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

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INSTRUCTORโS
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 ยฉ 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. Fermatโs and Eulerโs 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. Grรถbner 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, โ 3}
2. {2, โ3}.
3. {1, โ1, 2, โ2, 3, โ3, 4, โ4, 5, โ5, 6, โ6, 10, โ10, 12, โ12, 15, โ15, 20, โ20, 30, โ30,
60, โ60}
4. {2, 3, 4, 5, 6, 7, 8}
5. It is not a well-defined set. (Some may argue that no element of ๏ข + 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 โ
.)
6. โ
7. The set is โ
because 33 = 27 and 43 = 64.
8. { r โ ๏ค r = 2an for some a a โ ๏ข + and some integer n โฅ 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 ร 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. ฯ : [ 0,1] โ [ 0, 2] where ฯ ( x ) = 2 x
b. ฯ : [1, 3] โ [5, 25] where ฯ ( x ) = 2 x + 3
c.
ฯ : [ a, b ] โ [ c, d ] where ฯ ( x ) = c +
d โc
( x โ a)
bโa
15. Let ฯ : S โ ๏ก be defined by ฯ ( x ) = tan(ฯ (x โ 12 )).
16. a. โ
; cardinality 1
b. โ
, {a}; cardinality 2
c.
โ
, {a}, {b}, {a, b}; cardinality 4
d. โ
, {a}, {b},{c}, {a, b},{a, c}, {b, c}, {a, b, c}; cardinality 8
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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, ยท ยท ยท , s โ 1} and A = {1, 2,
3, ยท ยท ยท , 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(โ
)| = 1, so if |A| = s, then |P(A)| = 2s.
18. We define a one-to-one map ฯ of BA onto P(A). Let f โ BA, and let
ฯ ( f ) = {x โ A | f ( x ) = 1}. Suppose ฯ (f ) = ฯ (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 โ A so f = g and ฯ is
one to one. To show that ฯ is onto P(A), let S โ A, and let h : A โ {0, 1} be
defined by h(x) = 1 if x โ S and h(x) = 0 otherwise. Clearly ฯ (h) = S, showing that
ฯ is indeed onto P(A).
19. Picking up from the hint, let Z = {x โ A | x โ ฯ ( x )}. We claim that for any
a โ A, ฯ ( a ) = Z . Either a โ ฯ ( a ) , in which case a โ Z , or a โ ฯ ( a ) , in which
case a โ Z . Thus Z and ฯ (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 ๏ข, 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 doesnโt, 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 = {โ2, โ1, 0} and B = {1, 2, 3, ยท ยท ยท} = ๏ข + . Then |A| = 3 and |B| = โต0, and A and
B have no elements in common. The set C containing all elements in either A or B is
C = {โ2, โ1, 0, 1, 2, 3, ยท ยท ยท}. The map ฯ : C โ B defined by ฯ (x) = x + 3 is one to
one and onto B, so |C| = |B| = โต0. Thus we consider 3 + โต0 = โต0.
ii) Let A = {1, 2, 3, ยท ยท ยท} and B = {1/2, 3/2, 5/2, ยท ยท ยท}. Then |A| = |B| = โต0 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, ยท ยท ยท}. The map ฯ : C โ A defined by ฯ (x) = 2x is one to one
and onto A, so |C| = |A| = โต0. Thus we consider โต0 + โต0 = โต0
b. We leave the plotting of the points in A ร B to you. Figure 0.15 in the text, where
there are โต0 rows each having โต0 entries, illustrates that we would consider that
โต0 ยท โต0 = โต0.
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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 .##### ยท ยท ยท, we
expect 10โต0 sequences representing all numbers x โ ๏ก such that 0 โค x โค 1, but a
sequence trailing off in 0โs may represent the same x โ ๏ก as a sequence trailing of
in 9โs. At any rate, we should have 10โต0 โฅ |[0, 1]| = ๏ก ; see Exercise 15. On the
other hand, we can represent numbers in ๏ก using any integer base n > 1, and these
same 10โต0 sequences using digits from 0 to 9 in base n = 12 would not represent all
x โ [0, 1], so we have 10โต0 โค ๏ก . Thus we consider the value of 10โต0 to be ๏ก . We
could make the same argument using any other integer base n > 1, and thus
consider nโต0 = ๏ก for n โ ๏ข + , n > 1. In particular, 12โต0 = 12โต0 = ๏ก .
๏ก
22. โต0 , | ๏ก |, 2|๏ก| , 2( 2 ) , 2(2
(2
๏ก
)
)
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 โฅ 2 but 2 โฅ 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,ยท ยท ยท}, 2 = {2, 12, 22, 32,ยท ยท ยท}, ยท ยท ยท, 10 = {10, 20, 30, 40,ยท ยท ยท}.
35. a. { . . . , โ3, 0, 3, . . . }, { . . . , โ2, 1, 4, . . .}, { . . . , โ1, 2, 5, . . .}
b. { . . . , โ4, 0, 4, . . . }, { . . . , โ3, 1, 4, . . .}, { . . . , โ6, โ2, 2, . . .}, { . . . , โ5, โ1, 3, . . .}
c. { . . . , โ5, 0, 5, . . . }, {. . . , โ4, 1, 6, . . . }, { . . . , โ3, 2, 7, . . .}, { . . . , โ2, 3, 8, . . . }.
{ . . . , โ1, 4, 9, . . . }
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4
0. Sets and Relations
36. a. {0, 1, 2} b. {0, 1, 2, 3} c. {0, 1, 2, 3, 4}
37. 1 = { x โ ๏ข x รท 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 โ h = n0 so h โผ h.
Symmetric: If h โผ k so that h โ k = ns for some s โ ๏ข, then k โ h = n(โs) so k โผ h.
Transitive: If h โผ k and k โผ m, then for some s, t โ ๏ข, we have h โ k = ns and k โ m
= nt. Then h โ m = (h โ k) + (k โ m) = ns + nt = n(s + t), so h โผ m.
b. Let h, k โ ๏ข. In the sense of this exercise, h โผ k if and only if h โ k = nq for some
q โ ๏ข. In the sense of Example 0.19, h โก 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 โค r1 < n and 0 โค
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) ยท 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 ร n identity matrix I n has determinant 1 and is upper triangular. โ1 โ1 Inverse: The product property 1 = det(In ) = det( A A) = det( A ) โ det( A) shows that if det(A) = 1, then det(Aโ1) = 1 also. Copyright ยฉ 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 โ 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 โ, that is, that a + b + ab =/ โ1 for a, b โ S. Now a + b + ab = โ1 if and only if 0 = ab + a + b + 1 = ( a + 1)(b + 1). This is the case if and only if either a = โ1 or b = โ1, which is not the case for a, b โ S. b. Associative: We have a โ (b โ c ) = a โ (b + c + bc ) = a + (b + c + bc ) + a (b + c + bc ) = a + b + c + ab + ac + bc + abc and ( a โ b ) โ c = ( a + b + ab ) โ c = ( a + b + ab ) + c + ( a + b + ab ) c = a + b + c + ab + ac + bc + abc. Identity: 0 acts as identity element for โ, for 0 โ a = a โ 0 = a. โa Inverses: a+1 acts as inverse of a, for aโ โa โa โa a(a +1) โ a โ a2 0 =a+ +a = = = 0. a +1 a + 1 a +1 a +1 a +1 c. Because the operation is commutative, 2 โ x โ 3 = 2 โ 3 โ x = 11โ x. Now the inverse of 11 is โ11/12 by Part(b). From 11 โ x = 7, we obtain x= โ 11 โ 11 โ 11 โ 11 + 84 โ 77 โ 4 1 โ7 = +7+ 7= = =โ . 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 ยฉ 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. ๏ฉ0 โ1๏น ๏ฉโ1 0๏น ๏ฉ 0 1๏น ๏ฉ1 0๏น as e, ๏ช as a, ๏ช as b, ๏ช ๏บ ๏บ ๏บ as c to obtain ๏บ ๏ซ0 1 ๏ป ๏ซ1 0๏ป ๏ซ 0 โ1๏ป ๏ซโ1 0๏ป Relabel ๏ช Table III. c. Take n = 2. There are four 2 ร 2 diagonal matrices with entries ยฑ1, namely ๏ฉ1 0๏น ๏ฉโ1 0๏น ๏ฉ1 0๏น ๏ฉโ1 0๏น E=๏ช ,A= ๏ช ,B = ๏ช , and C = ๏ช ๏บ ๏บ ๏บ ๏บ. ๏ซ0 1 ๏ป ๏ซ 0 1๏ป ๏ซ0 โ1๏ป ๏ซ 0 โ1๏ป 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 โx = identityโ 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 โan inverse existsโ 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 โsuch that for all a, b โ G โ by โif for all a โ G โ. Delete โunder additionโ in line 2. The element should be e, not {e}. Replace โ= eโ by โ= aโ in line 3. Copyright ยฉ 2021 Pearson Education, Inc. 2. Groups 24. a. โ e a b e e a a a b b b. 11 โ 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 โ b = a โ c on the left by the inverse of a, and simplify, using the axioms for a group. 27. Show that x = aโฒโ b is a solution of a โ 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 โ x = b on the left by aโฒ and simplifying, using the axioms for a group. 28. First check that b โฒ โ a โ b =/ c. Use group properties to show (bโฒab)(bโฒab) = c. 29. Let S = { x โ G | x โฒ =/ x}. Then S has an even number of elements, because its elements can be grouped in pairs x, xโฒ. Because G has an even number of elements, the number of elements in G but not in S (the set G โ S) must be even. The set G โ S is nonempty because it contains e. Thus there is at least one element of G โ S other than e, that is, at least one element other than e that is its own inverse. 30. a. We have ( a โ b ) โ c = ( | a | b ) โ c | (| a | b ) | c = | ab | c. We also have a โ (b โ c ) = a โ (| b | c ) = | a || b | c = | ab | c , so * is associative. b. We have 1 โ a = | 1 | a = a for a โ๏ * so 1 is a left identity element. For a โ ๏*, 1/|a| is a right inverse. c. It is not a group because both 1/2 and โ1/2 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, โ be a group and let x โ G such that x โ x = x. Then x โ x = x โ e , and by left cancellation, x = e, so e is the only idempotent element in a group. 32. We have e = ( a โ b ) โ ( a โ b ) , and ( a โ a ) โ ( b โ b ) = e โ e = e also. Thus a โ b โ a โ b = a โ a โ b โ b. Using left and right cancellation, we have b โ a = a โ b . 33. Let P ( n ) be the statement ( a โ b ) n = a n โ b n . Since ( a โ a )1 = a โ b = a 1 โ b1 , we see P(1) is true. Suppose P(k) is true. Then ( a โ b ) k +1 = ( a โ b ) k โ ( a โ b ) = ( a k โ b k ) โ ( a โ b ) = [ a k โ (b k โ a )] โ b = [ a k โ ( a โ b k )] โ b = [( a k โ a ) โ b k ] โ b = ( a k +1 โ b k ) โ b = a k +1 โ (b k โ b ) = a k +1 โ b k +1 . This completes the induction argument. 34. Start with a โ b = b โ aโฒ and conclude aโฒ โ a โ b โ a = aโฒ โ b โ aโฒ โ a and simplify. 35. b 2 a 12 36. The elements e, a, a2 , a3 ,๏, am arenโt 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 = ajโi. Copyright ยฉ 2021 Pearson Education, Inc. 12 2. Groups 37. We have ( a โ b ) โ ( a โ b ) = ( a โ a ) โ (b โ b ) , so a โ [b โ ( a โ b )] = a โ [ a โ (b โ b )] and left cancellation yields b โ ( a โ b ) = a โ (b โ b ) . Then (b โ a ) โ b = ( a โ b ) โ b and right cancellation yields b โ a = a โ b . 38. Let a โ b = b โ a . Then ( a โ b )โฒ = (b โ a )โฒ = a โฒ โ b โฒ by Corollary 2.19. Conversely, if bโฒ โ aโฒ = aโฒ โ bโฒ . then Then so ( a โ b )โฒ = a โฒ โ b โฒ ( b โฒ โ a โฒ ) โฒ = ( a โฒ โ b โฒ) ( a โฒ) โฒ โ ( b โฒ) โฒ = ( b โฒ) โฒ โ ( a โฒ) โฒ and a โ b = b โ a . 39. We have a โ b โ c = a โ (b โ c ) = e , which implies that b*c is the inverse of a. Therefore ( b โ c ) โ a = b โ c โ 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 โฒ โ x ) โ e = x โฒ โ x so ( x โฒ)โฒ โ ( x โฒ โ x ) โ e = ( x โฒ)โฒ โ ( x โฒ โ x ). Using associativity, [( x โฒ) โฒ โ x โฒ] โ x โ e = [( x โฒ) โฒ โ xโฒ] โ x. Thus ( e โ x ) โ e = e โ x so x โ e = x and e is a right identity element also. If aโฒโ a = e , then ( a โฒ โ a ) โ a โฒ = e โ a โฒ = a โฒ . Multiplication of aโฒ โ a โ aโฒ = aโฒ on the left by (aโฒ)โฒ and associativity yield a * aโฒ = e, so aโฒ 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 โ G ; we are given that G is nonempty. Let e be a solution of y โ a = a. We show e โ b = b for any b โG. Let c be a solution of the equation a โ x = b. Then e โ b = e โ ( a โ c ) = ( e โ a ) โ c = a โ c = b. Thus e is a left identity. Now for each a โG. let aโฒ be a solution of y โ a = e. Then aโฒ is a left inverse of a. By Exercise 38, G is a group. 42. a and (aโฒ)โฒ both satisfy aโฒ โ x = e. So a = (aโฒ)โฒ by Theorem 2.17. 43. a) Let P, Qโ๏2 with P =/ Q . Then the distance from P to Q is positive which implies the distance from ฯ ( P ) and ฯ ( Q ) is positive. Therefore ฯ ( P ) =/ ฯ ( Q ). b) Let Qโ๏2 . We need to find a point T โ ๏ 2 with ฯ (T ) = Q . Let ฯ (0, 0) = C . If Q = C we are done. Otherwise, let the distance between Q and C be r. Then ฯ maps S1 the circle centered at (0, 0) with radius r to S2 the circle centered at C with radius r. Let P = ฯ ( r , 0) , W = ฯ (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 โ S 1 whose distance from (r, 0) is d1 and whose distance from (0, r) is d2. Then ฯ (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 = ฯ (T ). 44. Since f : G 1 โ G 2 is one-to-one and onto, f โ 1 : G 2 โ G 1 exists and f โ 1 is also one-to-one and onto. We only need to verify Condition 2 in the definition isomorphism. Let y1, y 2 โ G 2 be arbitrary, since f is onto, there exists x1, x 2 โ G 1 such that f ( x1 ) = y 1 and f ( x 2 ) = y 2 . Since f ( x1 โ1 x 2 ) = f ( x1 ) โ 2 f ( x 2 ) = y1 โ 2 y 2 , Thus f โ 1 ( f ( x1 โ 1 x 2 )) = f โ 1 ( y 1 โ 2 y 2 )). f โ 1 ( y 1 ) โ 1 f โ 1 ( y 2 ) = x1 โ 1 x 2 = f โ 1 ( f ( x1 โ x 2 ) = f โ 1 ( y 1 โ 2 y ) which is Condition 2. Copyright ยฉ 2021 Pearson Education, Inc. 2. Groups 13 45. Let g โ G and define B = { a โฒ โ g | a โ A} . Note the the function f : A โ B given by f(x) = xโฒg 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 โ A such that a โฒ โ g = b โ B which implies that g = a โ b . Copyright ยฉ 2021 Pearson Education, Inc. 14 3. Abelian Examples 3. Abelian Examples 1. i 3 = i 2 โ i = โ1 โ i = โi 2. i 4 = (i 2 )2 = (โ1)2 = 1 3. โ1 4. โi 5. 20 โ 9i 6. (8 + 2i)(3 โ i) = 24 โ 8i + 6i โ 2i 2 = 24 โ 2i โ 2(โ1) = 26 โ 2i 7. (2 โ 3i )(4 + i ) + (6 โ 5i ) = 8 + 2i โ 12i โ 3i 2 + 6 โ 5i = 14 โ 15i โ 3(โ1) = 17 โ 15i 8. (1 + i )3 = (1 + i)2 (1 + i ) = (1 + 2i โ 1)(1 + i ) = 2i (1 + i) = 2i 2 + 2i = โ2 + 2i 9. (1 โ i)5 = 15 + 1514 (โi) + 52โ โ 4113 (โi ) 2 + 52โ โ 41 12 (โi )3 + 1511 (โi ) 4 + (โi )5 = 1 โ 5i + 10i 2 โ 10i 3 + 5i 4 โ i 5 = 1 โ 5i โ 10 + 10i + 5 โ i = โ4 + 4i 10. 13 11. ฯ 2 + e2 12. | 3 โ 4i | = 32 + (โ4)2 = 25 = 5 and 3 โ 4i = 5 ( 53 โ 54 i ) 13. | z | = 2 and โ1 โ i = 2 ( โ 12 โ 12 i ) 5 14. |12 + 5i | = 122 + 52 = 169 and 12 + 5i = 13 (12 13 + 13 i ) 15. | โ3 + 5i | = (โ3)2 + 52 = 34 and โ3 + 5i = 34(โ 334 + 534 i) 16. | z |4 (cos 4ฮธ + i sin 4ฮธ ) = 1(1 + 0i ) so |z| = and cos 4ฮธ = 1 and sin 4ฮธ = 0 . Thus 4ฮธ = 0 + n (2ฯ) so ฮธ = n ฯ2 which yields values 0, ฯ2 , ฯ , and 3ฯ2 less than 2ฯ . The solutions are ฯ ฯ z1 = cos 0 + i sin 0 = 1, z2 = cos + i sin = i, 2 2 z3 = cos ฯ + i sin ฯ = โ1, and z4 = cos 3ฯ 3ฯ + i sin = โi. 2 2 17. | z |4 (cos 4ฮธ + i sin 4ฮธ ) = 1(โ1 + 0i ) so |z| = 1 and cos 4ฮธ = โ1 and sin 4ฮธ = 0 . Thus 4ฮธ = ฯ + n(2ฯ) so ฮธ = ฯ4 + n 2ฯ which yields values ฯ4 , 3ฯ4 , 5ฯ4 , and 7ฯ4 less than 2ฯ . The solutions are ฯ ฯ 1 1 3ฯ 3ฯ 1 1 z1 = cos + i sin = + i, z2 = cos + i sin = โ + i, 4 4 4 4 2 2 2 2 z3 = cos 5ฯ 5ฯ 1 1 + i sin = โ โ i, and 4 4 2 2 z4 = cos 7ฯ 7ฯ 1 1 + i sin = โ i. 4 4 2 2 18. | z |3 (cos 3ฮธ + i sin 3ฮธ ) = 125(โ1 + 0i ), so |z| = 5, cos 3ฮธ = โ 1 and sin 3ฮธ = 0. Thus 3ฮธ = ฯ + n(2ฯ) , so ฮธ = ฯ3 + n 2ฯ3 which yields values ฯ3 , ฯ and 5ฯ3 less than 2ฯ . The Copyright ยฉ 2021 Pearson Education, Inc.*

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