Solution Manual for Contemporary Abstract Algebra, 9th Edition

9 CHAPTER 2 Groups 1. c, d 2. c, d 3. none 4. a, c 1 3+2i 3 2 1 = 3−2i 5. 7; 13; n − 1; 3−2i 3+2i = 13 + 13 i 1 2 −3 2 6. a. −31 − i b. 5 c. d. 6 4 12 −8 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. 1 1 1 0 1 0 1 1 10. 6= . 0 1 1 1 1 1 0 1 11. a11 , a6 , a4 , a1 12. 5, 4, 8 13. (a) 2a + 3b; (b) −2a + 2(−b + c); (c) −3(a + 2b) + 2c = 0 14. (ab)3 = ababab and (ab−2 c)−2 = ((ab−2 c)−1 )2 = (c−1 b2 a−1 )2 = c−1 b2 a−1 c−1 b2 a−1 . 15. Observe that a5 = e implies that a−2 = a3 and b7 = e implies that b14 = e and therefore b−11 = b3 . Thus, a−2 b−11 = a3 b3 . Moreover, (a2 b4 )−2 = ((a2 b4 )−1 )2 = (b−4 a−2 )2 = (b3 a3 )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 = {R0 , R180 }; L = {R0 , R180 , H, V, D, D0 }. 2/Groups 10 19. The set is closed because det (AB) =(det A)(det B). Matrix 1 0 multiplication is associative. is the identity. 0 1 −1 a b d −b Since = its determinant is ad − bc = 1. c d −c a 20. 12 = (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 = an bn ab = an+1 bn+1 . 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 bn 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 . −1 −1 −1 28. (a1 a2 · · · an )(a−1 n an−1 · · · a2 a1 ) = e 29. By closure we have {1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45}. 30. Z105 ; Z44 and D22 . 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 1 1 5 5 7 7 11 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1 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 = a2 b2 ⇔ 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 x3 = 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 x3 = e and b is a nonidentity element such that b 6= a and b 6= a−1 and b3 = 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 x3 = e. To prove the second statement note that if x2 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 x2 6= e come in distinct pairs. 38. In D4 , HR90 V = DR90 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 R270 D0 H. 41. If F1 F2 = R0 then F1 F2 = F1 F1 and by cancellation F1 = F2 . 42. Observe that F1 F2 = F2 F1 implies that (F1 F2 )(F1 F2 ) = R0 . Since F1 and F2 are distinct and F1 F2 is a rotation it must be R180 . 43. Since F Rk is a reflection we have (F Rk )(F Rk ) = R0 . Multiplying on the left by F gives Rk F Rk = F . 44. Since F Rk is a reflection we have (F Rk )(F Rk ) = R0 . Multiplying on the right by R−k gives F Rk F = R−k . If Dn were Abelian, then F R360◦ /n F = R360◦ /n . But (R360◦ /n )−1 = R360◦ (n−1)/n 6= R360◦ /n when n ≥ 3. 2/Groups 45. a. R3 12 b. R c. R5 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 + a0 = 0, b0 + ac0 + b = 0, c0 + c = 0 for a0 , b0 , c0 . 47. Since a2 = b2 = (ab)2 = e, we have aabb = abab. Now cancel on left and right. 48. If a satisfies x5 = e and a 6= e, then so does a2 , a3 , a4 . Now, using cancellation we have that a2 , a3 , a4 are not the identity and are distinct from each other and distinct from a. If these are all of the nonidentity solutions of x5 = e we are done. If b is another solution that is not a power of a, then by the same argument b, b2 , b3 and b4 are four distinct nonidentity solutions. We must further show that b2 , b3 and b4 are distinct from a, a2 , a3 , a4 . If b2 = ai for some i, then cubing both sides we have b = b6 = a3i , which is a contradiction. A similar argument applies to b3 and b4 . Continuing in this fashion we have that the number of nonidentity solutions to x5 = e is a multiple of 4. In the general case, the number of solutions is a multiple of 4 or is infinite. a b 49. The matrix is in GL(2, Z2 ) if and only if ad 6= bc. This happens c d 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 1 1 1 0 1 1 0 1 0 1 . 0 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 and do not commute. 0 1 1 1 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 b0 be the element in G with the property that bb0 = e. Then observe that ba = (ba)e = ba(bb0 ) = b(ab)b0 = beb0 = (be)b0 = bb0 = e.