52, , 96 97 98 99 100, , 1 230, , 1, , 5, , 2, , 16, , Figure 24.2 The number of groups of a given order up to 100., , EXAMPLE 6 Determination of the Groups of Order 99, Suppose that G is a group of order 99. Let H be a Sylow 3-subgroup, of G and let K be a Sylow 11-subgroup of G. Since 1 is the only positive divisor of 99 that is equal to 1 modulo 11, we know from Sylow’s, Third Theorem and its corollary that K is normal in G. Similarly,, H is normal in G. It follows, by the argument used in the proof of, Theorem 24.6, that elements from H and K commute, and therefore, G 5 H 3 K. Since both H and K are Abelian, G is also Abelian. Thus,, G is isomorphic to Z99 or Z3 % Z33., EXAMPLE 7 Determination of the Groups of Order 66, Suppose that G is a group of order 66. Let H be a Sylow 3-subgroup of, G and let K be a Sylow 11-subgroup of G. Since 1 is the only positive, divisor of 66 that is equal to 1 modulo 11, we know that K is normal in, G. Thus, HK is a subgroup of G of order 33 (see Example 5 in Chapter 9, and Theorem 7.2). Since any group of order 33 is cyclic (Theorem, 24.6), we may write HK 5 kxl. Next, let y [ G and |y| 5 2. Since kxl, has index 2 in G, we know it is normal. So yxy21 5 xi for some i from, 1 to 32. Then, yx 5 xiy and, since every member of G is of the form, xsyt, the structure of G is completely determined by the value of i. We, claim that there are only four possibilities for i. To prove this, observe, that |xi| 5 |x| (Exercise 5, Supplementary Exercises for Chapters 1–4)., Thus, i and 33 are relatively prime. But also, since y has order 2,, 2, , x 5 y21( yxy21)y 5 y21xiy 5 yxiy21 5 (yxy21)i 5 (xi)i 5 x i ., 2, , So xi 21 5 e and therefore 33 divides i2 2 1. From this it follows that, 11 divides i 6 1, and therefore i 5 0 6 1, i 5 11 6 1, i 5 22 6 1, or, i 5 33 6 1. Putting this together with the other information we have, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 421, , about i, we see that i 5 1, 10, 23, or 32. This proves that there are at, most four groups of order 66., To prove that there are exactly four such groups, we simply observe, that Z66, D33, D11 % Z3, and D3 % Z11 each has order 66 and that no two, are isomorphic. For example, D11 % Z3 has 11 elements of order 2,, whereas D3 % Z11 has only three elements of order 2. (See Exercises, 27–30 of the Supplementary Exercises for Chapters 5–8.), EXAMPLE 8 The Only Group of Order 255 is Z255, Let G be a group of order 255 5 3 ? 5 ? 17, and let H be a Sylow 17-subgroup of G. By Sylow’s Third Theorem, H is the only Sylow 17-subgroup, of G, so N(H) 5 G. By Example 15 in Chapter 10, |N(H)/C(H)| divides, |Aut(H)| 5 |Aut(Z17)|. By Theorem 6.5, |Aut(Z17)| 5 |U(17)| 5 16. Since, |N(H)/C(H)| must divide 255 and 16, we have |N(H)/C(H)| 5 1. Thus,, C(H) 5 G. This means that every element of G commutes with every element of H, and, therefore, H # Z(G). Thus, 17 divides |Z(G)|, which in, turn divides 255. So |Z(G)| is equal to 17, 51, 85, or 255 and |G/Z(G)| is, equal to 15, 5, 3, or 1. But the only groups of order 15, 5, 3, or 1 are the, cyclic ones, so we know that G/Z(G) is cyclic. Now the G/Z Theorem, (Theorem 9.3) shows that G is Abelian, and the Fundamental Theorem of, Finite Abelian Groups tells us that G is cyclic., , Exercises, I have always grown from my problems and challenges, from the things that, don’t work out. That’s when I’ve really learned., carol burnett, , 1. Show that conjugacy is an equivalence relation on a group., 2. Calculate all conjugacy classes for the quaternions (see Exercise 4,, Supplementary Exercises for Chapters 1– 4)., 3. Show that the function T defined in the proof of Theorem 24.1 is, well-defined, is one-to-one, and maps the set of left cosets onto the, conjugacy class of a., 4. Show that cl(a) 5 {a} if and only if a [ Z(G)., 5. Let H be a subgroup of a group G. Prove that the number of conjugates of H in G is |G:N(H)|. (This exercise is referred to in this, chapter.), 6. Let H be a proper subgroup of a finite group G. Show that G is not, the union of all conjugates of H., 7. If G is a group of odd order and x [ G, show that x21 is not in cl(x)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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422, , Special Topics, , 8. Determine the class equation for non-Abelian groups of orders 39, and 55., 9. Determine which of the equations below could be the class equation given in the proof of Theorem 24.2. For each part, provide, your reasoning., a. 9 5 3 1 3 1 3, b. 21 5 1 1 1 1 3 1 3 1 3 1 3 1 7, c. 10 5 1 1 2 1 2 1 5, d. 18 5 1 1 3 1 6 1 8, 10. Exhibit a Sylow 2-subgroup of S4. Describe an isomorphism from, this group to D4., 11. Suppose that G is a group of order 48. Show that the intersection, of any two distinct Sylow 2-subgroups of G has order 8., 12. Find all the Sylow 3-subgroups of S4., 13. Let K be a Sylow p-subgroup of a finite group G. Prove that if x [, N(K) and the order of x is a power of p, then x [ K. (This exercise, is referred to in this chapter.), 14. Suppose that G is a group of order pnm, where p is prime and p does, not divide m. Show that the number of Sylow p-subgroups divides m., 15. Suppose that G is a group and UGU 5 pnm, where p is prime and, p 7 m. Prove that a Sylow p-subgroup of G must be normal in G., 16. Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow, p-subgroup of G contained in N(H)., 17. Suppose that G is a group of order 168. If G has more than one, Sylow 7-subgroup, exactly how many does it have?, 18. Show that every group of order 56 has a proper nontrivial normal, subgroup., 19. What is the smallest composite (that is, nonprime and greater than 1), integer n such that there is a unique group of order n?, 20. Let G be a noncyclic group of order 21. How many Sylow 3subgroups does G have?, 21. Prove that a noncyclic group of order 21 must have 14 elements of, order 3., 22. How many Sylow 5-subgroups of S5 are there? Exhibit two., 23. How many Sylow 3-subgroups of S5 are there? Exhibit five., 24. What are the possibilities for the number of elements of order 5 in, a group of order 100?, 25. What do the Sylow theorems tell you about any group of order 100?, 26. Prove that a group of order 175 is Abelian., 27. Let G be a group with UGU 5 pnm, where p is a prime that does not divide m and p $ m. Prove that the Sylow p-subgroup of G is normal., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 423, , 28. Determine the number of Sylow 2-subgroups of D2m, where m is an, odd integer at least 3., 29. Let K be a Sylow 2-subgroup of D2m, where m is an odd integer at, least 3. Prove that N(K) 5 K., 30. Generalize the argument given in Example 6 to obtain a theorem, about groups of order p2q, where p and q are distinct primes., 31. What is the smallest possible odd integer that can be the order of a, non-Abelian group?, 32. Prove that a group of order 375 has a subgroup of order 15., 33. Without using Theorem 24.6, prove that a group of order 15 is, cyclic. (This exercise is referred to in the discussion about groups, of order 30.), 34. Prove that a group of order 105 contains a subgroup of order 35., 35. Prove that a group of order 595 has a normal Sylow 17-subgroup., 36. Let G be a group of order 60. Show that G has exactly four elements of order 5 or exactly 24 elements of order 5. Which of these, cases holds for A5?, 37. Show that the center of a group of order 60 cannot have order 4., 38. Suppose that G is a group of order 60 and G has a normal subgroup N of order 2. Show that, a. G has normal subgroups of orders 6, 10, and 30., b. G has subgroups of orders 12 and 20., c. G has a cyclic subgroup of order 30., 39. Let G be a group of order 60. If the Sylow 3-subgroup is normal,, show that the Sylow 5-subgroup is normal., 40. Show that if G is a group of order 168 that has a normal subgroup, of order 4, then G has a normal subgroup of order 28., 41. Suppose that p is prime and |G| 5 pn. Show that G has normal subgroups of order pk for all k between 1 and n (inclusive)., 42. Suppose that G is a group of order pn, where p is prime, and G has, exactly one subgroup for each divisor of pn. Show that G is cyclic., 43. Suppose that p is prime and |G| 5 pn. If H is a proper subgroup of G,, prove that N(H) . H. (This exercise is referred to in Chapter 25.), 44. If H is a finite subgroup of a group G and x [ G, prove that, |N(H)| 5 |N(xHx21)|., 45. Let H be a Sylow 3-subgroup of a finite group G and let K be a, Sylow 5-subgroup of G. If 3 divides |N(K)|, prove that 5 divides, |N(H)|., 46. If H is a normal subgroup of a finite group G and |H| 5 pk, for some prime p, show that H is contained in every Sylow, p-subgroup of G., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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424, , Special Topics, , 47. Suppose that G is a finite group and G has a unique Sylow p-subgroup, for each prime p. Prove that G is the internal direct product of its, nontrivial Sylow p-subgroups. If each Sylow p-subgroup is cyclic, is, G cyclic? If each Sylow p-subgroup is Abelian, is G Abelian?, 48. If Gp is a Sylow p-subgroup of a group G and Hp is a Sylow psubgroup of a group H, prove that Gp % Hp is a Sylow p-subgroup, of G % H., 49. Let G be a finite group and let H be a normal Sylow p-subgroup, of G. Show that a(H) 5 H for all automorphisms a of G., 50. If H is a Sylow p-subgroup of a group, prove that N(N(H)) = N(H)., 51. Let p be a prime and H and K be Sylow p-subgroups of a group G., Prove that |N(H)| 5 |N(K)|., 52. Let G be a group of order p2q2, where p and q are distinct primes,, q B p2 2 1, and p B q2 2 1. Prove that G is Abelian. List three pairs, of primes that satisfy these conditions., 53. Let H be a normal subgroup of a group G. Show that H is the union, of the conjugacy classes in G of the elements of H. Is this true, when H is not normal in G?, 54. Let p be prime. If the order of every element of a finite group G is, a power of p, prove that |G| is a power of p., 55. For each prime p, prove that all Sylow p-subgroups of a finite, group are isomorphic., 56. Suppose that K is a normal subgroup of a finite group G and S, is a Sylow p-subgroup of G. Prove that K > S is a Sylow psubgroup of K., 57. Show that a group of order 12 cannot have nine elements of order 2., 58. If |G| 5 36 and G is non-Abelian, prove that G has more than one, Sylow 2-subgroup or more than one Sylow 3-subgroup., 59. Suppose G is a finite group and p is a prime that divides |G|. Let n, denote the number of elements of G that have order p. If the Sylow, p-subgroup of G is normal, prove that p divides n 1 1., 60. Determine the groups of order 45., 61. Show that there are at most three nonisomorphic groups of order 21., 62. Prove that if H is a normal subgroup of index p2 where p is prime,, then G9 # H (see Exercise 3 in the Supplementary Exercises for, Chapters 5–8 for a description of G9)., 63. Show that Z2 is the only group that has exactly two conjugacy, classes., 64. What can you say about the number of elements of order 7 in a, group of order 168 5 8 ? 3 ? 7?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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24 | Sylow Theorems, , 425, , 65. Explain why a group of order 4m where m is odd must have a subgroup isomorphic to Z4 or Z2 % Z2 but cannot have both a subgroup, isomorphic to Z4 and a subgroup isomorphic to Z2 % Z2. Show that, S4 has a subgroup isomorphic to Z4 and a subgroup isomorphic to, Z2 % Z2., 66. Let p be the smallest prime that divides the order of a finite group, G. If H is a Sylow p-subgroup of G and is cyclic, prove that N(H) 5, C(H)., 67. Let G be a group of order 715 5 5 ? 11 ? 13. Let H be a Sylow, 13-subgroup of G and K be a Sylow 11-subgroup of G. Prove that, H is contained in Z(G). Can the argument you used to prove that H, is contained in Z(G) also be used to show that K is contained in, Z(G)?, 68. Let G be a group of order 1925 5 52 ? 7 ? 11 and H be a subgroup of, order 7. Prove that |C(H)| is divisible by 385. What can you say, about Z(G) if the Sylow 5-subgroup is not cyclic?, 69. Let G be a group with UGU 5 595 5 5 ? 7 ? 17. Show that the Sylow, 5-subgroup of G is normal in G and is contained in Z(G)., 70. What is the probability that a randomly selected element from D4, commutes with the vertical reflection V?, 71. Prove that if x and y are in the same conjugacy class of a group,, then |C(x)| 5 |C(y)|. (This exercise is referred to in the discussion, on the probability that two elements from a group commute.), 72. Let G be a finite group and let a [ G. Express the probability that, a randomly selected element from G commutes with a in terms of, orders of subgroups of G., 73. Find Pr(D4), Pr(S3), and Pr(A4)., 74. Prove that Pr(Dn) 5 (n 1 3)/4n if n is odd and Pr(Dn) 5 n (n +, 6)/4n if n is even., 75. Prove that Pr(G % H) 5 Pr(G) ? Pr(H)., 76. Let R be a finite noncommutative ring. Show that the probability, that two randomly chosen elements from R commute is at most 5/8., [Hint: Mimic the group case and use the fact that the additive, group R/C(R) is not cyclic.], , Computer Exercises, Software for the computer exercises in this chapter is available at the, website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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426, , Special Topics, , References, 1. W. H. Gustafson, “What Is the Probability That Two Group Elements, Commute?” The American Mathematical Monthly 80 (1973):, 1031–1034., 2. Desmond MacHale, “How Commutative Can a Non-Commutative Group, Be?” The Mathematical Gazette 58 (1974): 199–202., , Suggested Readings, C. Clifton, D. Guichard, and P. Keef, “How Commutative Are Direct, Products of Dihedral Groups?” Mathematics Magazine 84 (2011): 137–140., The authors show that for any positive integer m, there is a group G, which is a direct product of dihedral groups such that Pr(G) 5 1/m., They also show that for any positive integer m, there is a dihedral, group G such that Pr(G) 5 m/m9, where m9 is relatively prime to m., J. A. Gallian and D. Moulton, “When Is Zn the Only Group of Order n?”, Elemente der Mathematik 48 (1993): 118–120., It is shown that Zn is the only group of order n if and only if n and f(n), are relatively prime. The article can be downloaded at http://www, .d.umn.edu/~jgallian/pq.pdf., W. H. Gustafson, “What Is the Probability That Two Group Elements, Commute?” The American Mathematical Monthly 80 (1973): 1031–1034., This paper is concerned with the problem posed in the title. It is shown, that for all finite non-Abelian groups and certain infinite non-Abelian, groups, the probability that two elements from a group commute is at, most 5/8. The paper concludes with several exercises., Desmond MacHale, “Commutativity in Finite Rings,” The American, Mathematical Monthly 83 (1976): 30–32., In this easy-to-read paper, it is shown that the probability that two, elements from a finite noncommutative ring commute is at most 5/8., Several properties of Pr(G) when G is a finite group are stated. For, example, if H # G, then Pr(G) # Pr(H). Also, there is no group G such, that 7/16 , Pr(G) , 1/2., T. Langley, D. Levitt, and J. Rower, “Two Generalizations of the 5/8, Bound on Commutativity in Nonabelian Finite Groups,” Mathematics, Magazine 84 (2011):128–136., This article provides a generalization of the notion of the probability of, two group elements commuting. It investigates the probability that for, a fixed positive integer n and n randomly chosen elements a1, a2, ), an, from a group, it is true that a1a2 ) an 5 an ) a2a1., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Ludwig Sylow, , Matematisk Unistitutt/Universitete I Oslo, , Sylow’s Theorem is 100 years old. In the, course of a century this remarkable theorem has been the basis for the construction of numerous theories., l. a. shemetkov, , Ludwig Sylow (pronounced “SEE-loe”), was born on December 12, 1832, in Christiania (now Oslo), Norway. While a student, at Christiania University, Sylow won a gold, medal for competitive problem solving. In, 1855, he became a high school teacher; despite the long hours required by his teaching, duties, Sylow found time to study the papers, of Abel. During the school year 1862–1863,, Sylow received a temporary appointment at, Christiania University and gave lectures, on Galois theory and permutation groups., Among his students that year was the great, mathematician Sophus Lie (pronounced, “Lee”), after whom Lie algebras and Lie, groups are named. From 1873 to 1881,, Sylow, with some help from Lie, prepared a, new edition of Abel’s works. In 1902, Sylow, and Elling Holst published Abel’s correspondence., , Sylow’s spectacular theorems came in, 1872. Upon learning of Sylow’s discovery,, C. Jordan called it “one of the essential, points in the theory of permutations.” The, results took on greater importance when the, theory of abstract groups flowered in the late, 19th century and early 20th century., In 1869, Sylow was offered a professorship at Christiania University but turned it, down. Upon Sylow’s retirement from high, school teaching at age 65, Lie mounted a, successful campaign to establish a chair for, Sylow at Christiania University. Sylow held, this position until his death on September 7,, 1918., To find more information about Sylow,, visit:, http://www-groups.dcs.st-and, .ac.uk/~history, , 427, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 Finite Simple Groups, , It is a widely held opinion that the problem of classifying finite simple, groups is close to a complete solution. This will certainly be one of the great, achievements of mathematics of this century., nathan jacobson, , Historical Background, We now come to the El Dorado of finite group theory—the simple, groups. Simple group theory is a vast and difficult subject; we call it, the El Dorado of group theory because of the enormous effort put forth, by hundreds of mathematicians over many years to discover and, classify all finite simple groups. Let’s begin our discussion with the, definition of a simple group and some historical background., Definition Simple Group, , A group is simple if its only normal subgroups are the identity, subgroup and the group itself., , The notion of a simple group was introduced by Galois about 180 years, ago. The simplicity of A5, the group of even permutations on five, symbols, played a crucial role in his proof that there is not a solution by, radicals of the general fifth-degree polynomial (that is, there is no “quintic, formula”). But what makes simple groups important in the theory of, groups? They are important because they play a role in group theory, somewhat analogous to that of primes in number theory or the elements, in chemistry; that is, they serve as the building blocks for all groups., These building blocks may be determined in the following way. Given a, finite group G, choose a proper normal subgroup G1 of G 5 G0 of largest, order. Then the factor group G0/G1 is simple, and we next choose a proper, normal subgroup G2 of G1 of largest order. Then G1/G2 is also simple, and, we continue in this fashion until we arrive at Gn 5 {e}. The simple groups, G0 /G1, G1/G2, . . . , Gn21/Gn are called the composition factors of, G. More than 100 years ago, Jordan and Hölder proved that these, 428, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 429, , factors are independent of the choices of the normal subgroups made in, the process described. In a certain sense, a group can be reconstructed, from its composition factors, and many of the properties of a group are, determined by the nature of its composition factors. This and the fact that, many questions about finite groups can be reduced (by induction) to questions about simple groups make clear the importance of determining all, finite simple groups., Just which groups are the simple ones? The Abelian simple groups, are precisely Zn, where n 5 1 or n is prime. This follows directly from the, corollary in Chapter 11. In contrast, it is extremely difficult to describe the, non-Abelian simple groups. The best we can do here is to give a few, examples and mention a few words about their discovery. It was Galois in, 1831 who first observed that An is simple for all n $ 5. The next, discoveries were made by Jordan in 1870, when he found four infinite, families of simple matrix groups over the field Zp, where p is prime. One, such family is the factor group SL(n, Zp)/Z(SL(n, Zp)), except when n 5 2, and p 5 2 or p 5 3. Between the years 1892 and 1905, the American, mathematician Leonard Dickson (see Chapter 22 for a biography) generalized Jordan’s results to arbitrary finite fields and discovered several new, infinite families of simple groups. About the same time, it was shown by, G. A. Miller and F. N. Cole that a family of five groups first described by, E. Mathieu in 1861 were in fact simple groups. Since these five groups, were constructed by ad hoc methods that did not yield infinitely many, possibilities, like An or the matrix groups over finite fields, they were, called “sporadic.”, The next important discoveries came in the 1950s. In that decade,, many new infinite families of simple groups were found, and the initial, steps down the long and winding road that led to the complete classification of all finite simple groups were taken. The first step was Richard, Brauer’s observation that the centralizer of an element of order 2 was an, important tool for studying simple groups. A few years later, John, Thompson, in his Ph.D. dissertation, introduced the crucial idea of, studying the normalizers of various subgroups of prime-power order., In the early 1960s came the momentous Feit–Thompson Theorem,, which says that a non-Abelian simple group must have even order. This, property was first conjectured around 1900 by one of the pioneers of, modern group theoretic methods, the Englishman William Burnside, (see Chapter 29 for a biography). The proof of the Feit–Thompson, Theorem filled an entire issue of a journal [1], 255 pages in all (see, Figure 25.1). This result provided the impetus to classify the finite simple groups—that is, a program to discover all finite simple groups and, prove that there are no more to be found. Throughout the 1960s, the, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , With permission of the Pacific Journal of Mathematics, , 430, , Oh, what are the orders of all simple, groups?, I speak of the honest ones, not of the loops., It seems that old Burnside their orders has, guessed, Except for the cyclic ones, even the rest., CHORUS: Finding all groups that are simple is no simple task., Groups made up with permutes will, produce some more:, For An is simple, if n exceeds 4., Then, there was Sir Matthew who came into, view, Exhibiting groups of an order quite new., Still others have come on to study this thing., Of Artin and Chevalley now we shall sing., With matrices finite they made quite a list, The question is: Could there be others, they’ve missed?, Suzuki and Ree then maintained it’s the, case, , That these methods had not reached the end, of the chase., They wrote down some matrices, just four by, four., That made up a simple group. Why not make, more?, And then came the opus of Thompson and, Feit, Which shed on the problem remarkable light., A group, when the order won’t factor by two,, Is cyclic or solvable. That’s what is true., Suzuki and Ree had caused eyebrows to raise,, But the theoreticians they just couldn’t faze., Their groups were not new: if you added a, twist,, You could get them from old ones with a, flick of the wrist., Still, some hardy souls felt a thorn in their, side., , Figure 25.1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , For the five groups of Mathieu all reason, defied;, Not An, not twisted, and not Chevalley,, They called them sporadic and filed them, away., Are Mathieu groups creatures of heaven or, hell?, Zvonimir Janko determined to tell., He found out [a new sporadic simple group], that nobody wanted to know:, The masters had missed 1 7 5 5 6 0., , 431, , (And twelve or more sprouted, to greet the, new age.), By Janko and Conway and Fischer and Held,, McLaughlin, Suzuki, and Higman, and Sims., No doubt you noted the last lines don’t, rhyme., Well, that is, quite simply, a sign of the time., There’s chaos, not order, among simple, groups;, And maybe we’d better go back to the loops., , The floodgates were opened! New groups, were the rage!, , methods introduced in the Feit–Thompson proof were generalized and, improved with great success by many mathematicians. Moreover, between 1966 and 1975, 19 new sporadic simple groups were discovered., Despite many spectacular achievements, research in simple group theory in the 1960s was haphazard, and the decade ended with many people believing that the classification would never be completed. (The, pessimists feared that the sporadic simple groups would foil all attempts. The anonymously written “song” in Figure 25.1 captures the, spirit of the times.) Others, more optimistic, were predicting that it, would be accomplished in the 1990s., The 1970s began with Thompson receiving the Fields Medal for his, fundamental contributions to simple group theory. This honor is among, the highest forms of recognition that a mathematician can receive, (more information about the Fields Medal is given near the end of this, chapter). Within a few years, three major events took place that ultimately led to the classification. First, Thompson published what is regarded as the single most important paper in simple group theory—the, N-group paper. Here, Thompson introduced many fundamental techniques and supplied a model for the classification of a broad family of, simple groups. Second, Daniel Gorenstein produced an elaborate outline for the classification, which he delivered in a series of lectures at, the University of Chicago in 1972. Here a program for the overall, proof was laid out. The army of researchers now had a battle plan and, a commander-in-chief. But this army still needed more and better, weapons. Thus came the third critical development: the involvement of, Michael Aschbacher. In a dazzling series of papers, Aschbacher combined his own insight with the methods of Thompson, which had been, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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432, , Special Topics, , generalized throughout the 1960s, and a geometric approach pioneered, by Bernd Fischer to achieve one brilliant result after another in rapid, succession. In fact, so much progress was made by Aschbacher, and others that by 1976, it was clear to nearly everyone involved that, enough techniques had been developed to complete the classification., Only details remained., The 1980s were ushered in with Aschbacher following in the footsteps of Feit and Thompson by winning the American Mathematical, Society’s Cole Prize in algebra (see the last section of this chapter)., A week later, Robert L. Griess made the spectacular announcement, that he had constructed the “Monster.”† The Monster is the largest of the, sporadic simple groups. In fact, it has vastly more elements than there, are atoms on the earth! Its order is, 808,017,424,794,512,875,886,459,904,961,710,757,005,754,, 368,000,000,000, (hence, the name). This is approximately 8 3 1053. The Monster is a, group of rotations in 196,883 dimensions. Thus, each element can be, expressed as a 196,883 3 196,883 matrix., At the annual meeting of the American Mathematical Society in 1981,, Gorenstein announced that the “Twenty-Five Years’ War” to classify all the, finite simple groups was over. Group theorists at long last had a list of all, finite simple groups and a proof that the list was complete. The proof was, spread out over hundreds of papers—both published and unpublished—, and ran more than 10,000 pages in length. Because of the proof’s extreme, length and complexity, and the fact that some key parts of it had not been, published, there was some concern in the mathematics community that the, classification was not a certainty. By the end of the decade, group theorists, had concluded that there was indeed a gap in the unpublished work that, would be difficult to rectify. In the mid-1990s, Aschbacher and Stephen, Smith began work on this problem. In 2004, at the annual meeting of the, American Mathematical Society, Aschbacher announced that he and Smith, had completed the classification. Their monograph is over 1200 pages, in length. Ronald Solomon, writing in Mathematical Reviews, called it, “an amazing tour de force” and a “major milestone in the history of finite group theory.” Aschbacher concluded his remarks by saying that, he would not bet his house that the proof is now error free., Several people who played a central role in the classification are working on a “second generation” proof that will be much shorter and more, comprehensible., †The, , name was coined by John H. Conway. Griess called the group the “Friendly Giant.” In 2010 the American Mathematical Society awarded Griess the Leroy P. Steele, Seminal Contribution to Research Prize for his construction of the Monster., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 433, , Nonsimplicity Tests, In view of the fact that simple groups are the building blocks for all, groups, it is surprising how scarce the non-Abelian simple groups are., For example, A5 is the only one whose order is less than 168; there are, only five non-Abelian simple groups of order less than 1000 and only, 56 of order less than 1,000,000. In this section, we give a few theorems, that are useful in proving that a particular integer is not the order of a, non-Abelian simple group. Our first such result is an easy arithmetic, test that comes from combining Sylow’s Third Theorem and the fact, that groups of prime-power order have nontrivial centers., Theorem 25.1 Sylow Test for Nonsimplicity, Let n be a positive integer that is not prime, and let p be a prime, divisor of n. If 1 is the only divisor of n that is equal to 1 modulo p,, then there does not exist a simple group of order n., , PROOF If n is a prime-power, then a group of order n has a nontrivial, center and, therefore, is not simple. If n is not a prime-power, then, every Sylow subgroup is proper, and, by Sylow’s Third Theorem, we, know that the number of Sylow p-subgroups of a group of order n is, equal to 1 modulo p and divides n. Since 1 is the only such number, the, Sylow p-subgroup is unique, and therefore, by the corollary to Sylow’s, Third Theorem, it is normal., How good is this test? Well, applying this criterion to all the, nonprime integers between 1 and 200 would leave only the following, integers as possible orders of finite non-Abelian simple groups: 12, 24,, 30, 36, 48, 56, 60, 72, 80, 90, 96, 105, 108, 112, 120, 132, 144, 150,, 160, 168, 180, and 192. (In fact, computer experiments have revealed, that for large intervals, say, 500 or more, this test eliminates more than, 90% of the nonprime integers as possible orders of simple groups. See [2], for more on this.), Our next test rules out 30, 90, and 150., Theorem 25.2 2 ? Odd Test, An integer of the form 2 ? n, where n is an odd number greater than 1,, is not the order of a simple group., , PROOF Let G be a group of order 2n, where n is odd and greater, than 1. Recall from the proof of Cayley’s Theorem (Theorem 6.1), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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434, , Special Topics, , that the mapping g S Tg is an isomorphism from G to a permutation, group on the elements of G [where Tg(x) 5 gx for all x in G]. Since, |G| 5 2n, Cauchy’s Theorem guarantees that there is an element g in, G of order 2. Then, when the permutation Tg is written in disjoint, cycle form, each cycle must have length 1 or 2; otherwise, |g| 2 2., But Tg can contain no 1-cycles, because the 1-cycle (x) would mean x 5, Tg(x) 5 gx, so g 5 e. Thus, in cycle form, Tg consists of exactly n, transpositions, where n is odd. Therefore, Tg is an odd permutation., This means that the set of even permutations in the image of G is a, normal subgroup of index 2. (See Exercise 23 in Chapter 5 and Exercise 9, in Chapter 9.) Hence, G is not simple., The next theorem is a broad generalization of Cayley’s Theorem., We will make heavy use of its two corollaries., Theorem 25.3 Generalized Cayley Theorem, Let G be a group and let H be a subgroup of G. Let S be the group, of all permutations of the left cosets of H in G. Then there is a, homomorphism from G into S whose kernel lies in H and contains, every normal subgroup of G that is contained in H., , PROOF For each g [ G, define a permutation Tg of the left cosets, of H by Tg(xH) 5 gxH. As in the proof of Cayley’s Theorem, it is easy to, verify that the mapping of a: g S Tg is a homomorphism from G into S., Now, if g [ Ker a, then Tg is the identity map, so H 5 Tg(H) 5 gH,, and, therefore, g belongs to H. Thus, Ker a # H. On the other hand, if, K is normal in G and K # H, then for any k [ K and any x in G, there, is an element k9 in K such that kx 5 xk9. Thus,, Tk(xH) 5 kxH 5 xk9H 5 xH, and, therefore, Tk is the identity permutation. This means that k [ Ker a., We have proved, then, that every normal subgroup of G contained in H, is also contained in Ker a., As a consequence of Theorem 25.3, we obtain the following very, powerful arithmetic test for nonsimplicity., Corollary 1 Index Theorem, If G is a finite group and H is a proper subgroup of G such that |G|, does not divide |G:H|!, then H contains a nontrivial normal subgroup, of G. In particular, G is not simple., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 435, , PROOF Let a be the homomorphism given in Theorem 25.3. Then, Ker a is a normal subgroup of G contained in H, and G/Ker a is, isomorphic to a subgroup of S. Thus, |G/Ker a| 5 |G|/|Ker a| divides, |S| 5 |G:H|!. Since |G| does not divide |G:H|!, the order of Ker a must, be greater than 1., Corollary 2 Embedding Theorem, If a finite non-Abelian simple group G has a subgroup of index n,, then G is isomorphic to a subgroup of An., , PROOF Let H be the subgroup of index n, and let Sn be the group, of all permutations of the n left cosets of H in G. By the Generalized, Cayley Theorem, there is a nontrivial homomorphism from G into Sn., Since G is simple and the kernel of a homomorphism is a normal subgroup of G, we see that the mapping from G into Sn is one-to-one, so, that G is isomorphic to some subgroup of Sn. Recall from Exercise 23, in Chapter 5 that any subgroup of Sn consists of even permutations only, or half even and half odd. If G were isomorphic to a subgroup of the, latter type, the even permutations would be a normal subgroup of index 2 (see Exercise 9 in Chapter 9), which would contradict the fact, that G is simple. Thus, G is isomorphic to a subgroup of An., Using the Index Theorem with the largest Sylow subgroup for H, reduces our list of possible orders of non-Abelian simple groups still, further. For example, let G be any group of order 80 5 16 ? 5. We may, choose H to be a subgroup of order 16. Since 80 is not a divisor of 5!,, there is no simple group of order 80. The same argument applies to 12,, 24, 36, 48, 96, 108, 160, and 192, leaving only 56, 60, 72, 105, 112,, 120, 132, 144, 168, and 180 as possible orders of non-Abelian simple, groups up to 200. Let’s consider these orders. Quite often we may use, a counting argument to eliminate an integer. Consider 56. By Sylow’s, Third Theorem, we know that a simple group of order 56 5 8 ? 7 would, contain eight Sylow 7-subgroups and seven Sylow 2-subgroups. Now,, any two Sylow p-subgroups that have order p must intersect in only the, identity. So the union of the eight Sylow 7-subgroups yields 48 elements of order 7, and the union of any two Sylow 2-subgroups gives at, least 8 1 8 2 4 5 12 new elements. But there are only 56 elements in, all. This contradiction shows that there is not a simple group of order 56., An analogous argument also eliminates the integers 105 and 132., So, our list of possible orders of non-Abelian simple groups up to, 200 is down to 60, 72, 112, 120, 144, 168, and 180. Of these, 60 and, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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436, , Special Topics, , 168 do correspond to simple groups. The others can be eliminated with, a bit of razzle-dazzle., The easiest case to handle is 112 5 24 ? 7. Suppose there were a simple group G of order 112. A Sylow 2-subgroup of G must have index 7., So, by the Embedding Theorem, G is isomorphic to a subgroup of A7., But 112 does not divide |A7|, which is a contradiction., Another easy case is 72. This case was done in Example 5 in, Chapter 24 but we eliminate it using the Index Theorem. Recall from, Exercise 5 in Chapter 24 that if we denote the number of Sylow, p-subgroups of a group G by np, then np 5 |G:N(H)|, where H is any, Sylow p-subgroup of G, and np mod p 5 1. It follows, then, that in a, simple group of order 72, we have n3 = 4, which is impossible, since 72, does not divide 4!., Next consider the possibility of a simple group G of order 144 5 9 ? 16., By the Sylow theorems, we know that n3 5 4 or 16 and n2 $ 3. The Index, Theorem rules out the case where n3 5 4, so we know that there are, 16 Sylow 3-subgroups. Now, if every pair of Sylow 3-subgroups had, only the identity in common, a straightforward counting argument would, produce more than 144 elements. So, let H and H9 be a pair of Sylow, 3-subgroups whose intersection has order 3. Then H > H9 is a subgroup, of both H and H9 and, by the corollary to Theorem 24.2 (or by Exercise 43, in Chapter 24), we see that N(H > H9) must contain both H and H9 and,, therefore, the set HH9. (HH9 need not be a subgroup.) Thus,, |N(H > H9)| $ |HH9| 5, , 0 H 0 0 H 0, 9?9, , 5 27., 0 H >H 0, 3, , Now, we have three arithmetic conditions on k 5 |N(H > H9)|. We, know that 9 divides k; k divides 144; and k $ 27. Clearly, then, k $ 36,, and so |G : N(H > H9)| # 4. The Index Theorem now gives us the desired contradiction., Finally, suppose that G is a non-Abelian simple group of order 180 5, 2, 2 ? 32 ? 5. Then n5 5 6 or 36 and n3 5 10 (n3 5 4 is ruled out by the, Index Theorem). First, assume that n5 5 36. Then G has 36 ? 4 5 144, elements of order 5. Now, if each pair of the Sylow 3-subgroups intersects in only the identity, then there are 80 more elements in the group,, which is a contradiction. So, we may assume that there are two Sylow, 3-subgroups L3 and L 39 whose intersection has order 3. Then, as was the, case for order 144, we have, |N(L3 > L 39)| $ |L3L 39| 5, , 9.9, 5 27., 3, , Thus,, |N(L3 > L 39)| 5 9 ? k,, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 437, , where k $ 3 and k divides 20. Clearly, then,, |N(L3 > L 39)| $ 36, and therefore, |G: N(L3 > L 39)| # 5., The Index Theorem now gives us another contradiction. Hence, we, may assume that n5 5 6. In this case, we let H be the normalizer of a, Sylow 5-subgroup of G. By Sylow’s Third Theorem, we have 6 5, |G:H|, so that |H| 5 30. In Chapter 24, we proved that every group of, order 30 has an element of order 15. On the other hand, since n5 5 6,, G has a subgroup of index 6 and the Embedding Theorem tells us that, G is isomorphic to a subgroup of A6. But A6 has no element of order 15., (See Exercise 9 in Chapter 5.), Unfortunately, the argument for 120 is fairly long and complicated., However, no new techniques are required to do it. We leave this as an, exercise (Exercise 17). Some hints are given in the answer section., , The Simplicity of A5, Once 120 has been disposed of, we will have shown that the only integers between 1 and 200 that can be the orders of non-Abelian simple, groups are 60 and 168. For completeness, we will now prove that A5,, which has order 60, is a simple group. A similar argument can be used, to show that the factor group SL(2, Z7)/Z(SL(2, Z7)) is a simple group, of order 168. [This group is denoted by PSL(2, Z7).], If A5 had a nontrivial proper normal subgroup H, then |H| would be, equal to 2, 3, 4, 5, 6, 10, 12, 15, 20, or 30. By Exercise 61 in Chapter 5,, A5 has 24 elements of order 5, 20 elements of order 3, and no elements, of order 15. Now, if |H| is equal to 3, 6, 12, or 15, then |A5/H| is relatively prime to 3, and by Exercise 61 in Chapter 9, H would have to, contain all 20 elements of order 3. If |H| is equal to 5, 10, or 20, then, |A5/H| is relatively prime to 5, and, therefore, H would have to contain the 24 elements of order 5. If |H| 5 30, then |A5 /H| is relatively, prime to both 3 and 5, and so H would have to contain all the elements, of orders 3 and 5. Finally, if |H| 5 2 or |H| 5 4, then |A5/H| 5 30 or, |A5/H| 5 15. But we know from our results in Chapter 24 that any group, of order 30 or 15 has an element of order 15. However, since A5 contains, no such element, neither does A5/H. This proves that A5 is simple., The simplicity of A5 was known to Galois in 1830, although the first, formal proof was done by Jordan in 1870. A few years later, Felix, Klein showed that the group of rotations of a regular icosahedron is, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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438, , Special Topics, , simple and, therefore, isomorphic to A5 (see Exercise 27). Since then it, has frequently been called the icosahedral group. Klein was the first to, prove that there is a simple group of order 168., The problem of determining which integers in a certain interval are, possible orders for finite simple groups goes back to 1892, when, Hölder went up to 200. His arguments for the integers 144 and 180, alone used up 10 pages. By 1975, this investigation had been pushed, to well beyond 1,000,000. See [3] for a detailed account of this endeavor. Of course, now that all finite simple groups have been classified, this problem is merely a historical curiosity., , The Fields Medal, Among the highest awards for mathematical achievement is the Fields, Medal. Two to four such awards are bestowed at the opening session of, the International Congress of Mathematicians, held once every four, years. Although the Fields Medal is considered by many mathematicians to be the equivalent of the Nobel Prize, there are great differences, between these awards. Besides the huge disparity in publicity and monetary value associated with the two honors, the Fields Medal is restricted to those under 40 years of age.† This tradition stems from John, Charles Fields’s stipulation, in his will establishing the medal, that the, awards should be “an encouragement for further achievement.” This restriction precluded Andrew Wiles from winning the Fields Medal for his, proof of Fermat’s Last Theorem., More details about the Fields Medal can be found at http://www, .wikipedia.com., , The Cole Prize, Approximately every five years, beginning in 1928, the American, Mathematical Society awards one or two Cole Prizes for research in, algebra and one or two Cole Prizes for research in algebraic number, theory. The prize was founded in honor of Frank Nelson Cole on the, occasion of his retirement as secretary of the American Mathematical, Society. In view of the fact that Cole was one of the first people interested in simple groups, it is interesting to note that no fewer than six, †“Take, , the sum of human achievement in action, in science, in art, in literature—, subtract the work of the men above forty, and while we should miss great treasures,, even priceless treasures, we would practically be where we are today. . . . The effective, moving, vitalizing work of the world is done between the ages of twenty-five and, forty.” Sir William Osler (1849–1919), Life of Sir William Osler, vol. I, chap. 24 (The, Fixed Period)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 439, , recipients of the prize—Dickson, Chevalley, Brauer, Feit, Thompson,, and Aschbacher—have made fundamental contributions to simple, group theory at some time in their careers. Recently the time between, Cole Prizes was reduced to three years., , Exercises, If you don’t learn from your mistakes, there’s no sense making them., herbert v. prochnow, , 1., 2., 3., 4., 5., 6., 7., 8., 9., 10., 11., 12., 13., , 14., 15., 16., 17., 18., 19., , 20., , Prove that there is no simple group of order 210 5 2 ? 3 ? 5 ? 7., Prove that there is no simple group of order 280 5 23 ? 5 ? 7., Prove that there is no simple group of order 216 5 23 ? 33., Prove that there is no simple group of order 300 5 22 ? 3 ? 52., Prove that there is no simple group of order 525 5 3 ? 52 ? 7., Prove that there is no simple group of order 540 5 22 ? 33 ? 5., Prove that there is no simple group of order 528 5 24 ? 3 ? 11., Prove that there is no simple group of order 315 5 32 ? 5 ? 7., Prove that there is no simple group of order 396 5 22 ? 32 ? 11., Prove that there is no simple group of order n, where 201 #, n # 235 and n is not prime., Without using the Generalized Cayley Theorem or its corollaries,, prove that there is no simple group of order 112., Without using the 2 ? Odd Test, prove that there is no simple group, of order 210., You may have noticed that all the “hard integers” are even. Choose, three odd integers between 200 and 1000. Show that none of these, is the order of a simple group unless it is prime., Show that there is no simple group of order pqr, where p, q, and r, are primes ( p, q, and r need not be distinct)., Show that A5 does not contain a subgroup of order 30, 20, or 15., Show that S5 does not contain a subgroup of order 40 or 30., Prove that there is no simple group of order 120 5 23 ? 3 ? 5. (This, exercise is referred to in this chapter.), Prove that if G is a finite group and H is a proper normal subgroup, of largest order, then G/H is simple., Suppose that H is a subgroup of a finite group G and that |H| and, (|G:H| 2 1)! are relatively prime. Prove that H is normal in G. What, does this tell you about a subgroup of index 2 in a finite group?, Suppose that p is the smallest prime that divides |G|. Show that, any subgroup of index p in G is normal in G., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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440, , Special Topics, , 21. Prove that the only nontrivial proper normal subgroup of S5 is A5., (This exercise is referred to in Chapter 32.), 22. Prove that a simple group of order 60 has a subgroup of order 6, and a subgroup of order 10., 23. Show that PSL(2, Z7) 5 SL(2, Z7)/Z(SL(2, Z7)), which has order, 168, is a simple group. (This exercise is referred to in this chapter.), 24. Show that the permutations (12) and (12345) generate S5., 25. Suppose that a subgroup H of S5 contains a 5-cycle and a 2-cycle., Show that H 5 S5. (This exercise is referred to in Chapter 32.), 26. Suppose that G is a finite simple group and contains subgroups H, and K such that |G:H| and |G:K| are prime. Show that |H| 5 |K|., 27. Show that (up to isomorphism) A5 is the only simple group of order, 60. (This exercise is referred to in this chapter.), 28. Prove that a simple group cannot have a subgroup of index 4., 29. Prove that there is no simple group of order p2q, where p and q are, odd primes and q 7 p., 30. If a simple group G has a subgroup K that is a normal subgroup of, two distinct maximal subgroups, prove that K 5 {e}., 31. Show that a finite group of even order that has a cyclic Sylow 2subgroup is not simple., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , References, 1. W. Feit and J. G. Thompson, “Solvability of Groups of Odd Order,” Pacific Journal of Mathematics 13 (1963): 775–1029., 2. J. A. Gallian, “Computers in Group Theory,” Mathematics Magazine 49, (1976): 69–73., 3. J. A. Gallian, “The Search for Finite Simple Groups,” Mathematics Magazine 49 (1976): 163–179., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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25 | Finite Simple Groups, , 441, , Suggested Readings, G. Cornell, N. Pele, and M. Wage, “Simple Groups of Orders Less Than, 1000,” Journal of Undergraduate Research 5 (1973): 77–86., In this charming article, three undergraduate students use slightly more, theory than was given in this chapter to show that the only integers less, than 1000 that could be orders of simple groups are 60, 168, 320, 504,, 660, and 720. All but the last one are orders of simple groups. The, proof that there is no simple group of order 720 is omitted because it is, significantly beyond most undergraduates., K. David, “Using Commutators to Prove A5 Is Simple,” The American, Mathematical Monthly 94 (1987): 775–776., This note gives an elementary proof that A5 is simple using commutators., J. A. Gallian, “The Search for Finite Simple Groups,” Mathematics Magazine 49 (1976): 163–179., A historical account is given of the search for finite simple groups., This article can be downloaded at http://www.d.umn.edu/~jgallian/, simple.pdf, Martin Gardner, “The Capture of the Monster: A Mathematical Group with a, Ridiculous Number of Elements,” Scientific American 242 (6) (1980): 20–32., This article gives an elementary introduction to groups and a discussion of simple groups, including the “Monster.”, Daniel Gorenstein, “The Enormous Theorem,” Scientific American 253, (6) (1985): 104–115., You won’t find an article on a complex subject better written for the, layperson than this one. Gorenstein, the driving force behind the classification, uses concrete examples, analogies, and nontechnical terms to, make the difficult subject matter of simple groups accessible., Sandra M. Lepsi, “PSL(2, Z7) Is Simple, by Counting,” Pi Mu Epsilon, Journal 9 (1993): 576–578., The author shows that the group SL(2, Z7)/Z(SL(2, Z7)) of order 168 is, simple using a counting argument., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Michael Aschbacher, , American Mathematical Society, , Fresh out of graduate school, he, [Aschbacher] had just entered the field,, and from that moment he became the, driving force behind my program. In rapid, succession he proved one astonishing, theorem after another. Although there, were many other major contributors to, this final assault, Aschbacher alone was, responsible for shrinking my projected, 30-year timetable to a mere 10 years., daniel gorenstein, Scientific American, , Michael Aschbacher was born on April 8,, 1944, in Little Rock, Arkansas. Shortly after, his birth, his family moved to Illinois, where, his father was a professor of accounting, and his mother was a high school English, teacher. When he was nine years old, his family moved to East Lansing, Michigan; six, years later, they moved to Los Angeles., After high school, Aschbacher enrolled at, the California Institute of Technology. In addition to his schoolwork, he passed the first, four actuary exams and was employed for a, few years as an actuary, full-time in the summers and part-time during the academic year., Two of the Caltech mathematicians who influenced him were Marshall Hall and Donald, Knuth. In his senior year, Aschbacher took, abstract algebra but showed little interest, in the course. Accordingly, he received a, grade of C., In 1966, Aschbacher went to the University of Wisconsin for a Ph.D. degree. He, completed his dissertation in 1969, and, after, spending one year as an assistant professor, at the University of Illinois, he returned to, Caltech and quickly moved up to the rank of, professor., , Aschbacher’s dissertation work in the, area of combinatorial geometries had led, him to consider certain group theoretic, questions. Gradually, he turned his attention more and more to purely group theoretic problems, particularly those bearing, on the classification of finite simple groups., The 1980 Cole Prize Selection Committee, said of one of his papers, “[It] lifted the, subject to a new plateau and brought the, classification within reach.” Aschbacher, has been elected to the National Academy, of Sciences, the American Academy of Sciences, and the vice presidency of the American Mathematical Society. In 2011, Aschbacher received the $75,000 Rolf Schock, Prize from the Royal Swedish Academy of, Sciences for “his fundamental contributions, to one of the largest mathematical projects, ever, the clasification of finite simple, groups.” In 2012, he shared the $100,000, Wolf Prize for his work in the theory of, finite groups and shared the American, Mathematical Society’s Steele Prize for, Exposition., , 442, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Daniel Gorenstein, Gorenstein was one of the most influential, mathematicians of the last few decades., , Daniel Gorenstein was born in Boston, on January 1, 1923. Upon graduating from, Harvard in 1943 during World War II,, Gorenstein was offered an instructorship at, Harvard to teach mathematics to army personnel. After the war ended, he began graduate, work at Harvard. He received his Ph.D. degree in 1951, working in algebraic geometry, under Oscar Zariski. It was in his dissertation, that he introduced the class of rings that is, now named after him. In 1951, Gorenstein, took a position at Clark University in, Worcester, Massachusetts, where he stayed, until moving to Northeastern University in, 1964. From 1969 until his death on August 26,, 1992, he was at Rutgers University., In 1957, Gorenstein switched from algebraic geometry to finite groups, learning the, basic material from I. N. Herstein while collaborating with him over the next few years. A, milestone in Gorenstein’s development as a, group theorist came during 1960–1961, when, he was invited to participate in a “Group, Theory Year” at the University of Chicago., , American Mathematical Society, , michael aschbacher,, Notices of the American Mathematical, Society, , It was there that Gorenstein, assimilating the, revolutionary techniques then being developed, by John Thompson, began his fundamental, work that contributed to the classification of, finite simple groups., Through his pioneering research papers,, his dynamic lectures, his numerous personal, contacts, and his influential book on finite, groups, Gorenstein became the leader in the, 25-year effort, by hundreds of mathematicians, that led to the classification of the, finite simple groups., Among the honors received by Gorenstein, are the Steele Prize from the American, Mathematical Society and election to membership in the National Academy of Sciences, and the American Academy of Arts and, Sciences., To find more information about Gorenstein, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , 443, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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John Thompson, There seemed to be no limit to his power., , John Thompson, , daniel gorenstein, , John G. Thompson was born on October 13,, 1932, in Ottawa, Kansas. In 1951, he entered, Yale University as a divinity student, but he, switched to mathematics in his sophomore, year. In 1955, he began graduate school at the, University of Chicago, he obtained his Ph.D., degree four years later. After one year on the, faculty at Harvard, Thompson returned to, Chicago. He remained there until 1968, when, he moved to Cambridge University in, England. In 1993, Thompson accepted an appointment at the University of Florida., Thompson’s brilliance was evident early., In his dissertation, he verified a 50-year-old, conjecture about finite groups possessing a, certain kind of automorphism. (An article, about his achievement appeared in The New, York Times.) The novel methods Thompson, used in his dissertation foreshadowed the, revolutionary ideas he would later introduce, in the Feit–Thompson paper and the classification of minimal simple groups (simple, groups that contain no proper non-Abelian, simple subgroups). The assimilation and extension of Thompson’s methods by others, throughout the 1960s and 1970s ultimately, , led to the classification of finite simple, groups., In the late 1970s, Thompson made significant contributions to coding theory, the, theory of finite projective planes, and the, theory of modular functions. His recent, work on Galois groups is considered the, most important in the field in the last half of, the 20th century., Among Thompson’s many honors are the, Cole Prize in algebra and the Fields Medal., He was elected to the National Academy of, Sciences in 1967, the Royal Society of, London in 1979, and the American Academy, of Arts and Sciences in 1998. In 2000,, President Clinton presented Thompson the, National Medal of Science. In 2008, he was a, cowinner of the $1,000,000 Abel Prize given, by the Norwegian Academy of Science and, Letters., To find more information about Thompson,, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , 444, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26, , Generators, and Relations, One cannot escape the feeling that these mathematical formulae have an, independent existence and an intelligence of their own, that they are, wiser than we are, wiser even than their discoverers, that we get more, out of them than we originally put into them., heinrich hertz, , Motivation, In this chapter, we present a convenient way to define a group with certain prescribed properties. Simply put, we begin with a set of elements, that we want to generate the group, and a set of equations (called relations) that specify the conditions that these generators are to satisfy., Among all such possible groups, we will select one that is as large as, possible. This will uniquely determine the group up to isomorphism., To provide motivation for the theory involved, we begin with a concrete, example. Consider D4, the group of symmetries of a square. Recall that, R 5 R90 and H, a reflection across a horizontal axis, generate the group., Observe that R and H are related in the following ways:, R4 5 H 2 5 (RH)2 5 R0, , (the identity)., , (1), , Other relations between R and H, such as HR 5 R3H and RHR 5 H,, also exist, but they can be derived from those given in Equation (1). For, example, (RH)2 5 R0 yields HR 5 R21H21, and R4 5 H 2 5 R0 yields, R21 5 R3 and H21 5 H. So, HR 5 R3H. In fact, every relation between, R and H can be derived from those given in Equation (1)., Thus, D4 is a group that is generated by a pair of elements a and b, subject to the relations a4 5 b2 5 (ab)2 5 e and such that all other relations between a and b can be derived from these relations. This last, stipulation is necessary because the subgroup {R0, R180, H, V} of D4 is, generated by the two elements a 5 R180 and b 5 H that satisfy the relations a4 5 b2 5 (ab)2 5 e. However, the “extra” relation a2 5 e satisfied by, this subgroup cannot be derived from the original ones (since R902 2 R0). It, is natural to ask whether this description of D4 applies to some other group, 445, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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446, , Special Topics, , as well. The answer is no. Any other group generated by two elements a, and b satisfying only the relations a4 5 b2 5 (ab)2 5 e, and those that, can be derived from these relations, is isomorphic to D4., Similarly, one can show that the group Z4 % Z2 is generated by two elements a and b such that a4 5 b2 5 e and ab 5 ba, and any other relation between a and b can be derived from these relations. The purpose of, this chapter is to show that this procedure can be reversed; that is, we, can begin with any set of generators and relations among the generators, and construct a group that is uniquely described by these generators and, relations, subject to the stipulation that all other relations among the, generators can be derived from the original ones., , Definitions and Notation, We begin with some definitions and notation. For any set S 5 {a, b, c, . . .}, of distinct symbols, we create a new set S21 5 {a21, b21, c21, . . .} by replacing each x in S by x21. Define the set W(S) to be the collection of all, formal finite strings of the form x1x2 ? ? ? xk, where each xi [ S < S21., The elements of W(S) are called words from S. We also permit the string, with no elements to be in W(S). This word is called the empty word and, is denoted by e., We may define a binary operation on the set W(S) by juxtaposition;, that is, if x1x2 ? ? ? xk and y1y2 ? ? ? yt belong to W(S), then so does x1x2, ? ? ? xk y1y2 ? ? ? yt. Observe that this operation is associative and the, empty word is the identity. Also, notice that a word such as aa21 is not, the identity, because we are treating the elements of W(S) as formal, symbols with no implied meaning., At this stage we have everything we need to make a group out of, W(S) except inverses. Here a difficulty arises, since it seems reasonable, that the inverse of the word ab, say, should be b21a21. But abb21a21 is, not the empty word! You may recall that we faced a similar obstacle, long ago when we carried out the construction of the field of quotients, of an integral domain. There we had formal symbols of the form a/b, and we wanted the inverse of a/b to be b/a. But their product, ab/(ba),, was a formal symbol that was not the same as the formal symbol 1/1,, the identity. So, we proceed here as we did there—by way of equivalence classes., Definition Equivalence Classes of Words, , For any pair of elements u and v of W(S), we say that u is related to v if v, can be obtained from u by a finite sequence of insertions or deletions of, words of the form xx21 or x21x, where x [ S., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26 | Generators and Relations, , 447, , We leave it as an exercise to show that this relation is an equivalence, relation on W(S). (See Exercise 1.), EXAMPLE 1 Let S 5 {a, b, c}. Then acc21b is equivalent to ab;, aab 21bbaccc 21 is equivalent to aabac; the word a 21aabb 21a 21 is, equivalent to the empty word; and the word ca 21b is equivalent to, cc21caa21a21bbca21ac21b21. Note, however, that cac21b is not equivalent to ab., , Free Group, Theorem 26.1 Equivalence Classes Form a Group, Let S be a set of distinct symbols. For any word u in W(S), let u, denote the set of all words in W(S) equivalent to u (that is, u is the, equivalence class containing u). Then the set of all equivalence, classes of elements of W(S) is a group under the operation, u ? v 5 uv., , PROOF This proof is left to the reader., The group defined in Theorem 26.1 is called a free group on S., Theorem 26.2 shows why free groups are important., Theorem 26.2 Universal Mapping Property, Every group is a homomorphic image of a free group., , PROOF Let G be a group and let S be a set of generators for G. (Such, a set exists, because we may take S to be G itself.) Now let F be the free, group on S. Unfortunately, since our notation for any word in W(S) also, denotes an element of G, we have created a notational problem for ourselves. So, to distinguish between these two cases, we will denote the, word x1x2 ? ? ? xn in W(S) by (x1x2 ? ? ? xn)F and the product x1x2 ? ? ? xn in, G by (x1x2 ? ? ? xn)G. As before, x1x2 . . . xn denotes the equivalence class, in F containing the word (x1x2 ? ? ? xn)F in W(S). Notice that x1x2 . . . xn, and (x1x2 ? ? ? xn)G are entirely different elements, since the operations, on F and G are different., Now consider the mapping from F into G given by, f( x1x2 . . . xn) 5 (x1x2 ? ? ? xn)G., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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448, , Special Topics, , [All we are doing is taking a product in F and viewing it as a product in, G. For example, if G is the cyclic group of order 4 generated by a, then, f(aaaaa) 5 (aaaaa)G 5 a.], Clearly, f is well-defined, for inserting or deleting expressions of the, form xx21 or x21x in elements of W(S) corresponds to inserting or deleting the identity in G. To check that f is operation-preserving, observe, that, f(x1x2 . . . xn)(y1y2 . . . ym) 5 f(x1x2 . . . xny1y2 . . . ym), 5 (x1x2 ? ? ? xny1y2 ? ? ? ym)G, 5 (x1x2 ? ? ? xn)G(y1y2 ? ? ? ym)G., Finally, f is onto G because S generates G., The following corollary is an immediate consequence of Theorem 26.2, and the First Isomorphism Theorem for Groups., Corollary Universal Factor Group Property, Every group is isomorphic to a factor group of a free group., , Generators and Relations, We have now laid the foundation for defining a group by way of generators and relations. Before giving the definition, we will illustrate the, basic idea with an example., EXAMPLE 2 Let F be the free group on the set {a, b} and let N be, the smallest normal subgroup of F containing the set {a4, b2, (ab)2}. We, will show that F/N is isomorphic to D4. We begin by observing that the, mapping f from F onto D4, which takes a to R90 and b to H (horizontal, reflection), defines a homomorphism whose kernel contains N. Thus,, F/Ker f is isomorphic to D4. On the other hand, we claim that the set, K 5 {N, aN, a2N, a3N, bN, abN, a2bN, a3bN}, of left cosets of N is F/N itself. To see this, notice that every member, of F/N can be generated by starting with N and successively multiplying on the left by various combinations of a’s and b’s. So, it suffices, to show that K is closed under multiplication on the left by a and b. It, is trivial that K is closed under left multiplication by a. For b, we will, do only one of the eight cases. The others can be done in a similar, fashion. Consider b(aN). Since b2, abab, a 4 [ N and Nb 5 bN, we, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26 | Generators and Relations, , 449, , have baN 5 baNb2 5 babNb 5 a21(abab)Nb 5 a21Nb = a21a4Nb 5, a3Nb 5 a3bN. Upon completion of the other cases (Exercise 3), we, know that F/N has at most eight elements. At the same time, we know, that F/Ker f has exactly eight elements. Since F/Ker f is a factor, group of F/N [indeed, F/Ker f < (F/N)/(Ker f/N)], it follows that F/N, also has eight elements and F/N 5 F/Ker f < D4., Definition Generators and Relations, , Let G be a group generated by some subset A 5 {a1, a2, . . . , an} and let, F be the free group on A. Let W 5 {w1, w2, . . . , wt} be a subset of F and, let N be the smallest normal subgroup of F containing W. We say that G, is given by the generators a1, a2, . . . , an and the relations w1 5 w2 5 ? ? ? 5, wt 5 e if there is an isomorphism from F/N onto G that carries ai N to ai ., , The notation for this situation is, G 5 ka1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 el., As a matter of convenience, we have restricted the number of generators and relations in our definition to be finite. This restriction is, not necessary, however. Also, it is often more convenient to write a, relation in implicit form. For example, the relation a21b23ab 5 e is, often written as ab 5 b3a. In practice, one does not bother writing, down the normal subgroup N that contains the relations. Instead, one, just manipulates the generators and treats anything in N as the identity, as our notation suggests. Rather than saying that G is given by, ka1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 el,, many authors prefer to say that G has the presentation, ka1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 el., Notice that a free group is “free” of relations; that is, the equivalence, class containing the empty word is the only relation. We mention in, passing the fact that a subgroup of a free group is also a free group., Free groups are of fundamental importance in a branch of algebra, known as combinatorial group theory., EXAMPLE 3 The discussion in Example 2 can now be summed up, by writing, D4 5 ka, b | a4 5 b2 5 (ab)2 5 el., EXAMPLE 4 The group of integers is the free group on one letter; that, is, Z < kal. (This is the only nontrivial Abelian group that is free.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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450, , Special Topics, , The next theorem formalizes the argument used in Example 2 to, prove that the group defined there has eight elements., Theorem 26.3 Dyck's Theorem (1882), Let, , G 5 ka1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 el, and let, , G 5 ka1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5, wt11 5 ? ? ? 5 wt1k 5 el., Then G is a homomorphic image of G., , PROOF See Exercise 5., In words, Theorem 26.3 says that if you start with generators and relations for a group G and create a group G by imposing additional, relations, then G is a homomorphic image of G., Corollary Largest Group Satisfying Defining Relations, If K is a group satisfying the defining relations of a finite group G, and |K| $ |G|, then K is isomorphic to G., , PROOF See Exercise 5., EXAMPLE 5 Quaternions Consider the group G 5 ka, b | a2 5, 5 (ab)2l. What does G look like? Formally, of course, G is isomorphic to F/N, where F is free on {a, b} and N is the smallest normal subgroup of F containing b22a2 and (ab)22a2. Now, let H 5 kbl and S 5, {H, aH}. Then, just as in Example 2, it follows that S is closed under, multiplication by a and b from the left. So, as in Example 2, we have, G 5 H < aH. Thus, we can determine the elements of G once we know, exactly how many elements there are in H. (Here again, the three relations come in.) To do this, first observe that b2 5 (ab)2 5 abab implies, b 5 aba. Then a2 5 b2 5 (aba)(aba) 5 aba2ba 5 ab4a and therefore, b4 5 e. Hence, H has at most four elements, and therefore G has at most, eight—namely, e, b, b2, b3, a, ab, ab2, and ab3. It is conceivable, however, that not all of these eight elements are distinct. For example, Z2 % Z2, satisfies the defining relations and has only four elements. Perhaps it is, b2, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26 | Generators and Relations, , 451, , the largest group satisfying the relations. How can we show that the eight, elements listed above are distinct? Well, consider the group G generated, by the matrices, A c, , 0 1, d, 1 0, , and, , B c, , 0 i, d,, i 0, , where i 5 21. Direct calculations show that in G, the elements e, B, B2,, B3, A, AB, AB2, and AB3 are distinct and that G satisfies the relations, A2 5 B2 5 (AB)2. So, it follows from the corollary to Dyck’s Theorem, that G is isomorphic to G and therefore G has order 8., The next example illustrates why, in Examples 2 and 5, it is necessary to show that the eight elements listed for the group are distinct., EXAMPLE 6 Let, G 5 ka, b | a3 5 b9 5 e, a21ba 5 b21l., Once again, we let H 5 kbl and observe that G 5 H < aH < a2H. Thus,, G 5 {aib j | 0 # i # 2, 0 # j # 8},, and therefore G has at most 27 elements. But this time we will not be, able to find some concrete group of order 27 satisfying the same relations that G does, for notice that b21 5 a21ba implies, b 5 (a21ba)21 5 a21b21a., Hence,, b 5 ebe 5 a23ba3 5 a22(a21ba)a2 5 a22b21a2, 5 a21(a21b21a)a 5 a21ba 5 b21., So, the original three relations imply the additional relation b2 5 e. But, b2 5 e 5 b9 further implies b 5 e. It follows, then, that G has at most, three distinct elements—namely, e, a, and a2. But Z3 satisfies the defining relations with a 5 1 and b 5 0. So, |G| 5 3., We hope Example 6 convinces you of the fact that, once a list of the, elements of the group given by a set of generators and relations has, been obtained, one must further verify that this list has no duplications., Typically, this is accomplished by exhibiting a specific group that satisfies the given set of generators and relations and that has the same size, as the list. Obviously, experience plays a role here., Here is a fun example adapted from [1]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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452, , Special Topics, , EXAMPLE 7 Let G be the group with the 26 letters of the alphabet, as generators. For relations we take strings A 5 B, where A and B are, words in some fixed reference, say [2], and have the same pronunciation but different meanings (such words are called homophones). For, example, buy 5 by 5 bye, hour 5 our, lead 5 led, whole 5 hole. From, these strings and cancellation, we obtain u 5 e 5 h 5 a 5 w 5 0/ (/0 is, the identity string). With these examples in mind, we ask, What is the, group given by these generators and relations? Surprisingly, the answer, is the infinite cyclic group generated by v. To verify this, one must show, that every letter except v is equivalent to 0/ and that there are no two homophones that contain a different number of v’s. The former can easily, be done with common words. For example, from inn 5 in, plumb 5, plum, and knot 5 not, we see that n 5 b 5 k 5 0., / From too 5 to, we have o 5 0., / That there are no two homophones in [2] that have, a different number of v’s can be verified by simply checking all, cases. In contrast, the reference Handbook of Homophones by W. C., Townsend (see http://members.peak.org/~jeremy/dictionaryclassic/, chapters/homophones.php) lists felt/veldt as homophones. Of course,, including these makes the group trivial., , Classification of Groups, of Order Up to 15, The next theorem illustrates the utility of the ideas presented in this, chapter., Theorem 26.4 Classification of Groups of Order 8 (Cayley, 1859), Up to isomorphism, there are only five groups of order 8: Z8, Z4 % Z2,, Z2 % Z2 % Z2, D4, and the quaternions., , PROOF The Fundamental Theorem of Finite Abelian Groups takes, care of the Abelian cases. Now, let G be a non-Abelian group of order, 8. Also, let G1 5 ka, b | a4 5 b2 5 (ab)2 5 el and let G2 5 ka, b | a2 5, b2 5 (ab)2l. We know from the preceding examples that G1 is isomorphic to D4 and G2 is isomorphic to the quaternions. Thus, it suffices to, show that G must satisfy the defining relations for G1 or G2. It follows, from Exercise 47 in Chapter 2 and Lagrange’s Theorem that G has an, element of order 4; call it a. Then, if b is any element of G not in kal, we, know that, G 5 kal < kalb 5 {e, a, a2, a3, b, ab, a2b, a3b}., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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454, , Special Topics, , Characterization of Dihedral Groups, As another nice application of generators and relations, we will now, give a characterization of the dihedral groups that has been known for, more than 100 years. For n $ 3, we have used Dn to denote the group of, symmetries of a regular n-gon. Imitating Example 2, one can show that, Dn < ka, b | an 5 b2 5 (ab)2 5 el (see Exercise 9). By analogy, these, generators and relations serve to define D1 and D2 also. (These are also, called dihedral groups.) Finally, we define the infinite dihedral group, D` as ka, b | a2 5 b2 5 el. The elements of D` can be listed as e, a, b, ab,, ba, (ab)a, (ba)b, (ab)2, (ba)2, (ab)2a, (ba)2b, (ab)3, (ba)3, . . . ., Theorem 26.5 Characterization of Dihedral Groups, Any group generated by a pair of elements of order 2 is dihedral., , PROOF Let G be a group generated by a pair of distinct elements of, order 2, say, a and b. We consider the order of ab. If |ab| 5 `, then G is, infinite and satisfies the relations of D`. We will show that G is isomorphic to D`. By Dyck’s Theorem, G is isomorphic to some factor group, of D`, say, D`/H. Now, suppose h [ H and h 2 e. Since every element, of D` has one of the forms (ab)i, (ba)i, (ab)ia, or (ba)ib, by symmetry,, we may assume that h 5 (ab)i or h 5 (ab)ia. If h 5 (ab)i, we will show, that D`/H satisfies the relations for Di given in Exercise 9. Since (ab)i is, in H, we have, H 5 (ab)iH 5 (abH)i,, so that (abH)21 5 (abH)i21. But, (ab)21H 5 b21a21H 5 baH,, and it follows that, , aHabHaH a2HbHaH eHbaH baH 1abH2 1., , Thus,, D`/H 5 kaH, bHl 5 kaH, abHl, (see Exercise 7), and D`/H satisfies the defining relations for Di (use, Exercise 9 with x 5 aH and y 5 abH). In particular, G is finite—an, impossibility., If h 5 (ab)ia, then, H 5 (ab)iaH 5 (ab)iHaH,, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26 | Generators and Relations, , 455, , and therefore, (abH)i 5 (ab)iH 5 (aH)21 5 a21H 5 aH., It follows that, kaH, bHl 5 kaH, abHl # kabHl., However,, (abH)2i 5 (aH)2 5 a2H 5 H,, so that D`/H is again finite. This contradiction forces H 5 {e} and G to, be isomorphic to D`., Finally, suppose that |ab| 5 n. Since G 5 ka, bl 5 ka, abl, we can, show that G is isomorphic to Dn by proving that b(ab)b 5 (ab)21, which, is the same as ba 5 (ab)21 (see Exercise 9). But (ab)21 5 b21a21 5 ba,, since a and b have order 2., , Realizing the Dihedral Groups, with Mirrors, A geometric realization of D` can be obtained by placing two mirrors, facing each other in a parallel position, as shown in Figure 26.1. If we, let a and b denote reflections in mirrors A and B, respectively, then ab,, viewed as the composition of a and b, represents a translation through, twice the distance between the two mirrors to the left, and ba is the, translation through the same distance to the right., A, aba ab, , FF, , a, , e, , FF, , B, b ba, , bab baba, , FF, , FF, , Figure 26.1 The group D`—reflections in parallel mirrors, , The finite dihedral groups can also be realized with a pair of mirrors., For example, if we place a pair of mirrors facing each other at a 45°, angle, we obtain the group D4. Notice that in Figure 26.2, the effect of, reflecting an object in mirror A, then mirror B, is a rotation of twice the, angle between the two mirrors (that is, 90°)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , F, , bab, , F, , e, , F, F, , F, aba, , A, a, , F, , baba = abab, , B, , b, , F, , ba, , F, , 456, , ab, , Figure 26.2 The group D4—reflections in mirrors at a 45° angle, , In Figure 26.3, we see a portion of the pattern produced by reflections, in a pair of mirrors set at a 1° angle. The corresponding group is D180. In, general, reflections in a pair of mirrors set at the angle 180°/n correspond, to the group Dn. As n becomes larger and larger, the mirrors approach a, parallel position. In the limiting case, we have the group D`., Β, , Α, , aba, , ab, , a, , e, , b ba, , bab ba, , ba, F, F, F, F, F, F, FF, , Figure 26.3 The group D180—reflections in mirrors at a 1° angle, , We conclude this chapter by commenting on the advantages and disadvantages of using generators and relations to define groups. The principal advantage is that in many situations—particularly in knot theory,, algebraic topology, and geometry—groups defined by way of generators and relations arise in a natural way. Within group theory itself, it is, often convenient to construct examples and counterexamples with generators and relations. Among the disadvantages of defining a group by, generators and relations is the fact that it is often difficult to decide, whether or not the group is finite, or even whether or not a particular, element is the identity. Furthermore, the same group can be defined with, entirely different sets of generators and relations, and, given two groups, defined by generators and relations, it is often extremely difficult to, decide whether or not these two groups are isomorphic. Nowadays,, these questions are frequently tackled with the aid of a computer., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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457, , 26 | Generators and Relations, , Exercises, It don’t come easy., ringo starr, “It Don't Come Easy,” single, , 1. Let S be a set of distinct symbols. Show that the relation defined on, W(S) in this chapter is an equivalence relation., 2. Let n be an even integer. Prove that Dn /Z(Dn) is isomorphic to Dn/2., 3. Verify that the set K in Example 2 is closed under multiplication, on the left by b., 4. Show that ka, b | a5 5 b2 5 e, ba 5 a2bl is isomorphic to Z2., 5. Prove Theorem 26.3 and its corollary., 6. Let G be the group {61, 6i, 6j, 6k} with multiplication defined, as in Exercise 54 in Chapter 9. Show that G is isomorphic to ka, b |, a2 5 b2 5 (ab)2l. (Hence, the name “quaternions.”), 7. In any group, show that ka, bl 5 ka, abl. (This exercise is referred, to in the proof of Theorem 26.5.), 8. Let a 5 (12)(34) and b 5 (24). Show that the group generated by, a and b is isomorphic to D4., 9. Prove that G 5 kx, y | x2 5 yn 5 e, xyx 5 y21l is isomorphic to Dn., (This exercise is referred to in the proof of Theorem 26.5.), 10. What is the minimum number of generators needed for Z2 % Z2 %, Z2? Find a set of generators and relations for this group., 11. Suppose that x2 5 y2 5 e and yz 5 zxy. Show that xy 5 yx., 12. Let G 5 ka, b | a2 5 b4 5 e, ab 5 b3al., a. Express a 3b 2abab 3 in the form b ia j, where 0 # i # 1 and, 0 # j # 3., b. Express b3abab3a in the form bia j, where 0 # i # 1 and 0 # j # 3., 13. Let G 5 ka, b | a2 5 b2 5 (ab)2l., a. Express b2abab3 in the form bia j., b. Express b3abab3a in the form bia j., 14. Let G be the group defined by the following table. Show that G is, isomorphic to Dn., 1, 2, 3, 4, 5, 6, :, 2n, , 1, , 2, , 3, , 4, , 5, , 6, , ???, , 2n, , 1, 2, 3, 4, 5, 6, :, 2n, , 2, 1, 4, 3, 6, 5, :, 2n 2 1, , 3, 2n, 5, 2, 7, 4, :, 2n 2 2, , 4, 2n 2 1, 6, 1, 8, 3, :, 2n 2 3, , 5, 2n 2 2, 7, 2n, 9, 2, :, 2n 2 4, , 6, 2n 2 3, 8, 2n 2 1, 10, 1, :, 2n 2 5, , ???, ???, ???, ???, ???, ???, :, ???, , 2n, 3, 2, 5, 4, 7, :, 1, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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458, , Special Topics, , 15. Let G 5 kx, y | x8 5 y2 5 e, yxyx3 5 el. Show that |G| # 16. Assuming that |G| 5 16, find the center of G and the order of xy., 16. Confirm the classification given in Table 26.1 of all groups of, orders 1 to 11., 17. Let G be defined by some set of generators and relations. Show, that every factor group of G satisfies the relations defining G., 18. Let G 5 ks, t | sts 5 tstl. Show that the permutations (23) and (13), satisfy the defining relations of G. Explain why this proves that G, is non-Abelian., 19. In D12 5 kx, y | x2 5 y12 5 e, xyx 5 y21l, prove that the subgroup, H 5 kx, y3l (which is isomorphic to D4) is not a normal subgroup., 20. Let G 5 kx, y | x2n 5 e, xn 5 y2, y21xy 5 x21l. Show that Z(G) 5, {e, xn}. Assuming that |G| 5 4n, show that G/Z(G) is isomorphic, to Dn. (The group G is called the dicyclic group of order 4n.), 21. Let G 5 ka, b | a6 5 b3 5 e, b21ab 5 a3l. How many elements, does G have? To what familiar group is G isomorphic?, 22. Let G 5 kx, y | x4 5 y4 5 e, xyxy21 5 el. Show that |G| # 16. Assuming that |G| 5 16, find the center of G and show that G/ky2l is, isomorphic to D4., 23. Determine the orders of the elements of D`., 1 a b, 24. Let G • £ 0 1 c § † a, b, c [ Z2 ¶ . Prove that G is isomorphic, 0 0 1, to D4., 25. Let G 5 ka, b, c, d | ab 5 c, bc 5 d, cd 5 a, da 5 bl. Determine |G|., 26. Let G ka, b 冟 a2 e, b2 e, aba babl. To what familiar, group is G isomorphic?, 27. Let G ka, b 冟 a3 e, b2 e, aba1b1 el. To what familiar, group is G isomorphic?, 28. Give an example of a non-Abelian group that has exactly three, elements of finite order., 29. Referring to Example 7 in this chapter, show as many letters as you, can that are equivalent to ~., 30. Suppose that a group of order 8 has exactly five elements of order 2., Identify the group., , References, 1. J.-F. Mestre, R. Schoof, L. Washington, D. Zagier, “Quotient homophones des groupes libres [Homophonic Quotients of Free Groups],” Experimental Mathematics 2 (1993): 153–155., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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26 | Generators and Relations, , 459, , 2. H. C. Whitford, A Dictionary of American Homophones and Homographs,, New York: Teachers College Press, 1966., , Suggested Readings, Alexander H. Frey, Jr., and David Singmaster, Handbook of Cubik Math,, Hillside, N.J.: Enslow, 1982., This book is replete with the group theoretic aspects of the Magic, Cube. It uses permutation group theory and generators and relations to, discuss the solutions to the cube and related results. The book has numerous challenging exercises stated in group theoretic terms., Lee Neuwirth, “The Theory of Knots,” Scientific American 240 (1979):, 110–124., This article shows how a group can be associated with a knotted string., Mathematically, a knot is just a one-dimensional curve situated in, three-dimensional space. The theory of knots—a branch of topology—, seeks to classify and analyze the different ways of embedding such a, curve. Around the beginning of the 20th century, Henri Poincaré observed that important geometric characteristics of knots could be described in terms of group generators and relations—the so-called knot, group. Among other knots, Neuwirth describes the construction of the, knot group for the trefoil knot pictured. One set of generators and relations for this group is kx, y, z | xy 5 yz, zx 5 yzl., , The trefoil knot, , David Peifer, “An Introduction to Combinatorial Group Theory and the, Word Problem,” Mathematics Magazine 70 (1997): 3–10., This article discusses some fundamental ideas and problems regarding, groups given by presentations., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Marshall Hall, Jr., Professor Hall was a mathematician in the, broadest sense of the word but with a, predilection for group theory, geometry, and combinatorics., , Jonathan Hall, , hans zassenhaus, Notices of, the American Mathematical Society, , Marshall Hall, Jr., was born on September, 17, 1910, in St. Louis, Missouri. He demonstrated interest in mathematics at the age of, 11 when he constructed a seven-place table of, logarithms for the positive integers up to, 1000. He completed a B.A. degree in 1932 at, Yale. After spending a year at Cambridge, University, where he worked with Philip, Hall, Harold Davenport, and G. H. Hardy, he, returned to Yale for his Ph.D. degree. At the, outbreak of World War II, he joined Naval, Intelligence and had significant success in deciphering both the Japanese codes and the, German Enigma messages. These successes, helped to turn the tide of the war. After the, war, Hall had faculty appointments at the, Ohio State University, Caltech, and Emory, University. He died on July 4, 1990., Hall’s highly regarded books on group, theory and combinatorial theory are classics., His mathematical legacy includes more than, , 120 research papers on group theory, coding, theory, and design theory. His 1943 paper on, projective planes ranks among the most cited, papers in mathematics. Several fundamental, concepts as well as a sporadic simple group, are identified with Hall’s name. One of Hall’s, most celebrated results is his solution to the, “Burnside Problem” for exponent 6—that is,, his proof that a finitely generated group in, which the order of every element divides 6, must be finite. Hall influenced both John, Thompson and Michael Aschbacher, two of finite group theory’s greatest contributors. It was, Hall who suggested Thompson’s Ph.D. dissertation problem. Hall’s Ph.D. students at Caltech, included Donald Knuth and Robert McEliece., To find more information about Hall,, visit:, http://www–groups.dcs.st–and, .ac.uk/~history/, , 460, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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27 Symmetry Groups, , I’m not good at math, but I do know that the universe is formed with, mathematical principles whether I understand them or not, and I am, going to let that guide me., bob dylan, Chronicles, Volume One, , Isometries, In the early chapters of this book, we briefly discussed symmetry, groups. In this chapter and the next, we examine this fundamentally, important concept in some detail. It is convenient to begin such a discussion with the definition of an isometry (from the Greek isometros,, meaning “equal measure”) in Rn., Definition Isometry, , An isometry of n-dimensional space Rn is a function from Rn onto Rn, that preserves distance., , In other words, a function T from Rn onto Rn is an isometry if, for, every pair of points p and q in Rn, the distance from T( p) to T(q) is the, same as the distance from p to q. With this definition, we may now, make precise the definition of the symmetry group of an n-dimensional figure., Definition Symmetry Group of a Figure in Rn, , Let F be a set of points in Rn. The symmetry group of F in Rn is the set, of all isometries of Rn that carry F onto itself. The group operation is, function composition., , It is important to realize that the symmetry group of an object depends not only on the object, but also on the space in which we view it., For example, the symmetry group of a line segment in R1 has order 2,, the symmetry group of a line segment considered as a set of points in, R2 has order 4, and the symmetry group of a line segment viewed as a, set of points in R3 has infinite order (see Exercise 9)., 461, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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462, , Special Topics, , Although we have formulated our definitions for all finite dimensions, our chief interest will be the two-dimensional case. It has been, known since 1831 that every isometry of R2 is one of four types:, rotation, reflection, translation, and glide-reflection (see [1, p. 46])., Rotation about a point in a plane needs no explanation. A reflection, across a line L is that transformation that leaves every point of L fixed, and takes every point Q, not on L, to the point Q9 so that L is the perpendicular bisector of the line segment from Q to Q9 (see Figure 27.1)., The line L is called the axis of reflection. In an xy-coordinate plane, the, transformation (x, y) S (x, 2y) is a reflection across the x-axis,, whereas (x, y) S ( y, x) is a reflection across the line y 5 x. Some authors call an axis of reflective symmetry L a mirror because L acts like, a two-sided mirror; that is, the image of a point Q in a mirror placed on, the line L is, in fact, the image of Q under the reflection across the line, L. Reflections are called opposite isometries because they reverse orientation. For example, the reflected image of a clockwise spiral is a, counterclockwise spiral. Similarly, the reflected image of a right hand, is a left hand. (See Figure 27.1.), Q, Q', L, Axis of reflection, , Axis of reflection, , Figure 27.1 Reflected images, , A translation is simply a function that carries all points the same distance in the same direction. For example, if p and q are points in a plane, and T is a translation, then the two directed line segments joining p to, T( p) and q to T(q) have the same length and direction. A glide-reflection, is the product of a translation and a reflection across the line containing, the translation line segment. This line is called the glide-axis. In, Figure 27.2, the arrow gives the direction and length of the translation,, and is contained in the axis of reflection. A glide-reflection is also an, opposite isometry. Successive footprints in wet sand are related by a, glide-reflection., , p, , T(p), , Figure 27.2 Glide-reflection, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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27 | Symmetry Groups, , 463, , Classification of Finite Plane, Symmetry Groups, , Symmetry in Science and Art by A. V. Shubnikov & V/A. Kopstik © 1974 Plenum, Publishing Company, , Our first goal in this chapter is to classify all finite plane symmetry, groups. As we have seen in earlier chapters, the dihedral group Dn is, the plane symmetry group of a regular n-gon. (For convenience, call, D2 the plane symmetry group of a nonsquare rectangle and D1 the, plane symmetry group of the letter “V.” In particular, D2 < Z2 % Z2 and, D1 < Z2.) The cyclic groups Zn are easily seen to be plane symmetry, groups also. Figure 27.3 is an illustration of an organism whose plane, symmetry group consists of four rotations and is isomorphic to Z4. The, surprising fact is that the cyclic groups and dihedral groups are, the only finite plane symmetry groups. The famous mathematician, Hermann Weyl attributes the following theorem to Leonardo da Vinci, (1452–1519)., , Figure 27.3 Aurelia insulinda, an organism, whose plane symmetry group is Z4, , Theorem 27.1 Finite Symmetry Groups in the Plane, The only finite plane symmetry groups are Zn and Dn., , PROOF Let G be a finite plane symmetry group of some figure. We, first observe that G cannot contain a translation or a glide-reflection,, because in either case G would be infinite. Now observing that the, composition of two reflections preserves orientation, we know that, such a composition is a translation or rotation. When the two reflections, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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464, , Special Topics, , have parallel axes of reflection, there is no fixed point (see Exercise 12, in the Supplementary Exercises for Chapters 1– 4), so the composition, is a translation. Thus, every two reflections in G have reflection axes, that intersect in some point. We claim that all reflections intersect in, the same point. Suppose that f and f 9 are two distinct reflections in G., Then because f f9 preserves orientation, we know that f f9 is a rotation., We use the fact from geometry [2, p. 366] that a finite group of rotations must have a common center, say P. This means that any two, reflections must intersect at point P. So, we have shown that all the elements of G have the common fixed point P., For convenience, let us denote a rotation about P of s degrees, by Rs. Now, among all rotations in G, let b be the smallest positive, angle of rotation. (Such an angle exists, since G is finite and R360 belongs to G.) We claim that every rotation in G is some power of Rb., To see this, suppose that Rs is in G. We may assume 0° , s # 360°., Then, b # s and there is some integer t such that tb # s ,, (t 1 1)b. But, then Rs2tb 5 Rs(Rb)2t is in G and 0 # s 2 tb , b., Since b represents the smallest positive angle of rotation among the, elements of G, we must have s 2 tb 5 0, and therefore, Rs 5 (Rb)t., This verifies the claim., For convenience, let us say that |Rb| 5 n. Now, if G has no reflections, we have proved that G 5 kRbl < Zn. If G has at least one reflection, say f, then, f, fRb, f (Rb)2, . . . , f (Rb)n21, are also reflections. Furthermore, this is the entire set of reflections of G., For if g is any reflection in G, then fg is a rotation, and so fg 5 (Rb)k for, some k. Thus, g 5 f21(Rb)k 5 f(Rb)k. So, G 5 {R0, Rb, (Rb)2, . . . , (Rb)n21, f, fRb, f (Rb)2, . . . , f(Rb)n21},, and G is generated by the pair of reflections f and fRb. Hence, by our, characterization of the dihedral groups (Theorem 26.5), G is the dihedral group Dn., , Classification of Finite Groups, of Rotations in R3, One might think that the set of all possible finite symmetry groups in, three dimensions would be much more diverse than is the case for two, dimensions. Surprisingly, this is not the case. For example, moving to, three dimensions introduces only three new groups of rotations. This, observation was first made by the physicist and mineralogist Auguste, Bravais in 1849, in his study of possible structures of crystals., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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27 | Symmetry Groups, , 465, , Theorem 27.2 Finite Groups of Rotations in R3, Up to isomorphism, the finite groups of rotations in R3 are Zn, Dn,, A4 , S4 , and A5 ., , Theorem 27.2, together with the Orbit-Stabilizer Theorem (Theorem 7.3), makes easy work of determining the group of rotations of an, object in R3., EXAMPLE 1 We determine the group G of rotations of the solid in, Figure 27.4, which is composed of six congruent squares and eight congruent equilateral triangles. We begin by singling out any one of the, squares. Obviously, there are four rotations that map this square to itself,, and the designated square can be rotated to the location of any of the, other five. So, by the Orbit-Stabilizer Theorem (Theorem 7.3), the rotation group has order 4 ? 6 5 24. By Theorem 27.2, G is one of Z24, D12,, and S4. But each of the first two groups has exactly two elements of, order 4, whereas G has more than two. So, G is isomorphic to S4., , Figure 27.4, , The group of rotations of a tetrahedron (the tetrahedral group) is, isomorphic to A4; the group of rotations of a cube or an octahedron (the, octahedral group) is isomorphic to S4; the group of rotations of a dodecahedron or an icosahedron (the icosahedral group) is isomorphic to, A5. (Coxeter [1, pp. 271–273] specifies which portions of the polyhedra, are being permuted in each case.) These five solids are illustrated in, Figure 27.5., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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466, , Special Topics, , Figure 27.5 The five regular solids as depicted by Johannes Kepler, in Harmonices Mundi, Book II (1619), , Exercises, Perhaps the most valuable result of all education is the ability to make, yourself do the thing you have to do, when it ought to be done, whether, you like it or not., thomas henry huxley, “Technical Education”, , 1., 2., 3., 4., , Show that an isometry of Rn is one-to-one., Show that the translations of Rn form a group., Exhibit a plane figure whose plane symmetry group is Z5., Show that the group of rotations in R3 of a 3-prism (that is, a prism, with equilateral ends, as in the following figure) is isomorphic to D3., , 5. What is the order of the (entire) symmetry group in R3 of a 3-prism?, 6. What is the order of the symmetry group in R3 of a 4-prism (a box, with square ends that is not a cube)?, 7. What is the order of the symmetry group in R3 of an n-prism?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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27 | Symmetry Groups, , 467, , 8. Show that the symmetry group in R3 of a box of dimensions 20 3, 30 3 40 is isomorphic to Z2 % Z2 % Z2., 9. Describe the symmetry group of a line segment viewed as, a. a subset of R1., b. a subset of R2., c. a subset of R3., (This exercise is referred to in this chapter.), 10. (From the “Ask Marilyn” column in Parade Magazine, December 11,, 1994.)* The letters of the alphabet can be sorted into the following, categories:, 1. FGJLNPQRSZ, 2. BCDEK, 3. AMTUVWY, 4. HIOX, What defines the categories?, 11. Exactly how many elements of order 4 does the group in Example 1, have?, 12. Why is inversion [that is, f 1x, y2 1x, y2] not listed as one of, the four kinds of isometries in R2?, 13. Explain why inversion through a point in R3 cannot be realized by, a rotation in R3., 14. Reflection across a line L in R3 is the isometry that takes each, point Q to the point Q9 with the property that L is a perpendicular, bisector of the line segment joining Q and Q9. Describe a rotation, that has this same effect., 15. In R2, a rotation fixes a point; in R3, a rotation fixes a line. In R4,, what does a rotation fix? Generalize these observations to Rn., 16. Show that an isometry of a plane preserves angles., 17. Show that an isometry of a plane is completely determined by the, image of three noncollinear points., 18. Suppose that an isometry of a plane leaves three noncollinear, points fixed. Which isometry is it?, 19. Suppose that an isometry of a plane fixes exactly one point. What, type of isometry must it be?, 20. Suppose that A and B are rotations of 180° about the points a and b,, respectively. What is A followed by B? How is the composite motion related to the points a and b?, *Copyright © 1994. Reprinted with permission of the author and the publisher from, PARADE, December 11, 1994., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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468, , Special Topics, , References, 1. H. S. M. Coxeter, Introduction to Geometry, 2nd ed., New York: Wiley,, 1969., 2. Marvin Jay Greenberg, Euclidean and Non-Euclidean Geometries:, Development and History, 3rd ed., New York: W. H. Freeman, 1993., , Suggested Readings†, Lorraine Foster, “On the Symmetry Group of the Dodecahedron,” Mathematics Magazine 63 (1990): 106–107., It is shown that the group of rotations of a dodecahedron and the, group of rotations of an icosahedron are both A5., J. Rosen, Symmetry Discovered, Cambridge: Cambridge University Press,, 1975., This excellent book was written for first- or second-year college students. It includes sections on group theory, spatial symmetry, temporal, symmetry, and color symmetry, and chapters on symmetry in nature, and the uses of symmetry in science., Andrew Watson, “The Mathematics of Symmetry,” New Scientist, (October 1990): 45–50., This article discusses how chemists use group theory to understand, molecular structure and how physicists use it to study the fundamental, forces and particles., , Suggested Website, http://en.wikipedia.org/wiki/Symmetry, This website has a wealth of material about symmetry. Included are, essays, photos, links, and references., , †See, , also the Suggested Readings for Chapter 1., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Frieze Groups, 28 and Crystallographic, Groups, Symmetry, considered as a law of regular composition of structural objects,, is similar to harmony. More precisely, symmetry is one of its components,, while the other component is dissymmetry. In our opinion the whole, esthetics of scientific and artistic creativity lies in the ability to feel this, where others fail to perceive it., a. v. shubnikov and v. a. koptsik,, Symmetry in Science and Art, , The Frieze Groups, In this chapter, we discuss an interesting collection of infinite symmetry groups that arise from periodic designs in a plane. There are two, types of such groups. The discrete frieze groups are the plane symmetry, groups of patterns whose subgroups of translations are isomorphic, to Z. These kinds of designs are the ones used for decorative strips and, for patterns on jewelry, as illustrated in Figure 28.1. In mathematics,, familiar examples include the graphs of y 5 sin x, y 5 tan x, y 5 |sin x|,, and |y| 5 sin x. After we analyze the discrete frieze groups, we examine the discrete symmetry groups of plane patterns whose subgroups of, translations are isomorphic to Z % Z., In previous chapters, it was our custom to view two isomorphic, groups as the same group, since we could not distinguish between them, algebraically. In the case of the frieze groups, we will soon see that, although some of them are isomorphic as groups (that is, algebraically, the same), geometrically they are quite different. To emphasize this, difference, we will treat them separately. In each of the following, cases, the given pattern extends infinitely far in both directions., A proof that there are exactly seven types of frieze patterns is given in, the appendix to [6]., , 469, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , Symmetry in Science and Art by A. V. Shubnikov & V/A. Kopstik © 1974 Plenum Publishing Company, , 470, , Figure 28.1 Frieze patterns, , The symmetry group of pattern I (Figure 28.2) consists of translations only. Letting x denote a translation to the right of one unit (that, is, the distance between two consecutive R’s), we may write the symmetry group of pattern I as, F1 5 {xn | n [ Z}., , R, , R, , R, , R, , Figure 28.2 Pattern I, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , 471, , The group for pattern II (Figure 28.3), like that of pattern I, is infinitely cyclic. Letting x denote a glide-reflection, we may write the, symmetry group of pattern II as, F2 5 {xn | n [ Z}., , R, , R, , R, , R, , R R, , R, , Figure 28.3 Pattern II, , Notice that the translation subgroup of pattern II is just kx2l., The symmetry group for pattern III (Figure 28.4) is generated by a, translation x and a reflection y across the dashed vertical line. (There, are infinitely many axes of reflective symmetry, including those midway between consecutive pairs of opposite-facing R’s. Any one will, do.) The entire group (the operation is function composition) is, F3 5 {xnym | n [ Z, m 5 0 or 1}., , RR, , RR, , RR, , RR, , RR, , Figure 28.4 Pattern III, , Note that the two elements xy and y have order 2, they generate F3,, and their product (xy)y 5 x has infinite order. Thus, by Theorem 26.5,, F3 is the infinite dihedral group. A geometric fact about pattern III, worth mentioning is that the distance between consecutive pairs of vertical reflection axes is half the length of the smallest translation vector., In pattern IV (Figure 28.5), the symmetry group F4 is generated by a, translation x and a rotation y of 180° about a point p midway between, consecutive R’s (such a rotation is often called a half-turn). This group,, like F3, is also infinite dihedral. (Another rotation point lies between a, top and bottom R. As in pattern III, the distance between consecutive, points of rotational symmetry is half the length of the smallest translation vector.) Therefore,, F4 5 {xnym | n [ Z, m 5 0 or m 5 1}., , R, , p, , R, , R, , R, , R, , R, , R, , R, , Figure 28.5 Pattern IV, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , RR, , R, , R, , RR, , R, , RR, , R, , R, , R, , p, , Figure 28.6 Pattern V, , The symmetry group F5 for pattern V (Figure 28.6) is yet another, infinite dihedral group generated by a glide-reflection x and a rotation y, of 180° about the point p. Notice that pattern V has vertical reflection, symmetry xy. The rotation points are midway between the vertical reflection axes. Thus,, F5 5 {xnym | n [ Z, m 5 0 or m 5 1}., The symmetry group F6 for pattern VI (Figure 28.7) is generated by, a translation x and a horizontal reflection y. The group is, F6 5 {xnym | n [ Z, m 5 0 or m 5 1}., Note that, since x and y commute, F6 is not infinite dihedral. In fact, F6, is isomorphic to Z % Z2. Pattern VI is invariant under a glide-reflection, also, but in this case the glide-reflection is called trivial, since the axis, of the glide-reflection is also an axis of reflection. (Conversely, a glidereflection is nontrivial if its glide-axis is not an axis of reflective symmetry for the pattern.), , R, R, , R, R, , R, R, , R, R, , Figure 28.7 Pattern VI, , The symmetry group F7 of pattern VII (Figure 28.8) is generated by, a translation x, a horizontal reflection y, and a vertical reflection z. It is, isomorphic to the direct product of the infinite dihedral group and Z2., The product of y and z is a 180° rotation. Therefore,, F7 5 {x ny mz k | n [ Z, m 5 0 or m 5 1, k 5 0 or k 5 1}., , RR, R, , RR, R, R, , R, , RR, R, , RR, R, R, , R, , 472, , Figure 28.8 Pattern VII, , The preceding discussion is summarized in Figure 28.9. Figure 28.10, provides an identification algorithm for the frieze patterns., In describing the seven frieze groups, we have not explicitly said, how multiplication is done algebraically. However, each group element, corresponds to some isometry, so multiplication is the same as function, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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473, , 28 | Frieze Groups and Crystallographic Groups, , Pattern, x21, R, I, , Generators, e, R, , x22, R, , II, , R, , e, R, , 21, , x, x yx, RR, , x21, R, , IV, , R, y, , y e, RR, , xy x, RR, , e, R, , x, R, , xy, , 2, , 21, , x ye, RR, , R, , x y, xy x2, RR, , R, , V, , yx, x21, R, R, , e, R, R, , x, R, R, , 21, , y, , xy, , x21z x21, RR, R, , ze, RR, R, R, , xz x, RR, R, , 21, , yz y, , xyz xy, , x y, , R, , VII, , 21, , x yz x y, , R, , VI, , x = translation, , Z, , x = glide-reflection, , Z, , x = translation, y = vertical reflection, , D, , x = translation, y = rotation of 180°, , D, , x = glide-reflection, y = rotation of 180°, , D, , x = translation, y = horizontal reflection, , Z, , x, , R, , III, , 21, , R, , x2, R, , R, , 21, , x2, R, , x, R, , Group, isomorphism, class, , x = translation, y = horizontal reflection, z = vertical reflection, , D, , Z2, , Z2, , Figure 28.9 The seven frieze patterns and their groups of symmetries, , composition. Thus, we can always use the geometry to determine the, product of any particular string of elements., For example, we know that every element of F7 can be written in the, form x ny mz k. So, just for fun, let’s determine the appropriate values for, n, m, and k for the element g 5 x21yzxz. We may do this simply by, looking at the effect that g has on pattern VII. For convenience, we will, pick out a particular R in the pattern and trace the action of g one step, at a time. To distinguish this R, we enclose it in a shaded box. Also, we, draw the axis of the vertical reflection z as a dashed line segment. See, Figure 28.11., Now, comparing the starting position of the shaded R with its final, position, we see that x21yzxz 5 x22y. Exercise 7 suggests how one may, arrive at the same result through purely algebraic manipulation., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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I, R, yes, , R R, R R, R R, , VII, , R R, , R R, R R, , R R, , yes, , yes, , V, , R R, , Is there a, half-turn?, , III, , no, , R R, , R R, , VI, , R, , II, , no, Is there a, horizontal, reflection?, , yes, no, , R R, , Figure 28.10 Recognition chart for frieze patterns., Adapted from [6, p. 83].*, , R, R, R, R, , R, , R, , yes, , IV, , R, , R, , Is there a, half-turn?, , no, , no, , R, , Is there a horizontal reflection, or glide-reflection?, , R, , R R, , Is there a horizontal reflection?, , R, , R, , no, , R, , R, , yes, , R, , R, , R, , R, , Special Topics, , Is there a vertical reflection?, , 474, , *Adaptation of figure from Dorothy K. Washburn and Donald W. Crowe. Symmetries, of Culture: Theory and Practice of Plane Pattern Analysis. Copyright © 1988 by the, University of Washington Press. Used by permission., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , RR, R, , RR, R, R, , RR, R, , RR, R, R, , RR, R, , RR, R, R, , RR, R, , RR, R, R, , RR, R, , RR, R, , RR, R, , R, , R, , RR, R, , 475, , R, , z, , RR, R, , R, , R, , RR, R, R, , x, , RR, R, , R, , R, , RR, R, R, , z, , RR, R, , R, , R, , RR, R, R, , y, , RR, R, , R, , R, , R, , RR, R, R, , x⫺1, , RR, R, R, , RR, R, , RR, R, R, , R, , RR, R, R, , Figure 28.11, , The Crystallographic Groups, The seven frieze groups catalog all symmetry groups that leave a, design invariant under all multiples of just one translation. However,, there are 17 additional kinds of discrete plane symmetry groups that, arise from infinitely repeating designs in a plane. These groups are the, symmetry groups of plane patterns whose subgroups of translations are, isomorphic to Z % Z. Consequently, the patterns are invariant under, linear combinations of two linearly independent translations. These, 17 groups were first studied by 19th-century crystallographers and are, often called the plane crystallographic groups. Another term occasionally used for these groups is wallpaper groups., Our approach to the crystallographic groups will be geometric. It, is adapted from the excellent article by Schattschneider [5] and the, monograph by Crowe [1]. Our goal is to enable the reader to determine, which of the 17 plane symmetry groups corresponds to a given periodic pattern. We begin with some examples., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , Makoto Nakamura and Kevin Lee, , 476, , Figure 28.12 Fish3 by Makoto Nakamura, adapted by Kevin Lee. Design with, symmetry group p1 (disregarding shading). The inserted arrows are translation, vectors., , The simplest of the 17 crystallographic groups contains translations, only. In Figure 28.12, we present an illustration of a representative, pattern for this group (imagine the pattern repeated to fill the entire, plane). The crystallographic notation for it is p1. (This notation is explained in [5].), The symmetry group of the pattern in Figure 28.13 contains translations and glide-reflections. This group has no (nonzero) rotational or, reflective symmetry. The crystallographic notation for it is pg., Figure 28.14 has translational symmetry and threefold rotational, symmetry (that is, the figure can be rotated 120° about certain points, and be brought into coincidence with itself). The notation for this, group is p3., Representative patterns for all 17 plane crystallographic groups,, together with their notations, are given in Figures 28.15 and 28.16., Figure 28.17 uses a triangle motif to illustrate the 17 classes of symmetry patterns., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , 477, , Makoto Nakamura and Kevin Lee, , Figure 28.13 Fish5 by Makoto, Nakamura, adapted by Kevin, Lee. Design with symmetry, group pg (disregarding shading)., The solid arrow is the translation vector. The dashed arrows, are the glide-reflection vectors., , Makoto Nakamura and Kevin Lee, , Figure 28.14, Horses1 by Makoto Nakamura,, adapted by Kevin, Lee. Design with, symmetry group, p3 (disregarding, shading). The inserted arrows are, translation vectors., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , Symmetry in Science and Art by A. V. Shubnikov & V/A., Kopstik © 1974 Plenum Publishing Company, , 478, , Figure 28.15 The plane symmetry groups, , All designs in Figures 28.15 and 28.16 except pm, p3, and pg are, found in [2]. The designs for p3 and pg are based on elements of Chinese, lattice designs found in [2]; the design for pm is based on a weaving pattern from Hawaii, found in [3]., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Symmetry in Science and Art by A. V. Shubnikov & V/A., Kopstik © 1974 Plenum Publishing Company, , 28 | Frieze Groups and Crystallographic Groups, , Figure 28.16 The plane symmetry groups, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , 479
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480, , Special Topics, , Figure 28.17 The 17 plane periodic patterns formed using a triangle motif, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , 481, , Identification of Plane, Periodic Patterns, To decide which of the 17 classes any particular plane periodic pattern, belongs to, we may use the flowchart presented in Figure 28.18. This is, done by determining the rotational symmetry and whether or not the, pattern has reflection symmetry or nontrivial glide-reflection symmetry., These three pieces of information will narrow the list of candidates to at, most two. The final test, if necessary, is to determine the locations of the, centers of rotation., For example, consider the two patterns in Figure 28.19 generated in a, hockey stick motif. Both patterns have a smallest positive rotational symmetry of 120°; both have reflectional and nontrivial glide-reflectional, symmetry. Now, according to Figure 28.18, these patterns must be of, type p3m1 or p31m. But notice that the pattern on the left has all its threefold centers of rotation on the reflection axis, whereas in the pattern on, the right the points where the three blades meet are not on a reflection, axis. Thus, the left pattern is p3m1, and the right pattern is p31m., Table 28.1 (reproduced from [5, p. 443]) can also be used to determine the type of periodic pattern and contains two other features that, are often useful. A lattice of points of a pattern is a set of images of any, particular point acted on by the translation group of the pattern. A lattice unit of a pattern whose translation subgroup is generated by u and, v is a parallelogram formed by a point of the pattern and its image, under u, v, and u 1 v. The possible lattices for periodic patterns in a, plane, together with lattice units, are shown in Figure 28.20. A generating region (or fundamental region) of a periodic pattern is the smallest, portion of the lattice unit whose images under the full symmetry group of, the pattern cover the plane. Examples of generating regions for the, patterns represented in Figures 28.12, 28.13, and 28.14 are given in, Figure 28.21. In Figure 28.21, the portion of the lattice unit with vertical, bars is the generating region. The only symmetry pattern in which the, lattice unit and the generating region coincide is the p1 pattern illustrated, in Figure 28.12. Table 28.1 tells what proportion of the lattice unit constitutes the generating region of each plane periodic pattern., Notice that Table 28.1 reveals that the only possible n-fold rotational, symmetries occur when n 5 1, 2, 3, 4, and 6. This fact is commonly, called the crystallographic restriction. The first proof of this was given, by the Englishman W. Barlow over 100 years ago. The information in, Table 28.1 can also be used in reverse to create patterns with a specific, symmetry group. The patterns in Figure 28.19 were made in this way., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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What is the, smallest angle, of rotation?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , (two of these), Is there a, reflection?, , (three of these), Is there a, reflection?, , (three of these), Is there a, reflection?, , (five of these), Is there a, reflection?, , (four of these), Is there a, reflection?, , no, , yes, , no, , yes, , no, , yes, , no, , yes, , no, , yes, , p6, , p6m, , p3, , Are all threefold, centers on reflection, axes?, , p4, , no, , yes, , no, , yes, , yes pgg, no p2, , Is there a glide-reflection?, Are there reflection axes in, four directions?, , no, , yes, , no, , yes, , no, , yes, , Are there reflection axes in, two directions?, , Is there a glide-reflection?, , Is there a glide-reflection, whose glide-axis is not a, reflection axis?, , Are all, rotation, centers on, reflection, axes?, , p31m, , p3m1, , p4m, p4g, , pmg, , p1, , pg, , pm, , cm, , Figure 28.18 Identification flowchart for symmetries of plane periodic patterns, , 60°, , 120°, , 90°, , 180°, , none, , no, , yes, , cmm, , pmm, , 482, Special Topics
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28 | Frieze Groups and Crystallographic Groups, , 483, , p31m, , p3m1, , Figure 28.19 Patterns generated in a hockey stick motif, , Rectangular, , Parallelogram, , Square, , Rhombic, , Hexagonal, (Equilateral triangles), , Figure 28.20 Possible lattices for plane periodic patterns, , In sharp contrast to the situation for finite symmetry groups, the transition from two-dimensional crystallographic groups to three-dimensional, crystallographic groups introduces a great many more possibilities, since, the motif is repeated indefinitely by three independent translations. Indeed,, there are 230 three-dimensional crystallographic groups (often called space, groups). These were independently determined by Fedorov, Schönflies, and, Barlow in the 1890s. David Hilbert, one of the leading mathematicians of, the 20th century, focused attention on the crystallographic groups in his, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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484, , Special Topics, , Table 28.1 Identification Chart for Plane Periodic Patternsa, , Type, , Lattice, , Highest, Order of, Rotation, , Reflections, , Nontrivial, GlideGenerating, Reflections, Region, , p1, p2, pm, pg, cm, pmm, pmg, , Parallelogram, Parallelogram, Rectangular, Rectangular, Rhombic, Rectangular, Rectangular, , 1, 2, 1, 1, 1, 2, 2, , No, No, Yes, No, Yes, Yes, Yes, , No, No, No, Yes, Yes, No, Yes, , 1 unit, 1, 2 unit, 1, 2 unit, 1, 2 unit, 1, 2 unit, 1, 4 unit, 1, 4 unit, , pgg, cmm, , Rectangular, Rhombic, , 2, 2, , No, Yes, , Yes, Yes, , 1, 4, 1, 4, , unit, unit, , p4, p4m, , Square, Square, , 4, 4, , No, Yes, , No, Yes, , 1, 4, 1, 8, , unit, unit, , p4g, , Square, , 4, , Yes, , Yes, , 1, 8, , unit, , p3, p3m1, , Hexagonal, Hexagonal, , 3, 3, , No, Yes, , No, Yes, , 1, 3, 1, 6, , unit, unit, , p31m, , Hexagonal, , 3, , Yes, , Yes, , 1, 6, , unit, , p6, p6m, , Hexagonal, Hexagonal, , 6, 6, , No, Yes, , No, Yes, , 1, 6 unit, 1, 12 unit, , Helpful, Distinguishing, Properties, , Parallel reflection, axes, Perpendicular, reflection axes, Fourfold centers, on reflection, axes, Fourfold centers, not on, reflection axes, All threefold, centers on, reflection axes, Not all threefold, centers on, reflection axes, , aA, , rotation through an angle of 360°/n is said to have order n. A glide-reflection is nontrivial if its glide-axis is not, an axis of reflective symmetry for the pattern., , famous lecture in 1900 at the International Congress of Mathematicians in, Paris. One of 23 problems he posed was whether or not the number of, crystallographic groups in n dimensions is always finite. This was answered affirmatively by L. Bieberbach in 1910. We mention in passing that, in four dimensions, there are 4783 symmetry groups for infinitely repeating patterns., As one might expect, the crystallographic groups are fundamentally, important in the study of crystals. In fact, a crystal is defined as a rigid, body in which the component particles are arranged in a pattern that, repeats in three directions (the repetition is caused by the chemical, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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485, , Makoto Nakamura and Kevin Lee, , Makoto Nakamura and Kevin Lee, , Makoto Nakamura and Kevin Lee, , 28 | Frieze Groups and Crystallographic Groups, , Figure 28.21 A lattice unit and generating region for the patterns in, Figures 28.12, 28.13, and 28.14. Generating regions are shaded with bars., , bonding). A grain of salt and a grain of sugar are two examples of, common crystals. In crystalline materials, the motif units are atoms,, ions, ionic groups, clusters of ions, or molecules., Perhaps it is fitting to conclude this chapter by recounting two, episodes in the history of science in which an understanding of symmetry groups was crucial to a great discovery. In 1912, Max von Laue, a, young German physicist, hypothesized that a narrow beam of x-rays directed onto a crystal with a photographic film behind it would be, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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486, , Special Topics, , deflected (the technical term is “diffracted”) by the unit cell (made up of, atoms or ions) and would show up on the film as spots. (See Figure 1.3.), Shortly thereafter, two British scientists, Sir William Henry Bragg and, his 22-year-old son William Lawrence Bragg, who was a student, noted, that von Laue’s diffraction spots, together with the known information, about crystallographic space groups, could be used to calculate the shape, of the internal array of atoms. This discovery marked the birth of modern mineralogy. From the first crystal structures deduced by the Braggs, to the present, x-ray diffraction has been the means by which the internal, structures of crystals are determined. Von Laue was awarded the Nobel, Prize in physics in 1914, and the Braggs were jointly awarded the, Nobel Prize in physics in 1915., Our second episode took place in the early 1950s, when a handful of, scientists were attempting to learn the structure of the DNA molecule—, the basic genetic material. One of these was a graduate student named, Francis Crick; another was an x-ray crystallographer, Rosalind Franklin., On one occasion, Crick was shown one of Franklin’s research reports, and an x-ray diffraction photograph of DNA. At this point, we let Horace, Judson [4, pp. 165–166], our source, continue the story., Crick saw in Franklin’s words and numbers something just as important,, indeed eventually just as visualizable. There was drama, too: Crick’s, insight began with an extraordinary coincidence. Crystallographers distinguish 230 different space groups, of which the face-centered monoclinic, cell with its curious properties of symmetry is only one—though in biological substances a fairly common one. The principal experimental subject of, Crick’s dissertation, however, was the x-ray diffraction of the crystals of a, protein that was of exactly the same space group as DNA. So Crick saw at, once the symmetry that neither Franklin nor Wilkins had comprehended,, that Perutz, for that matter, hadn’t noticed, that had escaped the theoretical, crystallographer in Wilkins’ lab, Alexander Stokes—namely, that the, molecule of DNA, rotated a half turn, came back to congruence with itself., The structure was dyadic, one half matching the other half in reverse., , This was a crucial fact. Shortly thereafter, James Watson and Crick, built an accurate model of DNA. In 1962, Watson, Crick, and Maurice, Wilkins received the Nobel Prize in medicine and physiology for their, discovery. The opinion has been expressed that, had Franklin correctly, recognized the symmetry of the DNA molecule, she might have been, the one to unravel the mystery and receive the Nobel Prize [4, p. 172]., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , 487, , Exercises, You can see a lot just by looking., yogi berra, , 1., 2., 3., 4., 5., 6., 7., , 8., 9., 10., 11., , Show that the frieze group F6 is isomorphic to Z % Z2., How many nonisomorphic frieze groups are there?, In the frieze group F7, write x2yzxz in the form x ny mz k., In the frieze group F7, write x23zxyz in the form x ny mz k., In the frieze group F7, show that yz 5 zy and xy 5 yx., In the frieze group F7, show that zxz 5 x21., Use the results of Exercises 5 and 6 to do Exercises 3 and 4, through symbol manipulation only (that is, without referring to the, pattern). (This exercise is referred to in this chapter.), Prove that in F7 the cyclic subgroup generated by x is a normal, subgroup., Quote a previous result that tells why the subgroups kx, yl and, kx, zl must be normal in F7., Look up the word frieze in an ordinary dictionary. Explain why the, frieze groups are appropriately named., Determine which of the seven frieze groups is the symmetry group, of each of the following patterns., , b., , c., , d., , Symmetry in Science and Art by A. V. Shubnikov & V/A. Kopstik © 1974, Plenum Publishing Company, , a., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Special Topics, , e., , Symmetry in Science and Art by A. V., Shubnikov & V/A. Kopstik © 1974 Plenum, Publishing Company, , f., , 12. Determine the frieze group corresponding to each of the following, patterns., a. y 5 sin x, b. y 5 |sin x|, c. |y| 5 sin x, d. y 5 tan x, e. y 5 csc x, 13. Determine the symmetry group of the tessellation of the plane exemplified by the brickwork shown., , 14. Determine the plane symmetry group for each of the patterns in, Figure 28.17., 15. Determine which of the 17 crystallographic groups is the symmetry group of each of the following patterns., a., , b., , c., , d., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , Symmetry in Science and Art by A. V. Shubnikov & V/A. Kopstik © 1974, Plenum Publishing Company, , 488
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28 | Frieze Groups and Crystallographic Groups, , 489, , 16. In the following figure, there is a point labeled 1. Let a be the, translation of the plane that carries the point labeled 1 to the point, labeled a, and let b be the translation of the plane that carries the, point labeled 1 to the point labeled b. The image of 1 under the, composition of a and b is labeled ab. In the corresponding fashion, label the remaining points in the figure in the form aib j., β, 1, , αβ, α, , 17. The patterns made by automobile tire treads in the snow are frieze, patterns. An extensive study of automobile tires revealed that only, five of the seven frieze patterns occur. Speculate on which two patterns do not occur and give a possible reason why they do not., 18. Locate a nontrivial glide-reflection axis of symmetry in the cm pattern in Figure 28.16., 19. Determine which of the frieze groups is the symmetry group of, each of the following patterns., a. ? ? ? D V, DDD, V???, b. ? ? ? V V ? ? ?, c. ? ? ? L L L L ? ? ?, d. ? ? ? V V V V ? ? ?, e. ? ? ? N N N N ? ? ?, f. ? ? ? H H H H ? ? ?, g. ? ? ? L L ? ? ?, 20. Locate a nontrivial glide-reflection axis of symmetry in the pattern, third from the left in the bottom row in Figure 28.17., L, , L, , References, 1. Donald Crowe, Symmetry, Rigid Motions, and Patterns, Arlington, Va.:, COMAP, 1986., 2. Daniel S. Dye, A Grammar of Chinese Lattice, Harvard-Yenching Institute, Monograph Series, vol. VI, Cambridge, Mass.: Harvard University Press,, 1937. (Reprinted as Chinese Lattice Designs, New York: Dover, 1974.), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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490, , Special Topics, , 3. Owen Jones, The Grammar of Ornament, New York: Van Nostrand Reinhold, 1972. (Reproduction of the same title, first published in 1856 and, reprinted in 1910 and 1928.), 4. Horace Freeland Judson, The Eighth Day of Creation, New York: Simon, and Schuster, 1979., 5. D. Schattschneider, “The Plane Symmetry Groups: Their Recognition, and Notation,” The American Mathematical Monthly 85 (1978): 439–450., 6. D. K. Washburn and D. W. Crowe, Symmetries of Culture: Theory and, Practice of Plane Pattern Analysis, Seattle: University of Washington, Press, 1988., , Suggested Readings, S. Garfunkel et al., For All Practical Purposes, 9th ed., New York: W. H., Freeman, 2012., This book has a well-written, richly illustrated chapter on symmetry in, art and nature., W. G. Jackson, “Symmetry in Automobile Tires and the Left-Right Problem,” Journal of Chemical Education 69 (1992): 624–626., This article uses automobile tires as a tool for introducing and explaining the symmetry terms and concepts important in chemistry., C. MacGillivray, Fantasy and Symmetry—The Periodic Drawings of, M. C. Escher, New York: Harry N. Abrams, 1976., This is a collection of Escher’s periodic drawings together with a mathematical discussion of each one., D. Schattschneider, Visions of Symmetry, New York: Harry Abrams, 2004., A loving, lavish, encyclopedic book on the drawings of M. C. Escher., H. von Baeyer, “Impossible Crystals,” Discover 11 (2) (1990): 69–78., This article tells how the discovery of nonperiodic tilings of the plane led, to the discovery of quasicrystals. The x-ray diffraction patterns of quasicrystals exhibit fivefold symmetry—something that had been thought to, be impossible., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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28 | Frieze Groups and Crystallographic Groups, , 491, , Suggested Websites, http://www.mcescher.com/, This is the official website for the artist M. C. Escher. It features many of, his prints and most of his 136 symmetry drawings., , http://britton.disted.camosun.bc.ca/jbsymteslk.htm, This spectacular website on symmetry and tessellations has numerous activities and links to many other sites on related topics. It is a wonderful, website for K–12 teachers and students., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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M. C. Escher, , The Granger Collection, New York, , I never got a pass mark in math. The funny, thing is I seem to latch on to mathematical, theories without realizing what is happening., m. c. escher, , M. C. Escher was born on June 17, 1898, in, the Netherlands. His artistic work prior to, 1937 was dominated by the representation, of visible reality, such as landscapes and, buildings. Gradually, he became less and, less interested in the visible world and became increasingly absorbed in an inventive, approach to space. He studied the abstract, space-filling patterns used in the Moorish, mosaics in the Alhambra in Spain. He also, studied the mathematician George Pólya’s, paper on the 17 plane crystallographic, groups. Instead of the geometric motifs used, by the Moors and Pólya, Escher preferred to, use animals, plants, or people in his spacefilling prints., Escher was fond of incorporating various, mathematical ideas into his works. Among, these are infinity, Möbius bands, stellations,, , deformations, reflections, Platonic solids,, spirals, and the hyperbolic plane., Although Escher originals are now quite, expensive, it was not until 1951 that he derived a significant portion of his income, from his prints. Today, Escher is widely, known and appreciated as a graphic artist., His prints have been used to illustrate ideas, in hundreds of scientific works. Despite this, popularity among scientists, however,, Escher has never been held in high esteem, in traditional art circles. Escher died on, March 27, 1972, in the Netherlands., To find more information about Escher, and his art, visit the official website of M. C., Escher:, http://www.mcescher.com/, , 492, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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George Pólya, , G. L. Alexanderson/Santa Clara University, , Thank you, Professor Pólya, for all your, beautiful contributions to mathematics, to, science, to education, and to humanity., A toast from frank harary on the, occasion of Pólya’s 90th birthday, , George Pólya was born in Budapest,, Hungary, on December 13, 1887. He received, a teaching certificate from the University of, Budapest in languages before turning to philosophy, mathematics, and physics., In 1912, he was awarded a Ph.D. in mathematics. Horrified by Hitler and World War, II, Pólya came to the United States in 1940., After two years at Brown University, he went, to Stanford University, where he remained, until his death in 1985 at the age of 97., In 1924, Pólya published a paper in a crystallography journal in which he classified the, plane symmetry groups and provided a fullpage illustration of the corresponding 17 periodic patterns. B. G. Escher, a geologist, sent a, copy of the paper to his artist brother, M. C., Escher, who used Pólya’s black-and-white, geometric patterns as a guide for making his, own interlocking colored patterns featuring, birds, reptiles, and fish., , Pólya contributed to many branches of, mathematics, and his collected papers fill four, large volumes. Pólya is also famous for his, books on problem solving and for his teaching. One of his books has sold more than, 1,000,000 copies. The Society for Industrial, and Applied Mathematics, the London Mathematical Society, and the Mathematical Association of America have prizes named after, Pólya., Pólya taught courses and lectured around, the country into his 90s. He never learned to, drive a car and took his first plane trip at, age 75. He was married for 67 years and had, no children., For more information about Pólya, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 493, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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John H. Conway, He’s definitely world class, yet he has this, kind of childlike enthusiasm., , Princeton University, , ronald graham, , John H. Conway ranks among the most, original and versatile contemporary mathematicians. Conway was born in Liverpool,, England, on December 26, 1937, and grew, up in a rough neighborhood. As a youngster,, he was often beaten up by older boys and, did not do well in high school. Nevertheless,, his mathematical ability earned him a scholarship to Cambridge University, where he, excelled., A pattern that uses repeated shapes to, cover a flat surface without gaps or overlaps, is called a tiling. In 1975, Oxford physicist, Roger Penrose invented an important new, way of tiling the plane with two shapes., Unlike patterns whose symmetry group is, one of the 17 plane crystallographic groups,, Penrose patterns can be neither translated nor, rotated to coincide with themselves. Many of, the remarkable properties of the Penrose patterns were discovered by Conway. In 1993,, , Conway discovered a new prism that can be, used to fill three-dimensional space without, gaps or overlaps., Conway has made many significant contributions to number theory, group theory,, game theory, knot theory, and combinatorics. Among his most important discoveries, are three simple groups, which are now, named after him. (Simple groups are the, basic building blocks of all groups.) Conway, is fascinated by games and puzzles. He invented the game Life and the game Sprouts., Conway has received numerous prestigious, honors. In 1987 he joined the faculty at, Princeton University, where his title is John, von Neumann Distinguished Professor of, Mathematics., For more information about Conway,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 494, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Symmetry, 29, and Counting, Let us pause to slake our thirst one last time at symmetry’s bubbling spring., timothy ferris, Coming of Age in the Milky Way, , Motivation, Permutation groups naturally arise in many situations involving symmetric designs or arrangements. Consider, for example, the task of coloring the six vertices of a regular hexagon so that three are black, 6, and three are white. Figure 29.1 shows the a b 5 20 possibilities., 3, However, if these designs appeared on one side of hexagonal ceramic, tiles, it would be nonsensical to count the designs shown in Figure, 29.1(a) as different, since all six designs shown there can be obtained, from one of them by rotating. (A manufacturer would make only one of, the six.) In this case, we say that the designs in Figure 29.1(a) are, equivalent under the group of rotations of the hexagon. Similarly, the, designs in Figure 29.1(b) are equivalent under the group of rotations, as, are the designs in Figure 29.1(c) and those in Figure 29.1(d). And, since, no design from Figure 29.1(a)–(d) can be obtained from a design in a, different part by rotation, we see that the designs within each part of the, figure are equivalent to each other but nonequivalent to any design in, another part of the figure. However, the designs in Figure 29.1(b) and, (c) are equivalent under the dihedral group D6, since the designs in, Figure 29.1(b) can be reflected to yield the designs in Figure 29.1(c)., For example, for purposes of arranging three black beads and three, white beads to form a necklace, the designs shown in Figure 29.1(b) and, (c) would be considered equivalent., In general, we say that two designs (arrangements of beads) A and B, are equivalent under a group G of permutations of the arrangements if, there is an element f in G such that f(A) 5 B. That is, two designs are, equivalent under G if they are in the same orbit of G. It follows, then,, , 495, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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496, , Special Topics, , (a), , (c), , (d), , (b), , Figure 29.1, , that the number of nonequivalent designs under G is simply the number, of orbits of designs under G. (The set being permuted is the set of all, possible designs or arrangements.), Notice that the designs in Figure 29.1 divide into four orbits under, the group of rotations but only three orbits under the group D6, since, the designs in Figure 29.1(b) and (c) form a single orbit under D6. Thus,, we could obtain all 20 tile designs from just four tiles, but we could, obtain all 20 necklaces from just three of them., , Burnside’s Theorem, Although the problems we have just posed are simple enough to, solve by observation, more complicated ones require a more sophisticated approach. Such an approach was provided by Georg Frobenius, in 1887. Frobenius’s theorem did not become widely known until it, appeared in the classic book on group theory by William Burnside, in 1911. By an accident of history, Frobenius’s theorem has come to, be known as Burnside’s Theorem. Before stating this theorem, we, recall some notation introduced in Chapter 7 and introduce new, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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29 | Symmetry and Counting, , 497, , notation. If G is a group of permutations on a set S and i [ S, then, s t a b G( i ) 5 { f [ G | f ( i ) 5 i } a n d o r b G( i ) 5 { f ( i ) |, f [ G}. For any set X, we use |X| to denote the number of elements in X., Definition Elements Fixed by F, , For any group G of permutations on a set S and any f in G, we let, fix(f) 5 {i [ S | f(i) 5 i}. This set is called the elements fixed by f (or, more simply, “fix of f”)., , Theorem 29.1 Burnside's Theorem, If G is a finite group of permutations on a set S, then the number, of orbits of elements of S under G is, , 1, a 0 fix1f2 0 ., 0 G 0 f[G, PROOF Let n denote the number of pairs (f, i), with f [ G, i [ S,, and f(i) 5 i. We begin by counting these pairs in two ways. First, for, each particular f in G, the number of such pairs is exactly |fix(f)|. So,, n a 0 fix1f2 0 ., , (1), , f[G, , Second, for each particular i in S, observe that |stabG(i)| is exactly the, number of pairs (f, i) for which f(i) 5 i. So,, n a 0 stabG 1i2 0 ., , (2), , i[S, , It follows from Exercise 43 in Chapter 7 that if s and t are in the same, orbit of G, then orbG(s) 5 orbG(t), and thus by the Orbit-Stabilizer Theorem (Theorem 7.3) we have |stabG(s)| 5 |G|/|orbG(s)| 5 |G|/|orbG(t)| 5, |stabG(t)|. So, if we choose s [ S and sum over orbG(s), we have, a 0 stabG 1t2 0 0 orbG 1s2 0 0 stabG 1s2 0 0 G 0 ., , t[orbG1s2, , (3), , Finally, by summing over all the elements of G, one orbit at a time, it, follows from Equations (1), (2), and (3) that, ., a 0 fix1f2 0 a 0 stab1i2 0 0 G 0 1number of orbits2,, , f[G, , i[S, , and the result follows., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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498, , Special Topics, , Applications, To illustrate how to apply Burnside’s Theorem, let us return to the ceramic, tile and necklace problems. In the case of counting hexagonal tiles with, three black vertices and three white vertices, the objects being permuted, are the 20 possible designs, whereas the group of permutations is the, group of six rotational symmetries of a hexagon. Obviously, the identity, fixes all 20 designs. We see from Figure 29.1 that rotations of 60°, 180°,, or 300° fix none of the 20 designs. Finally, Figure 29.2 shows fix(f) for, the rotations of 120° and 240°. These data are collected in Table 29.1., , Figure 29.2 Tile designs fixed by 120°, rotation and 240° rotation, Table 29.1, Number of Designs, Fixed by Element, , Element, Identity, Rotation of 60°, Rotation of 120°, Rotation of 180°, Rotation of 240°, Rotation of 300°, , 20, 0, 2, 0, 2, 0, , So, applying Burnside’s Theorem, we obtain the number of orbits, under the group of rotations as, 1, (20 1 0 1 2 1 0 1 2 1 0) 5 4., 6, Now let’s use Burnside’s Theorem to count the number of necklace, arrangements consisting of three black beads and three white beads. (For, the purposes of analysis, we may arrange the beads in the shape of a regular hexagon.) For this problem, two arrangements are equivalent if they, are in the same orbit under D6. Figure 29.3 shows the arrangements fixed, , Figure 29.3 Bead arrangements fixed, by the reflection across a diagonal, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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499, , 29 | Symmetry and Counting, , Table 29.2, , Type of Element, Identity, Rotation of order 2 (180°), Rotation of order 3 (120° or 240°), Rotation of order 6 (60° or 300°), Reflection across diagonal, Reflection across side bisector, , Number of, Elements, of This, Type, , Number of, Arrangements, Fixed by Type, of Element, , 1, 1, 2, 2, 3, 3, , 20, 0, 2, 0, 4, 0, , by a reflection across a diagonal. Table 29.2 summarizes the information, needed to apply Burnside’s Theorem., So, there are, 1, (1 ? 20 1 1 ? 0 1 2 ? 2 1 2 ? 0 1 3 ? 4 1 3 ? 0) 5 3, 12, nonequivalent ways to string three black beads and three white beads, on a necklace., Now that we have gotten our feet wet on a few easy problems, let’s, try a more difficult one. Suppose that we have the colors red (R), white, (W), and blue (B) that can be used to color the edges of a regular tetrahedron (see Figure 5.1). First, observe that there are 36 5 729 colorings, without regard to equivalence. How shall we decide when two colorings, of the tetrahedron are nonequivalent? Certainly, if we were to pick up a, tetrahedron colored in a certain manner, rotate it, and put it back down,, we would think of the tetrahedron as being positioned differently rather, than as being colored differently (just as if we picked up a die labeled in, the usual way and rolled it, we would not say that the die is now differently labeled). So, our permutation group for this problem is just the, group of 12 rotations of the tetrahedron shown in Figure 5.1 and is isomorphic to A4. (The group consists of the identity; eight elements of, order 3, each of which fixes one vertex; and three elements of order 2,, each of which fixes no vertex.) Every rotation permutes the 729 colorings, and to apply Burnside’s Theorem we must determine the size of, fix(f) for each of the 12 rotations of the group., Clearly, the identity fixes all 729 colorings. Next, consider the element (234) of order 3, shown in the bottom row, second from the left in, Figure 5.1. Suppose that a specific coloring is fixed by this element, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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500, , Special Topics, , Table 29.3 Nine Colorings Fixed by (234), , Edge, 12, 13, 14, 23, 34, 24, , Colorings, R, R, R, R, R, R, , R, R, R, W, W, W, , R, R, R, B, B, B, , W, W, W, W, W, W, , W, W, W, R, R, R, , W, W, W, B, B, B, , B, B, B, B, B, B, , B, B, B, R, R, R, , B, B, B, W, W, W, , (that is, the tetrahedron appears to be colored the same before and after, this rotation). Since (234) carries edge 12 to edge 13, edge 13 to edge, 14, and edge 14 to edge 12, these three edges must agree in color (edge, ij is the edge joining vertex i and vertex j). The same argument shows, that the three edges 23, 34, and 42 also must agree in color. So,, |fix(234)| 5 32, since there are three choices for each of these two sets, of three edges. The nine columns in Table 29.3 show the possible colorings of the two sets of three edges. The analogous analysis applies to, the other seven elements of order 3., Now consider the rotation (12)(34) of order 2. (See the second tetrahedron in the top row in Figure 5.1.) Since edges 12 and 34 are fixed,, they may be colored in any way and will appear the same after the rotation (12)(34). This yields 3 ? 3 choices for those two edges. Since edge, 13 is carried to edge 24, these two edges must agree in color. Similarly,, edges 23 and 14 must agree. So, we have three choices for the pair of, edges 13 and 24 and three choices for the pair of edges 23 and 14. This, means that we have 3 ? 3 ? 3 ? 3 ways to color the tetrahedron that will, be equivalent under (12)(34). (Table 29.4 gives the complete list of 81, colorings.) So, |fix((12)(34))| 5 34, and the other two elements of order, 2 yield the same results., Now that we have analyzed the three types of group elements, we, can apply Burnside’s Theorem. In particular, the number of distinct, Table 29.4 81 Colorings Fixed by (12)(34) (X and Y can be any of R, W, and B), , Edge, 12, 34, 13, 24, 23, 14, , Colorings, X, Y, R, R, R, R, , X, Y, R, R, W, W, , X, Y, R, R, B, B, , X, Y, W, W, W, W, , X, Y, W, W, R, R, , X, Y, W, W, B, B, , X, Y, B, B, B, B, , X, Y, B, B, R, R, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , X, Y, B, B, W, W
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29 | Symmetry and Counting, , 501, , colorings of the edges of a tetrahedron with three colors is, 1, (1 ? 36 1 8 ? 32 1 3 ? 34) 5 87., 12, Surely it would be a difficult task to solve this problem without Burnside’s Theorem., Just as surely, you are wondering who besides mathematicians are interested in counting problems such as the ones we have discussed. Well,, chemists are. Indeed, one set of benzene derivatives can be viewed as, six carbon atoms arranged in a hexagon with one of the three radicals, NH2, COOH, or OH attached at each carbon atom. See Figure 29.4 for, one example., OH, , C, , COOH, C, , C, COOH, , COOH, C, , C, C, , OH, , COOH, , Figure 29.4 A benzene derivative, , So Burnside’s Theorem enables a chemist to determine the number of, benzene molecules (see Exercise 4). Another kind of molecule considered by chemists is visualized as a regular tetrahedron with a carbon, atom at the center and any of the four radicals HOCH2 (hydroxymethyl),, C2H5 (ethyl), Cl (chlorine), or H (hydrogen) at the four vertices. Again,, the number of such molecules can be easily counted using Burnside’s, Theorem., , Group Action, Our informal approach to counting the number of objects that are considered nonequivalent can be made formal as follows. If G is a group, and S is a set of objects, we say that G acts on S if there is a homomorphism g from G to sym(S), the group of all permutations on S. (The, homomorphism is sometimes called the group action.) For convenience, we denote the image of g under g as gg. Then two objects x and, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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502, , Special Topics, , y in S are viewed as equivalent under the action of G if and only if, gg(x) 5 y for some g in G. Notice that when g is one-to-one, the elements of G may be regarded as permutations on S. On the other hand,, when g is not one-to-one, the elements of G may still be regarded as, permutations on S, but there are distinct elements g and h in G such, that gg and gh induce the same permutation on S [that is, gg(x) 5 gh(x), for all x in S]. Thus, a group acting on a set is a natural generalization, of the permutation group concept., As an example of group action, let S be the two diagonals of a square, and let G be D4, the group of symmetries of the square. Then gR0, gR180,, gD, gD9 are the identity; gR90, gR270, gH, gV interchange the two diagonals;, and the mapping g S gg from D4 to sym(S) is a group homomorphism., As a second example, note that GL(n, F), the group of invertible n 3 n, matrices with entries from a field F, acts on the set S of n 3 1 column, vectors with entries from F by multiplying the vectors on the left by the, matrices. In this case, the mapping g S gg from GL(n, F) to sym(S) is a, one-to-one homomorphism., We have used group actions several times in this text without calling, them that. The proof of Cayley’s Theorem ( Theorem 6.1) has a group G, acting on the elements of G; the proofs of Sylow’s Second Theorem and, Third Theorem ( Theorems 24.4 and 24.5) have a group acting on the set, of conjugates of a Sylow p-subgroup; and the proof of the Generalized, Cayley Theorem ( Theorem 25.3) has G acting on the left cosets of a, subgroup H., , Exercises, The greater the difficulty, the more glory in surmounting it., epicurus, , 1. Determine the number of ways in which the four corners of a, square can be colored with two colors. (It is permissible to use a, single color on all four corners.), 2. Determine the number of different necklaces that can be made using 13 white beads and 3 black beads., 3. Determine the number of ways in which the vertices of an equilateral triangle can be colored with five colors so that at least two colors are used., 4. A benzene molecule can be modeled as six carbon atoms arranged, in a regular hexagon in a plane. At each carbon atom, one of three, radicals NH 2, COOH, or OH can be attached. How many such, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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29 | Symmetry and Counting, , 5., 6., , 7., , 8., , 9., 10., 11., , 12., 13., , 14., , 15., , 503, , compounds are possible? (Make no distinction between single and, double bonds between the atoms.), Suppose that in Exercise 4 we permit only NH2 and COOH for the, radicals. How many compounds are possible?, Determine the number of ways in which the faces of a regular, dodecahedron (regular 12-sided solid) can be colored with three, colors., Determine the number of ways in which the edges of a square can, be colored with six colors so that no color is used on more than, one edge., Determine the number of ways in which the edges of a square can, be colored with six colors with no restriction placed on the number, of times a color can be used., Determine the number of different 11-bead necklaces that can be, made using two colors., Determine the number of ways in which the faces of a cube can be, colored with three colors., Suppose a cake is cut into six identical pieces. How many ways can, we color the cake with n colors assuming that each piece receives, one color?, How many ways can the five points of a five-pointed crown be, painted if three colors of paint are available?, Let G be a finite group and let sym(G) be the group of all permutations on G. For each g in G, let fg denote the element of sym(G) defined by fg(x) 5 gxg21 for all x in G. Show that G acts on itself under, the action g S fg. Give an example in which the mapping g S fg is, not one-to-one., Let G be a finite group, let H be a subgroup of G, and let S be the, set of left cosets of H in G. For each g in G, let gg denote the element of sym(S) defined by gg(xH) 5 gxH. Show that G acts on S, under the action g S gg., For a fixed square, let L1 be the perpendicular bisector of the top, and bottom of the square and let L2 be the perpendicular bisector of, the left and right sides. Show that D4 acts on {L1, L2} and determine the kernel of the mapping g S gg., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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504, , Special Topics, , Suggested Readings, Norman Biggs, Discrete Mathematics, Oxford: Clarendon Press, 1989., Chapter 20 of this book presents a more detailed treatment of the, subject of symmetry and counting., Doris Schattschneider, “Escher’s Combinatorial Patterns,” Electronic, Journal of Combinatorics 4(2) (1997): R17., This article discusses a combinatorial problem concerning generating, periodic patterns that the artist M. C. Escher posed and solved in an, algorithmic way. The problem can also be solved by using Burnside’s, Theorem. The article can be downloaded free from the website http://, www.combinatorics.org/, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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By permission of the Masters & Fellows of Pembroke, College in the University of Oxford, , William Burnside, In one of the most abstract domains of, thought, he [Burnside] has systematized, and amplified its range so that, there, his, work stands as a landmark in the widening, expanse of knowledge. Whatever be the, estimate of Burnside made by posterity,, contemporaries salute him as a Master, among the mathematicians of his own, generation., a. r. forsyth, , William Burnside was born on July 2,, 1852, in London. After graduating from, Cambridge University in 1875, Burnside, was appointed lecturer at Cambridge,, where he stayed until 1885. He then accepted a position at the Royal Naval College at Greenwich and spent the rest of his, career in that post., Burnside wrote more than 150 research, papers in many fields. He is best remembered, however, for his pioneering work in, group theory and his classic book Theory of, Groups, which first appeared in 1897. Because of Burnside’s emphasis on the abstract, approach, many consider him to be the first, pure group theorist., One mark of greatness in a mathematician is the ability to pose important and, challenging problems—problems that open, up new areas of research for future generations. Here, Burnside excelled. It was he, , who first conjectured that a group G of odd, order has a series of normal subgroups,, G 5 G 0 $ G 1 $ G 2 $ ? ? ? $ G n 5 {e},, such that Gi /Gi11 is Abelian. This extremely, important conjecture was finally proved, more than 50 years later by Feit and Thompson in a 255-page paper (see Chapter 25 for, more on this). In 1994, Efim Zelmanov, received the Fields Medal for his work on a, variation of one of Burnside’s conjectures., Burnside was elected a Fellow of the, Royal Society and awarded two Royal, Medals. He served as president of the Council of the London Mathematical Society and, received its De Morgan Medal. Burnside, died on August 21, 1927., To find more information about Burnside, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 505, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Cayley Digraphs, 30 of Groups, The important thing in science is not so much to obtain new facts as to, discover new ways of thinking about them., sir william lawrence bragg, Beyond Reductionism, , Motivation, In this chapter, we introduce a graphical representation of a group given, by a set of generators and relations. The idea was originated by Cayley, in 1878. Although this topic is not usually covered in an abstract algebra, book, we include it for five reasons: It provides a method of visualizing, a group; it connects two important branches of modern mathematics—, groups and graphs; it has many applications to computer science; it, gives a review of some of our old friends—cyclic groups, dihedral, groups, direct products, and generators and relations; and, most importantly, it is fun!, Intuitively, a directed graph (or digraph) is a finite set of points,, called vertices, and a set of arrows, called arcs, connecting some of the, vertices. Although there is a rich and important general theory of directed graphs with many applications, we are interested only in those, that arise from groups., , The Cayley Digraph of a Group, Definition Cayley Digraph of a Group, , Let G be a finite group and let S be a set of generators for G. We define, a digraph Cay(S:G), called the Cayley digraph of G with generating set, S, as follows., 1. Each element of G is a vertex of Cay(S:G)., 2. For x and y in G, there is an arc from x to y if and only if xs 5 y for, some s [ S., , To tell from the digraph which particular generator connects two vertices, Cayley proposed that each generator be assigned a color, and that the, arrow joining x to xs be colored with the color assigned to s. He called the, 506, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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507, , 30 | Cayley Digraphs of Groups, , resulting figure the color graph of the group. This terminology is still used, occasionally. Rather than use colors to distinguish the different generators,, we will use solid arrows, dashed arrows, and dotted arrows. In general, if, there is an arc from x to y, there need not be an arc from y to x. An arrow, emanating from x and pointing to y indicates that there is an arc from x to y., Following are numerous examples of Cayley digraphs. Note that, there are several ways to draw the digraph of a group given by a particular generating set. However, it is not the appearance of the digraph that, is relevant but the manner in which the vertices are connected. These, connections are uniquely determined by the generating set. Thus, distances between vertices and angles formed by the arcs have no significance. (In the digraphs below, a headless arrow joining two vertices x, and y indicates that there is an arc from x to y and an arc from y to x., This occurs when the generating set contains both an element and its, inverse. For example, a generator of order 2 is its own inverse.), EXAMPLE 1 Z6 5 k1l., 1, 55, , 00, , 0, 1, , 44, , 2, , 11, , 3, 4, 33, , 22, , 5, , Cay({1}:Z 6 ), , Cay({1}:Z 6 ), , EXAMPLE 2 Z3 % Z2 5 k(1, 0), (0, 1)l., (0, 1), , (1, 0), (0, 0), , (0, 0), , (0, 1), (0, 1), , (1, 0), , (1, 1), , (2, 0), , (2, 1), , (2, 1), , Cay({(1, 0), (0, 1)}:Z 3 ⊕ Z 2), , (1, 1), , (2, 0), , (1, 0), , Cay({(1, 0), (0, 1)}:Z 3 ⊕ Z 2), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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508, , Special Topics, , EXAMPLE 3 D4 5 kR90, Hl., R90, , H, R00, R, , R0, , H, , R90H, , R90, , R180 H, , R180, , R270 H, , R270, , R90, H, , R90H, , R270 H, , R180 H, , R270, , R180, , Cay({R90, H}:D 4), , Cay({R90, H}:D4 ), , EXAMPLE 4 S3 5 k(12), (123)l., (12), , (123), , (12), , (132), , (1), , (23), (13), , (123), , (23), , (132), (132), , (13), , (12), (1), , Cay({(12), (123)}:S3), , (123), Cay({(12), (123)}:S3), , EXAMPLE 5 S3 5 k(12), (13)l., (12), , (13), , (1), , (12), , (132), , (13), , (123), , (23), , Cay({(12), (13)}:S 3 ), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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509, , 30 | Cayley Digraphs of Groups, , EXAMPLE 6 A4 5 k(12)(34), (123)l., (123), , (12)(34), (124), , (14)(23), , (134), (123), (1), , (132), , (12)(34), , (234), , (143), , (243), , (13)(24), , (142), Cay({(12)(34), (123)}:A 4 ), , EXAMPLE 7 Q4 5 ka, b | a4 5 e, a2 5 b2, b21ab 5 a3l., a, , b, , b, , e, , ab, , a, , a2b, , a2, , a3b, , a3, Cay({a, b}:Q4), , EXAMPLE 8 D` 5 ka, b | a2 5 b2 5 el., a, bab, , ba, , b, , b, e, , a, , ab, , aba, , Cay({a, b}:D `), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , abab
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510, , Special Topics, , The Cayley digraph provides a quick and easy way to determine the, value of any product of the generators and their inverses. Consider, for, example, the product ab3ab22 from the group given in Example 7. To reduce this to one of the eight elements used to label the vertices, we need, only begin at the vertex e and follow the arcs from each vertex to the next, as specified in the given product. Of course, b21 means traverse the b arc, in reverse. (Observations such as b23 5 b also help.) Tracing the product, through, we obtain b. Similarly, one can verify or discover other relations, among the generators., , Hamiltonian Circuits and Paths, Now that we have these directed graphs, what is it that we care to know, about them? One question about directed graphs that has been the object, of much research was popularized by the Irish mathematician, Sir William Hamilton in 1859, when he invented a puzzle called, “Around the World.” His idea was to label the 20 vertices of a regular, dodecahedron with the names of famous cities. One solves this puzzle, by starting at any particular city (vertex) and traveling “around the, world,” moving along the arcs in such a way that each other city is, visited exactly once before returning to the original starting point. One, solution to this puzzle is given in Figure 30.1, where the vertices are, visited in the order indicated., Obviously, this idea can be applied to any digraph; that is, one starts, at some vertex and attempts to traverse the digraph by moving along, 20, 1, , 2, , 8, 19, , 15, , 9, , 4, , 6, , 10, , 16, , 3, , 7, , 14, , 5, 12, 11, 18, , 13, 17, , Figure 30.1 Around the World., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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30 | Cayley Digraphs of Groups, , 511, , arcs in such a way that each vertex is visited exactly once before, returning to the starting vertex. (To go from x to y, there must be an arc, from x to y.) Such a sequence of arcs is called a Hamiltonian circuit in, the digraph. A sequence of arcs that passes through each vertex exactly, once without returning to the starting point is called a Hamiltonian, path. In the rest of this chapter, we concern ourselves with the existence, of Hamiltonian circuits and paths in Cayley digraphs., Figures 30.2 and 30.3 show a Hamiltonian path for the digraph given, in Example 2 and a Hamiltonian circuit for the digraph given in, Example 7, respectively., Is there a Hamiltonian circuit in, Cay({(1, 0), (0, 1)}:Z3 % Z2)?, More generally, let us investigate the existence of Hamiltonian circuits in, Cay({(1, 0), (0, 1)}:Zm % Zn),, where m and n are relatively prime and both are greater than 1. Visualize, the Cayley digraph as a rectangular grid coordinatized with Zm % Zn, as, (0, 0), , (0, 1), , (1, 0), , (1, 1), , (2, 0), , (2, 1), , Figure 30.2 Hamiltonian path in Cay({(1, 0), (0, 1)}:Z3 % Z2), from (0, 0) to (2, 1)., b, , e, , ab, , a, , a 2b, , a2, , a 3b, , a3, , Figure 30.3 Hamiltonian circuit in Cay({a, b}:Q4)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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512, , Special Topics, , in Figure 30.4. Suppose there is a Hamiltonian circuit in the digraph and, (a, b) is some vertex from which the circuit exits horizontally. (Clearly,, such a vertex exists.) Then the circuit must exit (a 2 1, b 1 1) horizontally also, for otherwise the circuit passes through (a, b 1 1) twice—see, Figure 30.5. Repeating this argument again and again, we see that the, circuit exits horizontally from each of the vertices (a, b), (a 2 1, b 1 1),, (a 2 2, b 1 2), . . . , which is just the coset (a, b) 1 k(21, 1)l. But when, m and n are relatively prime, k(21, 1)l is the entire group. Obviously,, there cannot be a Hamiltonian circuit consisting entirely of horizontal, moves. Let us record what we have just proved., , (0,, (0, 0), 0), , (0,, (0, 1), 1), , (0,, (0, 2), 2), , (0, n – 1), , (1, 0), (1,, 0), , (1, 1), (1,, 1), , (1, 2), , (1, n – 1), (0, 1), (1, 0), (1,, 0), , (m – 1, 0), , (m – 1, 1), , (m – 1, 2), , (m – 1, n – 1), , Figure 30.4 Cay({(1, 0), (0, 1)}:Zm % Zn)., (a 2 1, b 1 1), , (a, b), , Figure 30.5, , Theorem 30.1 A Necessary Condition, Cay({(1, 0), (0, 1)}:Zm % Zn ) does not have a Hamiltonian circuit, when m and n are relatively prime and greater than 1., , What about when m and n are not relatively prime? In general, the, answer is somewhat complicated, but the following special case is easy, to prove., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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513, , 30 | Cayley Digraphs of Groups, , Theorem 30.2 A Sufficient Condition, Cay({(1, 0), (0, 1)}:Zm % Zn ) has a Hamiltonian circuit when n, divides m., , (0, 0 ), , (1, 0 ), , (0, 1), , (0, 2 ), , (1, 1), , (1, 2 ), , First 3 3 3 block, , (0, 1), , (2, 0 ), , (2, 1), (2, 2 ), (1, 0), , (3, 0), , (3, 1), , (3, 2 ), Repeat path used, in first block, , (4, 0), , (4, 1), (4, 2 ), , (5, 0), , (5, 1), (5, 2 ), , Repeat path used, in first block, kth 3 3 3 block, , (3k – 1, 0), , (3k – 1, 1), , (3k – 1, 2), , Figure 30.6 Cay({(1, 0), (0, 1)}:Z3k % Z3)., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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514, , Special Topics, , PROOF Say m 5 kn. Then we may think of Zm % Zn as k blocks of, size n 3 n. (See Figure 30.6 for an example.) Start at (0, 0) and cover, the vertices of the top block as follows. Use the generator (0, 1) to move, horizontally across the first row to the end. Then use the generator (1, 0), to move vertically to the point below, and cover the remaining points in, the second row by moving horizontally. Keep this process up until the, point (n 2 1, 0)—the lower left-hand corner of the first block—has, been reached. Next, move vertically to the second block and repeat the, process used in the first block. Keep this up until the bottom block is, covered. Complete the circuit by moving vertically back to (0, 0)., Notice that the circuit given in the proof of Theorem 30.2 is easy to, visualize but somewhat cumbersome to describe in words. A much, more convenient way to describe a Hamiltonian path or circuit is to, specify the starting vertex and the sequence of generators in the order, in which they are to be applied. In Example 5, for instance, we may, start at (1) and alternate the generators (12) and (13) until we return to, (1). In Example 3, we may start at R0 and successively apply R90, R90,, R90, H, R90, R90, R90, H. When k is a positive integer and a, b, . . . , c is, a sequence of group elements, we use k p (a, b, . . . , c) to denote the, concatenation of k copies of the sequence (a, b, . . . , c). Thus, 2 p (R90,, R90, R90, H) and 2 p (3 p R90, H) both mean R90, R90, R90, H, R90, R90,, R90, H. With this notation, we may conveniently denote the Hamiltonian circuit given in Theorem 30.2 as, m p ((n 2 1) p (0, 1), (1, 0))., We leave it as an exercise (Exercise 11) to show that if x1, x2, . . . , xn, is a sequence of generators determining a Hamiltonian circuit starting, at some vertex, then the same sequence determines a Hamiltonian circuit for any starting vertex., From Theorem 30.1, we know that there are some Cayley digraphs, of Abelian groups that do not have any Hamiltonian circuits. But Theorem, 30.3 shows that each of these Cayley digraphs does have a Hamiltonian, path. There are some Cayley digraphs for non-Abelian groups that do not, even have Hamiltonian paths, but we will not discuss them here., Theorem 30.3 Abelian Groups Have Hamiltonian Paths, Let G be a finite Abelian group, and let S be any (nonempty†) generating set for G. Then Cay(S:G) has a Hamiltonian path., S is the empty set, it is customary to define kSl as the identity group. We prefer to, ignore this trivial case., , †If, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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515, , 30 | Cayley Digraphs of Groups, , PROOF We use induction on |S|. If |S| 5 1, say, S 5 {a}, then the digraph is just a circle labeled with e, a, a2, . . . , am21, where |a| 5 m., Obviously, there is a Hamiltonian path for this case. Now assume that, |S| . 1. Choose some s [ S. Let T 5 S 2 {s}—that is, T is S with s, removed—and set H 5 ks1, s2, . . . , sn21l where S 5 {s1, s2, . . . , sn} and, s 5 sn. (Notice that H may be equal to G.), Because |T| , |S| and H is a finite Abelian group, the induction hypothesis guarantees that there is a Hamiltonian path (a1, a2, . . . , ak) in, Cay(T:H). We will show that, (a1, a2, . . . , ak, s, a1, a2, . . . , ak, s, . . . , a1, a2, . . . , ak, s, a1, a2, . . . , ak),, where a1, a2, . . . , ak occurs |G|/|H| times and s occurs |G|/|H| 2 1, times, is a Hamiltonian path in Cay(S:G)., Because S 5 T < {s} and T generates H, the coset Hs generates the, factor group G/H. (Since G is Abelian, this group exists.) Hence, the, cosets of H are H, Hs, Hs2, . . . , Hsn, where n 5 |G|/|H| 2 1. Starting, from the identity element of G, the path given by (a1, a2, . . . , ak) visits, each element of H exactly once [because (a1, a2, . . . , ak) is a, Hamiltonian path in Cay(T:H)]. The generator s then moves us to some, element of the coset Hs. Starting from there, the path (a1, a2, . . . , ak), visits each element of Hs exactly once. Then, s moves us to the coset, Hs2, and we visit each element of this coset exactly once. Continuing, this process, we successively move to Hs3, Hs4, . . . , Hsn, visiting each, vertex in each of these cosets exactly once. Because each vertex of, Cay(S:G) is in exactly one coset Hsi, this implies that we visit each vertex of Cay(S:G) exactly once. Thus, we have a Hamiltonian path., We next look at Cayley digraphs with three generators., EXAMPLE 9 Let, D3 5 kr, f | r 3 5 f 2 5 e, rf 5 fr 2l., Then a Hamiltonian circuit in, Cay({(r, 0), ( f, 0), (e, 1)}:D3 % Z6), is given in Figure 30.7., (f, 0), , (e, 0), , (f, 1), , (e, 1), , (f, 2), , (e, 2), , (f, 3), , (e, 3), , (f, 4), , (e, 4), , (f, 5), , (e, 5), , (rf, 0), , (r, 5), , (r2f, 0), , (r 2, 5), , Figure 30.7, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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516, , Special Topics, , Although it is not easy to prove, it is true that, Cay({(r, 0), ( f, 0), (e, 1)}:Dn % Zm), has a Hamiltonian circuit for all n and m. (See [3].) Example 10 shows, the circuit for this digraph when m is even., EXAMPLE 10 Let, Dn 5 kr, f | r n 5 f 2 5 e, rf 5 fr21l., Then a Hamiltonian circuit in, Cay({(r, 0), ( f, 0), (e, 1)}:Dn % Zm), with m even is traced in Figure 30.8. The sequence of generators that, traces the circuit is, m p [(n 2 1) p (r, 0), ( f, 0), (n 2 1) p (r, 0), (e, 1)]., (f, 0), , (e, 0), , (rf, 0), , (r, 0), , (f, 1), , (e, 1), , (r, 1), , ...iterate, (r2 f,, , 0), , (r n – 1f, 0), , (r 2,, , (r 2,, , 0), , (r n –1, 0), , (r n –1f, 1), , 1), , (r n–1, 1), , Figure 30.8, , Some Applications, Cayley digraphs are natural models for interconnection networks in, computer designs, and Hamiltonicity is an important property in relation to sorting algorithms on such networks. One particular Cayley digraph that is used to design and analyze interconnection networks, of parallel machines is the symmetric group Sn with the set of all transpositions as the generating set. Hamiltonian paths and circuits in Cayley, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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30 | Cayley Digraphs of Groups, , 517, , Don Crowe and Dorothy Washburn’s book “Symmetries of Culture”, , digraphs arise in a variety of group theory contexts. A Hamiltonian path, in a Cayley digraph of a group is simply an ordered listing of the group, elements without repetition. The vertices of the digraph are the group elements, and the arcs of the path are generators of the group. In 1948, R. A., Rankin used these ideas (although not the terminology) to prove that certain bell-ringing exercises could not be done by the traditional methods, employed by bell ringers. (See [1, Chap. 22] for the group theoretic, aspects of bell ringing.) In 1981, Hamiltonian paths in Cayley digraphs, were used in an algorithm for creating computer graphics of Escher-type, repeating patterns in the hyperbolic plane [2]. This program can produce, repeating hyperbolic patterns in color from among various infinite classes, of symmetry groups. The program has now been improved so that the user, may choose from many kinds of color symmetry. The 2003 Mathematics, Awareness Month poster featured one such image (see http://www, .mathaware.org/mam/03/index.html). Two Escher drawings and their, computer-drawn counterparts are given in Figures 30.9 through 30.12., In this chapter, we have shown how one may construct a directed, graph from a group. It is also possible to associate a group—called, the automorphism group—with every directed graph. In fact, several, of the 26 sporadic simple groups were first constructed in this way., , Figure 30.9 M. C. Escher’s Circle Limit I., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Douglas Dunham, , Special Topics, , Figure 30.10 A computer duplication of the pattern of M. C. Escher’s Circle, Limit I [2]. The program used a Hamiltonian path in a Cayley digraph of the, underlying symmetry group., , Douglas Dunham, , 518, , Figure 30.11 M. C. Escher’s Circle Limit IV., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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519, , Douglas Dunham, , 30 | Cayley Digraphs of Groups, , Figure 30.12 A computer drawing inspired by the pattern of, M. C. Escher’s Circle Limit IV [2]. The program used a Hamiltonian, path in a Cayley digraph of the underlying symmetry group., , Exercises, A mathematician is a machine for turning coffee into theorems., '' s, paul erdo, , 1. Find a Hamiltonian circuit in the digraph given in Example 7 different from the one in Figure 30.3., 2. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q4 % Z2)., 3. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q4 % Zm), where m is even., 4. Write the sequence of generators for each of the circuits found in, Exercises 1, 2, and 3., 5. Use the Cayley digraph in Example 7 to evaluate the product, a3ba21ba3b21., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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520, , Special Topics, , 6. Let x and y be two vertices of a Cayley digraph. Explain why two, paths from x to y in the digraph yield a group relation—that is, an, equation of the form a1a2 ? ? ? am 5 b1b2 ? ? ? bn, where the ai’s and, bj’s are generators of the Cayley digraph., 7. Use the Cayley digraph in Example 7 to verify the relation, aba21b21a21b21 5 a2ba3., 8. Identify the following Cayley digraph of a familiar group., , 9. Let D4 5 kr, f | r 4 5 e 5 f 2, rf 5 fr21l. Verify that, 6 p [3 p (r, 0), ( f, 0), 3 p (r, 0), (e, 1)], is a Hamiltonian circuit in, Cay({(r, 0), ( f, 0), (e, 1)}:D4 % Z6)., 10. Draw a picture of Cay({2, 5}:Z8)., 11. If s1, s2, . . . , sn is a sequence of generators that determines a, Hamiltonian circuit beginning at some vertex, explain why the same, sequence determines a Hamiltonian circuit beginning at any point., (This exercise is referred to in this chapter.), 12. Show that the Cayley digraph given in Example 7 has a, Hamiltonian path from e to a., 13. Show that there is no Hamiltonian path in, Cay({(1, 0), (0, 1)}:Z3 % Z2), from (0, 0) to (2, 0)., 14. Draw Cay({2, 3}:Z6). Is there a Hamiltonian circuit in this digraph?, 15. a. Let G be a group of order n generated by a set S. Show that a sequence s1, s2, . . . , sn21 of elements of S is a Hamiltonian path in, Cay(S:G) if and only if, for all i and j with 1 # i # j , n, we, have si si11 ? ? ? sj 2 e., b. Show that the sequence s1s2 ? ? ? sn is a Hamiltonian circuit if, and only if s1s2 ? ? ? sn 5 e, and that whenever 1 # i # j , n, we, have si si11 ? ? ? sj 2 e., 16. Let D4 5 ka, b | a2 5 b2 5 (ab)4 5 el. Draw Cay({a, b}:D4). Why, is it reasonable to say that this digraph is undirected?, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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30 | Cayley Digraphs of Groups, , 521, , 17. Let Dn be as in Example 10. Show that 2 p [(n 2 1) p r, f ] is a, Hamiltonian circuit in Cay({r, f}:Dn)., 18. Let Q8 5 ka, b | a8 5 e, a4 5 b2, b21ab 5 a21l. Find a Hamiltonian, circuit in Cay({a, b}:Q8)., 19. Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q8 % Z5)., 20. Prove that the Cayley digraph given in Example 6 does not have a, Hamiltonian circuit. Does it have a Hamiltonian path?, 21. Find a Hamiltonian circuit in, Cay({(R90, 0), (H, 0), (R0, 1)}:D4 % Z3)., Does this circuit generalize to the case Dn11 % Zn for all n $ 3?, 22. Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q8 % Zm) for all even m., 23. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q4 % Z3)., 24. Find a Hamiltonian circuit in, Cay({(a, 0), (b, 0), (e, 1)}:Q4 % Zm) for all odd m $ 3., 25. Write the sequence of generators that describes the Hamiltonian, circuit in Example 9., 26. Let Dn be as in Example 10. Find a Hamiltonian circuit in, Cay({(r, 0), ( f, 0), (e, 1)}:D4 % Z5)., 27., 28., , 29., 30., 31., , Does your circuit generalize to the case Dn % Zn11 for all n $ 4?, Prove that Cay({(0, 1), (1, 1)}:Zm % Zn) has a Hamiltonian circuit, for all m and n greater than 1., Suppose that a Hamiltonian circuit exists for Cay({(1, 0), (0, 1)}:, Zm % Zn) and that this circuit exits from vertex (a, b) vertically., Show that the circuit exits from every member of the coset, (a, b) 1 k(1, 21)l vertically., Let D2 5 kr, f | r2 5 f 2 5 e, rf 5 fr21l. Find a Hamiltonian circuit, in Cay({(r, 0), ( f, 0), (e, 1)}:D2 % Z3)., Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in Cay({(a, 0),, (b, 0), (e, 1)}:Q8 % Z3)., In Cay({(1, 0), (0, 1)}:Z4 % Z5), find a sequence of generators that, visits exactly one vertex twice and all others exactly once and returns to the starting vertex., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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522, , Special Topics, , 32. In Cay({(1, 0), (0, 1)}:Z4 % Z5), find a sequence of generators that, visits exactly two vertices twice and all others exactly once and returns to the starting vertex., 33. Find a Hamiltonian circuit in Cay({(1, 0), (0, 1)}:Z4 % Z6)., 34. (Factor Group Lemma) Let S be a generating set for a group G, let, N be a cyclic normal subgroup of G, and let, S 5 {sN | s [ S}., If (a1N, . . . , ar N) is a Hamiltonian circuit in Cay(S:G/N) and the, product a1 ? ? ? ar generates N, prove that, |N| p (a1, . . . , ar), is a Hamiltonian circuit in Cay(S:G)., 35. A finite group is called Hamiltonian if all of its subgroups are normal., (One non-Abelian example is Q4.) Show that Theorem 30.3 can be, generalized to include all Hamiltonian groups., , References, 1. F. J. Budden, The Fascination of Groups, Cambridge: Cambridge University Press, 1972., 2. Douglas Dunham, John Lindgren, and David Witte, “Creating Repeating, Hyperbolic Patterns,” Computer Graphics 15 (1981): 215–223., 3. David Witte, Gail Letzter, and Joseph A. Gallian, “On Hamiltonian Circuits in Cartesian Products of Cayley Digraphs,” Discrete Mathematics, 43 (1983): 297–307., , Suggested Readings, Frank Budden, “Cayley Graphs for Some Well-Known Groups,” The, Mathematical Gazette 69 (1985): 271–278., This article contains the Cayley graphs of A4, Q4, and S4 using a variety, of generators and relations., E. L. Burrows and M. J. Clark, “Pictures of Point Groups,” Journal of, Chemical Education 51 (1974): 87–90., Chemistry students may be interested in reading this article. It gives a, comprehensive collection of the Cayley digraphs of groups important, to chemists., Douglas Dunham, John Lindgren, and David Witte, “Creating Repeating, Hyperbolic Patterns,” Computer Graphics 15 (1981): 215–223., In this beautifully illustrated paper, a process for creating repeating, patterns of the hyperbolic plane is described. The paper is a blend of, group theory, geometry, and art., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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30 | Cayley Digraphs of Groups, , 523, , Joseph A. Gallian, “Circuits in Directed Grids,” The Mathematical Intelligencer 13 (1991): 40–43., This article surveys research done on variations of the themes discussed in this chapter., Joseph A. Gallian and David Witte, “Hamiltonian Checkerboards,” Mathematics Magazine 57 (1984): 291–294., This paper gives some additional examples of Hamiltonian circuits in, Cayley digraphs. It is available at http://www.d.umn.edu/~jgallian/, checker.pdf, Paul Hoffman, “The Man Who Loves Only Numbers,” The Atlantic, Monthly 260 (1987): 60–74., '' s, a prolific and eccentric mathematiA charming portrait of Paul Erdo, cian who contributed greatly to graph theory., Henry Levinson, “Cayley Diagrams,” in Mathematical Vistas: Papers, from the Mathematics Section, New York Academy of Sciences, J. Malkevitch and D. McCarthy, eds., 1990: 62–68., This richly illustrated article presents Cayley digraphs of many of the, groups that appear in this text., A. T. White, “Ringing the Cosets,” The American Mathematical Monthly, 94 (1987): 721–746., This article analyzes the practice of bell ringing by way of Cayley, digraphs., , Suggested Website, http://www.d.umn.edu/~ddunham/, This website has copies of several articles that describe the mathematics, involved in creating Escher-like repeating patterns in the hyperbolic plane, as shown in Figure 30.10., , Suggested DVD, N is a Number, Mathematical Association of America, 58 minutes., '' s discusses politics, death, and mathematics., In this documentary, Erdo, '' s’s collaborators and friends comment on his work and, Many of Erdo, life. It is available for purchase at http://www.amazon.com, , Suggested Software, Group Explorer is mathematical visualization software that allows, users to explore dozens of Cayley digraphs of finite groups visually, and interactively. This free software is available at http://, sourceforge.net/projects/groupexplorer, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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William Rowan, Hamilton, After Isaac Newton, the greatest mathematician of the English-speaking peoples is, William Rowan Hamilton., , William Rowan Hamilton was born on, August 3, 1805, in Dublin, Ireland. At, three, he was skilled at reading and arithmetic. At five, he read and translated Latin,, Greek, and Hebrew; at 14, he had mastered, 14 languages, including Arabic, Sanskrit,, Hindustani, Malay, and Bengali., In 1833, Hamilton provided the first, modern treatment of complex numbers. In, 1843, he made what he considered his greatest discovery—the algebra of quaternions., The quaternions represent a natural generalization of the complex numbers with three, numbers i, j, and k whose squares are 21., , This stamp featuring the quaternions was, issued in 1983., , Stock Montage, , sir edmund whittaker,, Scientific American, , With these, rotations in three and four dimensions can be algebraically treated. Of, greater significance, however, is the fact that, the quaternions are noncommutative under, multiplication. This was the first ring to be, discovered in which the commutative property does not hold. The essential idea for the, quaternions suddenly came to Hamilton after, 15 years of fruitless thought!, Today Hamilton’s name is attached to several concepts, such as the Hamiltonian function, which represents the total energy in a, physical system; the Hamilton–Jacobi differential equations; and the Cayley–Hamilton, Theorem from linear algebra. He also coined, the terms vector, scalar, and tensor., In his later years, Hamilton was plagued, by alcoholism. He died on September 2,, 1865, at the age of 60., For more information about Hamilton,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 524, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Paul Erd'O' s, , Seymour Schuster, Carleton College, , Paul Erdős is a socially helpless Hungarian, who has thought about more mathematical problems than anyone else in history., The Atlantic Monthly, , '' s (pronounced AIR-dish) was, Paul Erdo, one of the best-known and most highly respected mathematicians of the 20th century., Unlike most of his contemporaries, who, '' s, have concentrated on theory building, Erdo, focused on problem solving and problem, posing. The problems and methods of solu'' s—like those of Euler, whose, tion of Erdo, solutions to special problems pointed the, way to much of the mathematical theory, we have today—have helped pioneer new, theories, such as combinatorial and probabilistic number theory, combinatorial geometry,, probabilistic and transfinite combinatorics,, and graph theory., '' s was born on March 26, 1913, in, Erdo, Hungary. Both of his parents were high, '' s, a Jew,, school mathematics teachers. Erdo, left Hungary in 1934 at the age of 21 because of the rapid rise of anti-Semitism in, Europe. For the rest of his life he traveled, incessantly, rarely pausing more than a, month in any one place, giving lectures for, , small honoraria and staying with fellow, mathematicians. He had little property and, no fixed address. All that he owned he carried with him in a medium-sized suitcase,, frequently visiting as many as 15 places in a, month. His motto was, “Another roof, another proof.” Even in his eighties, he put in, 19-hour days doing mathematics., '' s wrote more than 1500 research paErdo, pers. He coauthored papers with more than, 500 people. These people are said to have, '' s number 1. People who do not have, Erdo, '' s number 1, but who have written a, Erdo, paper with someone who does, are said to, ' ' s number 2, and so on, induchave Erdo, '' s died of a heart attack on Septively. Erdo, tember 20, 1996, in Warsaw, Poland., '' s, visit:, For more information about Erdo, http://www-groups.dcs.st-and, .ac.uk/~history/, http://www.oakland.edu/enp, , 525, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Introduction, 31 to Algebraic, Coding Theory, Damn it, if the machine can detect an error, why can’t it locate the position, of the error and correct it?, richard w. hamming, , Motivation, One of the most interesting and important applications of finite fields, has been the development of algebraic coding theory. This theory,, which originated in the late 1940s, was created in response to practical, communication problems. (Algebraic coding has nothing to do with, secret codes.) Algebraic codes are now used in compact disc and DVD, players, fax machines, digital televisions, and bar code scanners, and, are essential to computer maintenance., To motivate this theory, imagine that we wish to transmit one of two, possible signals to a spacecraft approaching Mars. If the proposed, landing site appears unfavorable, we will command the craft to orbit, the planet; otherwise, we will command the craft to land. The signal for, orbiting will be a 0, and the signal for landing will be a 1. But it is possible that some sort of interference (called noise) could cause an incorrect message to be received. To decrease the chance of this happening,, redundancy is built into the transmission process. For example, if we, wish the craft to orbit Mars, we could send five 0s. The craft’s onboard, computer is programmed to take any five-digit message received and, decode the result by majority rule. So, if 00000 is sent and 10001 is received, the computer decides that 0 was the intended message. Notice, that, for the computer to make the wrong decision, at least three errors, must occur during transmission. If we assume that errors occur, independently, it is less likely that three errors will occur than that two, or fewer errors will occur. For this reason, this decision process is frequently called the maximum-likelihood decoding procedure. Our particular situation is illustrated in Figure 31.1. The general coding procedure is illustrated in Figure 31.2., 526, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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527, , 31 | Introduction to Algebraic Coding Theory, , Original, message, 0, earth, , Encoded, message, 00000, encoder, , Received, message, 10001, , Noise, transmitter, , spacecraft, , Decoded, message, 0, , decoder, , Figure 31.1 Encoding and decoding by fivefold repetition., , Encoded, message, Original, message, , encoder, , Received, message, , Noisy, channel, transmitter, , receiver, , decoder, , Decoded, message, , Figure 31.2 General encoding–decoding., , In practice, the means of transmission are telephone, radiowave,, microwave, or even a magnetic disk. The noise might be human error,, crosstalk, lightning, thermal noise, or deterioration of a disk. Throughout this chapter, we assume that errors in transmission occur independently. Different methods are needed when this is not the case., Now, let’s consider a more complicated situation. This time, assume, that we wish to send a sequence of 0s and 1s of length 500. Further,, suppose that the probability that an error will be made in the transmission of any particular digit is .01. If we send this message directly, without any redundancy, the probability that it will be received error-free is, (.99)500, or approximately .0066., On the other hand, if we adopt a threefold repetition scheme by, sending each digit three times and decoding each block of three digits, received by majority rule, we can do much better. For example, the sequence 1011 is encoded as 111000111111. If the received message is, 011000001110, the decoded message is 1001. Now, what is the probability that our 500-digit message will be error-free? Well, if a 1, say, is, sent, it will be decoded as a 0 if and only if the block received is 001,, 010, 100, or 000. The probability that this will occur is, (.01)(.01)(.99) 1 (.01)(.99)(.01) 1 (.99)(.01)(.01) 1 (.01)(.01)(.01), 5 (.01)2[3(.99) 1 .01], 5 .000298 , .0003., Thus, the probability that any particular digit in the sequence will be, decoded correctly is greater than .9997, and it follows that the probability that the entire 500-digit message will be decoded correctly is, greater than (.9997)500, or approximately .86—a dramatic improvement, over .0066., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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528, , Special Topics, , This example illustrates the three basic features of a code. There is a, set of messages, a method of encoding these messages, and a method of, decoding the received messages. The encoding procedure builds some, redundancy into the original messages; the decoding procedure corrects, or detects certain prescribed errors. Repetition codes have the advantage, of simplicity of encoding and decoding, but they are too inefficient. In a, fivefold repetition code, 80% of all transmitted information is redundant. The goal of coding theory is to devise message encoding and, decoding methods that are reliable, efficient, and reasonably easy to, implement., Before plunging into the formal theory, it is instructive to look at a, sophisticated example., EXAMPLE 1 Hamming (7, 4) Code, This time, our message set consists of all possible 4-tuples of 0’s and 1’s, (that is, we wish to send a sequence of 0’s and 1’s of length 4). Encoding will be done by viewing these messages as 1 3 4 matrices with entries from Z2 and multiplying each of the 16 messages on the right by, the matrix, 1, 0, G ≥, 0, 0, , 0, 1, 0, 0, , 0, 0, 1, 0, , 0, 0, 0, 1, , 1, 1, 1, 0, , 1, 0, 1, 1, , 0, 1, ¥., 1, 1, , (All arithmetic is done modulo 2.) The resulting 7-tuples are called, code words. (See Table 31.1.), Table 31.1, Message, , Encoder G, , Code Word, , 0000, 0001, 0010, 0100, 1000, 1100, 1010, 1001, , →, →, →, →, →, →, →, →, , 0000000, 0001011, 0010111, 0100101, 1000110, 1100011, 1010001, 1001101, , Message, 0110, 0101, 0011, 1110, 1101, 1011, 0111, 1111, , Encoder G, , Code Word, , →, →, →, →, →, →, →, →, , 0110010, 0101110, 0011100, 1110100, 1101000, 1011010, 0111001, 1111111, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 529, , Notice that the first four digits of each code word constitute just the, original message corresponding to the code word. The last three digits, of the code word constitute the redundancy features. For this code, we, use the nearest-neighbor decoding method (which, in the case that the, errors occur independently, is the same as the maximum-likelihood decoding procedure). For any received word v, we assume that the word, sent is the code word v9 that differs from v in the fewest number of positions. If the choice of v9 is not unique, we can decide not to decode or, arbitrarily choose one of the code words closest to v. (The first option, is usually selected when retransmission is practical.), Once we have decoded the received word, we can obtain the message, by deleting the last three digits of v9. For instance, suppose that 1000, were the intended message. It would be encoded and transmitted as u 5, 1000110. If the received word were v 5 1100110 (an error in the second, position), it would still be decoded as u, since v and u differ in only one, position, whereas v and any other code word would differ in at least two, positions. Similarly, the intended message 1111 would be encoded as, 1111111. If, instead of this, the word 0111111 were received, our decoding procedure would still give us the intended message 1111., The code in Example 1 is one of an infinite class of important codes, discovered by Richard Hamming in 1948. The Hamming codes are the, most widely used codes., The Hamming (7, 4) encoding scheme can be conveniently illustrated with the use of a Venn diagram, as shown in Figure 31.3. Begin, by placing the four message digits in the four overlapping regions I, II,, A, , I, , V, , B, , A, , VI, , 1, , B, , 1, 0, 0, , III, IV, , II, , 1, , 0, , VII, , 1, , C, , C, , Figure 31.3 Venn diagram of the message 1001 and the encoded, message 1001101., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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530, , Special Topics, , III, and IV, with the digit in position 1 in region I, the digit in position, 2 in region II, and so on. For regions V, VI, and VII, assign 0 or 1 so that, the total number of 1s in each circle is even., Consider the Venn diagram of the received word 0001101:, A, , B, , 0, , 1, , 0, , 0, 1, , 0, 1, , C, , How may we detect and correct an error? Well, observe that each of the, circles A and B has an odd number of 1s. This tells us that something is, wrong. At the same time, we note that circle C has an even number of 1s., Thus, the portion of the diagram that is in both A and B but not in C is, the source of the error. See Figure 31.4., Quite often, codes are used to detect errors rather than correct them., This is especially appropriate when it is easy to retransmit a message., If a received word is not a code word, we have detected an error. For, example, computers are designed to use a parity check for numbers. Inside the computer, each number is represented by a string of 0’s and 1’s., If there is an even number of 1’s in this representation, a 0 is attached to, the string; if there is an odd number of 1’s in the representation, a 1, is attached to the string. Thus, each number stored in the computer, memory has an even number of 1’s. Now, when the computer reads a, A, , B, , 0, , 1, , 0, , 0, 1, , 0, 1, , C, , Figure 31.4 Circles A and B but not C have wrong parity., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 531, , number from memory, it performs a parity check. If the read number, has an odd number of 1’s, the computer will know that an error has been, made, and it will reread the number. Note that an even number of errors, will not be detected by a parity check., The methods of error detection introduced in Chapters 0 and 5 are, based on the same principle. An extra character is appended to a string, of numbers so that a particular condition is satisfied. If we find that, such a string does not satisfy that condition, we know that an error has, occurred., , Linear Codes, We now formalize some of the ideas introduced in the preceding, discussion., Definition Linear Code, , An (n, k) linear code over a finite field F is a k-dimensional subspace V, of the vector space, , Fn 5 F % F % ? ? ? % F, n copies, over F. The members of V are called the code words. When F is Z2, the, code is called binary., , One should think of an (n, k) linear code over F as a set of n-tuples, from F, where each n-tuple has two parts: the message part, consisting, of k digits; and the redundancy part, consisting of the remaining n 2 k, digits. Note that an (n, k) linear code over a finite field F of order q has, q k code words, since every member of the code is uniquely expressible, as a linear combination of the k basis vectors with coefficients from F., The set of qk code words is closed under addition and scalar multiplication by members of F. Also, since errors in transmission may occur, in any of the n positions, there are qn possible vectors that can be, received. Where there is no possibility of confusion, it is customary to, denote an n-tuple (a1, a2, . . . , an) more simply as a1a2 ? ? ? an, as we, did in Example 1., EXAMPLE 2 The set, {0000000, 0010111, 0101011, 1001101,, 1100110, 1011010, 0111100, 1110001}, is a (7, 3) binary code., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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532, , Special Topics, , EXAMPLE 3 The set {0000, 0101, 1010, 1111} is a (4, 2) binary, code., Although binary codes are by far the most important ones, other, codes are occasionally used., EXAMPLE 4 The set, {0000, 0121, 0212, 1022, 1110, 1201, 2011, 2102, 2220}, is a (4, 2) linear code over Z3. A linear code over Z3 is called a ternary, code., To facilitate our discussion of the error-correcting and errordetecting capability of a code, we introduce the following terminology., Definitions Hamming Distance, Hamming Weight, , The Hamming distance between two vectors in Fn is the number of components in which they differ. The Hamming weight of a vector is the, number of nonzero components of the vector. The Hamming weight of a, linear code is the minimum weight of any nonzero vector in the code., , We will use d(u, v) to denote the Hamming distance between the, vectors u and v, and wt(u) for the Hamming weight of the vector u., EXAMPLE 5 Let s 5 0010111, t 5 0101011, u 5 1001101, and v 5, 1101101. Then, d(s, t) 5 4, d(s, u) 5 4, d(s, v) 5 5, d(u, v) 5 1; and, wt(s) 5 4, wt(t) 5 4, wt(u) 5 4, wt(v) 5 5., The Hamming distance and Hamming weight have the following, important properties., Theorem 31.1 Properties of Hamming Distance and Hamming Weight, For any vectors u, v, and w, d(u, v) # d(u, w) 1 d(w, v) and d(u, v) 5, wt(u 2 v)., , PROOF To prove that d(u, v) 5 wt(u 2 v), simply observe that both, d(u, v) and wt(u 2 v) equal the number of positions in which u and v, differ. To prove that d(u, v) # d(u, w) 1 d(w, v), note that if u and v differ, in the ith position and u and w agree in the ith position, then w and v, differ in the ith position., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 533, , With the preceding definitions and Theorem 31.1, we can now, explain why the codes given in Examples 1, 2, and 4 will correct any, single error, but the code in Example 3 will not., Theorem 31.2 Correcting Capability of a Linear Code, If the Hamming weight of a linear code is at least 2t 1 1, then the, code can correct any t or fewer errors. Alternatively, the same code, can detect any 2t or fewer errors., , PROOF We will use nearest-neighbor decoding; that is, for any received vector v, we will assume that the corresponding code word sent is, a code word v9 such that the Hamming distance d(v, v9) is a minimum., (If there is more than one such v9, we do not decode.) Now, suppose that, a transmitted code word u is received as the vector v and that at most t, errors have been made in transmission. Then, by the definition of distance between u and v, we have d(u, v) # t. If w is any code word other, than u, then w 2 u is a nonzero code word. Thus, by assumption,, 2t 1 1 # wt(w 2 u) 5 d(w, u) # d(w, v) 1 d(v, u) # d(w, v) 1 t,, and it follows that t 1 1 # d(w, v). So, the code word closest to the received vector v is u, and therefore v is correctly decoded as u., To show that the code can detect 2t errors, we suppose that a transmitted code word u is received as the vector v and that at least one, error, but no more than 2t errors, was made in transmission. Because, only code words are transmitted, an error will be detected whenever a, received word is not a code word. But v cannot be a code word, since, d(v, u) # 2t, whereas we know that the minimum distance between distinct code words is at least 2t 1 1., Theorem 31.2 is often misinterpreted to mean that a linear code with, Hamming weight 2t 1 1 can correct any t errors and detect any 2t or, fewer errors simultaneously. This is not the case. The user must choose, one or the other role for the code. Consider, for example, the Hamming, (7, 4) code given in Table 31.1. By inspection, the Hamming weight of, the code is 3 5 2 ? 1 1 1, so we may elect either to correct any single, error or to detect any one or two errors. To understand why we can’t do, both, consider the received word 0001010. The intended message, could have been 0000000, in which case two errors were made (likewise for the intended messages 1011010 and 0101110), or the intended, message could have been 0001011, in which case one error was made., But there is no way for us to know which of these possibilities occurred. If our choice were error correction, we would assume—perhaps, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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534, , Special Topics, , mistakenly—that 0001011 was the intended message. If our choice, were error detection, we simply would not decode. (Typically, one, would request retransmission.), On the other hand, if we write the Hamming weight of a linear code, in the form 2t 1 s 1 1, we can correct any t errors and detect any t 1 s, or fewer errors. Thus, for a code with Hamming weight 5, our options, include the following:, 1. Detect any four errors (t 5 0, s 5 4)., 2. Correct any one error and detect any two or three errors (t 5 1,, s 5 2)., 3. Correct any two errors (t 5 2, s 5 0)., EXAMPLE 6 Since the Hamming weight of the linear code given in, Example 2 is 4, it will correct any single error and detect any two errors, (t 5 1, s 5 1) or detect any three errors (t 5 0, s 5 3)., It is natural to wonder how the matrix G used to produce the Hamming code in Example 1 was chosen. Better yet, in general, how can, one find a matrix G that carries a subspace V of Fk to a subspace of Fn, in such a way that for any k-tuple v in V, the vector vG will agree with, v in the first k components and build in some redundancy in the last, n 2 k components? Such a matrix is a k 3 n matrix of the form, 1, 0, ?, ?, ?, 0, , 0, 1, ?, ?, ?, 0, , ???, ???, , ???, , 0, 0, ?, ?, ?, 1, , a11, ?, ?, ?, ?, ak1, , ???, , ???, , a1n 2 k, ?, ?, ?, ?, akn 2 k, , where the aij’s belong to F. A matrix of this form is called the standard, generator matrix (or standard encoding matrix) for the resulting code., Any k 3 n matrix whose rows are linearly independent will transform Fk to a k-dimensional subspace of Fn that could be used to build, redundancy, but using the standard generator matrix has the advantage, that the original message constitutes the first k components of the, transformed vectors. An (n, k) linear code in which the k information, digits occur at the beginning of each code word is called a systematic, code. Schematically, we have the following., Encoder, , message, |←k digits→|, , message redundant digits, |, , k, , |, , n2k, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , |
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31 | Introduction to Algebraic Coding Theory, , 535, , Notice that, by definition, a standard generator matrix produces a systematic code., EXAMPLE 7 From the set of messages, {000, 001, 010, 100, 110, 101, 011, 111},, we may construct a (6, 3) linear code over Z2 with the standard generator matrix, 1 0 0 1 1 0, G £0 1 0 1 0 1§ ., 0 0 1 1 1 1, The resulting code words are given in Table 31.2. Since the minimum, weight of any nonzero code word is 3, this code will correct any single, error or detect any double error., Table 31.2, Message, , Encoder G, , Code Word, , 000, 001, 010, 100, 110, 101, 011, 111, , →, →, →, →, →, →, →, →, , 000000, 001111, 010101, 100110, 110011, 101001, 011010, 111100, , EXAMPLE 8 Here we take a set of messages as, {00, 01, 02, 10, 11, 12, 20, 21, 22},, and we construct a (4, 2) linear code over Z3 with the standard generator matrix, G c, , 1 0 2 1, d., 0 1 2 2, , The resulting code words are given in Table 31.3. Since the minimum, weight of the code is 3, it will correct any single error or detect any, double error., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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536, , Special Topics, , Table 31.3, Message, , Encoder G, , Code Word, , 00, 01, 02, 10, 11, 12, 20, 21, 22, , →, →, →, →, →, →, →, →, →, , 0000, 0122, 0211, 1021, 1110, 1202, 2012, 2101, 2220, , Parity-Check Matrix Decoding, Now that we can conveniently encode messages with a standard generator matrix, we need a convenient method for decoding the received, messages. Unfortunately, this is not as easy to do; however, in the case, where at most one error per code word has occurred, there is a fairly, simple method for decoding. (When more than one error occurs in a, code word, our decoding method fails.), To describe this method, suppose that V is a systematic linear, code over the field F given by the standard generator matrix G 5, [Ik | A], where Ik represents the k 3 k identity matrix and A is the k 3, (n 2 k) matrix obtained from G by deleting the first k columns of G., Then, the n 3 (n 2 k) matrix, H c, , A, d,, Ink, , where 2A is the negative of A and In2k is the (n 2 k) 3 (n 2 k) identity matrix, is called the parity-check matrix for V. (In the literature, the, transpose of H is called the parity-check matrix, but H is much more, convenient for our purposes.) The decoding procedure is:, 1. For any received word w, compute wH., 2. If wH is the zero vector, assume that no error was made., 3. If there is exactly one instance of a nonzero element s [ F and a, row i of H such that wH is s times row i, assume that the sent word, was w 2 (0 . . . s . . . 0), where s occurs in the ith component. If, there is more than one such instance, do not decode., 39. When the code is binary, category 3 reduces to the following: If, wH is the ith row of H for exactly one i, assume that an error was, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 537, , made in the ith component of w. If wH is more than one row of H,, do not decode., 4. If wH does not fit into either category 2 or category 3, we know that, at least two errors occurred in transmission and we do not decode., EXAMPLE 9 Consider the Hamming (7, 4) code given in Example, 1. The generator matrix is, 1, 0, G≥, 0, 0, , 0, 1, 0, 0, , 0, 0, 1, 0, , 0, 0, 0, 1, , 1, 1, 1, 0, , 1, 0, 1, 1, , 0, 1, ¥, 1, 1, , and the corresponding parity-check matrix is, 1, 1, 1, H 5 G0, 1, 0, 0, , 1, 0, 1, 1, 0, 1, 0, , 0, 1, 1, 1W., 0, 0, 1, , Now, if the received vector is v 5 0000110, we find vH 5 110. Since this, is the first row of H and no other row, we assume that an error has been, made in the first position of v. Thus, the transmitted code word is assumed to be 1000110, and the corresponding message is assumed to be, 1000. Similarly, if w 5 1011111 is the received word, then wH 5 101,, and we assume that an error has been made in the second position. So,, we assume that 1111111 was sent and that 1111 was the intended message. If the encoded message 1001101 is received as z 5 1001011 (with, errors in the fifth and sixth positions), we find zH 5 110. Since this, matches the first row of H, we decode z as 0001011 and incorrectly, assume that the message 0001 was intended. On the other hand, nearestneighbor decoding would yield the same incorrect result., Notice that when only one error was made in transmission, the, parity-check decoding procedure gave us the originally intended message. We will soon see under what conditions this is true, but first we, need an important fact relating a code given by a generator matrix and, its parity-check matrix., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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538, , Special Topics, , Lemma Orthogonality Relation, Let C be a systematic (n, k) linear code over F with a standard, generator matrix G and parity-check matrix H. Then, for any vector v, in F n, we have vH 5 0 (the zero vector) if and only if v belongs to C., , PROOF First note that, since H has rank n 2 k, we may think of H as, a linear transformation from F n onto F n2k. Therefore, it follows from, the dimension theorem for linear transformations that n 5, n 2 k 1 dim (Ker H), so that Ker H has dimension k. (Alternatively,, one can use a group theory argument to show that |Ker H| 5 |F|k.), Then, since the dimension of C is also k, it suffices to show that, A, C # Ker H. To do this, let G 5 [Ik | A], so that H 5 c, d . Then,, Ink, A, d 5 2A 1 A 5 [0], (the zero matrix)., GH 5 [Ik | A] c, Ink, Now, by definition, any vector v in C has the form mG, where m is a, message vector. Thus, vH 5 (mG)H 5 m(GH) 5 m[0] 5 0 (the zero, vector)., Because of the way H was defined, the parity-check matrix method, correctly decodes any received word in which no error has been made., But it will do more., Theorem 31.3 Parity-Check Matrix Decoding, Parity-check matrix decoding will correct any single error if and only, if the rows of the parity-check matrix are nonzero and no one row is, a scalar multiple of any other row., , PROOF For simplicity’s sake, we prove only the binary case. In this, special situation, the condition on the rows is that they are nonzero and, distinct. So, let H be the parity-check matrix, and let’s assume that this, condition holds for the rows. Suppose that the transmitted code word w, was received with only one error, and that this error occurred in the ith, position. Denoting the vector that has a 1 in the ith position and 0’s elsewhere by ei, we may write the received word as w 1 ei. Now, using the, Orthogonality Lemma, we obtain, (w 1 ei)H 5 wH 1 eiH 5 0 1 eiH 5 eiH., But this last vector is precisely the ith row of H. Thus, if there was exactly one error in transmission, we can use the rows of the parity-check, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 539, , matrix to identify the location of the error, provided that these rows are, distinct. (If two rows, say, the ith and jth, are the same, we know that, the error occurred in either the ith position or the jth position, but we, do not know in which.), Conversely, suppose that the parity-check matrix method correctly, decodes all received words in which at most one error has been made, in transmission. If the ith row of the parity-check matrix H were, the zero vector and if the code word u 5 0 ? ? ? 0 were received as ei,, we would find eiH 5 0 ? ? ? 0, and we would erroneously assume that, the vector ei was sent. Thus, no row of H is the zero vector. Now, suppose that the ith row of H and the jth row of H are equal and i 2 j., Then, if some code word w is transmitted and the received word is, w 1 ei (that is, there is a single error in the ith position), we find, (w 1 ei)H 5 wH 1 eiH 5 ith row of H 5 jth row of H., Thus, our decoding procedure tells us not to decode. This contradicts, our assumption that the method correctly decodes all received words in, which at most one error has been made., , Coset Decoding, There is another convenient decoding method that utilizes the fact that, an (n, k) linear code C over a finite field F is a subgroup of the additive, group of V 5 Fn. This method was devised by David Slepian in 1956, and is called coset decoding (or standard decoding). To use this, method, we proceed by constructing a table, called a standard array., The first row of the table is the set C of code words, beginning in column 1 with the identity 0 ? ? ? 0. To form additional rows of the table,, choose an element v of V not listed in the table thus far. Among all the, elements of the coset v 1 C, choose one of minimum weight, say, v9., Complete the next row of the table by placing under the column headed, by the code word c the vector v9 1 c. Continue this process until all the, vectors in V have been listed in the table. [Note that an (n, k) linear code, over a field with q elements will have |V:C| 5 qn2k rows.] The words, in the first column are called the coset leaders. The decoding procedure, is simply to decode any received word w as the code word at the head, of the column containing w., EXAMPLE 10 Consider the (6, 3) binary linear code, C 5 {000000, 100110, 010101, 001011, 110011, 101101, 011110, 111000}., The first row of a standard array is just the elements of C. Obviously,, 100000 is not in C and has minimum weight among the elements of, 100000 1 C, so it can be used to lead the second row. Table 31.4 is the, completed table., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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540, , Special Topics, , Table 31.4 A Standard Array for a (6, 3) Linear Code, Words, Coset, Leaders, 000000, 100000, 010000, 001000, 000100, 000010, 000001, 100001, , 100110, 000110, 110110, 101110, 100010, 100100, 100111, 000111, , 010101, 110101, 000101, 011101, 010001, 010111, 010100, 110100, , 001011, 101011, 011011, 000011, 001111, 001001, 001010, 101010, , 110011, 010011, 100011, 111011, 110111, 110001, 110010, 010010, , 101101, 001101, 111101, 100101, 101001, 101111, 101100, 001100, , 011110, 111110, 001110, 010110, 011010, 011100, 011111, 111111, , 111000, 011000, 101000, 110000, 111100, 111010, 111001, 011001, , If the word 101001 is received, it is decoded as 101101, since, 101001 lies in the column headed by 101101. Similarly, the received, word 011001 is decoded as 111000., Recall that the first method of decoding that we introduced was the, nearest-neighbor method; that is, any received word w is decoded as, the code word c such that d(w, c) is a minimum, provided that there is, only one code word c such that d(w, c) is a minimum. The next result, shows that in this situation, coset decoding is the same as nearestneighbor decoding., Theorem 31.4 Coset Decoding Is Nearest-Neighbor Decoding, In coset decoding, a received word w is decoded as a code word c such, that d(w, c) is a minimum., , PROOF Let C be a linear code, and let w be any received word. Suppose, that v is the coset leader for the coset w 1 C. Then, w 1 C 5 v 1 C, so, w 5 v 1 c for some c in C. Thus, using coset decoding, w is decoded, as c. Now, if c9 is any code word, then w 2 c9 [ w 1 C 5 v 1 C, so, that wt(w 2 c9) $ wt(v), since the coset leader v was chosen as a vector, of minimum weight among the members of v 1 C., Therefore,, d(w, c9) 5 wt(w 2 c9) $ wt(v) 5 wt(w 2 c) 5 d(w, c)., So, using coset decoding, w is decoded as a code word c such that, d(w, c) is a minimum., Recall that in our description of nearest-neighbor decoding, we, stated that if the choice for the nearest neighbor of a received word v is, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 541, , not unique, then we can decide not to decode or to decode v arbitrarily, from among those words closest to v. In the case of coset decoding, the, decoded value of v is always uniquely determined by the coset leader, of the row containing the received word. Of course, this decoded value, may not be the word that was sent., When we know a parity-check matrix for a linear code, coset decoding can be considerably simplified., Definition Syndrome, , If an (n, k) linear code over F has parity-check matrix H, then, for any, vector u in Fn, the vector uH is called the syndrome† of u., , The importance of syndromes stems from the following property., Theorem 31.5 Same Coset—Same Syndrome, Let C be an (n, k) linear code over F with a parity-check matrix H., Then, two vectors of Fn are in the same coset of C if and only if they, have the same syndrome., , PROOF Two vectors u and v are in the same coset of C if and only if, u 2 v is in C. So, by the Orthogonality Lemma, u and v are in the same, coset if and only if 0 5 (u 2 v)H 5 uH 2 vH., We may now use syndromes for decoding any received word w:, 1. Calculate wH, the syndrome of w., 2. Find the coset leader v such that wH 5 vH., 3. Assume that the vector sent was w 2 v., With this method, we can decode any received word with a table that, has only two rows—one row of coset leaders and another row with the, corresponding syndromes., EXAMPLE 11 Consider the code given in Example 10. The paritycheck matrix for this code is, 1, 1, 0, H5F, 1, 0, 0, †This, , 1, 0, 1, 0, 1, 0, , 0, 1, 1, V., 0, 0, 1, , term was coined by D. Hagelbarger in 1959., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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542, , Special Topics, , The list of coset leaders and corresponding syndromes is the following., Coset leader, , 000000 100000, , 010000 001000, , 000100 000010 000001 100001, , Syndromes, , 000, , 101, , 100, , 110, , 011, , 010, , 001, , 111, , So, to decode the received word w 101001, we compute wH 100., Since the coset leader v 000100 has 100 as its syndrome, we assume that, w 000100 101101 was sent. If the received word is w 011001,, we compute wH 111 and assume w 100001 111000 was, sent because 100001 is the coset leader with syndrome 111. Notice that, these answers are in agreement with those obtained by using the standardarray method of Example 10., The term syndrome is a descriptive term. In medicine, it is used to, designate a collection of symptoms that typify a disorder. In coset decoding, the syndrome typifies an error pattern., In this chapter, we have presented algebraic coding theory in, its simplest form. A more sophisticated treatment would make substantially greater use of group theory, ring theory, and especially finite-field, theory. For example, Gorenstein (see Chapter 25 for a biography) and, Zierler, in 1961, made use of the fact that the multiplicative subgroup, of a finite field is cyclic. They associated each digit of certain codes, with a field element in such a way that an algebraic equation would be, derived whose zeros determined the locations of the errors., In some instances, two error-correcting codes are employed. The, European Space Agency space probe Giotto, which came within, 370 miles of the nucleus of Halley’s Comet in 1986, had two errorcorrecting codes built into its electronics. One code checked for, indepen dently occurring errors, and another—a so-called Reed–, Solomon code—checked for bursts of errors. Giotto achieved an errordetection rate of 0.999999. Reed–Solomon codes are also used on, compact discs. They can correct thousands of consecutive errors., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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HISTORICAL NOTE, , SIAM News (January 1993, Volume 25, Number 1,, pages 1 and 11), , The Ubiquitous, Reed–Solomon, Codes*, Irving Reed and Gustave Solomon, monitor the encounter of Voyager II, with Neptune at the Jet Propulsion, Laboratory in 1989., , We conclude this chapter with an adapted version of an article by Barry A. Cipra about, the Reed–Solomon codes [1]. It was the first in a series of articles called “Mathematics, That Counts” in SIAM News, the news journal of the Society for Industrial and Applied, Mathematics. The articles highlight developments in mathematics that have led to, products and processes of substantial benefit to industry and the public., In this “Age of Information,” no one need be, reminded of the importance not only of, speed but also of accuracy in the storage, retrieval, and transmission of data. Machines, do make errors, and their non-man-made, mistakes can turn otherwise flawless programming into worthless, even dangerous,, trash. Just as architects design buildings that, will remain standing even through an earthquake, their computer counterparts have, come up with sophisticated techniques capable of counteracting digital disasters., The idea for the current error-correcting, techniques for everything from computer, hard disk drives to CD players was first introduced in 1960 by Irving Reed and, Gustave Solomon, then staff members at, MIT’s Lincoln Laboratory. . . ., “When you talk about CD players and digital audio tape and now digital television, and, various other digital imaging systems that are, coming—all of those need Reed–Solomon, [codes] as an integral part of the system,” says, Robert McEliece, a coding theorist in the, electrical engineering department at Caltech., , Why? Because digital information, virtually by definition, consists of strings of, “bits”—0s and 1s—and a physical device,, no matter how capably manufactured, may, occasionally confuse the two. Voyager II,, for example, was transmitting data at incredibly low power—barely a whisper—, over tens of millions of miles. Disk drives, pack data so densely that a read/write head, can (almost) be excused if it can’t tell where, one bit stops and the next 1 (or 0) begins., Careful engineering can reduce the error, rate to what may sound like a negligible, level—the industry standard for hard disk, drives is 1 in 10 billion—but given the volume of information processing done these, days, that “negligible” level is an invitation, to daily disaster. Error-correcting codes are, a kind of safety net—mathematical insurance against the vagaries of an imperfect, material world., In 1960, the theory of error-correcting, codes was only about a decade old. The, basic theory of reliable digital communication had been set forth by Claude Shannon, , *Adapted version of an article called, “The Ubiquitous Reed-Solomon Codes” in SIAM News, the, news journal of the Society for Industrial and Applied Mathematics, by Barry A. Cipra. Reprinted from, SIAM News, Volume 26-1, January 1993., 543, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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in the late 1940s. At the same time, Richard, Hamming introduced an elegant approach to, single-error correction and double-error, detection. Through the 1950s, a number of, researchers began experimenting with a, variety of error-correcting codes. But with, their SIAM journal paper, McEliece says,, Reed and Solomon “hit the jackpot.”, The payoff was a coding system based on, groups of bits—such as bytes—rather than, individual 0s and 1s. That feature makes, Reed–Solomon codes particularly good at, dealing with “bursts” of errors: six consecutive bit errors, for example, can affect at, most two bytes. Thus, even a double-errorcorrection version of a Reed–Solomon code, can provide a comfortable safety factor. . . ., Mathematically, Reed–Solomon codes, are based on the arithmetic of finite fields., Indeed, the 1960 paper begins by defining a, code as “a mapping from a vector space of, dimension m over a finite field K into a vector space of higher dimension over the same, field.” Starting from a “message” (a 0, a 1,, . . . , am21), where each ak is an element of, the field K, a Reed–Solomon code produces, (P(0), P(g), P(g2), . . . , P(gN21)), where N is, the number of elements in K, g is a generator of the (cyclic) group of nonzero elements in K, and P(x) is the polynomial a0 1, a1x 1 ? ? ? 1 am21x m21. If N is greater than, m, then the values of P overdetermine the, polynomial, and the properties of finite, fields guarantee that the coefficients of, P—i.e., the original message—can be recovered from any m of the values . . . ., In today’s byte-sized world, for example,, it might make sense to let K be the field of, order 28, so that each element of K corresponds to a single byte (in computerese, there, are four bits to a nibble and two nibbles to a, byte). In that case, N 5 28 5 256, and hence, , messages up to 251 bytes long can be recovered even if two errors occur in transmitting, the values P(0), P(g), . . . , P(g255). That’s a, lot better than the 1255 bytes required by the, say-everything-five-times approach., Despite their advantages, Reed–Solomon, codes did not go into use immediately—, they had to wait for the hardware technology to catch up. “In 1960, there was no such, thing as fast digital electronics”—at least, not by today’s standards, says McEliece., The Reed–Solomon paper “suggested some, nice ways to process data, but nobody knew, if it was practical or not, and in 1960 it, probably wasn’t practical.”, But technology did catch up, and numerous researchers began to work on implementing the codes. . . . Many other bells, and whistles (some of fundamental theoretic significance) have also been added., Compact discs, for example, use a version, of a Reed–Solomon code., Reed was among the first to recognize, the significance of abstract algebra as the, basis for error-correcting codes. “In hindsight it seems obvious,” he told SIAM News., However, he added, “coding theory was not, a subject when we published that paper.”, The two authors knew they had a nice result;, they didn’t know what impact the paper, would have., Three decades later, the impact is clear., The vast array of applications, both current, and pending, has settled the question of the, practicality and significance of Reed–, Solomon codes. “It’s clear they’re practical,, because everybody’s using them now,” says, Elwyn Berkekamp. Billions of dollars in, modern technology depend on ideas that, stem from Reed and Solomon’s original, work. In short, says McEliece, “it’s been an, extraordinarily influential paper.”, , 544, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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545, , 31 | Introduction to Algebraic Coding Theory, , Exercises, The New Testament offers the basis for modern computer coding theory, in, the form of an affirmation of the binary number system., “But let your communication be yea, yea; nay, nay: for whatsoever is, more than these cometh of evil.”, anonymous, , 1. Find the Hamming weight of each code word in Table 31.1., 2. Find the Hamming distance between the following pairs of vectors: {1101, 0111}, {0220, 1122}, {11101, 00111}., 3. Referring to Example 1, use the nearest-neighbor method to decode the received words 0000110 and 1110100., 4. For any vector space V and any u, v, w in F n, prove that the, Hamming distance has the following properties., a. d(u, v) 5 d(v, u) (symmetry)., b. d(u, v) 5 0 if and only if u 5 v., c. d(u, v) 5 d(u 1 w, v 1 w) (translation invariance)., 5. Determine the (6, 3) binary linear code with generator matrix, 1 0 0 0 1 1, G £ 0 1 0 1 0 1§ ., 0 0 1 1 1 0, 6. Show that for binary vectors, wt(u 1 v) $ wt(u) 2 wt(v) and, equality occurs if and only if for all i the ith component of u is 1, whenever the ith component of v is 1., 7. If the minimum weight of any nonzero code word is 2, what can, we say about the error-detecting capability of the code?, 8. Suppose that C is a linear code with Hamming weight 3 and that, C9 is one with Hamming weight 4. What can C9 do that C can’t?, 9. Let C be a binary linear code. Show that the code words of even, weight form a subcode of C. (A subcode of a code is a subset of, the code that is itself a code.), 10. Let, C 5 {0000000, 1110100, 0111010, 0011101, 1001110,, 0100111, 1010011, 1101001}., What is the error-correcting capability of C? What is the errordetecting capability of C?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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546, , Special Topics, , 11. Suppose that the parity-check matrix of a binary linear code is, 1, 0, H 5 E1, 1, 0, , 0, 1, 1U ., 0, 1, , Can the code correct any single error?, 12. Use the generator matrix, Gc, , 1 0 1 1, d, 0 1 2 1, , to construct a (4, 2) ternary linear code. What is the parity-check, matrix for this code? What is the error-correcting capability of this, code? What is the error-detecting capability of this code? Use paritycheck decoding to decode the received word 1201., 13. Find all code words of the (7, 4) binary linear code whose generator matrix is, 1, 0, G ≥, 0, 0, , 0, 1, 0, 0, , 0, 0, 1, 0, , 0, 0, 0, 1, , 1, 1, 1, 0, , 1, 0, 1, 1, , 1, 1, ¥., 0, 1, , Find the parity-check matrix of this code. Will this code correct, any single error?, 14. Show that in a binary linear code, either all the code words end with, 0, or exactly half end with 0. What about the other components?, 15. Suppose that a code word v is received as the vector u. Show that, coset decoding will decode u as the code word v if and only if u 2 v, is a coset leader., 16. Consider the binary linear code, C 5 {00000, 10011, 01010, 11001, 00101, 10110, 01111, 11100}., Construct a standard array for C. Use nearest-neighbor decoding to, decode 11101 and 01100. If the received word 11101 has exactly, one error, can we determine the intended code word? If the received word 01100 has exactly one error, can we determine the intended code word?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 547, , 17. Construct a (6, 3) binary linear code with generator matrix, 1 0 0 1 1 0, G £0 1 0 0 1 1§., 0 0 1 1 0 1, Decode each of the received words, 001001, 011000, 000110, 100001, by the following methods:, a. Nearest-neighbor method., b. Parity-check matrix method., c. Coset decoding using a standard array., d. Coset decoding using the syndrome method., 18. Suppose that the minimum weight of any nonzero code word in a, linear code is 6. Discuss the possible options for error correction, and error detection., 19. Using the code and the parity-check matrix given in Example 9,, show that parity-check matrix decoding cannot detect any multiple, errors (that is, two or more errors)., 20. Suppose that the last row of a standard array for a binary linear, code is, 10000 00011 11010 01001 10101 00110 11111 01100., Determine the code., 21. How many code words are there in a (6, 4) ternary linear code?, How many possible received words are there for this code?, 22. If the parity-check matrix for a binary linear code is, 1, 0, 1, H5F, 1, 0, 0, , 1, 1, 0, 0, 1, 0, , 0, 1, 1, V,, 0, 0, 1, , will the code correct any single error? Why?, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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548, , Special Topics, , 23. Suppose that the parity-check matrix for a ternary code is, 2, 2, H 5 E1, 1, 0, , 1, 2, 2U., 0, 1, , Can the code correct all single errors? Give a reason for your, answer., 24. Prove that for nearest-neighbor decoding, the converse of Theorem 31.2 is true., 25. Can a (6, 3) binary linear code be double-error-correcting using the, nearest-neighbor method? Do not assume that the code is systematic., 26. Prove that there is no 2 3 5 standard generator matrix G that will, produce a (5, 2) linear code over Z3 capable of detecting all possible triple errors., 27. Why can’t the nearest-neighbor method with a (4, 2) binary linear, code correct all single errors?, 28. Suppose that one row of a standard array for a binary code is, 000100 110000 011110 111101 101010 001001 100111 010011., Determine the row that contains 100001., 29. Use the field F 5 Z2[x]/kx2 1 x 1 1l to construct a (5, 2) linear, code that will correct any single error., 30. Find the standard generator matrix for a (4, 2) linear code over Z3, that encodes 20 as 2012 and 11 as 1100. Determine the entire code, and the parity-check matrix for the code. Will the code correct all, single errors?, 31. Assume that C is an (n, k) binary linear code and that, for each position i 5 1, 2, . . . , n, the code C has at least one vector with a 1 in the, ith position. Show that the average weight of a code word is n/2., 32. Let C be an (n, k) linear code over F such that the minimum weight, of any nonzero code word is 2t 1 1. Show that not every vector of, weight t 1 1 in Fn can occur as a coset leader., 33. Let C be an (n, k) binary linear code over F 5 Z 2. If v [ Fn but, v o C, show that C < (v 1 C) is a linear code., 34. Let C be a binary linear code. Show that either every member of C, has even weight or exactly half the members of C have even, weight. (Compare with Exercise 23 in Chapter 5.), 35. Let C be an (n, k) linear code. For each i with 1 # i # n, let Ci 5, {v [ C | the ith component of v is 0}. Show that Ci is a subcode of C., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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31 | Introduction to Algebraic Coding Theory, , 549, , Reference, 1. Barry A. Cipra, “The Ubiquitous Reed-Solomon Codes,” SIAM News, 26 (January 1993): 1, 11., , Suggested Readings, Norman Levinson, “Coding Theory: A Counterexample to G. H. Hardy’s, Conception of Applied Mathematics,” The American Mathematical, Monthly 77 (1970): 249–258., The eminent mathematician G. H. Hardy insisted that “real” mathematics was almost wholly useless. In this article, the author argues that, coding theory refutes Hardy’s notion. Levinson uses the finite field of, order 16 to construct a linear code that can correct any three errors., T. M. Thompson, From Error-Correcting Codes Through Sphere Packings,, to Simple Groups, Washington, D.C.: The Mathematical Association of, America, 1983., Chapter 1 of this award-winning book gives a fascinating historical, account of the origins of error-correcting codes., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Richard W. Hamming, For introduction of error-correcting codes,, pioneering work in operating systems and, programming languages, and the advancement of numerical computation., , Richard W. Hamming, , Citation for the Piore, Award, 1979, , Richard W. Hamming was born in Chicago,, Illinois, on February 11, 1915. He graduated, from the University of Chicago with a B.S. degree in mathematics. In 1939, he received an, M.A. degree in mathematics from the University of Nebraska and, in 1942, a Ph.D. in mathematics from the University of Illinois., During the latter part of World War II,, Hamming was at Los Alamos, where he was, involved in computing atomic-bomb designs., In 1946, he joined Bell Telephone Laboratories, where he worked in mathematics, computing, engineering, and science., In 1950, Hamming published his famous, paper on error-detecting and error-correcting, codes. This work started a branch of information theory. The Hamming codes are used in, many modern computers. Hamming’s work, in the field of numerical analysis has also, been of fundamental importance., , Hamming received numerous prestigious awards, including the Turing Prize, from the Association for Computing Machinery, the Piore Award from the Institute, of Electrical and Electronics Engineers, (IEEE), and the Oender Award from the, University of Pennsylvania. In 1986 the IEEE, Board of Directors established the Richard, W. Hamming Medal “for exceptional contributions to information sciences, systems, and technology” and named Hamming as, its first recipient. Hamming died of a heart, attack on January 7, 1998, at age 82., To find more information about Hamming, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , 550, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Jessie MacWilliams, , Walter MacWilliams, , She was a mathematician who was instrumental in developing the mathematical, theory of error-correcting codes from its, early development and whose Ph.D. thesis, includes one of the most powerful theorems in coding theory., vera pless, SIAM News, , An important contributor to coding theory, was Jessie MacWilliams. She was born in, 1917 in England. After studying at Cambridge University, MacWilliams came to, the United States in 1939 to attend Johns, Hopkins University. After one year at Johns, Hopkins, she went to Harvard for a year., In 1955, MacWilliams became a programmer at Bell Labs, where she learned, about coding theory. Although she made a, major discovery about codes while a programmer, she could not obtain a promotion, to a math research position without a Ph.D., degree. She completed some of the requirements for the Ph.D. while working full-time, at Bell Labs and looking after her family., She then returned to Harvard for a year, (1961–1962), where she finished her degree., Interestingly, both MacWilliams and her, , daughter Ann were studying mathematics at, Harvard at the same time., MacWilliams returned to Bell Labs,, where she remained until her retirement in, 1983. While at Bell Labs, she made many, contributions to the subject of error-correcting, codes, including The Theory of ErrorCorrecting Codes, written jointly with Neil, Sloane. One of her results of great theoretical, importance is known as the MacWilliams, Identity. She died on May 27, 1990, at the, age of 73., To find more information about, MacWilliams, visit:, http://www.awm-math.org/, noetherbrochure/, MacWilliams80.html, , 551, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Vera Pless, , Vera Pless, , Vera Pless is a leader in the field of coding, theory., , Vera Pless was born on March 5, 1931, to, Russian immigrants on the West Side of, Chicago. She accepted a scholarship to attend, the University of Chicago at age 15. The program at Chicago emphasized great literature, but paid little attention to physics and mathematics. At age 18, with no more than one precalculus course in mathematics, she entered, the prestigious graduate program in mathematics at Chicago, where, at that time, there, were no women on the mathematics faculty, or even women colloquium speakers. After, passing her master’s exam, she took a job as a, research associate at Northwestern University while pursuing a Ph.D. there. In 1957,, she obtained her degree., Over the next several years, Pless stayed, at home to raise her children while teaching, , part-time at Boston University. When she, decided to work full-time, she found that, women were not welcome at most colleges, and universities. One person told her outright, “I would never hire a woman.” Fortunately, there was an Air Force Lab in the area, that had a group working on error-correcting, codes. Although she had never even heard of, coding theory, she was hired because of her, background in algebra. When the lab discontinued basic research, she took a position as, a research associate at MIT in 1972. In 1975,, she went to the University of Illinois–Chicago,, where she remained until her retirement., During her career, Pless wrote more than, 100 research papers, authored a widely used, textbook on coding theory, and had 11 Ph. D., students., , 552, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32, , An Introduction to, Galois Theory, Galois theory is a showpiece of mathematical unification, bringing together, several different branches of the subject and creating a powerful machine, for the study of problems of considerable historical and mathematical, importance., ian stewart, Galois Theory, , Fundamental Theorem of Galois Theory, The Fundamental Theorem of Galois Theory is one of the most elegant, theorems in mathematics. Look at Figures 32.1 and 32.2. Figure 32.1, depicts the lattice of subgroups of the group of field automorphisms of, 4, 2 , i). The integer along an upward lattice line from a group H1 to, Q(2, a group H2 is the index of H1 in H2. Figure 32.2 shows the lattice of, 4, 2, i). The integer along an upward line from a field, subfields of Q(2, K1 to a field K2 is the degree of K2 over K1. Notice that the lattice in, Figure 32.2 is the lattice of Figure 32.1 turned upside down. This is, only one of many relationships between these two lattices. The Fundamental Theorem of Galois Theory relates the lattice of subfields of an, algebraic extension E of a field F to the subgroup structure of the group, {e, a, a2, a3, b, ab, a2b, a3b}, 2, , 2, , {e, a, a 2, a 3}, , {e, a 2 , b, a2b }, 2, , {e, b}, , 2, , 2, , 2, , 2, , 2, , {e, a 2}, , {e, a 2b}, 2, , {e, a 2 , ab, a 3b }, , 2, , 2, , 2, , {e, ab }, 2, , 2, , {e, a3b }, , 2, , { e}, , Figure 32.1 Lattice of subgroups of the group of field automorphisms of, 4, 4, 4, 4, 4, Q(22, i), where a: i S i and 22 S 2i 22, b: i S 2i, and 22 S 22., 553, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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554, , Special Topics, 4, , Q(√2, i), 2, 4, , Q(√2), 2, , 2, , 2, , 4, , 2, , 2, , 2, , 2, , 2, , Q(√2), , Q((12 i)√2), 2, , Q((11 i)√2), 2, , 2, , Q(i√2), , Q(i), 2, , 2, , 4, , 4, , Q(√2, i), , Q(i√2), , 2, , Q, 4, , Figure 32.2 Lattice of subfields of Q(22, i)., , of automorphisms of E that send each element of F to itself. This relationship was discovered in the process of attempting to solve a polynomial equation f(x) 5 0 by radicals., Before we can give a precise statement of the Fundamental Theorem, of Galois Theory, we need some terminology and notation., Definitions Automorphism, Galois Group, Fixed Field of H, , Let E be an extension field of the field F. An automorphism of E is a, ring isomorphism from E onto E. The Galois group of E over F,, Gal(E/F), is the set of all automorphisms of E that take every element, of F to itself. If H is a subgroup of Gal(E/F ), the set, EH 5 {x [ E | f(x) 5 x for all f [ H}, is called the fixed field of H., , It is easy to show that the set of automorphisms of E forms a group, under composition. We leave as exercises (Exercises 3 and 5) the verifications that the automorphism group of E fixing F is a subgroup of, the automorphism group of E and that, for any subgroup H of, Gal(E/F), the fixed field EH of H is a subfield of E. Be careful not to, misinterpret Gal(E/F) as something that has to do with factor rings or, factor groups. It does not., The following examples will help you assimilate these definitions. In, each example, we simply indicate how the automorphisms are defined., We leave to the reader the verifications that the mappings are indeed, automorphisms., EXAMPLE 1 Consider the extension Q(22) of Q. Since, Q(22) 5 {a 1 b22 | a, b [ Q}, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 555, , and any automorphism of a field containing Q must act as the identity, on Q (Exercise 1), an automorphism f of Q(22) is completely determined by f(22). Thus,, 2 5 f(2) 5 f(2222) 5 (f(22))2,, and therefore f(22) 5 622. This proves that the group Gal(Q(22)/Q), has two elements, the identity mapping and the mapping that sends a 1, b22 to a 2 b 22., 3, EXAMPLE 2 Consider the extension Q(2, 2 ) of Q. An automor3, 3, 2 ). By an arguphism f of Q(22 ) is completely determined by f(2, 3, 2 ) must be a, ment analogous to that in Example 1, we see that f(2, 3, 3, 2 is, cube root of 2. Since Q(22 ) is a subset of the real numbers and 2, 3, 3, 2) 5 2, 2 . Thus, f is, the only real cube root of 2, we must have f(2, 3, 2 )/Q) has only one element., the identity automorphism and Gal(Q(2, 3, 3, 2 )., Obviously, the fixed field of Gal(Q( 22 )/Q) is Q (2, 4, EXAMPLE 3 Consider the extension Q(2, 2 , i) of Q(i). Any auto4, morphism f of Q(22 , i) fixing Q(i) is completely determined by, 4, 2 ). Since, f(2, 4, 4, 2 )4) 5 (f(2, 2 ))4,, 2 5 f(2) 5 f((2, 4, 2 ) must be a fourth root of 2. Thus, there are at most, we see that f(2, 4, 2 , i) fixing Q(i). If we define an, four possible automorphisms of Q(2, 4, 4, 2 ) 5 i2, 2 , then a [, automorphism a such that a(i) 5 i and a(2, 4, 4, Gal(Q(22 , i)/Q(i)) and a has order 4. Thus, Gal(Q(22 , i)/Q(i)) is a, cyclic group of order 4. The fixed field of {e, a2} (where e is the identity, 4, 2 , i)/, automorphism) is Q(22, i). The lattice of subgroups of Gal(Q(2, 4, Q(i)) and the lattice of subfields of Q(22 , i) containing Q(i) are shown, in Figure 32.3. As in Figures 32.1 and 32.2, the integers along the lines, , {e, a, a 2, a 3}, 2, , {e, a 2}, , 4, , Q(√2, i), 2, Q(√2, i), , 2, , 2, , {e}, , Q(i), 4, , Figure 32.3 Lattice of subgroups of Gal (Q(22, i)/Q(i)), 4, and lattice of subfields of Q(22, i) containing Q(i)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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556, , Special Topics, , of the group lattice represent the index of a subgroup in the group, above it, and the integers along the lines of the field lattice represent, the degree of the extension of a field over the field below it., EXAMPLE 4 Consider the extension Q(23, 25) of Q. Since, Q(23, 25) 5 {a 1 b23 1 c25 1 d2325 | a, b, c, d [ Q},, any automorphism f of Q(23, 25) is completely determined by the, two values f(23) and f(25). This time there are four automorphisms., e, , a, , b, , ab, , 23 S 23, , 2 3 S 22 3, , 23 S 23, , 23 S 223, , 25 S 25, , 25 S 25, , 25 S 225, , 25 S 225, , Obviously, Gal(Q(23, 25)/Q) is isomorphic to Z2 % Z2. The fixed, field of {e, a} is Q( 25), the fixed field of {e, b} is Q( 23), and, the fixed field of {e, ab} is Q(2325). The lattice of subgroups of, Gal(Q(23, 25)/Q) and the lattice of subfields of Q(23, 25) are shown, in Figure 32.4., {e, a, b, ab}, , 2, , Q(√3,√5 ), , {e, b }, , {e, a }, , {e, ab }, , 2, , 2, , 2, , 2, , 2, , 2, , Q(√5 ), , 2, , 2, , Q (√3 ), , 2, , 2, , {e}, , Q(√3√5 ), , 2, , Q, , Figure 32.4 Lattice of subgroups of Gal(Q( 13, 15)/Q) and lattice of subfields, of Q ( 13, 15)., , Example 5 is a bit more complicated than our previous examples. In, particular, the automorphism group is non-Abelian., EXAMPLE 5 Direct calculations show that v 5 21/2 1 i23/2 satisfies the equations v3 5 1 and v2 1 v 1 1 5 0. Now, consider the, 3, 2 ) of Q. We may describe the automorphisms of, extension Q(v, 2, 3, 3, 2 . There are six in all., Q(v, 22 ) by specifying how they act on v and 2, e, , a, , vSv, 3, 12 S 1, 2, , v S v2, 3, 12 S 1, 2, , 3, , 3, , b2, , b, vSv, 3, 1 2 S v1, 2, 3, , vSv, 3, 1 2 S v2 1, 2, 3, , ab2, , ab, v S v2, 3, 1 2 S v2 1, 2, 3, , v S v2, 3, 1 2 S v1, 2, 3, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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557, , 32 | An Introduction to Galois Theory, , 3, Since ab 2 ba, we know that Gal(Q(v, 2, 2 )/Q) is isomorphic to S3., (See Theorem 7.2.) The lattices of subgroups and subfields are shown, in Figure 32.5., , {e, a, b, b 2,ab, ab 2}, 2, , 3, , {e, b, b 2}, , 3, , {e, a }, 3, , 3, , {e, ab }, , 2, , 2, , {e, ab 2}, 2, , {e}, 3, , Q(ω ,√2), 3, Q(ω ), , 3, , 3, , 3, , 2 3, , Q(ω √2), , Q(√2), 2, , 2, , 2, , 2, , 3, , Q(ω √2), 3, , Q, 3, , Figure 32.5 Lattice of subgroups of Gal(Q(v, 22 )/Q) and lattice, 3, of subfields of Q(v, 22 ), where v 5 21/2 1 i 23/2., , The lattices in Figure 32.5 have been arranged so that each nontrivial, proper field occupying the same position as some group is the fixed field, 3, 2 ) is the fixed field of {e, ab}., of that group. For instance, Q(v2, The preceding examples show that, in certain cases, there is an intimate connection between the lattice of subfields between E and F and, the lattice of subgroups of Gal(E/F). In general, if E is an extension of F,, and we let ^ be the lattice of subfields of E containing F and let & be, the lattice of subgroups of Gal(E/F), then for each K in ^, the group, Gal(E/K) is in &, and for each H in &, the field EH is in ^. Thus, we, may define a mapping g: ^ S & by g(K) 5 Gal(E/K) and a mapping, f : & S ^ by f (H) 5 EH. It is easy to show that if K and L belong to ^, and K # L, then g(K) $ g(L). Similarly, if G and H belong to & and, G # H, then f (G) $ f (H). Thus, f and g are inclusion-reversing mappings between ^ and &. We leave it to the reader to show that for any K, in ^, we have (fg)(K) $ K, and for any G in &, we have (gf )(G) $ G., When E is an arbitrary extension of F, these inclusions may be strict., However, when E is a suitably chosen extension of F, the Fundamental, Theorem of Galois Theory, Theorem 32.1, says that f and g are inverses, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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558, , Special Topics, , of each other, so that the inclusions are equalities. In particular, f and g, are inclusion-reversing isomorphisms between the lattices ^ and &., A stronger result than that given in Theorem 32.1 is true, but our theorem illustrates the fundamental principles involved. The student is, referred to [1, p. 285] for additional details and proofs., Theorem 32.1 Fundamental Theorem of Galois Theory, Let F be a field of characteristic 0 or a finite field. If E is the splitting, field over F for some polynomial in F[x], then the mapping from the, set of subfields of E containing F to the set of subgroups of Gal(E/F), given by K S Gal(E/K) is a one-to-one correspondence. Furthermore, for any subfield K of E containing F,, 1. [E:K] 5 |Gal(E/K)| and [K:F] 5 |Gal(E/F)| / |Gal(E/K)|. [The, index of Gal(E/K) in Gal(E/F) equals the degree of K over F.], 2. If K is the splitting field of some polynomial in F[x],, then Gal(E/K) is a normal subgroup of Gal(E/F) and Gal(K/F), is isomorphic to Gal(E/F)/Gal(E/K)., 3. K 5 EGal(E/K). [The fixed field of Gal(E/K) is K.], 4. If H is a subgroup of Gal(E/F), then H 5 Gal(E/EH). [The, automorphism group of E fixing EH is H.], , Generally speaking, it is much easier to determine a lattice of subgroups than a lattice of subfields. For example, it is usually quite, difficult to determine, directly, how many subfields a given field has,, and it is often difficult to decide whether or not two extensions are, the same. The corresponding questions about groups are much more, tractable. Hence, the Fundamental Theorem of Galois Theory can be, a great labor-saving device. Here is an illustration. [Recall from, Chapter 20 that if f (x) [ F[x] and the zeros of f (x) in some extension, of F are a1, a2, . . . , an, then F(a1, a2, . . . , an) is the splitting field of, f (x) over F.], EXAMPLE 6 Let v 5 cos(2p/7) 1 i sin(2p/7), so that v7 5 1, and, consider the field Q(v). How many subfields does it have and what are, they? First, observe that Q(v) is the splitting field of x7 2 1 over Q, so, that we may apply the Fundamental Theorem of Galois Theory. A simple calculation shows that the automorphism f that sends v to v3 has, order 6. Thus,, [Q(v):Q] 5 |Gal(Q(v)/Q)| $ 6., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 559, , Also, since, x7 2 1 5 (x 2 1)(x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1), and v is a zero of x7 2 1, we see that, |Gal(Q(v)/Q)| 5 [Q(v):Q] # 6., Thus, Gal(Q(v)/Q) is a cyclic group of order 6. So, the lattice of subgroups of Gal(Q(v)/Q) is trivial to compute. See Figure 32.6., kφ l, 2, , 3, , kφ 2 l, , kφ 3 l, 3, , 2, {e}, , Figure 32.6 Lattice of subgroups of Gal(Q(v)/Q),, where v 5 cos(2p/7) 1 i sin(2p/7)., , This means that Q(v) contains exactly two proper extensions of Q:, one of degree 3 corresponding to the fixed field of kf3l and one of degree 2 corresponding to the fixed field of kf2l. To find the fixed field of, kf3l, we must find a member of Q(v) that is not in Q and that is fixed, by f3. Experimenting with various possibilities leads us to discover, that v 1 v 21 is fixed by f 3 (see Exercise 9), and it follows that, Q , Q(v 1 v21) # Q(v)kf 3l. Since [Q(v)kf 3l:Q] 5 3 and [Q(v 1, v21):Q] divides [Q(v)kf 3l:Q], we see that Q(v 1 v21) 5 Q(v)kf 3l., A similar argument shows that Q(v3 1 v5 1 v6) is the fixed field, of kf2l. Thus, we have found all subfields of Q(v)., EXAMPLE 7 Consider the extension E 5 GF(pn) of F 5 GF(p). Let, us determine Gal(E / F). By Corollary 2 of Theorem 22.2, E has the form, F(b) for some b where b is the zero of an irreducible polynomial p(x) of, the form x n 1 an21x n21 1 ? ? ? 1 a1x 1 a0, where an21, an22, . . . , a0 belong to F. Since any field automorphism f of E must take 1 to itself, it, follows that f acts as the identity on F. Thus, p(b) 5 0 implies that, p(f(b)) 5 0. And because p(x) has at most n zeros, we know that there, are at most n possibilities for f(b). On the other hand, by Exercise 49 in, Chapter 13, we know that the mapping s (a) 5 a p for all a [ E is an, automorphism of E, and it follows from the fact that E* is cyclic, (Theorem 22.2) that the group ksl has order n (see Exercise 11 in, Chapter 22). Thus, Gal(GF(p n )/GF(p)) < Zn., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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560, , Special Topics, , Solvability of Polynomials by Radicals, For Galois, the elegant correspondence between groups and fields, given by Theorem 32.1 was only a means to an end. Galois sought to, solve a problem that had stymied mathematicians for centuries., Methods for solving linear and quadratic equations were known thousands of years ago (the quadratic formula). In the 16th century, Italian mathematicians developed formulas for solving any third- or, fourth-degree equation. Their formulas involved only the operations, of addition, subtraction, multiplication, division, and extraction of, roots (radicals). For example, the equation, x3 1 bx 1 c 5 0, has the three solutions, A 1 B,, 2(A 1 B)/2 1 (A 2 B)23 /2,, 2(A 1 B)/2 2 (A 2 B)23 /2,, where, c, b3, c2, , , A3, B 2, A 27, 4, , and, , c, b3, c2, B3, , ., B 2, A 27, 4, , The formulas for the general cubic x3 1 ax2 1 bx 1 c 5 0 and the general quartic (fourth-degree polynomial) are even more complicated, but, nevertheless can be given in terms of radicals of rational expressions of, the coefficients., Both Abel and Galois proved that there is no general solution of a, fifth-degree equation by radicals. In particular, there is no “quintic formula.” Before discussing Galois’s method, which provided a group, theoretic criterion for the solution of an equation by radicals and led to, the modern-day Galois theory, we need a few definitions., Definition Solvable by Radicals, , Let F be a field, and let f (x) [ F[x]. We say that f (x) is solvable by radicals over F if f (x) splits in some extension F(a1, a2, . . . , an) of F and, there exist positive integers k1, . . . , kn such that a 1k1 [ F and ai ki [, F(a1, . . . , ai21) for i 5 2, . . . , n., , So, a polynomial in F[x] is solvable by radicals if we can obtain all, of its zeros by adjoining nth roots (for various n) to F. In other words,, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 561, , each zero of the polynomial can be written as an expression (usually a, messy one) involving elements of F combined by the operations of addition, subtraction, multiplication, division, and extraction of roots., EXAMPLE 8 Let v 5 cos( 2p/8) 1 i sin(2p/8) 522/2 1 i22/2., 8, 8, 3 ), v8 [ Q, and (2, 3 )8 [ Q , Q(v)., Then x8 2 3 splits in Q(v, 2, 8, Thus, x 2 3 is solvable by radicals over Q. Although the zeros of, 8, 8, 8, 3, 2, 3 v, 2, 3, x8 2 3 are most conveniently written in the form 2, 8, v2, . . . , 23 v7, the notion of solvable by radicals is best illustrated by, writing them in the form, 8, 8, 8, 3 , 6 2 12, 3, 6 2, 3 (222 1 21222),, 62, 8, 62, 3 (222 2 21222)., , Thus, the problem of solving a polynomial equation for its zeros can, be transformed into a problem about field extensions. At the same time,, we can use the Fundamental Theorem of Galois Theory to transform a, problem about field extensions into a problem about groups. This is exactly how Galois showed that there are fifth-degree polynomials that, cannot be solved by radicals, and this is exactly how we will do it. Before giving an example of such a polynomial, we need some additional, group theory., Definition Solvable Group, , We say that a group G is solvable if G has a series of subgroups, {e} 5 H0 , H1 , H2 , ? ? ? , Hk 5 G,, where, for each 0 # i , k, Hi is normal in Hi11 and Hi11/Hi is Abelian., , Obviously, Abelian groups are solvable. So are the dihedral groups, and any group whose order has the form pn, where p is a prime (see, Exercises 28 and 29). The monumental Feit–Thompson Theorem (see, Chapter 25) says that every group of odd order is solvable. In a certain, sense, solvable groups are almost Abelian. On the other hand, it, follows directly from the definitions that any non-Abelian simple, group is not solvable. In particular, A5 is not solvable. It follows from, Exercise 21 in Chapter 25 that S5 is not solvable. Our goal is to connect the notion of solvability of polynomials by radicals to that of, solvable groups. The next theorem is a step in this direction., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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562, , Special Topics, , Theorem 32.2 Splitting Field of xn 2 a, Let F be a field of characteristic 0 and let a [ F. If E is the splitting, field of xn 2 a over F, then the Galois group Gal(E/F) is solvable., , PROOF We first handle the case where F contains a primitive nth root, of unity v. Let b be a zero of xn 2 a in E. Then the zeros of xn 2 a are, b, vb, v2b, . . . , vn21b, and therefore E 5 F(b). In this case, we claim, that Gal(E/F) is Abelian and hence solvable. To see this, observe that, any automorphism in Gal(E/F) is completely determined by its action, on b. Also, since b is a zero of xn 2 a, we know that any element of, Gal(E/F) sends b to another zero of x n 2 a. That is, any element, of Gal(E/F) takes b to vib for some i. Let f and s be two elements, of Gal(E/F). Then, since v [ F, f and s fix v and f(b) 5 v jb and, s(b) 5 v kb for some j and k. Thus,, (sf)(b) 5 s (f(b)) 5 s (v jb) 5 s (v j)s (b) 5 v jv kb 5 v j1k b,, whereas, (fs)(b) 5 f(s(b)) 5 f(v kb) 5 f(v k)f(b) 5 v kv jb 5 v k1jb,, so that sf and fs agree on b and fix the elements of F. This shows, that sf 5 fs, and therefore Gal(E/F) is Abelian., Now suppose that F does not contain a primitive nth root of unity., Let v be a primitive nth root of unity and let b be a zero of xn 2 a in E., The case where a 5 0 is trivial, so we may assume that b ? 0. Since, vb is also a zero of xn 2 a, we know that both b and vb belong to E,, and therefore v 5 vb/b is in E as well. Thus, F(v) is contained in E,, and F(v) is the splitting field of xn 2 1 over F. Analogously to the, case above, for any automorphisms f and s in Gal(F(v)/F) we have, f(v) 5 v j for some j and s (v) 5 v k for some k. Then,, (sf)(v) 5 s (f(v)) 5 s (v j) 5 (s (v)) j 5 (v k) j, 5 (v j ) k 5 (f(v)) k 5 f(v k) 5 f(s (v)) 5 (fs)(v)., Since elements of Gal(F(v)/F) are completely determined by their action on v, this shows that Gal(F(v)/F) is Abelian., Because E is the splitting field of xn 2 a over F(v) and F(v) contains a primitive nth root of unity, we know from the case we have already done that Gal(E/F(v)) is Abelian and, by Part 2 of Theorem 32.1,, the series, {e} # Gal(E/F(v)) # Gal(E/F), , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 563, , is a normal series. Finally, since both Gal(E/F(v)) and, Gal(E/F )/Gal(E/F(v)) < Gal(F(v)/F), are Abelian, Gal(E/F) is solvable., To reach our main result about polynomials that are solvable by radicals, we need two important facts about solvable groups., Theorem 32.3 Factor Group of a Solvable Group Is Solvable, A factor group of a solvable group is solvable., , PROOF Suppose that G has a series of subgroups, {e} 5 H0 , H1 , H2 , ? ? ? , Hk 5 G,, where, for each 0 # i , k, Hi is normal in Hi11 and Hi11/Hi is Abelian., If N is any normal subgroup of G, then, {e} 5 H0N/N , H1N/N , H2N/N , ? ? ? , HkN/N 5 G/N, is the requisite series of subgroups that guarantees that G/N is solvable., (See Exercise 31.), Theorem 32.4 N and G/N Solvable Implies G Is Solvable, Let N be a normal subgroup of a group G. If both N and G/N are, solvable, then G is solvable., , PROOF Let a series of subgroups of N with Abelian factors be, N0 , N1 , ? ? ? , Nt 5 N, and let a series of subgroups of G/N with Abelian factors be, N/N 5 H0 /N , H1/N , ? ? ? , Hs /N 5 G/N., Then the series, N0 , N1 , ? ? ? , Nt 5 H0 , H1 , ? ? ? , Hs 5 G, has Abelian factors (see Exercise 33)., We are now able to make the critical connection between solvability, of polynomials by radicals and solvable groups., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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564, , Special Topics, , Theorem 32.5 (Galois) Solvable by Radicals Implies Solvable Group, Let F be a field of characteristic 0 and let f(x) [ F[x]. Suppose that, f(x) splits in F(a1, a2, . . . , at), where a1n1 [ F and aini [ F(a1, . . . ,, ai21) for i 5 2, . . . , t. Let E be the splitting field for f(x) over F in, F(a1, a2, . . . , at ). Then the Galois group Gal(E/F) is solvable., , PROOF We use induction on t. For the case t 5 1, we have F # E #, F(a1). Let a 5 a1n1 and let L be a splitting field of xn1 2 a over F. Then F, # E # L, and both E and L are splitting fields of polynomials over F., By part 2 of Theorem 32.1, Gal(E/F) < Gal(L/F)/Gal(L/E). It follows, from Theorem 32.2 that Gal(L/F) is solvable, and from Theorem 32.3, we know that Gal(L/F)/Gal(L/E) is solvable. Thus, Gal(E/F) is solvable., Now suppose t . 1. Let a 5 a1n1 [ F, let L be a splitting field of, n, x 1 2 a over E, and let K # L be the splitting field of xn1 2 a over F., Then L is a splitting field of (xn1 2 a) f (x) over F, and L is a splitting, field of f (x) over K. Since F(a 1) # K, we know that f (x) splits in, K(a2, . . . , at), so the induction hypothesis implies that Gal(L/K) is, solvable. Also, Theorem 32.2 asserts that Gal(K/F) is solvable, which,, from Theorem 32.1, tells us that Gal(L/F)/Gal(L/K) is solvable. Hence,, Theorem 32.4 implies that Gal(L/F) is solvable. So, by part 2 of Theorem 32.1 and Theorem 32.3, we know that the factor group, Gal(L/F)/Gal(L/E) < Gal(E/F) is solvable., It is worth remarking that the converse of Theorem 32.3 is true also;, that is, if E is the splitting field of a polynomial f (x) over a field F of, characteristic 0 and Gal(E/F) is solvable, then f (x) is solvable by radicals over F., It is known that every finite group is a Galois group over some field., However, one of the major unsolved problems in algebra, first posed, by Emmy Noether, is determining which finite groups can occur as, Galois groups over Q. Many people suspect that the answer is “all of, them.” It is known that every solvable group is a Galois group over Q., John Thompson has recently proved that certain kinds of simple, groups, including the Monster, are Galois groups over Q. The article by, Ian Stewart listed among this chapter’s suggested readings provides, more information on this topic., , Insolvability of a Quintic, We will finish our introduction to Galois theory by explicitly exhibiting a polynomial that has integer coefficients and that is not solvable, by radicals over Q., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 565, , Consider g(x) 5 3x5 2 15x 1 5. By Eisenstein’s Criterion (Theorem, 17.4), g(x) is irreducible over Q. Since g(x) is continuous and g(22) 5, 261 and g(21) 5 17, we know that g(x) has a real zero between 22, and 21. A similar analysis shows that g(x) also has real zeros between, 0 and 1 and between 1 and 2., Each of these real zeros has multiplicity 1, as can be verified by long, division or by appealing to Theorem 20.6. Furthermore, g(x) has no, more than three real zeros, because Rolle’s Theorem from calculus, guarantees that between each pair of real zeros of g(x) there must be a, zero of g9(x) 5 15x4 2 15. So, for g(x) to have four real zeros, g9(x), would have to have three real zeros, and it does not. Thus, the other two, zeros of g(x) are nonreal complex numbers, say, a 1 bi and a 2 bi., (See Exercise 65 in Chapter 15.), Now, let’s denote the five zeros of g(x) by a1, a2, a3, a4, a5. Since any, automorphism of K 5 Q(a1, a2, a3, a4, a5) is completely determined by its, action on the a’s and must permute the a’s, we know that Gal(K/Q) is isomorphic to a subgroup of S5, the symmetric group on five symbols. Since, a1 is a zero of an irreducible polynomial of degree 5 over Q, we know that, [Q(a1):Q] 5 5, and therefore 5 divides [K:Q]. Thus, the Fundamental, Theorem of Galois Theory tells us that 5 also divides |Gal(K/Q)|. So, by, Cauchy’s Theorem (corollary to Theorem 24.3), we may conclude that, Gal(K/Q) has an element of order 5. Since the only elements in S5 of order 5 are the 5-cycles, we know that Gal(K/Q) contains a 5-cycle. The, mapping from C to C, sending a 1 bi to a 2 bi, is also an element of, Gal(K/Q). Since this mapping fixes the three real zeros and interchanges, the two complex zeros of g(x), we know that Gal(K/Q) contains a 2-cycle., But, the only subgroup of S5 that contains both a 5-cycle and a 2-cycle is, S5. (See Exercise 25 in Chapter 25.) So, Gal(K/Q) is isomorphic to S5., Finally, since S5 is not solvable (see Exercise 27), we have succeeded in, exhibiting a fifth-degree polynomial that is not solvable by radicals., , Exercises, Seeing much, suffering much, and studying much are the three pillars, of learning., benjamin disraeli, , 1. Let E be an extension field of Q. Show that any automorphism of E, acts as the identity on Q. (This exercise is referred to in this chapter.), 2. Determine the group of field automorphisms of GF(4)., 3. Let E be an extension field of the field F. Show that the automorphism group of E fixing F is indeed a group. (This exercise is referred, to in this chapter.), Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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566, , Special Topics, , 4. Given that the automorphism group of Q(22, 25, 27) is isomorphic to Z2 % Z2 % Z2, determine the number of subfields of Q(22,, 25, 27) that have degree 4 over Q., 5. Let E be an extension field of a field F and let H be a subgroup of, Gal(E/F). Show that the fixed field of H is indeed a field. (This, exercise is referred to in this chapter.), 6. Let E be the splitting field of x4 1 1 over Q. Find Gal(E/Q). Find, all subfields of E. Find the automorphisms of E that have fixed, fields Q(22), Q(2 2 ), and Q(i). Is there an automorphism of E, whose fixed field is Q?, 7. Let f (x) [ F[x] and let the zeros of f (x) be a1, a2, . . . , an. If K 5, F(a1, a2, . . . , an), show that Gal(K/F ) is isomorphic to a group of, permutations of the ai’s. [When K is the splitting field of f (x) over F,, the group Gal(K/F ) is called the Galois group of f (x).], 8. Show that the Galois group of a polynomial of degree n has order, dividing n!., 9. Referring to Example 6, show that the automorphism f has order 6., Show that v 1 v21 is fixed by f3 and v3 1 v5 1 v6 is fixed by f2., (This exercise is referred to in this chapter.), 10. Let E 5 Q(22, 25). What is the order of the group Gal(E/Q)?, What is the order of Gal(Q(210)/Q)?, 11. Suppose that F is a field of characteristic 0 and E is the splitting, field for some polynomial over F. If Gal(E/F ) is isomorphic to, Z20 % Z2, determine the number of subfields L of E there are such, that L contains F and, a. [L:F ] 5 4., b. [L:F ] 5 25., c. Gal(E/L) is isomorphic to Z5., 12. Determine the Galois group of x2 2 10x 1 21 over Q. (See Exercise 7, for the definition)., 13. Determine the Galois group of x2 1 9 over R. (See Exercise 7 for, the definition)., 14. Suppose that F is a field of characteristic 0 and E is the splitting, field for some polynomial over F. If Gal(E/F ) is isomorphic to D6,, prove that there are exactly three fields L such that E $ L $ F and, [E:L] 5 6., 15. Suppose that E is the splitting field for some polynomial over, GF(p). If Gal(E/GF(p)) 5 p6, how many fields are there strictly, between E and GF(p)?, 16. Let p be a prime. Suppose that |Gal(E/F )| 5 p2. Draw all possible, subfield lattices for fields between E and F., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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32 | An Introduction to Galois Theory, , 567, , 17. Suppose that F is a field of characteristic 0 and E is the splitting, field for some polynomial over F. If Gal(E/F ) is isomorphic to A4,, show that there is no subfield K of E such that [K:F ] 5 2., 18. Determine the Galois group of x3 2 1 over Q and x3 2 2 over Q., (See Exercise 7 for the definition.), 19. Suppose that K is the splitting field of some polynomial over a field, F of characteristic 0. If [K:F ] 5 p2q, where p and q are distinct, primes, show that K has subfields L1, L2, and L3 such that [K:L1] 5 p,, [K:L2] 5 p2, and [K:L3] 5 q., 20. Suppose that E is the splitting field of some polynomial over a field F, of characteristic 0. If Gal(E/F ) is isomorphic to D5, draw the subfield, lattice for the fields between E and F., 21. Suppose that F , K , E are fields and E is the splitting field of, some polynomial in F[x]. Show, by means of an example, that K, need not be the splitting field of some polynomial in F[x]., 22. Suppose that E is the splitting field of some polynomial over a field, F of characteristic 0. If [E:F ] is finite, show that there is only a, finite number of fields between E and F., 23. Suppose that E is the splitting field of some polynomial over a field, F of characteristic 0. If Gal(E/F ) is an Abelian group of order 10,, draw the subfield lattice for the fields between E and F., 24. Let v be a nonreal complex number such that v5 5 1. If f is the, automorphism of Q(v) that carries v to v4, find the fixed field of kfl., 25. Determine the isomorphism class of the group Gal(GF(64)/GF(2))., 26. Determine the isomorphism class of the group Gal(GF(729)/GF(9))., Exercises 27, 28, and 29 are referred to in this chapter., 27., 28., 29., 30., 31., , Show that S5 is not solvable., Show that the dihedral groups are solvable., Show that a group of order pn, where p is prime, is solvable., Show that Sn is solvable when n # 4., Complete the proof of Theorem 32.3 by showing that the given, series of groups satisfies the definition for solvability., 32. Show that a subgroup of a solvable group is solvable., 33. Let N be a normal subgroup of G and let K/N be a normal subgroup of G/N. Prove that K is a normal subgroup of G. (This exercise is referred to in this chapter.), 34. Show that any automorphism of GF( pn) acts as the identity on, GF( p)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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568, , Special Topics, , Reference, 1. G. Ehrlich, Fundamental Concepts of Abstract Algebra, Boston:, PWS-Kent, 1991., , Suggested Readings, Tony Rothman, “The Short Life of Évariste Galois,” Scientific American,, April (1982): 136–149., This article gives an elementary discussion of Galois’s proof that the general fifth-degree equation cannot be solved by radicals. The article also, goes into detail about Galois’s controversial life and death. In this regard,, Rothman refutes several accounts given by other Galois biographers., Ian Stewart, “The Duellist and the Monster,” Nature 317 (1985): 12–13., This nontechnical article discusses recent work of John Thompson pertaining to the question of “which groups can occur as Galois groups.”, , Suggested Website, http://www-groups.dcs.st-and.ac.uk/~history/, Find more information about the history of quadratic, cubic, and quartic, equations at this site., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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By permission of the Masters & Fellows of Pembroke, College in the University of Oxford, , Philip Hall, He [Hall] was preeminent as a group theorist and made many fundamental discoveries; the conspicuous growth of interest, in group theory in the 20th century owes, much to him., j. e. roseblade, , Philip Hall was born on April 11, 1904, in, London. Abandoned by his father shortly, after birth, Hall was raised by his mother, a, dressmaker. He demonstrated academic, prowess early by winning a scholarship to, Christ’s Hospital, where he had several outstanding mathematics teachers. At Christ’s, Hospital, Hall won a medal for the best, English essay, the gold medal in mathematics, and a scholarship to King’s College,, Cambridge., Although abstract algebra was a field neglected at King’s College, Hall studied, Burnside’s book Theory of Groups and some, of Burnside’s later papers. After graduating in, 1925, he stayed on at King’s College for further study and was elected to a fellowship in, 1927. That same year, Hall discovered a, major “Sylow-like” theorem about solvable, groups: If a solvable group has order mn,, where m and n are relatively prime, then, every subgroup whose order divides m is contained in a group of order m and all subgroups, of order m are conjugate. Over the next three, decades, Hall developed a general theory of, , finite solvable groups that had a profound influence on John Thompson’s spectacular, achievements of the 1960s. In the 1930s, Hall, also developed a general theory of groups of, prime-power order that has become a foundation of modern finite group theory. In addition to his fundamental contributions to finite, groups, Hall wrote many seminal papers on, infinite groups., Among the concepts that have Hall’s name, attached to them are Hall subgroups, Hall, algebras, Hall–Littlewood polynomials, Hall, divisors, the marriage theorem from graph, theory, and the Hall commutator collecting, process. Beyond his own discoveries, Hall, had an enormous influence on algebra, through his research students. No fewer than, one dozen have become eminent mathematicians in their own right. Hall died on, December 30, 1982., To find more information about Hall,, visit:, http://www-groups.dcs.st-and, .ac.uk/~history/, , 569, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33 Cyclotomic Extensions, “. . . To regard old problems from a new angle requires creative imagination, and marks real advances in science.”, albert einstein, , Motivation, For the culminating chapter of this book, it is fitting to choose a topic, that ties together results about groups, rings, fields, geometric constructions, and the history of mathematics. The so-called cyclotomic extensions is such a topic. We begin with the history., The ancient Greeks knew how to construct regular polygons of 3, 4,, 5, 6, 8, 10, 12, 15, and 16 sides with a straightedge and compass. And,, given a construction of a regular n-gon, it is easy to construct a regular, 2n-gon. The Greeks attempted to fill in the gaps (7, 9, 11, 13, 14,, 17, . . .) but failed. More than 2200 years passed before anyone was, able to advance our knowledge of this problem beyond that of the, Greeks. Incredibly, Gauss, at age 19, showed that a regular 17-gon is, constructible, and shortly thereafter he completely solved the problem, of exactly which n-gons are constructible. It was this discovery of the, constructibility of the 17-sided regular polygon that induced Gauss to, dedicate his life to the study of mathematics. Gauss was so proud of, this accomplishment that he requested that a regular 17-sided polygon, be engraved on his tombstone., Gauss was led to his discovery of the constructible polygons through, his investigation of the factorization of polynomials of the form x n 2 1, over Q. In this chapter, we examine the factors of x n 2 1 and show how, Galois theory can be used to determine which regular n-gons are constructible with a straightedge and compass. The irreducible factors of, x n 2 1 are important in number theory and combinatorics., , 570, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33 | Cyclotomic Extensions, , 571, , Cyclotomic Polynomials, Recall from Example 2 in Chapter 16 that the complex zeros of xn 2 1, are 1, v 5 cos(2p/n) 1 i sin(2p/n), v2, v3, . . . , vn21. Thus, the splitting field of x n 2 1 over Q is Q(v). This field is called the nth cyclotomic extension of Q, and the irreducible factors of x n 2 1 over Q are, called the cyclotomic polynomials., Since v 5 cos(2p/n)1 i sin(2p/n) generates a cyclic group of order n, under multiplication, we know from Corollary 3 of Theorem 4.2 that the, generators of kvl are the elements of the form vk, where 1 # k # n and, gcd(n, k) 5 1. These generators are called the primitive nth roots of, unity. Recalling that we use f(n) to denote the number of positive integers less than or equal to n and relatively prime to n, we see that for, each positive integer n there are precisely f(n) primitive nth roots, of unity. The polynomials whose zeros are the f(n) primitive nth roots, of unity have a special name., Definition Cyclotomic Polynomial, , For any positive integer n, let v1, v2, . . . , vf(n) denote the primitive nth, roots of unity. The nth cyclotomic polynomial over Q is the polynomial, Fn(x) 5 (x 2 v1)(x 2 v2) ? ? ? (x 2 vf(n))., , In particular, note that F n(x) is monic and has degree f(n). In, Theorem 33.2 we will prove that Fn(x) has integer coefficients, and in, Theorem 33.3 we will prove that Fn(x) is irreducible over Z., EXAMPLE 1 F1(x) 5 x 2 1, since 1 is the only zero of x 2 1. F2(x) 5, x 1 1, since the zeros of x 2 2 1 are 1 and 21, and 21 is the only primitive, root. F3(x) 5 (x 2 v)(x 2 v2), where v 5 cos(2p/3) 1 i sin(2p/3) 5 (21 1, i23)/2, and direct calculations show that F3(x) 5 x2 1 x 1 1. Since, the zeros of x4 2 1 are 61 and 6i and only i and 2i are primitive, F4(x) 5, (x 2 i)(x 1 i) 5 x2 1 1., In practice, one does not use the definition of Fn(x) to compute it., Instead, one uses the formulas given in the exercises and makes recursive use of the following result., Theorem 33.1, For every positive integer n, xn 2 1 5 Pd|nFd (x), where the product, runs over all positive divisors d of n., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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572, , Special Topics, , Before proving this theorem, let us be sure that the statement, is clear. For n 5 6, for instance, the theorem asserts that x6 2 1 5, F1(x)F2(x)F3(x)F6(x), since 1, 2, 3, and 6 are the positive divisors of 6., PROOF Since both polynomials in the statement are monic, it suffices, to show that they have the same zeros and that all zeros have multiplicity 1. Let v 5 cos(2p/n) 1 i sin(2p/n). Then kvl is a cyclic group of, order n, and kvl contains all the nth roots of unity. From Theorem 4.3 we, know that for each j, |v j| divides n so that (x 2 v j) appears as a factor in, F|v j |(x). On the other hand, if x 2 a is a linear factor of Fd (x) for some, divisor d of n, then a d 5 1, and therefore a n 5 1. Thus, x 2 a is a factor, of x n 2 1. Finally, since no zero of x n 2 1 can be a zero of Fd (x) for two, different d’s, the result is proved., Before we illustrate how Theorem 33.1 can be used to calculate, Fn(x) recursively, we state an important consequence of the theorem., Theorem 33.2, For every positive integer n, Fn(x) has integer coefficients., , PROOF The case n 5 1 is trivial. By induction, we may assume that, g(x) 5 P d|n Fd (x) has integer coefficients. From Theorem 33.1 we, d6n, , know that x n 2 1 5 Fn(x)g(x), and, because g(x) is monic, we may, carry out the division in Z[x] (see Exercise 49 in Chapter 16). Thus,, Fn(x) [ Z[x]., Now let us do some calculations. If p is a prime, we have from Theorem 33.1 that x p 2 1 5 F 1 (x)F p (x) 5 (x 2 1)F p (x), so that, Fp(x) 5 (x p 2 1)/(x 2 1) 5 x p21 1 x p22 1 ? ? ? 1 x 1 1. From Theorem 33.1 we have, x6 2 1 5 F1(x)F2(x)F3(x)F6(x),, so that F6(x) 5 (x6 2 1)/((x 2 1)(x 1 1)(x2 1 x 1 1)). So, by long, division, F 6 (x) 5 x 2 2 x 1 1. Similarly, F 10 (x) 5 (x 10 2 1)/, ((x 2 1)(x 1 1)(x4 1 x3 1 x2 1 x 1 1)) 5 x4 2 x3 1 x2 2 x 1 1., The exercises provide shortcuts that often make long division unnecessary. The values of Fn(x) for all n up to 15 are shown in Table 33.1., The software for the computer exercises provides the values for Fn(x), for all values of n up to 1000. Judging from Table 33.1, one might be, led to conjecture that 1 and 21 are the only nonzero coefficients of the, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33 | Cyclotomic Extensions, , 573, , Table 33.1 The Cyclotomic Polynomials Fn(x) up to n 5 15, n, , Fn(x), , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, , x21, x11, x2 1 x 1 1, x2 1 1, x4 1 x3 1 x2 1 x 1 1, x2 2 x 1 1, x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1, x4 1 1, x6 1 x3 1 1, x4 2 x3 1 x2 2 x 1 1, x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1, x4 2 x2 1 1, x12 1 x11 1 x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1, x6 2 x5 1 x4 2 x3 1 x2 2 x 1 1, x8 2 x7 1 x5 2 x4 1 x3 2 x 1 1, , cyclotomic polynomials. However, it has been shown that every integer, is a coefficient of some cyclotomic polynomial., The next theorem reveals why the cyclotomic polynomials are, important., Theorem 33.3 (Gauss), The cyclotomic polynomials Fn(x) are irreducible over Z., , PROOF Let f (x) [ Z[x] be a monic irreducible factor of F n(x)., Because Fn(x) is monic and has no multiple zeros, it suffices to show, that every zero of Fn(x) is a zero of f(x)., Since Fn(x) divides xn 2 1 in Z[x], we may write xn 2 1 5 f (x)g(x),, where g(x) [ Z[x]. Let v be a primitive nth root of unity that is a zero, of f (x). Then f (x) is the minimal polynomial for v over Q. Let p be any, prime that does not divide n. Then, by Corollary 3 of Theorem 4.2,, v p is also a primitive nth root of unity, and therefore 0 5 (v p)n 2 1 5, f (v p)g(v p), and so f (v p) 5 0 or g(v p) 5 0. Suppose f (v p) 2 0. Then, g(v p) 5 0, and so v is a zero of g(x p). Thus, from Theorem 21.3, f (x), divides g(x p) in Q[x]. Since f (x) is monic, f (x) actually divides g(x p), in Z[x] (see Exercise 49 in Chapter 16). Say g(x p) 5 f (x)h(x), where, h(x) [ Z[x]. Now let g(x), f (x), and h(x) denote the polynomials in, Z p[x] obtained from g(x), f (x), and h(x), respectively, by reducing, each coefficient modulo p. Since this reduction process is a ring homomorphism from Z[x] to Z p[x] (see Exercise 11 in Chapter 16), we, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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574, , Special Topics, , have g(x p) 5 f (x)h(x) in Z p[x]. From Exercise 33 in Chapter 16 and, Corollary 5 of Theorem 7.1, we then have (g(x)) p 5 g(x p) 5 f (x)h(x),, and since Z p[x] is a unique factorization domain, it follows that g(x), and f (x) have an irreducible factor in Zp[x] in common; call it m(x)., Thus, we may write f (x) 5 k1(x)m(x) and g(x) 5 k2(x)m(x), where, k1(x), k2(x) [ Zp[x]. Then, viewing xn 2 1 as a member of Zp[x],, we have xn 2 1 5 f (x)g(x) 5 k1(x)k2(x)(m(x))2. In particular, x n 2 1, has a multiple zero in some extension of Z p. But because p does not, divide n, the derivative nx n21 of x n 2 1 is not 0, and so nx n21 and x n 2 1, do not have a common factor of positive degree in Zp[x]. Since this, contradicts Theorem 20.5, we must have f (v p) 5 0., We reformulate what we have thus far proved as follows: If b is, any primitive nth root of unity that is a zero of f (x) and p is any prime, that does not divide n, then b p is a zero of f (x). Now let k be any integer between 1 and n that is relatively prime to n. Then we can write, k 5 p1 p2 ? ? ? pt, where each pi is a prime that does not divide n (repetitions are permitted). It follows then that each of v, v p1, (v p1) p2, . . . ,, (v p1 p2???pt–1) pt 5 vk is a zero of f (x). Since every zero of Fn(x) has the, form v k, where k is between 1 and n and is relatively prime to n, we, have proved that every zero of Fn(x) is a zero of f (x). This completes, the proof., Of course, Theorems 33.3 and 33.1 give us the factorization of, x n 2 1 as a product of irreducible polynomials over Q. But Theorem 33.1, is also useful for finding the irreducible factorization of x n 2 1 over Zp., The next example provides an illustration. Irreducible factors of x n 2 1, over Zp are used to construct error-correcting codes., EXAMPLE 2 We determine the irreducible factorization of x6 2 1 over, Z2 and Z3. From Table 33.1, we have x6 2 1 5 (x 2 1)(x 1 1)(x 2 1, x 1 1)(x 2 2 x 1 1). Taking all the coefficients on both sides mod 2, we, obtain the same expression, but we must check that these factors are irreducible over Z2. Since x 2 1 x 1 1 has no zeros in Z2, it is irreducible, over Z2 (see Theorem 17.1). Finally, since 21 5 1 in Z2, we have the, irreducible factorization x6 2 1 5 (x 1 1)2 (x 2 1 x 1 1)2. Over Z3,, we again start with the factorization x6 2 1 5 (x 2 1)(x 1 1)(x 2 1, x 1 1)(x 2 2 x 1 1) over Z and view the coefficients mod 3. Then 1 is, a zero of x 2 1 x 1 1 in Z 3, and by long division we obtain x 2 1 x 1 1 5, (x 2 1)(x 1 2) 5 (x 1 2)2. Similarly, x 2 2 x 1 1 5 (x 2 2)(x 1 1) 5, (x 1 1)2. So, the irreducible factorization of x 6 2 1 over Z 3 is (x 1 1)3 ?, (x 1 2)3., We next determine the Galois group of the cyclotomic extensions of Q., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33 | Cyclotomic Extensions, , 575, , Theorem 33.4, Let v be a primitive nth root of unity. Then Gal(Q(v)/Q) < U(n)., , PROOF Since 1, v, v2, . . . , v n21 are all the nth roots of unity, Q(v), is the splitting field of x n 2 1 over Q. For each k in U(n), v k is a primitive nth root of unity, and by the lemma preceding Theorem 20.4, there, is a field automorphism of Q(v), which we denote by fk, that carries v, to v k and acts as the identity on Q. Moreover, these are all the automorphisms of Q(v), since any automorphism must map a primitive nth, root of unity to a primitive nth root of unity. Next, observe that for, every r, s [ U(n),, (fr fs )(v) 5 fr (v s ) 5 (fr (v)) s 5 (v r ) s 5 v rs 5 f rs (v)., This shows that the mapping from U(n) onto Gal(Q(v)/Q) given by, k S f k is a group homomorphism. Clearly, the mapping is an isomorphism, since v r 2 v s when r, s [ U(n) and r 2 s., The next example uses Theorem 33.4 and the results of Chapter 8 to, demonstrate how to determine the Galois group of cyclotomic extensions., EXAMPLE 3 Let a 5 cos(2p/9) 1 i sin(2p/9) and let b 5 cos(2p/15), 1 i sin(2p/15). Then, Gal(Q(a)/Q) < U(9) < Z6, and, Gal(Q(b)/Q) < U(15) < U(5) % U(3) < Z4 % Z2., , The Constructible Regular n-gons, As an application of the theory of cyclotomic extensions and Galois theory, we determine exactly which regular n-gons are constructible with a, straightedge and compass. But first we prove a technical lemma., Lemma, Let n be a positive integer and let v 5 cosA2p/nB 1 i sinA2p/nB. Then, QAcosA2p/nBB # Q(v)., , PROOF Observe that from (cos(2p/n) 1 i sin(2p/n))(cos(2p/n) 2, i sin(2p/n)) 5 cos2(2p/n) 1 sin2(2p/n) 5 1, we have cos(2p/n) 2, i sin(2p/n) 5 1/v. Moreover, (v 1 1/v)/2 5 (2cos( 2p/n))/2 5, cos(2p/n). Thus, cos(2p/n) [ Q(v)., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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576, , Special Topics, , Theorem 33.5 (Gauss, 1796), It is possible to construct the regular n-gon with a straightedge and, compass if and only if n has the form 2kp1 p2 ? ? ? pt, where k $ 0 and, the pi’s are distinct primes of the form 2m 1 1., , PROOF If it is possible to construct a regular n-gon, then we can construct the angle 2p/n and therefore the number cos(2p/n). By the results, of Chapter 23, we know that cos(2p/n) is constructible only if [Q, (cos(2p/n)):Q] is a power of 2. To determine when this is so, we will use, Galois theory., Let v 5 cos(2p/n) 1 i sin(2p/n). Then |Gal(Q(v)/Q)| 5 [Q(v):Q] 5, f(n). By the lemma on the preceding page, Q(cos(2p/n)) # Q(v), and, by Theorem 32.1 we know that, [Q(cos(2p/n)):Q] 5 |Gal(Q(v)/Q)|/|Gal(Q(v)/Q(cos(2p/n)))|, 5 f(n)/|Gal(Q(v)/Q(cos(2p/n)))|., Recall that the elements s of Gal(Q(v)/Q) have the property that, s(v) 5 vk for 1 # k # n. That is, s(cos(2p/n) 1 i sin(2p/n)) 5 cos(2pk/n), 1 i sin(2pk/n). If such a s belongs to Gal(Q(v)/Q(cos(2p/n))), then, we must have cos(2pk/n) 5 cos(2p/n). Clearly, this holds only when, k 5 1 and k 5 n 2 1. So, |Gal(Q(v)/Q(cos(2p/n)))| 5 2, and therefore, [Q(cos(2p/n)):Q] 5 f(n)/2. Thus, if an n-gon is constructible, then, f(n)/2 must be a power of 2. Of course, this implies that f(n) is a, power of 2., Write n 5 2kp1n 1p2 n 2 ? ? ? pt n t, where k $ 0, the pi’s are distinct, odd primes, and the n i ’s are positive. Then f(n) 5 |U(n)| 5, |U(2 k)||U( p1n1)||U( p 2n 2)| ? ? ? |U( pt n t)| 5 2 k21p 1n 121( p 1 2 1)p 2n221, ( p2 2 1)? ? ? pt n t21( pt 2 1) must be a power of 2. Clearly, this implies, that each ni 5 1 and each pi 2 1 is a power of 2. This completes the, proof that the condition in the statement is necessary., To prove that the condition given in Theorem 33.5 is also sufficient,, suppose that n has the form 2k p1p2 ? ? ? pt , where the pi’s are distinct, odd primes of the form 2m 1 1, and let v 5 cos(2p/n) 1 i sin(2p/n)., By Theorem 33.3, Q(v) is a splitting field of an irreducible polynomial over Q, and therefore, by the Fundamental Theorem of Galois, Theory, f(n) 5 [Q(v):Q] 5 |Gal(Q(v)/Q)|. Since f(n) is a power of 2, and Gal(Q(v)/Q) is an Abelian group, it follows by induction (see Exercise 15) that there is a series of subgroups, H0 , H1 , ? ? ? , Ht 5 Gal(Q(v)/Q),, where H0 is the identity, H1 is the subgroup of Gal(Q(v)/Q) of order 2, that fixes cos(2p/n), and |Hi11:Hi| 5 2 for i 5 0, 1, 2, . . . , t 2 1. By the, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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33 | Cyclotomic Extensions, , 577, , Fundamental Theorem of Galois Theory, we then have a series of subfields of the real numbers, Q 5 EH , EH, t, , t–1, , , ? ? ? , EH 5 Q(cos(2p/n)),, 1, , where [EH :EH ] 5 2. So, for each i, we may choose bi [ EH such, i21, i, i21, that EH 5 EH (bi). Then bi is a zero of a polynomial of the form, i21, i, x 2 1 bi x 1 ci [ EH [x], and it follows that EH 5 EH (2b2i 4ci)., i, i21, i, Thus, it follows from Exercise 3 in Chapter 23 that every element of, Q(cos(2p/n)) is constructible., It is interesting to note that Gauss did not use Galois theory in his, proof. In fact, Gauss gave his proof 15 years before Galois was born., , Exercises, Difficulties should act as a tonic. They should spur us to greater exertion., b. c. forbes, , 1. Determine the minimal polynomial for cos(p/3) 1 i sin(p/3) over Q., 2. Factor x12 2 1 as a product of irreducible polynomials over Z., 3. Factor x 8 2 1 as a product of irreducible polynomials over Z2, Z3,, and Z5., 4. For any n . 1, prove that the sum of all the nth roots of unity is 0., 5. For any n . 1, prove that the product of the nth roots of unity is, (21)n11., 6. Let v be a primitive 12th root of unity over Q. Find the minimal, polynomial for v4 over Q., 7. Let F be a finite extension of Q. Prove that there are only a finite, number of roots of unity in F., 8. For any n . 1, prove that the irreducible factorization over Z of, x n21 1 xn22 1 ? ? ? 1 x 1 1 is PFd (x), where the product runs, over all positive divisors d of n greater than 1., 9. If 2n 1 1 is prime for some n $ 1, prove that n is a power of 2., (Primes of the form 2n 1 1 are called Fermat primes.), 10. Prove that Fn(0) 5 1 for all n . 1., 11. Prove that if a field contains the nth roots of unity for n odd, then it, also contains the 2nth roots of unity., 12. Let m and n be relatively prime positive integers. Prove that the, splitting field of x mn 2 1 over Q is the same as the splitting field of, (x m 2 1)(x n 2 1) over Q., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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578, , Special Topics, , 13. Prove that F2n(x) 5 Fn(2x) for all odd integers n . 1., 14. Prove that if p is a prime and k is a positive integer, then Fp k (x) 5, Fp(x p k 21). Use this to find F8(x) and F27(x)., 15. Prove the assertion made in the proof of Theorem 33.5 that there exists a series of subgroups H0 , H1 , ? ? ? , Ht with |Hi11:Hi| 5 2, for i 5 0, 1, 2, . . . , t 2 1. (This exercise is referred to in this, chapter.), 16. Prove that x 9 2 1 and x 7 2 1 have isomorphic Galois groups over Q., (See Exercise 7 in Chapter 32 for the definition.), 17. Let p be a prime that does not divide n. Prove that F pn(x) 5, Fn(x p)/Fn(x)., 18. Prove that the Galois groups of x 10 2 1 and x 8 2 1 over Q are not, isomorphic., 19. Let E be the splitting field of x 5 2 1 over Q. Show that there is a, unique field K with the property that Q , K , E., 20. Let E be the splitting field of x 6 2 1 over Q. Show that there is no, field K with the property that Q , K , E., 21. Let v 5 cos(2p/15) 2 i sin(2p/15). Find the three elements of, Gal(Q(v)/Q) of order 2., , Computer Exercises, Computer exercises for this chapter are available at the website:, http://www.d.umn.edu/~jgallian, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Carl Friedrich Gauss, He [Gauss] lives everywhere in, mathematics., , The Granger Collection, New York, , e. t. bell, Men of Mathematics, , Carl Friedrich Gauss, considered by many, to be the greatest mathematician who has ever, lived, was born in Brunswick, Germany, on, April 30, 1777. While still a teenager, he made, many fundamental discoveries. Among these, were the method of “least squares” for handling statistical data, and a proof that a, 17-sided regular polygon can be constructed, with a straightedge and compass (this result, , This stamp was issued by East Germany in, 1977. It commemorates Gauss’s construction of a regular 17-sided polygon with a, straightedge and compass., , was the first of its kind since discoveries by the, Greeks 2000 years earlier). In his Ph.D. dissertation in 1799, he proved the Fundamental, Theorem of Algebra., Throughout his life, Gauss largely ignored the work of his contemporaries and, in, fact, made enemies of many of them. Young, mathematicians who sought encouragement, from him were usually rebuffed. Despite this, fact, Gauss had many outstanding students,, including Eisenstein, Riemann, Kummer,, Dirichlet, and Dedekind., Gauss died in Göttingen at the age of 77 on, February 23, 1855. At Brunswick, there is a, statue of him. Appropriately, the base is in the, shape of a 17-point star. In 1989, Germany issued a bank note (see page 117) depicting, Gauss and the Gaussian distribution., To find more information about Gauss,, visit:, http://www-groups.dcs, .st-and.ac.uk/~history/, , 579, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Trustees of Princeton University/Denise Applewhite;, bank notes and postage stamps on pages 8, 41, 41,, 117, 126, 161, 207, 279, 346, 524, 579: free domain, , Manjul Bhargava, We are watching him [Bhargava] very, closely., He is going to be a superstar., He's amazingly mature mathematically., He is changing the subject in a, fundamental way., peter sarnak, , Manjul Bhargava was born in Canada on, August 8, 1974, and grew up in Long Island,, New York. After graduating from Harvard in, 1996, Bhargava went to Princeton to pursue, his Ph.D. under the direction of Andrew, Wiles (see biography after Chapter 18)., Bhargava investigated a “composition law”, first formulated by Gauss in 1801 for combining two quadratic equations (equations in, a form such as x2 1 3xy 1 6y2 5 0) in a way, that was very different from normal addition, and revealed a lot of information about number systems. Bhargava tackled an aspect of, the problem in which no progress had been, made in more than 200 years. He not only, broke new ground in that area but also discovered 13 more composition laws and developed a coherent mathematical framework, to explain them. He then applied his theory, of composition to solve a number of fundamental problems concerning the distribution of extension fields of the rational numbers and of other, related algebraic objects., What made Bhargava’s work especially, remarkable is that he was able to explain, all his revolutionary ideas using only elementary mathematics. In commenting on, Bhargava’s results, Wiles said, “He did it in, , a way that Gauss himself could have understood and appreciated.”, Despite his youth, Bhargava already has, won many awards, including a Clay Research, Fellowship, the Clay Research Award, the, Blumenthal Award for the Advancement of, Research in Pure Mathematics, the SASTRA, Ramanujan Prize, the 2008 Cole Prize in number theory (see page 438), and the 2011 Fermat, Prize. In 2002 he was named one of Popular, Science magazine’s “Brilliant 10,” in celebration of scientists who are shaking up their fields., In 2003, Bhargava accepted a full professorship, with tenure at Princeton at the age of 28., In addition to doing mathematics,, Bhargava is an accomplished tabla player, who has studied with the world’s most, distinguished tabla masters. He performs, extensively in the New York and Boston, areas. To hear him play the tabla, visit, http://www.npr.org/templates/story/, story.php?storyId=4111253, To find more information about Bhargava,, visit, http://www.wikipedia.org, and, http:// www.d.umn.edu/~jgallian/, manjulMH4.pdf, , 580, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 24–33, , 581, , Supplementary Exercises for Chapters 24–33, The road to wisdom?—Well it’s plain and simple to express:, Err, and err, and err again, but less, and less, and less., piet hein, “The Road to Wisdom,” Grooks (1966)*, , True/false questions for Chapters 24–33 are available on the Web at:, http://www.d.umn.edu/~jgallian/TF, 1. Let G 5 kx, y | x 5 (xy)3, y 5 (xy)4l. To what familiar group is G, isomorphic?, 2. Let G 5 kz | z6 5 1l and H 5 kx, y | x2 5 y3 5 1, xy 5 yxl. Show, that G and H are isomorphic., 3. Show that a group of order 315 5 32 ? 5 ? 7 has a subgroup of, order 45., 4. Let G be a group of order p 2q 2, where p and q are primes and p . q., If |G| 2 36, prove that G has a normal Sylow p-subgroup., 5. Let H denote a Sylow 7-subgroup of a group G and K a Sylow, 5-subgroup of G. Assume that |H| 5 49, |K| 5 5, and K is a subgroup of N(H). Show that H is a subgroup of N(K)., 6. Prove that no finite group of order greater than 6 has exactly three, conjugacy classes., 7. Suppose that K is a normal Sylow p-subgroup of H and that H is a, normal subgroup of G. Prove that K is a normal subgroup of G., (Compare this with Exercise 55 in Chapter 9.), 8. Show that the polynomial x 5 2 6x 1 3 over Q is not solvable by, radicals., 9. Let H and K be subgroups of G. Prove that HK is a subgroup of G, if H # N(K)., 10. Suppose that H is a subgroup of a finite group G and that H contains N(P), where P is some Sylow p-subgroup of G. Prove that, N(H) 5 H., 11. Prove that a simple group G of order 168 cannot contain an element of order 21., 12. Prove that the only group of order 561 is Z561., *Piet Hein, “The Road to Wisdom,” Grooks (1966) Copyright © Piet Hein Grooks., Reprinted with kind permission from Piet Hein a/s, DK-5500 Middelfart, Denmark., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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582, , Special Topics, , 13. Prove that the center of a non-Abelian group of order 105 has, order 5., 14. Let n be an odd integer that is at least 3. Prove that every Sylow, subgroup of Dn is cyclic., 15. Let G be the digraph obtained from Cay({(1, 0), (0, 1)}:Z3 % Z5), by deleting the vertex (0, 0). [Also, delete each arc to or from, (0, 0).] Prove that G has a Hamiltonian circuit., 16. Prove that the digraph obtained from Cay({(1, 0), (0, 1)}:Z4 % Z7), by deleting the vertex (0, 0) has a Hamiltonian circuit., 17. Let G be a finite group generated by a and b. Let s1, s2, . . . , sn be, the arcs of a Hamiltonian circuit in the digraph Cay({a, b}:G). We, say that the vertex s1s2 ? ? ? si travels by a if si11 5 a. Show that if, a vertex x travels by a, then every vertex in the coset xkab21l travels by a., 18. Recall that the dot product u ? v of two vectors u 5 (u1, u2, . . . , un), and v 5 (v1, v2, . . . , vn) from F n is, u1v1 1 u2v2 1 ? ? ? 1 unvn, (where the addition and multiplication are those of F). Let C be an, (n, k) linear code. Show that, C > 5 {v [ Fn | v ? u 5 0 for all u [ C}, is an (n, n 2 k) linear code. This code is called the dual of C., 19. Find the dual of each of the following binary codes., a. {00, 11}, b. {000, 011, 101, 110}, c. {0000, 1111}, d. {0000, 1100, 0011, 1111}, 20. Let C be a binary linear code such that C # C >. Show that wt(v) is, even for all v in C., 21. Let C be an (n, k) binary linear code. If v is a binary n-tuple, but, v o C >, show that v ? u 5 0 for exactly half of the elements u in C., 22. Suppose that C is an (n, k) binary linear code and the vector, 11 ? ? ? 1 [ C >. Show that wt(v) is even for every v in C., 23. Suppose that C is an (n, k) binary linear code and C 5 C >. (Such a, code is called self-dual.) Prove that n is even. Prove that 11 ? ? ? 1, is a code word., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Supplementary Exercises for Chapters 24–33, , 583, , 24. If G is a finite solvable group, show that there exist subgroups of G, {e} 5 H0 , H1 , H2 , ? ? ? , Hn 5 G, such that Hi11/Hi has prime order., , The end., john lennon and paul mccartney, "The End,", Abbey Road, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, Don’t wait for answers, Just take your chances, Don’t ask me why, billy joel, “Don’t Ask Me Why,” Glass Houses*, , For some exercises only partial answers are provided., Many of the proofs given below are merely sketches. In these, cases, the student should supply the complete proof., Chapter 0, Things that hurt, instruct., benjamin franklin, , 1. {1, 2, 3, 4}; {1, 3, 5, 7}; {1, 5, 7, 11}; {1, 3, 7, 9, 11, 13, 17, 19}; {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13,, 14, 16, 17, 18, 19, 21, 22, 23, 24}, 3. 12, 2, 2, 10, 1, 0, 4, 5, 5. By using 0 as an exponent if necessary, we may write a 5 p1m1 . . . pkmk and b 5 p1n1 . . . pknk,, where the p’s are distinct primes and the m’s and n’s are nonnegative. Then lcm(a, b) 5 p1s1 . . . pksk,, where si 5 max(mi, ni), and gcd(a, b) 5 p1t1 . . . pktk, where ti 5 min(mi, ni). Then lcm(a, b) ?, gcd(a, b) 5 p1m11n1 . . . pkmk1nk 5 ab., 7. Write a 5 nq1 1 r1 and b 5 nq2 1 r2, where 0 # r1, r2 , n. We may assume that r1 $ r2. Then, a 2 b 5 n(q1 2 q2) 1 (r1 2 r2), where r1 2 r2 $ 0. If a mod n 5 b mod n, then r1 5 r2 and n, divides a 2 b. If n divides a 2 b, then by the uniqueness of the remainder, we have r1 2 r2 5 0., 9. Use Exercise 7., 11. Use Theorem 0.2., 13. By Theorem 0.2 there are integers s and t such that ms 1 nt 5 1. Then m(sr) 1 n(tr) 5 r., 15. Let p be a prime greater than 3. By the division algorithm, we can write p in the form 6n 1 r,, where r satisfies 0 # r , 6. Now observe that 6n, 6n 1 2, 6n 1 3, and 6n 1 4 are not prime., 17. Since st divides a 2 b, both s and t divide a 2 b. The converse is true when gcd(s, t) 5 1., 19. Use Euclid’s Lemma and the Fundamental Theorem of Arithmetic., 21. Use proof by contradiction., 30, 17, 23., 41 i, 41, 25. Observe that cos u + i sin u 5 cos n1 nu 2 1 i sin n1 nu 2 5 1cos nu i sin nu 2 n., 27. Let S be a set with n 1 1 elements and pick some a in S. By induction, S has 2n subsets that do not, contain a. But there is a one-to-one correspondence between the subsets of S that do not contain a, and those that do. So, there are 2 ? 2n 5 2n11 subsets in all., 29. Consider n 5 200! 1 2., 31. Say p1p2 . . . pr 5 q1q2 . . . qs, where the p’s and q’s are primes. By the Generalized Euclid’s, Lemma, p1 divides some qi, say q1 (we may relabel the q’s if necessary). Then p1 5 q1 and p2 . . ., pr 5 q2 . . . qs. Repeating this argument at each step, we obtain p2 5 q2, . . . , pr 5 qr and r 5 s., 33. Suppose that S is a set that contains a and whenever n $ a belongs to S, then n 1 1 [ S. We must, prove that S contains all integers greater than or equal to a. Let T be the set of all integers greater, *“Don’t Ask Me Why,” by Billy Joel. © 1980 Impulsive Music., A1, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A2, , 35., , 37., 39., 41., , 43., , 45., , 47., 51., 53., 55., 57., , 59., 61., , 63., , Selected Answers, , than a that are not in S and suppose that T is not empty. Let b be the smallest integer in T (if T has, no negative integers, b exists because of the Well Ordering Principle; if T has negative integers, it, can have only a finite number of them so that there is a smallest one). Then b 2 1 [ S, and therefore b 5 (b 2 1) 1 1 [ S., For n 5 1, observe that 13 1 23 1 33 5 36. Assume that n3 1 (n 1 1)3 + (n 1 2)3 5 9m for, some integer m. We must prove that (n 1 1)3 1 (n 1 2)3 1 (n 1 3)3 is a multiple of 9. Using the, induction hypothesis we have that (n 1 1)3 1 (n 1 2)3 1 (n 1 3)3 5 9m 2 n3 1 (n 1 3)3 5 9m 2, n3 1 n3 1 3 ? n2 ? 3 1 3 ? n ? 9 1 33 5 9m 1 9n2 1 27n 1 27., The statement is true for any divisor of 83 2 4 5 508., 6 p.m., Observe that the number with the decimal representation a9a8 . . . a1a0 is a9 ? 109 1 a8 ? 108 1 . . . 1, a1 ? 10 1 a0. Then use Exercise 9 and the fact that ai10i mod 9 5 ai mod 9 to deduce that the check, digit is (a9 1 a8 1 . . . 1 a1 1 a0) mod 9., For the case in which the check digit is not involved, see the answer to Exercise 41. If a transposition involving the check digit c 5 (a1 1 a2 1 . . . 1 a10) mod 9 goes undetected, then a10 5, (a1 1 a2 1 . . . 1 a 9 1 c) mod 9. Substitution yields 2(a1 1 a2 1 . . . 1 a9 ) mod 9 5 0. Therefore,, modulo 9, we have 10(a1 1 a2 1 . . . 1 a 9) 5 a1 1 a2 1 . . . 1 a9 5 0. It follows that c 5 a10., In this case the transposition does not yield an error., Say the number is a8a7 . . . a1a0 5 a8 ? 108 1 a7 ? 107 1 . . . 1 a1 ? 10 1 a0. Then the error is, undetected if and only if (ai10i 2 ai910i) mod 7 5 0. Multiplying both sides by 5i and noting that, 50 mod 75 1, we obtain (ai 2 ai9) mod 7 5 0., 4, Cases where (2a 2 b 2 c) mod 11 5 0 are undetected., The check digit would be the same., 4302311568, 2. Since b is one-to-one, b(a (a1)) 5 b(a(a2)) implies that a(a1) 5 a(a2); and since a is, one-to-one, a1 5 a2., 3. Let c [ C. There is a b in B such that b(b) 5 c and an a in A such that a(a) 5 b. Thus, (ba)(a) 5, b(a(a)) 5 b(b) 5 c., 4. Since a is one-to-one and onto, we may define a21(x) 5 y if and only if a( y) 5 x. Then, a21(a(a)) 5 a and a(a21(b)) 5 b., No. (1, 0) [ R and (0, 21) [ R, but (1, 21) o R., a belongs to the same subset as a. If a and b belong to the subset A, then b and a also belong to, A. If a and b belong to the subset A and b and c belong to the subset B, then A 5 B, since the, distinct subsets of P are disjoint. So, a and c belong to A., The last digit of 3100 is the value of 3100 mod 10. Observe that 3100 mod 10 is the same as, ((34 mod 10)25 mod 10 and 34 mod 10 5 1. Similarly, the last digit of 2100 is the value of 2100, mod 10. Observe that 25 mod 10 5 2 so that 2100 mod 10 is the same as (25 mod 10)20 mod, 10 5 220 mod 10 5 (25)4 mod 10 5 24 mod 10 5 6., , 65. Apply g 1 to both sides of ag bg., , Chapter 1, Think of what you’re saying, you can get it wrong and still think that it’s all right., john lennon and paul mccartney,, “We Can Work It Out,” single*, , 1. Three rotations—08, 1208, 2408—and three reflections across lines from vertices to midpoints of, opposite sides. See the back inside cover for a picture., 3. a. V, b. R270, c. R0, d. R180, H, V, D, D9, e. none, 5. Dn has n rotations of the form k(3608/n), where k 5 0, . . . , n 2 1. In addition, Dn has n reflections., When n is odd, the axes of reflection are the lines from the vertices to the midpoints of the opposite, *Copyright © 1965 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music, Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , 7., 9., 11., 13., 15., 17., 19., 21., 23., , A3, , sides. When n is even, half of the axes of reflection are obtained by joining opposite vertices; the, other half, by joining midpoints of opposite sides., A rotation followed by a rotation either fixes every point (and so is the identity) or fixes only the, center of rotation. However, a reflection fixes a line., Observe that 1 ? 1 5 1; 1(21) 5 21; (21)1 5 21; (21)(21) 5 1. These relationships also hold, when 1 is replaced by “rotation” and 21 is replaced by “reflection.”, In D4, HD 5 DV but H 2 V., R0, R180, H, V, See answer for Exercise 13., In each case, the group is D6., cyclic, Their only symmetry is the identity., 1808 rotational symmetry, , Chapter 2, There are no secrets to success. It is the result of preparation, hard work,, and learning from failure., colin powell, , 1. c, d, 3. none, 3, 2, 5. 17; 13; n 2 1;, , i, 13, 13, 7. Does not contain the identity; closure fails., 9. Under multiplication modulo 4, 2 does not have an inverse. Under multiplication modulo 5, each, element has an inverse., 11. c 9 9 d, 10 8, 13. a. 2a 1 3b b. 22a 1 2(2b 1 c) c. 23(a 1 2b) 1 2c 5 0, 15. Since the inverse of an element in G is in G, H 8 G. Let g belong to G. Then g21 belongs to G, and therefore (g21)21 5 g belongs to G. So, G 8 H., 17. e, 19. Use the fact that det (AB) 5 (det A)(det B)., 21. 29, 23. For n $ 0, use induction. For n , 0, note that e 5 (ab)0 5 (ab)n(ab)2n 5 (ab)na2nb2n so that, anbn 5 (ab)n. In a non-Abelian group (ab)n need not equal anbn., 25. Use the Socks–Shoes Property., 27. For the case n . 0, use induction. For n , 0, note that e 5 (a21ba)n(a21ba)2n 5 (a21ba)n, (a21b2na) and solve for (a21ba)n., 29. {1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45}, 31. Suppose x appears in a row labeled with a twice; say, x 5 ab and x 5 ac. Then cancellation, yields b 5 c. But we use distinct elements to label the columns., 33. Use Exercise 31., 35. a21cb21; aca21, 37. If x3 5 e and x 2 e, then (x21)3 5 e and x 2 x21. So nonidentity solutions come in pairs., If x2 2 e, then x21 2 x and (x21)2 2 e. So solutions to x2 2 e come in pairs., 39. Observe that aa21b 5 ba21a., 41. If F1F2 5 R0, then F1F1 5 F1F2 and by cancellation F1 5 F2., 43. Since FRk is a reflection we know that (FRk)(FRk) 5 R0. So Rk FRk 5 F21 5 F., 45. a. R3 b. R c. R5F, 47. Since a2 5 b2 5 (ab)2 5 e, we have aabb 5 abab. Now cancel on the left and right., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A4, , Selected Answers, , 49. If n is not prime, the set is not closed under multiplication modulo n. If n is prime, the set is, closed and for every r in the set there are integers s and t such that 1 5 rs 1 nt 5 rs modulo n., a b, 51. The matrix c, d is in GL(2, Z2) if and only if ad Z bc. This happens when a and d are 1 and at, c d, 1 0, 1 1, d do, least 1 of b and c is 0, and when b and c are 1 and at least 1 of a and d is 0. c, d and c, 1 1, 0, 1, not commute., 53. Let a be any element in G and write x 5 ea. Then a21x 5 a21(ea) 5 (a21e)a 5 a21a 5 e. Then, solving for x we obtain x 5 ae 5 a., , Chapter 3, Success is the ability to go from one failure to another with no loss of enthusiasm., sir winston churchill, , 1. |Z12| 5 12; |U(10)| 5 4; |U(12)| 5 4; |U(20)| 5 8; |D4| 5 8, In Z12, |0| 5 1; |1| 5 |5| 5 |7| 5 |11| 5 12; |2| 5 |10| 5 6; |3| 5 |9| 5 4; |4| 5 |8| 5 3; |6| 5 2., In U(10), |1| 5 1; |3| 5 |7| 5 4; |9| 5 2., In U(12), |1| 5 1; |5| 5 2; |7| 5 2; |11| 5 2., In U(20), |1| 5 1; |3| 5 |7| 5 |13| 5 |17| 5 4; |9| 5 |11| 5 |19| 5 2., In D4, |R0| 5 1; |R90| 5 |R270| 5 4; |R180| 5 |H| 5 |V| 5 |D| 5 |D9| 5 2., In each case, notice that the order of the element divides the order of the group., 3. In Q, |0| 5 1 and all other elements have infinite order. In Q*, |1| 5 1, |21| 5 2, and all other, elements have infinite order., 5. Each is the inverse of the other., 7. (a4c22b4)21 5 b24c2a24 5 b3c2a2, 9. D4; D4; it contains {R0, R180, H, V}, 11. If n is a positive integer, the real solutions of xn 5 1 are 1 when n is odd and 61 when n is even., So, the only elements of finite order in R* are 61., 13. H is a subgroup. To prove this we need only show that if a [ H then a21 [ H. But if a21 o H,, then the given property says that a 5 (a 21)21 o H., 15. Since |a| 5 7 we have a 5 a14a 5 a15 5 (a5)3., 17. If a and b are distinct elements of order 2, then ab has order 2 and is distinct from a and b. If c, is a fourth element of order 2, then ac, bc, and abc make at least 7 elements of order 2. D4 has, exactly five elements of order 2., 19. Suppose that m 6 n and am an. Then e ana m anm. This contradicts the assumption that, a has infinite order., 21. If a has infinite order, then e, a, a2, . . . are all distinct and belong to G, so G is infinite., If |a| 5 n, then e, a, a2, . . . , an 2 1 are distinct and belong to G., 23. By brute force, show that k 4 5 1 for all k., 25. Suppose that K is a subgroup of Dn that has at least one reflection F. Denote the rotations of K, by R1 R2, . . . , Rm. Then R1F, R2 F, . . . , RmF are distinct reflections in K. If F' is any reflection, in K, then F'F 5 Ri for some i. But then F' 5 RiF. Thus, K has exactly m reflections., 27. Since n is even, Dn contains R180. Let F be any reflection in Dn. Then the set {R0, R180, F, R180F}, is closed and therefore is a subgroup of Dn., 29. k2l, k3l, k6l, 31. U4(20) 5 {1, 9, 13, 17}; U5(20) 5 {1, 11}; U5(30) 5 {1, 11}; U10(30) 5 {1, 11}. To prove that, Uk(n) is a subgroup, it suffices to show that it is closed. Suppose that a and b belong to Uk(n). We, must show that in U(n), ab mod k 5 1. That is, (ab mod n) mod k 5 1. Let n 5 kt and ab 5 qn 1 r, where 0 # r , n. Then (ab mod n) mod k 5 r mod k 5 (ab 2 qn) mod k 5 (ab 2 qkt) mod k 5 ab, mod k 5 (a mod k)(b mod k) 5 1 ? 1 5 1. H is not a subgroup because 7 [ H but 7 ? 7 5 9 is not, 1 mod 3., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A5, , 33. If x [ Z(G), then x [ C(a) for all a, so x [ >a[G C(a). If x [ >a[G C(a), then xa 5 ax for all a, in G, so x [ Z(G)., 35. The case that k 0 is trivial. Let x [ C1a2. If k is positive, then by induction on k, xak1 xaak , axak aakx ak1x. The case where k is negative now follows from Exercise 34. In a group, if, x commutes with a, then x commutes with all powers of a. If x commutes with ak for some k,, then x need not commute with a., 37. a. C(5) 5 G; C(7) 5 {1, 3, 5, 7}, b. Z(G) 5 {1, 5}, c. |2| 5 2; |3| 5 4. They divide the order of the group., 39. a. First observe that because kSl is a subgroup of G containing S, it is a member of the intersection. So, H 8 kSl. On the other hand, since H is a subgroup of G and H contains S, by definition kSl 8 H., b. Let K 5 {s1n1 s2n2 … smnm | m $ 1, si [ S, ni [ Z}. Then because K satisfies the subgroup test, and contains S, we have kSl 8 K. On the other hand, if L is any subgroup of G that contains S,, then L also contains K by closure. Thus, by part a, H 5 kSl contains K., 41. Mimic the proof of Theorem 3.5., 43. No. In D4, C(R180) 5 D4., 45. For the first part, see Example 4. For the second part, use D4., 47. Note that for any polynomial f (x) 5 anxn 1 an21xn21 1 . . . 1 a0 in G we have 4f (x) 5 4anxn 1, 4an21xn21 1 . . . 1 4a0 5 0. Thus, the orders of elements of G are at most 4. No element has, order 3 because 3ai 5 0 mod 4 if and only if ai is 0. Since 2ai 5 0 mod 4 if and only if ai 5 0, or 2, we have 2f (x) 5 2anxn 1 2an21xn21 1 . . . 1 2a0 5 0 if and only if a1 5 0 or 2 for all i., Excluding the identity, this condition is necessary and sufficient for an element to have order 2., 49. 2, 51. First observe that (a d)n/d 5 an 5 e, so 0 ad 0 is at most n/d. Moreover, there is no positive integer, t , n/d such that 1ad 2 t adt e, for otherwise |a| Z n., 1 1 n, 1 n, 53. Note that c, d c, d., 0 1, 0 1, 55. For any positive integer n, a rotation of 3608/n has order n. A rotation of 22° has infinite order., 57. kR0l, kR90l, kR180l, kDl, kD9l, kH l, kV l (Note that kR90l 5 kR270l.), 59. Nonidentity elements of odd order come in pairs. So, there must be some element a of even order,, say |a| 5 2m. Then |am| 5 2., 61. Let |g| 5 m and write m 5 nq 1 r, where 0 # r , n. Then gr 5 gm2nq 5 gm(gn)2q 5 (gn)2q belongs to H. So, r 5 0., 63. 1 [ H. Let a, b [ H. Then (ab21)2 5 a2(b2)21, which is the product of two rationals. 2 can be, replaced by any positive integer., 65. |k3l| 5 4, 67. Let c, , a b, a b, d and c, d belong to H. It suffices to show that a 2 a9 1 b 2 b9 1 c 2 c9 1 d 2, c d, c d, , d9 5 0. This follows from a 1 b 1 c 1 d 5 0 5 a9 1 b9 1 c9 1 d9. If 0 is replaced by 1, H is, not a subgroup., 69. If 2a and 2b [ K, then 2a(2b)21 5 2a2b [ K, since a 2 b [ H., 71. c, , 1, 2 0 1, 0, d c 2 1 d is not in H., 0 2, 0 2, , 73. If a 1 bi and c 1 di [ H, then (a 1 bi)(c 1 di)21 5 (ac 1 bd) 1 (bc 2 ad)i and (ac 1 bd)2 1, (bc 2 ad)2 5 1, so that H is a subgroup. H is the unit circle in the complex plane., 75. Since ee 5 e is in HZ(G), it is nonempty. Let h1z1 and h2z2 belong to HZ(G). Then h1z1(h2z2)21 5, h1z1z221h221 5 h1h221z1z221 [ HZ (G)., 77. Use Theorem 0.2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A6, , Selected Answers, , 79. a. e c, , a, b, a, b. e c, b, a, c. e c, 0, , b, d 0 where a2 ab b2 ? 0; a, b [ R f, ab, b, d 0 a2 ? b2; a, b [ R f, a, 0, d 0 a ? 0; a [ R f, a, , Chapter 4, There will be an answer, let it be., john lennon and paul mccartney, “Let It Be,” single*, , 1. For Z6, generators are 1 and 5; for Z8, generators are 1, 3, 5, and 7; for Z20, generators are 1, 3, 7,, 9, 11, 13, 17, and 19., 3. k20l = {20, 10, 0}; k10l = {10, 20, 0}; ka20l = {a20, a10, a0}; ka10l = {a10, a20, a0], 5. k3l 5 {3, 9, 7, 1}; k7l 5 {7, 9, 3, 1}, 7. U(8) or D3, 9. Six subgroups; generators are the divisors of 20. Six subgroups; generators are ak, where k is a, divisor of 20., 11. By definition, a1 [ kal. So, ka 1l 8 kal. By definition, a 1a 1 2 1 [ ka 1l. So, kal 8 ka 1l., 13. k21l > k10l k18l k6l. In the general case kaml > kanl kakl, where, k 5 lcm (m, n) mod 24., 15. |g| divides 12 is equivalent to g12 5 e. So, if a12 5 e and b12 5 e, then (ab21)12 5 a12(b12)21 5, ee21 5 e. The general result is given in Exercise 45 of Chapter 3., 17. is odd or infinite, 19. k1l, k7l, k11l, k17l, k19l, k29l, 21. a. |a| divides 12. b. |a| divides m. c. By Theorem 4.3, |a| 5 1, 2, 3, 4, 6, 8, 12, or 24. If |a| 5 2,, then a8 5 (a2)4 5 e4 5 e. A similar argument eliminates all other possibilities except 24., 23. Yes, by Theorem 4.3. The subgroups of Z are of the form knl {0, 6n, 62n, 63n, . . .}, n 5 0,, 1, 2, 3, . . . . The subgroups of kal are of the form kanl for n 5 0, 1, 2, 3, . . . ., 25. For the first part, apply Theorem 4.3 to the subgroup of rotations; Dn has n elements of order 2, when n is odd and n 1 1 elements of order 2 when n is even., 27. See Example 15 of Chapter 2., 29. 1000000, 3000000, 5000000, 7000000; by Theorem 4.3, k1000000l is the unique subgroup of order 8, and only those on the list are generators. a1000000, a3000000, a5000000, a7000000; by Theorem 4.3,, ka1000000l is the unique subgroup of order 8, and only those on the list are generators., 31. Let G 5 {a1, a2, . . . , ak}. Now let |ai| 5 ni. Consider n 5 n1n2 . . . nk., 33. Mimic the lattice in Figure 14.1., 35. The lattice is a vertical line with successive terms from top to bottom kp0l, kp1l, kp2l, . . ., kpn21l, k0l., 37. Suppose a and b are relatively prime positive integers and ka/bl 5 Q1. Then there is some positive, integer n such that (a/b)n 5 2. Clearly, n 2 0, 1, or 21. If n . 1, an 5 2bn, so that 2 divides a. But, then 2 divides b as well. A similar contradiction occurs if n , 21., 39. For 6, use Z25. For n, use Z 2n21., 41. Let t 5 lcm(m, n) and |ab| 5 s. Then (ab)t 5 atbt 5 e, and therefore s divides t. Also, e 5, (ab)s 5 asbs, so that as 5 b2s, and therefore as and b2s belong to kal > kbl 5 {e}. Thus, m divides, s and n divides s, and, therefore, t divides s. This proves that s 5 t. For the second part, try D3., 43. Let |a| 5 m, b 5 n and d 5 gcd(m, n). Then lcm(m, n) 5 mnYd, |ad| 5 mYd, and |b| 5 n. Then, by Exercise 41, |adb| 5 lcm(m, n)., 45. all divisors of 60, 47. The argument given in the proof of the corollary to Theorem 4.4 shows that in an infinite group,, the number of elements of finite order n is a multiple of f(n) or there is an infinite number of, elements of order n., *Copyright © 1970 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music, Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A7, , 49. It follows from Example 15 in Chapter 2 and Example 10 in Chapter 0 that the group, H 5 kcos(3608Yn) 1 i sin(3608Yn)l is a cyclic group of order n and every member of this, group satisfies xn 2 1 5 0. Moreover, since every element of order n satisfies xn 2 1 5 0, and there can be at most n such elements, all complex numbers of order n are in H. Thus,, by Theorem 4.4, C* has exactly f(n) elements of order n., 51. Let x [ Z(G) and |x| 5 p where p is prime. Say y [ G with |y| 5 q where q is prime. Then, (xy)pq 5 e and therefore |xy| 5 1, p, or q. If |xy| 5 1, then p 5 q. If |xy| 5 p, then e 5 (xy)p 5 y p, and q divides p. Thus, q 5 p. A similar argument applies if |xy| 5 q., 53. An infinite cyclic group does not have an element of prime order. A finite cyclic group can have, only one subgroup for each divisor of its order. A subgroup of order p has exactly p 2 1 elements, of order p. Another element of order p would give another subgroup of order p., 55. 1 ? 4, 3 ? 4, 7 ? 4, 9 ? 4; x4, 1x4 2 3, 1x4 2 7, 1x4 2 9, 57. 1 of order 1; 33 of order 2; 2 of order 3; 10 of order 11; 20 of order 33, 59. 1, 2, 10, 20. In general, if an Abelian group contains cyclic subgroups of order m and n where m, and n are relatively prime, then it contains subgroups of order d for each divisor d of mn., 61. Say a and b are distinct elements of order 2. If a and b commute, then ab is a third element, of order 2. If a and b do not commute, then aba is a third element of order 2., 63. Use Exercise 32 of Chapter 3 and Theorem 4.3., 65. 1 and 2, 67. In a cyclic group there are at most n solutions to the equation xn 5 e., 69. 12 or 60; 48, 71. Observe that a280 e a440. Thus 0 a 0 is a common divisor of 280 and 440, and therefore 0 a 0, divides gcd(280, 440) 5 40., 73. Say b is a generator of the group. Since p and pn 2 1 are relatively prime, we know by Corollary, 3 of Theorem 4.2 that bp also generates the group. Finally, observe that (b p)k 5 (b k) p., 75. Use the fact that a cyclic group of even order has a unique element of order 2., 77. Since reflections have order 2, any cyclic subgroup of order 4 must be generated by a rotation. So,, by Theorem 4.3 there is exactly one cyclic subgroup of order 4 if 4 divides n and 0 otherwise., 79. Observe that Dn has nY2 noncyclic subgroups of order 4 of the form {R0, R180, F, R180 F}, where, F is a reflection. By Exercise 25 of Chapter 3, these are the only noncyclic subgroups of order 4., 81. First observe that the set of all rotations is the only cyclic subgroup of Dn of order n. It follows, from Exercise 25 of Chapter 3 that when n is odd, Dn has no noncyclic subgroup of order n, and, when n = 2m, the only noncyclic subgroups of Dn are of the form K 5 {R0, R, R2, . . . Rm21, F,, RF, R2F, . . ., Rm21F} where |R| 5 m and F is a reflection. (That K is closed follows from Exercises 43, and 44 in Chapter 2.) To find a second such subgroup let F9 be any reflection not in K. Then K9 5, {R0, R, R2, . . . Rm 21, F9, RF9, R2F9, . . ., Rm 21F9} is also a subgroup. (For example, in D4, {R0, R180,, H, R180H} and {R0, R180, D, R180D}.) Finally, by observing that for any reflection F0 in Dn we have, K0 5 {R0, R, R2, . . . Rm 21, F0, RF0, R2F0, . . ., Rm 21F0} 5 K if F0 is in K and K 0 5 K9 if F0 is in, K9 we know that K and K9 have no reflections in common. Since K and K9 account for all n reflections, we have proved that when n is even there are exactly three subgroups of order n., 83. Since m and n are relatively prime, it suffices to show both m and n divide k. By Corollary 2 of, Theorem 4.1, it is enough to show that ak e. Note that ak [ kal > kbl, and since kal > kbl is a, subgroup of both kal and kbl, we know that 0 kal > kbl 0 must divide both 0 kal 0 and 0 kbl 0. Thus,, 0 kal > kbl 0 1., 85. Observe that among the integers from 1 to pn, the pn21 integers p, 2p, 3p, . . . , pn21p are exactly the, ones that are not relatively prime to p., , Supplementary Exercises for Chapters 1–4, Four short words sum up what has lifted most successful individuals above the crowd: a little bit, more. They did all that was expected of them and a little bit more., a. lou vickery, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A8, , Selected Answers, , 1. a. Let xh1x21 and xh2x21 belong to xHx21. Then (xh1x21)(xh2x21)21 5 xh1h221x21 [ xHx21 also., b. Let khl 5 H. Then kxhx21l 5 xHx21. c. (xh1x21)(xh2x21) 5 xh1h2x21 5 xh2h1x21 5 (xh2x21), (xh1x21), 3. Suppose cl(a) > cl(b) 2 f. Say xax21 5 yby21. Then (y21x)a(y21x)21 5 b. Thus, for any ubu21, in cl(b), we have ubu21 5 (uy21x)a(uy21x)21 [ cl(a). This shows that cl(b) # cl(a). By symmetry, cl(a) # cl(b). Because a 5 eae21 [ cl(a), the union of the conjugacy classes is G., 5. Observe that (xax21)k 5 xakx21. Thus, (xax21)k 5 e if and only if ak 5 e., 7. By Exercise 4 of Chapter 3, Exercise 24 of Chapter 2, and the previous exercise, we have |ab| 5, |(ab)21| 5 |b21a21| 5 |a21b21|., 9. Try D4., 11. By Exercise 5, for every x in G, |xax21| 5 |a|, so that xax21 5 a or xa 5 ax., 13. 1 of order 1, 15 of order 2, 8 of order 15, 4 of order 5, 2 of order 3, 15. Let |G| 5 5. Let a 2 e belong to G. If |a| 5 5, we are done. If |a| 5 3, then {e, a, a2} is a subgroup of G. Let b be either of the remaining two elements of G. Then the set {e, a, a2, b, ab, a2b}, consists of six different elements, a contradiction. Thus, |a| 2 3. Similarly, |a| 2 4. We may now, assume that every nonidentity element of G has order 2. Pick a 2 e and b 2 e in G with a 2 b., Then {e, a, b, ab} is a subgroup of G. Let c be the remaining element of G. Then {e, a, b, ab, c,, ac, bc, abc} is a set of eight distinct elements of G, a contradiction. It now follows that if a [ G, and a 2 e, then |a| 5 5., 17. an(bn)21 5 (ab21)n, so Gn is a subgroup. For the non-Abelian group, try D3., 19. Suppose G 5 H < K. Pick h [ H with h o K. Pick k [ K, but k o H. Then, hk [ G, but hk o H, and hk o K. U(8) is the union of the three subgroups., 21. If |a| 5 pk and |b| 5 pr with k # r, say, then |ab21| divides pr., 23. Note that ba2 5 ab and a3 5 b2 5 e imply ba 5 a2b. Thus, every member of the group can be, written in the form a i b j. Therefore, the group is {e, a, a2, b, ab, a2b}. D3 satisfies these conditions., The Cayley table can be obtained from the one for the dihedral group of order 6 shown in the, inside back cover of the book by replacing R0 by e, R120 by a, R240 by a2, F by b, F9 by ab and F0, by a2b., 25. xy 5 yx if and only if xyx21y21 5 e. But, (xy)x21y21 5 x21(xy)y21 5 ee 5 e., 27. Let x [ N(gHg21). Then x(gHg21)x21 5 gHg21. Thus, g21xgHg21x21g 5 g21xgH(g21xg)21 5 H., This means that g21xg [ N(H). So x [ gN(H)g21. Reverse the argument to show gN(H)g21 #, N(gHg21)., 29. Look at D11., 31. Solution from Mathematics Magazine.† “Yes. Let a be an arbitrary element of S. The set {an | n 5, 1, 2, 3, . . .} is finite, and therefore a m 5 a n for some m, n with m . n 9 1. By cancellation we, have ar(a) 5 a, where r(a) 5 m 2 n 1 1 . 1. If x is any element of S, then aar(a)21x 5 ar(a)x 5 ax,, and this implies that ar(a)21x 5 x. Similarly, we see that xar(a)21 5 x, and the element e 5 ar(a)21 is, an identity. The identity element is unique, for if e9 is another identity, then e 5 ee9 5 e9. If r(a) . 2, then ar(a)22 is an inverse of a, and if r(a) 5 2 then a2 5 a 5 e is its own inverse. Thus S is a group.”, 33. {1, 2n 2 1, 2n 1 1, 4n 2 1}, 35. Use det (AB) 5 (det A)(det B) to prove H is a subgroup. H is not a subgroup when det A is an integer,, since det A21 need not be an integer., 37. Choose x 2 e and y o kxl. Then G 5 kxl < kyl. But then xy [ kyl, so that kxl # kyl and therefore, G 5 kyl. To prove that |G| 5 pq or p3, use Theorem 4.3., 39. If T and U are not closed, then there are elements x and y in T and w and z in U such that xy is not, in T and wz is not in U. It follows that xy [ U and wz [ T. Then xywz 5 (xy)wz [ U and xywz 5, xy(wz) [ T, a contradiction., 41. Let G be the group of all polynomials with integer coefficients under addition. Let Hk be the subgroup of polynomials of degree at most k together with the zero polynomial (the zero polynomial, does not have a degree)., †Mathematics, , Magazine 63 (April 1990): 136., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A9, , 43. Take g 5 a., 45. Let S 5 {s1, s2, s3, . . . , sk} and let g be any element in G. Then the set {gs121, gs221, gs321, . . . ,, gsk21} and S have at least one element in common. Say gsi21 5 sj. Then g 5 sj si., 47. Let K 5 {x [ G | |x| divides d}. By Exercise 15 of Chapter 4, K is a subgroup. Let x [ H. By, Theorem 4.3, |x| divides d. So, H # K. Let y [ K, |y| 5 t, and d 5 tq. By Theorem 4.3, H has a, subgroup of order t and G has only one subgroup of order t. So, kyl # H., 49. To check associativity, note (a * b) * c 5 ((a 1 b) 2 1) * c 5 a 1 b 2 1 1 c 2 1 5 a 1 b 1, c 2 2 and a * (b * c) 5 a * (b 1 c 2 1) 5 a 1 (b 1 c 2 1) 2 1 5 a 1 b 1 c 2 2. To determine, the identity e, we observe that a * e 5 a if and only if a 1 e 2 1 5 a. Thus, 1 is the identity (it is, obvious that the operation is commutative). If a21 exists, we have must a * a21 5 a 1, a21 2 1 5 1, and therefore a21 is a 2. To find a generator, observe that for any positive, integer k, ak 5 ka 2 (k 2 1). So, for positive k and a 5 2, we have 2k 5 k 1 1. One can also check, that 2k 5 k 1 1 when k 5 0 or negative. Thus, 2 generates all integers., , Chapter 5, Mistakes are often the best teachers., james a. froude, , 1. a., c., 3., 5., 7., 9., 11., 13., 15., 17., 19., , 21., 23., 25., 27., 29., 31., 33., 35., 37., 39., 41., 43., 45., 47., , 1, 5 c, 2, 1, ab c, 6, (15)(234), 3 b. 12, a21, , 2 3 4 5 6, 1 2 3 4 5 6, d b. ba c, d, 1 3 5 4 6, 1 6 2 3 4 5, 2 3 4 5 6, d, 2 1 5 3 4, b. (124)(35)(6) c. (1423), c. 6 d. 6 e. 12 f. 2, , a., a., 12, For S6, the possible orders are 1, 2, 3, 4, 5, 6; for A6, 1, 2, 3, 4, 5; for A7, 1, 2, 3, 4, 5, 6, 7., a. even b. odd c. even d. odd e. even, Let a(x1) 5 a(x2). Then x1 5 a(a(x1)) 5 a(a(x2)) 5 x2. For any s in S, we have a(a(s)) 5 s., even; odd, An even number of 2-cycles followed by an even number of 2-cycles gives an even number of, 2-cycles in all. So the Finite Subgroup Test is verified., Suppose that a can be written as a product of m 2-cycles and b can be written as a product of n, 2-cycles. Then ab can be written as a product of m 1 n 2-cycles. Now observe that m 1 n is, even if and only if m and n are both even or both odd., (AHMPRS) (BDGC) (EJLNF) (I) (KO) (QU) (TWV) (XZY), Suppose H contains at least one odd permutation, say, s. Imitate the proof of Theorem 5.7 with, s in place of (12)., The identity is even; the set is not closed., a. C(a3) 5 {a1, a2, a3, a4}; b. C(a12) 5 {a1, a7, a12}, 180; 75, In S7, b 5 (2457136). In S9, b 5 (2457136) or b 5 (2457136)(89)., (124586739), (142568793), (214856379), Let a, b [ stab(a). Then ab(a) 5 a(b(a)) 5 a(a) 5 a. Also, a(a) 5 a implies a21(a(a)) 5, a21(a) or a 5 a21(a)., m is a multiple of 6 but not a multiple of 30., 6!/5 5 144, 3, 7, 9, Let a 5 (123) and b 5 (145)., (123)(12) 2 (12)(123) in Sn (n $ 3)., The Finite Subgroup Test shows that H is a subgroup. |H| 5 2(n 2 2)!., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A10, , Selected Answers, , 49. Theorem 5.2 shows that disjoint cycles commute. For the other half, observe that (ab)(ac) 5 (acb), whereas (ac)(ab) 5 (abc)., 51. Let b 5 b1b2 . . . bk, where each bi is a cycle. Because aba21 5 ab1b2 . . . bka21 5 (ab1a21), (ab2a21) . . . (abka21), the problem is reduced to showing that if g is a t-cycle for some t, then, aga21 is a t-cycle. Now write a 5 a1a2 . . . as, where each ai is a 2-cycle. Then aga21 5 a1a2 . . ., (asgas) . . . a2a1 since each ai has order 2. Now by Exercise 50 we can replace as gas by a cycle, the same length as g. Repeated use of this argument finishes the proof., 53. S9 has an element of order 20 and n 5 9 is the smallest integer that has the desired property., 55. The cases where n 5 1, 3, 5, 7, 9, or 11 are done by examining cases as in Example 4 of this, chapter. For n $ 13, observe that (1, 2, . . . , n 2 6) (n 2 5, n 2 4, n 2 3, n 2 2) (n 2 1, n) is in, An and has order 4(n 2 6), which is greater than 2n when n $ 13., 57. R0, R180, H, V, 59. The permutation corresponding to the rotation of 360/n degrees, (1, 2, . . . , n), is an even permutation so all rotations are even., 61. Cycle decomposition shows that any nonidentity element of A5 is a 5-cycle, a 3-cycle, or a product of, a pair of disjoint 2-cycles. Then, observe that there are (5 ? 4 ? 3 ? 2 ? 1)/5 5 24 group elements of the, form (abcde), (5 ? 4 ? 3)/3 5 20 group elements of the form (abc), and (5 ? 4 ? 3 ? 2)/(2 ? 2 ? 2) 5, 15 group elements of the form (ab)(cd)., 63. One possibility is {(1), (12)(34), (56)(78), (12)(34)(56)(78)}., 65. Hint: (13)(12) 5 (123) and (12)(34) 5 (324)(132)., 67. Verifying that a * s(b) 2 b * s(a) is done by examining all cases. To prove the general case,, observe that si(a) * s i11(b) 2 s i(b) * s i11(a) can be written in the form si(a) * s(si(b)) 2, si(b) * s(si(a)), which is the case already done. If a transposition were not detected, then, s(a1) * . . . * si(ai) * si11(ai11) * . . . * sn(an) 5 s(a1) * . . . * si(ai11) * si11(ai) * . . . * sn(an),, which implies si(ai) * s i11(ai11) 5 si(ai11) * s i11(ai)., 69. By Theorem 5.4 it is enough to prove that every 2-cycle can be expressed as a product of elements, of the form (1k). To this end, observe that if a ? 1, b ? 1, then (ab) 5 (1a)(1b)(1a)., 71. If a has odd order k and a is an odd permutation, then e 5 ak would be odd., 73. Observe that every element of H is an even permutation, so H # A4. Elements of A4 have order 3,, 2, or 1. For any a in A4 of order 3, kal 5 ka2l , H. The only elements in A4 of order 2 are the, products of two disjoint 2-cycles—(ab)(cd) 5 (acbd)2, which belong to H. A similar argument, applies to K., 75. By case-by-case analysis, H is a subgroup for n 5 1, 2, 3, and 4. For n $ 5, observe that (12)(34), and (12)(35) belong to H but their product does not., 77. The product of an element of Z(A4) of order 2 and an element of A4 of order 3 would have order 6., The product of an element of Z(A4) of order 3 and an element of A4 of order 2 would have order 6., 79. In Exercise 35, let G be A6. Then, stab(1) is the subgroup of A6 consisting of the 60 even permutations of the set {2,3,4,5,6}. Similarly, stab(2), stab(3), stab(4), stab(5), stab(6) are subgroups, of order 60., 81. Labeling the four tires 1, 2, 3, and 4 in clockwise order starting with 1 being the tire in the, upper left-hand corner, we may represent the four patterns as, a 5 (1324) top left-hand pattern,, b 5 (1423) top right-hand pattern,, g 5 (14)(23) bottom right-hand pattern,, d 5 (13)(24) bottom left-hand pattern., Notice that a21 5 b and that d 5 a2g. Thus, we need only find the smallest subgroup of S4, containing a and g. To this end, observe that the set {e, a, a2, a3, g, ag, a2g, a3g} is closed under, multiplication on the left and right by both a and g. This implies that the set is closed under multiplication and is therefore a group. Since ag 2 ga, the subgroup is non-Abelian., 83. Then a permutation such as (23) in Sn for n $ 3 could also be written as (11)(23) so it would be, both even and odd. In fact, (11) could be appended to every permutation written in cycle form,, making them both even and odd., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A11, , Chapter 6, Think and you won’t sink., b. c. forbes, Epigrams, , 1., 3., 5., 7., 9., 11., 13., 15., , 17., 19., , 21., 23., , 25., 27., 29., 31., , 33., 35., 37., 39., 41., 43., 45., 47., 49., 51., , Try n → 2n., f(xy) 5 2xy 2x 2y f(x)f(y)., Try 1 → 1, 3 → 5, 5 → 7, 7 → 11., D12 has elements of order 12 and S4 does not., Since Te(x) 5 ex 5 x for all x, Te is the identity. For the second part, observe that Tg 8 (Tg)21 5 Te 5, Tgg21 5 Tg 8 Tg21 and cancel., For any x in the group, we have (fgfh)(x) 5 fg(fh(x)) 5 fg(hxh21) 5 ghxh21g21 5 (gh)x(gh)21 5, fgh(x)., fR and fR disagree on H; fR and fH disagree on R90; fR and fD disagree on R90. The remain90, 0, 90, 90, ing cases are similar., 21, Let a [ Aut(G). We show that a is operation-preserving: a21(xy) 5 a21(x)a21(y) if and only, if a(a21(xy)) 5 a(a21(x)a21(y)), that is, if and only if xy 5 a(a21(x))a(a21(y)) 5 xy. So a21 is, operation-preserving. That Inn(G) is a group follows from the equation fgfh 5 fgh., Since b 5 f(a) 5 af(1), it follows that f(1) 5 a21b and therefore f(x) 5 a21bx. [Here a21 is the, multiplicative inverse of a mod n, which exists because a [ U(n).], Note that both H and K are isomorphic to the group of all permutations on four symbols,, which is isomorphic to S4. The same is true when 5 is replaced by n, since both H and K are, isomorphic to Sn21., Recall that, when n is even, Z(Dn) 5{R0, R180}. Since R180 and f(R180) are not the identity and, belong to Z(Dn), they must be equal., Z60 contains cyclic subgroups of orders 12 and 20, and any cyclic group that has subgroups or orders, 12 and 20 must be divisible by 12 and 20. So, 60 is the smallest order of any cyclic group that has, subgroups isomorphic to Z12 and Z20., See Example 15 of Chapter 2., That a is one-to-one follows from the fact that r21 exists modulo n. The operation-preserving condition is Exercise 9 in Chapter 0., Use property 2 of Theorem 6.2., The inverse of a one-to-one function is one-to-one. For any g [ G, we have f21(f(g)) 5 g, and, therefore f21 is onto. To verify that f21 is operation-preserving, see the answer to Exercise 15 of, this chapter., Tg(x) 5 Tg(y) if and only if gx 5 gy or x 5 y. This shows that Tg is a one-to-one function. Let y [, G. Then Tg(g21y) 5 y, so that Tg is onto., Apply the appropriate definitions., Show that Q is not cyclic., a b, Try a 1 bi → c, d., b, a, Yes, by Cayley’s Theorem., Observe that fg(y) 5 gyg21 and fzg(y) 5 zgy(zg)21 5 zgyg21z21 5 gyg21 since z [ Z(G). So,, fg 5 fzg., fg 5 fh implies gxg21 5 hxh21 for all x. This implies h21gx(h21g)21 5 x, and therefore h21g [ Z(G)., By Exercise 45, f x 5 f y implies y21 x is in Z(Sn); and by Exercise 66 in Chapter 5, Z(Sn) 5 {e}., Since both f and g take e to itself, H is not empty. Assume a and b belong to H. Then f(ab21), 5 f(a)f(b21) 5 f(a)f(b)21 5 g(a)g(b)21 5 g(a)g(b 21) 5 g(ab21). Thus, ab21 is in H., Since f(e) 5 e 5 e21, H is not empty. Assume that a and b belong to H. Then f(ab) 5 f(a)f(b), 5 a21b21 5 b21a21 5 (ab)21, and H is closed under multiplication. Moreover, because f(a21), 5 f(a)21 5 (a21)21, we have that H is closed under inverses., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A12, , Selected Answers, , 53. Say |a| 5 n. Then f a n(x) 5 anxa2n 5 x, so that f an is the identity. For the example, take, a 5 R90 in D4., 55. Observe that D 5 R90V and H 5 R90D., 57. (R0R90R180R270)(HD9VD)., 59. Consider the mapping f(x) 5 x2 and note that 2 is not in the image., 61. Use the fact that if a . 0, then a 5 2a2a. For the second part, use the first part together with, the fact that the inverse of an automorphism is an automorphism., 63. Say f is an isomorphism from Q to R1 and f takes 1 to a. It follows that the integer r maps to ar, and the rational r/s maps to ar/s. But ar/s 2 ap for any r/s., , Chapter 7, Use missteps as stepping stones to deeper understanding and greater achievement., susan taylor, , 1., 3., 5., 7., 9., 11., , 13., 15., 17., 19., 21., 23., , 25., 27., , 29., 31., 33., 35., , 37., 39., 41., , H 5 {a1, a2, a3, a4}, a5H 5 {a5, a8, a6, a7}, a9H 5 {a9, a11, a12, a10}, H, 1 1 H, 2 1 H, a. yes b. yes c. no, 8/2 5 4, so there are four cosets. Let H 5 {1, 11}. The cosets are H, 7H, 13H, 19H., Since |a4| 5 15, there are two cosets: ka4l and aka4l., Let ga belong to g(H d K), where a is in H d K. Then by definition ga is in gH d gK. Now let, x [ gH d gK. Then x 5 gh for some h [ H, and x 5 gk for some k [ K. Cancellation then gives, h 5 k. Thus, x [ g(H d K)., Suppose that h [ H and h , 0. Then hR1 # hH 5 H. But hR1 is the set of all negative real, numbers. Thus, H 5 R*., 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, Use Lagrange’s Theorem (Theorem 7.1) and Corollary 3., By Exercise 18, we have 56 mod 7 5 1 . So, using mod 7, we have 515 5 56 ? 56 ? 52 ? 5 5 1 ? 1 ?, 4 ? 5 5 6; 713 mod 11 5 2., Use Corollary 4 of Lagrange’s Theorem (Theorem 7.1) together with Theorem 0.2., By closure, (234)(12) 5 (1342) belongs to H so that |H| is divisible by 3 and 4 and divides 24., But if |H| 5 12 then the even permutations in H would be a subgroup of A4 of order 6, which, does not exist (see Example 5)., Since G has odd order, no element can have order 2. Thus, for each x 2 e, we know that x 2 x21., So, we can write the product of all the elements in the form ea1a121a2 a221 . . . an a n21 5 e., Let H be the subgroup of order p and K be the subgroup of order q. Then H < K has p 1 q 2 1 ,, pq elements. Let a be any element in G that is not in H < K. By Lagrange’s Theorem, |a| 5 p, q,, or pq. But |a| 2 p, for if so, then kal 5 H. Similarly, |a| 2 q., 1, 3, 11, 33. If |x| 5 33, then |x11| 5 3. Elements of order 11 occur in multiples of 10., No. Observe that by Lagrange’s Theorem, the elements of a group of order 55 must have orders, 1, 5, 11, or 55; then use the corollary of Theorem 4.4., Observe that |G:H| = |G| / |H|, |G:K| = |G| / |K|, and |K:H| = |K| / |H|., Since the reflections in a dihedral group have order 2, the generators of the subgroups of orders, 12 and 20 must be rotations. The smallest rotation subgroup of a dihedral group that contains, rotations of orders 12 and 20 must have order divisible by 12 and 20 and therefore must be a, multiple of 60. So, D60 is the smallest such dihedral group., Let a have order 3 and b be an element of order 3 not in kal. Then kal kbl is a subgroup of G of, order 9. Now use Lagrange's Theorem., Look at D3., Let a [ G and |a| 5 5. Then the set kalH has exactly 5?|H|/|kal d H| elements and |kal d H|, divides |kal| 5 5. It follows that |kal d H| 5 5 and therefore kal d H 5 kal., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A13, , 43. Certainly, a [ orbG(a). Now suppose that c [ orbG(a) > orbG(b). Then c 5 a(a) and c 5 b(b) for, some a and b, and therefore (b21a)(a) 5 b. So, if x [ orbG(b), then x 5 g(b) 5 (gb21a)(a) for, some g. This proves that orbG(b) # orbG(a). By symmetry, orbG(a) # orbG(b)., 45. a. stabG(1) 5 {(1), (24)(56)}; orbG(1) 5 {1, 2, 3, 4}, b. stabG(3) 5 {(1), (24)(56)}; orbG(3) 5 {3, 4, 1, 2}, c. stabG(5) 5 {(1), (12)(34), (13)(24), (14)(23)}; orbG(5) 5 {5, 6}, 47. Consider the mapping from G to G defined by f(x) 5 x2 and let |G| 5 2k 1 1. Use the observation that x 5 xe 5 xx2k11 5 x2k12 5 (x2)k11 to prove that f is one-to-one and Exercise 12 of, Chapter 5 to show that f is onto., 49. By Corollary 3 of Lagrange's Theorem, a group of order 5 is cyclic. Suppose G is a group with, distinct subgroups kal and kbl of order 5. Because 5 is prime, the identity is the only element, common to the two subgroups. This implies that the 25 elements of the form aib j where i,, j [ {0, 1, 2, 3, 4}are distinct., 51. Suppose that H is a subgroup of A5 of order 30. We claim that H contains all 20 elements of A5, that have order 3. To verify this, assume that there is some a in A5 of order 3 that is not in H., Then A5 5 H < aH. It follows that a2H 5 H or a2 5 aH. Since the latter implies that a [ H,, we have that a2H 5 H, which implies that a2 [ H. But then kal 5 ka2l # H, which is a, contradiction of our assumption that a is not in H. The same argument, shows that H must, contain all 24 elements of order 5. Since uHu 5 30, we have a contradiction., 53. Observe that a(ai) 5 ai11, a2(ai) 5 ai12, . . . , ak(ai) 5 ai, where all subscripts are taken mod k., 55. Suppose that H is a subgroup of S5 of order 60. An argument analogous to that given in Exercise, 51 in this chapter shows that H must contain all 24 elements in S5 of order 5 and all 20 elements, in S5 of order 3. Since these 44 elements are also in A5, we know that Z A5 > HZ divides 60 and is, greater than 30. So, H 5 A5., 57. Suppose that B [ G and det (B) 5 2. Then det (A21B) 5 1, so that A21B [ H and therefore B [, AH. Conversely, for any Ah [ AH we have det (Ah) 5 det (A)det (h) 5 2 ? 1 5 2., 59. It is the set of all permutations that carry face 2 to face 1., 61. aH 5 bH if and only if det (a) 5 6det (b)., 63. Closure of the set follows from using ab2 5 b2a3., 65. 50, , Chapter 8, Practice isn't the thing you do when you're good. It's the thing you do that makes you good., malcolm gladwell, , 1. Closure and associativity in the product follow from the closure and associativity in each component. The identity in the product is the n-tuple with the identity in each component. The inverse, of (g1, g2, . . . , gn) is (g121, g221, . . . , gn21)., 3. Use g → (g, eH) and h → (eG, h)., 5. To show that Z % Z is not cyclic, note that (a, b 1 1) o k(a, b)l., 7. Use (g1, g2) → (g2, g1). In general, G1 % G2 . . . % Gn is isomorphic to the external direct product of, any rearrangement of G1, G2, . . . , Gn., 9. Yes, by Theorem 8.2., 11. There are 12 elements of order 4. Observe by Theorem 4.4 that as long as d divides n, the number, of elements of order d in a cyclic group depends only on d. So, in both Z8000000 and Z4 there are, f(4) 5 2 elements of order 4 and f(2) 5 1 element of order 2. Similarly for Zm % Zn., 13. Zn2 and Zn % Zn, 15. Try a 1 bi → (a, b)., 17. Use Exercise 3 and Theorem 4.3., 19. km/rl % kn/sl, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A14, , Selected Answers, , 21. Since k(g, h)l # kgl % khl, a necessary and sufficient condition for equality is that, lcm (|g|,|h|) 5 |(g, h)| 5 |kgl % khl)| 5 |g||h|. This is equivalent to gcd (|g|,|h|) 5 1., 23. |(a, b, c)| 5 lcm (|a|,|b|,|c|) 5 3, unless a b c e. In general, the order of every, nonidentity element of Zp % Zp % . . . % Zp, where p is prime, is p., a b, 25. Map c, d to (a, b, c, d). Let Rk denote R % R % ? ? ? % R (k factors). Then the group of m 3, c d, n matrices under addition is isomorphic to Rmn., 27. (g, g)(h, h)21 5 (gh21, gh21). When G 5 R, G % G is the plane and H is the line y 5 x., 29. k(3, 0)l, k(3, 1)l, k(3, 2)l, k(0, 1)l, 31. 60, 33. k(1, 1, 1)l, k(5, 1, 1)l, k(1, 3, 1)l. (Others exist.), 35. {0, 400} % {0, 50, 100, 150}, 37. Compare the number of elements of order 2 in each group., 39. The mapping f13m6n 2 1m, n2 is an isomorphism. The mapping f13m9n 2 1m, n2, is not well-defined, since f13290 2 ? f130912., 41. Compare the number of elements of order 6 in each group., 43. U(35) L U(5)%U(7) L U7(35)3U5(35) 5 {1, 8, 22, 29}3{1, 6, 11, 16, 26, 31}, 45. C* has only one element of order 2, whereas Z2 % Z2 has three elements of order 2., 47. 12, 49. Aut(U(25)) < Aut(Z20) < U(20) < U(4) % U(5) < Z2 % Z4, 51. 2k 2 1; 2t 2 1, where t is the number of the integers n1, n2, . . . , nk that are even., 53. No. Z10 % Z12 % Z6 has seven elements of order 2, whereas Z15 % Z4 % Z12 has only three., 55. Using the fact that an isomorphism from Z12 is determined by the image of 1 and the fact that a, generator must map to a generator, we determine that there are four isomorphisms., 57. Since (2, 0) has order 2, it must map to an element in Z12 of order 2. The only such element in Z12, is 6. The isomorphism defined by (1, 1) x → 5x with x 5 6 takes (2, 0) to 6. Since (1, 0) has order, 4, it must map to an element in Z12 of order 4. The only such elements in Z12 are 3 and 9. The first, case occurs for the isomorphism defined by (1, 1) x → 7x with x 5 9 [recall that (1, 1) is a generator of Z4 % Z3]; the second case occurs for the isomorphism defined by (1, 1) x → 5x with x 5 9., 59. Since a [ Zm and b [ Zn, we know that |a| divides m and |b| divides n. So, |(a, b)| 5 lcm(|a|,, |b|) divides lcm(m, n)., 61. Z, Z3, Z4, Z6, 63. Observe that every nonidentity element of Zp % Zp has order p and each subgroup of order p, contains p 2 1 of them. So, there are exactly (p2 2 1)/(p 2 1) 5 p 1 1 subgroups of order p., 65. Look at Z % Z2., 67. U(165) < U(11) % U(15) < U(5) % U(33) < U(3) % U(55) < U(3) % U(5) % U(11), 69. Mimic the analysis for elements of order 12 in U(720) in this chapter., 71. 60, 73. They are both isomorphic to Z10 % Z4., 75. That U(n)2 is a subgroup follows from Exercise 17 of the Supplementary Exercises, for Chapters 1–4. 12 5 (n 2 1)2 shows that it is a proper subgroup., 77. 275, 79. U(117) < U(9) % U(13) < Z6 % Z12, which contains k(2, 0)l % k(0, 4)l., 81. Since U(pq) < U(p) % U(q) < Zp21 % Zq21, it follows that k 5 lcm(p21, q21)., 83. In the first case there are 2k 21; in the second case there are 2k12 21., 85. Consider U(49)., 87. Consider U(65)., 89. NO., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Page 619 : Selected Answers, , A15, , Supplementary Exercises for Chapters 5–8, All things are difficult before they are easy., thomas fuller, , 1. Consider the finite and infinite cases separately. In the finite case, note that |H| 5 |f(H)|. Now, use Theorem 4.3. For the infinite case, use Exercise 2 in Chapter 6., 3. Observe that f(x21y21xy) 5 (f(x))21(f(y))21f(x)f(y), so f carries the generators of G9 to the, generators of G9., 5. All nonidentity elements of G and H have order 3. G 2 H., 7. Let a and b be distinct nonidentity elements in G of order 2. Let H1 5 {e, a, b, ab}. Note that H1, is a subgroup of G and the product of all its elements is e. If H1 5 G, we are done. If not, then let, c be an element of G not in H1. Then H2 5 H1 < cH1 is a subgroup of G and the product of all, the elements in H2 is c4 5 e. If H2 5 G, we are done. If not, then let d be an element of G not in, H2. Then H3 5 H2 < dH2 is a subgroup of G and the product of all the elements in H3 is d 8 5 e., Continuing in this way finishes the proof., 9. Let x and y belong to H. Then x 5 aiz1 and y 5 a jz2, where 0 # i, j , 4, and z1 and z2 are in, Z(G). So, xy21 5 aiz1(a jz2)21 5 ai2jz1z221, where i2j mod 4 [ {0, 1, 2, 3} and z1z221 [ Z(G)., In the general case, H 5 Z(G) < aZ(G) < a2Z(G) < . . . <ak21 Z(G) is a subgroup of G., 11. U(n), where n 5 4, 8, 3, 6, 12, 24., 3, 2, 13. Hint: 3 2i 213 a, , ib, 213, 213, 15. Suppose f: Q S R is an isomorphism. Let f(1) 5 x0. Show that f(a/b) 5 (a/b)x0 for all integers, a, b with b 2 0., 17. In Q, the equation 2x 5 a has a solution for all a. The corresponding equation x2 5 b in Q1 does, not have a solution for all b., 19. Suppose x p22 5 1. Since |U(p)| 5 p 2 1, we have that x p21 5 1 for all x [ U(p). So, by cancellation, x 5 1., 21. k3l % k4l, 23. Z18, Z2 % Z3 % Z3, D9, D3 % Z3, 25. Say a 5 a1a2 . . . an and b 5 b1 . . . bm, where the a’s and b’s are cycles. Then ab21 5 a1a2 . . ., a b 21 . . . b 21 is a finite number of cycles., n m, , 27., 29., 31., 33., 35., 37., 39., 41., 43., 45., 47., 49., 51., 53., 55., , 1, , Count elements of order 2., Count elements of order 2., x 5 fa(x) 5 axa21, so that xa 5 ax. Conversely, if G is Abelian, fa is the identity., U50(450), (4, 10), Count elements of order 2., 20; (8, 7, (3251)), Let H 5x [ Zp2 % Zp2 0 xp 10, 026. Then 冟H冟 p2 and every nonidentity element of, 1Zp2 % Zp2 2 /H has order p., (12)(34)(56789), 1260, b 5 (17395)(286), Say the points in H lie on the line y 5 mx. Then (a, b) 1 H 5 {(a 1 x, b 1 mx) | x [ R}. This, set is the line y 2 b 5 m(x 2 a)., aH 5 bH implies a21b [ H. So (a21b)21 5 b21a [ H. Thus, Hb21a 5 H or Hb21 5 Ha21., These steps are reversible., p2 1, p1p 12, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A16, , Selected Answers, , 57. By Theorem 8.3, U(pq) < U(p) % U(q), so an element xn in U1pq2 corresponds to an element, 1xn1, xn2 2[U1p2 % U1q2. It follows from Corollary 4 of Theorem 7.1 that 1xn1, xn2 2 11, 12, the, identity of U1p2 % U1q2., 59. First observe that (n, n21, . . . 2, 1)(12)(123 . . . n) 5 (1n). Also, (1n)(123 . . . n) 5 (123 . . . n 2 1)., So, by induction, 1122 and 1123 . . . n2 generate Sn1. This means that every 2-cycle not involving n, can be generated. Now note that (1k)(1n)(1k) 5 (kn), so all 2-cycles are generated., 61. Let b have order 2. In disjoint cycle form, b is a product of transpositions, so there must be some i, missing from this product. Thus, b(i) 5 i. Pick j such that b( j) 2 j. Since s is an n-cycle, some, power of s, say st, takes i to j. If b commutes with s, it commutes with st as well. Then, (stb)(i) 5 st(b(i)) 5 st(i) 5 j, whereas (bst)(i) 5 b(st(i)) 5 b( j) 2 j. This proves that, stb 2 bst., 63. Write s as (a1a2 . . . an). Then for each entry ai in the cycle, sm takes ai to ai1m, where the subscript i 1 m is taken mod n. So, the cycle decomposition of sm has the cycle (a1, a11m, a112m, . . .,, a11(r21)m), where r is the smallest positive integer such that 1 5 (11rm) mod n. Thus, r is the, smallest positive integer with the property that rm 5 0 mod n. But n / gcd (m, n) is the smallest, such integer. The same argument applies to all the other cycles in sm. Since the sum of the lengths, of the cycles in the decomposition of sm is n and each cycle in the decomposition of sm has, length n / gcd (m, n), there must be gcd (m, n) cycles., , Chapter 9, There’s a mighty big difference between good, sound reasons and reasons that sound good., burton hillis, , 1., 3., 5., 7., 9., 11., 13., 15., 17., 19., 21., 23., 25., 27., 29., 31., 33., 35., 37., 39., , No., HR90 5 R270 H; DR270 5 R90D; R90V 5 VR270, Say i , j and let h [ Hi > Hj. Then h [ H1H2 . . . Hi . . . Hj21 > Hj 5 {e}., Recall that if A and B are matrices, then det (ABA21) 5 (det A)(det B)(det A)21., Let x [ G. If x [ H, then xH 5 H 5 Hx. If x o H, then xH is the set of elements in G, not in H., But Hx is also the set of elements in G, not in H., G/H < Z4, but G/K < Z2 % Z2., Observe that aHbH 5 abH 5 baH 5 bHaH., 2, H 5 {0 1 k20l, 4 1 k20l, 8 1 k20l, 12 1 k20l, 16 1 k20l}; G/H 5 {0 1 k20l 1 H, 1 1 k20l 1 H,, 2 1 k20l 1 H, 3 1 k20l 1 H}, 40/10 5 4, By Theorem 9.5, the group has an element a of order 3 and an element b of order 11. Then, |ab| 5 33., `; no, (6, 3) 1 k(4, 2)l has order 2., Z8, Yes; no, Mimic the argument given in Example 16 in this chapter., Certainly, every nonzero real number is of the form 6r, where r is a positive real number. Real, numbers commute, and R1 > {1, 21} 5 {1}., No. If G 5 H 3 K, then |g| 5 lcm(|h|, |k|), provided that |h| and |k| are finite. If |h| or |k| is infinite, so is |g|., For the first question, note that k3l > k6l 5 {1} and k3lk6l > k10l 5 {1}. For the second question, observe that 12 5 32162., Say |g| 5 n. Then (gH)n 5 gnH 5 eH 5 H. Now use Corollary 2 to Theorem 4.1., Let x [ C(H), g [ G, and h [ H. We must show that gxg21h 5 hgxg21. Note that in the expression (gxg21)h(gxg21)21 5 gxg21hgx21g21, the terms x and x21 cancel since g21hg [ H and x, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A17, , Selected Answers, , 41., , 43., 45., 47., 49., 51., , 53., 55., 57., 59., 61., 63., 65., 67., 69., 71., 73., 75., 77., , 79., , commutes with every element of H. Then we have (gxg21)h(gxg21)21 5 gxg21hgx21g21 5, gg21hgg21 5 h. So, gxg21 [ C(H)., Suppose that H is a proper subgroup of Q of index n. Then Q/H is a finite group of order n. By, Corollary 4 of Theorem 7.1, we know that for every x in Q we have nx is in H. Now observe, that the function f(x) 5 nx maps Q onto Q. So, Q # H., Take G 5 Z6, H 5 {0, 3}, a 5 1, and b 5 9., Use Lagrange’s Theorem and Exercise 9 of this chapter., Since N # NH # G, we have |G:N| 5 |G:NH| |NH:N|. Thus, |G:H| 5 1 or |NH:N| 5 1. It follows, that G 5 NH or NH 5 N., Use the G/Z Theorem., If H is normal in G, then xNhN(xN)21 5 xhx21N [ H/N, so H/N is normal in G/N. Now assume, H/N is normal in G/N. Then xhx21N 5 xNhN(xN)21 [ H/N. Thus, xhx21N 5 h9N for h9 [ H. So,, xhx21 5 h9n for some n [ N., Say H has index n. Then (R*)n 5 {x n | x [ R*} # H. If n is odd, then (R*)n 5 R*; if n is even,, then (R*)n 5 R1. So, H 5 R* or H 5 R1., Use Exercise 9 and observe that VK 2 KV., Look at S3., Let N 5 kal, H 5 kakl, and x [ G. Then, x(ak)mx21 5 (xamx21)k 5 (ar)k 5 (ak)r [ H., gcd(|x|, |G/H|) 5 1 implies gcd(|xH|, |G/H|) 5 1. But |xH| divides |G/H|. Thus |xH| 5 1 and, therefore xH 5 H., Note that G/N is a group and use Corollary 4 of Theorem 7.1., Use Theorems 9.4 and 9.3., Say |gH| 5 n. Then |g| 5 nt (by Exercise 37) and |gt| 5 n. For the second part, consider Z/kkl., It is not a group table. No, because _ is not normal in D4., Use Theorem 9.3 and Theorem 7.3., By Exercise 72, A5 would have an element of the form (ab)(cd) that commutes with every element of A5. Try (abc)., To see that H is normal, observe that xgpx1 1xgx1 2 p. To verify the second part, note that, 1gH2 p gpH H., Since H has index 2 in G, it is a normal subgroup of G and |G/H| 5 2. It follows that for every a, in G, we have 1aH2 2 H. If a is an element of G of order 2n 1, then H a2n1H , 1 1aH2 2 2 naH aH. Thus, a is in H., Let H be unique subgroup of order 5. By Example 7 H is normal. Now observe that G exactly 4, elements of order 5 and elements of order 25 come in multiples of f(25) 5 20 (see the Corollary, of Theorem 4.4). So, elements of order 1, 5 or 25 account for at most 85 elements of G. By, Lagrange’s Theorem every other element of G must have order 2, 4, 10, 20, 50 or 100. It follows, from Theorem 4.3 that G has a subgroup K of order 2. By Example 5 HK is a subgroup and by, Theorem 7.2 |HK| 5 10., , Chapter 10, It’s always helpful to learn from your mistakes, because then your mistakes seem worthwhile., garry marshall, , 1. Note that det(AB) 5 (det A)(det B)., 3. Note that ( f 1 g)9 5 f9 1 g9., 5. Observe that (xy)r 5 xryr. Odd values of r yield an isomorphism. For even values of r the kernel, is {1, 21}., 7. (sf)(g1g2) 5 s(f(g1g2)) 5 s(f(g1)f(g2)) 5 s(f(g1))s(f(g2)) 5 (sf)(g1)(sf)(g2)., Ker f is a normal subgroup of Ker sf. |H|/|K| 5 [Ker sf:Ker f]., 9. f((g, h) (g9, h9)) 5 f((gg9, hh9)) 5 gg9 5 f((g, h))f((g9, h9)). The kernel is {(e, h) | h [ H}., 11. Consider f: Z % Z S Za % Zb given by f((x, y)) 5 (x mod a, y mod b) and use Theorem 10.3., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A18, , Selected Answers, , 13. (a, b) S b is a homomorphism from A % B onto B with kernel A % {e}., 15. 3, 13, 23, 17. Suppose f is such a homomorphism. By Theorem 10.3, Ker f 5 k(8, 1)l, k(0, 1)l, or k(8, 0)l. In, these cases, (1, 0) 1 Ker f has order either 16 or 8. So, (Z16 % Z2) / Ker f is not isomorphic to, Z4 % Z4., 19. Since |Ker f| is not 1 and divides 17, f is the trivial map., 21. k5l, 23. a. The possible images are isomorphic to Z1, Z2, Z3, Z4, Z6, and Z12., b. k1l < Z36, k2l < Z18, k3l < Z12, k4l < Z9, k6l < Z6, and k12l < Z3., 25. 4 onto; 10 to, 27. For each k with 0 # k # n 2 1, the mapping 1 S k determines a homomorphism., 29. Use Theorem 10.3 and properties 5, 7, and 8 of Theorem 10.2., 31. f21(7) 5 7 Ker f 5 {7, 17}, 33. 11 Ker f, 35. f((a, b) 1 (c, d)) 5 f((a 1 c, b 1 d)) 5 (a 1 c) 2 (b 1 d) 5 a 2 b 1 c 2 d 5 f((a, b)) 1, f((c, d)). Ker f 5 {(a, a) | a [ Z}. f21(3) 5 {(a 1 3, a) | a [ Z}., 37. Use the property of complex numbers that |xy| 5 |x||y| and the First Isomorphism Theorem., 39. f(xy) 5 (xy)6 5 x6y6 5 f(x)f(y). Ker f 5 kcos 60° 1 i sin 60°l., 41. Show that the mapping from K to KN/N given by k S kN is an onto homomorphism with kernel, K > N., 43. For each divisor d of k there is a unique subgroup of Zk of order d, and this subgroup is generated by f(d) elements. A homomorphism from Zn to a subgroup of Zk must carry 1 to a generator of the subgroup. Furthermore, the order of the image of 1 must divide n, so we need consider only those divisors d of k that also divide n., 45. D4, {e}, Z2, Z2 % Z2, 47. It is divisible by 10. 10 can be replaced by any positive integer., 49. It is infinite. Look at Z., 51. Let g be the natural homomorphism from G onto G/N. Let H be a subgroup of G/N and let, g21(H) 5 H. Then H is a subgroup of G and H/N 5 g(H) 5 g(g21(H)) 5 H., 53. The mapping g S fg is a homomorphism with kernel Z(G)., 55. ( f 1 g)(3) 5 f(3) 1 g(3). The kernel is the set of elements in Z[x] whose graphs pass through the, point (3, 0). 3 can be replaced by any integer., 57. Let g belong to G. Since f1g2 belongs to Z2 % Z2 5 k(1, 0)l x k(0, 1)l x k(1, 1)l, it follows that, G 5 f21(k(1, 0)l) x f21(k(0, 1)l) x f21(k(1, 1)l). Moreover, each of these three subgroups is, proper and by property 8 of Theorem 10.2 normal., 59. Use Exercise 56 in Chapter 9 and Exercise 41 above to prove the first assertion. To verify that, G/(H > K) is not cyclic, observe that it has two subgroups of order 2., 61. Mimic Example 16., 63. Let f be a homomorphism from S3 to G. Since |f(S3)| must divide 6, we have that |f(S3)| 5 1,, 2, 3, or 6. In the first case, f maps every element to 0. If |f(S3)| 5 2, then n is even and f maps, the even permutations to 0 and the odd permutations to an element of order 2. The case that, |f(S3)| 5 3 cannot occur, because it implies that Ker f is a normal subgroup of order 2, whereas, S3 has no normal subgroup of order 2. The case that |f(S3)| 5 6 cannot occur, because it implies, that f is an isomorphism from a non-Abelian group to an Abelian group., 65. f(zw) 5 z2w2 5 f(z) f(w). Ker f 5 {1, 21} and, because f is onto C* we have by Theorem, 10.3, C*/{1, 21} is isomorphic to C*., 67. Suppose that H is a proper subgroup of G that is not properly contained in a proper subgroup of, G. Then G/H has no nontrivial, proper subgroup. It follows from Exercise 26 in Chapter 7 that, G/H is isomorphic to Zp for some prime p. But then for every coset g 1 H we have p(g 1 H) 5 H,, so that pg [ H for all g [ G. But then G 5 pG # H. Both Q and R satisfy the hypothesis., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A19, , Chapter 11, Ever tried. Ever failed. No matter. Try again. Fail again. Fail better., Samuel Beckett, , 1. n 5 4; Z4, Z2 % Z2, 3. n 5 36; Z9 % Z4, Z3 % Z3 % Z4, Z9 % Z2 % Z2, Z3 % Z3 % Z2 % Z2, 5. The only Abelian groups of order 45 are Z45 and Z3 % Z3 % Z5. In the first group, |3| 5 15; in the, second one, |(1, 1, 1)| 5 15. Z3 % Z3 % Z5 does not have an element of order 9., 7. Z9 % Z3 % Z4; Z9 % Z3 % Z2 % Z2, 9. Z4 % Z2 % Z3 % Z5, 11. By the Fundamental Theorem, any finite Abelian group G is isomorphic to some direct product of, cyclic groups of prime-power order. Now go across the direct product and, for each distinct prime, you have, pick off the largest factor of the prime power. Next, combine all of these into one factor (you can do this, since the subscripts are relatively prime). Let us call the order of this new, factor n1. Now repeat this process with the remaining original factors and call the order of the, resulting factor n2. Then n2 divides n1, since each prime-power divisor of n2 is also a primepower divisor of n1. Continue in this fashion. Example: If, G < Z27 % Z3 % Z125 % Z25 % Z4 % Z2 % Z2,, then, G < Z27 ? 125 ? 4 % Z3 ? 25 ? 2 % Z2., Now note that 2 divides 3 ? 25 ? 2 and 3 ? 25 ? 2 divides 27 ? 125 ? 4., 13. Z2 % Z2, 15. a. 1 b. 1 c. 1 d. 1 e. 1 f. There is a unique Abelian group of order n if and only if n is, not divisible by the square of any prime., 17. This is equivalent to asking how many Abelian groups of order 16 have no elements of order 8., From the Fundamental Theorem of Finite Abelian Groups the only choices are Z4 % Z4, Z4 % Z2, % Z2, and Z2 % Z2 % Z2 % Z2., 19. Z2 % Z2, 21. Z3 % Z3, 23. n is square-free (no prime factor of n occurs more than once)., 25. Among the first 11 elements in the table, there are nine elements of order 4. None of the other, isomorphism classes has this many., 27. Z4 % Z2 % Z2; one internal direct product is k7l 3 k101l 3 k199l., 29. 3; 6; 12, 31. Z4 % Z4, 33. Use Theorems 11.1, 8.1, and 4.3., 35. |kalK| 5 |a||K|/|kal > K| 5 |a||K| 5 |a||K|p 5 |G|p 5 |G|, 37. By the Fundamental Theorem of Finite Abelian Groups, it suffices to show that every group of the, form Zp n1 % Zp n2 % . . . % Zp nk is a subgroup of a U-group. Consider first a group of the form, 1, 2, k, Zp n1 % Zp n2 ( p1 and p2 need not be distinct). By Dirichlet’s Theorem, for some s and t there are, 1, 2, distinct primes q and r such that q 5 tp1n1 1 1 and r 5 sp2n2 1 1. Then U(qr) 5 U(q) % U(r) <, Ztp n1 % Zsp n2, and this latter group contains a subgroup isomorphic to Zp n1 % Zp n2. The general, 1, 2, 1, 2, case follows in the same way., 39. Look at D4., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A20, , Selected Answers, , Supplementary Exercises for Chapters 9–11, You cannot have success without the failures., h. g. hasler, The Observer, , 1. Say aH 5 Hb. Then a 5 hb for some h in H. Then Ha 5 Hhb 5 Hb 5 aH., 3. Suppose diag(G) is normal. Then (e, a)(b, b)(e, a)21 5 (b, aba21) [ diag(G). Thus b 5 aba21. If, G is Abelian, (g, h)(b, b)(g, h)21 5 (gbg21, hbh21) 5 (b, b). The index of diag(G) is |G|., 5. Let a [ Aut(G) and fa [ Inn(G). Then (afaa21)(x) 5 (afa)(a21(x)) 5 a(aa21(x)a21) 5, a(a)x(a(a))21 5 fa(a)(x)., 7. R* (See Example 2 in Chapter 10.), 1 0 b, 9. a. Z(H) 5 • £ 0 1 0 § ` b [ Q ¶, 0 0 1, b. The mapping, 1 0 b, £0 1 0 § S b, 0 0 1, is an isomorphism., c. The mapping, 1 a b, £ 0 1 c § S 1a, c2, 0 0 1, is a homomorphism with Z(H) as the kernel., d. The proofs are valid with R and Zp., 11. b(a/b 1 Z) 5 a 1 Z 5 Z, 13. Use Exercise 5 of the Supplementary Exercises for Chapters 1– 4. Such a set is possible only when n is, prime. For the first example, consider Dp, where p is a prime. For the second example, try D4., 15. Observe that hkh21k21 5 (hkh21)k21 [ K and hkh21k21 5 h(kh21k21) [ H., 17. Use Theorem 7.4 and Exercise 9 of Chapter 9., 19. First observe that f((4, 0, 0)) 5 f(4(1, 0, 0)) 5 4f(1, 0, 0) 5 (0, 0), so that Ker f 5 {(0, 0, 0),, (4, 0, 0)}. But then (Z8 % Z2 % Z2)/Ker f has more than three elements of order 2, whereas Z4 % Z4, has only three., 21. Use Theorem 7.3 together with the fact that S4 has no element of order 6., 23. The number is m in all cases., 25. The mapping g → gn is a homomorphism from G onto Gn with kernel Gn., 27. Let |H| 5 p. Theorem 7.2 shows that H is the only subgroup of order p. But xHx21 is also a, subgroup of order p. So, xHx21 5 H., 29. Say a and b are integers and a/b 1 Z has order n in Q/Z. Then na/b 5 m for some integer m., Thus, a/b 1 Z 5 m/n 1 Z 5 m(1/n 1 Z) [ k1/n 1 Zl., 31. If (1, 0) → a and (0, 1) → b, then (x, 0) → ax and (0, y) → by., 33. First note that by Exercise 11, every element in Q/Z has finite order. For each positive integer n, let, Bn denote the set of elements of order n, and suppose that f is an isomorphism from Q/Z to itself., Then, by property 5 of Theorem 6.2, f(Bn) # Bn. By Exercise 29 we know that Bn is finite, and, since f preserves orders and is one-to-one, we must have f(Bn) 5 Bn. Since it follows from Exercise 11 and Exercise 29 that Q/Z 5 < Bn, where the union is taken over all positive integers n, we, have f(Q/Z) 5 Q/Z., 35. If the group is not Abelian, for any element a not in the center, the inner automorphism induced by, a is not the identity; if the group is Abelian and contains an element a with |a| . 2, then x → x21, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A21, , works; if every nonidentity element has order 2, then G is isomorphic to a group of the form, Z2 % Z2 % . . . % Z2 . In this case, the automorphism that takes (a1, a2, a3, . . . , ak) to (a2, a1, a3, . . . ,, ak) is not the identity., 37. G/H is isomorphic to Z2 % Z2. G/H is not isomorphic to a subgroup of G, since G has only one, element of order 2., 1 x 1 y, 1 xy, 1 x 1, 1 x, d , so H is closed. Also, c, d c, 39. Observe that c, dc, d c, d,, 0 1 0 1, 0, 1, 0 1, 0, 1, which is in H. Thus, H is a subgroup of G. Since c, c, , 1, 0, , ab 1, 1, d, c, b 1 5 0, , 1, 0, , a 1, dc, b 0, , x 1, dc, 1 0, , 1 a 1 x, a 1, dc, d, d c, 0 b 0 1, b, , b1x, d belongs to H, we have that H is normal in G., 1, , 41. Let g belong to G. Since gKg1 K, conjugation is an automorphism of K. Thus gHg1 H., , Chapter 12, Mistakes are the portals of discovery., james joyce, , 1. For any n . 1, the ring M2(Zn ) of 2 3 2 matrices with entries from Zn is a finite noncommutative, ring. The set M2(2Z ) of 2 3 2 matrices with even integer entries is an infinite noncommutative, ring that does not have a unity., 3. In R, consider 5n12 0 n [ Z 6, 5. The proofs given for a group apply to a ring as well., 7. In Zp, nonzero elements have multiplicative inverses. Use them., 9. If a and b belong to the intersection, then they belong to each member of the intersection. Thus,, a 2 b and ab belong to each member of the intersection. So, a 2 b and ab belong to the, intersection., 11. Rule 3: 0 5 0(2b) 5 (a 1 (2a))(2b) 5 a(2b) 1 (2a)(2b) 5 2(ab) 1 (2a)(2b)., So, ab 5 (2a)(2b)., Rule 4: a(b 2 c) 5 a(b 1 (2c)) 5 ab 1 a(2c) 5 ab 1 (2(ac)) 5 ab 2 ac., Rule 5: Use rule 2., Rule 6: Use rule 3., 13. Hint: Z is a cyclic group under addition, and every subgroup of a cyclic group is cyclic., 15. For positive m and n, observe that (m ? a)(n ? b) 5 (a 1 a 1 . . . 1 a)(b 1 b 1 . . . 1 b) 5, (ab 1 ab 1 . . . 1 ab), where the last term has mn summands. Similar arguments apply in the, remaining cases., 17. From Exercise 15, we have (n ? a)(m ? a) 5 (nm) ? a2 5 (mn) ? a2 5 (m ? a)(n ? a)., 19. Let a, b belong to the center. Then (a 2 b)x 5 ax 2 bx 5 xa 2 xb 5 x(a 2 b). Also,, (ab)x 5 a(bx) 5 a(xb) 5 (ax)b 5 (xa)b 5 x(ab)., 21. (x1, . . . , xn)(a1, . . . , an) 5 (x1, . . . , xn) for all xi in Ri if and only if xiai 5 xi for all xi in Ri and, i 5 1, . . . , n., 23. {1, 21, i, 2i}, 25. f (x) 5 1 and g(x) 5 21., 27. If a is a unit, then b 5 a(a21b)., 29. Consider a21 2 a22b., 31. Try the ring M2(Z)., 33. Note that 2x 5 (2x)3 5 8x3 5 8x., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A22, , Selected Answers, , 35. For Z 6, use n 5 3. For Z10, use n 5 5. Say m 5 p2t, where p is a prime. Then (pt)n 5 0 in Z m,, since m divides (pt)n., 37. Every subgroup of Zn is closed under multiplication., 39. ara 2 asa 5 a(r 2 s)a. (ara)(asa) 5 ara2sa 5 arsa. a1a 5 a2 5 1, so 1 [ S., 41. The Subring Test is satisfied., 43. Look at (1, 0, 1) and (0, 1, 1)., 45. Observe that n ? 1 2 m ? 1 5 (n 2 m) ? 1. Also, (n ? 1)(m ? 1) 5 (nm) ? ((1)(1)) 5 (nm) ? 1., 47. {m/2n | m [ Z, n [ Z1}, 49. (a 1 b)(a 2 b) 5 a2 1 ba 2 ab 2 b2 5 a2 2 b2 if and only if ba 2 ab 5 0., 51. Z2 % Z2; Z2 % Z2 % . . . (infinitely many copies), , Chapter 13, Work now or wince later., b. c. forbes, Epigrams, , 1. The verifications for Examples 1– 6 follow from elementary properties of real and complex numbers. For Example 7, note that, 1 0 0 0, 0 0 ., d, c, dc, d c, 0 0 0 1, 0 0, For Example 8, note that (1, 0)(0, 1) 5 (0, 0)., 3. Let ab 5 0 and a 2 0. Then ab 5 a ? 0, so b 5 0., 5. Let k [ Zn. If gcd(k, n) 5 1, then k is a unit. If gcd(k, n) 5 d . 1, write k 5 sd. Then k(n/d) 5, sd(n/d) 5 sn 5 0., 7. Let s [ R, s 2 0. Consider the set S 5 {sr | r [ R}. If S 5 R, then sr 5 1 (the unity) for some r., If S 2 R, then there are distinct r1 and r2 such that sr1 5 sr2. In this case, s(r1 2 r2) 5 0. To see, what happens when the “finite” condition is dropped, consider Z., 9. Take a 5 (1, 1, 0), b 5 (1, 0, 1), and c 5 (0, 1, 1)., 11. (a1 1 b12d) 2 (a2 1 b22d) 5 (a1 2 a2) 1 (b1 2 b2)2d; (a1 1 b12d)(a2 1 b22d) 5, (a1a2 1 b1b2d) 1 (a1b2 1 a2b1)2d. Thus, the set is a ring. Since Z[2d] is a subring of the, ring of complex numbers, it has no zero-divisors., 13. The even integers., 15. (1 2 a)(1 1 a 1 a2 1 . . . 1 an21 ) 5 1 1 a 1 a2 1 . . . 1 an21 2 a 2 a2 2 . . . 2 an 5 1, 2 an 5 1 2 0 5 1., 17. Suppose a 2 0 and an 5 0 (where we take n to be as small as possible). Then a ? 0 5 0 5 an 5, a ? an21, so by cancellation, an21 5 0., 19. If a2 a and b2 b, then 1ab2 2 a2b2 ab. The other cases are similar., 21. Let f(x) 5 x on [21, 0], f(x) 5 0 on (0, 1], g(x) 5 0 on [21, 0], and g(x) 5 x on (0, 1]. Then f(x), and g(x) are in R and f(x)g(x) 5 0 on [21, 1]., 23. Suppose that a is an idempotent and an 0. By the previous exercise, a 0., 25. (3 1 4i)2 5 3 1 4i., 27. a2 5 a implies a(a 2 1) 5 0. So if a is a unit, a 2 1 5 0 and a 5 1., 29. See Theorems 3.1 and 12.3., 31. Note that ab 5 1 implies aba 5 a. Thus 0 5 aba 2 a 5 a(ba 2 1). So, ba 2 1 5 0., 33. A subdomain of an integral domain D is a subset of D that is an integral domain under the operations of D. To show that P is a subdomain, show that it is a subring and contains 1. Every, subdomain contains 1 and is closed under addition and subtraction, so every subdomain contains P. |P| 5 char D when char D is prime and |P| is infinite when char D is 0., 35. Use Theorems 13.3, 13.4, and 7.1 (Lagrange’s Theorem)., 37. By Exercise 36, 1 is the only element of an integral domain that is its own multiplicative inverse, if and only if 1 5 21. This is true only for fields of characteristic 2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A23, , Selected Answers, , 39. a. Since a3 5 b3, a6 5 b6. Then a 5 b because we can cancel a5 from both sides (since a5 5 b5)., b. Use the fact that there exist integers s and t such that 1 5 sn 1 tm, but remember that you, cannot use negative exponents in a ring., 41. (1 2 a)2 5 1 2 2a 1 a2 5 1 2 2a 1 a 5 1 2 a., 43. Z8, 45. Let S 5 {a1, a2, . . . , an} be the nonzero elements of the ring. First show that S 5 {a1a1, a1a2, . . . ,, a1an}. Thus, a1 5 a1ai for some i. Then ai is the unity, for if ak is any element of S, we have, a1ak 5 a1ai a k, so that a1(ak 2 aiak) 5 0., 47. Say |x| 5 n and |y| 5 m with n , m. Consider (nx)y 5 x(ny)., 49. a. Use the Binomial Theorem., b. Use part a and induction., c. Look at Z4., 51. Use Theorems 13.4 and 9.5 and Exercise 47., 53. n c, 55., 57., 59., 61., 63., 65., 67., 69., , 0 0, a b, d 5 c, d for all members of M2(R) if and only if na 5 0 for all a in R., c d, 0 0, , Use Exercise 54., a. 2 b. 2, 3 c. 2, 3, 6, 11 d. 2, 3, 9, 10, 2, See Example 10., Use Exercise 29 and part a of Exercise 49., Choose a 2 0 and a 2 1 and consider 1 1 a., f(x) 5 f(x . 1) 5 f(x) . f(1), so f(1) 5 1. Also, 1 5 f(1) 5 f(xx21) 5 f(x) f(x21)., Since a field of order 27 has characteristic 3, we have 3a 5 0 for all a. From this, we have, 6a 5 0 and 5a 5 2a., , Chapter 14, The paradox of excellence is that it is built upon the foundations of necessary failure., matthew syed, , 1. Let r1a and r2a belong to kal. Then r1a 2 r2a 5 (r1 2 r2)a [ kal. If r [ R and r1a [ kal, then, r(r1a) 5 (rr1)a [ kal., 3. Clearly, I is not empty. Now observe that (r1a1 1 . . . 1 rnan) 2 (s1a1 1 . . . 1 snan) 5 (r1 2 s1)a1, 1 . . . 1 (rn 2 sn)an [ I. Also, if r [ R, then r(r1a1 1 . . . 1 rnan) 5 (rr1)a1 1 . . . 1 (rrn)an [ I., That I # J follows from closure under addition and multiplication by elements from R., 5. Let a 1 bi, c 1 di [ S. Then (a 1 bi) 2 (c 1 di) 5 a 2 c 1 (b 2 d)i and b 2 d is even. Also,, (a 1 bi)(c 1 di) 5 ac 2 bd 1 (ad 1 cb)i and ad 1 cb is even. Finally, (1 1 2i)(1 1 i) 5, 21 1 3i o S., 7. Since ar1 2 ar2 5 a(r1 2 r2) and (ar1)r 5 a(r1r), 4R 5 {. . . , 216, 28, 0, 8, 16, . . .}., 9. If n is prime, use Euclid’s Lemma (Chapter 0). If n is not prime, say n 5 st where s , n and t , n;, then st belongs to nZ but s and t do not., 11. a. a 5 1 b. a 5 2 c. a 5 gcd(m, n), 13. a. a 5 12, b. a 5 48. To see this, note that every element of k6lk8l has the form 6t18k1 1 6t28k2 1 . . . 1, 6tn8kn 5 48s [ k48l. So, k6lk8l # k48l. Also, since 48 [ k6lk8l, we have k48l # k6lk8l., c. a 5 mn, 15. Let r [ R. Then r 5 1r [ A., 17. Let u [ I be a unit and let r [ R. Then r 5 r(u21u) 5 (ru21)u [ I., 19. Observe that k2l and k3l are the only nontrivial ideals of Z6, so both are maximal. More generally, Zpq , where p and q are distinct primes, has exactly two maximal ideals., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A24, , Selected Answers, , 21. Clearly, I is closed under subtraction. Also, if b1, b2, b3, and b4 are even, then every, a1 a2 b1 b2, dc, d is even., entry of c, a3 a4 b3 b4, 2q1 r1 2q2 r2, d., 23. Use the observation that every member of R can be written in the form c, 2q3 r3 2q4 r4, r 1 r2, 2q1 r1 2q2 r2, d I c, d I., Then note that c, 2q3 r3 2q4 r4, r3 r4, 25. (br1 1 a1) 2 (br2 1 a2) 5 b(r1 2 r2) 1 (a1 2 a2) [ B; r9(br 1 a) 5 b(r9r) 1 r9a [ B since r9a, is in A., 27. Use Exercise 17., 29. Since every element of kxl has the form xg(x), we have kxl # I. If f(x) [ I, then f(x) 5 anxn 1 . . . 1, a1x 5 x(anxn21 1 . . . 1 a1) [ kxl., 31. Suppose f(x) 1 A 2 A. Then f(x) 1 A 5 f(0) 1 A and f(0) 2 0. Thus,, 1, A., 1 f1x2 A2 1 , f 102, This shows that R/A is a field. Now use Theorem 14.4., 33. Since (3 1 i)(3 2 i) 5 10, 10 1 k3 1 il 5 0 1 k3 1 il. Also, i 1 k3 1 il 5 23 1 k3 1 il 5 7 1, k3 1 il. So, Z[i]/k3 1 il 5 {k 1 k3 1 il | k 5 0, 1, . . . , 9}, since 1 1 k3 1 il has additive order 10., 35. Use Theorems 14.3 and 14.4., 37. Since every f(x) in kx, 2l has the form f(x) 5 xg(x) 1 2h(x), we have f(0) 5 2h(0), so that f(x) [ I., If f(x) [ I, then f(x) 5 anxn 1 ? ? ? 1 a1x 1 2k 5 x(anxn21 1 ? ? ? 1 a1) 1 2k [ kx, 2l. I is prime, and maximal. Z[x]/I has two elements., 39. 3x 1 1 1 I, 41. Every ideal is a subgroup. Every subgroup of a cyclic group is cyclic., 43. Let I be any ideal of R % S and let IR 5 {r [ R | (r, s) [ I for some s [ S} and IS 5 {s [ S | (r, s) [ I, for some r [ R}. Then IR is an ideal of R and IS is an ideal of S. Let IR 5 krl and IS 5 ksl. Since,, for any (a, b) [ I there are elements a9 [ R and b9 [ S such that (a, b) 5 (a9r, b9s) 5 (a9, b9)(r, s),, we have that I = k(r, s)l., 45. Say b, c [ Ann(A). Then (b 2 c)a 5 ba 2 ca 5 0 2 0 5 0. Also, (rb)a 5 r(ba) 5 r ? 0 5 0., 47. a. k3l b. k3l c. k3l, 49. Suppose (x 1 N(k0l))n 5 0 1 N(k0l). We must show that x [ N(k0l). We know that xn 1, N(k0l) 5 0 1 N(k0l), so that xn [ N(k0l). Then, for some m, (xn)m 5 0, and therefore x [ N(k0l)., 51. The set Z2[x]/kx2 1 x 1 1l has only four elements and each of the nonzero ones has a multiplicative inverse. For example,, (x 1 kx2 1 x 1 1l)(x 1 1 1 kx2 1 x 1 1l) 5 1 1 kx2 1 x 1 1l., 53. x 1 2 1 kx2 1 x 1 1l is not zero, but its square is., 55. If f and g [ A, then ( f 2 g)(0) 5 f (0) 2 g(0) is even and ( f ? g)(0) 5 f (0) ? g(0) is even., f(x) 5 12 [ R and g(x) 5 2 [ A, but f(x)g(x) o A., 57. Hint: Any ideal of R/I has the form A/I, where A is an ideal of R., 59. Use the fact that R/I is an integral domain to show that R/I 5 {I, 1 1 I}., 61. kxl , kx, 2nl , kx, 2n21l , . . . , kx, 2l, 63. Taking r 5 1 and s 5 0 shows that a [ I. Taking r 5 0 and s 5 1 shows that b [ I. If J is any, ideal that contains a and b, then it contains I because of the closure conditions., , Supplementary Exercises for Chapters 12–14, One problem after another presents itself and in the solving of them we can find our greatest, pleasure., karl menninger, , 1. In Z10, they are 0, 1, 5, and 6. In Z20, they are 0, 1, 5, and 16. In Z30, they are 0, 1, 6, 10, 15, 16,, 21, and 25., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A25, , 3. We must show that an 5 0 implies a 5 0. First show this for the case when n is a power of 2., If n is not a power of 2, say 13, for example, note that a13 5 0 implies a16 5 0., 5. Suppose A s C and B s C. Pick a [ A and b [ B such that a, b o C. But ab [ C and C is, prime., 7. {0} % {0}, R % R, R % {0}, and {0} % R. The ideals of F % F are {0} % {0}, F % F,, F % {0}, and {0} % F., 9. Suppose that am mod n 5 0. Since n divides am, every prime divisor p of n divides am. By Euclid’s, Lemma (Chapter 0), p divides a, and since n is square-free, if follows that n divides a., 11. Suppose a1, a2 [ A but a1 o B and a2 o C. Use a1 1 a2 to derive a contradiction., 13. Clearly kal contains the right-hand side. Now show that the right-hand side contains a and is an, ideal., 15. Since A is an ideal, ab [ A. Since B is an ideal, ab [ B. So ab [ A > B 5 {0}., 17. 6, 19. Use Exercise 4., 21. Since 2x 1 1 1 k2x 1 1l = 0 1 k2x 1 1l, we have 22x 1 k2x 1 1l 5 1 1 k2x 1 1l., So, (22 + k2x 1 1l) (x 1 k2x 1 1l) 5 1 1 k2x 1 1l., 23. Consider x2 1 1 1 kx4 1 x2l., 25. Consider Z8., 27. Say char R 5 p (remember p must be prime). Then char R/A 5 the additive order of 1 1 A., But |1 1 A| divides |1| 5 p., 29. Use Theorems 13.2, 14.3, and 14.4., 1 1 1 0, 1 0, a b, 31. Observe that A e c, dc, d c, d is not in A., d ` a, b [ Z2 f but c, 1 1 0 0, 1 0, 0 0, 33. Z[i]/A has two elements. (From this it follows that A is maximal. See Theorem 14.4.), 35. A finite subset of a field is a subfield if it contains a nonzero element and is closed under addition and multiplication., 37. Observe that (a 1 bi) (a 2 bi) 5 a2 1 b2. In Z13[i], 2 1 3i is a zero-divisor., 39. 5, 41. The inverse is 2x 1 3., 43. Observe that Z5[x, y]/kx, yl < Z5 and use Theorem 14.4., 45. Say (a,b)n 5 (0, 0). Then an 5 0 and bn 5 0. If am 5 0 and bn 5 0, then (a, b)mn 5, ((am)n, (bn)m) 5 (0, 0)., 47. If a2 5 a, then pk | a(a 2 1). Since a and a 2 1 are relatively prime, pk | a or pk | (a 2 1)., So, a 5 0 or a 5 1., 49. In Z3 3 224, (a 1 b22)21 5 (a 2 b22)/(a2 2 2b2) 5 (a 2 b22)/(a2 1 b2). In Z7 3 224 , (1 1 222), (1 1 522) 5 0., 51. If xn 5 0, then (rx)n 5 rnxn 5 0., , Chapter 15, For every problem there is a solution which is simple, clean and wrong., h. l. mencken, , 1. Property 3: f(A) is a subgroup because f is a group homomorphism. Let s [ S and f(r) 5 s., Then sf(a) 5 f(r)f(a) 5 f(ra) and f(a)s 5 f(a)f(r) 5 f(ar)., Property 4: Let a and b belong to f21(B) and r belong to R. Then f(a) and f(b) are in B. So,, f(a) 2 f(b) 5 f(a) 1 f(2b) 5 f(a 2 b) [ B. Thus, a 2 b [ B. Also, f(ra) 5 f(r)f(a) [ B, and f(ar) 5 f(a)f(r) [ B. So, ra and ar [ f21(B)., 3. We already know the mapping is an isomorphism of groups. Let F(x 1 Ker f) 5 f(x). Note that, F((r 1 Ker f)(s 1 Ker f)) 5 F(rs 1 Ker f) 5 f(rs) 5 f(r)f(s) 5 F(r 1 Ker f)F(s 1 Ker f)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A26, 5., 7., 9., 11., 13., 15., 17., 19., 21., 23., 25., 27., 29., 31., 33., 35., 37., 39., 41., 43., 45., 47., 49., , 51., 53., 55., 57., 59., 61., 63., 65., 67., , 69., , Selected Answers, , f(2 1 4) 5 f(1) 5 5, whereas f(2) 1 f(4) 5 0 1 0 5 0., Observe that (x 1 y)/1 5 x/1 1 y/1 and (xy)/1 5 (x/1)(y/1)., a 5 f(1) 5 f(1·1) 5 f(1)f(1) 5 aa 5 a2. For the example look at Z6., If a and b (b 2 0) belong to every member of the collection, then so do a 2 b and ab21. Thus,, by Exercise 29 in Chapter 13, the intersection is a subfield., Apply the definition., Multiplication is not preserved., Yes., The set of all polynomials passing through the point (1, 0)., For Z6 to Z6, 1 S 0, 1 S 1, 1 S 3, and 1 S 4 each define a homomorphism. For Z20 to Z30, 1 S 0,, 1 S 6, 1 S 15, and 1 S 21 each define a homomorphism., The zero map and the identity map., Use Exercise 24., Say 1 is the unity of R. Let s 5 f(r) be any element of S. Then f(1)s 5 f(1)f(r) 5 f(1r) 5, f(r) 5 s. Similarly, sf(1) 5 s., Observe that an idempotent must map to an idempotent. So, (1, 0) and (0, 1) must map to 0 or 1., It follows that (a, b) S a, (a, b) S b, and (a, b) S 0 are the only ring homomorphisms., Say m 5 akak21 . . . a1a0 and n 5 bkbk21 . . . b1b0. Then m 2 n 5 (ak 2 bk)10k 1, (ak21 2 bk21)10k21 1 . . . 1 (a1 2 b1)10 1 (a0 2 b0). Now use the test for divisibility by 9., Use the appropriate divisibility tests., Mimic Example 8., Use Exercise 35., Look at both sides mod 2., Observe that (2 ? 1075 1 2) mod 3 5 1 and (10100 1 1) mod 3 5 2 5 21 mod 3., This follows directly from Theorem 13.3 and Theorem 10.1, part 3., No. The kernel must be an ideal., a. Suppose ab [ f21(A). Then f(a)f(b) [ A, so that a [ f21(A) or b [ f21(A)., b. Consider the natural homomorphism from R to S/A. Then use Theorems 15.3 and 14.4., a. f((a, b) 1 (a9, b9)) 5 f((a 1 a9, b 1 b9)) 5 a 1 a9 5 f((a, b)) 1 f((a9, b9)), so f preserves, addition. Also, f((a, b)(a9, b9)) 5 f((aa9, bb9)) 5 aa9 5 f((a, b))f((a9, b9))., b. f(a) 5 f(b) implies that (a, 0) 5 (b, 0), which implies that a 5 b. f(a 1 b) 5 (a 1 b, 0) 5, (a, 0) 1 (b, 0) 5 f(a) 1 f(b). Also, f(ab) 5 (ab, 0) 5 (a, 0)(b, 0) 5 f(a)f(b)., c. Use (r, s) S (s, r)., Observe that x4 5 1 has two solutions in R but four in C., Use Exercises 46 and 52., If a / b 5 a9/ b9 and c / d 5 c9 / d9, then ab9 5 ba9 and cd9 5 dc9. So, acb9d9 5 (ab9)(cd9) 5, (ba9)(dc9) 5 bda9c9. Thus, ac / bd 5 a9c9/ b9d9 and therefore (a / b)(c / d) 5 (a9/ b9)(c9/ d9)., First note that any field containing Z and i must contain Q[i]. Then prove (a 1 bi)/(c 1 di) [ Q[i]., The subfield of E is {ab21 | a, b [ D, b 2 0}., Reflexive and symmetric properties follow from the commutativity of D. For transitivity, assume, a/b ; c/d and c/d ; e/f. Then adf 5 (bc)f 5 b(cf ) 5 bde, and cancellation yields af 5 be., Try ab21 S a/b., The mapping a 1 bi S a 2 bi is a ring isomorphism of C., Certainly the unity 1 is contained in every subfield. So, if a field has characteristic p, the subfield, {0, 1, . . . , p 2 1} is contained in every subfield. If a field has characteristic 0, then {(m ? 1)(n ? 1)21, | m, n [ Z, n 2 0} is a subfield contained in every subfield. This subfield is isomorphic to Q, [map (m ? 1)(n ? 1)21 to m/n]., The mapping f(x) 5 (x mod m, x mod n) from Zmn to Zm { Zn is a ring isomorphism., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A27, , Chapter 16, You know my methods. Apply them!, sherlock holmes,, The Hound of the Baskervilles*, , 1. f 1 g 5 3x4 1 2x3 1 2x 1 2; f ? g 5 2x7 1 3x6 1 x5 1 2x4 1 3x2 1 2x 1 2., 3. 1, 2, 4, 5, 5. Write f(x) 5 (x 2 a)q(x) 1 r(x). Since deg (x 2 a) 5 1, deg r(x) 5 0 or r(x) 5 0. So r(x) is a, constant. Also, f(a) 5 r(a)., 7. x3 1 1 and x3 1 x2 1x 11, 9. Use Corollary 1 of Theorem 16.2., 11. Let f(x), g(x) [ R[x]. By inserting terms with the coefficient 0, we may write, f(x) 5 anxn 1 . . . 1 a0, and, g(x) 5 bnxn 1 . . . 1 b0., Then, f( f(x) 1 g(x)) 5 f(an 1 bn)xn 1 . . . 1 f(a0 1 b0), 5 (f(an) 1 f(bn))xn 1 . . . 1 f(a0) 1 f(b0), 5 (f(a )xn 1 . . . 1 f(a )) 1 (f(b )xn 1 . . . 1 f(b )), n, , 5 f( f(x)) 1 f(g(x))., 13., 15., 17., 19., 21., , 23., 25., 27., 29., 31., 33., 35., 37., 39., 41., 43., 45., 47., 49., , 0, , n, , 0, , Multiplication is done similarly., 4x2 1 3x 1 6 is the quotient and 6x 1 2 is the remainder., It is its own inverse., No. See Exercise 19., If f(x) 5 anxn 1 . . . 1 a0 and g(x) 5 bmxm 1 . . . 1 b0, then f(x) ? g(x) 5 anbmxm1n 1 . . . 1 a0b0., Let m be the multiplicity of b in q(x). Then we may write f(x) 5 (x 2 a)n (x 2 b)m q9(x), where, q9(x) is in F[x] and q9(b) Z 0. This means that b is a zero of f(x) of multiplicity at least m., If b is a zero of f(x) greater than m, then b is a zero of g(x) 5 f(x)/(x 2 b)m 5 (x 2 a)nq9(x)., But then 0 5 g(b) 5 (b 2 a)n q9(b), and therefore q9(b) 5 0., Use Corollary 3 of Theorem 16.2., If f(x) 2 g(x), then deg[f(x) 2 g(x)] , deg p(x). But the minimum degree of any member of, kp(x)l is deg p(x)., Start with (x 2 1/2)(x 1 1/3) and clear fractions., “Long divide” x 2 a into f(x) and induct on deg f(x)., By Theorem 16.4, I 5 kx 2 1l., Use Corollary 2 of Theorem 15.5 and Exercise 11 in this chapter., For any a in U(p), ap21 5 1, so every member of U(p) is a zero of x p21 2 1. Now use the Factor, Theorem and a degree argument., C(x) (field of quotients of C[x]), Use Exercise 38., Observe that, modulo 101, (50!)2 5 (50!)(21)(22) . . . (250) 5 (50!)(100)(99) . . . (51) 5 100!, and use Exercise 36., Take R 5 Z and I 5 k2l., Hint: F[x] is a PID. So kf(x), g(x)l 5 ka(x)l for some a(x) [ F[x]. Thus, a(x) divides both f(x) and, g(x). This means that a(x) is a constant., Write f(x) 5 (x 2 a)g(x). Use the product rule to compute f9(x)., Say deg g(x) 5 m, deg h(x) 5 n, and g(x) has leading coefficient a. Let k(x) 5 g(x) 2 axm2nh(x)., Then deg k(x) , deg g(x) and h(x) divides k(x) in Z[x] by induction. So, h(x) divides k(x) 1, axm2nh(x) 5 g(x) in Z[x]., , *Copyright © 1968 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music, Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A28, , Selected Answers, , 51. If f(x) takes on only finitely many values, then there is at least one a in Z with the property that, f(x) 5 a for infinitely many x in Z. But then g(x) 5 f(x) 2 a has infinitely many zeros. This, contradicts Corollary 3 of Theorem 16.2., 53. Observe that (1 1 tmx) (1 2 tmx) 5 1., 55. Let f(x) = anxn 1 an21xn21 1 ) 1 a1x 1 a0 and assume that p/q is a zero of f(x), where p and q, are integers and n is even. We may assume that p and q are relatively prime. Substituting p/q for, x and clearing fractions, we have anpn 1 an21pn21 q 1 ) 1 a1pq n21 52a0qn. If p is even, then, the left side is even. If p is odd, then each summand on the left side is odd and since there is an, even number of summands, the left side is still even. Because a0 is odd, we then have that q is, even. It follows that anpn 52 (an21pn21 q 1 ) 1 a1pq n21 1 a0qn) is even, since the right side is, divisible by q. This implies that p is even. This contradicts the assumption that p and q are relatively prime., 57. Consider the remainder when x43 is divided by x2 1 x 1 1., 59. Observe that every term of f(a) has the form ci ai and ci ai mod m 5 ci bi mod m. To prove the second statement, assume that there is some integer k such that f(k) 5 0. If k is even, then because, k mod 2 5 0, we have by the first statement 0 5 f(k) mod 2 5 f(0) mod 2 so that f(0) is even., This shows that k is not even. If k is odd, then k mod 2 5 1, so by the first statement f(k) 5 0 is, odd. This contradiction completes the proof., 61. A solution to x25 2 1 5 0 in Z37 is a solution to x25 5 1 in U(37). So, by Corollary 2 of, Theorem 4.1, |x| divides 25. Moreover, we must also have that |x| divides |U(37)| 5 36., , Chapter 17, Experience enables you to recognize a mistake when you make it again., franklin p. jones*, , 1. Use Theorem 17.1., 3. If f(x) is not primitive, then f(x) 5 ag(x), where a is an integer greater than 1. Then a is not a unit, in Z[x] and f(x) is reducible., 5. a. If f(x) 5 g(x)h(x), then af(x) 5 ag(x)h(x)., b. If f(x) 5 g(x)h(x), then f(ax) 5 g(ax)h(ax)., c. If f(x) 5 g(x)h(x), then f(x 1 a) 5 g(x 1 a)h(x 1 a)., d. Try a 5 1., 7. Use part a Exercise 5 and clear fractions., 9. Find an irreducible polynomial p(x) of degree 2 over Z5. Then Z5[x]/kp(x)l is a field of order 25., 11. Note that 21 is a zero. No, since 4 is not a prime., 13. Let f(x) 5 x4 1 1 and g(x) 5 f(x 1 1) 5 x4 1 4x3 1 6x2 1 4x 1 2. Then f(x) is irreducible over, Q if g(x) is. Eisenstein’s Criterion shows that g(x) is irreducible over Q. To see that x4 1 1 is, reducible over R, observe that, x8 2 1 5 (x4 1 1)(x4 2 1),, , 15., 17., 19., 21., , so any complex zero of x4 1 1 is a complex zero of x8 2 1. Also note that the complex zeros, of x4 1 1 must have order 8 (when considered as an element of C). Let v 5 22/2 1 i 22/2., Then Example 2 in Chapter 16 tells us that the complex zeros of x4 1 1 are v, v3, v5, and v7,, so x4 1 1 5 (x 2 v)(x 2 v3)(x 2 v5)(x 2 v7). But we may pair these factors up as ((x 2 v), (x 2 v7)) ((x 2 v3)(x 2 v5)) 5 (x2 2 22 x 1 1) . (x2 1 22 x 11) to factor using reals (see, DeMoivre’s Theorem, Example 7 in Chapter 0)., (x 1 3)(x 1 5)(x 1 6), a. Consider the number of distinct expressions of the form (x 2 c)(x 2 d)., b. Reduce the problem to the case considered in part a., Use Exercise 18, and imitate Example 10., Map Z3[x] onto Z3[i] by f(x) → f(i). This is a ring homomorphism with kernel kx2 1 1l., , *Copyright © 1968 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music, Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A29, , 23. x2 1 1, x2 1 x 1 2, x2 1 2x 1 2, 25. 1 has multiplicity 1, 3 has multiplicity 2., 27. We know that an(r/s)n 1 an21(r/s)n21 1 . . . 1 a0 5 0. So anr n 1 san21r n21 1 . . . 1 sna0 5 0., This shows that s | anrn and r | sna0. Now use Euclid’s Lemma and the fact that r and s are relatively prime., 29. If there is an a in Zp such that a2 5 21, then x4 1 1 5 (x2 1 a)(x2 2 a)., If there is an a in Zp such that a2 5 2, then x4 1 1 5 (x2 1 ax 1 1)(x2 2 ax 1 1)., If there is an a in Zp such that a2 5 22, then x4 1 1 5 (x2 1 ax 2 1)(x2 2 ax 2 1)., To show that one of these three cases must occur, consider the group homomorphism from Z *p to, itself given by x → x2. Since the kernel is {1, 21}, the image H has index 2 (we may assume that, p 2 2). Suppose that neither 21 nor 2 belongs to H. Then, since there is only one coset other, than H, we have 21H 5 2H. Thus, H 5 (21H)(21H) 5 (21H)(2H) 5 22H, so that 22 is in H., 31. Since ( f 1 g)(a) 5 f(a) 1 g(a) and ( f ? g)(a) 5 f(a)g(a), the mapping is a homomorphism., Clearly, p(x) belongs to the kernel. By Theorem 17.5, kp(x)l is a maximal ideal, so the kernel, is kp(x)l., 33. The mapping a → a 1 kp(x)l is an isomorphism., 35. f(x) is primitive., 37. Although the probability of rolling any particular sum is the same with either pair of dice, the, probability of rolling doubles is different (1/6 with ordinary dice, 1/9 with Sicherman dice). Thus,, the probability of going to jail is different. Other probabilities are also affected. For example, if, in jail one cannot land on Virginia by rolling a pair of 2’s with Sicherman dice, but one is twice, as likely to land on St. James with a pair of 3’s with the Sicherman dice as with ordinary dice., 39. The analysis is identical except that 0 # q, r, t, u # n. Now, just as when n 5 2, we have q 5 r 5, t 5 1, but this time 0 # u # n. However, when u . 2, P(x) 5 x(x 1 1)(x2 1 x 1 1)(x2 2 x 1 1)u, has (2u 1 2)x2u13 as one of its terms. Since the coefficient of x2u13 represents the number of dice, with the label 2u 1 3, the coefficient cannot be negative. Thus, u # 2, as before., , Chapter 18, If you have great talents, industry will improve them; if you have but moderate abilities, industry, will supply their deficiency., sir joshua reynolds, , 1. 1. |a2 2 db2| 5 0 implies a2 5 db2. Thus, a 5 0 5 b, since otherwise d 5 1 or d is divisible by, the square of a prime., 2. N((a 1 b2d)(a9 1 b92d)) 5 N(aa9 1 dbb9 1 (ab9 1 a9b)2d) 5 |(aa9 1 dbb9)2 2, d(ab9 1 a9b)2| 5 |a2a92 1 d 2b2b92 2 da2b92 2 da92b2| 5 |a2 2 db2||a92 2 db92| 5, N(a 1 b2d)N(a9 1 b92d)., 3. If xy 5 1, then 1 5 N(1) 5 N(xy) 5 N(x)N(y) and N(x) 5 1 5 N(y). If N(a 1 b2d) 5 1, then, 61 5 a2 2 db2 5 (a 1 b2d)(a 2 b2d) and a 1 b2d is a unit., 4. This property follows directly from properties 2 and 3., 3. Let I 5 <Ii. Let a, b [ I and r [ R. Then a [ Ii for some i and b [ Ij for some j. Thus, a, b [ Ik,, where k 5 max{i, j}. So, a 2 b [ Ik # I and ra, ar [ Ik # I., 5. Clearly, kabl # kbl. If kabl 5 kbl, then b 5 rab, so that 1 5 ra and a is a unit., 7. Say x 5 a 1 bi and y 5 c 1 di. Then, xy 5 (ac 2 bd) 1 (bc 1 ad)i., So, d(xy) 5 (ac 2 bd)2 1 (bc 1 ad)2 5 (ac)2 1 (bd)2 1 (bc)2 1 (ad)2., On the other hand,, d(x)d(y) 5 (a2 1 b2)(c 2 1 d 2) 5 a2c2 1 b2d 2 1 b2c2 1 a2d 2., 9. Suppose a 5 bu, where u is a unit. Then d(b) # d(bu) 5 d(a). Also, d(a) # d(au21) 5 d(b)., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A30, , Selected Answers, , 11. m 5 0 and n 5 21 give q 5 2i, r 5 22 2 2i., 13. 3 ? 7 and 11 2252 11 2252. Mimic Example 8 to show that these are irreducible., 15. Observe that 10 5 2 ? 5 and 10 5 12 262 12 262 and mimic Example 8. A PID, is a UFD., 17. Suppose 3 5 ab, where a, b [ Z[i] and neither is a unit. Then 9 5 d(3) 5 d(a)d(b), so that, d(a) 5 3. But there are no integers such that a2 1 b2 5 3. Observe that 2 5 2i(1 1 i)2 and, 5 5 (1 1 2i)(1 2 2i)., 19. Use Exercise 1 with d 5 21. 5 and 1 1 2i; 13 and 3 1 2i; 17 and 4 1 i., 21. Mimic Example 1., 23. 11 252 11 252 5 4 5 2 ? 2. Now use Exercise 22., 25. Use the fact that x is a unit if and only if N(x) 5 1., 27. 1 5 N (ab) 5 N(a)N(b), so that N(a) 5 1 5 N(b)., 29. Suppose that bc 5 pt in Zn. Then there exists an integer k such that bc 5 pt 1 kn. This implies, that p divides bc in Z, and by Euclid's Lemma we know that p divides b or p divides c., 31. See Example 3., 33. p | (a1a2 . . . an21)an implies that p | a1a2 . . . an21 or p | an. Thus, by induction, p divides some ai., 35. Use Exercise 10 and Theorem 14.4., 37. Suppose R satisfies the ascending chain condition and there is an ideal I of R that is not finitely, generated. Then pick a1 [ I. Since I is not finitely generated, ka1l is a proper subset of I, so, we may choose a2 [ I but a2 o ka1l. As before, ka1, a2l is proper, so we may choose a3 [ I but, a3 o ka1, a2l. Continuing in this fashion, we obtain a chain of infinite length ka1l , ka1, a2l ,, ka1, a2, a3l , . . .., Now suppose every ideal of R is finitely generated and there is a chain I1 , I2 , I3 , . . .., Let I 5 <Ii. Then I 5 ka1, a2, . . . , anl. Since I 5 <Ii, each ai belongs to some member of the, union, say Ii9. Letting k 5 max {i9 | i 5 1, . . . , n}, we see that all ai [ Ik. Thus, I # Ik and the, chain has length at most k., 39. Say I 5 ka 1 bil. Then a2 1 b2 1 I 5 (a 1 bi)(a 2 bi) 1 I 5 I and a2 1 b2 [ I. For any, c, d [ Z, let c 5 q1(a2 1 b2) 1 r1 and d 5 q2(a2 1 b2) 1 r2, where 0 # r1, r2 , a2 1 b2. Then, c 1 di 1 I 5 r1 1 r2i 1 I., 41. N16 2272 64 N11 3272. For the other part, use Exercise 25., 43. Theorem 18.1 shows that primes are irreducible. So, assume that a is an irreducible in a UFD R, and that a | bc in R. We must show that a | b or a | c. Since a | bc, there is an element d in R such, that bc 5 ad. Now replace b, c, and d by their factorizations as a product of irreducibles and use, uniqueness., 45. See Exercise 21 in Chapter 0., , Supplementary Exercises for Chapters 15–18, Errors, like straws, upon the surface flow;, He who would search for pearls must dive below., john dryden, , 1. Use Theorem 15.3, Supplementary Exercise 8 for Chapters 12–14, Theorem 14.4, and Example 13 in Chapter 14., 3. To show the isomorphism, use the First Isomorphism Theorem., 5. Use the First Isomorphism Theorem., 7. Consider the obvious homomorphism from Z[x] onto Z2[x]. Then use the First Isomorphism, Theorem and Theorem 14.3., 9. As in Example 7 in Chapter 6, the mapping is onto, is one-to-one, and preserves multiplication., Also, a(x 1 y)a21 5 axa21 1 aya21, so that it preserves addition as well., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A31, , 11. Z[i]/k2 1 il 5 {0 1 k2 1 il, 1 1 k2 1 il, 2 1 k2 1 il, 3 1 k2 1 il, 4 1 k2 1 il}. Note that, 5 1 k2 1 il 5 (2 1 i)(2 2 i) 1 k2 1 il, 5 0 1 k2 1 il., 13. Use the fact that ap 5 a for all a in Zp and Exercise 49 in Chapter 13., 15. In Zn we are given (k 1 1)2 5 k 1 1. So, k2 1 2k 1 1 5 k 1 1 or k2 5 2k 5 n 2 k., Also, (n 2 k)2 5 n2 2 2nk 1 k2 5 k2, so (n 2 k)2 5 n 2 k., 17. Observe that for any integer a, a2 mod 4 5 0 or 1., 19. Use the Mod 2 Irreducibility Test., 21. Use Theorem 14.4. The factor ring has two elements., 23. Use Theorem 14.4., 25. Say a/b, c/d [ R. Then (ad 2 bc)/(bd) and ac/(bd) [ R by Euclid’s Lemma. The field of, quotients is Q., 27. Z[i]/k3l is a field and Z3 % Z3 is not., 29. Consider the mapping from R[x] to (R/I)[x] given by anxn 1 . . . 1 a0 → (an 1 I)x n 1 . . . 1, (a0 1 I )., 31. Let I 5 k2l[x]. Then Z8[x]/I is isomorphic to Z2[x]., , Chapter 19, When I was young I observed that nine out of every ten things I did were failures, so I did ten times, more work., george bernard shaw, , 1., , Rn, , has basis {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1)}; M2(Q) has basis, ec, , 3., 5., 7., 9., , 11., 13., 15., , 0 0, 1 0, 0 1, 0 0, d, c, d, c, d, c, d f;, 0 0, 0 0, 1 0, 0 1, , Zp[x] has basis {1, x, x2, . . .}; C has basis {1, i}., (a2x2 1 a1x 1 a0) 1 (a29x2 1 a19x 1 a09) 5 (a2 1 a29)x2 1 (a1 1 a19)x 1 (a0 1 a09) and, a(a2x2 1 a1x 1 a0) 5 aa2x2 1 aa1x 1 aa0. A basis is {1, x, x2}. Yes., Linearly dependent, since 23(2, 21, 0) 2 (1, 2, 5) 1 (7, 21, 5) 5 (0, 0, 0)., Suppose au 1 b(u 1 v) 1 c(u 1 v 1 w) 5 0. Then (a 1 b 1 c)u 1 (b 1 c)v 1 cw 5 0. Since {u, v, w}, are linearly independent, we obtain c 5 0, b 1 c 5 0, and a 1 b 1 c 5 0. So, a 5 b 5 c 5 0., If the set is linearly independent, it is a basis. If not, then delete one of the vectors that is a linear, combination of the others (see Exercise 8). This new set still spans V. Repeat this process until you, obtain a linearly independent subset. Since the set is finite, you will eventually obtain a linearly, independent set that still spans V., Let u1, u2, u3 be a basis for U and w1, w2, w3 be a basis for W. Use the fact that u1, u2, u3, w1, w2,, w3 are linearly dependent over F. In general, if dim U 1 dim W . dim V, then U > W 2 {0}., no, yes; 2, , 17. c, , a, ab, a, a b, a a, a b a b, d c, d c, d and, ab, b, a b, b, a b a b, b b, , a, ab, ac, ac bc, d c, d., ab, a, ac bc, bc, 19. Suppose B is a basis. Then every member of V is some linear combination of elements of B. If, a1v1 1 . . . 1 anvn 5 a19v1 1 . . . 1 an9 vn, where vi [ B, then (a1 2 a19)v1 1 . . . 1 (an 2 an9)vn 5 0, and ai 2 a i 9 5 0 for all i. Conversely, if every member of V is a unique linear combination of, elements of B, certainly B spans V. Also, if a1v1 1 . . . 1 anvn 5 0, then a1v1 1 . . . 1 anvn 5, 0v1 1 . . . 1 0vn and ai 5 0 for all i., cc, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A32, , Selected Answers, , 21. Since w1 5 a1u1 1 a2u2 1 . . . 1 anun and a1 2 0, we have u1 5 a121(w1 2 a2u2 2 . . . 2 anun),, and therefore u1 [ kw1, u2, . . . , unl. Clearly, u2, . . . , un [ kw1, u2, . . . , unl. Hence every linear, combination of u1, . . . , un is in kw1, u2, . . . , unl., 23. {(1, 0, 1, 1), (0, 1, 0, 1)}, 25. Study the proof of Theorem 19.1., 27. If V and W are vector spaces over F, then the mapping must preserve addition and scalar multiplication. That is, T: V → W must satisfy T(u 1 v) 5 T(u) 1 T(v) for all vectors u and v in V, and, T(au) 5 aT(u) for all vectors u in V and scalars a in F. A vector space isomorphism from V to W, is a one-to-one linear transformation from V onto W., 29. Suppose v and u belong to the kernel and a is a scalar. Then T(v 1 u) 5 T(v) 1 T(u) 5 0 1 0 5 0, and T(av) 5 aT(u) 5 a ? 0 5 0., 31. Let {v1, v2, . . . , vn} be a basis for V. Map a1v1 1 a2v2 1 . . . 1 anvn to (a1, a2,. . . , an)., , Chapter 20, Well here’s another clue for you all., john lennon and paul mccartney,, “Glass Onion,” The White Album, , 1. Compare with Exercise 26 in the Supplementary Exercises for Chapters 12–14., 3. Q1 232, 5. Q1 232, 7. Note that x 5 21 25 implies x4 2 2x2 2 4 5 0., 9. a5 5 a2 1 a 1 1; a22 5 a2 1 a 1 1; a100 5 a2, 11. The set of all expressions of the form, (a pn 1 a pn21 1 . . . 1 a )/(b pm 1 b, n, , n21, , 0, , m, , m21, , m21p, , 1 . . . 1 b0),, , where bm 2 0., 13. x7 2 x 5 x(x6 2 1) 5 x(x3 1 1)(x3 2 1) 5 x(x 2 1)3(x 1 1)3 ; x10 2 x 5 x(x9 2 1) 5 x(x 2 1)9, (see Exercise 49 in Chapter 13)., 15. Hint: Use Exercise 49 in Chapter 13., 17. a 5 4/3, b 5 2/3, c 5 5/6, 19. Use the fact that 1 1 i 5 2(4 2 i) 1 5 and 4 2 i 5 5 2 (1 1 i)., 21. If the zeros of f(x) are a1, a2, . . . , an, then the zeros of f(x 1 a) are a1 2 a, a2 2 a, . . . , an 2 a. Now, use Exercise 20., 23. Q and Q(22), 25. Let F 5 Z3[x]/kx3 1 2x 1 1l and denote the cosets x 1 kx3 1 2x 1 1l by b and 2 1 kx3 1 2x 1 1l, by 2. Then x3 1 2x 1 1 5 (x 2 b)(x 2 b 2 1)(x 1 2b 1 1)., 27. Suppose that f: Q(23) → Q(23) is an isomorphism. Since f(1) 5 1, we have f(23) 5 23., Then 23 5 f(23) 5 f( 2323) 5 [f( 23)]2. This is impossible, since f( 23) is a real, number., 29. Use long division., 31. Use Theorem 20.5., 33. Use Theorem 20.5., 35. Let K be the intersection of all subfields of E that contain F and the set {a1, a2, . . ., an}. It follows from the subfield test given in Exercise 29 Chapter 13 that K is a subfield of E and, by the, definition, that K contains F and the set {a1, a2, . . ., an}. Since F (a1, a2, . . ., an) is the smallest, such field we have F(a1, a2, . . ., an) # K. Moreover, since the field F(a1, a2, . . ., an) is one member of the intersection we have K # F(a1, a2, . . ., an}. This proves that K 5 F(a1, a2, . . ., an}., 37. Since |(Z2[x]/kf(x)l)*| 5 31, every nonidentity is a generator., 39. Mimic the argument given in Example 9 of this chapter., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A33, , Chapter 21, A good proof is one which makes us wiser., yu. manin, , 1. It follows from Theorem 21.1 that if p(x) and q(x) are both monic irreducible polynomials in, F[x] with p(a) 5 q(a) 5 0, then deg p(x) 5 deg q(x). If p(x) 2 q(x), then ( p 2 q)(a) 5 p(a) 2, q(a) 5 0 and deg (p(x) 2 q(x)) , deg p(x), contradicting Theorem 21.1. To prove Theorem 21.3,, use the Division Algorithm for F[x] (Theorem 16.2)., n, 3. Note that 3Q1 2 22 :Q4 5 n and use Theorem 21.5., 5. Use Exercise 4., 7. Suppose Q(2a) 5 Q(2b). If 2b [ Q, then 2a [ Q and we may take c 5 2a/2b. If 2b o Q,, then 2a o Q. Write 2a 5 r 1 s2b. It follows that r 5 0 and a 5 bs2. The other direction, follows from Exercise 20 in Chapter 20., 9. Observe that [F(a):F] must divide [E:F]., 11. Note that [F(a, b):F] is divisible by both m 5 [F(a):F] and n 5 [F(b):F], and that [F(a,b):F] # mn., 13. Note that a is a zero of x3 2 a3 over F(a3)[x]. For the second part, take F 5 Q, a 5 1; F 5 Q, a 5, 3, 11 i232 /2; F Q, a 2, 2., 15. Suppose E1 > E2 2 F. Then [E1:E1 > E2][E1 > E2:F] 5 [E1:F] implies [E1:E1 > E2] 5 1, so that, E1 5 E1 > E2. Similarly, E2 5 E1 > E2., 17. E must be an algebraic extension of R, so that E # C. But then [C:E][E:R] 5 [C:R] 5 2., 19. Let a be a zero of p(x) in some extension of F. First note [E(a):E] # [F(a):F] 5 deg p(x). Then, observe that [E(a):F(a)][F(a):F] 5 [E(a):E][E:F]. This implies that deg p(x) divides [E(a):E],, so deg p(x) 5 [E(a):E]., 21. Hint: If a 1 b and ab are algebraic, then so is 2 1a b2 2 4ab., 23. 2b2 4ac, 25. Use the Factor Theorem., 27. Say a is a generator of F*. If char F 5 0, then the prime subfield of F is isomorphic to Q. Since Q*, is not cyclic, we have that F 5 Zp(a), and it suffices to show that a is algebraic over Zp. If a [ Zp,, we are done. Otherwise, 1 1 a 5 ak for some k 2 0. If k . 0, we are done. If k , 0, then a2k 1, a12k 5 1 and we are done., 29. If [K:F] 5 n, then there are elements v1, v2, . . . , vn in K that constitute a basis for K over F., The mapping a1v1 1 . . . 1 anvn → (a1, . . . , an) is a vector space isomorphism from K to F n. If, K is isomorphic to Fn, then the n elements in K corresponding to (1, 0, . . . , 0), (0, 1, . . . , 0), . . . ,, (0, 0, . . . , 1) in F n constitute a basis for K over F., 31. Observe that [F(a, b):F(a)] 5 [F(a)(b):F(a)] # [F(b):F] # [F(a)(b):F(b)] [F(b):F] 5, [F(a)(b):F] 5 [F(a, b):F]., 33. Mimic Example 5., 35. Mimic Example 6., 37. Observe that K 5 F(a1, a2, . . . , an), where a1, a2, . . . , an are the zeros of the polynomial. Now, use Theorem 21.5., 39. Elements of Q(p) have the form (amp m 1 am21p m21 1 . . . 1 a0)/(bnp n 1 bn21p n21 1 . . . 1 b0),, where the a’s and b’s are rational numbers. So, if 22 [ Q1p2, we have an expression of the form, 2(b p n 1 b p n21 1 . . . 1 b )2 5 (a p m 1 a p m21 1 . . . 1 a )2. Equating the lead terms, n, , n21, , 0, , m, , m21, , 0, , of both sides, we have 2b2np2n a2mp2m. But then we have m n, and 22 is equal to the rational, number am/bn., , 41. Observe that F1a2 F11 a1 2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A34, , Selected Answers, , Chapter 22, Tell me tell me tell me come on tell me the answer., john lennon and paul mccartney,, “Helter Skelter,” The White Album*, , 1. [GF(729):GF(9)] 5 3; [GF(64):GF(8)] 5 2, 3. The lattice of subfields of GF(64) looks like Figure 21.3 with GF(2) at the bottom, GF(64) at the, top, and GF(4) and GF(8) on the sides., 5. 2a 1 1, 7. Use Theorem 22.2., 9. The only possibilities for f(x) are x3 1 x 1 1 and x3 1 x2 1 1. See Exercise 8 in Chapter 20 for the, first case. See Example 2 in this chapter for the second case., 11. Use Exercise 44 in Chapter 15 and Corollary 4 of Lagrange’s Theorem (Theorem 7.1)., 13. Use the fact that if g(x) is an irreducible factor of x8 2 x over Z 2 and deg g(x) 5 m, then the field, Z 2[x]/kg(x)l has order 2m and is a subfield of GF(8). Now use Theorem 22.3., 15. Since GF(pn)* is a cyclic group of order pn 2 1, we seek the smallest n such that pn 2 1 is divisible, by 5. By observation, n 5 4 for p 5 2 or 3., 17. Direct calculations show that given x3 1 2x 1 1 5 0, we have x2 2 1 and x13 2 1., 19. Direct calculations show that x13 5 1, whereas (2x)2 2 1 and (2x)13 2 1. Thus, 2x is a generator., 21. First observe that for any field F, the set F* is a group under multiplication. Now use Theorem 22.2, and Theorem 4.3., 23. Find a quadratic irreducible polynomial p(x) over Z 3; then Z 3[x]/kp(x)l is a field of order 9., m, , m, , m, , 25. Let a, b [ K. Then, by Exercise 49b in Chapter 13, (a 2 b) p 5 a p 2 b p 5 a 2 b. Also,, m, m, m, (ab) p 5 a p b p 5 ab. So, K is a subfield., n21, p, 27. Consider x, 2 1 and use Corollary 4 of Lagrange’s Theorem (Theorem 7.1)., 29. Structurally identical, 31. Consider g(x) 5 x2 2 a. Note that |GF(p)[x]/kg(x)l| 5 p2, so that g(x) has a zero in GF(p2). Now, use Theorem 22.3., 33. Use Exercise 11., 35. Since F* is a cyclic group of order 124, it has a unique element of order 2., 37. See the solution for Exercise 25., 39. Consider the field of quotients of Zp[x]. The polynomial f(x) 5 x is not the image of any element., n, 41. Observe that p 2 1 5 2 1 has multiplicative order 2 and a(p 21)/2 is the unique element in kal of, order 2., 43. Since p mod 4 5 1, we have pn mod 4 5 1, and GF(pn)* is a cyclic group of order pn 2 1., , Chapter 23, Why, sometimes I’ve believed as many as six impossible things before breakfast., lewis carroll, , 1. To construct a 1 b, first construct a. Then use a straightedge and compass to extend a to the, right by marking off the length of b. To construct a 2 b, use the compass to mark off a length, of b from the right endpoint of a line of length a., 3. Let y denote the length of the hypotenuse of the right triangle with base 1, and let x denote the, length of the hypotenuse of the right triangle with base |c|. Then y2 5 1 1 d 2, y2 1 x2 5 (1 1 |c|)2,, and |c|2 1 d 2 5 x2. So, 1 1 2|c| 1 |c|2 5 1 1 d 2 1 |c|2 1 d 2, which simplifies to |c| 5 d 2., 5. Use sin2 u 1 cos2 u 5 1., 7. Use cos 2u 5 2 cos2 u 21., 9. Use sin(a 2 b) 5 sin a cos b 2 cos a sin b and Exercise 8., 11. Solving two linear equations with coefficients from F involves only the operations of F., *Quote from “While My Guitar Gently Weeps” by George Harrison. Copyright © 1968 Harrisong, Limited. Used by permission. All rights reserved., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A35, , 13. Use Theorem 17.1 and Exercise 27 in Chapter 17., 15. If so, then an angle of 40° is constructible. Now use Exercise 10., 17. This amounts to showing that 2p is not constructible. But if 2p is constructible, so is p., However, [Q(p):Q] is infinite., 3, 19. No, since [Q(2, 3 ):Q] 5 3., 3, 21. No, since [Q(2p):Q] is infinite., , Supplementary Exercises for Chapters 19–23, The things taught in colleges and schools are not an education, but the means of education., ralph waldo emerson, Journals, , 1. Use Theorem 20.5., 3. Suppose b is one solution of xn 5 a. Since F* is a cyclic group of order q 2 1, it has a cyclic, subgroup of order n, say kcl. Then each member of kcl is a solution to the equation xn 5 1. It, follows that bkcl is the solution set of xn 5 a., 5. (5a 2 1 2)/a 5 5a 1 2a21. Now observe that since a2 1 a 1 1 5 0, we know that, a(2a 2 1) 5 1, and so a21 5 2a 2 1. Thus, (5a2 1 2)/a 5 22 1 3a., 7. 5, 9. Since F(a) 5 F(a21), we have degree of a 5 [F(a):F] 5 [F(a21):F] 5 degree of a21., 11. If ab is a zero of cnx n 1 . . . 1 c1x 1 c0 [ F[x], then a is a zero of cnb nx n 1 . . . 1, c1bx 1 c0 [ F(b)[x]., 13. Every element of F(a) can be written in the form f(a)/g(a), where f(x), g(x) [ F[x]. If f(a)/g(a) is, algebraic and not in F, then there is some h(x) [ F[x] such that h( f (a)/g(a)) 5 0. By clearing, fractions and collecting like powers of a, we obtain a polynomial in a with coefficients from F, equal to 0. But then a would be algebraic over F., 15. Use Corollary 2 to Theorem 22.2., 17. If the basis elements commute, then so would any combination of basis elements. However, the, entire space is not commutative., 19. {x, x2, x3}, 21. Use Exercise 49 in Chapter 13., , Chapter 24, Difficulty, my brethren, is the nurse of greatness., william cullen bryant, , 1. a 5 eae21; cac21 5 b implies a 5 c21bc 5 c21b(c21)21; a 5 xbx21 and b 5 ycy21 imply, a 5 xycy21x21 5 xyc(xy)21., 3. Observe that T(xC(a)) 5 xax21 5 yay21 5 T(yC(a)) if and only if y21xa 5 ay21x, which is true, if and only if y21x [ C(a), which in turn is true if and only if yC(a) 5 xC(a). This proves that, T is well-defined and one-to-one. T is onto by definition., 5. Consider the correspondence T from the left cosets of N(H) in G to the conjugates of H in G, given by T(xN(H)) 5 xHx21., 7. Say cl(x) 5 {x, g1xg121, g2xg221, …, gkxgk21}. If x21 5 gi xgi21, then for each gj xgj21 in, cl(x), we have (gj xgj21)21 5 gj x21gj21 5 gj (gixgi21)gj21 [ cl(x). Because |G| has odd order,, gj xgj21 Z (gj xgj21)21. It follows that |cl(x)| is even. But |cl(x)| divides |G|., 9. Part a is not possible by the corollary of Theorem 24.2. Part b is not possible because it implies, that the center would have order 2, and 2 does not divide 21. Part c is the class equation for D5., Part d is not possible because of Corollary 1 of Theorem 24.1., 11. Use Theorem 7.2., 13. Use Example 5 of Chapter 9 and Theorem 7.2., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A36, , Selected Answers, , 15. By Lagrange we know that |H > K| 5 1, 3, 5 or 15. The first case is trivial. Corollary 3 of, Theorem 7.1 handles the next two cases. Theorem 24.6 takes care of the last case., 17. 8, 19. 15, 21. By Exercise 20, G has seven subgroups of order 3., 23. 10; k(123)l, k(234)l, k(134)l, k(345)l, k(245)l, 25. A group of order 100 has 1, 5, or 25 subgroups of order 4; exactly one subgroup of order 25, (which is normal); at least one subgroup of order 5; and at least one subgroup of order 2., 27. By the corollary of Theorem 24.5, it suffices to show that there is only one Sylow p-subgroup., By Sylow's Third Theorem, np, the number of Sylow p-subgroups of G, has the form 1 1 kp and, divides m. But if k . 0, 1 1 kp . p $ m. So k 5 0., 29. By Sylow's Third Theorem and Exercises 5 and 28, we know |D2m: N(K)| 5 m. Also, |D2m:K| 5 m., Thus we have |N(K)| 5 |K|, and since K 8 N(K), we are done., 31. 21, 33. Sylow’s Third Theorem implies that the Sylow 3- and Sylow 5-subgroups are unique. Pick any, x not in the union of these. Then |x| 5 15., 35. By Sylow’s Third Theorem, n17 5 1 or 35. Assume n17 5 35. Then the union of the Sylow, 17-subgroups has 561 elements. By Sylow's Third Theorem, n5 5 1. Thus, we may form a, cyclic subgroup of order 85 (Exercise 57 in Chapter 9 and Theorem 24.6). But then there are, 64 elements of order 85. This gives too many elements., 37. Use the G/Z Theorem (Theorem 9.3)., 39 . Let H be the Sylow 3-subgroup and suppose that the Sylow 5-subgroups are not normal. By, Sylow, there must be six Sylow 5-subgroups, call them K1, . . . , K6. These subgroups have, 24 elements of order 5. Also, each of the cyclic subgroups HK1, . . . , HK6 has eight generators., Thus, there are 48 elements of order 15, which results in more than 60 elements in the group., 41. By Theorem 24.2 and Theorem 9.5, Z(G) has an element x of order p. By induction, the group, G/kxl has normal subgroups of order pk for every k between 1 and n 2 1, inclusively. Now use, Exercise 51 in Chapter 9 and Exercise 51 of Chapter 10., 43. Pick x [ Z(G) such that |x| 5 p. If x [ H, by induction, N(H/kxl) . H/kxl, say ykxl [ N(H/kxl), but not H/kxl. Now show y [ N(H) but not H. If x o H, then x [ N(H), so that N(H) . H., 45. The hypotheses and G/Z(G) Theorem (Theorem 9.3) precludes all but |Z(G)| 5 1, p, q and r. If, |Z(G)| 5 p, then |G/Z(G)| 5 qr and by Theorems 24.6 and 9.3 G/Z(G) is cyclic and G is Abelian., The same contradiction occurs if |Z(G)| 5 r. So, |Z(G)| 5 1 or q. If also p does not divide r 2 1, then |Z(G)| 5 1., 47. Sylow's Third Theorem shows that all the Sylow subgroups are normal. Then Theorem 7.2 and, Example 5 of Chapter 9 ensure that G is the internal direct product of it Sylow subgroups. G is cyclic because of Theorem 9.6 and Corollary 1 of Theorem 8.2. G is Abelian because of Theorem 9.6, and Exercise 4 in Chapter 8., 49. Automorphisms preserve order., 51. That |N (H)| 5 |N (K )| follows directly from the last part of Sylow's Third Theorem and Exercise 5., 53. Normality of H implies cl(h) # H for h in H. Now observe that h [ cl(h). This is true only when, H is normal., 55. The mapping from H to xHx21 given by h → xhx21 is an isomorphism., 57. Suppose that G is a group of order 12 that has nine elements of order 2. By the Sylow theorems,, G has three Sylow 2-subgroups whose union contains the identity and the nine elements of order, 2. If H and K are both Sylow 2-subgroups, then by Theorem 7.2, |H y K| 5 2. Thus, the union, of the three Sylow 2-subgroups has at most seven elements of order 2, since there are three in H,, two more in K that are not in H, and at most two more that are in the third but not in H or K., 59. First note that p divides n 1 1 is equivalent to n mod p 5 2 1. Let H be the unique Sylow, p-subgroup of G and let |H| 5 pm. By Theorem 24.4, all the elements whose order is a power of, p are contained in H. Since every element of G whose order is a power of p lies in H, we have that, n 5 pm 2 (number of elements of order pm 1 number of elements of order pm21 1 . . . 1 number, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , 61., 63., , 65., , 67., , 69., , 71., 73., 75., , A37, , of elements of order p2 1 number of elements of order p0). By the corollary of Theorem 4.4 and, the fact that f(pk) is divisible by p for each k $ 2 (see page 167), each term on the right is divisible, by p except p0 5 1. Thus, n mod p 5 2 1., Mimic Example 7., Say cl(e) and cl(a) are the only two conjugacy classes of a group G of order n. Then cl(a) has n 2 1, elements all of the same order, say m. If m 5 2, then it follows from Exercise 47 in Chapter 2, that G is Abelian. But then cl(a) 5 {a} and so n 5 2. If m . 2, then cl(a) has at most n 2 2, elements, since conjugation of a by e, a, and a2 each yields a., Note that any subgroup of order 4 in a group of order 4m where m is odd is a Sylow 2-subgroup., By Sylow’s Third Theorem, the Sylow 2-subgroups are conjugate and therefore isomorphic. S4, contains both the subgroups k(1234)l and {(1), (12), (34), (12) (34)}., By Sylow’s Third Theorem, the number of Sylow 13-subgroups is equal to 1 mod 13 and, divides 55. This means that there is only one Sylow 13-subgroup, so it is normal in G. Thus, |N(H)/C(H)| 5 715/|C(H)| divides both 55 and 12. This forces 715/|C(H)| 51 and therefore, C(H) = G. This proves that H is contained in Z(G). Applying the same argument to K, we get, that K is normal in G and |N(K)/C(K)| 5 715/|C(K)| divides both 65 and 10. This forces, 715/|C(K)| 5 1 or 5. In the latter case, K is not contained in Z(G)., Let H be a Sylow 5-subgroup. Since the number of Sylow 5-subgroups is 1 modulo 5 and divides, 7 . 17, the only possibility is 1. So, H is normal in G. Then by the N/C Theorem (Example 15 of, Chapter 10), 冟G/C1H2冟 divides both 4 and 冟G冟. Thus C1H2 G., This follows directly from Theorem 24.1., Pr(D4) 5 5/8; Pr(S3) 5 1/2; Pr(A4) 5 1/3, By the discussion in this chapter, Pr(G % H) 5 m/(|G||H|), where m is the number of conjugacy, classes in G % H. Next prove that |C(g, h)| 5 |C(g)||C(h)|. Then use Theorem 24.1., , Chapter 25, Sweet are the uses of adversity., william shakespeare, As You Like It, , 1. Use the 2 ? Odd Test., 3. Use the Index Theorem., 5. Suppose G is a simple group of order 525. Let L7 be a Sylow 7-subgroup of G. It follows from, Sylow’s theorems that |N(L 7)| 5 35. Let L be a subgroup of N(L 7) of order 5. Since N(L 7) is, cyclic (Theorem 24.6), N(L) $ N(L 7), so that 35 divides |N(L)|. But L is contained in a Sylow, 5-subgroup (Theorem 24.4), which is Abelian (see the corollary to Theorem 24.2). Thus, 25, divides |N(L)| as well. It follows that 175 divides |N(L)|. The Index Theorem now yields a, contradiction., 7. n11 5 12. Use the N/C Theorem (Example 15 in Chapter 10) to show that there is an element of, order 22; then use the Embedding Theorem and observe that A12 has no element of order 22., 9. Suppose that there is a simple group of order 396 and L11 is a Sylow 11-subgroup. Use the N/C, Theorem given in Example 15 of Chapter 10 to show that C(L11) has an element of order 33,, whereas A12 does not., 11. If we can find a pair of distinct Sylow 2-subgroups A and B such that |A > B| 5 8, then, N(A > B) $ AB, so that N(A > B) 5 G. Now let H and K be any distinct pair of Sylow 2-subgroups., Then 16 ? 16/|H > K| 5 |HK| # 112 (Theorem 7.2), so that |H > K| is at least 4. If |H > K| 5 8,, we are done. So, assume |H > K| 5 4. Then N(H > K) picks up at least 8 elements from H and, at least 8 from K (see Exercise 43 in Chapter 24). Thus, |N(H > K)| $ 16 and is divisible by 8. So,, |N(H > K)| 5 16, 56, or 112. Since the latter two cases yield normal subgroups, we may assume, |N(H > K)| 5 16. If N(H > K) 5 H, then |H > K| 5 8, since N(H > K) contains at least 8 elements, from K. So, we may assume that N(H > K) 2 H. Then, we may take A 5 N(H > K) and B 5 H., 15. Use the Index Theorem., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A38, , Selected Answers, , 17. n5 5 6 and n3 5 10 or 40 (n3 5 4 is ruled out by the Index Theorem). If there are two Sylow, 2-subgroups L2 and L29 whose intersection has order 4, show that N(L2 > L29) has index at most 5., Now use the Embedding Theorem. If n35 40, the union of all the Sylow subgroups has more than, 120 elements. If n35 10, use the N/C Theorem to show that there is an element of order 6 and then, use the Embedding Theorem and observe that A6 has no element of order 6., 19. Let a be as in the proof of the Generalized Cayley Theorem. Then Ker a # H and |G/Ker a|, divides |G:H|!. Now show |Ker a| 5 |H|. A subgroup of index 2 is normal., 21. Since A5 is simple, if H is a proper normal subgroup of S5, then H > A5 5 A5 or {e}. But H > A5, 5 A5 implies H 5 A5, whereas H > A5 5 {e} implies H 5 {e} or |H| 5 2. (See Exercise 23 in, Chapter 5.) Now use Exercise 72 in Chapter 9 and Exercise 66 in Chapter 5., 23. By direct computation, show that PSL(2, Z7) has more than four Sylow 3-subgroups, more than, one Sylow 7-subgroup, and more than one Sylow 2-subgroup. Hint: Observe that c, , 25., 27., , 29., , 31., , 1 4, d has, 1 5, 5 5, order 3. Now use conjugation to find four other subgroups of order 3; observe that ` c, d ` 7, 1 4, 5 1, and use conjugation to find another subgroup of order 7; observe that ` c, d ` 4 and use, 3 5, conjugation to find six more elements of order 4 (which guarantees that more than one Sylow, 2-subgroup exists). Now argue as we did to show that A5 is simple. In the cases that the supposed, normal subgroup N has order 2 or 4, show that in G/N, the Sylow 7-subgroup is normal. But then,, G has a normal subgroup of order 14 or 28, which were already ruled out., Mimic Exercise 24., Suppose there is a simple group of order 60 that is not isomorphic to A5. The Index Theorem, implies n2 2 1 or 3, and the Embedding Theorem implies n2 2 5. Thus, n2 5 15. Counting shows, that there must be two Sylow 2-subgroups whose intersection has order 2. Now mimic the argument used in showing that there is no simple group of order 144 to show that the normalizer of, this intersection has index 5, 3, or 1, but the Embedding Theorem and the Index Theorem rule, these out., Suppose there is such a simple group G. Since the number of Sylow q-subgroups is 1 modulo q, and divides p2, it must be p2. Thus there are p2 1q 12 elements of order q in G. These elements,, together with the p2 elements in one Sylow p-subgroup, account for all p2q elements in G. Thus,, there cannot be another Sylow p-subgroup. But then the Sylow p-subgroup is normal in G., Consider the right regular representation of G. Let g be a generator of the Sylow 2-subgroup and, suppose that 冟G冟 2kn where n is odd. Then by Exercise 62 in Chapter 6, every cycle of the permutation Tg in the right regular representation of G has length 2k. This means that there are exactly n such cycles. Since each cycle is odd and there is an odd number of them, Tg is odd. This, means that the set of even permutations in the regular representations has index 2 and is therefore, normal. (See Exercise 23 in Chapter 5 and Exercise 9 in Chapter 9)., , Chapter 26, If you make a mistake, make amends., lou holtz, , 1. u is related to u because u is obtained from itself by no insertions; if v can be obtained from u, by inserting or deleting words of the form xx21 or x21x, then u can be obtained from v by reversing the procedure; if u can be obtained from v and v can be obtained from w, then u can be, obtained from w by obtaining first v from w and then u from v., 3. b(a2N) 5 b(aN)a 5 a3bNa 5 a3b(aN), 5 a3a3bN, 5 a6bN 5 a6Nb 5 a2Nb 5 a2bN, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , 5., , 7., 9., 11., 13., 15., , 17., 19., 21., 23., 25., , 27., , A39, , b(a3N) 5 b(a2N)a 5 a2bNa 5 a2b(aN), 5 a2a3bN, 5 a5bN 5 a5Nb 5 aNb 5 abN, b(bN) 5 b2N 5 N, b(abN) 5 baNb 5 a3bNb 5 a3b2N 5 a3N, b(a2bN) 5 ba2Nb 5 a2bNb 5 a2b2N 5 a2N, b(a3bN) 5 ba3Nb 5 abNb 5 ab2N 5 aN, Let F be the free group on {a1, a2, . . . , an}. Let N be the smallest normal group containing, {w1, w2, . . . , wt} and let M be the smallest normal subgroup containing {w1, w2, . . . , wt,, wt11, . . . , wt1k}. Then F/N < G and F/M < G. The homomorphism from F/N to F/M given by, aN → aM induces a homomorphism from G onto G. To prove the corollary, observe that the theorem shows that K is a homomorphic image of G, so |K| # |G|., Clearly, a and ab belong to ka, bl, so ka, abl # ka, bl. Now show that a and b belong to ka, abl., Show that |G| # 2n and that Dn satisfies the relations that define G., Since x2 5 y2 5 e, we have (xy)21 5 y21x21 5 yx. Also, xy 5 z21yz, so (xy)21 5 (z21yz)21 5, z21y21z 5 z21yz 5 xy., a. b6 b. b7a, Note that yxyx3 5 e implies that yxy21 5 x5 and therefore kxl is normal. So, G 5 kxl x ykxl and, |G| # 16. Use y2 5 e and yxyx3 5 e, to prove that x2 [ Z(G). Then prove G is not Abelian and, use Theorem 9.3 to show that |Z(G)| 2 8. Thus, Z(G) 5 kx2l. Finally, prove that (xy)2 5 x22,, so that |xy| 5 8., Use the fact that the mapping from G onto G/N given by x → xN is a homomorphism., For H to be a normal subgroup we must have yxy21 [ H 5 {e, y3, y6, y9, x, xy3, xy6, xy9}. But, yxy21 5 yxy11 5 (yxy)y10 5 xy10., 6; the given relations imply that a2 5 e. G is isomorphic to Z6., 1, 2, and `, ab 5 c 1 abc21 5 e, cd 5 a 1 (abc21)cd 5 ae 1 bd 5 e 1 d 5 b21, da 5 b 1 bda 5 b2 1 ea 5 b2 1 a 5 b2, ab 5 c 1 b3 5 c, So G 5 kbl., bc 5 d 1 bb3 5 b21 1 b5 5 e. So |G| 5 1 or 5., But Z5 satisfies the defining relations with a 5 1, b 5 3, c 5 4, and d 5 2., Z6, , Chapter 27, If at first you don’t succeed—that makes you about average., bradenton, [Florida] Herald, , 1. If T is a distance-preserving function and the distance between points a and b is positive, then the, distance between T(a) and T(b) is positive., 3. See Figure 1.5., 5. 12, 7. 4n, 9. a. Z2, b. Z2 % Z2, c. G % Z2, where G is the plane symmetry group of a circle (see Exercise, 55 of Chapter 3)., 11. 6, 13. An inversion in R3 leaves only a single point fixed, whereas a rotation leaves a line fixed., 15. In R4, a plane is fixed. In Rn, a hyperplane of dimension n 2 2 is fixed., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A40, , Selected Answers, , 17. Create a coordinate system for the plane. Let T be an isometry; p, q, and r the three noncollinear, points; and s any other point in the plane. Then the quadrilateral determined by T( p), T(q), T(r),, and T(s) is congruent to the one formed by p, q, r, and s. Thus, T(s) is uniquely determined by, T ( p), T(q), and T (r)., 19. a rotation, , Chapter 28, The thing that counts is not what we know but the ability to use what we know., leo l. spears, , Try, → (n, m)., xy, Use Figure 28.9., x2yzxz 5 x2yx21 5 x2x21y 5 xy; x23zxyz 5 x23x21y 5 x24y, A subgroup of index 2 is normal., a. V b. I c. II d. VI e. VII f. III, cmm, a. p4m b. p3 c. p31m d. p6m, The principal purpose of tire tread design is to carry water away from the tire. Patterns I and III do, not have horizontal reflective symmetry. Thus, these designs would not carry water away equally, on both halves of the tire., 19. a. VI b. V c. I d. III e. IV f. VII g. IV, 1., 3., 5., 7., 9., 11., 13., 15., 17., , xnym, , Chapter 29, With every mistake we must surely be learning., george harrison, “While My Guitar Gently Weeps,” The White Album, , 1., 3., 5., 7., 9., , 6, 30, 13, 45, 126, 1, 11. 1n6 2 ? n 2 ? n2 n3 2, 6, 13. For the first part, see Exercise 11 in Chapter 6. For the second part, try D4., 15. R0, R180, H, V act as the identity and R90, R270, D, D9 interchange L1 and L2., , Chapter 30, I am not bound to please thee with my answers., shakespeare, The Merchant of Venice, , 1., 3., 5., 7., 11., , 4 * (b, a), (m/2) * {3 * [(a, 0), (b, 0)], (a, 0), (e, 1), 3 * (a, 0), (b, 0), 3 * (a, 0), (e, 1)}, a3b, Both yield paths from e to a3b., Say we start at x. Then we know the vertices x, xs1, xs1s2, . . . , xs1s2 . . . sn21 are distinct and, x 5 xs1s2 . . . sn. So if we apply the same sequence beginning at y, then cancellation shows that, y, ys1, ys1s2, . . . , ys1s2 . . . sn21 are distinct and y 5 ys1s2 . . . sn., 13. If there were a Hamiltonian path from (0, 0) to (2, 0), there would be a Hamiltonian circuit in the, digraph, since (2, 0) 1 (1, 0) 5 (0, 0). This contradicts Theorem 30.1., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Selected Answers, , A41, , 15. a. If s1, s2, . . . , sn21 traces a Hamiltonian path and sisi11 . . . sj 5 e, then the vertex s1s2 . . . si21, appears twice. Conversely, if sisi11 . . . sj 2 e, then the sequence e, s1, s1s2, . . . , s1s2 . . . sn21, yields the n vertices (otherwise, cancellation gives a contradiction)., b. This follows directly from part a., 17. The sequence traces the digraph in a clockwise fashion., 19. Abbreviate (a, 0), (b, 0), and (e, 1) by a, b, and 1, respectively. A circuit is 4 * (4 * 1, a), 3 * a, b, 7, * a, 1, b, 3 * a, b, 6 * a, 1, a, b, 3 * a, b, 5 * a, 1, a, a, b, 3 * a, b, 4 * a, 1, 3 * a, b, 3 * a, b, 3 * a, b., 21. Abbreviate (R90, 0), (H, 0), and (R0, 1) by R, H, and 1, respectively. A circuit is 3 * (R, 1, 1), H,, 2 * (1, R, R), R, 1, R, R, 1, H, 1, 1., 23. Abbreviate (a, 0), (b, 0), and (e, 1) by a, b, and 1, respectively. A circuit is 2 * (1, 1, a), a, b,, 3 * a, 1, b, b, a, b, b, 1, 3 * a, b, a, a., 25. Abbreviate (r, 0), ( f, 0), and (e, 1) by r, f, and 1, respectively. Then the sequence is r, r, f, r, r, 1,, f, r, r, f, r, 1, r, f, r, r, f, 1, r, r, f, r, r, 1, f, r, r, f, r, 1, r, f, r, r, f, 1., 27. m * [(n 2 1) * (0, 1), (1, 1)], 29. Abbreviate (r, 0), ( f, 0), and (e, 1) by r, f, and 1, respectively. A circuit is 1, r, 1, 1, f, r, 1, r, 1, r, f, 1., 31. 5 * [3 * (1, 0), (0, 1)], (1, 0), 33. 12 * [(1, 0), (0, 1)], 35. In the proof of Theorem 30.3, we used the hypothesis that G is Abelian in two places: We needed, H to satisfy the induction hypothesis, and we needed to form the factor group G/H. Now, if we, assume only that G is Hamiltonian, then H also is Hamiltonian and G/H exists., , Chapter 31, We must view with profound respect the infinite capacity of the human mind to resist the, introduction of useful knowledge., thomas r. lounsbury, , 1., 3., 5., 7., 9., , wt(0001011) 5 3; wt(0010111) 5 4; wt(0100101) 5 3; etc., 1000110; 1110100, 000000, 100011, 010101, 001110, 110110, 101101, 011011, 111000, By using t = 1/2 in the proof of Theorem 31.2 we have that all single errors can be detected., Observe that a vector has even weight if and only if it can be written as a sum of an even number, of vectors of weight 1., 11. No, by Theorem 31.3., 13. 0000000, 1000111, 0100101, 0010110, 0001011, 1100010, 1010001, 1001100, 0110011,, 0101110, 0011101, 1110100, 1101001, 1011010, 0111000, 1111111;, , 1 1 1, 1 0 1, 1 1 0, H 5 G0 1 1 W ;, 1 0 0, 0 1 0, 0 0 1, yes., 15. Suppose that u is decoded as v and that x is the coset leader of the row containing u. Coset decoding, means v is at the head of the column containing u. So, x 1 v 5 u and x 5 u 2 v. Now suppose u 2 v, is a coset leader and u is decoded as y. Then y is at the head of the column containing u. Since v is a, code word, u 5 u 2 v 1 v is in the row containing u 2 v. Thus, u 2 v 1 y 5 u and y 5 v., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A42, , Selected Answers, , 17. 000000, 100110, 010011, 001101, 110101, 101011, 011110, 111000;, 1, 0, 1, HG, 1, 0, 0, , 19., , 21., 23., 25., 27., 29., , 1, 1, 0, 0, 1, 0, , 0, 1, 1, W., 0, 0, 1, , 001001 is decoded as 001101 by all four methods., 011000 is decoded as 111000 by all four methods., 000110 is decoded as 100110 by all four methods., Since there are no code words whose distance from 100001 is 1 and three whose distance is 2,, the nearest-neighbor method will not decode or will arbitrarily choose a code word; parity-check, matrix decoding does not decode 100001; the standard-array and syndrome methods decode, 100001 as 000000, 110101, or 101011, depending on which of 100001, 010100, or 001010 is a, coset leader., For any received word w, there are only eight possibilities for wH. But each of these eight possibilities satisfies condition 2 or the first portion of condition 39 of the decoding procedure, so, decoding assumes that no error was made or one error was made., There are 34 code words and 36 possible received words., No; row 3 is twice row 1., No. For if so, nonzero code words would be all words with weight at least 5. But this set is not, closed under addition., Use Exercise 24, together with the fact that the set of code words is closed under addition., Abbreviate the coset a 1 kx2 1 x 1 1l with a. The following generating matrix will produce the, desired code:, , c, , 1 0 1, 1, x, d., 0 1 x x1 x1, , 31. Use Exercise 14., 33. Let c, c9 [ C. Then, c 1 (v 1 c9) 5 v 1 c 1 c9 [ v 1 C and (v 1 c) 1, (v 1 c9) 5 c 1 c9 [ C, so the set C < (v 1 C) is closed under addition., 35. If the ith component of both u and v is 0, then so is the ith component of u 2 v and au, where a is, a scalar., , Chapter 32, Wisdom rises upon the ruins of folly., thomas fuller, Gnomologia, , 1. Note that f(1) 5 1. Thus f(n) 5 n. Also, 1 5 f(1) 5 f(nn21) 5 f(n)f(n21) 5 nf(n21), so that, 1/n 5 f(1/n)., 3. If a and b are automorphisms of E fixing F, so are a21 and ab., 5. If a and b are fixed by elements of H, so are a 1 b, a 2 b, a ? b, and a/b., 7. It suffices to show that each member of Gal(K/F) defines a permutation on the ai’s. Let, a [ Gal(K/F) and write, f(x) 5 cnxn 1 cn21xn21 1 . . . 1 c0, 5 cn(x 2 a1)(x 2 a2) . . . (x 2 an)., Then f(x) 5 a(f(x)) 5 cn(x 2 a(a1))(x 2 a(a2)) ? ? ? (x 2 a(an)). Thus, f(ai) 5 0 implies, ai 5 a(aj) for some j, so that a permutes the ai’s., Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A43, , Selected Answers, , 9. Observe that f6(v) 5 v729 5 v whereas f3(v) 5 v27 5 v21 and f2(v) 5 v9 5 v2. f3(v 1, v21) 5 v27 1 v227 5 v21 1 v; f2(v3 1 v5 1 v6) 5 v27 1 v45 1 v54 5 v6 1 v3 1 v5., 11. a. Z20 % Z2 has three subgroups of order 10. b. 25 does not divide 40, so there are none., c. Z20 % Z2 has one subgroup of order 5., 13. The splitting field over R is R (23 ). The Galois group is the identity and the mapping a 1 b, 23 S a 2 b23 ., 15. Use Theorem 23.3., 17. Recall that A4 has no subgroup of order 6. (See Example 16 in Chapter 9.), 19. Use Sylow’s First Theorem., 3, 3, 3, 21. Let v be a primitive cube root of 1. Then Q , Q(2 2) , Q(v,2 2 and Q(2 2) is not the splitting, field of a polynomial in Q[x]., 23. Use the lattice of Z10., 25. Z6 (Be sure you know why the group is cyclic.), 27. See Exercise 21 in Chapter 25., 29. Use Exercise 41 in Chapter 24., 31. Use Exercise 42 in Chapter 10., 33. Since K/N v G/N, for any x [ G and k [ K, there is a k9 [ K such that k9N 5 (xN)(kN)(xN)21 5, xNkNx21N 5 xkx21N. So, xkx21 5 k9n for some n [ N. And since N # K, we have k9n [ K., , Chapter 33, All wish to possess knowledge, but few, comparatively speaking, are willing to pay the price., juvenal, , 1. x 2 2 x 1 1, 3. Over Z, x8 2 1 5 (x 2 1)(x 1 1)(x2 1 1)(x4 1 1). Over Z2, x2 1 1 5 (x 1 1)2 and, x4 1 1 5 (x 1 1)4. So, over Z2, x8 2 1 5 (x 1 1)8. Over Z3, x2 1 1 is irreducible, but x4 1 1 factors into irreducibles as (x2 1 x 1 2)(x2 2 x 2 1). So, x8 2 1 5 (x 2 1)(x 1 1) (x2 1 1)(x2 1, x 1 2)(x2 2 x 2 1). Over Z5, x2 1 1 5 (x 2 2)(x 1 2), x4 1 1 5 (x2 1 2)(x2 2 2), and these last, two factors are irreducible. So, x8 2 1 5 (x 2 1)(x 1 1)(x 2 2)(x 1 2)(x2 1 2)(x2 2 2)., 5. Let v be a primitive nth root of unity. We must prove vv2 . . . vn 5 (21)n11. Observe that, vv2 . . . vn 5 vn(n11)/2. When n is odd, vn(n11)/2 5 (vn)(n11)/2 5 1(n11)/2 5 1. When n is even,, (vn/2)n11 5 (21)n11 5 21., 7. If [F:Q] 5 n and F has infinitely many roots of unity, then there is no finite bound on their, multiplicative orders. Let v be a primitive mth root of unity in F such that f(m) . n. Then, [Q(v):Q] 5 f(m). But F $ Q(v) $ Q implies [Q(v):Q] # n., 9. Let 2n 1 1 5 q. Then 2 [ U(q) and 2n 5 q 2 1 5 21 in U(q) implies that |2| 5 2n. So, by, Lagrange’s Theorem, 2n divides |U(q)| 5 q 2 1 5 2n., 11. Let v be a primitive nth root of unity. Then 2nth roots of unity are 61, 6v, . . . , 6vn21. These, are distinct, since 21 5 (2vi)n, whereas 1 5 (v i)n., 13. First observe that deg F2n(x) 5 f(2n) 5 f(n) and deg Fn(2x) 5 deg Fn(x) 5 f(n). Thus, it suffices to show that every zero of Fn(2x) is a zero of F2n(x). But the fact that v is a zero of, Fn(2x) means that |2v| 5 n, and because n is odd, this implies that |v| 5 2n., 15. Let G 5 Gal(Q(v)/Q) and H1 be the subgroup of G of order 2 that fixes cos(2p, n ). Then, by, induction, G/H1 has a series of subgroups H1/H1 , H2/H1 , . . . , Ht /H1 5 G/H1, so that, |Hi11/H1:Hi /H1| 5 2. Now observe that |Hi11/H1:Hi /H1| 5 |Hi11/Hi|., 17. Instead, prove that Fn(x)Fpn(x) 5 Fn(xp). Since both sides are monic and have degree pf(n), it, suffices to show that every zero of Fn(x)Fpn(x) is a zero of Fn(xp). If v is a zero of Fn(x), then, |v| 5 n. By Theorem 4.2, |vp| 5 n also. Thus, v is a zero of Fn(xp). If v is a zero of Fpn(x), then, |v| 5 pn and therefore |v p| 5 n., 19. Use Theorem 33.4 and Theorem 32.1., 21. v → v4, v → v21, v → v24, Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A44, , Selected Answers, , Supplementary Exercises for Chapters 24–33, For those who keep trying, failure is temporary., frank tyger, , 1. Z6, 3. Let |G| 5 315 and let H be a Sylow 3-subgroup and K a Sylow 5-subgroup. If H v G, then, HK 5 45. If H is not normal, then by Sylow’s Third Theorem, |G/N(H)| 5 7, so that |N(H)| 5 45., 5. Observe that K # N(H) implies that HK is a group of order 245. Now, use Sylow’s Third Theorem., 7. Note that gKg21 # gHg21 5 H. Now use the corollary to Sylow’s Third Theorem., 9. Use the same proof as for Example 5 in Chapter 9., 11. Since n7 5 8, we know by the Embedding Theorem (Chapter 25) that G # A8. But A8 does not, have an element of order 21., 13. Let G be a non-Abelian group of order 105. By Theorem 9.3, G/Z(G) is not cyclic. So |Z(G)| 2 3,, 7, 15, 21, or 35. This leaves only 1 or 5 for |Z(G)|. Let H, K, and L be Sylow 3-, Sylow 5-, and, Sylow 7-subgroups of G, respectively. Now, counting shows that K v G or L v G. Thus,, |KL| 5 35 and KL is a cyclic subgroup of G. But, KL has 24 elements of order 35 (since, |U(Z35)| 5 24). Thus, a counting argument shows that K v G and L v G. Now, |HK| 5 15 and HK, is a cyclic subgroup of G. Thus, HK # C(K ) and KL # C(K ). This means that 105 divides, |C(K )|. So K # Z(G)., 15., , 17. It suffices to show that x travels by a implies xab21 travels by a (for we may successively replace, x by xab21). If xab21 traveled by b, then the vertex xa would appear twice in the circuit., 19. a. {00, 11}, b. {000, 111}, c. {0000, 1100, 1010, 1001, 0101, 0110, 0011, 1111}, d. {0000, 1100, 0011, 1111}, 21. The mapping Tv: F n → {0, 1} given by Tv(u) 5 u ? v is an onto homomorphism. So |F n/Ker Tv| 5 2., 23. It follows from Exercise 18 that if C is an (n, k) linear code, then C> is an (n, n 2 k) linear code., Thus, in this problem, k 5 n 2 k. To prove the second claim, use Exercise 18, Exercise 21, the, definition of C>, and the hypothesis that C> 5 C., , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Index of Mathematicians, , (Biographies appear on pages in boldface.), , Abbati, Pietro, 147, Abel, Niels, 34, 41, 331, 427, 560, Adleman, Leonard, 170, 171, 172, 180, Allenby, R., 127, Artin, Emil, 327, 358, 359, Artin, Michael, 358, Aschbacher, Michael, 431, 432, 439,, 442, 460, Bell, E. T., 389, Berlekamp, Elwyn, 544, Bhargava, Manjul, 580, Bieberbach, L., 484, Birkhoff, Garrett, 310, Boole, George, 253, Brauer, Richard, 429, 439, Burnside, William, 429, 496, 505, 569, Cauchy, Augustin Louis, 102, 108,, 126, 195, 207, 271, 331, 360, 414, Cayley, Arthur, 33, 93, 94, 99, 131,, 143, 388, 452, 506, Chevalley, Claude, 439, Cole, Frank, 429, 438, Conway, John H., 432, 494, Courant, Richard, 359, Crowe, Donald, 475, Davenport, Harold, 460, Davidson, Morley, 155, da Vinci, Leonardo, 463, Dedekind, Richard, 245, 279, 389, 579, de Fermat, Pierre, 149, 331, 332, De Morgan, Augustus, 14, 93, de Séguier, J. A., 66, Dethridge, John, 155, Dickson, L. E., 398, 429, 439, Dirichlet, Peter, 331, 579, Dyck, Walther, 42, 450, , Eisenstein, Ferdinand, 315, 579, " s, Paul, 525, Erdo, Escher, M. C., 492, 493, 517–519, Euclid, 5, 399, Euler, Leonhard, 41, 46, 161, 331,, 345, 376, 525, Feit, Walter, 429, 432, 439, 444,, 505, 561, Fields, John, 438, Fischer, Bernd, 432, Fourier, Joseph, 207, Fraenkel, Abraham, 245, Frobenius, Georg, 181, 414, 496, Galois, Évariste, 42, 127, 144,, 161, 185, 207, 375, 388, 389,, 428, 429, 437, 560, 561, 564,, 575, 577, Gauss, Carl, 41, 117, 126, 167, 279,, 313, 315, 319, 331, 340, 345, 384,, 389, 400, 570, 573, 576, 577, 579,, 580, Germain, Sophie, 345, Gersonides, 287, Gorenstein, Daniel, 431, 432, 443,, 542, Griess, Robert, 432, Hall, Marshall, 442, 460, Hall, Philip, 460, 569, Hamilton, William Rowan, 203,, 510, 524, Hamming, Richard, 528, 529,, 544, 550, Hardy, G. H., 460, Hermite, Charles, 376, Herstein, I. N., 144, 254, 443, A45, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A46, , INDEX OF MATHEMATICIANS, , Hilbert, David, 245, 280, 331, 358,, 359, 483, Hölder, Otto, 136, 187, 225, 428, 438, Holst, Elling, 427, , Pascal, Blaise, 14, Pless, Vera, 552, Poincaré, Henri, 375, 459, Pólya, George, 492, 493, , Jacobson, Nathan, 266, Jordan, Camille, 187, 208, 215, 225,, 389, 427, 428, 429, 437, , Rankin, R. A., 517, Reed, Irving, 543, 544, Riemann, Bernhard, 579, Rivest, Ronald, 170, 171, 172, 180, Rokicki, Tomas, 155, Ruffini, Paolo, 106, 149, , Kaplansky, Irving, 387, Klein, Felix, 280, 437, 438, Kline, Morris, 279, Knuth, Donald, 442, 460, Kociemba, Herbert, 155, Kronecker, Leopold, 226, 331,, 360, 375, Kummer, Ernst, 331, 332, 375, 579, Lagrange, Joseph Louis, 41, 126,, 147, 161, 207, 345, Lamé, Gabriel, 331, 332, Landau, Edmund, 332, Lang, Serge, 327, Laplace, Pierre-Simon, 126, 161, Larson, Loren, 113, Legendre, Adrien-Marie, 207, 331, Lie, Sophus, 427, Lindemann, Ferdinand, 366, 375,, 376, 400, Liouville, Joseph, 376, Mac Lane, Saunders, 310, MacWilliams, Jessie, 551, Mathieu, Emile, 429, Maurolico, Francisco, 14, McEliece, Robert, 460, 543, 544, Miller, G. A., 144, 181, 429, Miyaoka, Yoichi, 332, Moore, E. H., 136, 389, Motzkin, T., 339, Netto, Eugen, 93, Newton, Isaac, 41, 161, 345, Noether, Emmy, 280, 336, 564, , Schattschneider, Doris, 475, Shamir, Adi, 170, 171, 172, 180, Shannon, Claude, 543, Singer, Richard, 337, Slepian, David, 539, Sloane, Neil, 551, Smith, Stephen, 432, Solomon, Gustave, 543, 544, Solomon, Ronald, 432, Steinitz, Ernst, 290, 381, 383, Sylow, Ludwig, 95, 412, 427, Sylvester, J. J., 93–94, 143, Taussky-Todd, Olga, 359, Taylor, Richard, 332, 346, Thompson, John G., 429, 431,, 432, 439, 443, 444, 460, 505,, 561, 564, 569, van der Waerden, B.L., 225, Verhoeff, J., 115, 116, Wantzel, Pierre, 400, Weber, Heinrich, 42, 46, 257, Weyl, Hermann, 280, 463, Wiles, Andrew, 332, 334, 346, 438,, 580, Zariski, Oscar, 443, Zelmanov, Efim, 505, Zierler, N., 542, Zorn, Max, 254, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Index of Terms, , Abel Prize, 41, 444, Abelian group, 34, 43, Addition modulo n, 7, Additive group of integers modulo n, 44, Algebraic, closure, 383, element, 376, extension, 376, Algebraically closed field, 383, 384, Alternating group, 110, Annihilator, 277, Arc, 506, Ascending chain condition, 335, 343, Associates, 328, Associativity, 34, 43, Automorphism(s), Frobenius, 395, group, 136, 517, group of E over F, 554, inner, 135, of a group, 135, Axioms, for a group, 43, for a ring, 245, for a vector space, 351, Basis for a vector space, 353, Binary, code, 531, operation, 42, strings, 168, Boolean ring, 253, Burnside’s Theorem, 497, Cancellation, property for groups, 50, property for integral domains, 256, Cauchy’s Theorem, 195, 414, Cayley digraph, 506, Cayley table, 33, Cayley’s Theorem, 131, generalized, 434, , Center, of a group, 66, of a ring, 251, Centralizer, of an element, 68, of a subgroup, 72, Characteristic of a ring, 258, Characteristic subgroup, 181, Check digit, 7, Check-digit scheme, 115, Chinese Remainder Theorem for, Rings, 347, Circle in F, 400, Class equation, 410, Closure, 33, 42, Code, binary, 531, dual of, 582, Hamming, 528, (n,k) linear, 531, self-dual, 582, systematic, 534, ternary, 532, word, 528, 531, Color graph, 507, Cole Prize, 327, 358, 398, 432, 438,, 439, 442, 444, 580, Commutative diagram, 216, Commutative operation, 34, Commutator subgroup, 181, Complex numbers, polar form, 13, standard form, 13, Composition factors, 428, Composition of functions, 21, Conjugacy class, 95, 409, Conjugate, elements, 409, subgroups, 95, 414, Conjugation, 131, Constant polynomial, 300, Constructible number, 400, A47, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A48, , INDEX OF TERMS, , Constructible regular n-gons, 575, Content of a polynomial, 312, Coset, decoding, 539, leader, 539, left, 144, representative, 144, right, 144, Crystallographic groups, 475, Crystallographic restriction, 481, Cube, rotation group of, 154, Cycle, m-, 102, notation, 102, Cyclic, group, 77, rotation group, 36, subgroup, 65, Cyclotomic, extension, 571, polynomial, 316, 571, Decoding, coset, 539, maximum-likelihood, 526, nearest-neighbor, 529, parity-check matrix, 536, Degree, of a over F, 378, of an extension, 378, of a polynomial, 300, rule, 306, DeMoivre’s Theorem, 15, Derivative, 368, Determinant, 45, Diagonal of G % G, 175, Digital signatures, 172, Dihedral groups, 33, 34, Dimension of a vector space, 355, Direct product of groups, external, 162, internal, 196, 197, Direct sum, of groups, 199, of rings, 247, Dirichlet’s Theorem, 236, Discrete frieze group, 469, Distance between vectors, 532, Divides, 246, 303, , Division algorithm, for F[x], 301, for Z, 3, Divisor, 3, Domain, Euclidean, 337, integral, 255, Noetherian, 336, unique factorization, 334, Doubling the cube, 399, 401, Dual code, 582, Eisenstein’s Criterion, 315, Element(s), algebraic, 376, conjugate, 409, degree of, 378, fixed by f, 497, idempotent, 262, identity, 33, 43, 246, inverse, 33, 43, nilpotent, 261, order of, 60, primitive, 382, square, 202, transcendental, 376, Embedding Theorem, 435, Empty word, 446, Equivalence class, 18, Equivalence relation, 18, Equivalent under group action,, 495, Euclidean domain, 337, Euclid’s Lemma, 5, generalization of, 25, Euler phi function, 84, Even permutation, 110, Exponent of a group, 182, Extension, algebraic, 376, cyclotomic, 571, degree, 378, field, 360, finite, 378, infinite, 378, simple, 376, transcendental, 376, External direct product, 162, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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INDEX OF TERMS, , Factor, group, 187, of a ring element, 246, ring, 268, Factor Theorem, 303, Feit–Thompson Theorem, 429, 431,, 444, 505, 561, Fermat prime, 577, Fermat’s Last Theorem, 331–333, Fermat’s Little Theorem, 149, Field, algebraic closure of, 383, algebraically closed, 383, definition of, 256, extension, 360, fixed, 554, Galois, 389, of quotients, 290, perfect, 370, splitting, 362, Fields Medal, 431, 438, 444, 505, Finite dimensional vector space,, 355, Finite extension, 378, First Isomorphism Theorem, for groups, 215, for rings, 288, Fixed field, 554, Free group, 447, Frieze pattern, 469, Frobenius map, 294, 395, Function, composition, 21, definition of, 20, domain, 20, image under, 20, one-to-one, 21, onto, 22, range, 20, Fundamental region, 481, Fundamental Theorem, of Algebra, 384, of Arithmetic, 6, of Cyclic Groups, 82, of Field Theory, 360, of Finite Abelian Groups, 226, of Galois Theory, 558, of Group Homomorphisms, 215, of Ring Homomorphisms, 289, , A49, , GAP, 113, G/Z Theorem, 194, Galois, field, 389, group, 554, 566, Gaussian integers, 249, 338, Gauss’s Lemma, 313, Generating region of a pattern, 481, Generator(s), of a cyclic group, 65, 77, of a group, 49, in a presentation, 449, Geometric constructions, 399, Glide-axis, 462, Glide-reflection, 462, nontrivial, 472, trivial, 472, Greatest common divisor, 4, Group, Abelian, 34, 43, action, 501, alternating, 110, automorphism, 136, 517, automorphism of, 135, center of, 66, color graph of a, 507, commutative, 34, composition factors, 428, crystallographic, 475, cyclic, 36, 65, 77, definition, 43, dicyclic, 453, 458, dihedral, 33, 34, discrete frieze, 469, factor, 187, finite, 60, free, 447, frieze, 469, Galois, 554, 566, general linear, 45, generator(s), 49, 65, 77, 449, Hamiltonian, 522, Heisenberg, 58, homomorphism of, 208, icosahedral, 438, 465, infinite dihedral, 454, inner automorphism, 136, integers mod n, 44, isomorphic, 128, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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A50, , INDEX OF TERMS, , isomorphism, 128, non-Abelian, 34, 43, octahedral, 465, order of, 60, p-, 410, permutation, 99, presentation, 449, quaternions, 95, 201, 203, 450, quotient, 187, representation, 219, simple, 428, solvable, 561, space, 483, special linear, 48, symmetric, 101, symmetry, 35, 36, 461, tetrahedral, 465, of units, 251, wallpaper, 475, Half-turn, 471, Hamiltonian, circuit, 511, group, 522, path, 511, Hamming, code, 528, distance, 532, weight of a code, 532, weight of a vector, 532, Homomorphism(s), Fundamental Theorem of, 215, 289, kernel of, 208, of a group, 208, natural, 218, 289, of a ring, 285, Ideal, annihilator, 277, definition of, 267, finitely generated, 342, generated by, 268, maximal, 272, nil radical of, 277, prime, 272, principal, 268, product of, 275, proper, 267, sum of, 275, , test, 267, trivial, 268, Idempotent, 261, Identity element, 33, 43, 246, Imaginary axis, 13, Index of a subgroup, 148, Index Theorem, 434, Induction, First Principle of, 15, Second Principle of, 16, Inner automorphism, 135, Integral domain, 255, Internal direct product, 196, 197, International standard book number,, 26, Inverse element, 33, 43, Inverse image, 212, Inversion, 140, Irreducibility tests, 312, 314, Irreducible element, 328, Irreducible polynomial, 311, ISBN, 26, Isometry, 461, Isomorphism(s), class, 226, First Theorem for groups, 215, First Theorem for rings, 288, of groups, 128, of rings, 285, Second Theorem for groups, 222, Second Theorem for rings, 347, Third Theorem for groups, 222, Third Theorem for rings, 347, Kernel, of a homomorphism, 208, of a linear transformation, 357, Key, 169, Kronecker’s Theorem, 360, Lagrange’s Theorem, 147, Latin square, 56, Lattice, diagram, 86, of points, 481, unit, 481, Leading coefficient, 300, Least common multiple, 6, Left regular representation, 132, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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INDEX OF TERMS, , Line in F, 400, Linear, code, 531, combination, 353, transformation, 357, Linearly dependent vectors, 353, Linearly independent vectors, 353, Mapping, 20, Mathematical induction, First Principle, 15, Second Principle, 16, Matrix, addition, 44, determinant of, 45, multiplication, 45, standard generator, 534, Maximal,, ideal, 272, subgroup, 205, 240, Maximum-likelihood decoding, 526, Measure, 337, Minimal polynomial, 377, Mirror, 462, Mod p Irreducibility Test, 314, Modular arithmetic, 6–7, Monic polynomial, 300, Monster, 432, 564, Multiple, 3, Multiple zeros, 369, Multiplication modulo n, 7, Multiplicity of a zero, 303, Natural homomorphism, 218, 286, 289, Natural mapping, 216, N/C Theorem, 217, Nearest-neighbor decoding, 529, Nil radical, 277, Nilpotent element, 261, Noetherian domain, 336, Norm, 329, Normal subgroup, 185, Normal Subgroup Test, 186, Normalizer, 95, Odd permutation, 110, Operation, associative, 43, binary, 42, commutative, 34, , preserving mapping, 129, table, 33, Opposite isometry, 462, Orbit of a point, 151, Orbit-Stabilizer Theorem, 152, Order, of a group, 60, of an element, 60, Orthogonality relation, 538, Parity-check matrix, 536, Partition, of a set, 19, of an integer, 227, Perfect field, 370, Permutation, definition of, 99, even, 110, group, 99, odd, 110, p-group, 410, Phi function, Euler, 84, PID, 304, Plane of F, 400, Plane symmetry, 35, Polynomial(s), alternating, 110, constant, 300, content of, 312, cyclotomic, 316, 571, degree of, 300, derivative of, 368, Galois group of, 566, irreducible, 311, leading coefficient of, 300, minimal, 377, monic, 300, primitive, 312, reducible, 311, relatively prime, 308, ring of, 298, splits, 362, symmetric, 110, zero of, 303, Prime, element of a domain, 328, ideal, 272, integer, 3, relatively, 4, 308, subfield, 290, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , A51
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A52, , INDEX OF TERMS, , Primitive, element, 382, Element Theorem, 381, nth root of unity, 304, 571, polynomial, 312, Principal ideal domain, 276, 304, Principal ideal ring, 295, Projection, 220, Proper ideal, 267, Proper subgroup, 61, Pullback, 212, Quaternions, 95, 201, 203, 450, Quotient, 4, 302, Quotient group, 187, Quotients, field of, 290, Range, 20, Rational Root Theorem, 324, Reducible polynomial, 311, Reflection, 36, 462, Relation, equivalence, 18, in a presentation, 449, Relatively prime, 4, 308, Remainder, 4, 302, Remainder Theorem, 303, Ring(s), Boolean, 253, center of, 251, characteristic of, 258, commutative, 245, definition of, 245, direct sum of, 247, factor, 268, homomorphism of, 285, isomorphism of, 285, of polynomials, 298, with unity, 246, RSA public encryption, 171, Rubik’s Cube, 114, 155, Scalar, 351, Scalar multiplication, 351, Self-dual code, 582, Sicherman dice, 321, Simple extension, 376, Simple group, 428, Socks–Shoes Property, 52, 56, Solvable by radicals, 560, , Solvable group, 561, Spanning set, 353, Splitting field, 362, Squaring the circle, 399, 402, Stabilizer of a point, 120, 151, Standard array, 539, Standard decoding, 539, Standard encoding matrix, 534, Standard generator matrix, 534, Subcode, 545, Subfield Test, 262, Subgroup(s), centralizer, 70, characteristic, 181, commutator, 181, conjugate, 95, 414, cyclic, 65, definition of, 61, diagonal, 175, Finite Test, 64, generated by a, 65, generated by S, 66, index of, 148, lattice, 86, maximal, 205, 240, nontrivial, 61, normal, 185, One-Step Test, 62, proper, 61, Sylow p-, 413, torsion, 96, trivial, 61, Two-Step Test, 63, Subring, definition of, 248, Test, 248, Trivial, 249, Subspace, 352, Subspace spanned by vectors,, 353, Subspace Test, 355, Sylow p-subgroup, 413, Sylow test for nonsimplicity, 433, Sylow Theorems, 412, 414, 415, Symmetric group, 101, Symmetries of a square, 31, Symmetry group, 35, 36, 461, Syndrome of a vector, 541, Systematic code, 534, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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INDEX OF TERMS, , Torsion subgroup, 96, Transcendental element, 376, Transcendental extension, 376, Translation, 47, 462, Transposition, 107, Trisecting an angle, 399, 400, 401, , dimension of, 355, finite dimensional, 355, infinite dimensional, 355, spanned by a set, 353, trivial, 355, Vertex of a graph, 506, , UFD, 334, Unique factorization domain, 334, Unique factorization theorem, for a PID, 335, for D[x], 340, for F[x], 337, for Z, 6, for Z[x], 319, in a Euclidean domain, 339, Unity, 246, Universal Factor Group Property, 448, Universal Mapping Property, 447, Universal Product Code, 9, , Wallpaper groups, 475, Weight of a vector, 532, Weighting vector, 9, Weird dice, 321, Well-defined function, 209, Well Ordering Principle, 3, Word, code, 528, 531, empty, 446, in a group, 446, , Vector, 351, Vector space, basis of, 353, definition of, 351, , Zero, multiple, 369, multiplicity of, 303, of a polynomial, 303, Zero-divisor, 255, , Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it., , A53
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
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Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially, affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
