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Solution Manual for Contemporary Abstract Algebra, 9th Edition

Solution Manual for Contemporary Abstract Algebra, 9th Edition makes textbook problem-solving easy with a comprehensive guide that explains every step.

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    Complete Solutions Manual to Accompany Contemporary Abstract Algebra NINTH EDITION Joseph Gallian University of Minnesota Duluth

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    i CONTEMPORARY ABSTRACT ALGEBRA 9TH EDITION SOLUTION MANUAL CONTENTS Integers and Equivalence Relations 0 Preliminaries 1 Groups 1 Introduction to Groups 7 2 Groups 9 3 Finite Groups; Subgroups 13 4 Cyclic Groups 20 5 Permutation Groups 27 6 Isomorphisms 34 7 Cosets and Lagrange’s Theorem 40 8 External Direct Products 46 9 Normal Subgroups and Factor Groups 53 10 Group Homomorphisms 59 11 Fundamental Theorem of Finite Abelian Groups 65 12 Introduction to Rings 69 13 Integral Domains 74 14 Ideals and Factor Rings 80 15 Ring Homomorphisms 87 16 Polynomial Rings 94 17 Factorization of Polynomials 100 18 Divisibility in Integral Domains 105

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    ii Fields 19 Vector Spaces 110 20 Extension Fields 114 21 Algebraic Extensions 118 22 Finite Fields 123 23 Geometric Constructions 127 Special Topics 24 Sylow Theorems 129 25 Finite Simple Groups 135 26 Generators and Relations 140 27 Symmetry Groups 144 28 Frieze Groups and Crystallographic Groups 146 29 Symmetry and Counting 148 30 Cayley Digraphs of Groups 151 31 Introduction to Algebraic Coding Theory 154 32 An Introduction to Galois Theory 158 33 Cyclotomic Extensions 161

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    1 CHAPTER 0 Preliminaries 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 } 2. a. 2; 10 b. 4; 40 c. 4: 120; d. 1; 1050 e. pq 2 ; p 2 q 3 3. 12, 2, 2, 10, 1, 0, 4, 5. 4. s = − 3, t = 2; s = 8, t = − 5 5. By using 0 as an exponent if necessary, we may write a = p m 1 1 · · · p m k k and b = p n 1 1 · · · p n k k , where the p ’s are distinct primes and the m ’s and n ’s are nonnegative. Then lcm( a, b ) = p s 1 1 · · · p s k k , where s i = max( m i , n i ) and gcd( a, b ) = p t 1 1 · · · p t k k , where t i = min( m i , n i ) Then lcm( a, b ) · gcd( a, b ) = p m 1 + n 1 1 · · · p m k + n k k = ab . 6. The first part follows from the Fundamental Theorem of Arithmetic; for the second part, take a = 4, b = 6, c = 12. 7. Write a = nq 1 + r 1 and b = nq 2 + r 2 , where 0 ≤ r 1 , r 2 < n . We may assume that r 1 ≥ r 2 . Then a − b = n ( q 1 − q 2 ) + ( r 1 − r 2 ), where r 1 − r 2 ≥ 0. If a mod n = b mod n , then r 1 = r 2 and n divides a − b . If n divides a − b , then by the uniqueness of the remainder, we then have r 1 − r 2 = 0. Thus, r 1 = r 2 and therefore a mod n = b mod n . 8. Write as + bt = d . Then a ′ s + b ′ t = ( a/d ) s + ( b/d ) t = 1. 9. By Exercise 7, to prove that ( a + b ) mod n = ( a ′ + b ′ ) mod n and ( ab ) mod n = ( a ′ b ′ ) mod n it suffices to show that n divides ( a + b ) − ( a ′ + b ′ ) and ab − a ′ b ′ . Since n divides both a − a ′ and n divides b − b ′ , it divides their difference. Because a = a ′ mod n and b = b ′ mod n there are integers s and t such that a = a ′ + ns and b = b ′ + nt . Thus ab = ( a ′ + ns )( b ′ + nt ) = a ′ b ′ + nsb ′ + a ′ nt + nsnt . Thus, ab − a ′ b ′ is divisible by n . 10. Write d = au + bv . Since t divides both a and b , it divides d . Write s = mq + r where 0 ≤ r < m . Then r = s − mq is a common multiple of both a and b so r = 0. 11. Suppose that there is an integer n such that ab mod n = 1. Then there is an integer q such that ab − nq = 1. Since d divides both a and n , d also divides 1. So, d = 1. On the other hand, if d = 1, then by the corollary of Theorem 0.2, there are integers s and t such that as + nt = 1. Thus, modulo n , as = 1.

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    0/Preliminaries 2 12. 7(5 n + 3) − 5(7 n + 4) = 1 13. By the GCD Theorem there are integers s and t such that ms + nt = 1. Then m ( sr ) + n ( tr ) = r . 14. It suffices to show that ( p 2 + q 2 + r 2 ) mod 3 = 0. Notice that for any integer a not divisible by 3, a mod 3 is 1 or 2 and therefore a 2 mod 3 = 1. So, ( p 2 + q 2 + r 2 ) mod 3 = p 2 mod 3 + q 2 mod 3 + r 2 mod 3 = 3 mod 3= 0. 15. Let p be a prime greater than 3. By the Division Algorithm, we can write p in the form 6 n + r , where r satisfies 0 ≤ r < 6. Now observe that 6 n, 6 n + 2 , 6 n + 3, and 6 n + 4 are not prime. 16. By properties of modular arithmetic we have (7 1000 ) mod 6 = (7 mod 6) 1000 = 1 1000 = 1. Similarly, (6 1001 ) mod 7 = (6 mod 7) 1001 = − 1 1001 mod 7 = − 1 = 6 mod 7. 17. Since st divides a − b , both s and t divide a − b . The converse is true when gcd( s, t ) = 1. 18. Observe that 8 402 mod 5 = 3 402 mod 5 and 3 4 mod 5 = 1. Thus, 8 402 mod 5 = (3 4 ) 100 3 2 mod 5 = 4. 19. If gcd( a, bc ) = 1, then there is no prime that divides both a and bc . By Euclid’s Lemma and unique factorization, this means that there is no prime that divides both a and b or both a and c . Conversely, if no prime divides both a and b or both a and c , then by Euclid’s Lemma, no prime divides both a and bc . 20. If one of the primes did divide k = p 1 p 2 · · · p n + 1, it would also divide 1. 21. Suppose that there are only a finite number of primes p 1 , p 2 , . . . , p n . Then, by Exercise 20, p 1 p 2 . . . p n + 1 is not divisible by any prime. This means that p 1 p 2 . . . p n + 1, which is larger than any of p 1 , p 2 , . . . , p n , is itself prime. This contradicts the assumption that p 1 , p 2 , . . . , p n is the list of all primes. 22. − 7 58 + 3 58 i 23. − 5+2 i 4 − 5 i = − 5+2 i 4 − 5 i 4+5 i 4+5 i = − 30 41 + − 17 41 i 24. Let z 1 = a + bi and z 2 = c + di . Then z 1 z 2 = ( ac − bd ) + ( ad + bc ); | z 1 | = √ a 2 + b 2 , | z 2 | = √ c 2 + d 2 , | z 1 z 2 | = √ a 2 c 2 + b 2 d 2 + a 2 d 2 + b 2 c 2 = | z 1 || z 2 | . 25. x NAND y is 1 if and only if both inputs are 0; x XNOR y is 1 if and only if both inputs are the same. 26. If x = 1, the output is y , else it is z .

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    0/Preliminaries 3 27. Let S be a set with n + 1 elements and pick some a in S . By induction, S has 2 n subsets that do not contain a . But there is one-to-one correspondence between the subsets of S that do not contain a and those that do. So, there are 2 · 2 n = 2 n +1 subsets in all. 28. Use induction and note that 2 n +1 3 2 n +2 − 1 = 18(2 n 3 2 n ) − 1 = 18(2 n 3 3 n − 1) + 17. 29. Consider n = 200! + 2. Then 2 divides n , 3 divides n + 1, 4 divides n + 2 , . . . , and 202 divides n + 200. 30. Use induction on n . 31. Say p 1 p 2 · · · p r = q 1 q 2 · · · q s , where the p ’s and the q ’s are primes. By the Generalized Euclid’s Lemma, p 1 divides some q i , say q 1 (we may relabel the q ’s if necessary). Then p 1 = q 1 and p 2 · · · p r = q 2 · · · q s . Repeating this argument at each step we obtain p 2 = q 2 , · · · , p r = q r and r = s . 32. 47. Mimic Example 12. 33. Suppose that S is a set that contains a and whenever n ≥ a belongs to S , then n + 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 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 − 1 ∈ S , and therefore b = ( b − 1) + 1 ∈ S . This contradicts our assumption that b is not in S . 34. By the Second Principle of Mathematical Induction, f n = f n − 1 + f n − 2 < 2 n − 1 + 2 n − 2 = 2 n − 2 (2 + 1) < 2 n . 35. For n = 1, observe that 1 3 + 2 3 + 3 3 = 36. Assume that n 3 + ( n + 1) 3 + ( n + 2) 3 = 9 m for some integer m . We must prove that ( n + 1) 3 + ( n + 2) 3 + ( n + 3) 3 is a multiple of 9. Using the induction hypothesis we have that ( n + 1) 3 + ( n + 2) 3 + ( n + 3) 3 = 9 m − n 3 + ( n + 3) 3 = 9 m − n 3 + n 3 +3 · n 2 · 3+3 · n · 9+3 3 = 9 m +9 n 2 +27 n +27 = 9( m + n 2 +3 n +3). 36. You must verify the cases n = 1 and n = 2. This situation arises in cases where the arguments that the statement is true for n implies that it is true for n + 2 is different when n is even and when n is odd. 37. The statement is true for any divisor of 8 3 − 4 = 508. 38. One need only verify the equation for n = 0 , 1 , 2 , 3 , 4 , 5 . Alternatively, observe that n 3 − n = n ( n − 1)( n + 1). 39. Since 3736 mod 24 = 16, it would be 6 p.m.

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    0/Preliminaries 4 40. 5 41. Observe that the number with the decimal representation a 9 a 8 . . . a 1 a 0 is a 9 10 9 + a 8 10 8 + · · · + a 1 10 + a 0 . From Exercise 9 and the fact that a i 10 i mod 9 = a i mod 9 we deduce that the check digit is ( a 9 + a 8 + · · · + a 1 + a 0 ) mod 9. So, substituting 0 for 9 or vice versa for any a i does not change the value of ( a 9 + a 8 + · · · + a 1 + a 0 ) mod 9. 42. No 43. For the case in which the check digit is not involved, the argument given Exercise 41 applies to transposition errors. Denote the money order number by a 9 a 8 . . . a 1 a 0 c where c is the check digit. For a transposition involving the check digit c = ( a 9 + a 8 + · · · + a 0 ) mod 9 to go undetected, we must have a 0 = ( a 9 + a 8 + · · · + a 1 + c ) mod 9. Substituting for c yields 2( a 9 + a 8 + · · · + a 0 ) mod 9 = a 0 . Then cancelling the a 0 , multiplying by sides by 5, and reducing module 9, we have 10( a 9 + a 8 + · · · + a 1 ) = a 9 + a 8 + · · · + a 1 = 0. It follows that c = a 9 + a 8 · · · + a 1 + a 0 = a 0 . In this case the transposition does not yield an error. 44. 4 45. Say the number is a 8 a 7 . . . a 1 a 0 = a 8 10 8 + a 7 10 7 + · · · + a 1 10 + a 0 . Then the error is undetected if and only if ( a i 10 i − a ′ i 10 i ) mod 7 = 0. Multiplying both sides by 5 i and noting that 50 mod 7 = 1, we obtain ( a i − a ′ i ) mod 7 = 0. 46. All except those involving a and b with | a − b | = 7. 47. 4 48. Observe that for any integer k between 0 and 8, k ÷ 9 = .kkk . . . . 50. 7 51. Say that the weight for a is i . Then an error is undetected if modulo 11, ai + b ( i − 1) + c ( i − 2) = bi + c ( i − 1) + a ( i − 2). This reduces to the cases where (2 a − b − c ) mod 11 = 0. 52. Say the valid number is a 1 a 2 . . . a 10 and a i and a i +1 were transposed. Then, modulo 11, 10 a 1 + 9 a 2 + · · · + a 10 = 0 and 10 a 1 + · · · + (11 − i ) a i +1 + (11 − ( i + 1)) a i + · · · + a 10 = 5. Thus, 5 = 5 − 0 = (10 a 1 + · · · + (11 − i ) a i +1 + (11 − ( i + 1)) a i + a 10 ) − (10 a 1 + 9 a 2 + · · · + a 10 ). It follows that ( a i +1 − a i ) mod 11 = 5. Now look for adjacent digits x and y in the invalid number so that ( x − y ) mod 11 = 5. Since the only pair is 39, the correct number is 0-669-09325-4.

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    0/Preliminaries 5 53. Since 10 a 1 + 9 a 2 + · · · + a 10 = 0 mod 11 if and only if 0 = ( − 10 a 1 − 9 a 2 − · · · − 10 a 10 ) mod 11 = ( a 1 + 2 a 2 + · · · + 10 a 10 ) mod 11, the check digit would be the same. 54. 7344586061 55. First note that the sum of the digits modulo 11 is 2. So, some digit is 2 too large. Say the error is in position i . Then 10 = (4 , 3 , 0 , 2 , 5 , 1 , 1 , 5 , 6 , 8) · (1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10) mod 11 = 2 i . Thus, the digit in position 5 to 2 too large. So, the correct number is 4302311568. 56. An error in an even numbered position changes the value of the sum by an even amount. However, (9 · 1 + 8 · 4 + 7 · 9 + 6 · 1 + 5 · 0 + 4 · 5 + 3 · 2 + 2 · 6 + 7) mod 10 = 5. 57. 2. Since β is one-to-one, β ( α ( a 1 )) = β ( α ( a 2 )) implies that α ( a 1 ) = α ( a 2 ) and since α is one-to-one, a 1 = a 2 . 3. Let c ∈ C . There is a b in B such that β ( b ) = c and an a in A such that α ( a ) = b . Thus, ( βα )( a ) = β ( α ( a )) = β ( b ) = c . 4. Since α is one-to-one and onto we may define α − 1 ( x ) = y if and only if α ( y ) = x . Then α − 1 ( α ( a )) = a and α ( α − 1 ( b )) = b . 58. a − a = 0; if a − b is an integer k then b − a is the integer − k ; if a − b is the integer n and b − c is the integer m , then a − c = ( a − b ) + ( b − c ) is the integer n + m . The set of equivalence classes is { [ k ] | 0 ≤ k < 1 , k is real } . The equivalence classes can be represented by the real numbers in the interval [0 , 1). For any real number a , [ a ] = { a + k | where k ranges over all integers } . 59. No. (1 , 0) ∈ R and (0 , − 1) ∈ R but (1 , − 1) 6 ∈ R . 60. Obviously, a + a = 2 a is even and a + b is even implies b + a is even. If a + b and b + c are even, then a + c = ( a + b ) + ( b + c ) − 2 b is also even. The equivalence classes are the set of even integers and the set of odd integers. 61. a belongs to the same subset as a . If a and b belong to the subset A and b and c belong to the subset B , then A = B , since the distinct subsets of P are disjoint. So, a and c belong to A . 62. Suppose that n is odd prime greater than 3 and n + 2 and n + 4 are also prime. Then n mod 3 = 1 or n mod 3 = 2. If n mod 3 = 1 then n + 2 mod 3 = 0 and so is not prime. If n mod 3 = 2 then n + 4 mod 3 = 0 and so is not prime.

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    0/Preliminaries 6 63. The last digit of 3 100 is the value of 3 100 mod 10. Observe that 3 100 mod 10 is the same as ((3 4 mod 10) 25 mod 10 and 3 4 mod 10 = 1. Similarly, the last digit of 2 100 is the value of 2 100 mod 10. Observe that 2 5 mod 10 = 2 so that 2 100 mod 10 is the same as (2 5 mod 10) 20 mod 10 = 2 20 mod 10 = (2 5 ) 4 mod 10 = 2 4 mod 10 = 6. 64. Suppose that there are integers a, b, c, and d with gcd( a, b ) = 1 and gcd( c, d ) = 1 such that a 2 /b 2 − c 2 /d 2 = 1002. Then a 2 d 2 − c 2 b 2 = 1002 b 2 d 2 . If both b and d are odd, then modulo 4, b 2 = d 2 = 1 and a 2 /b 2 − c 2 /d 2 = 1002 reduces to a 2 − c 2 = 2. This case is handled in Example 7. If 2 i ( i > 0) divides b , then a is odd and a 2 d 2 − c 2 b 2 = 1002 b 2 d 2 implies that 2 i divides d also. It follows that if 2 n is the highest power of 2 that divides one of b or d , then 2 n is the highest power of 2 that divides the other. So dividing both sides of a 2 d 2 − c 2 b 2 = 1002 b 2 d 2 by 2 n we get an equation of the same form where both b and d are odd. Taking both sides modulo 4 and recalling that for odd x, x 2 mod 4 = 1 we have that a 2 d 2 − c 2 b 2 = 1002 b 2 d 2 reduces a 2 − c 2 = 2, which was done in Example 7. 65. Apply γ − 1 to both sides of αγ = βγ .

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    7 CHAPTER 1 Introduction to Groups 1. Three rotations: 0 ◦ , 120 ◦ , 240 ◦ , and three reflections across lines from vertices to midpoints of opposite sides. 2. Let R = R 120 , R 2 = R 240 , F a reflection across a vertical axis, F ′ = RF and F ′′ = R 2 F R 0 R R 2 F F ′ F ′′ R 0 R 0 R R 2 F F ′ F ′′ R R R 2 R 0 F ′ F ′′ F R 2 R 2 R 0 R F ′′ F F ′ F F F ′′ F ′ R 0 R 2 R F ′ F ′ F F ′′ R R 0 R 2 F ′′ F ′′ F ′ F R 2 R R 0 3. a. V b. R 270 c. R 0 d. R 0 , R 180 , H, V, D, D ′ e. none 4. Five rotations: 0 ◦ , 72 ◦ , 144 ◦ , 216 ◦ , 288 ◦ , and five reflections across lines from vertices to midpoints of opposite sides. 5. D n has n rotations of the form k (360 ◦ /n ), where k = 0 , . . . , n − 1. In addition, D n has n reflections. When n is odd, the axes of reflection are the lines from the vertices to the midpoints of the opposite 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. 6. A nonidentity rotation leaves only one point fixed – the center of rotation. A reflection leaves the axis of reflection fixed. A reflection followed by a different reflection would leave only one point fixed (the intersection of the two axes of reflection) so it must be a rotation. 7. 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. 8. In either case, the set of points fixed is some axis of reflection. 9. Observe that 1 · 1 = 1; 1( − 1) = − 1; ( − 1)1 = − 1; ( − 1)( − 1) = 1. These relationships also hold when 1 is replaced by a “rotation” and − 1 is replaced by a “reflection.” 10. reflection.

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    1/Introduction to Groups 8 11. Thinking geometrically and observing that even powers of elements of a dihedral group do not change orentation we note that each of a, b and c appears an even number of times in the expression. So, there is no change in orentation. Thus, the expression is a rotation. Alternatively, as in Exercise 9, we associate each of a, b and c with 1 if they are rotations and − 1 if they are reflections and we observe that in the product a 2 b 4 ac 5 a 3 c the terms involving a represents six 1s or six − 1s, the term b 4 represents four 1s or four − 1s, and the terms involving c represents six 1s or six − 1s. Thus the product of all the 1s and − 1s is 1. So the expression is a rotation. 12. H, I, O, X. Rotations of 0 ◦ , 180 ◦ , horizontal reflection, and vertical reflection. 13. In D 4 , HD = DV but H 6 = V . 14. D n is not commutative. 15. R 0 , R 180 , H , V 16. Rotations of 0 ◦ and 180 ◦ ; Rotations of 0 ◦ and 180 ◦ and reflections about the diagonals. 17. R 0 , R 180 , H , V 18. Let the distance from a point on one H to the corresponding point on an adjacent H be one unit. Then translations of any number of units to the right or left are symmetries; reflection across the horizontal axis through the middle of the H ’s is a symmetry; reflection across any vertical axis midway between two H ’s or bisecting any H is a symmetry. All other symmetries are compositions of finitely many of those already described. The group is non-Abelian. 19. In each case the group is D 6 . 20. D 28 21. First observe that X 2 6 = R 0 . Since R 0 and R 180 are the only elements in D 4 that are squares we have X 2 = R 180 . Solving X 2 Y = R 90 for Y gives Y = R 270 . 22. X 2 = F has no solutions; the only solution to X 3 = F is F . 23. 180 ◦ rotational symmetry. 24. Z 4 , D 5 , D 4 , Z 2 D 4 , Z 3 , D 3 , D 16 D 7 , D 4 , D 5 , Z 10 25. Their only symmetry is the identity.

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    9 CHAPTER 2 Groups 1. c, d 2. c, d 3. none 4. a, c 5. 7; 13; n − 1; 1 3 − 2 i = 1 3 − 2 i 3+2 i 3+2 i = 3 13 + 2 13 i 6. a. − 31 − i b. 5 c. 1 12 [ 2 − 3 − 8 6 ] d. [ 2 4 4 6 ] . 7. The set does not contain the identity; closure fails. 8. 1, 3, 7, 9, 11, 13, 17, 19. 9. Under multiplication modulo 4, 2 does not have an inverse. Under multiplication modulo 5, { 1 , 2 , 3 , 4 } is closed, 1 is the identity, 1 and 4 are their own inverses, and 2 and 3 are inverses of each other. Modulo multiplication is associative. 10. [ 1 1 0 1 ] [ 1 0 1 1 ] 6 = [ 1 0 1 1 ] [ 1 1 0 1 ] . 11. a 11 , a 6 , a 4 , a 1 12. 5 , 4 , 8 13. (a) 2 a + 3 b ; (b) − 2 a + 2( − b + c ); (c) − 3( a + 2 b ) + 2 c = 0 14. ( ab ) 3 = ababab and ( ab − 2 c ) − 2 = (( ab − 2 c ) − 1 ) 2 = ( c − 1 b 2 a − 1 ) 2 = c − 1 b 2 a − 1 c − 1 b 2 a − 1 . 15. Observe that a 5 = e implies that a − 2 = a 3 and b 7 = e implies that b 14 = e and therefore b − 11 = b 3 . Thus, a − 2 b − 11 = a 3 b 3 . Moreover, ( a 2 b 4 ) − 2 = (( a 2 b 4 ) − 1 ) 2 = ( b − 4 a − 2 ) 2 = ( b 3 a 3 ) 2 . 16. The identity is 25. 17. Since the inverse of an element in G is in G , H ⊆ G . Let g belong to G . Then g − 1 belongs to G and therefore ( g − 1 ) − 1 = g belong to G . So, G ⊆ H . 18. K = { R 0 , R 180 } ; L = { R 0 , R 180 , H, V, D, D ′ } .

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    2/Groups 10 19. The set is closed because det ( AB ) = (det A )(det B ). Matrix multiplication is associative. [ 1 0 0 1 ] is the identity. Since [ a b c d ] − 1 = [ d − b − c a ] its determinant is ad − bc = 1. 20. 1 2 = ( n − 1) 2 = 1. 21. Using closure and trial and error, we discover that 9 · 74 = 29 and 29 is not on the list. 22. Consider xyx = xyx . 23. For n ≥ 0, we use induction. The case that n = 0 is trivial. Then note that ( ab ) n +1 = ( ab ) n ab = a n b n ab = a n +1 b n +1 . For n < 0, note that e = ( ab ) 0 = ( ab ) n ( ab ) − n = ( ab ) n a − n b − n so that a n b n = ( ab ) n . In a non-Abelian group ( ab ) n need not equal a n b n . 24. The “inverse” of putting on your socks and then putting on your shoes is taking off your shoes then taking off your socks. Use D 4 for the examples. (An appropriate name for the property ( abc ) − 1 = c − 1 b − 1 a − 1 is “Socks-Shoes-Boots Property.”) 25. Suppose that G is Abelian. Then by Exercise 24, ( ab ) − 1 = b − 1 a − 1 = a − 1 b − 1 . If ( ab ) − 1 = a − 1 b − 1 then by Exercise 24 e = aba − 1 b − 1 . Multiplying both sides on the right by ba yields ba = ab . 26. By definition, a − 1 ( a − 1 ) − 1 = e . Now multiply on the left by a . 27. The case where n = 0 is trivial. For n > 0, note that ( a − 1 ba ) n = ( a − 1 ba )( a − 1 ba ) · · · ( a − 1 ba ) ( n terms). So, cancelling the consecutive a and a − 1 terms gives a − 1 b n a . For n < 0, note that e = ( a − 1 ba ) n ( a − 1 ba ) − n = ( a − 1 ba ) n ( a − 1 b − n a ) and solve for ( a − 1 ba ) n . 28. ( a 1 a 2 · · · a n )( a − 1 n a − 1 n − 1 · · · a − 1 2 a − 1 1 ) = e 29. By closure we have { 1 , 3 , 5 , 9 , 13 , 15 , 19 , 23 , 25 , 27 , 39 , 45 } . 30. Z 105 ; Z 44 and D 22 . 31. Suppose x appears in a row labeled with a twice. Say x = ab and x = ac . Then cancellation gives b = c . But we use distinct elements to label the columns. 32. 1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1

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    2/Groups 11 33. Proceed as follows. By definition of the identity, we may complete the first row and column. Then complete row 3 and column 5 by using Exercise 31. In row 2 only c and d remain to be used. We cannot use d in position 3 in row 2 because there would then be two d ’s in column 3. This observation allows us to complete row 2. Then rows 3 and 4 may be completed by inserting the unused two elements. Finally, we complete the bottom row by inserting the unused column elements. 34. ( ab ) 2 = a 2 b 2 ⇔ abab = aabb ⇔ ba = ab . ( ab ) − 2 = b − 2 a − 2 ⇔ b − 1 a − 1 b − 1 a − 1 = b − 1 b − 1 a − 1 a − 1 ⇔ a − 1 b − 1 = b − 1 a − 1 ⇔ ba = ab . 35. axb = c implies that x = a − 1 ( axb ) b − 1 = a − 1 cb − 1 ; a − 1 xa = c implies that x = a ( a − 1 xa ) a − 1 = aca − 1 . 36. Observe that xabx − 1 = ba is equivalent to xab = bax and this is true for x = b . 37. Since e is one solution it suffices to show that nonidentity solutions come in distinct pairs. To this end note that if x 3 = e and x 6 = e , then ( x − 1 ) 3 = e and x 6 = x − 1 . So if we can find one nonidentity solution we can find a second one. Now suppose that a and a − 1 are nonidentity elements that satisfy x 3 = e and b is a nonidentity element such that b 6 = a and b 6 = a − 1 and b 3 = e . Then, as before, ( b − 1 ) 3 = e and b 6 = b − 1 . Moreover, b − 1 6 = a and b − 1 6 = a − 1 . Thus, finding a third nonidentity solution gives a fourth one. Continuing in this fashion we see that we always have an even number of nonidentity solutions to the equation x 3 = e . To prove the second statement note that if x 2 6 = e , then x − 1 6 = x and ( x − 1 ) 2 6 = e . So, arguing as in the preceding case we see that solutions to x 2 6 = e come in distinct pairs. 38. In D 4 , HR 90 V = DR 90 H but HV 6 = DH . 39. Observe that aa − 1 b = ba − 1 a . Cancelling the middle term a − 1 on both sides we obtain ab = ba . 40. X = V R 270 D ′ H . 41. If F 1 F 2 = R 0 then F 1 F 2 = F 1 F 1 and by cancellation F 1 = F 2 . 42. Observe that F 1 F 2 = F 2 F 1 implies that ( F 1 F 2 )( F 1 F 2 ) = R 0 . Since F 1 and F 2 are distinct and F 1 F 2 is a rotation it must be R 180 . 43. Since F R k is a reflection we have ( F R k )( F R k ) = R 0 . Multiplying on the left by F gives R k F R k = F . 44. Since F R k is a reflection we have ( F R k )( F R k ) = R 0 . Multiplying on the right by R − k gives F R k F = R − k . If D n were Abelian, then F R 360 ◦ /n F = R 360 ◦ /n . But ( R 360 ◦ /n ) − 1 = R 360 ◦ ( n − 1) /n 6 = R 360 ◦ /n when n ≥ 3 .

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    2/Groups 12 45. a. R 3 b. R c. R 5 F 46. Closure and associativity follow from the definition of multiplication; a = b = c = 0 gives the identity; we may find inverses by solving the equations a + a ′ = 0, b ′ + ac ′ + b = 0, c ′ + c = 0 for a ′ , b ′ , c ′ . 47. Since a 2 = b 2 = ( ab ) 2 = e , we have aabb = abab . Now cancel on left and right. 48. If a satisfies x 5 = e and a 6 = e , then so does a 2 , a 3 , a 4 . Now, using cancellation we have that a 2 , a 3 , a 4 are not the identity and are distinct from each other and distinct from a . If these are all of the nonidentity solutions of x 5 = e we are done. If b is another solution that is not a power of a , then by the same argument b, b 2 , b 3 and b 4 are four distinct nonidentity solutions. We must further show that b 2 , b 3 and b 4 are distinct from a, a 2 , a 3 , a 4 . If b 2 = a i for some i , then cubing both sides we have b = b 6 = a 3 i , which is a contradiction. A similar argument applies to b 3 and b 4 . Continuing in this fashion we have that the number of nonidentity solutions to x 5 = e is a multiple of 4. In the general case, the number of solutions is a multiple of 4 or is infinite. 49. The matrix [ a b c d ] is in GL(2 , Z 2 ) if and only if ad 6 = bc . This happens when a and d are 1 and at 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. So, the elements are [ 1 0 0 1 ] [ 1 1 0 1 ] [ 1 0 1 1 ] [ 1 1 1 0 ] [ 0 1 1 1 ] [ 0 1 1 0 ] . [ 1 1 0 1 ] and [ 1 0 1 1 ] do not commute. 50. If n is not prime, we can write n = ab , where 1 < a < n and 1 < b < n . Then a and b belong to the set { 1 , 2 , . . . , n − 1 } but 0 = ab mod n does not. 51. Let a be any element in G and write x = ea . Then a − 1 x = a − 1 ( ea ) = ( a − 1 e ) a = a − 1 a = e . Then solving for x we obtain x = ae = a . 52. Suppose that ab = e and let b ′ be the element in G with the property that bb ′ = e . Then observe that ba = ( ba ) e = ba ( bb ′ ) = b ( ab ) b ′ = beb ′ = ( be ) b ′ = bb ′ = e .

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