solutions (Acrobat; 948 kB)

141 Pages·2005·0.92 MB·English

by  Ben Hekster

Solutions
to
A First Course in
Abstract Algebra
John B. Fraleigh
sixth edition
ISBN 0-201-33596-4
Addison Wesley Longman
by
Ben Hekster
PO Box 391852
Mountain View, CA 94039-1852
[email protected]
http://www.hekster.org/Academic/Mathematics/
These are completely unofficial and unverified worked solutions by me. Corrections welcome.
Typeset in Galliard and Gill Sans using AppleWorks 5 on Apple Macintosh.
Copyright © 1999-2005 All Rights Reserved by: Ben Hekster.
1
Abstract Algebras
set
+ binary operation
binary algebraic structure
+ associative binary operation
semigroup
+ identity
monoid
+ inverse
group
+ finite order + commutative operation
finite group commutative group
+ prime order + generator
symmetric group cyclic group
+ prime order
alternating group prime order group
Glossary
: reads as “so that”
+ , ⋅ summation, multiplication over i
i i
( { ordered, unordered set over i
i i
∧,∨,<,> scalar operators

∩,∪,⊂,⊃ set operators

= congruent modulo n
n
< is normal to, is ideal to
( )
fx function application f x

commutative group abelian group
maximal p-group Sylow p-group
1
§0.1 Preliminaries
1. proving theorems
2. set
3. precision?
4. definition
5. A triangle with vertices P, Q, R is the collection of points X such that
• X is in the line segment PQ, or
• X is in the line segment QR, or
• X is in the line segment RP.
6. An equilateral triangle is a triangle with vertices P, Q, R such that the length of the line segment PQ equals both
the length of the line segment QR and the length of the line segment RP.
7. A right triangle is a triangle with vertices P, Q, R in which the two line segments through one of its vertices (say PQ
and PR) are such, that for any point X on PQ there is no point Y on PR such that the length of the line segment XY
is less than the length of the line segment XP.
8. The interior of a triangle is the collection of points X such that the line segments XP, XQ, XR from X to its vertices
P, Q, R have only the vertices in common with the triangle.
9. A circle with center C and radius r is the collection of points X such that the length of the line segment XC equals
r.
10. A disk with center C and radius r is the collection of points X such that the length of the line segment XC is less
than or equal to r.
11. Define the relationship between PQ and PR in 7. to be a right angle. Then, a rectangle with vertices P, Q, R, S is
the collection of points formed by the four line segments PQ, QR, RS, SP, where PQ is at a right angle to QR, QR to
RS, RS to SP, and SP to PQ.
12. Let n and m be even integers. Then by (2), there are integers p, q such that n = 2p, m = 2q. Then n + m = 2p + 2q
= 2(p + q), so n + m is even.
13. Let n, m, p, q as in 12. Then nm = 2p · 2q = 4pq. Since pq is an integer, 4pq is an integral multiple of 4.
14. Define an odd integer m to be an integer such that there exists another integer n such that m = 2n + 1.
Let r be an even integer and s an odd integer. Then there are integers p, q such that r = 2p, s = 2q + 1. So r + s =
2p + 2q + 1 = 2(p + q) + 1, so r + s is odd.
15. counterexample
16. A B F G M, C D J, E H K N, I, L, O.
17. 1, 2, 4, 8, 16, 31 (the conjecture is false).
18. Suppose that i is the square of an odd integer k. Then
∃ i∈(cid:2):k =2l +1⇒i =k2 =(2l +1)2 =4l2+4l +1
Since i is also even,
∃j ∈(cid:2): i =2j ⇒ 4l2+4l +1=2j ⇒ 2l2+2l + 1 = j ∉(cid:2)
2
which is a contradiction, so k cannot be odd. Since k must be even,
∃l∈(cid:3): k =2l⇒ i =k2 =(2l)2 = 4l2
so i is indeed an integral multiple of 4.
19. Let n = 0, then (n+3)2 =32 =9>/ 9.
20. Let n2+2=3⇒ n2 =1⇒ n=−1∨n=+1, so n is not unique.

21. Let n=2⇒ n2+4=22+4=8.
22. Let n=3⇒ n2+5=32+5=14.
23. Let n=−3⇒ n2+5=(−3)2+5=14. With 22., n is not unique.

24. Let n=0: n2 >n⇐ 02 >0, which is a contradiction.

25. Let n ∈(cid:3),n <0⇒ n2 > 0⇒ n2 >0 >n.
( )
2
26. Let x = 1 ⇒ x2 <x ⇐ 1 < 1 ⇐ 1 < 1, which is a contradiction.
2 2 2 4 2
2
27. Let n=2: n2 >n⇐ 22 >2⇐ 4>2.

n=0: n2 =n⇐ 02 =0⇐ 0=0
28. Let , so x is not unique.
n=1: n2 =n⇐ 12 =1⇐ 1=1

29. Let j be an odd integer, so
∃k∈(cid:2): j =2k+1
⇒ j 2 =(2k+1)2
=4k2 +4k+1
= k2 +k∈(cid:3)
( )
30. ∃m∈(cid:3): n =3m+1, so n2 =(3m+1)2 =9m2+6m+1=33m2+2m +1, and 3 m2+2m is integral.

31. Let n=−2: n3 <n⇐ (−2)3 <−2⇐ −8<−2.

( ) ( )
2 2
32. Let n=−2,m =1: n = −2 =(−2)2 =4</1.
m 1
( )
2
33. n < n ⇒(m =0) n2 <nm⇒(n<0) n≥m⇒ n</m.
m m
( )3 ( )2 m ≥0: n3 ≤mn2 ⇒(n≠0) n≤m⇒ n<m
34. n ≤ n ⇒ . So let m = –1 and n = –2:
m m m ≤0: n3≥mn2 ⇒(n≠0) n≥m⇒ n</m

( ) ( ) ( ) ( )
3 2 3 2
n ≤ n ⇒ −2 ≤ −2 ⇒ 8≤4, which is a contradiction.
m m −1 −1
§0.2 Sets and Relations
♥ 17. An equivalence relation ∼ extracts a property from the whole identity of its arguments and asserts the equality of just
this property: equivalence is property equality. For example, ‘congruence modulo’ ≡ asserts equality of the
remainder under division.
1. {x ∈(cid:4)|x2 =3}={− 3, + 3}
2. {m ∈(cid:2)|m2 =3}=∅
{ } { }
3. m ∈(cid:2)|mn =60 for some n ∈(cid:2) = ±1, 2,3,4,5,6,10,12,15, 20,30
{ }
4. m ∈(cid:2)|m2 −m <115 . Solve the inequality:
m2−m =115⇒ m2−m−115=0⇒

+1± (−1)2−4⋅1⋅−115 1± 1+460
m = =
2⋅1 2
= 1(1± 461)≈−10.2,11.2 1(1− 461) 0 1(1+ 461)
2 2 2
so m∈{−10,−9,…,10,11}.

5. not a set
6. ∅
7. ∅
8. (cid:5)
9. (cid:5)
{ }
10. m |m ∈(cid:2)
2
11. {(a,1), (a,2), (a,c), (b,1), (b,2), (b,c), (c,1), (c,2), (c,c)}
12. function one-to-one onto
a. yes no no
b. yes no no
c. no
3
d. yes yes yes
e. yes no no
f. no
13. Map x to y(x).
P
A B
x
C D
y(x)
14 a. f :[0,1]→[0,2]:x a2x
b. f:[1,3]→[5,25]:x a(x −1)20+5
2
d−c
c. f:[a,b]→[c,d]:x a(x −a) +c
b−a
( )
15. f : S→ (cid:4): x a tanxπ− 1π
2
16. a. P(∅)=∅, P(∅)=1
{ }
b. P({a})= ∅,{a}, P({a}) =2
{ }
c. P({a,b})= ∅,{a},{b},{a,b}, P({a,b}) =4
{ }
d. P({a,b,c})= ∅,{a},{b},{a,b},{c},{a,c},{b,c},{a,b,c}, P({a,b,c}) =8
17. Conjecture P(A) =2A.

Let A be a series of sets such that |A | = n, and A ⊂A .
n n n n+1
• P(A ) = P(∅) =1.
0
• Let P(A ) =2An .
n
There is s ∉A such that A =A ∪s . Consider the set
n+1 n n+1 n n+1
U ( )
A′ = P∪ P∪{s }
n P⊆A n+1
n
Since every element of A n′ is a subset of A n+1, A n′ ⊆P(An+1).
Every subset P of A either does or does not contain s :
n+1 n+1
s ∉P ⇒ P ⊆A ⇒ P ∈A′
n+1 n n
s ∈P ⇒ P \{s }⊆A ⇒ P ∈A′
n+1 n+1 n n
so P (An+1)⊆An′.
So P (An+1)=An′, and P(An+1) =2⋅P(An) =2⋅2An =2An +1=2An+1 .
( )
18. Let f:A→B ∈BA.
• For each subset P ⊆A, there is a corresponding function
a∉P:0
f :A→B:aa
P a∈P:1
Let there be two such subsets P,P′⊆A such that f = f . Then ∀a∈A:
P P′
a ∈P ⇒ fP(a)= fP′(a)=1⇒ a∈P′; a ∉P ⇒…⇒a∉P′
so P =P′.
4
• Conversely, for each function f ∈BA there is a corresponding subset P ⊆A:
f
{ }
P = a∈A| f (a)=1 .
f
Let there be two functions f ,f ′∈BA such that P f =Pf′. Then ∀ a∈A:
f (a)=0 a∉P f ′(a)=0
∨ ⇒ ∨ f ⇒ ∨
f (a)=1 a∈Pf f ′(a)=1
so f = f ′.

So, P :BA →P(A) is a bijection, and BA = P(A).
f
19. For every element of A there is a distinct singleton subset containing just that element, which is an element of P(A)

. ∅ is not such a singleton set, yet is an element of P(A). So P(A) > A .

( )
Let A be such that A =ℵ. Then the power set of A has P(A) >ℵ, and P P(A) > P(A), ad infinitum.

20. a. It is possible to define addition in (cid:3) in terms of the union of disjoint sets, so
2+3=5⇐ A =2, B =3, A∪B =5.

i. 3+ℵ = {0}∪ (cid:2)+ (=*) (cid:2)+ =ℵ , where (*): φ:(cid:2)+ →{0}∪ (cid:2)+ :ma m−1.
0 0
( ) (*) ( )  m odd: 1(m−1)+ 1
ii. ℵ +ℵ = (cid:2)+ − 1 ∪ (cid:2)+ = (cid:2)+ =ℵ , where (*) φ: (cid:2)+ → (cid:2)+ − 1 ∪ (cid:2)+ :m a  2 2.
0 0 2 0 2  m even: 1m
2
b. It is possible to define multiplication in (cid:3) in terms of a Cartesian product:
2⋅3={1,2}×{1,2,3}=6, so

fig 14
ℵ ⋅ℵ = (cid:2)+ × (cid:2)+ = (cid:2)+ =ℵ .
0 0 0
21. 102 digits, 105 digits. By extrapolation, 1 0ℵ0 would equal the number of digits of the form 0.###…, where ‘#’ is
{ }
repeated ℵ times— name this set R. Since any number in R ′ = x ∈(cid:4)|0 ≤x <1 can be expressed arbitrarily
0
precise by an element of R, R ′⊆R. Since R ⊆R′, R =R′. By Exercise 15, R′ =ℵ, so R =ℵ and 10ℵ0 =ℵ.

Similar arguments can be made in terms of duodecimal and binary expansions of numbers of R′, so 1 2ℵ0 =2ℵ0 =ℵ.
22. Since
P((cid:2))(1=7)2(cid:2) = 2ℵ0 (1=8)ℵ; P((cid:2))(1=9){0,1}(cid:2)
The next higher cardinals after ℵ are
0
{ }(cid:2) { }
ℵ= 0,1 = 0,1 exp (cid:2),
{0,1}{0, 1}(cid:2) ={0,1}exp {0,1} exp (cid:2)
et cetera.
28. xRy ⇒∃i:x,y ∈P ⇒y,x ∈P ⇒y Rx (symmetric)
i i
xRx ⇔∃i:x ∈P ⇔x ∈S (reflexive)
i
xRy ∧y Rz ⇒∃i:x,y ∈P ∧∃j:y,z ∈P
i j
⇒(P disjoint⇒i = j)∃i:x,y,z ∈P (transitive).
i i
⇒y Rz

29. not reflexive because 0⋅0>/ 0⇒0R/ 0

30. not symmetric because 2≥1,1≥/ 2⇒2R1,1R/ 2.

5
31. R is a relation, because
y
x = x xRx R
 
x = y ⇒ y = x ⇒xRy ⇒y Rx
  x
x = y ∧ y = x ⇒ x = z xRy ∧y Rz ⇒xRz

.
0−3 =3≤3, 3−6 =3≤3, 0−6 =6≤/ 3⇒
32. so R is not transitive.
0R3, 3R6, 0R/ 6

33. The number of digits of n ∈(cid:2)+ is base 10 notation is 1+10logn. Obviously R is reflexive and symmetric, and

transitive.
R
0 10 100
34. R is congruence modulo 10 on (cid:2)*.
35. a. {1, 3, 5, …}, {2, 4, 6, …}
b. {1, 4, 7, …}, {2, 5, 8, …}, {3, 6, 9, …}
c. {1, 6, 11, …}, {2, 7, 12, …}, {3, 8, 13, …}, {4, 9, 14, …}, {5, 10, 15, …}
36a.
∀r ∈(cid:2): r −r =0 =0⋅n ⇒r ~r
( ) ( )
∀r,s,r ~s : ∃q∈(cid:2): r −s =qn ⇒ s −r =− qn = −q n⇒ s ~r
∀r,s,t,r ~ s,s ~t: ∃p,q ∈(cid:2): r −s =pn, s −t=qn
r −s +s −t =pn +qn
r −t =(p +q)n⇒ r ~t
r −s r s
b. ∀r,s ∈(cid:2)+,r ~s : ∃q∈(cid:2): r −s =qn ⇒(n∈(cid:2)) = − =q
n n n
∃r ′ ,s ′ ∈(cid:2),r ′′ , s ′′ ∈(cid:3): r =r ′ n +r ′′ , s = s ′ n +s ′′ ,0 ≤r ′′ , s ′′ <n
n n n n n n n n n n
r −s =qn
r ′ n +r ′′ −s ′ n −s ′′ =qn
n n n n
( ) ( )
r ′ −s ′ n + r ′′ −s ′′ =qn
n n n n
( )
r ′′ −s ′′ r ′′ −s ′′
r ′ −s ′ + n n =q ∧ 0 ≤ n n <1
n n n n
r ′′ −s ′′
Since r ′ ,s ′ ∈(cid:3),q ∈(cid:2), n n =0 ⇒ r ′′ −s ′′ .
n n n n n
c.
{…, –2, 1, 3, …}, {…, –2, 0, 2, …}
{…, –2, 1, 4, …}, {…, –1, 2, 5, …}, {…, –3, 0, 3, …}
{…, –4, 1, 6, …}, {…, –3, 2, 7, …}, {…, –2, 3, 8, …}, {…, –1, 4, 9, …}, {…, –5, 0, 5, …}
§0.3 Mathematical Induction
( )( )
n n+1 2n+1
1. Prove that + i2 = .
i =1…n 6
( )( )
11+1 2⋅1+1 2⋅3
n=1: 12 = = =1
6 6
6
( )( ) ( )( ) ( )
( ) n n+1 2n+1 n n+1 2n+1 +6n2+2n+1
n+1: + i2 = + i2 + n+1 2 = +n2+2n+1=
i…n+1 i…(n ) 6 ( )( )( ) 6
=n 2n2+3n+1 +6n2+12n+6=…= n+1 n+2 2n+3

n2(n +1)2
2. Prove that + i3 = ,n ∈(cid:2)+.
i=1…n 4
( )
12 1+1 2 1⋅22
n=1: 13 = = =1
4 4
( )
( ) n2 n+1 2 ( )( )
n+1: + i3 = + i3 + n+13 = + n+1 n+1 2
1…n+1 i…n( ) ( 4)( )
n2 n2+2n+1 +4n+1 n2+2n+1 n4+2n3+n2+4n3+8n2+4n+4n2+8n+4
= =
4 4
( ) ( )
n4+6n3+13nn2+12n+4 n+1 2 n+2 2
= =…=
4 4
( )
3. Prove that + 2i−1 =n2.
i =1…n
n=1: 1=12

( ) ( ) ( ) ( )
n+1: + 2i−1 = + 2i−1 +2n+1 −1=n2+2n+1= n+1 2
i=1…n+1 i=i…n
4. Prove that + ( 1 ) = n ,n ∈(cid:2)+.
i=1…ni i +1 n +1
1 1
n=1: ( ) = 1 = = 1
11+1 2 1+1 2

1 1 1 n 1
n+1: + ( ) = + ( ) + ( )( ) = +( )( )
i=1…n+1i i+1 i=1…ni i+1 n+1 n+2 n+1 n+1 n+2
( ) ( )
n n+2 +1 n2+2n+1 n+1 2 nn+1
= ( )( ) = ( )( ) = ( )( ) =
n+1 n+2 n+1 n+2 n+1 n+2 n+2

( )
n a 1−rn+1
5. Prove that ∀a,r ∈(cid:4),r ≠1,n ∈(cid:4)+ : + ari = .
i=0 1−r
( )
( )( )
a 1−r2 a 1−r 1+r ( )
n=1: a+ar = = =a 1−r
1−r 1−r
( ) ( ) ( )
n+1 n a 1−rn+1 a 1−rn+1 + 1−r arn+1
n+1: + ari = + ari +arn+1= +arn+1=
i=0 i=0 1−r 1−r(( )
( )
1−rn+1+ 1−r rn+1 1−rn+1+rn+1−rn+2 a 1−rn+2
=a =a =
1−r 1−r 1−r
6. max is only defined on (cid:2)+, so max(i−1,j −1) is undefined.
7. the concept ‘interesting property’ is not well defined
§0.4 Complex and Matrix Algebra
( ) ( )
1. 2+3i + 4+5i =6+2i.
2. i +5−3i =5−2i.
7
( ) ( )
3. 5+7i − 3−2i =2+5i.

( ) ( )
4. 1−3i − −4+2i =5−5i.

5. i 3 =ii2 =−i.
6. i 4 =i2⋅i2 =−1⋅−1=+1.
7. i23 =i20i3 =(i4)5⋅i3 =15⋅i3 =−i.

( ) ( ) ( )
8. −i 35=−i35=− i32i3 =−1⋅−i =i.

( )( )
9. 4−i 5+3i =20+12i−5i−3i2 =23+7i.

( )( )
10. 8+2i 3−i =24+6i−8i−2i2 =26−2i.

( )( ) ( )
11. 2−3i 4+i + 6−5i =8+2i−12i−3i2+6−5i =17−15i.

( ) ( )( ) ( )( ) ( )
12. 1+i 3 = 1+i 1+i 2 = 1+i 1+2i+i2 = 1+i 2i =2i+2i2 =−2+2i.

( )( )
7−5i 7−5i 1−6i 7−42i−5i+30i2 ( )
14. = ( )( ) = = 1 −23−47i .
1+6i 1+6i 1−6i 1−36 35

( )
1 i 1−i i−i2 i+1
15. = ( )( ) = = .
1+i 1+i 1−i 1−i2 2

( )
1−i 1−i i ( )
16. = =− i−i2 =−1−i.
i i2
( ) ( )( ) ( )( )
i 3+i 1 i 3+i 1+2i 3i−1 1+2i ( ) ( )
17. = ⋅( )( ) = 1 = 1 −1+3i−2i+6i2 = 1 −7+i .
2−4i 2 1−2i 1+2i 2 12−4i2 10 10

( )( )( ) ( )( )
3+7i 3+7i 1−i 2+3i 3−3i+7i−7i2 2+3i
( )( ) = ( )( )( )( ) = ( )( )
1+i 2−3i 1+i 1−i 2−3i 2+3i 1−i2 4−9i2
18. .
( )( ) ( )( )
10+4i 2+3i 5+2i 2+3i 10+15i+4i+6i2 4+19i
= = = =
2⋅13 1133 13 13
( )( )
( )( ) ( ) ( )( )
1−i 2+i 1−i 2 2+i 1+2i 1−2i+i2 2+4i+i+2i2 −2i⋅5i −i2
19. ( )( ) = ( )( )( )( ) = ( )( ) = = =1.
1−2i 1+i 1−2i 1+2i 1+i 1−i 1−4i2 1−i2 5⋅2 1
20. 3−4i =5.
21. 6+4i =23+2i =2 9+4 =2 13.

( )
22. 3−4i =5 ⇒3−4i =5 3− 4i .
5 5
 
1 1  
23. −1+i = 2 ⇒−1+i = 2− +i  = 2−1 2+ 1i 2.
  2 2
2 2

( )
24. 12+5i = 144+25= 169 =13 ⇒12+5i =13 12+ 5 i .
13 13
( )
25. −3+5i = 9+25= 36 =6 ⇒−3+5i =6 −1+ 5i .
2 6
26. z1=r1eiθ1,z2 =r2eiθ2 ⇒ z1 z2 =z1z2−1=r1eiθ1 ⋅(r2eiθ2)−1= r1 ei(θ1−θ2). So z 1 z2 is the point in the complex
r
2
point at the end of a line from the origin with length r r and angle θ −θ from the positive x-axis.
1 2 1 2
8
( )4 { }
27. z4 =1 ⇒ reiθ =1e0i ⇒r =1,4θ= 0 ⇒r =1,θ= 0 ⇒z ∈ 1,i,−1,−i .
2π 1π
2
( )
4
28. z4 =−1 ⇒ reiθ =1eiπ ⇒r =41,4θ= iπ ⇒r =1,θ= 1π .
{ 2π 12π 4 }
⇒z ∈ 1 2+ 1i 2,− 1 2+i 1 2,− 1 2− 1i 2, 1 2− 1i 2
2 2 2 2 2 2 2 2
§1.1 Binary Operations
(( ) ) ( )
1. b∗d =e,c∗c =b, a∗c ∗e ∗a = c∗e ∗a =a∗a =a.

( )
a∗b ∗c =b∗c =a
2. ( ) , so * could be, but is not necessarily, associative.
a∗ b∗c =a∗a =a
( )
b∗d ∗c =e∗c =a
3. ( ) , so * is not associative.
b∗ d∗c =b∗b =c

4. no, because e∗b ≠b∗e.
* a b c d
a a b c d
b b d a c
c c a d b
5. d d c b a
* a b c d
a a b c d
b b a c d
c c d c d
6. d d c c d
( ) ( )
d∗a = c∗b ∗a =c∗ b∗a =c∗b =d,
( ) ( )
d∗b = c∗b ∗b =c∗ b∗b =c∗a =c,

( ) ( )
d∗c = c∗b ∗c =c∗ b∗c =c∗c =c,

( ) ( )
d∗d = c∗b ∗d =c∗ b∗d =c∗d =d.

( ) ( ) ( ) ( )
7. 1∗0=1−0=1, 0∗1=0−1=−1, a∗b ∗c = a−b −c =a−b−c, a∗ b∗c =a− b−c =a−b+c, so * is neither

commutative nor associative.
( ) ( ) ( ) ( )
8. Let ∀a,b ∈(cid:5): a∗b = ab+1=ba+1=b ∗a, 0∗0 ∗1= 0⋅0+1 ⋅1+1=2, 0∗ 0∗1 =0⋅ 0⋅1+1 +1=1, so *
is commutative, but not associative.
( ) ( ) ( ) ( )
9. ∀a,b ∈(cid:5): a∗b = 1ab= 1ba=b ∗a, ∀a,b,c ∈(cid:5): a∗b ∗c = 1 1abc = 1a 1bc = a∗ b ∗c , so * is
2 2 2 2 2 2
commutative and associative.
( ) ( )
10. Let ∀a,b ∈(cid:2)+ : a∗b = 2ab = 2ba=b ∗a, then 0∗ 0∗1 =20⋅20⋅1 =20 =1 and 0∗0 ∗1=220⋅0⋅1=21=2, so * is

commutative, but not associative.
11. 1∗2=12 =1; 2∗1=21=2 and 2∗(3∗2)=2(32) =29; (2∗3)∗2=(23)2 =26, so * is neither commutative nor

associative.
12. 1; 222 =24 =16; 333 =39 =19683; nnn; the table defining * has n2 entries, each having n possible values.
( ) ( ) ( ) ( )
13. 1; 212⋅22−1 =21=2; 312⋅33−1 =33 =27; n12nn−1; the table defining commutative * has 1n n−1 entries, each
2
having n possible values.
9