Solution Manual for Contemporary Abstract Algebra 9th Edition Gallian / All Chapters / Full Complete 2023
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Solutions Manual to Accompany Contemporary Abstract Algebra
NINTH EDITION Joseph Gallian University of Minnesota Duluth
CONTENTS
Integers and Equivalence Relations
0 Preliminaries 1
Groups
1 Introduction to Groups 8
2 Groups 10
3 Finite Groups; Subgroups 15
4 Cyclic Groups 25
Supplementary Exercises for Chapters 1-4 35
5 Permutation Groups 42
6 Isomorphisms 51
7 Cosets and Lagrange’s Theorem 58
8 External Direct Products 66
Supplementary Exercises for Chapters 5-8 74
9 Normal Subgroups and Factor Groups 81
10 Group Homomorphisms 89
11 Fundamental Theorem of Finite Abelian Groups 96
Supplementary Exercises for Chapters 9-11 101
12 Introduction to Rings 108
13 Integral Domains 113
14 Ideals and Factor Rings 120
Supplementary Exercises for Chapters 12-14 127
15 Ring Homomorphisms 132
16 Polynomial Rings 141
17 Factorization of Polynomials 148
18 Divisibility in Integral Domains 154
Supplementary Exercises for Chapters 15-18 160
24 Sylow Theorems 190
25 Finite Simple Groups 199
26 Generators and Relations 205
27 Symmetry Groups 209
28 Frieze Groups and Crystallographic Groups 211
29 Symmetry and Counting 213
30 Cayley Digraphs of Groups 216
31 Introduction to Algebraic Coding Theory 220
32 An Introduction to Galois Theory 225
33 Cyclotomic Extensions 228
Supplementary Exercises for Chapters 24-33 231
5. By using 0 as an exponent if necessary, we may write
a = pm1 · · · pmk and b = pn1 · · · pnk , where the p’s are distinct primes
1 k 1 k
and the m’s and n’s are nonnegative. Then lcm(a, b) = ps11 · · · psk ,
t t k
where si = max(mi, ni) and gcd(a, b) = p 11· · · p kk , where
1+n 1
ti = min(mi, ni) Then lcm(a, b) · gcd(a, b) = pm 1 ·· · pm
k
k +n k
= ab.
6. The first part follows from the Fundamental Theorem of
Arithmetic; for the second part, take a = 4, b = 6, c = 12.
7. 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.
8. Write as + bt = d. Then aJs + bJt = (a/d)s + (b/d)t = 1.
9. By Exercise 7, to prove that (a + b) mod n = (aJ + bJ) mod n and
(ab) mod n = (aJbJ) mod n it suffices to show that n divides
(a + b) − (aJ + bJ) and ab − aJbJ. Since n divides both a − aJ and n
divides b − bJ , it divides their difference. Because a = aJ mod n and
b = bJ mod n there are integers s and t such that a = aJ + ns and
b = bJ + nt. Thus ab = (aJ + ns)(bJ + nt) = aJbJ + nsbJ + aJnt + nsnt.
Thus, ab − aJbJ is divisible by n.
Test Bank A+ Page 1
, Contemporary Abstract Algebra
10. 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.
11. 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.
12. 7(5n + 3) − 5(7n + 4) = 1
13. By the GCD Theorem there are integers s and t such that
ms + nt = 1. Then m(sr) + n(tr) = r.
14. It suffices to show that (p2 + q2 + r2) mod 3 = 0. Notice that for
any integer a not divisible by 3, a mod 3 is 1 or 2 and therefore a2
mod 3 = 1. So, (p2 + q2 + r2) mod 3 = p2 mod 3 + q2 mod 3 + r2
mod 3 = 3 mod 3= 0.
15. 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.
16. By properties of modular arithmetic we have
(71000) mod 6 = (7 mod 6)1000 = 11000 = 1. Similarly,
(61001) mod 7 = (6 mod 7)1001 = −11001 mod 7 = −1 = 6 mod 7.
17. Since st divides a − b, both s and t divide a − b. The converse is
true when gcd(s, t) = 1.
18. Observe that 8402 mod 5 = 3402 mod 5 and 34 mod 5 = 1. Thus,
8402 mod 5 = (34)10032 mod 5 = 4.
19. 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.
20. If one of the primes did divide k = p1p2 · · · pn + 1, it would also
divide 1.
Test Bank A+ Page 2
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