SOLUTION MANUAL First Course in Abstract
Algebra A 8th Edition by John B. Fraleigh
All Chapters Full Complete
, CONTENTS
0. Sets and Relations 1
I. Groups and Subgroups
1. Introduction and Examples 4
2. Binary Operations 7
3. Isomorphic Binary Structures 9
4. Groups 13
5. Subgroups 17
6. Cyclic Groups 21
7. Generators and Cayley Digraphs 24
II. Permutations, Cosets, and Direct Products
8. Groups of Permutations 26
9. Orbits, Cycles, and the Alternating Groups 30
10. Cosets and the Theorem of Lagrange 34
11. Direct Products and Finitely Generated Abelian Groups 37
12. Plane Isometries 42
III. Homomorphisms and Factor Groups
13. Homomorphisms 44
14. Factor Groups 49
15. Factor-Group Computations and Simple Groups 53
16. Group Action on a Set 58
17. Applications of G-Sets to Counting 61
IV. Rings and Fields
18. Rings and Fields 63
19. Integral Domains 68
20. Fermat’s and Euler’s Theorems 72
21. The Field of Quotients of an Integral Domain 74
22. Rings of Polynomials 76
23. Factorization of Polynomials over a Field 79
24. Noncommutative Examples 85
25. Ordered Rings and Fields 87
V. Ideals and Factor Rings
26. Homomorphisms and Factor Rings 89
27. Prime and Maximal Ideals 94
28. Gröbner Bases for Ideals 99
, VI. Extension Fields
29. Introduction to Extension Fields 103
30. Vector Spaces 107
31. Algebraic Extensions 111
32. Geometric Constructions 115
33. Finite Fields 116
VII. Advanced Group Theory
34. Isomorphism Theorems 117
35. Series of Groups 119
36. Sylow Theorems 122
37. Applications of the Sylow Theory 124
38. Free Abelian Groups 128
39. Free Groups 130
40. Group Presentations 133
VIII. Groups in Topology
41. Simplicial Complexes and Homology Groups 136
42. Computations of Homology Groups 138
43. More Homology Computations and Applications 140
44. Homological Algebra 144
IX. Factorization
45. Unique Factorization Domains 148
46. Euclidean Domains 151
47. Gaussian Integers and Multiplicative Norms 154
X. Automorphisms and Galois Theory
48. Automorphisms of Fields 159
49. The Isomorphism Extension Theorem 164
50. Splitting Fields 165
51. Separable Extensions 167
52. Totally Inseparable Extensions 171
53. Galois Theory 173
54. Illustrations of Galois Theory 176
55. Cyclotomic Extensions 183
56. Insolvability of the Quintic 185
APPENDIX Matrix Algebra 187
iv
, 0. Sets and Relations 1
0. 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)
c. φ : [a, b] → [c, d] where φ(x) = c + d−c (x
− a)
b−a
15. Let φ : S → R be defined by φ(x) = tan(π(x − 1 )).
2
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)|s = 2 .
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 ∈ / φ(a), in which case a ∈ Z. Thus Z and φ(a) are certainly
/ Z, or a ∈
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 10ℵ0 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 10ℵ0 ≥ |[0, 1]| = |R|; see
Exercise
15. On the other hand, we can represent numbers in R 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
≤ |R|. Thus we consider the value of 10ℵ0 to be |R|. We could make the same argument using any
other integer base n > 1, and thus consider nℵ0 = |R| for n ∈ Z+, n > 1. In particular, 12ℵ0 = 2ℵ0 =
|R|.
(2|R|)
22. ℵ0, |R|, 2|R|, 2(2
|R|
), 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}}.
, 0. Sets and Relations 3
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. It is an equivalence relation; 0 = {0} and a = {a, −a} for a ∈ R, a /= 0.
32. It is not an equivalence relation; 1 R 3 and 3 R 5 but we do not have 1 R 5 because |1 − 5| = 4 > 3.
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. (See the answer in the text.)
36. 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 ∈ Z, then k − h = n(−s) so k ∼ h.
Transitive: If h ∼ k and k ∼ m, then for some s, t ∈ Z, 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 ∈ Z+. In the sense of this exercise, h ∼ k if and only if h − k = nq for some q ∈ Z. 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 < n. Then
h − k = n(q1 − q2) + (r1 − r2)
and we see that h − k is a multiple of n if and only if r1 = r2. Thus the conditions are the same.
c. a. 0 = {· · · , −2, 0, 2, · · ·}, 1 = {· · · , −3, −1, 1, 3, ···}
b. 0 = {· · · , −3, 0, 3, · · ·}, 1 = {· · · , −5, −2, 1, 4, · · ·}, 2 = {· · · , −1, 2, 5, · }
c. 0 = {· · · , −5, 0, 5, · · ·}, 1 = {· · · , −9, −4, 1, 6, · · ·}, 2 = {· · · , −3, 2, 7, · },
3 = {· · · , −7, −2, 3, 8, · · ·}, 4 = {· · · , −1, 4, 9,··· }
,4 1. Introduction and Examples
37. The name two-to-two function suggests that such a function f should carry every pair of distinct points into two
distinct points. Such a function is one-to-one in the conventional sense. (If the domain has only one
element, the function cannot fail to be two-to-two, because the only way it can fail to be two-to-two is to
carry two points into one point, and the set does not have two points.) Conversely, every function that is
one-to-one in the conventional sense carries each pair of distinct points into two distinct points. Thus the
functions conventionally called one-to-one are precisely those that carry two points into two points, which
is a much more intuitive unidirectional way of regarding them. Also, the standard way of trying to
show that a function is one-to-one is precisely to show that it does not fail to be two-to-two. That is,
proving that a function is one-to-one becomes more natural in the two-to-two terminology.
1. Introduction and Examples
1. i3 = i2 · i = −1 · i = −i 2. i4 = (i2)2 = (−1)2 = 1 3. i23 = (i2)11 · i = (−1)11 · i = (−1)i = −i
4. (−i)35 = (i2)17(−i) = (−1)17(−i) = (−1)(−i) = i
5. (4 − i)(5 + 3i) = 20 + 12i − 5i − 3i2 = 20 + 7i + 3 = 23 + 7i
6. (8 + 2i)(3 − i) = 24 − 8i + 6i − 2i2 = 24 − 2i − 2(−1) = 26 − 2i
7. (2 − 3i)(4 + i) + (6 − 5i) = 8 + 2i − 12i − 3i2 + 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) = 2i2 + 2i = −2 + 2i
9. (1 − i)5 = 15 + 5 14(−i) + 5·4 13(−i)2 + 5·4 12(−i)3 + 5 11(−i)4 +(−i)5 = 1 − 5i + 10i2 − 10i3 + 5i4 − i5 =
1 2·1 2·1 1
1 − 5i − 10 + 10i + 5 — i = − 4 +
4i
√ √ √ √ √ √ √
10. |3−4i| = 32 + (−4)2 = 9 + 16 = 25 = 5 11. |6+4i| = 62 + 42 = 36 + 16 = 52 = 2 13
√ √
12. |3 − 4i| = 32 + (−4)2 = 25 = 5 and 3 − 4i = 5( 3 − 54 i) 5
√ √ √
13. | − 1 + i| = (−1)2 + 12 = 2 and − 1 + i = 2(− + √1 i)
2 2
√1
√ √
14. |12 + 5i| = 122 + 52 = 169 and 12 + 5i = 13(
13
12 + 5 i)
13
√ √ √
15. | − 3 + 5i| = (−3) + 5 = 34 and − 3 + 5i = 34(−
2 2 + √ 5 i)
34 34
√3
16. |z|4(cos 4θ + i sin 4θ) = 1(1 + 0i) so |z| = 1 and cos 4θ = 1 and sin 4θ = 0. Thus 4θ = 0 + n(2π) so
θ = n π which yields values 0, π , π, and 3π less than 2π. The solutions are
2 2 2
π π
z1 = cos 0 + i sin 0 = 1, z2 = cos + i sin = i,
2 2
3π 3π
z3 = cos π + i sin π = −1, and z4 = +i = −i.
cos sin 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= π + n π which yields values π , 3π , 5π , and 7π less than 2π. The solutions are
4 2 4 4 4 4
π π 1 1 3π 3π 1 1
z1 = cos + i sin = √ + √ i, z2 = +i = −√ + √ i,
4 4 2 2 sin 4 2 2
cos 4
5π 5π 1 1 7π 7π 1 1
z3 = +i 4 = −√ − √ i, and z4 = cos
cos sin 4 2 2
, + i sin =√
4 4 2
− √ i.
2
, 1. Introduction and Examples 5
18. |z|3(cos 3θ + i sin 3θ) = 8(−1 + 0i) so |z| = 2 and cos 3θ = −1 and sin 3θ = 0. Thus 3θ = π + n(2π) so
θ = π + n 2π which yields values π , π, and 5π less than 2π. The solutions are
3 3 3 √ 3
π π 1 3 √
z = 2(cos + i sin ) = 2( + i) = 1 + 3i, z = 2(cos π + i sin π) = 2(−1 + 0i) = −2,
1 2
3 3 2 2
and √
5π 5π 1 3 √
z3 = 2(cos +i ) = 2( i) = 1 3i.
sin 3 2 2
3 − −
19. |z|3(cos 3θ + i sin 3θ) = 27(0 − i) so |z| = 3 and cos 3θ = 0 and sin 3θ = −1. Thus 3θ = 3π/2 + n(2π)
so θ = π + n 2π which yields values π , 7π , and 11π less than 2π. The solutions are
2 3 2 6 6
√ √
π π 7π 7π 3 1 3 3 3
z1 = 3(cos + i sin ) = 3(0 + i) = 3i,
2 2 z2 = 3(cos 6 + i sin 6 ) = 3(− 2 — 2 i) = − 2 — 2 i
√ √
and 11π 11π 3 1 3 3 3
z3 = 3(cos +i )= − i) = − i.
sin 3( 2 2 2 2
6 6
20. |z|6(cos 6θ + i sin 6θ) = 1 + 0i so |z| = 1 and cos 6θ = 1 and sin 6θ = 0. Thus 6θ = 0 + n(2π)
so
θ = 0 + n 2π which yields values 0, π , 2π , π, 4π , and 5π less than 2π. The solutions are
6 3 3 3 3
√
π π 1 3
z1 = 1(cos 0 + i sin 0) = 1 + 0i = 1, z2 = 1(cos + i sin ) = + i,
3 3 2 2
√
2π 2π 1 3
z3 = +i )=− i, z4 = 1(cos π + i sin π) = −1 + 0i = −1,
1(cos sin 3 2 2
3 + √ √
4π 4π 1 3 5π 5π 1 3
z5 = +i )=− − i, z6 = +i ) = − i.
1(cos sin 3 2 2 1(cos sin 3 2 2
3 3
21. |z|6(cos 6θ + i sin 6θ) = 64(−1 + 0i) so |z| = 2 and cos 6θ = −1 and sin 6θ = 0. Thus 6θ = π + n(2π)
so θ = π + n 2π which yields values π , π , 5π , 7π , 3π and 11π6 less than 2π. The solutions are
6 6 6 2 6 6 2
√ √
z = 2(cos π π 3 1 3 + i,
1 + i sin ) = + i) =
2( 2 2
6 6
π π
z2 = 2(cos + i sin ) = 2(0 + i) = 2i,
2 2 √
5π 5π 3 1 √
z3 = 2(cos + i sin ) = 2(− + i) = − 3 + i,
6 6 √2 2 √
7π
z = 2(cos + i sin ) = 2(−
= 7π 3 1
− i) 3 − i,
−
4 6 6 2 2
3π 3π
z5 = 2(cos +i ) = 2(0 − i) = −2i,
2 sin 2
√
11π 11π 3 1 √
z6 = 2(cos + i sin ) = 2( − i) = 3 − i.
6 6 2 2
22. 10 + 16 = 26 > 17, so 10 +17 16 = 26 − 17 = 9. 23. 8 + 6 = 14 > 10, so 8 +10 6 = 14 − 10 = 4.
24. 20.5 + 19.3 = 39.8 > 25, so 20.5 +25 19.3 = 39.8 − 25 = 14.8.
, 25. 1 + 7 = 11 > 1, so 1 +1 7 = 11 − 1 = 3. 26. 3π + 3π = 9π > 2π, so 3π +2π 3π = 9π − 2π = π .
2 8 8 2 8 8 8 4 2 4 4 2 4 4
