Instructor's Solution Manual for Contemporary Abstract Algebra 11th Edition by Joseph Gallian, Chapter 1-33 | All Chapters

INSTRUCTOR SOLUTION
MANUAL
Contemporary Abstract Algebra (11th Edition)
by Joseph Gallian
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, Table of content
Integers and Equivalence Relations
0 Preliminaries
Introduction to Groups
2 Groups
3 Finite Groups; Subgroups
4 Cyclic Groups
5 Permutation Groups
6 Isomorphisms
7 Cosets and Lagrange’s Theorem
8 External Direct Products
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9 Normal Subgroups and Factor Groups
10 Group Homomorphisms
11 Fundamental Theorem of Finite Abelian Groups
12 Introduction to Rings
13 Integral Domains
14 Ideals and Factor Rings
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15 Ring Homomorphisms
16 Polynomial Rings
17 Factorization of Polynomials
18 Divisibility in Integral Domains
Fields
19 Vector Spaces
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20 Extension Fields
21 Algebraic Extensions
22 Finite Fields
23 Geometric Constructions
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Special Topics
24 Sylow Theorems
25 Finite Simple Groups
26 Generators and Relations
27 Symmetry Groups
28 Frieze Groups and Crystallographic Groups
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29 Symmetry and Counting
30 Cayley Digraphs of Groups
31 Introduction to Algebraic Coding Theory
32 An Introduction to Galois Theory
33 Cyclotomic Extensions
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, 1




CHAPTER 0
Preliminaries
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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
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5. By using 0 as an exponent if necessary, we may write a = p1m1 · · · pkmk and
b = p1n1 · · · pnkk , where the p’s are distinct primes and the m’s and n’s are
nonnegative. Then lcm(a, b) = p1s1 · · · pk sk , where si = max(mi, ni) and
gcd(a, b) = pt11 · · · ptkk , where ti = min(mi, ni) Then
lcm(a, b) · gcd(a, b) = p1m1 +n 1 · · · pm
k
k +n k
= ab.
6. The first part follows from the Fundamental Theorem of Arithmetic; for
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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
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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 a′s + b′t = (a/d)s + (b/d)t = 1.
9. 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
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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.
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.
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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.

, 0/Preliminaries 2


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.
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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.
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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
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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
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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.

21. 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
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prime. This contradicts the assumption that p1, p2, . . . , pn is the list of all
primes.
22. −7 + 3
i
58 58
−5+2i −5+2i 4+5i −30
23. 4−5i
= 4−5i 4+5i
= 41
+ −17
41
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L e t z1 = a + bi a√n d z2 = c + di. Th√en z1z2 = (ac — bd) + (ad + bc); |z1| =
24. √
a2 + b2, |z2| = c2 + d2, |z1z2| = a2c2 + b2d2 + a2d2 + b2c2 = |z1||z2|.
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25. 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.
26. If x = 1, the output is y, else it is z.