SOLUTION MANUAL First Course in Abstract Algebra A 8th Edition by John B. Fraleigh All Chapters Full Complete

SOLUTION MANUAL
First Course in Abstract
Algebra A 8th Edition by
John B. Fraleigh
All Chapters Full Complete

, CONTENTS
1. Sets and Relations 1

I. Groups and Subgroups

2. Introduction and Examples 4
3. Binary Operations 7
4. Isomorphic Binary Structures 9
5. Groups 13
6. Subgroups 17
7. Cyclic Groups 21
8. Generators and Cayley Digraphs 24

II. Permutations, Cosets, and Direct Products

9. Groups of Permutations 26
10. Orbits, Cycles, and the Alternating Groups
30
11. Cosets and the Theorem of Lagrange34
12. Direct Products and Finitely Generated Abelian Groups 37
13. Plane Isometries 42

III. Homomorphisms and Factor Groups

14. Homomorphisms 44
15. Factor Groups 49
16. Factor-Group Computations and Simple Groups 53
17. Group Action on a Set 58
18. Applications of G-Sets to Counting 61

IV. Rings and Fields

19. Rings and Fields 63
20. Integral Domains 68
21. Fermat’s and Euler’s Theorems 72
22. The Field of Quotients of an Integral Domain 74
23. Rings of Polynomials 76
24. Factorization of Polynomials over a Field 79
25. Noncommutative Examples 85
26. Ordered Rings and Fields 87

V. Ideals and Factor Rings

27. Homomorphisms and Factor Rings 89
28. Prime and Maximal Ideals 94

,29. Gröbner Bases for Ideals 99

, VI. Extension Fields

30. Introduction to Extension Fields 103
31. Vector Spaces 107
32. Algebraic Extensions 111
33. Geometric Constructions 115
34. Finite Fields 116

VII. Advanced Group Theory

35. Isomorphism Theorems 117
36. Series of Groups 119
37. Sylow Theorems 122
38. Applications of the Sylow Theory 124
39. Free Abelian Groups 128
40. Free Groups 130
41. Group Presentations 133

VIII. Groups in Topology

42. Simplicial Complexes and Homology Groups 136
43. Computations of Homology Groups 138
44. More Homology Computations and Applications 140
45. Homological Algebra 144

IX. Factorization
46. Unique Factorization Domains 148
47. Euclidean Domains 151
48. Gaussian Integers and Multiplicative Norms 154

X. Automorphisms and Galois Theory
49. Automorphisms of Fields 159
50. The Isomorphism Extension Theorem 164
51. Splitting Fields 165
52. Separable Extensions 167
53. Totally Inseparable Extensions 171
54. Galois Theory 173
55. Illustrations of Galois Theory 176
56. Cyclotomic Extensions 183
57. Insolvability of the Quintic 185

APPENDIX Matrix Algebra 187


iv

, 0. Sets and Relations 1

1. Sets and Relations
√ √
1. 3, 3} 2. The set is empty.
{ −

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. {−10, −9, −8, −7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

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. It is not a well-defined set. 9. Q

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. It is a function. It is not one-to-one since there are two pairs with second member 4. It is
not onto
B because there is no pair with second member 2.
b. (Same answer as Part(a).)
c. It is not a function because there are two pairs with first member 1.
d. It is a function. It is one-to-one. It is onto B because every element of B appears
as second member of some pair.
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. It is not a function because there are two pairs with first member 2.

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) = 2x b. φ : [1, 3] → [5, 25] where φ(x) = 5 + 10(x −
1)
d−c
c. φ : [a, b] → [c, d] where φ(x) = c + (x − a)
b−a

15. Let φ : S → R be defined by φ(x) = 1
)2).
tan(π(x −
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

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)|.

, 2 0. Sets and Relations

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)| = 2 .s

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 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| =
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 0.14 in the text, where there
are ℵ0
rows each having ℵ0 entries, illustrates that we would consider that ℵ0 · ℵ0 = ℵ0.

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 10ℵ 0 sequences representing all numbers x ∈ R such that 0 ≤ x ≤ 1, but a
sequence trailing off in 0’s may represent the same x ∈ R as a sequence trailing of in 9’s.
At any rate, we should have 10ℵ 0 ≥ |[0, 1]| = |R|; see Exercise
15. On the other hand, we can represent numbers in R using any integer base n > 1,