solutions manual to accompany introduction to abstract algebra fourth edition pdf

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Solutions Manual to
Accompany Introduction to
Abstract Algebra

www.TechnicalBooksPdf.com


www.TechnicalBooksPdf.com


Solutions Manual to
Accompany Introduction to
Abstract Algebra
Fourth Edition

W. Keith Nicholson
University of Calgary
Calgary, Alberta, Canada

www.TechnicalBooksPdf.com


Copyright 2012 by John Wiley & Sons, Inc. All rights reserved
Published by John Wiley & Sons, Inc., Hoboken, New Jersey
Published simultaneously in Canada
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Library of Congress Cataloging-in-Publication Data:
Nicholson, W. Keith.
Introduction to abstract algebra / W. Keith Nicholson. – 4th ed.
p. cm.
Includes bibliographical references and index.
ISBN 978-1-118-28815-3 (cloth)
1. Algebra, Abstract. I. Title.
QA162.N53 2012
512’.02–dc23

2011031416
Printed in the United States of America.
10 9 8 7 6 5 4 3 2 1

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Contents

0

Preliminaries
0.1
0.2
0.3
0.4

1

2

Proofs / 1
Sets / 2
Mappings / 3
Equivalences / 4

Integers and Permutations
1.1
1.2
1.3

1.4

6

Induction / 6
Divisors and Prime Factorization / 8
Integers Modulo n / 11
Permutations / 13

Groups
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11

1

17

Binary Operations / 17
Groups / 19
Subgroups / 21
Cyclic Groups and the Order of an Element / 24

Homomorphisms and Isomorphisms / 28
Cosets and Lagrange’s Theorem / 30
Groups of Motions and Symmetries / 32
Normal Subgroups / 34
Factor Groups / 36
The Isomorphism Theorem / 38
An Application to Binary Linear Codes / 43
v


vi

3

Contents

Rings
3.1
3.2
3.3
3.4
3.5

4

5

7

8

9

108

Products and Factors / 108
Cauchy’s Theorem / 111
Group Actions / 114
The Sylow Theorems / 116
Semidirect Products / 118
An Application to Combinatorics / 119

Series of Subgroups
9.1
9.2
9.3

102

Modules / 102
Modules over a Principal Ideal Domain / 105

p-Groups and the Sylow Theorems
8.1
8.2
8.3
8.4
8.5
8.6

88

Vector Spaces / 88
Algebraic Extensions / 90
Splitting Fields / 94
Finite Fields / 96
Geometric Constructions / 98
An Application to Cyclic and BCH Codes / 99

Modules over Principal Ideal Domains
7.1
7.2

81

Irreducibles and Unique Factorization / 81
Principal Ideal Domains / 84

Fields
6.1
6.2
6.3
6.4
6.5
6.7

64

Polynomials / 64
Factorization of Polynomials over a Field / 67

Factor Rings of Polynomials over a Field / 70
Partial Fractions / 76
Symmetric Polynomials / 76

Factorization in Integral Domains
5.1
5.2

6

Examples and Basic Properties / 47
Integral Domains and Fields / 52
Ideals and Factor Rings / 55
Homomorphisms / 59
Ordered Integral Domains / 62

Polynomials
4.1
4.2
4.3
4.4
4.5

47

The Jordan-Hă
older Theorem / 122
Solvable Groups / 124
Nilpotent Groups / 127

122


Contents

10 Galois Theory
10.1
10.2
10.3
10.4

vii

130

Galois Groups and Separability / 130
The Main Theorem of Galois Theory / 134
Insolvability of Polynomials / 138
Cyclotomic Polynomials and Wedderburn’s Theorem / 140

11 Finiteness Conditions for Rings and Modules

142

11.1 Wedderburn’s Theorem / 142
11.2 The Wedderburn-Artin Theorem / 143
Appendices
Appendix A: Complex Numbers / 147
Appendix B: Matrix Arithmetic / 148
Appendix C: Zorn’s Lemma / 149

147



Chapter 0

Preliminaries
0.1 PROOFS
1. (a) (1) If n = 2k, k an integer, then n2 = (2k)2 = 4k 2 is a multiple of 4.
(2) The converse is true: If n2 is a multiple of 4 then n must be even
because n2 is odd when n is odd (Example 1).
(c) (1) Verify: 23 − 6 · 22 + 11 · 2 − 6 = 0 and 33 − 6 · 32 + 11 · 3 − 6 = 0.
(2) The converse is false: x = 1 is a counterexample. because
13 − 6 12 + 11 · 1 − 6 =
/ 0.
2. (a) Either n = 2k or n = 2k + 1, for some integer k. In the first case n2 = 4k 2 ;
in the second n2 = 4(k 2 + k) + 1.
(c) If n = 3k, then n3 − n = 3(9k 3 − k); if n = 3k + 1, then
n3 − n = 3(9k 3 + 9k 2 + 2k);
if n = 3k + 2, then n3 − n = 3(9k 3 + 18k 2 + 11k + 2).
3. (a) (1) If n is not odd, then n = 2k, k an integer, k ≥ 1, so n is not a prime.
(2) The converse is false: n = 9 is a counterexample; it is odd but is not a
prime.
√
√
√
√
(c) (1) If a > b then ( a)2 > ( b)2 , that is a > b, contrary to the assumption.
√

√
√
√
(2) The converse is true: If a ≤ b then ( a)2 ≤ ( b)2 , that is a ≤ b.
√
√
√
4. (a) If x > 0 and y > 0 assume
x + y = x + y. Squaring gives
√
√
x + y = x + 2 xy + y, whence 2 xy = 0. This means xy = 0 so x = 0 or
y = 0, contradicting our assumption.

Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition.
W. Keith Nicholson.
© 2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc.

1


2

0. Preliminaries

(c) Assume all have birthdays in different months. Then there can be at most
12 people, one for each month, contrary to hypothesis.
5. (a) n = 11 is a counterexample because then n2 + n + 11 = 11 · 13 is not prime.
Note that n2 + n + 11 is prime if 1 ≤ n ≤ 9 as is readily verified, but n = 10
is also a counterexample as 102 + 10 + 11 = 112 .

(c) n = 6 is a counterexample because there are then 31 regions. Note that the
result holds if 2 ≤ n ≤ 5.

0.2 SETS
1. (a) A = {x | x = 5k, k ∈ Z, k ≥ 1}
2. (a) {1, 3, 5, 7, . . .}
(c) {−1, 1, 3}
(e) { } = ∅ is the empty set by Example 3.
3. (a) Not equal: −1 ∈ A but −1 ∈
/ B.
(c) Equal to {a, l, o, y}.
(e) Not equal: 0 ∈ A but 0 ∈
/ B.
(g) Equal to {−1, 0, 1}.
4. (a)

∅, {2}

(c) {1}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
5. (a) True. B ⊆ C means each element of B (in particular A) is an element of
C.
(c) False. For example, A = {1}, B = C = {{1}, 2}.
6. (a) Clearly A ∩ B ⊆ A and A ∩ B ⊆ B; If X ⊆ A and X ⊆ B, then x ∈ X implies x ∈ A and x ∈ B, that is x ∈ A ∩ B. Thus X ⊆ A ∩ B.
7. If x ∈ A ∪ (B1 ∩ B2 ∩ . . . ∩ Bn ), then x ∈ A or x ∈ Bi for all i. Thus x ∈ A ∪ Bi
for all i, that is x ∈ (A ∪ B1 ) ∩ (A ∪ B2 ) ∩ . . . ∩ (A ∪ Bn ). Thus
A ∪ (B1 ∩ B2 ∩ . . . ∩ Bn ) ⊆ (A ∪ B1 ) ∩ (A ∪ B2 ) ∩ . . . ∩ (A ∪ Bn ),
and the reverse argument proves equality. The other formula is proved similarly.
9. A = {1, 2}, B = {1, 3}, C = {2, 3}.
10. (a) Let A × B = B × A, and fix a ∈ A and b ∈ B (since these sets are
nonempty). If x ∈ A, then (x, b) ∈ A × B = B × A. This implies x ∈ B; so

A ⊆ B. Similarly B ⊆ A.
(c) If x ∈ A ∩ B, then x ∈ A and x ∈ B, so (x, x) ∈ A × B. If (x, x) ∈ A × B,
then x ∈ A and x ∈ B, so x ∈ A ∩ B.
11. (a) (x, y) ∈ A × (B ∩ C)
if and only if
if and only if
if and only if

x ∈ A and y ∈ (B ∩ C)
(x, y) ∈ A × B and (x, y) ∈ A × C
(x, y) ∈ (A × B) ∩ (A × C).


0.3. Mappings

3

(c) (x, y) ∈ (A ∩ B) × (A ∩ B )
if and only if x ∈ A ∩ B and y ∈ A ∩ B
if and only if (x, y) ∈ A × A and (x, y) ∈ (B × B )
if and only if (x, y) ∈ (A × A ) ∩ (B × B ).
0.3 MAPPINGS
1. (a) Not a mapping: α(1) = −1 is not in N.
√
(c) Not a mapping: α(−1) = −1 is not in R.
(e) Not a mapping: α(6) = α(2 · 3) = (2, 3) and α(6) = α(1 · 6) = (1, 6).
(g) Not a mapping: α(2) is not defined.
2. (a) Bijective. α(x) = α(x1 ) implies 3 − 4x = 3 − 4x1 , so x = x1 , and α is oneto-one. Given y ∈ R, y = α 14 (3 − y) , so α is onto.
(c) Onto: If m ∈ N , then m = α(2m − 1) = α(2m). Not one-to-one: In fact we
have α(1) = 1 = α(2).

(e) One-to-one: α(x) = α(x1 ) implies (x + 1, x − 1) = (x1 + 1, x1 − 1), whence
x = x1 . Not onto: (0, 0) =
/ α(x) for any x because (0, 0) = (x + 1, x − 1)
would give x = 1 and x = −1.
(g) One-to-one: α(a) = α(a1 ) implies (a, b0 ) = (a1 , b0 ) implies a = a1 . Not onto
if |B| ≥ 2 since no element (a, b) is in α(A) for b =
/ b0 .
3. (a) Given c ∈ C, let c = βα(a) with a ∈ A (because βα is onto). Hence
c = β(α(a)), where α(a) ∈ B, so β is onto.
(c) Let β(b) = β(b1 ). Write b = α(a) and b1 = α(a1 ) (since α is onto). Then
βα(a) = β(α(a)) = β(b) = β(b1 ) = β(α(a1 )) = βα(a1 ),
so a = a1 (because βα is one-to-one), and hence b = b1 as required.
(e) Let b ∈ B. As α is onto, let b = α(a), a ∈ A. Hence
β(b) = β(α(a)) = βα(a) = β1 α(a) = β1 (α(a)) = β1 (b).
Since b ∈ B was arbitrary, this shows that β = β1 .
5. (a) If α2 = α, let x ∈ α(A), say x = α(a). Then α(x) = α2 (a) = α(a) = x. Conversely, let α(x) = x for all x ∈ α(A). If a ∈ A, write α(a) = x. Then
α2 (a) = α(α(a)) = α(x) = x = α(a), so α2 = α.
(c) α2 = (βγ)(βγ) = β(γβ)γ = β(1A )γ = βγ = α.
7. (a) If y ∈ R, write α−1 (y) = x. Hence y = α(x), that is y = ax + b. Solving for
x gives α−1 (y) = x = a1 (y − b). As this is possible for all y ∈ R, this shows
that α−1 (y) = a1 (y − b) for all y ∈ R.
(c) First verify that α2 = 1N , that is αα = 1N . Hence α−1 = α by the definition
of the inverse of a function.
9. Let βα = 1A . Then α is one-to-one because α(a) = α(a1 ) implies that
a = βα(a) = βα(a1 ) = a1 ; and β is onto because if a ∈ A then a = βα(a)
= β(α(a)) and α(a) ∈ B. Hence both are bijections as |A| = |B| (Theorem 2),
and hence α−1 and β −1 exist. But then β −1 = β −1 1A = β −1 (βα) = α. Similarly
α−1 = β.

0. Preliminaries

4

11. Let ϕ(α) = ϕ(α1 ) where α and α1 are in M. Then (α(1), α(2)) = (α1 (1), α1 (2)),
so α(1) = α1 (1) and α(2) = α1 (2). Thus α = α1 (by Theorem 1), so ϕ is oneto-one. Conversely, let (x, y) ∈ B × B, and define α2 : {1, 2} → B by α2 (1) = x
and α2 (2) = y. Then α2 ∈ M, and ϕ(β) = (α2 (1), α2 (2)) = (x, y). Thus ϕ is
onto. Then ϕ−1 : B × B → M has action ϕ−1 (x, y) = α2 where α2 (1) = x and
α2 (2) = y.
13. For each a ∈ A there are m choices for α(a) ∈ B. Since |A| = n, there are mn
choices in all, and they all lead to different functions α because α is determined
by these choices.
15.

(a) ⇒ (b) Given b ∈ B, write Ab = {a ∈ A | α(a) = b}. Then Ab =
/ ∅ for each
b (α is onto), so choose ab ∈ Ab for each b ∈ B. Then define β : B → A by
β(b) = ab . Then αβ(a) = α(β(b)) = α(ab ) = b for each b; that is αβ = 1B .
(c) ⇒ (a) If b0 ∈ B − α(A), we deduce a contradiction. Choose a0 ∈ A, and
define β : B → B by:
β(b) =

b
if b =
/ b0
α(a0 ) if b = b0 .

Then α(a) =
/ b0 for all a ∈ A, so
βα(a) = β(α(a)) = α(a) = 1B (α(a)) = 1B α(a)

for all a ∈ A. Hence, βα = 1B α, so β = 1B by (c). Finally then
b0 = β(b0 ) = α(a0 ), a contradiction.

0.4 EQUIVALENCES
1. (a) It is an equivalence by Example 4.
[−1] = [0] = [1] = {−1, 0, 1}, [2] = {2}, [−2] = {−2}.
(c) Not an equivalence. x ≡ x only if x = 1, so the reflexive property fails.
(e) Not an equivalence. 1 ≡ 2 but 2 ≡
/ 1, so the symmetric property fails.
(g) Not an equivalence. x ≡ x is never true. Note that the transitive property
also fails.
(i) It is an equivalence by Example 4. [(a, b)] = {(x, y) | y − 3x = b − 3a} is
the line with slope 3 through (a, b).
2. In every case (a, b) ≡ (a1 , b1 ) if α(a, b) = α(a1 , b1 ) for an appropriate function
α : A → R. Hence ≡ is the kernel equivalence of α.
(a) The classes are indexed by the possible sums of elements of U .
Sum is 2: [(1, 1)] = {(1, 1)}
Sum is 3: [(1, 2)] = [(2, 1)] = {(1, 2), (2, 1)}
Sum is 4: [(1, 3)] = [(2, 2)] = [(3, 1)] = {(1, 3), (2, 2), (3, 1)}
Sum is 5: [(2, 3)] = [(3, 2)] = {(2, 3), (3, 2)}
Sum is 6: [(3, 3)] = {(3, 3)}.


0.4. Equivalences

5

(c) The classes are indexed by the first components.
First component is 1: [(1, 1)] = [(1, 2)] = [(1, 3)] = {(1, 1), (1, 2), (1, 3)}
First component is 2: [(2, 1)] = [(2, 2)] = [(3, 2)] = {(2, 1), (2, 2), (2, 3)}

First component is 3: [(3, 1)] = [(3, 2)] = [(3, 3)] = {(3, 1), (3, 2), (3, 3)}.
3. (a) It is the kernel equivalence of α : Z → Z where α(n) = n2 . Here
[n] = {−n, n} for each n. Define σ : Z≡ → B by σ[n] = |n|, where |n| is
the absolute value. Then [m] = [n] ⇔ m ≡ n ⇔ |m| = |n|. Thus σ is welldefined and one-to-one. It is clearly onto.
(c) It is the kernel equivalence of α : R → R where α(x, y) = y. Define
σ : (R × R)≡ → B by σ[(x, y)] = y. Then
[(x, y)] = [(x1 , y1 )] ⇔ (x, y) ≡ (x1 , y1 ) ⇔ y = y1 ,
so σ is well-defined and one-to-one. It is clearly onto.
(e) Reflexive: x ≡ x ∈ Z;
Symmetric: x ≡ y ⇒ x − y ∈ Z ⇒ y − x ∈ Z ⇒ y ≡ x;
Transitive: x ≡ y and y ≡ z gives x − y ∈ Z and y − z ∈ Z. Hence
x − z = (x − y) + (y − z) ∈ Z, that is x ≡ z.
Now define σ : R≡ → B by σ[x] = x − x where x denotes the
greatest integer ≤ x. Then [x] = [y] ⇒ x ≡ y ⇒ x − y = n, n ∈ Z. Thus
x = y + n, so x = y + n. Hence,
x − x = (y + n) − ( y + n) = y − y ,
and σ is well-defined. To see that σ is one-to-one, let σ[x] = σ[y], that is
x − x = y − y . Then x − y = y − x ∈ Z, so x ≡ y, that is x = y .
Finally, σ is onto because, if 0 ≤ x < 1, x = 0, so x = σ[x].
5. (a) If a ∈ A, then a ∈ Ci and a ∈ Dj for some i and j, so a ∈ Ci ∩ Dj . If
Ci ∩ Dj =
/ Ci ∩ Dj , then either i =
/ i or j =
/ j . Thus
(Ci ∩ Dj ) ∩ (Ci ∩ Dj ) = ∅
in either case.
7. (a) Not well defined: α(2) = α
(c) Not well defined: α

1

2

2
1

= 2 and α(2) = α

= 3 and α

1
2

=α

2
4

4
2

= 4.

= 6.

9. (a) [a] = [a1 ] ⇔ a ≡ a1 ⇔ α(a) = α(a1 ). The implication ⇒ proves σ is well
defined; the implication ⇐ shows it is one-to-one. If α is onto, so is σ.
(c) If we regard σ : A≡ → a(A), then σ is a bijection.


Chapter 1

Integers and Permutations
1.1 INDUCTION
1. In each case we give the equation that makes pk imply pk+1 .
(a) k(2k − 1) + (4k + 1) = 2k 2 + 3k + 1 = (k + 1)(2k + 1)
(c)
(e)
(g)

(i)

1 2
1
1
2
3
2 2
2
2
4 k (k + 1) + (k + 1) = 4 (k + 1) (k + 4k + 4) = 4 (k + 1) (k + 2)
1
2
12 k(k + 1)(k + 2)(3k + 5) + (k + 1)(k + 2)
1
1
2
= 12 (k + 1)(k + 2)(3k + 17k + 24) = 12 (k + 1)(k + 2)(k + 3)(3k + 8)
k
k
2

2
2
3 (4k − 1) + (2k + 1) = 3 (2k − 1)(2k + 1) + (2k + 1)
1
1
2
= 3 (2k + 1)[2k + 5k + 3] = 3 (2k + 1)(k + 1)(2k + 3)
= 13 (k + 1)[4(k + 1)2 − 1]
1
k+1
1
1
+ (k+2)!
= 1 − (k+2)!
[(k + 2) − (k + 1)] = 1 − (k+2)!
1 − (k+1)!

2. In each case we give the inequality that makes pk imply pk+1 .
(a) 2k+1 = 2 · 2k > 2 · k ≥ k + 1.
2

2

2

(c) If k! ≤ 2k , then (k + 1)! = (k + 1)k! ≤ (k + 1)2k ≤ 2(k+1) provided
k + 1 ≤ 22k+1 . This latter inequality follows, again by induction on k ≥ 1,
because 22k+3 = 4 · 22k+1 ≥ 4(k + 1) ≥ k + 2.
√
√

√
2 +k+1
1
1
(e) √11 + · · · + √1k + √k+1
≥ k + √k+1
= k√k+1
≥ √k+1
= k + 1.
k+1
3. In each case we give the calculation that makes pk imply pk+1 .
(a) If k 3 + (k + 1)3 + (k + 2)3 = 9m, then
(k + 1)3 + (k + 2)3 + (k + 3)3 = 9m − k 3 + (k + 3)3 = 9m + 9k 2 + 27k + 27.
(c) If 32k+1 + 2k+2 = 7m, then
32k+3 + 2k+3 = 9(7m − 2k+2 ) + 2k+3 = 9 · 7m − 2k+2 (9 − 2).

Student Solution Manual to Accompany Introduction to Abstract Algebra, Fourth Edition.
W. Keith Nicholson.
© 2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc.

6


1.1. Induction

7

5. If 33k + 1 = 7m where k is odd, then passing to k + 2,
33(k+2) + 1 = 36 (7m − 1) + 1 = 36 · 7m − (36 − 1)
= 36 · 7m − 728 = 7(36 · m − 104).

7. It is clear if n = 1. In general, such a (k + 1) digit number must end in 4, 5 or
6, and there are 3k of each by induction. We are done since 3 · 3k = 3k+1 .
9. It is clear if n = 1. Given k + 1 secants, remove one and color the result unambiguously by induction. Now reinsert the removed secant. On one side of this
secant, leave all regions the original color (including the new regions of that side
created by the new secant). On the other side, interchange colors everywhere
(including those regions newly created). This is an unambiguous coloring.
10. (a) If k ≥ 2 cents can be made up, there must be a 2-cent or a 3-cent stamp.
In the first case, replace a 2-cent stamp by a 3-cent stamp; in the second
case, replace a 3-cent stamp by two 2-cent stamps.
(c) If k ≥ 18 can be made up, either one 7-cent stamp is used (replace with
two 4-cent stamps) or five 4-cent stamps are used (replace with three 7-cent
stamps).
11. a0 = 0 , a1 = 7, a2 = 63 = 7.9, a3 = 511 = 7 · 73. The conjecture is that 23n − 1
is a multiple of 7 for all n ≥ 0. If 23k − 1 = 7x for some n ≥ 0, then we have
23(k+1) − 1 = 23 (7x + 1) − 1 = 7(23 + 1).
12. (a) If Sn is the statement “13 + 23 + 33 + · · · + n3 is a perfect square”, then
S1 is true. If k ≥ 1, assume that 13 + 23 + · · · + k 3 = x2 for some integer
x. Then 13 + 23 + · · · + (k + 1)3 = x2 + (k + 1)3 and it is not clear how
to deduce that this is a perfect square without some knowledge about
how x is dependent upon k. Thus induction fails for Sn . However, if we
2
strengthen the statement to 13 + 23 + · · · + n3 = 12 n(n + 1) , induction
does go through (see Exercise 1(c)). The reason is that now the inductive
hypothesis brings more information to the inductive step and so allows the
(stronger) conclusion to be deduced.
13.

n
r−1

14. (a)

+

n
r

=

n!
(r−1)!(n−r+1)!

+
+ ··· +
6 with x = 1).
n
0

n
1

n
n

+

n!
r!(n−r)!
n

=

n!
r!(n+1−r)! [r

+ (n + 1 − r)] =

n+1
r

.

n

= (1 + 1) = 2 by the binomial theorem (Example

15. We use the well-ordering principle to prove the principle of induction. Let
p1 , p2 , p3 , · · · be statements such that p1 is true and pk ⇒ pk+1 for every k ≥ 1.
We must show that pn is true for every n ≥ 1. To this end consider the set
X = {n ≥ 1 | pn is false}; we must show that X is empty. But if X is nonempty
it has a smallest member m by the well-ordering principle. Hence m =
/ 1
(because p1 is true), so m − 1 is a positive integer. But then pm−1 is true
(because m is the smallest member of X) and so pm is true (because
pm−1 ⇒ pm ). This contradiction shows that X must be empty, as required.
17. If pn is “n has a prime factor”, then p2 is true. Assume p2 , . . . , pk are all true.
If k + 1 is a prime, we are done. If k + 1 = ab write 2 ≤ a ≤ k and 2 ≤ b ≤ k,
then a (and b) has a prime factor by strong induction. Thus k + 1 has a prime
factor.

8

1. Integers and Permutations

18. (a) an = 2(−1)n

an+1 = −an = −2(−1)n = 2(−1)n+1

(c) an = 12 [1 + (−1)n ]
an+1 = 1 − an = 1 − 12 [1 + (−1)n ] = 12 [2 − 1 − (−1)n ] = 12 [1 + (−1)n+1 ]
19. Given n lines, another line intersects all existing lines (because no two are
parallel) at new intersection points (none of these are concurrent) and so enters n + 1 regions. Hence it creates n + 1 new regions; so an+1 = an + (n + 1).
Then a0 = 1, a1 = 1 + 1, a2 = 1 + 1 + 2, a3 = 1 + 1 + 2 + 3; and this suggests
an = 1 + (1 + 2 + · · · + n). Hence Gauss’ formula (Example 1) gives
an = 1 + 12 n(n + 1) = 12 (n2 + n + 2).
This is valid for n = 0; if it holds for n = k ≥ 1 then
ak+1 = ak + (k + 1) = 12 [(k 2 + k + 2) + 2(k + 1)] = 12 [(k + 1)2 + (k + 1) + 2].
21. (a) Let pn denote the statement an = (−1)n . Then p0 and p1 are
true by hypothesis. If pk and pk+1 are true for some k ≥ 0, then
ak = (−1)k , ak+1 = (−1)k+1 and so
ak+2 = ak+1 + 2ak = (−1)k+1 + 2(−1)k = (−1)k [−1 + 2] = (−1)k = (−1)k+2 .
Thus pk+2 is true and the principle applies.
23. p1 ⇒ p2 fails.
24. (a) Prove p1 and p2 are true.
25. If pk is true for some k, then pk−1 , pk−2 , . . . , p1 are all true by induction using
the first condition. Given m, the second condition implies that pk is true for
some k ≥ m, so pm is true.
27. (a) Apply the recursion theorem with s0 = a0 and sn = sn−1 + an .
1.2 DIVISORS AND PRIME FACTORIZATION

1. (a) 391 = 23 · 17 + 0

(c) −116 = (−9) · 13 + 1

2. (a) n/d = 51837/386 = 134.293, so q = 134. Thus r = n − qd = 113.
3. If d > 0, then |d| = d and this is the division algorithm. If d < 0, then
|d| = −d > 0 so n = q(−d) + r = (−q)d + r, 0 ≤ r ≤ |d|.
5. Write m = 2k + 1, n = 2j + 1. Then m2 − n2 = 4[k(k + 1) − j(j + 1)]. But
each of k(k + 1) and j(j + 1) is even, so 8 | (m2 − n2 ).
7. (a) 10(11k + 4) − 11(10k + 3) = 7, so d | 7. Thus d = 1 or d = 7.
9. (a) 72 = 42 + 30
42 = 30 + 12
30 = 2 · 12 + 6
12 = 2 · 6
Thus, gcd(72, 42) = 6 and
6 = 30 − 2(42 − 30)
= 3 · 30 − 2 · 42
= 3(72 − 42) − 2 · 42
= 3 · 72 − 5 · 42

(c) 327 = 6 · 54 + 3
54 = 3 · 18
Thus gcd(327 · 54) = 3 and
3 = 1 · 327 − 6 · 54


1.2. Divisors and Prime Factorization

(e) 377 = 13 · 29
Hence 29 | 377, so

gcd(29, 377) = 29. Thus

9

(g) 72 = 0 · (−176) + 72
−175 = (−3) · 72 + 41
72 = 41 + 31
41 = 31 + 10

29 = 0 · 377 + 1 · 29

31 = 3 · 10 + 1
Hence gcd(72, −175) = 1 and
1 = 31 − 3(41 − 31)
= 4(72 − 41) − 3 · 41
= 4 · 72 − 7(−175 + 3 · 72)
= (−17) · 72 − 7 · (−175)

d
d
m
11. If m = qd, then m
k = q k , so k | k . Similarly,
d
m
n
m
k = x k + y k , so any common divisor of k and

d

k
n
k

| nk . If d = xm + yn, then
is a divisor of kd .

13. It is prime for n = 1, 2, . . . , 9; but 102 + 10 + 11 = 121 = 112 .
15. If d = gcd(m, n) and d1 = gcd(m1 , n1 ), then d | m and d | n, so d | m1 and d | n1
by hypothesis. Thus d | d1 .
17. If 1 = xm + yn and 1 = x1 k + y1 n, then
1 = (xm + yn)(x1 k + y1 n) = (xx1 )mk + (xmy1 + yx1 k + yny1 )n.
Thus gcd(mk, n) = 1 by Theorem 4.
Alternatively, if d = gcd(mk, n) =
/ 1 let p | d, p a prime. Then p | n and
p | mk But then p | m or p | k, a contradiction either way because we have
gcd(m, n) = 1 = gcd(m, n).
19. Write d = gcd(m, n) and d = gcd(km, kn). We must show kd = d . First, d | m
and d | n, so kd | km and kd | kn. Hence, kd | d . On the other hand, write
km = qd and kn = pd . We have d = xm + yn, x, y ∈ Z, so
kd = xkm + ykn = xqd + ypd .
Thus d | kd. As k ≥ 1 it follows that d = kd.
21. If p is not a prime, then assume p = mn with m ≥ 2 and n ≥ 2. But then p | m
or p | n by hypothesis, so p ≤ m < p or p ≤ n < p, a clear contradiction.
23. No. If a = 18 and n = 12 then d = 6 so

a
d

= 3 is not relatively prime to n = 12.

25. Let them be 2k + 1, 2k + 3, 2k + 5. We have k = 3q + r, r = 0, 1, 2. If r = 0 then
3 | (2k + 3); if r = 1, then 3 | (2k + 1); and if r = 2, then 3 | (2k + 5). Thus one
of these primes is a multiple of 3, and so is 3.
27. Let d = gcd(m, pk ), then d | m and d | pk . Thus d = pj , j ≤ k. If j > 0, then
p | d, so (since d | m) p | m. This contradicts gcd(m, p) = 1. So j = 0 and d = 1.
29. We have a | a1 b1 and (a, b1 ) = 1. Hence a | a1 by Theorem 5. Similarly a1 | a,
so a = a1 because both are positive. Similarly b = b1 .
30. (a) 27783 = 34 · 73
(c) 2431 = 11 · 13 · 17
(e) 241 = 241 (a prime)


10

1. Integers and Permutations

31. (a) 735 = 20 · 31 · 51 · 72 · 110 and 110 = 21 · 30 · 51 · 70 · 111 . Hence
gcd(735, 110) = 20 · 30 · 51 · 72 · 110 = 5, and
lcm(735, 110) = 21 · 31 · 51 · 72 · 111 = 16170.
(c) 139 = 20 · 1391 and 278 = 21 · 1391 . Hence
gcd(139, 278) = 20 · 1391 = 139, and lcm(139, 278) = 21 · 1391 = 278.
33. (a) Use Theorem 8. In forming d = pd11 . . . pdr r , there are (n1 + 1) choices for
d1 among 0, 1, 2, . . . , ni ; then there are (n2 + 1) choices for d2 among
0, 1, 2, . . . , n2 ; and so on. Thus there are (n1 + 1)(n2 + 1) · · · (nr + 1)
choices in all, and each leads to a different divisor by the uniqueness in
the prime factorization theorem.
mr
1
35. Let m = pm

and n = q1n1 . . . qsns be the prime factorizations of m and
1 . . . pr
n. Since gcd(m, n) = 1, pi =
/ qj for all i and j, so the prime factorization of mn
mr n1
1
.
.
.
p
q
.
.
.
qsns . Since d | mn, we have d = pd11 . . . pdr r q1e1 . . . qses
is mn = pm
r
1
1
where 0 ≤ di ≤ mi for each i and 0 ≤ ej ≤ nj for each j. Take m1 = pd11 . . . pdr r
and n1 = q1e1 . . . qses .

37. Write a = pa1 1 pa2 2 . . . par r and b = pb11 pb22 . . . pbrr where the pi are distinct primes,
0 if ai < bi
bi if ai < bi
and vi =
, and then
ai ≥ 0 and bi ≥ 0. Let ui =
ai if ai ≥ bi
0 if ai ≥ bi

take u = pu1 1 pu2 2 . . . pur r and v = pv11 pv22 . . . pvrr . Then u | a, v | b and gcd(u, v)=1.
Moreover uv = lcm(a, b) by Theorem 9 because ui + vi = max(ai , bi ) for each i.
39. (a) By the division algorithm, p = 4k + r for r = 0, 1, 2 or 3. But r = 0 or 2 is
impossible since p is odd (being a prime greater than 2).
41. (a) 28665 = 32 · 51 · 72 · 110 · 131 and 22869 = 33 · 50 · 71 · 112 · 130 so,
gcd(28665, 22869) = 32 · 50 · 71 · 110 · 130 = 63
lcm(28665, 22869) = 33 · 51 · 72 · 112 · 131 = 10, 405, 395
43. Let X = {x1 a1 + · · · + xk ak | xi ∈ Z, x1 a1 + · · · + xk ak ≥ 1}. Then X =
/ ∅
because a21 · · · + a2k ∈ X, so let m be the smallest member of X. Then
m = x1 a1 + · · · + xk ak for integers ak , so we show d = m. Since d | ai for each
i, it is clear that d | m. We can show m | d, if we can show that m is a common divisor of the ai (by definition of d = gcd(a1 , · · · , ak )). Write a1 = qm + r,
0 ≤ r < m. Then
r = a1 − qm = (1 − qx1 )a1 + (−qx2 )a2 + · · · + (−qxk )ak ,
and this contradicts the minimality if r ≥ 1. So r = 0 and m | a1 . A similar
argument shows m | ai for each i.
45. (a) Let m = qn + r, 0 ≤ r < n. If m < n, then q = 0 and r = m. If m ≥ n, then
q ≥ 1. Thus q ≥ 0. We want x ∈ Z such that 2m − 1 = x(2n − 1) + (2r − 1).
Solving for x (possibly in Q):
x=

2 m − 2r
= 2r
2n − 1

2m−r − 1
2n − 1

= 2r

(2n )q − 1
2n − 1

.

If q = 0, take x = 2r = 2m ; if q > 0, take x = (2n )q−1 + · · · + 2n + 1.


1.3. Integers Modulo n

11

1.3 INTEGERS MODULO n
1. (a) True.
(c) True.

40 − 13 = 3 · 9
−29 − 6 = (−5)7

(e) True.

8 − 8 = 0 · n for any n.

(g) False.

84 ≡ (64)2 ≡ (−1)2 ≡ 1 (mod 13).

2. (a) 2k − 4 = 7q, so q is even. Thus k = 2 + 7x for some integer x; that is
k ≡ 2 (mod 7).
(c) 2k ≡ 0 (mod 9), so 2k = 9q. Thus 2 | q, so k = 9x for some integer x; that

is k ≡ 0 (mod 9).
3. (a) 10 ≡ 0 (mod k), so k | 10: k = 2, 5, 10.
(c) k 2 − 3 = qk, so k | 3. Thus k = 1, 3 so, (as k ≥ 2 by assumption) k = 3.
5. (a) a ≡ b (mod 0) means a − b = q · 0 for some q, that is a = b.
6. (a) a ≡ a for all a because n | (a − a). Hence if n | (a − b), then n | (b − a).
Hence if a − b = xn and b − c = yn, x, y ∈ Z, then a − c = (x + y)n.
7. If n = pm and a ≡ b(mod n), then a − b = qn = qpm. Thus a ≡ b(mod m).
2
3
6
8. (a) In Z7 : 10 = ¯
3, so 10 = ¯
9 = ¯2, 10 = ¯6 = −1, 10 = ¯1. Since
515
6
5
2
3
515 = 6 · 85 + 5 we get 10
= (10 )85 · 10 = ¯185 · 10 · 10 = ¯2 · (−1) = ¯5.
515
Hence 10
≡ 5(mod 7).
2
¯ =9
¯ = −1, so ¯
9. (a) In Z10 : 3
34 = ¯1. Since 1027 = 4 · 256 + 3, we get
1027
4

256
3
256
¯
3
= (¯
3 ) ·¯
3 =¯
1 · 27 = ¯
7. The unit decimal is 7.
¯ 1,
¯ 2,
¯ 3,
¯ 4,
¯ 5
¯ in Z6 . If p¯ = 0,
¯ 2,
¯ 4
¯ then 2 | p; if p¯ = ¯3, then 3 | p. So p¯ = ¯1 or
11. p¯ = 0,
p¯ = ¯
5.
¯2 = ¯
0, ¯
1, ¯
0, ¯
1 respectively.
12. (a) a
¯=¯
0, ¯

1, ¯
2, ¯
3 in Z4 , so a

13. a
¯=¯
0, ¯
1, . . . , 10 in Z11 . Taking each case separately:
¯
05 = ¯
0

¯65 = (−5)5 = −55 = −1

¯
1
15 = ¯

¯75 = (−4)5 = −45 = −1

¯
25 = 32 = −1

¯85 = (−3)5 = −35 = −1

¯
9 · 27 = ¯
9·¯
5=¯
1

35 = ¯

¯95 = (−2)5 = −25 = ¯1

¯
5·¯
9=¯
1
45 = 16 · 64 = ¯
5
¯ = 25 · 25 · ¯
5=¯
3·¯
3·¯
5=¯
1
5

10 = (−1) = −1

5

5

15. One of a, a + 1 must be even so 2 | a(a + 1)(a + 2); similarly, one of a,
a + 1, a + 2 is a multiple of 3 [in fact a ≡ 0 means 3 | a, a ≡ 1 means 3 | a + 2,
and a ≡ 2 means 3 | a + 1]. Hence 3 | a(a + 1)(a + 2). But 2 and 3 are relatively
prime so 2 · 3 = 6 also divides a(a + 1)(a + 2). Hence
¯ a + 2)
¯ = a(a + 1)(a + 2) = 0¯ in Z6 .

a
¯(¯
a + 1)(¯


12

1. Integers and Permutations

17. Since a
¯=¯
0, ¯
1, . . . , ¯
5 in Z6 , we examine every case.
¯
¯
03 = ¯
0
33 = 27 = ¯3
¯
1
13 = ¯

3
¯
43 = (−2) = −(¯2)3 = −2 = ¯4

¯
8=¯
2

23 = ¯

3
¯
53 = (−1) = −1 = ¯5

Hence a
¯3 = a
¯ in all cases.
18. (a) Since a
¯=¯
0, ¯
1, . . . , ¯
4 in Z5 , it suffices to show each of these is a cube
in Z5 . Look at the cubes in Z5 : ¯03 = ¯0, ¯13 = ¯1, ¯23 = ¯3, ¯33 = ¯2, and
3
¯
43 = (−1) = −¯
1=¯
4. Thus every residue ¯0, ¯1, ¯2, ¯3, ¯4 is a cube in Z5 .
¯
¯ 1,
¯ 2,
¯ 3,
¯ 4,
¯ 5,
¯ 6
¯ in Z7 , we get k¯2 + ¯1 = ¯1, ¯2, ¯5, ¯3, ¯3, ¯5, ¯2 respectively.
19. (a) Since k = 0,
2

¯
1=¯
0 does not occur in Z7 .
Clearly k + ¯
21. We have n = d0 + 10d1 + 102 d2 + · · · + 10k dk .
(a) 10 = ¯
1 in Z3 , so n
¯ = d0 + ¯
1 · d1 + ¯12 d2 + · · · + ¯1k dk = d0 + d1 + · · · + dk .
Thus n
¯ = d0 + d1 + · · · + dk (mod 3).
35 = 2 · 13 + 9

1 = 9 − 2(13 − 9)

22. (a) By the euclidean algorithm, 13 = 1 · 9 + 4 so = 3(35 − 2 · 13) − 2 · 13 .
9=2·4+1
= 3 · 35 − 8 · 13
Hence (−8) · 13 ≡ 1(mod 35), so −8 = 27 is the inverse of 13 in Z35 . Then
13 · x
¯=¯
9 gives x
¯ = 27 · 13 · x
¯ = 27 · ¯9 = −8 · ¯9 = −72 = −2 = 33.
1 = 9 − 4(11 − 9)

20 = 11 + 9
(c) Euclidean algorithm: 11 = 9 + 2 so
9=4·2+1

23. (a)
24. (a)

25. (a)

(c)

(e)

= 5 · 9 − 4 · 11
.
= 5(20 − 11) − 4 · 11

= 5 · 20 − 9 · 11
Hence the inverse of 11 is −9 = 11, so 11 · x
¯ = 16 gives x
¯ = 11 · 16 = 16.
Let d¯ be the inverse of a
¯ in Zn , so d¯ · a
¯ = ¯1 in Zn , then multiply a
¯ · ¯b = a
¯ · c¯
by d¯ to get d¯ · a
¯ · ¯b = d¯ · a
¯ · c¯, that is ¯1 · a
¯ = ¯1 · c¯, that is a
¯ = c¯.
If c¯ and d¯ are the inverses of a
¯ and ¯b respectively in Zn , then c¯ · a
¯ = ¯1 and

¯ a · ¯b) = ¯1. Hence
d¯ · ¯b = ¯
1. Multiplying, we find c¯ · a
¯ · d¯ · ¯b = ¯1, that is (¯
c · d)(¯
c¯ · d¯ is the inverse of a
¯ · ¯b = ab in Zn .
¯ to get 10x + 2y
¯ = 2.
¯ Subtract this from equation
Multiply equation 2 by 2
1: ¯
7x = ¯
1. But ¯
8·¯
7=¯
1 in Z11 , so x = ¯8 · ¯1 = ¯8. Then equation 2 gives
y=¯
1−¯
5·¯
8=¯
5.
¯ to get 3x
¯ + 2y
¯ = 2.
¯ Comparing this with the first
Multiply equation 2 by 2
equation gives ¯
1=¯
3x + ¯

2y = ¯
2, an impossibility. So there is no solution to
these equations in Z7 . (Compare with (a)).
¯ to get 3x
¯ + 2y
¯ = 1,
¯ which is just equation 1. Hence,
Multiply equation 2 by 2
we need only solve equation 2. If x = r¯ is arbitrary in Z7 (so r¯ = ¯0, ¯1, . . . , ¯6),
then y = ¯
4−¯
5x = 4 − 5r. Thus the solutions are:
x
y

¯
0
¯
4

¯
1
¯
6

¯
2
¯
1

¯
3
¯
3

¯
4
¯
5

¯
5
¯
0

¯
6
¯
2

.


1.4. Permutations

13

27. If an expression x2 + ax is given where a is a number, we can
2
2

complete the square by adding 12 a . Then x2 + ax + 12 a = (x + 12 a)2 . The
1
same thing works in Zn except 2 is replaced by the inverse of ¯2 if it exists.
(a) x2 + ¯
5x + ¯
4=¯
0 means x2 + ¯
5x = ¯3 in Z7 . The inverse of ¯2 is ¯4 in Z7 , so
2
the square is completed by adding (¯4 · ¯5) = ¯1 to both sides. The result is
(x + ¯
6)2 = x2 + ¯5x + ¯1 = ¯3 + ¯1 = ¯4.
The only members of Z7 which square to ¯4 are ¯2 and −2 = ¯5. (See Exercise
26.) Hence x + ¯
6=¯
2 or ¯
5; that is x = ¯3 or ¯6.
(c) x2 + x + ¯
2=¯
0 gives x2 + x = ¯
3 in Z5 . The inverse of ¯2 is ¯3 in Z5 , so add
2
¯
¯
3 = 4 to both sides
(x + ¯
3)2 = x2 + x + ¯4 = ¯3 + ¯4 = ¯2.
But ¯
2 is not a square in Z5 [¯
02 = ¯0, ¯12 = ¯42 = ¯1, ¯22 = ¯32 = ¯4], so there is

no solution.
(e) Since n is odd, gcd(2, n) = 1, so ¯2 has an inverse in Zn ; call it r¯. Now
x2 + a
¯x + ¯b = ¯
0 in Zn means x2 + a
¯x = −b. Complete the square by adding
2
2
(¯
r·a
¯) = ra to both sides. The result is
(x + ra)2 = x2 + a
¯ + ra2 = −b + ra2 = (¯
r2 a
¯2 − ¯b).
¯2 − ¯b) is a square in Zn .
Thus, there is a solution if and only if (¯
r2 a
29. (a) Let a
¯ · ¯b = ¯
0 in Zn . If gcd(a, n) = 1, then a has an inverse in Zn , say
c¯ · a
¯=¯
1. Then ¯b = ¯
1 ¯b = c¯ · a
¯ · ¯b = c¯ · ¯0 = ¯0.
31. (1) ⇒ (2). Assume (1) holds but n is not a power of a prime. Then n = pk a
where p is a prime, k ≥ 1, and a > 1 has p /| a. Then gcd(n, a) = a > 1, so a
¯ has
no inverse in Zn . But a

¯n =
/ ¯
0 too. In fact a
¯n = ¯0 means n | an whence p | an .
By Euclid’s lemma, this implies p | a, contrary to choice.
2
2
33. In Z223 , ¯
28 = 256 = 33. Thus ¯
216 = 33 = 197, ¯232 = 197 = ¯7, and finally
37
32 ¯5
¯
¯
¯
¯
2 = 2 · 2 = 7 · 32 = 224 = 1. Similarly, in Z641 ,

¯16 = 2562 = 154, 2¯32 = 1542 = 640 = −1.
¯
28 = 256, 2
34. (a) If ax ≡ b has a solution x in Zn , then b − ax = qn, q an integer, so
b = ax + qn. It follows that d = gcd(a, n) divides b. Conversely, if d | b write
b = qd, q an integer. Now d = ra + sn for integers r and s (Theorem 3 §1.2),
so b = qd = (qr)a + (qs)n. Thus, (qr)a ≡ b(mod n) and we have our
solution.
¯ means x2 − 1¯ = 0.
¯ Thus (x − 1)(x
¯
¯ = 0¯ in Zp ,

35. Working modulo p, x2 = 1
+ 1)
so x = ¯
1 or x = −¯
1 by Theorem 7.
37. (a) If n = p2 m and a = pm, then a ≡
/ 0(mod n) and a2 ≡ 0(mod n). Hence
n
a ≡
/ a.

1.4 PERMUTATIONS
1. (a) τ σ =

1

2

3

4

5

2

3

5

1

4

(c) τ −1 = τ =

1
3

2
2

3
1

4
5

5
4


14

1. Integers and Permutations

(e) μτ σ −1 =

1
4

3. (a) χ = σ −1 τ =

2
5

3
2

4 5
3 1

1
4

2
2

3 4
3 1

(e) χ = τ −1 εσ −1 = τ −1 σ −1 =

(c) χ = στ =
1 2 3

4

1 3 2

4

1
1

2
3

3
2

4
4

5. Solution 1. We must have σ1 = 1, 2, 3 or 4; in each case we find σ1 = σ3, a
contradiction.
If σ1 = 1 :
If σ1 = 2 :
τ 1 = τ σ1 = 2
τ 2 = τ σ1 = 2
σ2 = στ 1 = 2
τ 2 = τ σ2 = 3

σ2 = στ 2 = 1
τ 1 = τ σ2 = 3

σ3 = στ 2 = 1
If σ1 = 3 :
τ 3 = τ σ1 = 2

σ3 = στ 1 = 2
If σ1 = 4 :
τ 4 = τ σ1 = 2

σ2 = τ σ3 = 4

σ2 = στ 4 = 3

τ 4 = τ σ2 = 3
σ3 = στ 4 = 3

τ 3 = τ σ2 = 3
σ3 = στ 3 = 4

Solution 2. Let σ =

1

2 3 4

. Then we show στ = (a
a b c d
cycle, contrary to στ = (1 2)(3 4) :

b

c

d) is a

σ1 = a ⇒ τ a = τ σ1 = 2 ⇒ στ a = σ2 = b
σ2 = b ⇒ τ b = τ σ2 = 3 ⇒ στ b = σ3 = c
σ3 = c ⇒ vτ c = τ σ3 = 4 ⇒ στ c = σ4 = d
σ4 = d ⇒ τ d = τ σ4 = 1 ⇒ στ d = σ1 = a
6. If σk = k, then σ −1 k = σ −1 (σk) = k. If also τ k = k, then (τ σ)k = τ (σk)
= τ k = k.
1 2 3 4 5
7. (a) Here σ =
where a, b, c, d are 2, 3, 4, 5 in some order.
1 a b c d
Thus there are 4 choices for a, 3 for b, 2 for c, and 1 for d; and so we have
4 · 3 · 2 · 1 = 4! = 24 choices in all for σ.
1 2 3 4 5
(b) Now σ =
where a, b, c are 3, 4, 5 in some order. As in
1 2 a b c
(a), there are 3 · 2 · 1 = 3! = 6 choices in all for σ.
8. (a) If στ = ε, then σ = σε = σ(τ τ −1 ) = (στ )τ −1 = τ −1 .
9. If σ = τ , then στ −1 = τ τ −1 = ε; if στ −1 = ε, then
τ = ετ = (στ −1 )τ = σ(τ −1 τ ) = σε = σ.


1.4. Permutations

11. (a)

1
8

2

2

3
6

4
1

5
9

6
4

15

7 8 9
5 7 3

12. (a) ε, σ = (1 2 3), σ 2 = (1 3 2), τ = (1 2), στ = (1 3), σ 2 τ = (2 3).
These are all six elements of S3 . We have σ 3 = σσ 2 = ε, τ 2 = ε and hence
τ σ = (2 3) = σ 2 τ .
13. (a) σ = (1 4 8 3 9 5 2 7 6); σ −1 = (1 6 7 2 5 9 3 8 4)
(c) σ = (1 2 8)(3 6 7)(4 9 5); σ −1 = (1 8 2)(3 7 6)(4 5 9)
(e) σ = (1 3 8 7 2 5); σ −1 = (1 5 2 7 8 3)
15. (a) ε, (1 2 3 4 5), (1 2 3 4), (1 2 3), (1 2 3)(4 5), (1 2), (1 2)(3 4)
17. (a) σ −1 = (4 3 2 1)(7 6 5).
19. They are factored into disjoint cycles in the solution to Exercise 13, so the
parities are:
(a) even

(c) even + even + even = even
(e) odd
21. (a) We have γi2 = ε for all i because the γi are transpositions. Hence
(γ1 γ2 . . . γm )(γm γm−1 . . . γ2 γ2 ) = (γ1 γ2 . . . γm−1 )(γm−1 . . . γ2 γ1 ) = . . . = ε.
Now use Exercise 8(a).
(c) If σ and τ are products of k and m transpositions respectively, then τ −1 is
also a product of m transpositions (by (a)) so τ στ −1 is a product of k + 2m
transpositions. This has the same parity as k.
23. Let σk = 1 for some k =
/ 1. Then, as n ≥ 3, choose an m ∈
/ {k, 1}. Now let
γ = (k, m). This gives γσk = γ1 = 1, but σγk = σm =
/ 1, since if σm = 1 = σk,
then m = k as σ is one-to-one, contrary to assumption.
25. It suffices to show that any pair of transpositions is a product of
3-cycles. If k, l, m and n are distinct, this follows from
(k l)(m n) = (k m l)(k m n), (k l)(k m) = (k m l), and (k l)2 = ε.
27. (a) Both sides have the same effect on each ki , and both sides fix each
k∈
/ {k1 , k2 , . . . kr }.
(c) Using Exercise 26, we have for all a = 1, 2, . . . , n − 1:
(1 a + 1) = (1 a)(a a + 1)(1 a)

(*)

Now if σ ∈ Sn , write it as a product of factors (1 n). Use (*) to write each
(1 n) as a product of (1 2), . . . , (1 n − 1), and (n − 1 n). Then write each
(1, n − 1) in terms of (1 2), . . . , (1 n − 2) and (n − 2, n − 1). Continue. The
result is (c).
28. (a) σ = (1 2 3 4 . . . 2k − 1 2k) so σ 2 = (1 3 5 . . . 2k − 1)(2 4 6 . . . 2k).

(c) The action of σ is depicted in the diagram,
and carries k → k + 1 → k + 2 . . .. If k + m >
n, the correct location on the circle is given by
the remainder r when k + m is divided by n,
That is k + m ≡ 4(modn. Now the action of
σ m is σ m k = k + m, so σ m k ≡ k + m mod n.

1
2

n-1

3
n-2

k
k+1