CONTEMPORARY ABSTRACT ALGEBRA

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CONTEMPORARY ABSTRACT ALGEBRA

9TH EDITION INSTRUCTOR SOLUTION

MANUAL

CONTENTS

Integers and Equivalence Relations

• Preliminaries 1

Groups 1 Introduction to Groups 7

• Groups 9
• Finite Groups; Subgroups 13
• Cyclic Groups 20
• Permutation Groups 27
• Isomorphisms 34
• Cosets and Lagrange’s Theorem 40
• External Direct Products 46
• Normal Subgroups and Factor Groups 53

10 Group Homomorphisms 59 11 Fundamental Theorem of Finite Abelian Groups 65 12 Introduction to Rings 69 13 Integral Domains 74 14 Ideals and Factor Rings 80 15 Ring Homomorphisms 87 16 Polynomial Rings 94 17 Factorization of Polynomials 100 18 Divisibility in Integral Domains 105 1 / 4

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Fields 19 Vector Spaces 110 20 Extension Fields 114 21 Algebraic Extensions 118 22 Finite Fields 123 23 Geometric Constructions 127 Special Topics 24 Sylow Theorems 129 25 Finite Simple Groups 135 26 Generators and Relations 140 27 Symmetry Groups 144 28 Frieze Groups and Crystallographic Groups 146 29 Symmetry and Counting 148 30 Cayley Digraphs of Groups 151 31 Introduction to Algebraic Coding Theory 154 32 An Introduction to Galois Theory 158 33 Cyclotomic Extensions 161 2 / 4

1

− − −

≤ −

— − −

n n

• k
• k
• k

m +n m +n

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 ; p 2 q 3

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

• s = −3, t = 2; s = 8, t = −5
• By using 0 as an exponent if necessary, we may write a = p

m1 · · · p mk and b = p 1 1 · · · p k k , where the p’s are distinct primes and the m’s and n’s are nonnegative. Then lcm(a, b) = p s1 · · · p sk , where si = max(mi, ni) and

gcd(a, b) = p 1 · · · p k , where ti = min(mi, ni) Then lcm(a, b) · gcd(a, b) = p 1

• 1

· · · p k k k = ab.

• The first part follows from the Fundamental Theorem of Arithmetic; for

the second part, take a = 4, b = 6, c = 12.

• 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 then have r1 − r2 = 0. Thus, r1 = r2 and therefore a mod n = b mod n.

• Write as + bt = d. Then a

′ s + b ′ t = (a/d)s + (b/d)t = 1.

• By Exercise 7, 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.

• 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.

• 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.t t 3 / 4

0/Preliminaries 2

≤

— −

• 7(5n + 3) − 5(7n + 4) = 1
• By the GCD Theorem there are integers s and t such that ms + nt = 1.

Then m(sr) + n(tr) = r.

• It suffices to show that (p

2

• q

2

• r

2

• mod 3 = 0. Notice that for any

integer a not divisible by 3, a mod 3 is 1 or 2 and therefore a 2 mod 3 = 1.So, (p 2

• q

2

• r

2

• mod 3 = p

2 mod 3 + q 2 mod 3 + r 2 mod 3 = 3 mod 3= 0.

• Let p be a prime greater than 3. By the Division Algorithm, we can write

p in the form 6n + r, where r satisfies 0 r < 6. Now observe that 6n, 6n + 2, 6n + 3, and 6n + 4 are not prime.

• By properties of modular arithmetic we have

(7 1000

• mod 6 = (7 mod 6)

1000 = 1 1000 = 1. Similarly, (6 1001

• mod 7 = (6 mod 7)

1001

= −1

1001 mod 7 = −1 = 6 mod 7.

• Since st divides a b, both s and t divide a b. The converse is true when

gcd(s, t) = 1.

• Observe that 8

402 mod 5 = 3 402 mod 5 and 3 4 mod 5 = 1. Thus, 8 402 mod

5 = (3

4 ) 100 3 2 mod 5 = 4.

• If gcd(a, bc) = 1, then there is no prime that divides both a and bc. By

Euclid’s Lemma and unique factorization, this means that there is no prime that divides both a and b or both a and c. Conversely, if no prime divides both a and b or both a and c, then by Euclid’s Lemma, no prime divides both a and bc.

• If one of the primes did divide k = p1p2 · · · pn + 1, it would also divide 1.
• Suppose that there are only a finite number of primes p1, p2, . . . , pn. Then,

by Exercise 20, p1p2 . . . pn + 1 is not divisible by any prime. This means that p1p2 . . . pn + 1, which is larger than any of p1, p2, . . . , pn, is itself prime. This contradicts the assumption that p1, p2, . . . , pn is the list of all primes.

22.−7 + 3 i 58 58 23.−5+2i = −5+2i 4+5i = −30 + −17 i

4−5i 4−5i 4+5i 41 41

• L

√ et z1 = a + bi a √ nd z2 = c + di. Th √ en z1z2 = (ac − bd) + (ad + bc); |z1| = a 2

• b

2 , |z2| = c 2

• d

2 , |z1z2| = a 2 c 2

• b

2 d 2

• a

2 d 2

• b

2 c 2 = |z1||z2|.

• x NAND y is 1 if and only if both inputs are 0; x XNOR y is 1 if and only

if both inputs are the same.

• If x = 1, the output is y, else it is z.
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