Solution Manual for First Course in Abstract Algebra, A, 8th Edition

INSTRUCTOR’S SOLUTIONS MANUAL JOHN B. FRALEIGH AND NEAL BRAND A FIRST COURSE IN ABSTRACT ALGEBRA EIGHTH EDITION John B. Fraleigh University of Rhode Island Neal Brand University of North Texas The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2021, 2003 by Pearson Education, Inc. 221 River Street, Hoboken, NJ 07030. All rights reserved. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-32-139037-0 ISBN-10: 0-321-39037-7 Preface for Seventh Edition This manual contains solutions to all exercises in the text, except those odd-numbered exercises for which fairly lengthy complete solutions are given in the answers at the back of the text. Then reference is simply given to the text answers to save typing. I prepared these solutions myself. While I tried to be accurate, there are sure to be the inevitable mistakes and typos. An author reading proof tends to see what he or she wants to see. However, the instructor should find this manual adequate for the purpose for which it is intended. Morgan, Vermont J.B.F July, 2002 Preface for Eighth Edition In keeping with the seventh edition, this manual contains solutions to all exercises in the text except for some of the odd-numbered exercises whose solutions are in the back of the text book. I made few changes to solutions to exercises that were in the seventh edition. However, solutions to new exercises do not always include as much detail as would be found in the seventh edition. My thinking is that instructors teaching the class would use the solution manual to see the idea behind a solution and they would easily fill in the routine details. As in the seventh edition, I tried to be accurate. However, there are sure to be some errors. I hope instructors find the manual helpful. Denton, Texas March, 2020 N.B. CONTENTS 0. Sets and Relations 01 I. Groups and Subgroups 1. Binary Operations 2. Groups 05 08 3. Abelian Examples 14 4. Nonabelian Examples 5. Subgroups 19 22 6. Cyclic Groups 27 7. Generators and Cayley Digraphs 32 II. Structure of Groups 8. Groups of Permutations 34 9. Finitely Generated Abelian Groups 10. Cosets and the Theorem of Lagrange 11. Plane Isometries 40 45 50 III. Homomorphisms and Factor Groups 12. Factor Groups 53 13. Factor Group Computations and Simple Groups 14. Group Action on a Set 58 65 15. Applications of G-Sets to Counting 70 VI. Advanced Group Theory 16. Isomorphism Theorems 17. Sylow Theorems 75 18. Series of Groups 80 19. Free Abelian Groups 20. Free Groups 73 85 88 21. Group Presentations 91 V. Rings and Fields 22. Rings and Fields 95 23. Integral Domains 102 24. Fermat’s and Euler’s Theorems 25. RSA Encryption 106 109 VI. Constructing Rings and Fields 26. The Field of Quotients of an Integral Domain 27. Rings of Polynomials 110 112 28. Factorization of Polynomials over a Field 29. Algebraic Coding Theory 116 123 30. Homomorphisms and Factor Rings 31. Prime and Maximal Ideals 131 32. Noncommutative Examples 137 125 VII. Commutative Algebra 33. Vector Spaces 140 34. Unique Factorization Domains 35. Euclidean Domains 36. Number Theory 145 149 154 37. Algebraic Geometry 160 38. Gröbner Bases for Ideals 163 VIII. Extension Fields 39. Introduction to Extension Fields 40. Algebraic Extensions 174 41. Geometric Constructions 42. Finite Fields 168 179 182 IX. Galois Theory 43. Automorphisms of Fields 44. Splitting Fields 191 45. Separable Extensions 46. Galois Theory 199 195 185 47. Illustrations of Galois Theory 48. Cyclotomic Extensions 203 211 49. Insolvability of the Quintic 214 APPENDIX: Matrix Algebra 216 0. Sets and Relations 1 0. Sets and Relations 1. { 3, − 3} 2. {2, –3}. 3. {1, −1, 2, −2, 3, −3, 4, −4, 5, −5, 6, −6, 10, −10, 12, −12, 15, −15, 20, −20, 30, −30, 60, −60} 4. {2, 3, 4, 5, 6, 7, 8} 5. It is not a well-defined set. (Some may argue that no element of  + is large, because every element exceeds only a finite number of other elements but is exceeded by an infinite number of other elements. Such people might claim the answer should be ∅.) 6. ∅ 7. The set is ∅ because 33 = 27 and 43 = 64. 8. { r ∈  r = 2an for some a a ∈  + and some integer n ≥ 0}. 9. It is not a well-defined set. 10. The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or 1/3. 11. {(a, 1), (a, 2), (a, c), (b, 1), (b, 2), (b, c), (c, 1), (c, 2), (c, c)} 12. a. This is a function which is both one-to-one and onto B. b. This not a subset of A × B, and therefore not a function. c. It is not a function because there are two pairs with first member 1. d. This is a function which is neither one-to-one (6 appears twice in the second coordinate) nor onto B ( 4 is not in the second coordinate). e. It is a function. It is not one-to-one because there are two pairs with second member 6. It is not onto B because there is no pair with second member 2. f. This is not a function mapping A into B since 3 is not in the first coordinate of any ordered pair. 13. Draw the line through P and x, and let y be its point of intersection with the line segment CD. 14. a. φ : [ 0,1] → [ 0, 2] where φ ( x ) = 2 x b. φ : [1, 3] → [5, 25] where φ ( x ) = 2 x + 3 c. φ : [ a, b ] → [ c, d ] where φ ( x ) = c + d −c ( x − a) b−a 15. Let φ : S →  be defined by φ ( x ) = tan(π (x − 12 )). 16. a. ∅; cardinality 1 b. ∅, {a}; cardinality 2 c. ∅, {a}, {b}, {a, b}; cardinality 4 d. ∅, {a}, {b},{c}, {a, b},{a, c}, {b, c}, {a, b, c}; cardinality 8 Copyright © 2021 Pearson Education, Inc. 2 0. Sets and Relations 17. Conjecture: | P ( A ) |= 2 s = 2| A|. Proof The number of subsets of a set A depends only on the cardinality of A, not on what the elements of A actually are. Suppose B = {1, 2, 3, · · · , s − 1} and A = {1, 2, 3, · · · , s}. Then A has all the elements of B plus the one additional element s. All subsets of B are also subsets of A; these are precisely the subsets of A that do not contain s, so the number of subsets of A not containing s is |P(B)|. Any other subset of A must contain s, and removal of the s would produce a subset of B. Thus the number of subsets of A containing s is also |P(B)|. Because every subset of A either contains s or does not contain s (but not both), we see that the number of subsets of A is 2|P(B)|. We have shown that if A has one more element that B, then |P(A)| = 2|P(B)|. Now |P(∅)| = 1, so if |A| = s, then |P(A)| = 2s. 18. We define a one-to-one map φ of BA onto P(A). Let f ∈ BA, and let φ ( f ) = {x ∈ A | f ( x ) = 1}. Suppose φ (f ) = φ (g). Then f (x) = 1 if and only if g(x) = 1. Because the only possible values for f (x) and g(x) are 0 and 1, we see that f (x) = 0 if and only if g(x) = 0. Consequently f (x) = g(x) for all x ∈ A so f = g and φ is one to one. To show that φ is onto P(A), let S ⊆ A, and let h : A → {0, 1} be defined by h(x) = 1 if x ∈ S and h(x) = 0 otherwise. Clearly φ (h) = S, showing that φ is indeed onto P(A). 19. Picking up from the hint, let Z = {x ∈ A | x ∈ φ ( x )}. We claim that for any a ∈ A, φ ( a ) = Z . Either a ∈ φ ( a ) , in which case a ∈ Z , or a ∈ φ ( a ) , in which case a ∈ Z . Thus Z and φ (a) are certainly different subsets of A; one of them contains a and the other one does not. Based on what we just showed, we feel that the power set of A has cardinality greater than |A|. Proceeding naively, we can start with the infinite set , form its power set, then form the power set of that, and continue this process indefinitely. If there were only a finite number of infinite cardinal numbers, this process would have to terminate after a fixed finite number of steps. Since it doesn’t, it appears that there must be an infinite number of different infinite cardinal numbers. The set of everything is not logically acceptable, because the set of all subsets of the set of everything would be larger than the set of everything, which is a fallacy. 20. a. The set containing precisely the two elements of A and the three (different) elements of B is C = {1, 2, 3, 4, 5} which has 5 elements. i) Let A = {−2, −1, 0} and B = {1, 2, 3, · · ·} =  + . Then |A| = 3 and |B| = ℵ0, and A and B have no elements in common. The set C containing all elements in either A or B is C = {−2, −1, 0, 1, 2, 3, · · ·}. The map φ : C → B defined by φ (x) = x + 3 is one to one and onto B, so |C| = |B| = ℵ0. Thus we consider 3 + ℵ0 = ℵ0. ii) Let A = {1, 2, 3, · · ·} and B = {1/2, 3/2, 5/2, · · ·}. Then |A| = |B| = ℵ0 and A and B have no elements in common. The set C containing all elements in either A of B is C = {1/2, 1, 3/2, 2, 5/2, 3, · · ·}. The map φ : C → A defined by φ (x) = 2x is one to one and onto A, so |C| = |A| = ℵ0. Thus we consider ℵ0 + ℵ0 = ℵ0 b. We leave the plotting of the points in A × B to you. Figure 0.15 in the text, where there are ℵ0 rows each having ℵ0 entries, illustrates that we would consider that ℵ0 · ℵ0 = ℵ0. Copyright © 2021 Pearson Education, Inc. 0. Sets and Relations 3 21. There are 102 = 100 numbers (.00 through .99) of the form .##, and 105 = 100, 000 numbers (.00000 through .99999) of the form .#####. Thus for .##### · · ·, we expect 10ℵ0 sequences representing all numbers x ∈  such that 0 ≤ x ≤ 1, but a sequence trailing off in 0’s may represent the same x ∈  as a sequence trailing of in 9’s. At any rate, we should have 10ℵ0 ≥ |[0, 1]| =  ; see Exercise 15. On the other hand, we can represent numbers in  using any integer base n > 1, and these same 10ℵ0 sequences using digits from 0 to 9 in base n = 12 would not represent all x ∈ [0, 1], so we have 10ℵ0 ≤  . Thus we consider the value of 10ℵ0 to be  . We could make the same argument using any other integer base n > 1, and thus consider nℵ0 =  for n ∈  + , n > 1. In particular, 12ℵ0 = 12ℵ0 =  .  22. ℵ0 , |  |, 2|| , 2( 2 ) , 2(2 (2  ) ) 23. 1. There is only one partition {{a}} of a one-element set {a}. 24. There are two partitions of {a, b}, namely {{a, b}} and {{a}, {b}}. 25. There are five partitions of {a, b, c}, namely {{a, b, c}}, {{a}, {b, c}}, {{b}, {a, c}}, {{c}, {a, b}}, and {{a}, {b}, {c}}. 26. 15. The set {a, b, c, d} has 1 partition into one cell, 7 partitions into two cells (four with a 1,3 split and three with a 2,2 split), 6 partitions into three cells, and 1 partition into four cells for a total of 15 partitions. 27. 52. The set {a, b, c, d, e} has 1 partition into one cell, 15 into two cells, 25 into three cells, 10 into four cells, and 1 into five cells for a total of 52. (Do a combinatorics count for each possible case, such as a 1,2,2 split where there are 15 possible partitions.) 28. Reflexive: In order for x R x to be true, x must be in the same cell of the partition as the cell that contains x. This is certainly true. Transitive: Suppose that x R y and y R z. Then x is in the same cell as y so x = y , and y is in the same cell as z so that y = z . By the transitivity of the set equality relation on the collection of cells in the partition, we see that x = z so that x is in the same cell as z. Consequently, x R z. 29. Not an equivalence relation; 0 is not related to 0, so it is not reflexive. 30. Not an equivalence relation; 3 ≥ 2 but 2 ≥ 3, so it is not symmetric. 31. Not an equivalence relation since transitivity fails: 3R 15 and 15 R 5, but 3 R 5. Also not reflexive: 1 R 1. 32. 0 = (0, 0) and ( x, y ) is the circle centered at the origin with radius x2 + y2 . 33. (See the answer in the text.) 34. It is an equivalence relation; 1 = {1, 11, 21, 31,· · ·}, 2 = {2, 12, 22, 32,· · ·}, · · ·, 10 = {10, 20, 30, 40,· · ·}. 35. a. { . . . , –3, 0, 3, . . . }, { . . . , –2, 1, 4, . . .}, { . . . , –1, 2, 5, . . .} b. { . . . , –4, 0, 4, . . . }, { . . . , –3, 1, 4, . . .}, { . . . , –6, –2, 2, . . .}, { . . . , –5, –1, 3, . . .} c. { . . . , –5, 0, 5, . . . }, {. . . , –4, 1, 6, . . . }, { . . . , –3, 2, 7, . . .}, { . . . , –2, 3, 8, . . . }. { . . . , –1, 4, 9, . . . } Copyright © 2021 Pearson Education, Inc. 4 0. Sets and Relations 36. a. {0, 1, 2} b. {0, 1, 2, 3} c. {0, 1, 2, 3, 4} 37. 1 = { x ∈  x ÷ n has remainder 1} depends on the value of n. 38. a. Let h, k, and m be positive integers. We check the three criteria. Reflexive: h – h = n0 so h ∼ h. Symmetric: If h ∼ k so that h – k = ns for some s ∈ , then k – h = n(–s) so k ∼ h. Transitive: If h ∼ k and k ∼ m, then for some s, t ∈ , we have h – k = ns and k – m = nt. Then h – m = (h − k) + (k – m) = ns + nt = n(s + t), so h ∼ m. b. Let h, k ∈ . In the sense of this exercise, h ∼ k if and only if h – k = nq for some q ∈ . In the sense of Example 0.19, h ≡ k (mod n) if and only if h and k have the same remainder when divided by n. Write h = nq1 + r1 and k = nq2 + r2 where 0 ≤ r1 < n and 0 ≤ r2 j, because for each product aikbkj where i > j appearing in the computation of cij, either k j so that bkj = 0. Thus the product of two upper-triangular matrices is again upper triangular. The equation det(AB) = det(A) · det(B), shows that the product of two matrices of determinant 1 again has determinant 1. Associative: We know that matrix multiplication is associative. Identity: The n × n identity matrix I n has determinant 1 and is upper triangular. −1 −1 Inverse: The product property 1 = det(In ) = det( A A) = det( A ) ⋅ det( A) shows that if det(A) = 1, then det(A−1) = 1 also. Copyright © 2021 Pearson Education, Inc. 2. Groups 9 18. Matrix multiplication is associative, so it remains to show that G is closed under matrix multiplication, G has an identity and each element of G has an inverse. The ∗ e a b table for G is e e a b a a b e b b e a from which all of these properties are easily spotted. 19. a. We must show that S is closed under ∗, that is, that a + b + ab =/ −1 for a, b ∈ S. Now a + b + ab = −1 if and only if 0 = ab + a + b + 1 = ( a + 1)(b + 1). This is the case if and only if either a = −1 or b = −1, which is not the case for a, b ∈ S. b. Associative: We have a ∗ (b ∗ c ) = a ∗ (b + c + bc ) = a + (b + c + bc ) + a (b + c + bc ) = a + b + c + ab + ac + bc + abc and ( a ∗ b ) ∗ c = ( a + b + ab ) ∗ c = ( a + b + ab ) + c + ( a + b + ab ) c = a + b + c + ab + ac + bc + abc. Identity: 0 acts as identity element for ∗, for 0 ∗ a = a ∗ 0 = a. −a Inverses: a+1 acts as inverse of a, for a∗ −a −a −a a(a +1) − a − a2 0 =a+ +a = = = 0. a +1 a + 1 a +1 a +1 a +1 c. Because the operation is commutative, 2 ∗ x ∗ 3 = 2 ∗ 3 ∗ x = 11∗ x. Now the inverse of 11 is –11/12 by Part(b). From 11 ∗ x = 7, we obtain x= − 11 − 11 − 11 − 11 + 84 − 77 − 4 1 ∗7 = +7+ 7= = =− . 12 12 12 12 12 3 e a b c e e a b c a a e c b b c c c b 20. Table I e a b c e a b c e e a b c e e a b c b a a e c b a a b c e e a b b c a e b b c e a a e c c b e a c c e a b Table II Table III Table I is structurally different from the others because every element is its own inverse. Table II can be made to look just like Table III by interchanging the names a and b everywhere to obtain. Copyright © 2021 Pearson Education, Inc. 10 2. Groups e b a c e e b a c b b e c a a a c b c c c a e b and rewriting this table in the order e, a, b, c. a. The symmetry of each table in its main diagonal shows that all groups of order 4 are commutative. b. 0 −1 −1 0  0 1 1 0 as e,  as a,  as b,     as c to obtain  0 1  1 0  0 −1 −1 0 Relabel  Table III. c. Take n = 2. There are four 2 × 2 diagonal matrices with entries ±1, namely 1 0 −1 0 1 0 −1 0 E= ,A=  ,B =  , and C =     . 0 1   0 1 0 −1  0 −1 If we write the table for this group using the letters E, A, B, C in that order, we obtain Table I with the letters capitalized. 21. A binary operation on a set {x, y} of two elements that produces a group is completely determined by the choice of x or y to serve as identity element, so just 2 of the 16 possible tables give groups. For a set {x, y, z} of three elements, a group binary operation is again determined by the choice x, y, or z to serve as identity element, so there are just 3 of the 19,683 binary operations that give groups. (Recall that there is only one way to fill out a group table for {e, a} and for {e, a, b} if you require e to be the identity element.) 22. The orders G1G3G2, G3G1G2, and G3G2G1 are not acceptable. The identity element e occurs in the statement of G3, which must not come before e is defined in G2. 23. Ignoring spelling, punctuation and grammar, here are some of the mathematical errors. a. The statement “x = identity” is wrong. b. The identity element should be e, not (e). It would also be nice to give the properties satisfied by the identity element and by inverse elements. c. Associativity is missing. Logically, the identity element should be mentioned before inverses. The statement “an inverse exists” is not quantified correctly: for each element of the set, an inverse exists. Again, it would be nice to give the properties satisfied by the identity element and by inverse elements. d. Replace “such that for all a, b ∈ G ” by “if for all a ∈ G ”. Delete “under addition” in line 2. The element should be e, not {e}. Replace “= e” by “= a” in line 3. Copyright © 2021 Pearson Education, Inc. 2. Groups 24. a. ∗ e a b e e a a a b b b. 11 ∗ e a b c b e e a b c e b a a e b b b e b b b e b c c b b e 25. F T T F F T T T F T 26. Multiply both sides of the equation a ∗ b = a ∗ c on the left by the inverse of a, and simplify, using the axioms for a group. 27. Show that x = a′∗ b is a solution of a ∗ x = b by substitution and the axioms for a group. Then show that it is the only solution by multiplying both sides of the equation a ∗ x = b on the left by a′ and simplifying, using the axioms for a group. 28. First check that b ′ ∗ a ∗ b =/ c. Use group properties to show (b′ab)(b′ab) = c. 29. Let S = { x ∈ G | x ′ =/ x}. Then S has an even number of elements, because its elements can be grouped in pairs x, x′. Because G has an even number of elements, the number of elements in G but not in S (the set G – S) must be even. The set G – S is nonempty because it contains e. Thus there is at least one element of G – S other than e, that is, at least one element other than e that is its own inverse. 30. a. We have ( a ∗ b ) ∗ c = ( | a | b ) ∗ c | (| a | b ) | c = | ab | c. We also have a ∗ (b ∗ c ) = a ∗ (| b | c ) = | a || b | c = | ab | c , so * is associative. b. We have 1 ∗ a = | 1 | a = a for a ∈ * so 1 is a left identity element. For a ∈ *, 1/|a| is a right inverse. c. It is not a group because both 1/2 and –1/2 are right inverse of 2. d. The one-sided definition of a group, mentioned just before the exercises, must be all left sided or all right sided. We must not mix them. 31. Let G, ∗ be a group and let x ∈ G such that x ∗ x = x. Then x ∗ x = x ∗ e , and by left cancellation, x = e, so e is the only idempotent element in a group. 32. We have e = ( a ∗ b ) ∗ ( a ∗ b ) , and ( a ∗ a ) ∗ ( b ∗ b ) = e ∗ e = e also. Thus a ∗ b ∗ a ∗ b = a ∗ a ∗ b ∗ b. Using left and right cancellation, we have b ∗ a = a ∗ b . 33. Let P ( n ) be the statement ( a ∗ b ) n = a n ∗ b n . Since ( a ∗ a )1 = a ∗ b = a 1 ∗ b1 , we see P(1) is true. Suppose P(k) is true. Then ( a ∗ b ) k +1 = ( a ∗ b ) k ∗ ( a ∗ b ) = ( a k ∗ b k ) ∗ ( a ∗ b ) = [ a k ∗ (b k ∗ a )] ∗ b = [ a k ∗ ( a ∗ b k )] ∗ b = [( a k ∗ a ) ∗ b k ] ∗ b = ( a k +1 ∗ b k ) ∗ b = a k +1 ∗ (b k ∗ b ) = a k +1 ∗ b k +1 . This completes the induction argument. 34. Start with a ∗ b = b ∗ a′ and conclude a′ ∗ a ∗ b ∗ a = a′ ∗ b ∗ a′ ∗ a and simplify. 35. b 2 a 12 36. The elements e, a, a2 , a3 ,, am aren’t all different since G has only m elements. If one of a, a2 , a3 , , am is e, then we are done. If not, then we must have ai = aj where i < j. Repeated left cancellation of a yields e = aj–i. Copyright © 2021 Pearson Education, Inc. 12 2. Groups 37. We have ( a ∗ b ) ∗ ( a ∗ b ) = ( a ∗ a ) ∗ (b ∗ b ) , so a ∗ [b ∗ ( a ∗ b )] = a ∗ [ a ∗ (b ∗ b )] and left cancellation yields b ∗ ( a ∗ b ) = a ∗ (b ∗ b ) . Then (b ∗ a ) ∗ b = ( a ∗ b ) ∗ b and right cancellation yields b ∗ a = a ∗ b . 38. Let a ∗ b = b ∗ a . Then ( a ∗ b )′ = (b ∗ a )′ = a ′ ∗ b ′ by Corollary 2.19. Conversely, if b′ ∗ a′ = a′ ∗ b′ . then Then so ( a ∗ b )′ = a ′ ∗ b ′ ( b ′ ∗ a ′ ) ′ = ( a ′ ∗ b ′) ( a ′) ′ ∗ ( b ′) ′ = ( b ′) ′ ∗ ( a ′) ′ and a ∗ b = b ∗ a . 39. We have a ∗ b ∗ c = a ∗ (b ∗ c ) = e , which implies that b*c is the inverse of a. Therefore ( b ∗ c ) ∗ a = b ∗ c ∗ a = e also. 40. We need to show that a left identity element is a right identity element and that a left inverse is a right inverse. Note that e * e = e. Then ( x ′ ∗ x ) ∗ e = x ′ ∗ x so ( x ′)′ ∗ ( x ′ ∗ x ) ∗ e = ( x ′)′ ∗ ( x ′ ∗ x ). Using associativity, [( x ′) ′ ∗ x ′] ∗ x ∗ e = [( x ′) ′ ∗ x′] ∗ x. Thus ( e ∗ x ) ∗ e = e ∗ x so x ∗ e = x and e is a right identity element also. If a′∗ a = e , then ( a ′ ∗ a ) ∗ a ′ = e ∗ a ′ = a ′ . Multiplication of a′ ∗ a ∗ a′ = a′ on the left by (a′)′ and associativity yield a * a′ = e, so a′ is also a right inverse of a. 41. Using the hint, we show there is a left identity element and that each element has a left inverse. Let a ∈ G ; we are given that G is nonempty. Let e be a solution of y ∗ a = a. We show e ∗ b = b for any b ∈G. Let c be a solution of the equation a ∗ x = b. Then e ∗ b = e ∗ ( a ∗ c ) = ( e ∗ a ) ∗ c = a ∗ c = b. Thus e is a left identity. Now for each a ∈G. let a′ be a solution of y ∗ a = e. Then a′ is a left inverse of a. By Exercise 38, G is a group. 42. a and (a′)′ both satisfy a′ ∗ x = e. So a = (a′)′ by Theorem 2.17. 43. a) Let P, Q∈2 with P =/ Q . Then the distance from P to Q is positive which implies the distance from φ ( P ) and φ ( Q ) is positive. Therefore φ ( P ) =/ φ ( Q ). b) Let Q∈2 . We need to find a point T ∈  2 with φ (T ) = Q . Let φ (0, 0) = C . If Q = C we are done. Otherwise, let the distance between Q and C be r. Then φ maps S1 the circle centered at (0, 0) with radius r to S2 the circle centered at C with radius r. Let P = φ ( r , 0) , W = φ (0, r ) , d1 the distance from P to Q and d2 the distance from W to Q. Then Q is the unique point in S2 with distance d1 from P and distance d2 from W. Since S1 is congruent with S2 and under the congruence, (r, 0) and (0, r) correspond with P and W respectively, there is a unique point T ∈ S 1 whose distance from (r, 0) is d1 and whose distance from (0, r) is d2. Then φ (T ) is on the circle S2 and the distance from P and W are d1 and d2 respectively. Since the Q is the unique point with this property, Q = φ (T ). 44. Since f : G 1 → G 2 is one-to-one and onto, f − 1 : G 2 → G 1 exists and f − 1 is also one-to-one and onto. We only need to verify Condition 2 in the definition isomorphism. Let y1, y 2 ∈ G 2 be arbitrary, since f is onto, there exists x1, x 2 ∈ G 1 such that f ( x1 ) = y 1 and f ( x 2 ) = y 2 . Since f ( x1 ∗1 x 2 ) = f ( x1 ) ∗ 2 f ( x 2 ) = y1 ∗ 2 y 2 , Thus f − 1 ( f ( x1 ∗ 1 x 2 )) = f − 1 ( y 1 ∗ 2 y 2 )). f − 1 ( y 1 ) ∗ 1 f − 1 ( y 2 ) = x1 ∗ 1 x 2 = f − 1 ( f ( x1 ∗ x 2 ) = f − 1 ( y 1 ∗ 2 y ) which is Condition 2. Copyright © 2021 Pearson Education, Inc. 2. Groups 13 45. Let g ∈ G and define B = { a ′ ∗ g | a ∈ A} . Note the the function f : A → B given by f(x) = x′g is one-to-one. Therefore both A and B have more than half the elements of G which implies that A and B have a common element, say b. Thus there is an a ∈ A such that a ′ ∗ g = b ∈ B which implies that g = a ∗ b . Copyright © 2021 Pearson Education, Inc. 14 3. Abelian Examples 3. Abelian Examples 1. i 3 = i 2 ⋅ i = −1 ⋅ i = −i 2. i 4 = (i 2 )2 = (−1)2 = 1 3. –1 4. –i 5. 20 – 9i 6. (8 + 2i)(3 − i) = 24 − 8i + 6i − 2i 2 = 24 − 2i − 2(−1) = 26 − 2i 7. (2 − 3i )(4 + i ) + (6 − 5i ) = 8 + 2i − 12i − 3i 2 + 6 − 5i = 14 − 15i − 3(−1) = 17 − 15i 8. (1 + i )3 = (1 + i)2 (1 + i ) = (1 + 2i − 1)(1 + i ) = 2i (1 + i) = 2i 2 + 2i = −2 + 2i 9. (1 − i)5 = 15 + 1514 (−i) + 52⋅⋅4113 (−i ) 2 + 52⋅⋅41 12 (−i )3 + 1511 (−i ) 4 + (−i )5 = 1 − 5i + 10i 2 − 10i 3 + 5i 4 − i 5 = 1 − 5i − 10 + 10i + 5 − i = −4 + 4i 10. 13 11. π 2 + e2 12. | 3 − 4i | = 32 + (−4)2 = 25 = 5 and 3 − 4i = 5 ( 53 − 54 i ) 13. | z | = 2 and −1 − i = 2 ( − 12 − 12 i ) 5 14. |12 + 5i | = 122 + 52 = 169 and 12 + 5i = 13 (12 13 + 13 i ) 15. | −3 + 5i | = (−3)2 + 52 = 34 and −3 + 5i = 34(− 334 + 534 i) 16. | z |4 (cos 4θ + i sin 4θ ) = 1(1 + 0i ) so |z| = and cos 4θ = 1 and sin 4θ = 0 . Thus 4θ = 0 + n (2π) so θ = n π2 which yields values 0, π2 , π , and 3π2 less than 2π . The solutions are π π z1 = cos 0 + i sin 0 = 1, z2 = cos + i sin = i, 2 2 z3 = cos π + i sin π = −1, and z4 = cos 3π 3π + i sin = −i. 2 2 17. | z |4 (cos 4θ + i sin 4θ ) = 1(−1 + 0i ) so |z| = 1 and cos 4θ = −1 and sin 4θ = 0 . Thus 4θ = π + n(2π) so θ = π4 + n 2π which yields values π4 , 3π4 , 5π4 , and 7π4 less than 2π . The solutions are π π 1 1 3π 3π 1 1 z1 = cos + i sin = + i, z2 = cos + i sin = − + i, 4 4 4 4 2 2 2 2 z3 = cos 5π 5π 1 1 + i sin = − − i, and 4 4 2 2 z4 = cos 7π 7π 1 1 + i sin = − i. 4 4 2 2 18. | z |3 (cos 3θ + i sin 3θ ) = 125(−1 + 0i ), so |z| = 5, cos 3θ = − 1 and sin 3θ = 0. Thus 3θ = π + n(2π) , so θ = π3 + n 2π3 which yields values π3 , π and 5π3 less than 2π . The Copyright © 2021 Pearson Education, Inc.