Gallian's Contemporary Abstract Algebra 11th Edition – Complete Student Solutions Manual (All Chapters Included)

SOLUTION MANUAL
All Chapters Included

, CONTEMPORARY ABSTRACT ALGEBRA 11TH EDITION
INSTRUCTOR SOLUTIONS MANUAL

CONTENTS

Integers and Equivalence Relations

0 Preliminaries 1

Groups

1 Introduction to Groups 5
2 Groups 7
3 Finite Groups; Subgroups 11
4 Cyclic Groups 18
5 Permutation Groups 25
6 Isomorphisms 31
7 Cosets and Lagrange’s Theorem 37
8 External Direct Products 44
9 Normal Subgroups and Factor Groups 50
10 Group Homomorphisms 56
11 Fundamental Theorem of Finite Abelian Groups 62
12 Introduction to Rings 66
13 Integral Domains 71
14 Ideals and Factor Rings 77
15 Ring Homomorphisms 84
16 Polynomial Rings 91
17 Factorization of Polynomials 97
18 Divisibility in Integral Domains 102

,vi


Fields

19 Extension Fields 106
20 Algebraic Extensions 111
21 Finite Fields 116
22 Geometric Constructions 121

Special Topics

23 Sylow Theorems 123
24 Finite Simple Groups 129
25 Generators and Relations 133
26 Symmetry Groups 136
27 Symmetry and Counting 138
28 Cayley Digraphs of Groups 140
29 Introduction to Algebraic Coding Theory 143
30 An Introduction to Galois Theory 147
31 Cyclotomic Extensions 150

, 1




CHAPTER 0
Preliminaries
1. {1, 2, 3, 4}; {1, 3, 5, 7}; {1, 5, 7, 11}; {1, 3, 7, 9, 11, 13, 17, 19};
{1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}
2. a. 2; 10 b. 4; 40 c. 4: 120; d. 1; 1050 e. pq2; p2q3

3. 12, 2, 2, 10, 1, 0, 4, 5.

4. s = —3, t = 2; s = 8, t = —5

5. Let a be the least common multiple of every element of the set and b be any common
multiple of every element of the set. Write b = aq + r where 0 ≤ r < a. Then, for
any element c in the set, we have that c divides b — aq = r. This means that r is a
common multiple of every element of the set and therefore is greater than or equal to
a, which is a contradiction.

6. If n = 0 mod 3, we are done. If n = 1 mod 3, the n + 2 = 0 mod 3. If n = 2 mod 3,
then n + 1 = 0 mod 3.
7. By using 0 as an exponent if necessary, we may write a = p1m1 · · · pkmk and
b = p n11 · · · p nkk , where the p’s are distinct primes and the m’s and n’s are nonnegative.
Then lcm(a, b) = ps11 · · · pskk , where si = max(m i, ni) and gcd(a, b) = p t11 · · · p tkk , where
ti = min(m i, ni) Then lcm(a, b) · gcd(a, b) = p1m1 +n 1 · · · pm k
k +n k
= ab.
8. The first part follows from the Fundamental Theorem of Arithmetic; for the second
part, take a = 4, b = 6, c = 12.

9. Write a = nq1 + r1 and b = nq2 + r2, where 0 ≤ r1, r2 < n. We may assume that
r1 ≥ r2. Then a — b = n(q1 — q2) + (r1 — r2), where r1 — r2 ≥ 0. If a mod n = b mod
n, then r1 = r2 and n divides a — b. If n divides a — b, then by the uniqueness of the
remainder, we have r1 — r2 = 0. Thus, r1 = r2 and therefore a mod n = b mod n.

10. Write as + bt = d. Then, a′s + b′t = (a/d)s + (b/d)t = 1.

11. By Exercise 9, to prove that (a + b) mod n = (a′ + b′) mod n and (ab)
mod n = (a′b′) mod n it suffices to show that n divides (a + b) — (a′ + b′) and
ab — a′b′. Since n divides both a — a′ and n divides b — b′, it divides their difference.
Because a = a′ mod n and b = b′ mod n, there are integers s and t such that
a = a′ + ns and b = b′ + nt. Thus ab = (a′ + ns)(b′ + nt) = a′b′ + nsb′ + a′nt + nsnt.
Thus, ab — a′b′ is divisible by n.

12. Write d = au + bv. Since t divides both a and b, it divides d. Write s = mq + r,
where 0 ≤ r < m. Then, r = s — mq is a common multiple of both a and b so r = 0.

13. Suppose that there is an integer n such that ab mod n = 1. Then there is an integer
q such that ab — nq = 1. Since d divides both a and n, d also divides 1. So, d = 1.
On the other hand, if d = 1, then by the corollary of Theorem 0.2, there are integers
s and t such that as + nt = 1. Thus, modulo n, as = 1.

14. 7(5n + 3) — 5(7n + 4) = 1.