Solutions Manual for Gallians Contemporary Abstract Algebra 11th Edition

Student Solutions
Manual

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STUDENT SOLUTIONS MANUAL
CONTEMPORARY ABSTRACT ALGEBRA,
TENTH EDITION
SELECTED PROBLEMS

CONTENTS

Integers and Equivalence Relations

0 Preliminaries 1

Groups

1 Introduction to Groups 4
2 Groups 6
3 Finite Groups; Subgroups 9
4 Cyclic Groups 15
5 Permutation Groups 21
6 Isomorphisms 27
7 Cosets and Lagrange’s Theorem 32
8 External Direct Products 38
9 Normal Subgroups and Factor Groups 45
10 Group Homomorphisms 50
11 Fundamental Theorem of Finite Abelian Groups 55
12 Introduction to Rings 59
13 Integral Domains 63
14 Ideals and Factor Rings 69
15 Ring Homomorphisms 74
16 Polynomial Rings 80
17 Factorization of Polynomials 85
18 Divisibility in Integral Domains 89

Fields

19 Extension Fields 93
20 Algebraic Extensions 97
21 Finite Fields 101
22 Geometric Constructions 105

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Special Topics

23 Sylow Theorems 106
24 Finite Simple Groups 111
25 Generators and Relations 115
26 Symmetry Groups 118
27 Symmetry and Counting 120
28 Cayley Digraphs of Groups 123
29 Introduction to Algebraic Coding Theory 125
30 An Introduction to Galois Theory 128
31 Cyclotomic Extensions 130

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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. pq 2 ; p2 q 3
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.
7. By using 0 as an exponent if necessary, we may write
mk
a = pm
1 · · · pk
1
and b = pn1 1 · · · pnk k , 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(mi , ni ) and
gcd(a, b) = pt11 · · · ptkk , where ti = min(mi , ni ). Then
lcm(a, b) · gcd(a, b) = p1m1 +n1 · · · pm k
k +nk
= ab.
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.
11. By Exercise 9, to prove that (a + b) mod n = (a0 + b0 ) mod n
and (ab) mod n = (a0 b0 ) mod n it suffices to show that n divides
(a + b) − (a0 + b0 ) and ab − a0 b0 . Since n divides both a − a0 and n
divides b − b0 , it divides their difference. Because a = a0 mod n
and b = b0 mod n, there are integers s and t such that
a = a0 + ns and b = b0 + nt. Thus
ab = (a0 + ns)(b0 + nt) = a0 b0 + nsb0 + a0 nt + nsnt. Thus, ab − a0 b0
is divisible by n.
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.