Solution Manual for First Course in Abstract Algebra, 8th Edition by John B. Fraleigh| 9780135859759| All Chapters| LATEST

SOLUTION MANUAL
A FIRST COURSE IN ABSTRACT
ALGEBRA

, 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

, Contents

0. Sets And Relations 01


I. Groups And Subgroups

1. Binary Operations 05
2. Groups 08
3. Abelian Examples 14
4. Nonabelian Examples 19
5. Subgroups 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 40
10. Cosets And The Theorem Of Lagrange 45
11. Plane Isometries 50


Iii. Homomorphisms And Factor Groups

12. Factor Groups 53
13. Factor Group Computations And Simple Groups 58
14. Group Action On A Set 65
15. Applications Of G-Sets To Counting 70


Vi. Advanced Group Theory

16. Isomorphism Theorems 73
17. Sylow Theorems 75
18. Series Of Groups 80
19. Free Abelian Groups 85
20. Free Groups 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 106
25. Rsa Encryption 109


VI. Constructing Rings And Fields

26. The Field Of Quotients Of An Integral Domain 110
27. Rings Of Polynomials 112
28. Factorization Of Polynomials Over A Field 116
29. Algebraic Coding Theory 123
30. Homomorphisms And Factor Rings 125
31. Prime And Maximal Ideals 131
32. Noncommutative Examples 137


VII. Commutative Algebra

33. Vector Spaces 140
34. Unique Factorization Domains 145
35. Euclidean Domains 149
36. Number Theory 154
37. Algebraic Geometry 160
38. Gröbner Bases For Ideals 163


VIII. Extension Fields

39. Introduction To Extension Fields 168
40. Algebraic Extensions 174
41. Geometric Constructions 179
42. Finite Fields 182


IX. Galois Theory

43. Automorphisms Of Fields 185
44. Splitting Fields 191
45. Separable Extensions 195
46. Galois Theory 199

,47. Illustrations Of Galois Theory 203
48. Cyclotomic Extensions 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 Z 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 Z 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  : 0,1  0, 2 Where   X  2x
.
 : 1, 3   5, 25 Where   X  2x  3
B.


D
C.  : A, B  C, D  Where   X   X  A
CB
C 
A




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, 0. Sets and Relations 2
15. Let  : S  R Be Defined By   X  Tan( (X  )).
1


2

16. A. D.
B.
C.



;
Cardinali
Ty 1
,{A};
Cardinal
i Ty 2
,{A},{
B},{A,
B};
Cardinal
i Ty 4
,{A},{
B},{C},{
A,
B},{A,
C},{B,
C},{A, B,
C};
Cardinal
Ity 8




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, 0. Sets and Relations 3

17. Conjecture: | P  A | 2s  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  Of Ba Onto P(A). Let F ∈ Ba, And Let
  F   {X  A | F  X  1}. Suppose  (F ) =  (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  Is One To One. To Show That  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 
(H) = S, Showing That
 Is Indeed Onto P(A).
19. Picking Up From The Hint, Let Z  {X  A | X    X}. We Claim That

For Any A  A,  A  Z. Either A  A , In Which Case A  Z , Or A   A

, In Which Case A  Z. Thus Z And  (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
Z, 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, · · ·} = Z . Then |A| = 3 And |B| = 0, 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  : C → B Defined By  (X) = X + 3 Is One
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, 0. Sets and Relations 4
To One And Onto B, So |C| = |B| = 0. Thus We Consider 3 + 0 = 0.
ii) Let A = {1, 2, 3, · · ·} And B = {1/2, 3/2, 5/2, · · ·}. Then |A| = |B| = 0 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  : C → A Defined By  (X) = 2x Is One To
One And Onto A, So |C| = |A| = 0. Thus We Consider 0 + 0 = 0
B. We Leave The Plotting Of The Points In A × B To You. Figure 0.15 In The Text,
Where There Are 0 Rows Each Having 0 Entries, Illustrates That We Would
Consider That




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, 0. Sets and Relations 5




0 · 0 = 0.




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