Joseph Gallian Solutions Manual To Contemporary Abstract Algebra (2012)
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i
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
ii
Fields
19 Vector Spaces 165
20 Extension Fields 170
21 Algebraic Extensions 174
22 Finite Fields 180
23 Geometric Constructions 185
Supplementary Exercises for Chapters 19-23 187
Special Topics
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
1
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. 2 ·32·7; 23·33·5·7·11
3. 12, 2, 2, 10, 1, 0, 4, 5.
4. s=−3, t= 2; s= 8, t=−5
5. By using 0 as an exponent if necessary, we may write
a=pm1
1···pmk
kand b=pn1
1···pnk
k, where the p’s are distinct primes
and the m’s and n’s are nonnegative. Then lcm(a, b) = ps1
1···psk
k,
where si= max(mi, ni) and gcd(a, b) = pt1
1···ptk
k, where
ti= min(mi, ni) Then lcm(a, b)·gcd(a, b) = pm1+n1
1···pmk+nk
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+r1and 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 amod n=bmod n, then r1=r2and ndivides a−b.
If ndivides a−b, then by the uniqueness of the remainder, we then
have r1−r2= 0. Thus, r1=r2and therefore amod n=bmod n.
8. Write as +bt =d. Then a0s+b0t= (a/d)s+ (b/d)t= 1.
9. By Exercise 7, to prove that (a+b) mod n= (a0+b0) mod nand
(ab) mod n= (a0b0) mod nit suffices to show that ndivides
(a+b)−(a0+b0) and ab −a0b0. Since ndivides both a−a0and n
divides b−b0, it divides their difference. Because a=a0mod nand
b=b0mod nthere are integers sand tsuch that a=a0+ns and
b=b0+nt. Thus ab = (a0+ns)(b0+nt) = a0b0+nsb0+a0nt +nsnt.
Thus, ab −a0b0is divisible by n.
0/Preliminaries 2
10. Write d=au +bv. Since tdivides both aand b, it divides d. Write
s=mq +rwhere 0 ≤r < m. Then r=s−mq is a common
multiple of both aand bso r= 0.
11. Suppose that there is an integer nsuch that ab mod n= 1. Then
there is an integer qsuch that ab −nq = 1. Since ddivides both a
and n,dalso divides 1. So, d= 1. On the other hand, if d= 1,
then by the corollary of Theorem 0.2, there are integers sand 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 sand tsuch 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 anot divisible by 3, amod 3 is 1 or 2 and therefore a2
mod 3 = 1. So, (p2+q2+r2) mod 3 = p2mod 3 + q2mod 3 + r2
mod 3 = 3 mod 3= 0.
15. Let pbe a prime greater than 3. By the Division Algorithm, we can
write pin the form 6n+r, where rsatisfies 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 sand tdivide a−b. The converse is
true when gcd(s, t) = 1.
18. Observe that 8402 mod 5 = 3402 mod 5 and 34mod 5 = 1. Thus,
8402 mod 5 = (34)10032mod 5 = 4.
19. If gcd(a, bc) = 1, then there is no prime that divides both aand bc.
By Euclid’s Lemma and unique factorization, this means that there
is no prime that divides both aand bor both aand c. Conversely,
if no prime divides both aand bor both aand c, then by Euclid’s
Lemma, no prime divides both aand bc.
20. If one of the primes did divide k=p1p2···pn+ 1, it would also
divide 1.
0/Preliminaries 3
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 prime. This contradicts the assumption that
p1, p2, . . . , pnis the list of all primes.
22. −7
58 +3
58 i
23. −5+2i
4−5i=−5+2i
4−5i
4+5i
4+5i=−30
41 +−17
41 i
24. √2
2+√2
2i
25. Observe that cos θ+isin θ= cos( θ
n)n+isin( θ
n)n= (cos θ
n+isin θ
n)n
26. Observe that
1+2+···+n+(n+1) = n(n+1)
2+(n+1) = (n+1)( n
2+1) = (n+1)(n+2)
2.
27. Let Sbe a set with n+ 1 elements and pick some ain S. By
induction, Shas 2nsubsets that do not contain a. But there is
one-to-one correspondence between the subsets of Sthat do not
contain aand those that do. So, there are 2 ·2n= 2n+1 subsets in
all.
28. Use induction and note that
2n+132n+2 −1 = 18(2n32n)−1 = 18(2n33n−1) + 17.
29. Consider n= 200! + 2. Then 2 divides n, 3 divides n+ 1, 4 divides
n+ 2, . . ., and 202 divides n+ 200.
30. Use induction on n.
31. Say p1p2···pr=q1q2···qs, where the p’s and the q’s are primes.
By the Generalized Euclid’s Lemma, p1divides some qi, say q1(we
may relabel the q’s if necessary). Then p1=q1and
p2···pr=q2···qs. Repeating this argument at each step we obtain
p2=q2,···, pr=qrand r=s.
32. 47. Mimic Example 12.
33. Suppose that Sis a set that contains aand whenever n≥abelongs
to S, then n+ 1 ∈S. We must prove that Scontains all integers
greater than or equal to a. Let Tbe the set of all integers greater
than athat are not in Sand suppose that Tis not empty. Let bbe
the smallest integer in T(if Thas no negative integers, bexists
0/Preliminaries 4
because of the Well Ordering Principle; if Thas negative integers,
it can have only a finite number of them so that there is a smallest
one). Then b−1∈S, and therefore b= (b−1) + 1 ∈S. This
contradicts our assumption that bis not in S.
34. By the Second Principle of Mathematical Induction,
fn=fn−1+fn−2<2n−1+ 2n−2= 2n−2(2 + 1) <2n.
35. For n= 1, observe that 13+ 23+ 33= 36. Assume that
n3+ (n+ 1)3+ (n+ 2)3= 9mfor some integer m. We must prove
that (n+ 1)3+ (n+ 2)3+ (n+ 3)3is a multiple of 9. Using the
induction hypothesis we have that
(n+ 1)3+ (n+ 2)3+ (n+ 3)3= 9m−n3+ (n+ 3)3=
9m−n3+n3+ 3 ·n2·3+3·n·9+33= 9m+ 9n2+ 27n+ 27.
36. You must verify the cases n= 1 and n= 2. This situation arises in
cases where the arguments that the statement is true for nimplies
that it is true for n+ 2 is different when nis even and when nis
odd.
37. The statement is true for any divisor of 84−4 = 4092.
38. One need only verify the equation for n= 0,1,2,3,4,5.
Alternatively, observe that n3−n=n(n−1)(n+ 1).
39. Since 3736 mod 24 = 16, it would be 6 p.m.
40. 5
41. Observe that the number with the decimal representation
a9a8. . . a1a0is a9109+a8108+··· +a110 + a0. From Exercise 9
and the fact that ai10imod 9 = aimod 9 we deduce that the check
digit is (a9+a8+··· +a1+a0) mod 9. So, substituting 0 for 9 or
vice versa for any aidoes not change the value of
(a9+a8+··· +a1+a0) mod 9.
42. No
43. For the case in which the check digit is not involved, the argument
given Exercise 41 applies to transposition errors. Denote the money
order number by a9a8. . . a1a0cwhere cis the check digit. For a
transposition involving the check digit c= (a9+a8+···+a0) mod 9
to go undetected, we must have a0= (a9+a8+···+a1+c) mod 9.
Substituting for cyields 2(a9+a8+··· +a0) mod 9 = a0. Then
0/Preliminaries 5
cancelling the a0, multiplying by sides by 5, and reducing module 9,
we have 10(a9+a8+··· +a1) = a9+a8+··· +a1= 0. It follows
that c=a9+a8··· +a1+a0=a0. In this case the transposition
does not yield an error.
44. 4
45. Say the number is a8a7. . . a1a0=a8108+a7107+··· +a110 + a0.
Then the error is undetected if and only if
(ai10i−a0
i10i) mod 7 = 0. Multiplying both sides by 5iand noting
that 50 mod 7 = 1, we obtain (ai−a0
i) mod 7 = 0.
46. All except those involving aand bwith |a−b|= 7.
47. 4
48. Observe that for any integer kbetween 0 and 8, k÷9 = .kkk . . . .
49. If nis not prime, we can write n=ab, where 1 < a < n and
1< b < n. Then aand bbelong to the set {1,2, . . . , n}but
0 = ab mod ndoes not.
50. 7
51. Say that the weight for ais i. Then an error is undetected if
modulo 11, ai +b(i−1) + c(i−2) = bi +c(i−1) + a(i−2). This
reduces to the cases where (2a−b−c) mod 11 = 0.
52. Say the valid number is a1a2. . . a10 and aiand ai+1 were
transposed. Then, modulo 11, 10a1+ 9a2+··· +a10 = 0 and
10a1+··· + (11 −i)ai+1 + (11 −(i+ 1))ai+··· +a10 = 5. Thus,
5 = 5 −0 =
(10a1+···+(11−i)ai+1+(11−(i+1))ai+a10)−(10a1+9a2+···+a10).
It follows that (ai+1 −ai) mod 11 = 5. Now look for adjacent digits
xand yin the invalid number so that (x−y) mod 11 = 5. Since
the only pair is 39, the correct number is 0-669-09325-4.
53. Since 10a1+ 9a2+··· +a10 = 0 mod 11 if and only if
0=(−10a1−9a2− ··· − 10a10) mod 11 =
(a1+ 2a2+··· + 10a10) mod 11, the check digit would be the same.
54. 7344586061
0/Preliminaries 6
55. First note that the sum of the digits modulo 11 is 2. So, some digit
is 2 too large. Say the error is in position i. Then
10 = (4,3,0,2,5,1,1,5,6,8) ·(1,2,3,4,5,6,7,8,9,10) mod 11 = 2i.
Thus, the digit in position 5 to 2 too large. So, the correct number
is 4302311568.
56. An error in an even numbered position changes the value of the
sum by an even amount. However,
(9 ·1+8·4+7·9+6·1+5·0+4·5+3·2+2·6 + 7) mod 10 = 5.
57. 2. Since βis one-to-one, β(α(a1)) = β(α(a2)) implies that
α(a1) = α(a2) and since αis one-to-one, a1=a2.
3. Let c∈C. There is a bin Bsuch that β(b) = cand an ain A
such that α(a) = b. Thus, (βα)(a) = β(α(a)) = β(b) = c.
4. Since αis one-to-one and onto we may define α−1(x) = yif and
only if α(y) = x. Then α−1(α(a)) = aand α(α−1(b)) = b.
58. a−a= 0; if a−bis an integer kthen b−ais the integer −k; if
a−bis the integer nand b−cis the integer m, then
a−c= (a−b) + (b−c) is the integer n+m. The set of equivalence
classes is {[k]|0≤k < 1, k is real}. The equivalence classes can be
represented by the real numbers in the interval [0,1). For any real
number a, [a] = {a+k|where kranges over all integers}.
59. No. (1,0) ∈Rand (0,−1) ∈Rbut (1,−1) 6∈ R.
60. Obviously, a+a= 2ais even and a+bis even implies b+ais even.
If a+band b+care even, then a+c= (a+b)+(b+c)−2bis also
even. The equivalence classes are the set of even integers and the
set of odd integers.
61. abelongs to the same subset as a. If aand bbelong to the subset
Aand band cbelong to the subset B, then A=B, since the
distinct subsets of Pare disjoint. So, aand cbelong to A.
62. Suppose that nis odd prime greater than 3 and n+ 2 and n+ 4 are
also prime. Then nmod 3 = 1 or nmod 3 = 2. If nmod 3 = 1 then
n+ 2 mod 3 = 0 and so is not prime. If nmod 3 = 2 then
n+ 4 mod 3 = 0 and so is not prime.
0/Preliminaries 7
63. The last digit of 3100 is the value of 3100 mod 10. Observe that 3100
mod 10 is the same as ((34mod 10)25 mod 25 and 34mod 10 = 1.
Similarly, the last digit of 2100 is the value of 2100 mod 10. Observe
that 25mod 10 = 2 so that 2100 mod 10 is the same as
(25mod 10)20 mod 10 = 220 mod 10 = (25)4mod 10 =
24mod 10 = 6.
64. Write the numbers in the form
11,11 + 100,11 + 1100,11 + 11100, . . . and consider them modulo 4.
65. Apply γ−1to both sides of αγ =βγ.
8
CHAPTER 1
Introduction to Groups
1. Three rotations: 0◦, 120◦, 240◦, and three reflections across lines
from vertices to midpoints of opposite sides.
2. Let R=R120,R2=R240,Fa reflection across a vertical axis,
F0=RF and F00 =R2F
R0R R2F F 0F00
R0R0R R2F F 0F00
R R R2R0F0F00 F
R2R2R0R F 00 F F 0
F F F 00 F0R0R2R
F0F0F F 00 R R0R2
F00 F00 F0F R2R R0
3. a. Vb. R270 c. R0d. R180, H, V, D, D0e. none
4. Five rotations: 0◦, 72◦, 144◦, 216◦, 288◦, and five reflections across
lines from vertices to midpoints of opposite sides.
5. Dnhas nrotations of the form k(360◦/n), where k= 0, . . . , n −1.
In addition, Dnhas nreflections. When nis odd, the axes of
reflection are the lines from the vertices to the midpoints of the
opposite sides. When nis even, half of the axes of reflection are
obtained by joining opposite vertices; the other half, by joining
midpoints of opposite sides.
6. A nonidentity rotation leaves only one point fixed – the center of
rotation. A reflection leaves the axis of reflection fixed. A reflection
followed by a different reflection would leave only one point fixed
(the intersection of the two axes of reflection) so it must be a
rotation.
7. A rotation followed by a rotation either fixes every point (and so is
the identity) or fixes only the center of rotation. However, a
reflection fixes a line.
1/Introduction to Groups 9
8. In either case, the set of points fixed is some axis of reflection.
9. Observe that 1 ·1 = 1; 1(−1) = −1; (−1)1 = −1; (−1)(−1) = 1.
These relationships also hold when 1 is replaced by a “rotation”
and −1 is replaced by a “reflection.”
10. reflection.
11. In D4, HD =DV but H6=V.
12. Dnis not commutative.
13. R0,R180,H,V
14. Rotations of 0◦and 180◦; Rotations of 0◦and 180◦and reflections
about the diagonals.
15. R0,R180,H,V
16. Let the distance from a point on one Hto the corresponding point
on an adjacent Hbe one unit. Then translations of any number of
units to the right or left are symmetries; reflection across the
horizontal axis through the middle of the H’s is a symmetry;
reflection across any vertical axis midway between two H’s or
bisecting any His a symmetry. All other symmetries are
compositions of finitely many of those already described. The
group is non-Abelian.
17. In each case the group is D6.
18. D28
19. cyclic
20. D22
21. Their only symmetry is the identity.
22. It is symmetric under a horizontal reflection.
23. 180 degree rotational symmetry
24. Z4, D5, D4, Z2
10
CHAPTER 2
Groups
1. c, d
2. a, c, d
3. none
4. a, c
5. 17; 13; n−1; 1
3−2i=1
3−2i
3+2i
3+2i=3
13 +2
13 i
6. a. −31 −ib. 5c. 1
12 "2−3
−8 6 #d. "2 10
5 6 #
7. The set does not contain the identity; closure fails.
8. (3 −2) −1 = 0 while 3 −(2 −1) = 2.
9. Under multiplication modulo 4, 2 does not have an inverse. Under
multiplication modulo 5, {1,2,3,4}is closed, 1 is the identity, 1
and 4 are their own inverses, and 2 and 3 are inverses of each other.
Modulo multiplication is associative.
10. "1 1
0 1 #" 1 0
1 1 #6="1 0
1 1 #" 1 1
0 1 #.
11. First observe that taking the entries modulo 11
5−6
−3 2 =5 5
8 2 . Also, modulo 11, the determinant of 2 6
3 5
is −8 = 3. Finally, instead of dividing each entry of 5 5
8 2 by 3 we
must multiply each entry by 3−1mod 11 = 4 and reduce modulo 11
to obtain 9 9
10 8 .
12. Use D4.
13. (a) 2a+ 3b; (b) −2a+ 2(−b+c); (c) −3(a+ 2b)+2c= 0
2/Groups 11
14. (ab)3=ababab and
(ab−2c)−2= ((ab−2c)−1)2= (c−1b2a−1)2=c−1b2a−1c−1b2a−1.
15. Since the inverse of an element in Gis in G,H⊆G. Let gbelong
to G. Then g= (g−1)−1and g−1belong to G. So, G⊆H.
16. The identity is 25.
17. First note that if 1 is in Hthen by closure all five elements are in
H. Then, since gcd(pq, qp) = 1 and gcd(p+q, pq) = 1, The corollary
of Theorem 0.2 shows that 1 belongs to Hin all cases except case e.
18. H={R0, R180};K={R0, R180, H, V, D, D0}.
19. The set is closed because det (AB) = (det A)(det B). Matrix
multiplication is associative. "1 0
0 1 #is the identity.
Since "a b
c d #−1
="d−b
−c a #its determinant is ad −bc = 1.
20. 12= (n−1)2= 1.
21. Using closure and trial and error, we discover that 9 ·74 = 29 and
29 is not on the list.
22. Consider xyx =xyx.
23. For n≥0, we use induction. The case that n= 0 is trivial. Then
note that (ab)n+1 = (ab)nab =anbnab =an+1bn+1. For n < 0, note
that e= (ab)0= (ab)n(ab)−n= (ab)na−nb−nso that anbn= (ab)n.
In a non-Abelian group (ab)nneed not equal anbn.
24. The “inverse” of putting on your socks and then putting on your
shoes is taking off your shoes then taking off your socks. Use D4for
the examples. (An appropriate name for the property
(abc)−1=c−1b−1a−1is “Socks-Shoes-Boots Property.”)
25. Suppose that Gis Abelian. Then by Exercise 24,
(ab)−1=b−1a−1=a−1b−1. If (ab)−1=a−1b−1then by Exercise 24
e=aba−1b−1. Multiplying both sides on the right by ba yields
ba =ab.
26. By definition, a−1(a−1)−1=e. Now multiply on the left by a.
2/Groups 12
27. The case where n= 0 is trivial. For n > 0, note that
(a−1ba)n= (a−1ba)(a−1ba)···(a−1ba) (nterms). So, cancelling the
consecutive aand a−1terms gives a−1bna. For n < 0, note that
e= (a−1ba)n(a−1ba)−n= (a−1ba)n(a−1b−na) and solve for
(a−1ba)n.
28. (a1a2···an)(a−1
na−1
n−1···a−1
2a−1
1) = e
29. By closure we have {1,3,5,9,13,15,19,23,25,27,39,45}.
30. Z105;Z44 and D22.
31. Suppose xappears in a row labeled with atwice. Say x=ab and
x=ac. Then cancellation gives b=c. But we use distinct elements
to label the columns.
32. 1 5 7 11
1 1 5 7 11
5 5 1 11 7
7 7 11 1 5
11 11 7 5 1
33. Proceed as follows. By definition of the identity, we may complete
the first row and column. Then complete row 3 and column 5 by
using Exercise 31. In row 2 only cand dremain to be used. We
cannot use din position 3 in row 2 because there would then be
two d’s in column 3. This observation allows us to complete row 2.
Then rows 3 and 4 may be completed by inserting the unused two
elements. Finally, we complete the bottom row by inserting the
unused column elements.
34. Use cancellation.
35. axb =cimplies that x=a−1(axb)b−1=a−1cb−1;a−1xa =cimplies
that x=a(a−1xa)a−1=aca−1.
36. Observe that xabx−1=ba is equivalent to xab =bax and this is
true for x=b.
37. Since eis one solution it suffices to show that nonidentity solutions
come in distinct pairs. To this end note that if x3=eand x6=e,
then (x−1)3=eand x6=x−1. So if we can find one nonidentity
solution we can find a second one. Now suppose that aand a−1are
nonidentity elements that satisfy x3=eand bis a nonidentity
2/Groups 13
element such that b6=aand b6=a−1and b3=e. Then, as before,
(b−1)3=eand b6=b−1. Moreover, b−16=aand b−16=a−1. Thus,
finding a third nonidentity solution gives a fourth one. Continuing
in this fashion we see that we always have an even number of
nonidentity solutions to the equation x3=e.
To prove the second statement note that if x26=e, then x−16=x
and (x−1)26=e. So, arguing as in the preceding case we see that
solutions to x26=ecome in distinct pairs.
38. In D4, HR90V=DR90Hbut HV 6=DH.
39. Observe that aa−1b=ba−1a. Cancelling the middle term a−1on
both sides we obtain ab =ba.
40. X=V R270D0H.
41. If F1F2=R0then F1F2=F1F1and by cancellation F1=F2.
42. Observe that F1F2=F2F1implies that (F1F2)(F1F2) = R0. Since
F1and F2are distinct and F1F2is a rotation it must be R180.
43. Since F Rkis a reflection we have (F Rk)(F Rk) = R0. Multiplying
on the left by Fgives RkF Rk=F.
44. Since F Rkis a reflection we have (F Rk)(F Rk) = R0. Multiplying
on the right by R−kgives F RkF=R−k. If Dnwere Abelian, then
F R360◦/nF=R360◦/n. But (R360◦/n)−1=R360◦(n−1)/n 6=R360◦/n
when n≥3.
45. a. R3b. Rc. R5F
46. Closure: (3m6n)(3s6t)=3m+s6n+t; multiplication of real numbers
is associative; identity: 3060; (3m6n)−1= 3−m6−n.
47. Since a2=b2= (ab)2=e, we have aabb =abab. Now cancel on left
and right.
48. Closure and associativity follow from the definition of
multiplication; a=b=c= 0 gives the identity; we may find
inverses by solving the equations a+a0= 0, b0+ac0+b= 0,
c0+c= 0 for a0, b0, c0.
49. If nis not prime, we can write n=ab, where 1 < a < n and
1< b < n. Then aand bbelong to the set {1,2, . . . , n}but
0 = ab mod ndoes not.
2/Groups 14
50. If asatisfies x5=eand a6=e, then so does a2, a3, a4. Now, using
cancellation we have that a2, a3, a4are not the identity and are
distinct from each other and distinct from a. If these are all of the
nonidentity solutions of x5=ewe are done. If bis another solution
that is not a power of a, then by the same argument b, b2, b3and b4
are four distinct nonidentity solutions. We must further show that
b2, b3and b4are distinct from a, a2, a3, a4. If b2=aifor some i,
then cubing both sides we have b=b6=a3i,which is a
contradiction. A similar argument applies to b3and b4. Continuing
in this fashion we have that the number of nonidentity solutions to
x5=eis a multiple of 4. In the general case, the number of
solutions is a multiple of 4 or is infinite.
51. The matrix "a b
c d #is in GL(2, Z2) if and only if ad 6=bc. This
happens when aand dare 1 and at least 1 of band cis 0 and when
band care 1 and at least 1 of aand dis 0. So, the elements are
"1 0
0 1 #" 1 1
0 1 #" 1 0
1 1 #" 1 1
1 0 #" 0 1
1 1 #" 0 1
1 0 #.
"1 1
0 1 #and "1 0
1 1 #do not commute.
52. Since "a a
a a #" b b
b b #="2ab 2ab
2ab 2ab #
and 2ab 6= 0 we have closure; matrix multiplication is associative;
from the product above we observe that the identity is
"1/2 1/2
1/2 1/2#and that the inverse of "a a
a a #is
"1/(4a) 1/(4a)
1/(4a) 1/(4a)#.
The group GL(2,R) has a different identity than the group G.
53. Let abe any element in Gand write x=ea. Then
a−1x=a−1(ea)=(a−1e)a=a−1a=e. Then solving for xwe
obtain x=ae =a.
54. Suppose that ab =eand let b0be the element in Gwith the
property that bb0=e. Then observe that
ba = (ba)e=ba(bb0) = b(ab)b0=beb0= (be)b0=bb0=e.
15
CHAPTER 3
Finite Groups; Subgroups
1. |Z12|= 12; |U(10)|= 4; |U(12)|= 4; |U(20)|= 8; |D4|= 8.
In Z12,|0|= 1; |1|=|5|=|7|=|11|= 12; |2|=|10|= 6; |3|=|9|=
4; |4|=|8|= 3; |6|= 2.
In U(10), |1|= 1; |3|=|7|= 4; |9|= 2.
In U(20), |1|= 1; |3|=|7|=|13|=|17|= 4; |9|=|11|=|19|= 2.
In D4,|R0|= 1; |R90|=|R270|= 4;
|R180|=|H|=|V|=|D|=|D0|= 2.
In each case, notice that the order of the element divides the order
of the group.
2. In Q,h1/2i={n(1/2)|n∈Z}={0,±1/2,±1,±3/2, . . .}. In Q∗,
h1/2i={(1/2)n|n∈Z}={1,1/2,1/4,1/8, . . . ; 2,4,8, . . .}.
3. In Q,|0|= 1. All other elements have infinite order since
x+x+··· +x= 0 only when x= 0.
4. Suppose |a|=nand |a−1|=k. Then (a−1)n= (an)−1=e−1=e.
So k≤n. Now reverse the roles of aand a−1to obtain n≤k. The
infinite case follows from the finite case.
5. In Z30, 2 + 28 = 0 and 8 + 22 = 0. So, 2 and 28 are inverses of each
other and 8 and 22 are inverses of each other. In U(15), 2 ·8=1
and 7 ·13 = 1. So, 2 and 8 are inverses of each other and 7 and 13
are inverses of each other.
6. a. |6|= 2,|2|= 6,|8|= 3; b. |3|= 4,|8|= 5,|11|= 12 ;
c. |5|= 12,|4|= 3,|9|= 4. In each case |a+b|divides lcm(|a|,|b|).
7. (a4c−2b4)−1=b−4c2a−4=b3c2a2.
8. If a subgroup of D3contains R240 and Fit also contains
R0, R2
240 =R120, R240F, and R120F, which is all six elements of D3.
If Fand F0are distinct reflections in a subgroup of D3, then
F F 0=R240 is also in the subgroup. Thus the subgroup must be D3.
3/Finite Groups; Subgroups 16
9. If a subgroup of D4contains R270 and a reflection F, then it also
contains the six other elements
R0,(R270)2=R180,(R270)3=R90, R270F, R180Fand R90F. If a
subgroup of D4contains Hand D, then it also contains HD =R90
and DH =R270. But this implies that the subgroup contains every
element of D4. If it contains Hand Vthen it contains HV =R180
and R0.
10. {R0, R90, R180, R270},{R0, R180, H, V }, and {R0, R180, D, D0}.
11. If nis a positive integer, the real solutions of xn= 1 are 1 when n
is odd and ±1 when nis even. So, the only elements of finite order
in R∗are ±1.
12. If aba =e, then b=a−2so that bcommutes with a.
13. His a subgroup. To prove this we need only show that if a∈H
then a−1∈H. But if a−16∈ Hthen the given property says that
a= (a−1)−16∈ H.
14. |f(x)|= 10; |g(x)|= 5; |f(x) + g(x)|= 10; |h(x)|=
10/gcd(a1, a2, . . . , an).
15. Since |a|= 7 we have a=a14a=a15 = (a5)3.
16. No, let the group be Z3and H={1}. To see that Hhas the desired
property, observe that the only choice for aand bis a= 1 and b= 1
and (−1) + (−1) = −2 = 1 belongs to Hbut His not a subgroup.
17. If aand bare distinct elements of order 2 then ab has order 2 and is
distinct from aand b. If cis a fourth element of order 2 then ac, bc
and abc make at least 7 elements of order 2. D4has exactly five
elements of order 2.
18. The possibilities are 1, 2, 3 and 6. 5 is not possible for if a5=e,
then e=a6=aa5=a. 4 is not possible for if a4=e, then
e=a6=a2a4=a2.
19. Suppose that m<nand am=an. Then e=ana−m=an−m. This
contradicts the assumption that ahas infinite order.
20. If x4=e, then e= (x4)2=x8=x6·x2=x2. If x5=e, then
e= (x5)2=x10 =x6·x4=x4.|x|= 3 or 6.
3/Finite Groups; Subgroups 17
21. If ahas infinite order, then e, a, a2, . . . are all distinct and belong
to G, so Gis infinite. If |a|=n, then ai=ajwhere 0 < i < j < n
implies aj−i=e, which is a contradiction. Thus, e, a, a2, . . . , an−1
are all distinct and belong to G, so Ghas at least nelements.
22. h3i={3,32,33,34,35,36}={3,9,13,11,5,1}=U(14).h5i=
{5,52,53,54,55,56}={5,11,13,9,3,1}=U(14).h11i={11,9,1} 6=
U(14).
23. Since |U(20)|= 8, for U(20) = hkifor some kit must the case that
|k|= 8. But 11= 1, 34= 1, 74= 1, 92= 1, 112= 1, 134= 1,
174= 1, and 192= 1. So, the maximum order of any element is 4.
24. Let Abe the subset of even members of Znand Bthe subset of
odd members of Zn. If x∈B, then x+A={x+a|a∈A} ⊆ B, so
|A|≤|B|. Also, x+B={x+b|b∈B} ⊆ A, so |B|≤|A|.
25. Suppose that Kis a subgroup of Dnthat has at least one reflection
F. Denote the rotations of Kby R1, R2, . . . , Rm. Then
R1F, R2F, . . . , RmFare distinct reflections in K. If F0is any
reflection in K, then F0F=Rifor some i. But then F0=RiF.
Thus Khas exactly mreflections.
26. Suppose that aand bare two elements of order 2 that commute.
Then {e, a, b, ab}is closed and therefore a subgroup.
27. Observe that by Exercise 26 we have that for any reflection Fin
Dnthe set {R0, R180, F, R180F}is a subgroup of order 4.
28. h2i
29. Let H=hkiand observe that because 6 = 30 + 30 −54 belongs to
Hwe know 6 is a multiple of k. Thus the possibilities for Hare
h6i,h3iand h2i. None of these can be excluded because each
contains 12, 30 and 54.
30. Suppose that His a subgroup of D3of order 4. Since D3has only
two elements of order 2, Hmust contain R120 or R240. By closure,
it follows that Hmust contain R0, R120, and R240 as well as some
reflection F. But then Hmust also contain the reflection R120F.
3/Finite Groups; Subgroups 18
31. U4(20) = {1,9,13,17};U5(20) = {1,11};U5(30) = {1,11};
U10(30) = {1,11}.Uk(n) is closed because (ab) mod k= (amod
k)(bmod k) = 1 ·1 = 1 (here we used the fact that kdivides n). H
is not closed since 7 ∈Hbut 7 ·7 = 9 is not in H.
32. H∩K6=∅, since e∈H∩K. Now suppose that x, y ∈H∩K.
Then, since Hand Kare subgroups, we know xy−1∈Hand
xy−1∈K. That is, xy−1∈H∩K.
33. If x∈Z(G), then x∈C(a) for all a, so x∈T
a∈G
C(a). If
x∈T
a∈G
C(a), then xa =ax for all ain G, so x∈Z(G).
34. Suppose x∈C(a). Then xa =ax. So a−1(xa) = a−1(ax) = x.
Thus, (a−1x)a=xand therefore a−1x=xa−1. This shows
x∈C(a−1). The other half follows by symmetry.
35. The case that k= 0 is trivial. Let x∈C(a). If kis positive, then
by induction on k, xak+1 =xaak=axak=aakx=ak+1x. The case
where kis negative now follows from Exercise 34. The statement
“If for some integer k, x commutes ak, then xcommutes with a” is
false as can be seen in the group D4with x=H, a =R90 and k= 2.
36. Observe that 2 = (6 ∗6) ∗2=6∗(6 ∗2) = 6 ∗5. The remaining
cases are similar.
37. a. C(1) = C(5) = G;C(2) = C(6) = {1,2,5,6};C(3) = C(7) =
{1,3,5,7};C(4) = C(8) = {1,4,5,8}.
b. Z(G) = {1,5}
c. |1|= 1; |2|=|4|=|5|=|6|=|8|= 2; |3|=|7|= 4. They divide
the order of the group.
38. If a2=b2and a3=b3, then a2a=b2b. Now cancel a2and b2.
39. a. First observe that because hSiis a subgroup of Gcontaining S,
it is a member of the intersection. So, H⊆ hSi. On the other hand,
since His a subgroup of Gand Hcontains S, by definition
hSi ⊆ H.
b. Let K={sn1
1sn2
2. . . snm
m|m≥1, si∈S, ni∈Z}. Then because
Ksatisfies the subgroup test and contains Swe have hSi ⊆ K. On
the other hand, if Lis any subgroup of Gthat contains Sthen L
also contains Kby closure. Thus, by part a, H=hSicontains K.
3/Finite Groups; Subgroups 19
40. a. h2ib. h1ic. h3id. hgcd(m, n)ie. h3i.
41. Since ea =ae, C(a)6=∅. Suppose that xand yare in C(a). Then
xa =ax and ya =ay. Thus,
(xy)a=x(ya) = x(ay)=(xa)y= (ax)y=a(xy)
and therefore xy ∈C(a). Starting with xa =ax, we multiply both
sides by x−1on the right and left to obtain x−1xax−1=x−1axx−1
and so ax−1=x−1a. This proves that x−1∈C(a). By the
Two-Step Subgroup Test, C(a) is a subgroup of G.
42. Mimic the proof of Theorem 3.5.
43. No. In D4,C(R180) = D4.
44. Yes. Elements in the center commute with all elements.
45. Let H={x∈G|xn=e}. Since e1=e, H 6=∅. Now let a, b ∈H.
Then an=eand bn=e. So, (ab)n=anbn=ee =eand therefore
ab ∈H. Starting with an=eand taking the inverse of both sides,
we get (an)−1=e−1. This simplifies to (a−1)n=e. Thus, a−1∈H.
By the Two-Step test, His a subgroup of G. In D4,
{x|x2=e}={R0, R180, H, V, D, D0}. This set is not closed
because HD =R90.
46. The C(a)⊆C(a3) is easy. To prove the other inclusion, observe
that a6=aso if x∈C(a3), then
xa =xa6=x(a3a3) = (xa3)a3= (a3x)a3=a3(xa3)
=a3(a3x)=(a3a3)x=a6x=ax.
For the second part of the exercise, try D6.
47. Note that for any polynomial f(x) = anxn+an−1xn−1+···a0in G
we have 4f(x)=4anxn+ 4an−1xn−1+··· + 4a0= 0. Thus the
orders of elements of Gare at most 4. No element has order 3
because 3ai= 0 mod 4 if and only if aiis 0. Since 2ai= 0 mod 4 if
and only if ai= 0 or 2 we have
2f(x)=2anxn+ 2an−1xn−1+···+ 2a0in Gif and only if a1= 0 or
2 for all i. Excluding the identity, this condition is necessary and
sufficient for an element to have order 2.
3/Finite Groups; Subgroups 20
48. For any integer n≥3, observe that the rotation R360/n in Dnhas
order n. Now in Dnlet Fbe any reflection. Then F0=R360/nFis a
reflection in Dn. Also |F0|=|F|= 2 and F0F=R360/n has order n.
49. Note that (ab)2=abab =a(ba)b=a(a3b)b=a4b2=ee =e. So, |ab|
is 1 or 2. But |ab|= 1 implies that a=b−1=b, which is false.
50. Note that b4=abaaba =ab2a=a(aba)a=b. Thus, b3=e. On the
other hand, b26=efor if so then b3=b2implies that b=e.
51. First observe that (ad)n/d =an=e, so |ad|is at most n/d.
Moreover, there is no positive integer t < n/d such that
(ad)t=adt =e, for otherwise |a| 6=n.
52. |A|= 4, |B|= 3, |AB|=∞.
53. By induction we will prove that any positive integer nwe have
"1 1
0 1 #n
="1n
0 1 #.
The n= 1 case is true by definition. Now assume
"1 1
0 1 #k
="1k
0 1 #.
Then "1 1
0 1 #k+1
="1 1
0 1 #k"1 1
0 1 #=
"1k
0 1 #" 1 1
0 1 #="1k+ 1
0 1 #.
So, when the entries are from R,"1 1
0 1 #has infinite order.
When the entries are from Zp, the order is p.
54. For the first part use induction; 6, ∞.
55. For any positive integer n, a rotation of 360◦/n has order n. If we
let Rbe a rotation of √2 degrees then Rnis a rotation of √2n
degrees. This is never a multiple of 360◦, for if √2n= 360kthen
√2 = 360k/n, which is rational. So, Rhas infinite order.
3/Finite Groups; Subgroups 21
56. |x2|= 3; |x3|= 2; |x4|= 3; |x5|= 6;
|y2|=|y4|=|y5|=|y7|=|y8|= 9; |y3|=|y6|= 3.
57. hR0i,hR180i,hR90i,hDi,hD0i,hHi,hVi. (Note that hR90i=hR270i.)
58. h1i,h2i,h4i,h7i,h11i,h14i.
59. For every nonidentity element aof odd order, a−1is distinct from a
and has the same order as a. Thus nonidentity elements of odd
order come in pairs. So, there must be some element aof even
order, say |a|= 2m. Then |am|= 2.
60. If |a|= 3, then {e, a, a2}is a subgroup. So, 8 elements of order 3
produce 4 subgroups of order 3. To show that there are no other
subgroups of order 3 note that a subgroup of order 3 can’t have an
element of order greater than 3. So, the only other possibility is to
have H={e, a, b}where |a|=|b|= 2. But then ab is a fourth
element of H.
61. Let |g|=mand write m=nq +rwhere 0 ≤r < n. Then
gr=gm−nq =gm(gn)−qbelongs to H. So, r= 0.
62. a. 2, 2, 4 b. 4, 6, 24 c. 2, 4, 8 d. 2, 4, 8.
63. 1 ∈H, so H6=∅. Let a, b ∈H. Then (ab−1)2=a2(b2)−1, which is
the product of two rationals. The integer 2 can be replaced by any
positive integer.
64. 2, 4, 16. |U(rs)|=|U(r)||U(s)|when gcd(r, s)=1
65. |h3i| = 4
66. {1,9,11,19}
67. Let "a b
c d #and "a0b0
c0d0#belong to H. By the One-Step
Subgroup Test it suffices to show that
a−a0+b−b0+c−c0+d−d0= 0. This follows from
a+b+c+d= 0 = a0+b0+c0+d0. If 0 is replaced by 1, His not a
subgroup since it does not contain the identity.
68. Say det A= 2mand det B= 2n. Then det (AB) = 2m+nand
det A−1= 2−m.
3/Finite Groups; Subgroups 22
69. If 2aand 2b∈K, then 2a(2b)−1= 2a−b∈K, since a−b∈H.
70. Let f, g ∈H. Then (f·g−1)(2) = f(1)g−1(2) = 1 ·1 = 1. The 2 can
be replaced by any number.
71. "2 0
0 2 #−1
="1
20
01
2#is not in H.
72. His not closed.
73. If a+bi and c+di ∈H, then
(a+bi)(c+di)−1=a+bi
c+di
c−di
c−di =(ac+bd)+(bc−ad)i
c2+d2= (ac+bd)+(bc−ad)i.
Moreover,
(ac +bd)2+ (bc −ad)2=a2c2+ 2acbd +b2d2+b2c2−2bcad +a2d2.
Simplifying we obtain,
(a2+b2)c2+ (a2+b2)d2= (a2+b2)(c2+d2)=1·1 = 1. So, His a
subgroup. His the unit circle in the complex plane.
74. Since Gis Abelian the set is closed and therefore a subgroup.
|ha, bi| ≤ |a||b|.
75. Since ee =eis in HZ(G) it is non-empty. Let h1z1and h2z2belong
to HZ(G). Then
h1z1(h2z2)−1=h1z1z−1
2h−1
2=h1h−1
2z1z−1
2∈HZ(G).
76. Since eis in H∪gH,Kis not empty. Suppose the xand ybelong
to K. We will show that xy−1is in K. If xand ybelong to H,
then xy−1is in H. If xis in Hand yis in gH, then we can write
xy−1in the form h1(gh2)−1where h1is in Hand h2is in H. Thus,
since g=g−1we have
xy−1=h1(gh2)−1=h1h−1
2g−1=g−1h1h−1
2=g(h1h−1
2) is in gH. A
similar argument argument applies if xis in gH and yis in H.
Lastly, if both xand yare in gH then xy−1has the form
gh1(gh2)−1=gg−1h1h−1
2=h1h−1
2, which is in H. The argument is
valid for K=Z(G)∪gZ(G).
77. By the corollary of Theorem 0.2 there are integers sand tso that
1 = ms +nt. Then a1=ams+nt =amsant = (am)s(an)t= (at)n.
78. The argument given in Example 14 shows that the only rotations
that commute with Fare R0when nis odd and R0and R180 when
nis even. Now suppose that some reflection F0in Dncommutes
with F. Consider the rotation R=F F 0. Then F0=F R and by
3/Finite Groups; Subgroups 23
Exercise 44 of Chapter 2 we know that R−1=F RF . Then F0
commutes with Fif and only if F F 0=F0F. This is the same as
F(F R)=(F R)For R=R−1. And this holds if and only if R=R0
or R=R180. So, for nodd, C(F) = {R0, F }. For neven,
C(F) = {R0, R180, F, F R180}.C(R0) = C(R180) = Dn. If Ris
rotation in Dnother than R0or R180 then C(R) is the subgroup of
all rotations.
79. a. Suppose that "a b
c d #commutes with "1 1
1 0 #. Then
"a b
c d #" 1 1
1 0 #="1 1
1 0 #" a b
c d #. This implies that
"a+b a
c+d c #="a+c b +d
a b #. From this we have b=cand
a=b+d. For convenience, we solve the latter equality for din
terms of aand band get d=a−b. So, C " 1 1
1 0 #!=
(" a b
b a −b#|where a2−ab −b26= 0; a, b ∈R).
b. Suppose that "a b
c d #commutes with "0 1
1 0 #. Then
"a b
c d #" 0 1
1 0 #="0 1
1 0 #" a b
c d #.
This implies that "b a
d c #="c d
a b #.
From this we have b=cand a=d. So,
C " 0 1
1 0 #!=(" a b
b a #|where a2−b26= 0; a, b ∈R).
3/Finite Groups; Subgroups 24
c. Suppose that "a b
c d #commutes with every element of
GL(2,R). By part b we know that c=band d=a. So, any
element in the center of GL(2,R) has the form "a b
b a #.Also,
"a b
b a #" 0−1
1 0 #="0−1
1 0 #" a b
b a #.
This gives "b−a
a−b#="−b−a
a b #.
So, b= 0.Thus,
Z(G) = (" a0
0a#|a6= 0; a∈R).
80. Let g∈G,g6=e. If |g|=pm, then |gm|=p.
25
CHAPTER 4
Cyclic Groups
1. For Z6, generators are 1 and 5; for Z8generators are 1, 3, 5, and 7;
for Z20 generators are 1, 3, 7, 9, 11, 13, 17, and 19.
2. For hai, generators are aand a5; for hbi, generators are b,b3,b5,
and b7; for hci, generators are c,c3,c7,c9,c11,c13,c17,c19.
3. h20i={20,10,0};h10i={10,20,0}
ha20i={a20, a10, a0};ha10i={a10, a20, a0}
4. h3i={3,6,9,12,15,0};
h15i={15,12,9,6,3,0};ha3i={a3, a6, a9, a12, a15, a0};
ha15i={a15, a12, a9, a6, a3, a0}.
5. h3i={3,9,7,1}
h7i={7,9,3,1}
6. In any group, hai=ha−1i. See Exercise 11.
7. U(8) or D3.
8. (a) All have order 5. (b) Both have order 3. (c) All have order 15.
9. Six subgroups; generators are the divisors of 20.
Six subgroups; generators are ak, where kis a divisor of 20.
10. 3 ·1,3·3,3·5,3·7; a3,(a3)3,(a3)5,(a3)7.
11. By definition, a−1∈ hai. So, ha−1i ⊆ hai. By definition,
a= (a−1)−1∈ ha−1i. So, hai ⊆ ha−1i.
12. h3i,h3i;a3, a−3.
13. Observe that h21i={0,21,18,15,12,9,6,3}and
h10i={0,10,20,6,16,2,12,22,8,18,4,14}. Since the intersection of
two subgroups is a subgroup, according to the proof of Theorem
4.3, we can find a generator of the intersection by taking the
4/Cyclic Groups 26
smallest positive multiple of 1 that is in the intersection. So,
h21i∩h10i=h6i.
Similarly, ha21i={e, a21, a18, a15, a12, a9, a6, a3}and
ha10i={e, a10, a20, a6, a16, a2, a12, a22, a8, a18, a4, a14}. Then again
by the proof of Theorem 4.3, we can find a generator of the
intersection by taking the smallest positive power of athat is in the
intersection. So, ha21i∩ha10i=ha6i.
For the case hami∩hani, let k= lcm(m, n). Write k=ms and
k=nt. Then ak= (am)s∈ hamiand ak= (an)t∈ hani. So,
haki⊆hami∩hani. Now let arbe any element in hami∩hani. Then
ris a multiple of both mand n. It follows that ris a multiple of k
(see Exercise 10 of Chapter 0). So, ar∈ haki.
14. 49. First note that the group is not infinite since an infinite cyclic
group has infinitely many subgroups. Let |G|=n. Then 7 and n/7
are both divisors of n. If n/76= 7,then Ghas at least 4 divisors.
So, n/7 = 7. When 7 is replaced by p , |G|=p2.
15. |g|divides 12 is equivalent to g12 =e. So, if a12 =eand b12 =e,
then (ab−1)12 =a12(b12)−1=ee−1=e. The same argument works
when 12 is replaced by any integer (see Exercise 45 of Chapter 3).
16. h0i ⊂ h120i ⊂ h60i⊂h30i⊂h15i ⊂ h5i ⊂ h1i.
17. is odd or infinite. To see this note that if |a|=∞, then |a2|cannot
be finite and if |a|=n, by Theorem 4.2 we have
n=|a2|=n/gcd(n, 2) and therefore gcd(n, 2) = 1.
18. 1.
19. h1i,h7i,h11i,h17i,h19i,h29i
20. By Corollary 2 of Theorem 4.1 a nonidentity element of Gmust
have order 5, 7 or 35. We may assume that Ghas no element of
order 35. Since 34 is not a multiple of φ(5) = 4, not all of the
nonidentity elements can have order 5. Similarly, not all of them
can have order 7. So, Ghas elements of orders both 5 and 7. Say,
|a|= 5 and |b|= 7. Then, since (ab)5=b56=eand
(ab)7=a7=a26=e, we must have |ab|= 35, a contradiction.
21. a. |a|divides 12. b. |a|divides m. c. By Theorem 4.3,
|a|= 1,2,3,4,6,8,12, or 24. If |a|= 2, then a8= (a2)4=e4=e. A
similar argument eliminates all other possibilities except 24.
4/Cyclic Groups 27
22. Let G={e, a, b}. Cancellation shows ab must be e. Thus
G={e, a, a−1}.
23. Yes, by Theorem 4.3. The subgroups of Zare of the form
hni={0,±n, ±2n, ±3n, . . .}, for n= 0,1,2,3, . . .. The subgroups of
haiare of the form hanifor n= 0,1,2,3, . . ..
24. Certainly, a∈C(a). Thus, hai ⊆ C(a).
25. Dnhas nreflections each of which has order 2. Dnalso has n
rotations that form a cyclic group of order n. So, according to
Theorem 4.4, there are φ(d) rotations of order din Dn. If nis odd,
there are no rotations of order 2. If nis even, there is φ(2) = 1
rotation of order 2. (Namely, R180.) So, when nis odd Dnhas n
elements of order 2; when nis even, Dnhas n+ 1 elements of order
2.
26. 1 and −1 are the only generators of Z. Suppose that akgenerates
hai. Then there is an integer tso that (ak)t=a. By Theorem 4.1,
we conclude that kt = 1. So, k=±1.
27. See Example 15 of Chapter 2.
28. The case that i=−jis Exercise 4 of Chapter 3. If haii=haji, then
ai= (aj)k=ajk for some k. Since ahas infinite order this means
that i=jk and therefore jdivides i. Likewise, idivides j. So,
i=±j.
29. 1000000, 3000000, 5000000, 7000000. By Theorem 4.3, h1000000iis
the unique subgroup of order 8, and only those on the list are
generators; a1000000, a3000000, a5000000, a7000000. By Theorem 4.3,
ha1000000iis the unique subgroup of order 8, and only those on the
list are generators.
30. Observe that
a2ba−2=a(aba−1)a−1=ab−1a−1= (aba−1)−1= (b−1)−1=b. So,
a2∈C(b). Since ahas odd order ha2i=haiso that a∈C(b). Thus,
aba−1=band therefore b=b−1.
31. Let G={a1, a2, . . . , ak}. Now let |ai|=niand n=n1n2. . . nk.
Then an
i=efor all isince nis a multiple of ni.
4/Cyclic Groups 28
32.
@
@
@
@
@
@
h0i
h4i h6i
h2i h3i
h1i
33.
@
@
@
@
@
@
@
@
h0i
hpqihq2i
hpi hqi
h1i
34. h1i
|
h2i
|
h4i
|
h0i
4/Cyclic Groups 29
35.
.
.
.
h0i
hpn−1i
hpn−2i
hpn−3i
hpn−ni
36. First suppose that Gis the union of proper subgroups. If Gwere
cyclic, say G=hai, and Gwas the union of proper subgroups
H1, H2, . . . , Hn, then amust be in one of Hisince the union
contains every element. But if abelongs to Hithen G=haiis a
subgroup of the proper subgroup Hi. This is a contradiction.
Now suppose that Gis not cyclic. Let abe any element in G. Since
Gis not cyclic haiis a proper subgroup of G. So, we can take hai
as one of the members of the union. Continuing in this way we get
a union that contains all of G.
37. Suppose aand bare relatively prime positive integers and
ha/bi=Q+. Then there is some integer nsuch that
(a/b)n= 2.Clearly n6= 0,1, or −1. If n > 1, an= 2bn, so that 2
divides a. But then 2 divides bas well. If n < −1, then b−n= 2a−n
and as before 2 divides both aand b.
38. 4 8 12 16
4 16 12 8 4
8 12 4 16 8
12 8 16 4 12
16 4 8 12 16
The identity is 16. The group is generated by 8 and by 12.
39. For 6, use Z27. For n, use Z2n−1.
40. lcm(m, n)
4/Cyclic Groups 30
41. Let t= lcm(m, n) and let |ab|=s. Then (ab)t=atbt=e, and
therefore sdivides t. Also, e= (ab)s=asbs, so that as=b−s, and
therefore asand b−sbelong to hai∩hbi={e}. Thus, mdivides s
and ndivides s, and therefore tdivides s.
For the second part, let R120 denote a rotation of 120◦and Fbe a
reflection in D3. Then hR120i∩hFi={R0},|R120|= 3,|F|= 2
but D3has no element of order 6.
42. |ab|could be any divisor of lcm(|a|,|b|).
43. Let |a|=m, b =nand d= gcd(m, n). Then
lcm(m, n) = mn/d, |ad|=m/d, and |b|=n. Then by Exercise 41,
|adb|= lcm(m, n).
44. Since F F 0is a rotation other than the identity and the rotations of
D21 form a cyclic subgroup of order 21, we know by Theorem 4.3
that |F F 0|is a divisor of 21. Moreover, F F 0cannot be the identity
for then F F 0=F F , which implies that F0=F. So, |F F 0|= 3,7 or
21.
45. All divisors of 60.
46. Use the corollary to Theorem 4.4.
47. In an infinite group the number of elements of finite order nis a
multiple of φ(n) or there is an infinite number of elements of order
n.
48. Since Khas two reflections it has a non-trivial rotation. The order
of the subgroup of Kconsisting of rotations only must be 5 or 7
since it divides 35 and is proper. So, |K|= 10 or 14.
49. It follows from Example 15 in Chapter 2 and Example 10 in
Chapter 0 that the group H=hcos(360◦/n) + isin(360◦/n)iis a
cyclic group of order nand every member of this group satisfies
xn−1=0.Moreover, since every element of order nsatisfies
xn−1 = 0 and there can be at most nsuch elements, all complex
numbers of order nare in H. Thus, by Theorem 4.4, C∗has
exactly φ(n) elements of order n.
50. Clearly 0 is in H. If mand nare in Hthen mhas the forms 8m1
and 10m2and nhas the forms 8n1and 10n2. Then m−nhas the
forms 8m1−8n1and 10m1−10n1. So m−nis in H. So, His a
subgroup of Z.
4/Cyclic Groups 31
51. Let x∈Z(G) and |x|=pwhere pis prime. Say y∈Gwith |y|=q
where qis prime. Then (xy)pq =eand therefore |xy|= 1, p or q. If
|xy|= 1, then x=y−1and therefore p=q. If |xy|=p, then
e= (xy)p=ypand qdivides p. Thus, q=p. A similar argumernt
applies if |xy|=q.
52. If an infinite group had only a finite number of subgroups then it
would have only a finite number of cyclic subgroups. If each of
these cyclic subgroups is finite then the group would be finite since
every element is in the cyclic subgroup generated by itself. So the
group contains at least one infinite cyclic subgroup, call it hai.
Then the subgroups hai,ha2i,ha3i, . . . are distinct subgroups since
the least positive power of ain haiiis aiand aiis not the least
positive power of ain any subgroup hajifor j6=i.
53. An infinite cyclic group does not have element of prime order. A
finite cyclic group can have only one subgroup for each divisor of its
order. A subgroup of order phas exactly p−1 elements of order p.
Another element of order pwould give another subgroup of order p.
54. 2; 4; a3,a5,a7.
55. 1 ·4, 3 ·4, 7 ·4, 9 ·4; x4,(x4)3,(x4)7,(x4)9.
56. In group, the number of elements order dis divisible by φ(d) or
there are infinitely many elements of order d.
57. D33 has 33 reflections each of which has order 2 and 33 rotations
that form a cyclic group. So, according to Theorem 4.4, for each
divisor dof 33 there are φ(d) rotations of order din Dn. This gives
one element of order 1; φ(3) = 2 elements of order 3; φ(11) = 10
elements of order 11; and φ(33) = 20 elements of order 33.
58. The number of solutions to x15 =ein Gis 15. To see this let Hbe
the unique subgroup of Gof order 15. Then Halso contains the
unique subgroups of Gorders 1, 3, and 5. Since every element of G
that is a solution to x15 =ehas order 1,3, 5 or 15 we know that H
contains all solutions and every element of His a solution. The
same argument works when 15 is replaced by any positive integer n.
59. Let |hai| = 4 and |hbi| = 5. Since (ab)20 = (a4)5(b5)4=e·e=ewe
know that |ab|divides 20. Noting that (ab)4=b46=ewe know that
|ab| 6= 1,2 or 4. Likewise, (ab)10 =a26=eimplies that |ab| 6= 5 or
4/Cyclic Groups 32
10. So, |ab|= 20. Then, by Theorem 4.3, habihas subgroups of
orders 1, 2, 4, 5, 10 and 20. In general, if an Abelian group contains
cyclic subgroups of order mand nwhere mand nare relatively
prime, then it contains subgroups of order dfor each divisor dof
mn.
60. 1, 2, 3, 12. In general, if an Abelian group contains cyclic
subgroups of order mand n, then it contains subgroups of order d
for each divisor dof the least common multiple of mand n.
61. Say aand bare distinct elements of order 2. If aand bcommute,
then ab is a third element of order 2. If aand bdo not commute,
then aba is a third element of order 2.
62. 12
63. By Exercise 32 of Chapter 3, hai∩hbiis a subgroup. Also,
hai∩ hbi⊆haiand hai∩ hbi⊆hbi. So, by Theorem 4.3, |hai∩ hbi| is
a common divisor of 10 and 21. Thus, |hai∩hbi| = 1 and therefore
hai∩hbi={e}.
64. Mimic Exercise 53.
65. Similar to Exercise 63, |hai∩hbi| must divide both 24 and 10. So,
hai∩hbi| = 1 or 2.
66. From Theorem 4.4 we know that a finite cyclic group has at most 1
element of order 2. Now observe that 2n−1−1 and 2n−1 have
order 2.
67. If x8=e, then |x|divides 8. If Gis cyclic, then by Theorem 4.4, G
has exactly φ(8) + φ(4) + φ(2) + φ(1) = 4 + 2 + 1 + 1 = 8 elements
with orders that divide 8. So, Gis not cyclic. For the case x4=e,
if Gis cyclic, then Ghas φ(4) + φ(2) + φ(1) = 4 elements with
orders that divide 4. So, again Gis not cyclic. If Gis a finite cyclic
group and ndivides |G|, then there are exactly nelements in G
that are solutions of xn=e. So, if there are more than nelements
xsuch that xn=e, the group is not cyclic.
68. First note that if kis a generator then so is −k. Thus it suffices to
show that k6=−k. But k=−kimplies that 2k= 0 so that
n=|k|= 1 or 2.
4/Cyclic Groups 33
69. Observe that |a5|= 12 implies that e= (a5)12 =a60 so |a|divides
60. Since ha5i ⊆ haiwe know that |hai| is divisible by 12. So,
|hai| = 12 or 60.
If |a4|= 12, then |a|divides 48. Since ha4i⊆haiwe know that |hai|
is divisible by 12. So, |hai| = 12,24, or 48. But |a|= 12 implies
|a4|= 3 and |a|= 24 implies |a4|= 6. So, |a|= 48.
70. By Theorem 4.3, it suffices to find necessary and sufficient
conditions so that |xr|divides |xs|. By Theorem 4.2, we obtain
gcd(n, s) divides gcd(n, r).
71. Observe that a280 =e=a440. Thus |a|is common divisor of 280
and 440, and therefore |a|divides gcd(280,440) = 40. Since
|a22|= 20 implies that 20 divides |a|we have that |a|= 20 or 40.
But |a|= 20 implies that 20 = |a22|=|a2|= 10. So, |a|= 40.
72. gcd(48,21) = 3; gcd(48,14) = 2; gcd(48,18) = 6.
73. Say bis a generator of the group. Since pand pn−1 are relatively
prime, we know by Corollary 3 of Theorem 4.2 that bpalso
generates the group. Finally, observe that (bp)k= (bk)p.
74. "1 1
0 1 #n
="1n
0 1 #.
75. By Exercise 32 of Chapter 3, hai∩hbiis a subgroup haiand hbi. So,
|hai∩hbi| divides 12 and 22. If follows that |hai∩hbi| = 1 or 2 and
since hai∩hbi 6={e}we have that |hai∩hbi| = 2. Because ha6iis the
only subgroup of haiof order 2 and hb11iis the only subgroup of hbi
of order 2, we have hai∩hbi=ha6i=hb11iand therefore a6=b11.
because the addition is term wise and the set {0,1,2}under
modulo 3 addition is associative. The identity is the polynomial
0x2+ 0x+ 0. The inverse of ax2+bx +cis a0x2+b0x+c0where
a0, b0and c0are the inverses of a, b, and cmodulo 3. Since there are
3 choices for each of a, b, and c, the group has 27 elements. Noting
that adding any polynomial in the group to itself 3 times results in
a polynomial with coefficients 0 modulo 3, we see that the largest
order of any element is 3 so Gis not cyclic.
76. First note that x6=e. If x3=x5, then x2=e. By Corollary 2
Theorem 4.1 and Theorem 4.3 we then have |x|divides both 2 and
15. Thus |x|= 1 and x=e. If x3=x9, then x6=eand therefore
4/Cyclic Groups 34
|x|divides 6 and 15. This implies that |x|= 3. Then
|x13|=|x(x3)4|=|x|= 3. If x5=x9, then x4=eand |x|divides
both 4 and 15, and therefore x=e.
77. Since reflections have order 2, any cyclic subgroup of order 4 must
be generated by a rotation. So, by Theorem 4 there is exactly one
cyclic subgroup of order 4.
78. Suppose that Kis a subgroup of Dnof order 4 where nis odd. By
Exercise 5.1, Kconsists of 4 rotations or two rotations and two
reflections. Since the subgroup of Dnof all rotations is cyclic of
odd order, Kcannot consist of 4 rotations nor have a rotation of
order 2. This rules out both cases for K.
79. Observe that Dnhas n/2 non-cyclic subgroups of order 4 of the
form {R0, R180, F, R180F}where Fis a reflection. By Exercise 25 of
Chapter 3 these are the only non-cyclic subgroups of order 4.
80. By Exercise 78 there are exactly n/2 non-cyclic subgroups of order
4. By Theorem 4.3 there is no cyclic subgroup of order 4.
81. First observe that the set of all rotations is the only cyclic subgroup
of Dnof order n. It follows from Exercise 25 of Chapter 3 that
when nis odd Dnhas no non-cyclic subgroup of order nand when
n= 2mthe only non-cyclic subgroups of Dnare the two of then
form {R, R2,...Rm, RF, R2F, . . . , RmF}where |R|=mand Fis a
reflection.
82. Gis a group because it is closed. It is not cyclic becasue every
nonzero element has order 3.
83. Since mand nare relatively prime, it suffices to show both mand
ndivide k. By Corollary 2 of Theorem 4.1, it is enough to show
that ak=e. Note that ak∈ hai∩hbi, and since hai∩hbiis a
subgroup of both haiand hbi, we know that |hai∩hbi| must divide
both |hai| and |hbi|. Thus, |hai∩hbi| = 1.
84. Note that nand n2−2 are distinct elements of order 2 and appeal
to Theorem 4.4.
85. Note that among the integers from 1 to pnthe pn−1integers
p, 2p, . . . , pn−1pare exactly the ones not relatively prime to p.
86. In C∗the solutions of x4= 1 are ±1,±iand | ± i|= 4.
35
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1-4
1. a. Let xh1x−1and xh2x−1belong to xHx−1. Then
(xh1x−1)(xh2x−1)−1=xh1h−1
2x−1∈xHx−1also.
b. Let hhi=H. Then hxhx−1i=xHx−1.
c.
(xh1x−1)(xh2x−1) = xh1h2x−1=xh2h1x−1= (xh2x−1)(xh1x−1).
2. Let x, y ∈N(H). Then
(xy−1)H(xy−1)−1=x(y−1Hy)x−1=xHx−1=H. (Note that
yHy−1=Himplies y−1Hy =H.)
3. Suppose cl(a)∩cl(b)6=∅. Say xax−1=yby−1. Then
(y−1x)a(y−1x)−1=b. Thus, for any ubu−1in cl(b), we have
ubu−1=uy−1xa(y−1x)−1u−1= (uy−1x)a(uy−1x)−1∈cl(a). This
shows that cl(b)⊆cl(a). By symmetry, cl(a)⊆cl(b). Because
a=eae−1∈cl(a), the union of the conjugacy classes is G.
4. a. {e, a2}b. cl(a) = {a, a3}c. cl(b) = {b, ba2}
d. hei={e}
hai={e, a, a2, a3}=ha3i
ha2i={e, a2}
hbi={e, b, a2, ba2}=hba2i
hbai={e, ba, a2, ba3}=hba3i
5. Observe that e= (xax−1)k=xakx−1if and only if ak=e.
6. If both ab and ba have infinite order, we are done. Suppose that
|ab|=k. Then e= (ab)(ab)···(ab) [kfactors]. Thus,
ba =bea =b((ab)(ab)···(ab))a= (ba)k+1. This shows that
(ba)k=eso that |ba|≤|ab|. By symmetry, |ab| ≤ |ba|.
7. By Exercise 4 of Chapter 3, Exercise 24 of Chapter 2, and the
previous exercise we have |ab|=|(ab)−1|=|b−1a−1|=|a−1b−1|.
8. By Exercise 22 of Chapter 4 we may write Hin the form {e, a, a2}.
Let G={e, a, a2, b}. By cancellation, ab cannot be a, a2of b. So,
ab =eand b=a−1belongs to H.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 36
9. In D4, take a=R90,b=H, and c=D.
10. If there is an element xin Gsuch that xax =bthen axax =ab and
c=ax. If there is an element cin Gso that ab =c2, let x=a−1c.
11. By Exercise 5, for every xin G|xax−1|=|a|, so that xax−1=aor
xa =ax.
12. |x|=|y|= 2; |xy|=∞. No!
13. 1 of order 1, 15 of order 2, 8 of order 15, 4 of order 5, 2 of order 3.
14. Let |G|= 4. If Ghas an element of order 4, Gis cyclic. Gcannot
have an element of order 3 for if so then Ghas the form
{e, x, x2, y}. But cancellation shows xy 6∈ G. The remaining case is
handled by Exercise 35 of Chapter 2.
15. Let |G|= 5. Let a6=ebelong to G. If |a|= 5, we are done. If
|a|= 3, then {e, a, a2}is a subgroup of G. Let bbe either of the
remaining two elements of G. Then the set {e, a, a2, b, ab, a2b}
consists of six different elements, a contradiction. Thus |a| 6= 3.
Similarly |a| 6= 4. We may now assume that every nonidentity
element of Ghas order 2. Pick a6=eand b6=ein Gwith a6=b.
Then {e, a, b, ab}is a subgroup of G. Let cbe the remaining
element of G. Then {e, a, b, ab, c, ac, bc, abc}is a set of eight distinct
elements of G, a contradiction. It now follows that if a∈Gand
a6=e, then |a|= 5.
16. Suppose that Gis an Abelian group of order 6 but has no element
of order 6. Gcannot have an element aof order 5 for letting b
denote the element not in haiwe know that Gwould contain
e, a, a2, a3, a4, b and ab, which is too many. Moreover, Ghas no
element aof order 4 for then Gwould contain e, a, a2, a3, b, ab and
a2b, which is too many. So, we may assume that all nonidentity
elements of Ghave orders 2 or 3. Since the number of elements of
order 3 is divisible by φ(3) = 2 we know that Gmust have at least
one element of order 2. Now note that Gcannot have two elements,
say aand b, of order 2. For if so, then letting cbe any element not
in {e, a, b, ab}we have that Gwould contain the elements
e, a, b, ab, c, ac, bc, which is too many. So, Gmust have an element a
of order 2 and an element bof order 3. Then ab has order 6.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 37
17. an(bn)−1= (ab−1)n, so by the One-Step Subgroup Test Gnis a
subgroup. For the non-Abelian group, use D3with n= 3.
18. Let x=a+b√2 and y=c+d√2 belong to G. First note that
y−1=c
c2−2d2−d
c2−2d2√2∈G.
Now show that xy−1∈G. (How do we know c2−2d26= 0?)
19. Suppose G=H∪K. Pick h∈Hwith h6∈ K. Pick k∈Kwith
k6∈ H. Then, hk ∈Gbut hk 6∈ Hand hk 6∈ K.
U(8) = {1,3}∪{1,5}∪{1,7}.
20. If |a|=mand |b|=n, then (ab−1)mn =e. The elements of finite
order do not always form a subgroup in a non-Abelian group. (See
Exercise 8 of this set.)
21. If |a|=pkand |b|=prwith k≤r, say, then
(ab−1)pr=apr(bpr)−1=eso |ab−1|divides pr.
22. 1, 3, 5, and 15. To see this observe that
a16 = (a4)4= (bab)4=ba4b=bbabb =a. So, a15 =eand therefore
|a|divides 15.
23. Because ba2=ab and a3=b2=eimply that ba =a2b, every
member of the group can be written in the form aibj. Therefore,
the group is {e, a, a2, b, ab, a2b}.
e a a2b ab a2b
e e a a2b ab a2b
a a a2e ab a2b b
a2a2e a a2b b ab
b b a2b ab e a2a
ab ab b a2b a e a2
a2b a2b ab b a2a e
D3satisfies these conditions.
24. If xcommutes with athen abelongs to the subgroup C(x). By
closure, akbelongs to C(x) and therefore xand (ak) commute .
Now suppose that xand akcommute. Then akbelongs to C(x)
and by closure haki ⊆ C(x). By the Corollary to Theorem 0.2,
there are integers sand tsuch that 1 = ns +kt. Thus,
a=ans+kt = (an)s(ak)t= (ak)talso belongs to C(x).
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 38
25. xy =yx if and only if xyx−1y−1=e. But
(xy)x−1y−1=x−1(xy)y−1=ee =e.
26. If a/b ∈Hand c/d ∈K, then ac ∈H∩K. No, h√2i∩h1i={0}.
27. Let x∈N(gHg−1). Then x(gHg−1)x−1=gHg−1. Thus
g−1xgHg−1x−1g=g−1xgH(g−1xg)−1=H. This means that
g−1xg ∈N(H). So x∈gN(H)g−1. To show
gN(H)g−1⊆N(gHg−1) let x=gng−1where n∈N(H). Then
x(gHg−1)x−1= (gng−1)(gHg−1)(gn−1g−1) = gnHn−1g−1=
gHg−1. So, x∈N(gHg−1).
28. Let sand tbe such that ms +nt = 1. Then
g=gms+nt = (gm)s(gn)t= (gm)s.
29. In D11 let abe a reflection and let bbe a rotation other than the
identity. Then |a|= 2, |b|= 11 and |ab|= 2 because ab is a
reflection.
30. 2
31. Solution from Mathematics Magazine.1“Yes. Let abe an arbitrary
element of S. The set {an|n= 1,2,3, . . .}is finite, and therefore
am=anfor some m, n with m>n≥1. By cancellation we have
ar(a)=a, where r(a) = m−n+ 1 >1. If xis any element of S,
then aar(a)−1x=ar(a)x=ax, and this implies that ar(a)−1x=x.
Similarly, we see that xar(a)−1=x, and therefore the element
e=a(r(a)−1is an identity. The identity element is unique, for e0is
another identity, then e=ee0=e0. If r(a)>2 then ar(a)−2is an
inverse of a, and if r(a) = 2 then a2=a=eis its own inverse.
Thus Sis a subgroup.”
32. Let Hbe the union. If x, y ∈Hthen there are iand jso that
x∈Hiand y∈Hj. Say i≤j. Then x, y ∈Hjand therefore
xy−1∈Hj⊆H.
33. {1,2n−1,2n+ 1,4n−1}. Since the square of each of these is 1
they are relatively prime to 4n.
1Mathematics Magazine 63 (April 1990): 136.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 39
34. Since e1=e, H is not empty. Let aand bbelong to H. Then there
are odd integers mand nsuch that am=eand bn=e. Then mn is
odd and (ab−1)mn = (am)n(bn)−m=ene−m=ee =eso ab−1is in
H.His a subgroup when “odd” is replaced by “even” because
e2=eis in Hand the argument given for odd exponents remains
valid.
35. Hcontains the identity. If A, B ∈H, then
det (AB−1) = (det A)(det B)−1is rational. His not a subgroup
when det Ais an integer, since det A−1= (det A)−1is an integer
only when det A=±1.
36. Let Hbe the nontrivial proper subgroup of Gand pick g∈Gso
that g6∈ H. Then G=hgi. That |G|=p2follows from Theorem
4.3 and Exercise 42 of Chapter 4.
37. First suppose that Gis not cyclic. Choose x6=eand choose
y6∈ hxi. Then, since Ghas only two proper subgroups
G=hxi∪hyi. But then xy ∈ hyiso that hxi⊆hyiand therefore
G=hyiand Gis cyclic. To prove that |G|=pq, where pand qare
distinct primes or |G|=p3, where pis prime we first observe that
Gis not infinite since an infinite cyclic group has infinitely many
subgroups. If |G|is divisible by at least two distinct primes pand
q, then |G|=pqm. If m > 1, then Gwould have nontrivial proper
subgroups of order p,q, and m. So, m= 1. If |G|=pn,where pis
prime, then by Theorem 4.3, Ghas exactly n−1 nontrivial proper
subgroups and therefore n= 3.
38. In D6, take a=R60 and b=R120. Then |a2|=|b2|but |a| 6=|b|.
39. If Tand Uare not closed, then there are elements xand yin Tand
wand zin Usuch that xy is not in Tand wz is not in U. It
follows that xy ∈Uand wz ∈T. Then xywz = (xy)wz ∈Uand
xywz =xy(wz)∈T, a contradiction since Tand Uare disjoint.
40. If ahas order p, then haihas p−1 elements of order p. So, there is
an element bnot in haithat also has order p. Then hai∪hbihas
2(p−1) > p elements of order p.
41. Let Gbe the group of all polynomials with integer coefficients
under addition. Let Hkbe the subgroup of polynomials of degree at
most ktogether with the zero polynomial (the zero polynomial
does not have a degree).
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 40
42. Observe that
b=a−3ba3=a−2a−1baa2=a−2b2a2=a−1a−1b2aa =a−1b4a=b8.
In the general case we can say |b|divides 2k−1.
43. Take g=a.
44. We first prove that if a, b ∈Hthen ab =ba. Let |a|=mand
|b|=nand let ms1+ 3t1= 1 and ns2+ 3t2= 1. Then
a=ams1a3t1= (a3)t1and b=bns2+3t2=bns2b3t2= (b3)t2. So,
ab = (at1)3(bt2)3= (bt2)3(at1)3= (bt2)3bns2(at1)3ams1=ba. Now
that we know that elements from Hcommute it follows that |ab|
divides lcm(|a|,|b|) and from this we have that Hsatisfies the
subgroup test. This argument is valid for all positive integers.
45. Let S={s1, s2, s3, . . . , sk}and let gbe any element in G. Then the
set {gs−1
1, gs−1
2, gs−1
3, . . . , gs−1
k}consists of kdistinct elements so it
and Shave at least one element in common. Say gs−1
i=sj. Then
g=sjsi.
46. f(xe) = f(x) so His not empty. If a, b ∈H, the
f(xab) = f(xa) = f(x) so ab ∈H. Finally, if a∈H, then
f(x) = f(xa−1a) = f(xa−1) so that a−1∈H. When Gis the reals
H=h2πi.
47. Let K={x∈G| |x|divides d}. By Exercise 15 of Chapter 4 Kis
a subgroup. Let x∈H. By Theorem 4.3 |x|divides d. So, H⊆K.
Let y∈K,|y|=tand d=tq. By Theorem 4.3, Hhas a subgroup
of order tand Ghas only one subgroup of order t. So, hyi ⊆ H.
48. Suppose that there is an element bsuch that |b|does not divide |a|.
Then there is some prime-power divisor pnof |b|such that pndoes
not divide |a|. Let |a|=pkm, where gcd(p, m) = 1. Then k < n.
Let |b|=pnt. Consider a0=apkand b0=bt. Then |a0|=mand
|b0|=pn. Since gcd(p, m) = 1,|a0b0|=mpn>|a|,we have a
contradiction. For the non-Abelian example, consider D3.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 1–4 41
49. To check associativity, note
(a∗b)∗c= ((a+b)−1) ∗c=a+b−1 + c−1 = a+b+c−2 and
a∗(b∗c) = a∗(b+c−1) == a+ (b+c−1) −1 = a+b+c−2. To
determine the identity e, we observe that a∗e=aif and only if
a+e−1 = a. Thus 1 is the identity (it is obvious that the
operation is commutative). If a−1exists, we have must
a∗a−1=a+a−1−1 = 1, and therefore a−1is −a+ 2. To find a
generator, observe that for any positive integer
k, ak=ka −(k−1). So, for positive kand a= 2, we have
2k=k+ 1. One can also check that 2k=k+ 1 when k= 0 or
negative. Thus 2 generates all integers.
50. {1,2n−1,2n+ 1,4n−1}
42
CHAPTER 5
Permutation Groups
1. a. α−1=123456
213546
b. βα =123456
162345
c. αβ =123456
621534
2. α= (12345)(678) = (15)(14)(13)(12)(68)(67); β= (23847)(56) =
(27)(24)(28)(23)(56); αβ = (12485736) =
(16)(13)(17)(15)(18)(14)(12).
3. a. (15)(234) b. (124)(35)(6) c. (1423)
4. 2; 3; 5; k.
5. a. By Theorem 5.3 the order is lcm(3,3) = 3.
b. By Theorem 5.3 the order is lcm(3,4) = 12.
c. By Theorem 5.3 the order is lcm(3,2) = 6.
d. By Theorem 5.3 the order is lcm(3,6) = 6.
e. |(1235)(24567)|=|(124)(3567)|= lcm(3,4) = 12.
f. |(345)(245)|=|(25)(34)|= lcm(2,2) = 2.
6. 6; 12
7. By Theorem 5.3 the order is lcm(4,6) = 12.
8. |(123)(45678)|= 15
9. We find the orders by looking at the possible products of disjoint
cycle structures arranged by longest lengths left to right and denote
an n-cycle by (n).
(6) has order 6 and is odd;
(5)(1) has order 5 and is even;
(4)(2) has order 4 and is even;
5/Permutation Groups 43
(4)(1)(1) has order 4 and is odd;
(3)(3) has order 3 and is even;
(3)(2)(1) has order 6 and is odd;
(3)(1)(1)(1) has order 3 and is even;
(2)(2)(2) has order 2 and is odd;
(2)(2)(1)(1) has order 2 and is even;
(2)(1)(1)(1)(1) has order 2 and is odd.
So, for S6, the possible orders are 1, 2, 3, 4, 5, 6; for A6the possible
orders are 1, 2, 3, 4, 5.
We see from the cycle structure of S7shown in Example 4 that in
A7the possible orders are 1, 2, 3, 4, 5, 6, 7.
10. 21.
11. (135) =(15)(13) even; (1356) = (16)(15)(13) odd; (13567) =
(17)(16)(15)(13) even; (12)(134)(152) = (12)(14)(13)(12)(15) odd;
(1243)(3521) = (13)(14)(12)(31)(32)(35) even.
12. Say S={s1, . . . , sn}and φis one-to-one from Sto S. Then
φ(s1), . . . , φ(sn) are all distinct and all in Sso φ(S) = S. On the
other hand, if φ(si) = φ(sj) for some i6=j, then φ(S) has at most
n−1 members. The mapping from Zto Zthat takes xto 2xis
one-to-one but not onto.
13. To prove that αis 1 −1 assume α(x1) = α(x2). Then
x1=α(α(x1) = αα(x2) = x2. To prove that αis onto note that for
any sin S, we have α(α(s)) = s.
14. (1), (12), (34), (56), (12)(34), (12)(56), (34)(56), (12)(34)(56).
These eight elements form a subgroup.
15. An n-cycle is even when nis odd since we can write it as a product
of n−1 2-cycles by successively pairing up the first element of the
cycle with each of the other cycle elements starting from the last
element of the cycle and working towards the front. The same
process shows that when nis odd we get an even permutation.
16. If α1, α2, . . . , αnare 2-cycles and α=α1···αnthen
α−1=αnαn−1···α2α1.
17. An even number of two cycles followed by an even number of two
cycles gives an even number of two cycles in all. So the Finite
Subgroup Test is verified.
5/Permutation Groups 44
18. even; odd.
19. Suppose that αcan be written as a product on m2-cycles and β
can be written as product of n2-cycles. Then juxtaposing these
2-cycles we can write αβ as a product of m+n2-cycles. Now
observe that m+nis even if and only if mand nare even or both
odd.
20. (+1) ·(+1) = (+1) (−1) ·(−1) = +1
even ·even =even odd ·odd =even
(+1) ·(−1) = (−1) (−1) ·(+1) = (−1)
even ·odd =odd odd ·even =odd
21. (AHMPRS)(BDGC)(EJLNF)(I)(KO)(QU)(TWV)(XZY)
22. If αand βare disjoint 2-cycles, then |αβ|= lcm(2,2) = 2. If αand
βhave exactly one symbol in common we can write α= (ab) and
β= (ac). Then αβ = (ab)(ac) = (acb) and |αβ|= 3.
23. If all members of Hare even we are done. So, suppose that Hhas
at least one odd permutation σ. For each odd permutation βin H
observe that σβ is even and, by cancellation, different βs give
different σβs. Thus, there are at least as many even permutations
are there are odd ones. Conversely, for each even permutation βin
Hobserve that σβ is odd and, by cancellation, different βs give
different σβs. Thus, there are at least as many odd permutations
are there are even ones.
24. By Exercise 23, either every element of His even or half are even
and half are odd. In the latter case, Hwould have even order.
25. The identity is even; the set is not closed.
26. If αcan written as the product of m2-cycles and βcan be written
as a product n2-cycles, then α−1β−1αβ can be written as the
product of n+m+n+m= 2(m+n) 2-cycles.
27. C(α3) = {α1, α2, α3, α4},C(α12) = {α1, α7, α12}.
28. 7 ·6·5·4·3/5
5/Permutation Groups 45
29. An odd permutation of order 4 must be of the form (a1a2a3a4).
There are 6 choices for a1, 5 for a4, 4 for a3, and 3 for a4. This
gives 6 ·5·4·3 choices. But since for each of these choices the
cycles (a1a2a3a4)=(a2a3a4a1)=(a3a4a1a2)=(a4a3a2a1) give the
same group element we must divide 6 ·5·4·3 by 4 to obtain 90. An
even permutation of order 4 must be of the form (a1a2a3a4)(a5a6).
As before, there are 90 choices (a1a2a3a4). Since (a5a6)=(a6a5)
there are 90 elements of the form (a1a2a3a4)(a5a6). This gives 180
elements of order 4 in S6.
A permutation in S6of order 2 has three possible disjoint cycle
forms: (a1a2),(a1a2)(a3a4) and (a1a2)(a3a4)(a5a6). For (a1a2)
there are 6 ·5/2 =15 distinct elements; for (a1a2)(a3a4) there are
6·5·4·3 choices for the four entries but we must divide by 2 ·2·2
since (a1a2) = (a2a1),(a3a4)=(a4a3) and
(a1a2)(a3a4) = (a4a3)(a1a2). This gives 45 distinct elements. For
(a1a2)(a3a4)(a5a6) there are 6! choices for the six entries but we
must divide by 2 ·2·2·3! since each of the three 2-cycles can be
written 2 ways and the three 2-cycles can be permuted 3! ways.
This gives 15 elements. So, the total number of elements of order 2
is 75.
30. Any product of 3-cycles is even whereas (1234) is odd.
31. Since β28 = (β4)7=, we know that |β|divides 28. But β46=so
|β| 6= 1,2, or 4. If |β|= 14, then βwritten in disjoint cycle form
would need at least one 7-cycle and one 2-cycle. But that requires
at least 9 symbols and we have only 7. Likewise, |β|= 28 requires
at least one 7-cycle and one 4-cycle. So, |β|= 7. Thus,
β=β8= (β4)2= (2457136).In S9,β= (2457136) or
β= (2457136)(89).
32. Observe that β= (123)(145) = (14523) so that
β99 =β4=β−1= (13254).
33. Observe that if we start with a 9-cycle (a1a2a3a4a5a6a7a8a9) and
cube it we get (a1a4a7)(a2a5a8)(a3a6a9). So, we can let
σ= (124586739). But
(157)(283)(469) = (157)(469)(283) = (283)(157)(469) so these give
us (124586739), (142568793) and (214856379).
34. anan−1···a2a1
5/Permutation Groups 46
35. Let α, β ∈stab(a). Then (αβ)(a) = α(β(a)) = α(a) = a. Also,
α(a) = aimplies α−1(α(a)) = α−1(a) or a=α−1(a).
36. Since |β|= 21, we have n= 16.
37. Since αm= (1,3,5,7,9)m(2,4,6)m(8,10)mand the result is a
5-cycle we deduce that (2,4,6)m=and (8,10)m=. So, 3 and 2
divide m. Since (1,3,5,7,9)m6=we know that 5 does not divide
m. Thus, we can say that mis a multiple of 6 but not a multiple of
30.
38. Let β, γ ∈H. Then (βγ)(1) = β(γ(1)) = β(1) = 1;
(βγ)(3) = β(γ(3)) = β(3) = 3.So, by Theorem 3.3, His a
subgroup. |H|= 6. The proof is valid for all n≥3. In the general
case, |H|= (n−2)!. When Snis replaced by An,|H|= (n−2)!/2.
39. An element of order 5 in A6must be a 5-cycle. There are
6·5·4·3·2 = 720 ways to create a 5-cycle. But the same 5-cycle
can be written in 5 ways so we must divide 720 by 5 to obtain 144.
40. h(1234)i;{(1),(12),(34),(12)(34)}
41. 3, 7, 9
42. Let α= (12) and β= (13).
43. Let α= (123) and β= (145).
44. R0= (1)(2)(3); R120 = (123); R240 = (132); (12); (13); (23).
45. Observe that (12) and (123) belong to Snfor all n≥3 and they do
not commute.
46. Observe that (123)(124) 6= (124)(123).
47. Let αand βbelong to H. Then α(1) is 1 or 2, and α(2) is the
unused choice between 1 and 2 for α(1). Likewise, β(1) is 1 or 2,
and β(2) is the unused choice between 1 and 2 not used β(1). Thus
αβ(1) = α(1) or αβ(1) = α(2). This means that αβ(1) = α(1) or
α(2) and these are 1 or 2. The same argument applies to αβ(2). To
find |H|observe that in matrix form we have 2 choices (1 or 2) for
the image of 1, the second entry must be the choice of 1 or 2 not
used as the image of 1, and (n−2)! choices for the remaining n−2
images. So, |H|= 2(n−2)!
5/Permutation Groups 47
48. For any permutation αin S7, α2is even whereas (1234) is odd. For
the second part we can take x= (1432),(1432)(567),(1432)(576)
49. Theorem 5.2 shows that disjoint cycles commute. For the other
half, observe that (ab)(ac) = (acb) whereas (ac)(ab)=(abc).
50. If αand βhave two symbols in common write α= (b1bi) and
β= (b1b2. . . bt). Then αβα = (b1bi+1bi+2 . . . btbib2b3. . . bi−1) (b1
and biswitch places) which is a t-cycle. If αand βhave exactly one
symbol in common write α= (abi) and β= (b1b2. . . bt). Then
αβα = (b1b2. . . bi−1ab1+1 . . . bt) (biis replaced by a) . If αand β
are disjoint, then αβα =β.
51. Let β=β1β2. . . βkwhere each βiis a cycle. Because
αβα−1=αβ1β2. . . βkα−1= (αβ1α−1)(αβ2α−1). . . (αβkα−1) the
problem is reduced to showing that if γis a t-cycle for some tthen
αγα−1is a t-cycle. Now write α=α1α2. . . αswhere each αiis a
2-cycle. Then αγα−1=α1α2. . . (αsγαs). . . α2α1since each αihas
order 2. Now by Exercises 50 we can replace αsγαsby a cycle the
same length as γ. Repeated use of this argument finishes the proof.
52. Say βcan be written with m2-cycles and αwith n2-cycles.
Then β−1αβ can be written with 2m+n2-cycles.
53. Examining Snfor each nup to 8 as in Example 4 we see that Sn
has no element of order greater than 2n. For n= 9 we have
|(12345)(6789)|= 20.
54. The case where n= 2,4,or 6 is done by examining cases as in
Example 4 of this chapter. Note that the permutation (12345)(678)
is in both A8and A10 and has order 15. For n≥12 observe that
(1,2,···, n −5)(n−4, n −3)(n−2, n −1) is in Anand has order
2(n−5), which is greater than n.
55. The cases where n= 1,3,5,7,9, or 11 are done by examining cases
as in Example 4 of this chapter. For n≥13 observe that
(1,2, . . . , n −6)(n−5, n −4, n −3, n −2)(n−1, n) is in Anand has
order 4(n−6), which is greater than 2nwhen n≥13.
56. The cases where n < 16 are done by examining cases as in Example
4 of this chapter. For n≥16 observe that
(1,2, . . . , n −7)(n−6, n −5, n −4, n −3)(n−2, n −1) is in Anand
has order 4(n−6), which is greater than 2nwhen n≥13.
5/Permutation Groups 48
57. R0, R180, H, V .
58. 216◦rotation; reflection about the axis joining vertex 1 to the
midpoint of the opposite side.
59. The permutation corresponding to the rotation of 360/ndegrees,
(1,2, . . . , n), is an even permutation so all rotations are even.
60. nmod 4 = 1.
61. Cycle decomposition shows any nonidentity element of A5is a
5-cycle, a 3-cycle, or a product of a pair of disjoint 2-cycles. Then,
observe there are (5 ·4·3·2·1)/5 = 24 group elements of the form
(abcde),(5 ·4·3)/3 = 20 group elements of the form (abc) and
(5 ·4·3·2)/8 = 15 group elements of the form (ab)(cd). In this last
case we must divide by 8 because there are 8 ways to write the
same group element (ab)(cd)=(ba)(cd)=(ab)(dc)=(ba)(dc) =
(cd)(ab)=(cd)(ba) = (dc)(ab)=(dc)(ba).
62. One possibility is h(1234)(5678)i.
63. One possibility is {(1),(12)(34),(56)(78),(12)(34)(56)(78)}.
64. α2,α3, and α4have order 2; α5, α6, . . . , α12 have order 3. The
orders of the elements divide the order of the group.
65. Any element from Anis expressible as a product of an even number
of 2-cycles. For each pair of 2-cycles there are two cases. One is
that they share an element in common (ab)(ac) and the other is
that they are disjoint (ab)(cd). Now observe that (ab)(ac) = (abc)
and (ab)(cd)=(cbd)(acb).
66. Suppose α6=εand α∈Z(Sn). Write αin disjoint cycle form
(a1a2. . .). . . where a16=a2. Let a3be different from a1and a2and
let β= (a1)(a2a3). Then (αβ)(a1) = a2while (βα)(a1) = a3.
67. That a∗σ(b)6=b∗σ(a) is done by examining all cases. To prove
the general case, observe that σi(a)∗σi+1(b)6=σi(b)∗σi+1(a) can
be written in the form σi(a)∗σ(σi(b)) 6=σi(b)∗σ(σi(a)), which is
the case already done. If a transposition were not detected, then
σ(a1)∗ ··· ∗ σi(ai)∗σi+1(ai+1)∗ ··· ∗ σn(an) =
σ(a1)∗ ··· ∗ σi(ai+1)∗σi+1(ai)∗ ··· ∗ σn(an), which implies
σi(ai)∗σi+1(ai+1) = σi(ai+1)∗σi+1(ai).
5/Permutation Groups 49
68. 5
69. By Theorem 5.4 it is enough to prove that every 2-cycle can be
expressed as a product of elements of the form (1k). To this end
observe that if a6= 1, b 6= 1, then (ab) = (1a)(1b)(1a).
70. Let αdenote the permutation of positions induced by a shuffling.
Label the positions ace to king as 1 through 13. We are given that
α2="1 2 3 4 5 6 7 8 9 10 11 12 13
8 12 6 7 9 11 13 4 2 1 10 3 5 #=
(1,8,4,7,13,5,9,2,12,3,6,11,10).
Since |α2|= 13 we know that |α|= 13 or 26. But S13 has no
elements of order 26. So, |α|= 13. Thus,
α=α14 = (1,2,8,12,4,3,7,6,13,11,5,10,9).
71. If αhas odd order kand αis an odd permutation then =αk
would be odd.
72. From the elements on the diagonal in Table 5.1 we see that when
G=A4,Hcontains α8and α12 but their product.
73. Observe that every element of His an even permutation so
H⊆A4. Elements of A4have order 3, 2 or 1. For any αin A4of
order 3, hαi=hα2i ⊂ H. The only elements in A4of order 2 are
the products of two disjoint 2-cycles–(ab)(cd) = (acbd)2, which
belong to H. A similar argument applies to K.
74. Observe that (1234)(56) belongs to A6but is not in H. For if
(1234)(56) = α2, then |α|= 12 and S6has no element of order 12.
75. By case-by-case analysis His a subgroup for n= 1,2,3 and 4. For
n≥5, observe that (12)(34) and (12)(35) belong to Hbut their
product does not.
76. β= (1423); γ= (234). To see this we compute (γβ)(1) two ways.
Using γβ = (1243) we obtain (γβ)(1) = 2. From we have
(γβ)(1) = γ(β(1)) = γ(4). So, γ(4) = 2. Now continue in this way.
Another correct answer is β= (14)(23); γ= (24).
77. The product of an element from Z(A4) of order 2 and an element of
A4of order 3 would have order 6. The product of an element from
Z(A4) of order 3 and an element of A4of order 2 would have order
6. But A4has no element of order 6.
5/Permutation Groups 50
78. See Exercise 35.
79. See Exercise 35.
80. By definition, H⊆An. To prove An⊆H, let α=α1α2···α2k.
Then Then α=α1α2···α2kα1α1α2α2···αkαk.
81. Labeling the four tires 1, 2, 3, 4 in clockwise order starting with 1
being the tire in the upper left-hand corner, we may represent the
four patterns as
α= (1324) top left-hand pattern
β= (1423) top right-hand pattern
γ= (14)(23) bottom right-hand pattern
δ= (13)(24) bottom left-hand pattern
Notice that α−1=βand that δ=α2γ. Thus, we need only find the
smallest subgroup of S4containing αand γ. To this end, observe
that the set {ε, α, α2, α3, γ, αγ, α2γ, α3γ}is closed under
multiplication on the left and right by both αand γ. (Here we have
used the fact that γα =α3γ.) This implies that the set is closed
under multiplication and is therefore a group. Since αγ 6=γα, the
subgroup is non-Abelian.
82. 12.
83. Then a permutation such as (23) in Snfor n≥3 could also be
written as (11)(23) so it would be both even and odd. In fact, (11)
could be appended to every permutation written in cycle form
making it both even and odd.
51
CHAPTER 6
Isomorphisms
1. Let φ(n) = 2n. Then φis onto since the even integer 2nis the
image of n.φis one-to-one since 2m= 2nimplies that m=n.
φ(m+n) = 2(m+n)=2m+ 2nso φis operation preserving.
2. An automorphism of a cyclic group must carry a generator to a
generator. Thus 1 →1 and 1 → −1 are the only two choices for the
image of 1. So let α:n→nand β:n→ −n. Then
Aut(Z) = {α, β}.
3. φis onto since any positive real number ris the image of √r.φis
one-to-one since √a=√bimplies that a=b. Finally,
φ(xy) = √xy =√x√y=φ(x)φ(y).
4. U(8) is not cyclic while U(10) is.
5. Define φfrom U(8) to U(12) by φ(1) = 1; φ(3) = 5; φ(5) = 7;
φ(7) = 11. To see that φis operation preserving we observe that
φ(1a) = φ(a) = φ(a)·1 = φ(a)φ(1) for all a;
φ(3 ·5) = φ(7) = 11 = 5 ·7 = φ(3)φ(5);
φ(3 ·7) = φ(5) = 7 = 5 ·11 = φ(3)φ(7);
φ(5 ·7) = φ(3) = 5 = 7 ·11 = φ(5)φ(7).
6. If βis an isomorphism from Gonto H, and αis an isomorphism
from Honto K, then αβ is an isomorphism from Gonto K. That
αβ is one-to-one and onto is done in Theorem 0.7. If a, b ∈G, then
(αβ)(ab) = α(β(ab)) = α(β(a)β(b)) = α(β(a))α(β(b)) =
(αβ)(a)(αβ)(b).
7. D12 has an element of order 12 and S4does not.
8. Properties of real numbers assure that the mapping is one-to-one
and onto Rand that log10(ab) = log10(a)+ log10(b) is a property of
logs.
6/Isomorphisms 52
9. Since Te(x) = ex =xfor all x,Teis the identity. For the second
part, observe that Tg◦(Tg)−1=Te=Tgg−1=Tg◦Tg−1and cancel.
10. Suppose αis an automorphism of G. Then α(ab)=(ab)−1and
α(ab) = α(a)α(b) = a−1b−1. So a−1b−1= (ab)−1=b−1a−1for all a
and bin G. Taking the inverse of both sides proves that Gis
Abelian.
If Gis Abelian, then for all aand bin G, we have
(ab)−1= (ba)−1=a−1b−1. Thus α(ab) = α(a)α(b).
That αis one-to-one and onto follows directly from the definitions.
11. For any xin the group, we have (φgφh)(x) = φg(φh(x)) =
φg(hxh−1) = ghxh−1g−1= (gh)x(gh)−1=φgh(x).
12. Aut(Z2)≈Aut(Z1)≈Z1;
Aut(Z6)≈Aut(Z4)≈Aut(Z3)≈Z2;
Aut(Z10)≈Aut(Z5)≈Z4(see Example 4 and Theorem 6.5);
Aut(Z12)≈Aut(Z8) (see Exercise 5 and Theorem 6.5).
13. φR0and φR90 disagree on H;φR0and φHdisagree on R90;φR0and
φDdisagree on R90;φR90 and φHdisagree on R90;φR90 and φD
disagree on R90;φHand φDdisagree on D.
14. By Theorem 6.5, we know |Aut(Z6)|=|U(6)|= 2. So α:n→n
and β:n→ −nare the only two automorphisms of Z6.
15. Let α∈Aut(G). We show that α−1is operation preserving:
α−1(xy) = α−1(x)α−1(y) if and only if
α(α−1(xy)) = α(α−1(x)α−1(y)). That is, if and only if
xy =α(α−1(x))α(α−1(y)) = xy. So α−1is operation preserving.
That Inn(G) is a group follows from the equation φgφh=φgh.
16. Let φbe an isomorphism from Gto H. For any βin Aut(G) define
a mapping from Aut(G) to Aut(H) by Γ(β) = φβφ−1. Then Γ is
1-1 and operation preserving. (See Theorem 0.8 and Exercise 6). To
see that Γ is onto observe that for any γin Aut(H), Γ(φ−1γφ) = γ.
17. Since b=φ(a) = aφ(1) it follows that φ(1) = a−1band therefore
φ(x) = a−1bx. (Here a−1is the multiplicative inverse of amod n,
which exists because a∈U(n).)
6/Isomorphisms 53
18. Note that φmust take R360◦/n to an element of order n. Since the
reflections have order 2 we know φ(R360◦/n) is a rotation. Thus,
φ(H) = φ(hR360◦/ni)⊆K.
19. Note that both Hand Kare isomorphic to the group of all
permutations on the four symbols, which is isomorphic to S4. The
same is true when 5 is replaced by nsince both Hand Kare
isomorphic to Sn−1.
20. Observe that h2i,h3i, . . . are distinct and each is isomorphic to Z.
21. Recall when nis even, Z(Dn) = {R0, R180}. Since R180 and φ(R180)
are not the identity and belong to Z(Dn) they must be equal.
22. This follows directly from the subgroup tests.
23. Z60 contains cyclic subgroups of orders 12 and 20 and any cyclic
group that has subgroups or orders 12 and 20 must be divisible by
12 and 20. So, 60 is the smallest order of any cyclic group that
subgroups isomorphic to Z12 and Z60.
24. φ(x) = x;φ(x)=9x;φ(x) = 13x;φ(x) = 17x.
25. See Example 15 of Chapter 2.
26. It is enough to prove that the mapping is one-to-one. If a3=b3,
then a9=b9. Now use the fact that x4= 1 for all xin U(16). In
general, x→xnis an automorphism of U(16) when nis odd.
27. That αis a one-to-one follows from the fact that r−1exists module
n. The operation preserving condition is Exercise 9 of Chapter 0.
28. The mapping "1a
0 1 #→ais an isomorphism to Zwhen a∈Z
and to Rwhen a∈R.
29. By Part 2 of Theorem 6.2, we have φ(an) = φ(a)n=γ(a)n=γ(an)
thus φand γagree on all elements of hai.
30. Observe that φ(11) = 11φ(1) = 13 and since 11 is relatively prime
to 50, 11−1exists modulo 50. Thus, we have
φ(1) = 11−1·13 = 41 ·13 = 33 and φ(x) = φ(x·1) = xφ(1) = 33x.
6/Isomorphisms 54
31. The inverse of a one-to-one function is one-to-one. For any g∈G
we have φ−1(φ(g)) = gand therefore φ−1is onto. To verify that
φ−1is operation preserving see the answer to Exercise 15 of this
chapter.
32. Since φ(K) contains φ(e), φ(K)6=∅. Also,
φ(a)(φ(b))−1=φ(a)φ(b−1) = φ(ab−1)∈φ(K).
33. Tg(x) = Tg(y) if and only if gx =gy or x=y. This shows that Tgis
a one-to-one function. Let y∈G. Then Tg(g−1y) = y, so that Tgis
onto.
35. To prove that φis 1-1 observe that φ(a+bi) = φ(c+di) implies
that a−bi =c−di. From properties of complex numbers this gives
that a=cand b=d. Thus a+bi =c+di. To prove φis onto let
a+bi be any complex number. Then φ(a−bi) = a+bi. To prove
that φpreserves addition and multiplication note that
φ((a+bi)+(c+di)) = φ((a+c)+(b+d)i)=(a+c)−(b+d)i=
(a−bi)+(c−di) = φ(a+bi) + φ(c+di). Also,
φ((a+bi)(c+di) = φ((ac −bd) + (ad +bc)i) = (ac −bd)−(ad +bc)i
and φ(a+bi)φ(c+di) = (a−bi)(c−di)=(ac −bd)−(ad +bc)i.
34. U(20) has three elements of order 2 whereas U(24) has seven.
35. To prove that φis 1-1 observe that φ(a+bi) = φ(c+di) implies
that a−bi =c−di. From properties of complex numbers this gives
that a=cand b=d. Thus a+bi =c+di. To prove φis onto let
a+bi be any complex number. Then φ(a−bi) = a+bi. To prove
that φpreserves addition and multiplication note that
φ((a+bi)+(c+di)) = φ((a+c)+(b+d)i)=(a+c)−(b+d)i=
(a−bi)+(c−di) = φ(a+bi) + φ(c+di). Also,
φ((a+bi)(c+di) = φ((ac −bd) + (ad +bc)i) = (ac −bd)−(ad +bc)i
and φ(a+bi)φ(c+di) = (a−bi)(c−di)=(ac −bd)−(ad +bc)i.
36. Map a+b√2→"a2b
b a #. This map preserves both addition and
multiplication.
37. First observe the Zis a cyclic group generated by 1. By property 3
of Theorem 6.3, it suffices to show that Qis not cyclic under
addition. By way of contradiction suppose that Q=hp/qi. But
then p/2qis a rational number that is not in hp/qi.
6/Isomorphisms 55
38. D4has 5 elements of order 2, the quaternion group has only 1.
39. The notation itself suggests that
φ(a+bi) = "a−b
b a #
is the appropriate isomorphism. To verify this note that
φ((a+bi)+(c+di)) = "a+c−(b+d)
(b+d)a+c#=
"a−b
b a #+"c−d
d c #=φ(a+bi) + φ(c+di).
Also, φ((a+bi)(c+di)) = φ((ac −bd)+(ad +bc)i) =
"(ac −bd)−(ad +bc))
(ad +bc)ac −bd)#="a−b
b a #" c−d
d c #=
φ(a+bi)φ(c+di).
40. φ((a1, . . . , an)+(b1, . . . , bn)) = (−a1,...,−an)=(−b1,...,−bn)
implies (a1, . . . , an)=(b1, . . . , bn)) so that φis 1-1. For any
(a1, . . . , an),we have φ(−a1,...,−an)=(a1, . . . , an) so φis onto.
φ((a1+b1, . . . , an+bn)) = (−(a1+b1),...,−(an+bn)) =
(−a1,...,−an)(−b1,...,−bn) = φ((a1, . . . , an)) + φ((b1, . . . , bn)) .
φreflects each point through the origin.
41. Yes, by Cayley’s Theorem.
42. (ab)2=a2b2shows that the mapping is O.P. To show it is 1-1, note
that a2=b2, implies e=a2b−2= (ab−1)2so that |ab−1|= 1 or 2.
Thus, a=b. Since Gis finite, 1-1 implies onto. For Zunder
addition, g→2gis not onto.
43. Observe that φg(y) = gyg−1and
φzg(y) = zgy(zg)−1=zgyg−1z−1=gyg−1, since z∈Z(G). So,
φg=φzg.
44. For any x∈C(a) let φ(x) = gxg−1. To verify that
gxg−1∈C(gag−1) observe that (gxg−1)(gag−1) = (gag−1)(gxg−1)
if and only if gxag−1=gaxg−1if and only if xa =ax. So, φis a
function from C(a) to C(gag−1). To see that φis onto we let
6/Isomorphisms 56
y∈C(gag−1) and observe that φ(g−1yg) = g(g−1yg)g−1=yand
that (g−1yg)a=a(g−1yg) if and only if
g(g−1yg)a)g−1=g(a(g−1yg)g−1if and only if ygag−1=gag−1yif
and only if y(gag−1)=(gag−1)yif and only if y∈C(gag−1). This
proves that φis onto. Finally
φ(xy) = gxyg−1=gxg−1gyg−1=φ(x)φ(y) so φis an isomorphism.
45. φg=φhimplies gxg−1=hxh−1for all x. This implies
h−1gx(h−1g)−1) = x, and therefore h−1g∈Z(G).
46. φg=φhif and only if h−1g∈Z(G).
47. By Exercise 45 φx=φyimplies y−1xis in Z(Sn) and by Exercise
66 in Chapter 5, Z(Sn) = {}.
48. This follows directly from property 3 of Theorem 6.2.
49. Since both φand γboth take eto itself, His not empty. Assume a
and bbelongs to H. Then φ(ab−1) = φ(a)φ(b−1) = φ(a)φ(b)−1=
γ(a)γ(b)−1=γ(a)γ(b−1) = γ(ab−1). Thus ab−1is in H.
50. For any integer n, H ={g∈G|βn(g) = g}is a subgroup of G. This
is a special case of Exercise 22 where φ=βnand γis the identity
automorphism.
51. Since φ(e) = e=e−1, H is not empty. Assume that aand bbelong
to H. Then φ(ab) = φ(a)φ(b) = a−1b−1=b−1a−1= (ab)−1, and H
is closed under multiplication. Moreover, because
φ(a−1) = φ(a)−1= (a−1)−1we have that His closed under inverses.
52. Clearly φis 1-1 and onto. Observe that φ(ab)=(ab)−1=b−1a−1
and φ(a)∗φ(b) = a−1∗b−1=b−1a−1. So, φ(ab) = φ(a)∗φ(b).
53. Say |a|=n. Then φn
a(x) = anxa−n=x, so that φn
ais the identity.
For example, take a=R90 in D4. Then |φa|= 2.
54. The mapping φ(x) = (3/2)xis an isomorphism from Gonto H.
Multiplication is not preserved. When G=hmiand H=hnithe
mapping φ(x)=(n/m)xis an isomorphism from Gonto H.
55. Observe that φ(D) = φ(R90V) = φ(R90)φ(V) = R270V=D0and
φ(H) = φ(R90D) = φ(R90)φ(D) = R270D0=H.
56. α5= (0)(157842)(36); α8= (0)(18)(27)(36)(45).
6/Isomorphisms 57
57. (R0R90R180R270)(HD0V D).
58. By part 2 of Theorem 6.2, nφ(1/n) = φ(1) so that
φ(1/n) = (1/n)φ(1). Also, by Part 2 of Theorem 6.2,
φ(m/n) = mφ(1/n) = m(1/n)φ(1) = (m/n)φ(1).
59. The mapping φ(x) = x2is one-to-one from Q+to Q+since a2=b2
implies a=bwhen both aand bare positive. Moreover,
φ(ab) = φ(a)φ(b) for all aand b. However, φis not onto since there
is no rational whose square is 2. So, the image of φis a proper
subgroup of Q+.
60. The argument given in Exercise 40 shows that an isomorphic image
of Qhas the form aQ where ais a nonzero rational. But aQ =Q.
61. Suppose that φis an automorphism of R∗and ais positive. Then
φ(a) = φ(√a√a) = φ(√a)φ(√a) = φ(√a)2>0. Now suppose that
ais negative but φ(a) = bis positive. Then by the case we just did
a=φ−1(φ(a)) = φ−1(b) is positive. This is a contradiction.
62. Consider first the cycle that begins with e. That cycle is
(e, eg, eg2, . . . , egn−1) Now if xis some element of Gnot appearing
in the first cycle then we have the cycle (x, xg, xg2, . . . , xgn−1).
Continue in this fashion.
63. Say φis an isomorphism from Qto R+and φtakes 1 to a. It
follows that the integer rmaps to ar. Then
a=φ(1) = φ(s1
s) = φ(1
s+··· +1
s) = φ(1
s)sand therefore
a1
s=φ(1
s). Thus, the rational r/s maps to ar/s. But ar/s 6=aπfor
any rational number r/s.
58
CHAPTER 7
Cosets and Lagrange’s Theorem
1. H={α1, α2, α3, α4},α5H={α5, α8, α6, α7},
α9H={α9, α11, α12, α10}.
2. 24/4=6
3. H, 1 + H, 2 + H. To see that there are no others notice that for any
integer nwe can write n= 3q+rwhere 0 ≤r < 3. So,
n+H=r+ 3q+H=r+H, where r= 0,1 or 2.
4. b−a∈H
5. a. 11 + H= 17 + Hbecause 17 −11 = 6 is in H;
b. −1 + H= 5 + Hbecause 5 −(−1) = 6 is in H;
c. 7 + H6= 23 + Hbecause 23 −7 = 16 is not in H.
6. 0 + hni,1 + hni, . . . , n −1 + hni;n
7. Since 8/2 = 4, there are four cosets. Let H={1,11}. The cosets
are H, 7H, 13H, 19H.
8. |ha5i| = 3 so there are 15/3 = 5 cosets. They are ha5i,aha5i,a2ha5i,
a3ha5i,a4ha5i.
9. Since |a4|= 15, there are two cosets: ha4iand aha4i.
10. Let Fand F0be distinct reflections in D3. Then take H={R0, F }
and K={R0, F 0}.
11. Let ga belong to g(H∩K) where ais in H∩K. Then by definition
ga is in gH ∩gK. Now let x∈gH ∩gK. Then x=gh for some
h∈Hand x=gk for some k∈K. Cancellation then gives h=k.
Thus x∈g(H∩K).
12. By Lagrange’s Theorem |H|= 5,31 or 155. But |H|= 5 implies
that all non-identity elements in Hhave order 5 and |H|= 31
implies that all non-identity elements in Hhave order 31.
7/Cosets and Lagrange’s Theorem 59
13. Suppose that h∈Hand h < 0. Then hR+⊆hH =H. But hR+is
the set of all negative real numbers. Thus H=R∗.
14. The coset containing c+di is the circle with center at the origin
and radius √c2+d2.
15. By Lagrange’s Theorem the possible orders are 1, 2, 3, 4, 5, 6, 10,
12, 15, 20, 30, 60.
16. 84 or 210.
17. By Lagrange’s Theorem, the only possible orders for the subgroups
are 1, pand q. By Corollary 3 of Lagrange’s Theorem, groups of
prime order are cyclic. The subgroup of order 1 is hei.
18. Note that φ(n) = |U(n)|then use Corollary 4 of Lagrange’s
Theorem and mimic the proof of Corollary 5 of Lagrange’s
Theorem.
19. By Exercise 18 we have 56mod 7 = 1. So, using mod 7 we have
515 = 56·56·52·5=1·1·4·5 = 6; 713 mod 11 = 2.
20. Note that n−1∈U(n) and has order 2.
21. By Corollary 4 of Theorem 7.1 gn=e. Then since we know that |g|
is a common divisor of both mand n. So, |g|= 1.
22. Since |H∩K|must divide 12 and 35, |H∩K|= 1.
23. By closure (234)(12) = (1342) belongs to Hso that |H|is divisible
by 3 and 4 and divides 24. But if |H|= 12 then the even
permutations in Hwould be a subgroup of A4of order 6, which is
does not exist (see Example 5).
24. Let hbe any element in H. By assumption there is a k∈Kso that
ah =bk. From aH ⊆bK we have that b−1aH ⊆K. In particular,
b−1a=b−1ae ∈K. Since Kis a group we have that
a−1b= (b−1a)−1∈K. Then h=a−1bk ∈K.
25. Since Ghas odd order, no element can have order 2. Thus, for each
x6=e, we know that x6=x−1. So, we can write the product of all
elements in the form ea1a−1
1a2a−1
2···ana−1
n=e.
7/Cosets and Lagrange’s Theorem 60
26. First suppose Gis infinite. If x∈Gand |x|=∞, then hx2i
contradicts the hypothesis. So we may assume all elements have
finite order. Then hxicontradicts the hypothesis for any x6=e.
Next assume Gis finite and e6=x∈G. Then hxi=G(otherwise
hxiis nontrivial and proper). Now use Theorem 4.3.
27. Let Hbe the subgroup of order pand Kbe the subgroup of order
q. Then H∪Khas p+q−1< pq elements. Let abe any element
in Gthat is not in H∪K. By Lagrange’s Theorem, |a|=p, q, or
pq. But |a| 6=p, for if so, then hai=H. Similarly, |a| 6=q.
28. Let e6=g∈G. Then |g|= 5 or 25. If |g|= 25 for some g, then Gis
cyclic. If there is no such g, then |g|= 1 or 5 for all g.
29. The possible orders are 1, 3, 11, 33. If |x|= 33, then |x11|= 3 so we
may assume that there is no element of order 33. By the Corollary
of Theorem 4.4, the number of elements of order 11 is a multiple of
10 so they account for 0, 10, 20, or 30 elements of the group. The
identity accounts for one more. So, at most we have accounted for
31 elements. By Corollary 2 of Lagrange’s Theorem, the elements
unaccounted for have order 3.
30. Let e6=g∈G. If |g|= 8, |g4|= 2. If |g|= 4, |g2|= 2.
31. No. By Lagrange’s Theorem, the elements of a group of order 55
must have orders 1, 5, 11, or 55. By Theorem 4.4 a cyclic group of
order 55 cannot have exactly 20 elements of order 11. So, a group
with exactly 20 elements of order 11 must have exactly 34 elements
of order 5. This contradicts the Corollary to Theorem 4.4.
32. For any positive integer nlet ωn= cos 2π
n+isin 2π
n. The finite
subgroups of C∗are those of the form hωni. To verify this, let H
denote any finite subgroup of C∗of order n. Then every element of
His a solution to xn= 1. But the solution set of xn= 1 in C∗is
hωni.
33. Observe that |G:H|=|G|/|H|,|G:K|=|G|/|K|,
|K:H|=|K|/|H|. So, |G:K||K:H|=|G|/|H|=|G:H|.
34. 2520
7/Cosets and Lagrange’s Theorem 61
35. Since the reflections in a dihedral group have order 2, the
generators of the subgroups of orders 12 and 20 must be rotations.
The smallest rotation subgroup of a dihedral group that contains
rotations of orders 12 and 20 must have order divisible by 12 and
20 and therefore must be a multiple of 60. So, D60 is the smallest
such dihedral group.
36. If some element does not have prime-power order, then by
Lagrange’s theorem it must have order 10, 20, 50 or 100. But in
each of the last three cases Theorem 4.4 guarantees the existence of
an element of order 10.
37. Let ahave order 3 and bbe an element of order 3 not in hai. Then
haihbi={aibj|i= 0,1,2, j = 0,1,2}is a subgroup of Gof order 9.
Now use Lagrange’s Theorem.
38. Suppose that |G|=pm where pis prime and |Z(G)|=m. Let
a∈G, but a6∈ Z(G). Since C(a) contains both aand Z(G) we
know that mis proper divisor of |C(a)|and |C(a)|divides pm. It
follows that C(a) = Gand a∈Z(G).
39. In D3, let H={R0, F }, and a=b=R120. Then
R120H={R120, F 0}, HR120 ={R120, F 00}, and
R120H∩HR120 ={R120}. (See the Cayley Table for D3on the
inside back cover.)
40. Suppose Gis a group of order 63. Let abe any non-identity
element in G. By Lagrange’s Theorem, |a|= 63,21,9,7,or 3. If
|a|= 3, we are done. If |a|= 63, then |a21|= 3; if |a|= 21, then
|a7|= 3; and if |a|= 9, then |a3|= 3. So, if any of these cases occur
we are done. Thus we may assume that all 62 non-identity elements
in Ghave order 7. But by the Corollary to Theorem 4.4, the
number of elements of order 7 must be a multiple of 6.
41. Let a∈Gand |a|= 5. Then by Theorem 7.2 we know that the set
haiHhas exactly 5 · |H|/|hai ∩ H|elements and |hai ∩ H|divides
|hai| = 5. It follows that |hai ∩ H|= 5 and therefore hai ∩ H=hai.
42. Suppose the ak=bk. By the Corollary to Theorem 0.2 there are
integers sand tsuch that 1 = ns +kt. Then by Corollary 4 of
Lagrange’s Theorem we have
a=ans+kt = (an)s(ak)t= (ak)t= (bn)s(bk)t=bns+kt =b. To prove
that the mapping is an automorphism when the group is also
7/Cosets and Lagrange’s Theorem 62
Abelian note that by Exercise 10 of Chapter 5 a 1-1 mapping from
a finite set to itself is onto. Lastly, observe that (ab)k=akbk.
43. Certainly, a∈orbG(a). Now suppose c∈orbG(a)∩orbG(b). Then
c=α(a) and c=β(b) for some αand β, and therefore
(β−1α)(a) = β−1(α(a)) = β−1(c) = b. So, if x∈orbG(b), then
x=γ(b) = γ(β−1α)(a)) = (γβ−1α)(a). This proves
orbG(b)⊆orbG(a). By symmetry, orbG(a)⊆orbG(b).
44. Since reflections have order 2 the subgroup must consist entirely of
rotations and the subgroup of all rotations is cyclic.
45. a. stabG(1) = {(1),(24)(56)}; orbG(1) = {1,2,3,4}
b. stabG(3) = {(1),(24)(56)}; orbG(3) = {3,4,1,2}
c. stabG(5) = {(1),(12)(34),(13)(24),(14)(23)}; orbG(5) = {5,6}
46. Let |G|= 12 and let a∈Gbe a nonidentity element. By
Lagrange’s Theorem, |a|= 12,6,4,3,or 2. If |a|= 12, then |a6|= 2.
Similarly, if ahas order 6 or 4 then there is an element of order 2.
So, we may assume that all 11 nonidentity elements have order 3.
But elements of order 3 come in pairs (if |a|= 3, then |a2|= 3).
Since this is a contradiction, one of the earlier cases must occur.
47. Consider the mapping from Gto Gdefined by φ(x) = x2. To prove
that it is one-to-one assume that x2=y2and let |G|= 2k+ 1.
Then x=xe =xx2k+1 =x2k+2 = (x2)k+1 = (y2)k+1 =y2k+2 =
yy2k+1 =ye =y. By Exercise 12 of Chapter 5, φis also onto.
48. It follows from Lagrange’s Theorem that |H∩K|= 1 or q. Since
|K|=qr and |G:H|=r, there must be distinct elements k1and k2
in Ksuch that k1H=k2H. Thus k−1
2k1H=Hand
k−1
2k1∈H∩K. So, |H∩K| 6= 1.
49. By Corollary 3 of Lagrange’s Theorem a group of order 5 is cyclic.
Suppose Gis a group with distinct subgroups haiand hbiof order
5. Because 5 is prime, the identity is the only element common to
the two subgroups groups. This implies that the 25 elements of the
form aibjwhere i, j ∈ {0,1,2,3,4}are distinct.
50. Observe that |orbA5(5)|= 5. Now use the Orbit-Stabilizer Theorem
to show that |stabA5(5)|= 12. Note that the same argument
applies to stabA5(i) for i= 1,2,3,and 4.
7/Cosets and Lagrange’s Theorem 63
51. Suppose that His a subgroup of A5of order 30. We claim that H
contains all 20 elements of A5that have order 3. To verify this
assume that there is some αin A5of order 3 that is not in H.
Then A5=H∪αH. It follows that α2H=Hor α2=αH. Since
the latter implies that α∈H, we have that α2H=H, which
implies that α2∈H. But then hαi=hα2i ⊆ H, which is a
contradiction of our assumption that αis not in H. The same
argument shows that Hmust contain all 24 elements of order 5.
Since |H|= 30 we have a contradiction.
52. Suppose that His a subgroup of A5of order 20. We claim that H
contains all 24 elements of A5that have order 5. To verify this
assume that there is some αin A5of order 5 that is not in H.
Then A5=H∪αH ∪α2H. To see that the coset α2His not the
same as Hnote that α2H=Himplies that hα2i ⊂ Hand
hαi=hα2i.) Moreover, α2His not the same as αH for then α∈H.
It follows that α3His equal to one of the cosets H, αH or α2H. If
α3H=Hthen α3∈Hand therefore hαi=hα3i ⊆ H, which
contradicts the assumption that αis not in H. If α3H=αH then
α2∈Hand therefore hαi=hα2i ⊆ H, which contradicts the
assumption that αis not in H. If α3H=α2Hthen α∈Hwhich
contradicts the assumption that αis not in H. The same argument
shows that Hmust contain all 24 elements of order 5. Since
|H|= 20 we have a contradiction. An analogous argument shows
that A5has no subgroup of order 15.
53. Observe that α(ai) = ai+1, α2(ai) = ai+2, . . . , αk(ai) = aiwhere
all subscripts are taken mod k.
54. Since eis in H,Kis not empty. Note that every element of Kcan
be written in the from aihwhere i= 0 or 1 and h∈H. To verify
that the One-Step Subgroup test is fulfilled observe that
aih1(ajh2)−1where iand jare 0 or 1 and h1and h2are in Hcan
be written in the form aih1h−1
2a−j=ai(h1h−1
2)a−j=ai−jh3for
some h3∈H. Since i−j=−1,0 , or 1 and a−1=a, aih1(ajh2)−1
is in K. In the general case, K=H∪aH ∪a2H∪ ··· ∪ ak−1His a
subgroup.
55. Suppose that His a subgroup of S5of order 60. An argument
analogous to that given in Exercise 51 in this chapter shows that H
must contain all 24 elements in S5of order 5 and all 20 elements in
7/Cosets and Lagrange’s Theorem 64
S5of order 3. Since these 44 elements are also in A5we know that
|A5∩H|divides 60 and is greater than 30. So, H=A5.
56. If |Z(A4)|>1, then A4would have an element of order 2 or order 3
that commutes with every element. But any subgroup generated by
an element of order 2 and an element of order 3 that commute has
order 6. This contradicts the fact shown in Example 5 that A4has
no subgroup of order 6.
57. Suppose that B∈Gand det B= 2. Then det (A−1B) = 1, so that
A−1B∈Hand therefore B∈AH. Conversely, for any Ah ∈AH
we have det Ah = (det A)(det h) = 2 ·1 = 2.
58. The circle passing through Q, with center at P.
59. It is the set of all permutations that carry face 2 to face 1.
60.
@@@@@
@
@@@@@
@
@@@@@
@
@@@@@
@
@@@@@
@
@@@@@
@
s
s
s s
s
s
s s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s s
s s
s
s
s s
{R0, H};{R0, D0};{R0, H}
{R0};{R0};{R0}.
61. If aH =bH, then b−1a∈H. So det (b−1a) = (det b−1)(det a) =
(det b−1)(det a) = (det b)−1(det a) = 1. Thus det a= det b.
Conversely, we can read this argument backwards to get that det a
= det bimplies aH =bH.
62. a 12 b 24 c 60 d 60
7/Cosets and Lagrange’s Theorem 65
63. To prove that the set is closed note that αβ2= (13) = β2α3,
α2β2= (14)(23) = β2α2, and α3β2= (24) = β2α.
64. The order of the symmetry group would have to be 6 ·20 = 120.
65. Since the order of Gis divisible by both 10 and 25 it must be
divisible by 50. But the only number less than 100 that is divisible
by 50 is 50.
66
CHAPTER 8
External Direct Products
1. Closure and associativity in the product follows from the closure
and associativity in each component. The identity in the product is
the n-tuple with the identity in each component. The inverse of
(g1, g2, . . . , gn) is (g−1
1, g−1
2, . . . , g−1
n).
2. Every nonidentity element in the group has order 2. Each of these
generates a distinct subgroup of order 2.
3. The mapping φ(g) = (g, eH) is an isomorphism from Gto
G⊕ {eH}. To verify that φis one-to-one, we note that φ(g) = φ(g0)
implies (g, eH) = (g0, eH) which means that g=g0. The element
(g, eH)∈G⊕ {eH}is the image of g. Finally, φ((g, eH)(g0, eH)) =
φ((gg0, eHeH)) = φ((gg0, eH)) = gg0=φ((g, eH))φ((g0, eH)). A
similar argument shows that φ(h)=(eG, h) is an isomorphism from
Honto {eG} ⊕ H.
4. (g, h)(g0, h0)=(g0, h0)(g, h) for all g, g0, h, h0if and only if gg0=g0g
and hh0=h0h, that is, if and only if Gand Hare Abelian. A
corresponding statement holds for the external direct product of
any number of groups.
5. To show that Z⊕Zis not cyclic note that (a, b + 1) 6∈ h(a, b)i.
6. Z8⊕Z2contains elements of order 8, while Z4⊕Z4does not.
7. Define a mapping from G1⊕G2to G2⊕G1by φ(g1, g2)=(g2, g1).
To verify that φis one-to-one, we note that φ((g1, g2)) = φ((g0
1, g0
2))
implies (g2, g1)=(g0
2, g0
1). From this we obtain that g1=g0
1and
g2=g0
2. The element (g2, g1) is the image on (g1, g2) so φis onto.
Finally, φ((g1, g2)(g0
1, g0
2)) = φ((g1g0
1, g2g0
2)) = (g2g0
2, g1g0
1) =
(g2, g1)(g0
2, g0
1) = φ((g1, g2))φ((g0
1, g0
2)). In general, the external
direct product of any number of groups is isomorphic to the
external direct product of any rearrangement of those groups.
8/External Direct Products 67
8. No, Z3⊕Z9does not have an element of order 27. See also
Theorem 8.2.
9. Yes, By Theorem 8.2.
10. Z9has 6 elements of order 9 (the members of U(9)). Any of these
together with any element of Z3gives an ordered pair whose order
is 9. So Z3⊕Z9has 18 elements of order 9.
11. In both Z4⊕Z4and Z8000000 ⊕Z400000,|(a, b)|= 4 if and only if
|a|= 4 and |b|= 1,2 or 4 or if |b|= 4 and |a|= 1 or 2 (we have
already counted the case that |a|= 4). For the first case, we have
φ(4) = 2 choices for aand φ(4) = φ(2) + φ(1) = 4 choices for bto
give us 8 in all. For the second case, we have φ(4) = 2 choices for b
and φ(2) + φ(1) = 2 choices for b. This gives us a total of 12.
In the general case observe that by Theorem 4.4 that as long as d
divides nthe number of elements of order din a cyclic group
depends only on d.
12. Z12, Z6⊕Z2, D6, A4.The first two are Abelian and the second two
are not. Z12 is cyclic and Z6⊕Z2is not. D6has an element of
order 6 and A4does not.
13. Zn2and Zn⊕Zn.
14. The group of rotations is Abelian and a group of order 2 is Abelian;
now use Exercise 4.
15. Define a mapping φfrom Cto R⊕Rby φ(a+bi)=(a, b). To
verify that φis one-to-one note that φ(a+bi) = φ(a0+b0i) implies
that (a, b)=(a0, b0). So, a=a0and b=b0and therefore
a+bi =a0+b0i. The element (a, b) in R⊕Ris the image of a+bi
so φis onto. Finally,
φ((a+bi)+(a0+b0i)) = φ((a+a0)+(b+b0)i) = (a+a0, b +b0) =
(a, b)+(a0, b0) = φ(a+bi) + φ(a0+b0i).
16. Let α:G1→G2and β:H1→H2be isomorphisms. Then
γ:G1⊕H1→G2⊕H2given by γ((g1, h1)) = (α(g1), β(h1)) is an
isomorphism. A corresponding statement holds for the external
direct product of any number of groups.
8/External Direct Products 68
17. By Exercise 3 in this chapter Gis isomorphic to G⊕{eH}and His
isomorphic to {eG} ⊕ H. Since subgroups of cyclic groups are
cyclic, we know that G⊕{eH}and {eG} ⊕Hare cyclic. In general,
if the external direct product of any number of groups is cyclic,
each of the factors is cyclic.
18. h10i⊕h10i;h20i⊕h5i.
19. hm/ri⊕hn/si.
20. Observe that Z9⊕Z4≈Z4⊕Z9≈ h3i⊕h2i.
21. Since h(g, h)i ⊆ hgi⊕hhi, a necessary and sufficient condition for
equality is that lcm(|g|,|h|) = |(g, h)|=|hgi⊕hhi| =|g||h|. This is
equivalent to gcd(|g|,|h|) = 1.
22. 48; 6
23. |(a, b, c)|= lcm{|a|,|b|,|c|} = 3, unless a=b=c=e. In general,
the order of every nonidentity element of Zp⊕Zp⊕ ··· ⊕ Zp, where
pis prime, is p.
24. Observe that |(a, b)|= 2 if and only if |a|= 1 or 2 and |b|= 1 or 2
but not both |a|= 1 and |b|= 1. So, there are
(m+ 2)(n+ 1) −1 = mn +m+ 2n+ 1 elements of order 2.
25. Define a mapping φfrom Mto Nby φ " a b
c d #!= (a, b, c, d).
To verify that φis one-to-one we note that
φ " a b
c d #!=φ " a0b0
c0d0#!implies (a, b, c, d) = (a0, b0, c0, d0).
Thus a=a0, b =b0, c =c0, and d=d0. This proves that φis
one-to-one. The element (a, b, c, d) is the image of "a b
c d #so φis
onto. Finally, φ " a b
c d #+"a0b0
c0d0#!=
φ " a+a0b+b0
c+c0d+d0#!= (a+a0, b +b0, c +c0, d +d0) =
(a, b, c, d)+(a0, b0, c0, d0) = φ " a b
c d #!+φ " a0b0
c0d0#!.
Let Rkdenote R⊕R⊕ ··· ⊕ R(kfactors). Then the group of
m×nmatrices under addition is isomorphic to Rmn.
8/External Direct Products 69
26. D6. Since S3⊕Z2is non-Abelian, it must be isomorphic to A4or
D6. But S3⊕Z2contains an element of order 6 and A4does not.
27. Since (g, g)(h, h)−1= (gh−1, gh−1), H is a subgroup. When
G=R,G⊕Gis the plane and His the line y=x.
28. {(0,0),(2,1)}.
29. h(3,0)i,h(3,1)i,h(3,2)i,h(0,1)i
30. h(1,0)i,h(1,1)i,h(1,2)i,h(1,3)i,h(0,1)i,h(2,1)i,
{(0,0),(2,0),(0,2),(2,2)}
31. |(1,1)|= lcm{30,20}= 60.
32. lcm(6,10,15) = 30; lcm(n1, n2, . . . , nk).
33. h(1,1,1)i,h(5,1,1)i,h(1,3,1)i
34. In general, if mand nare even, then Zm⊕Znhas exactly 3
elements of order 2. For if |(a, b)|= 2, then |a|= 1 or 2 and |b|= 1
or 2 but not both aand bhave order 1. Since any cycle group of
even order has exactly 1 element of order 2 and 1 of order 1 there
are only 3 choices for (a, b).
35. h400i⊕h50i={0,400}⊕{0,50,100,150}
36. h4i⊕h0i⊕h5i
37. In R∗⊕R∗(1,−1), (−1,1) and (−1,−1) have order 2 whereas in
C∗the only element of order 2 is −1. But isomorphisms preserve
order.
38. Z3⊕Z3
39. Define the mapping from Gto Z⊕Zby φ(3m6n)=(m, n). To
verify that φis one-to-one note that φ(3m6n) = φ(3s6t) implies that
(m, n)=(s, t), which in turn implies that m=sand n=t. So,
3m6n= 3s6t. The element (m, n) is the image of 3m6nso φis onto.
Finally, φ((3m6n)(3s6t)) = φ(3m+s6n+t) = (m+s, n +t) =
(m, n)+(s, t) = φ(3m6n)φ(3s6t) shows that φis operation
preserving.
When G={3m9n|m, n ∈Z}the correspondence from Gto Z⊕Z
given by φ(3m9n)=(m, n) is not well-defined since
φ(3290)6=φ(3091) and 3290= 9 = 3091.
8/External Direct Products 70
40. |ai|=∞for some i.
41. D3⊕D4has 10 elements of order 6 whereas D12 ⊕Z2has only 4.
42. 4
43. U(35) ≈U(5) ⊕U(7) ≈U7(35) ×U5(35) =
{1,8,22,29}×{1,6,11,16,26,31}.
44. Observe that U(40) ⊕Z6≈U(8) ⊕U(5) ⊕Z6≈Z2⊕Z2⊕Z4⊕Z6
and U(72) ⊕Z4≈U(9) ⊕U(8) ⊕Z4≈Z6⊕Z2⊕Z2⊕Z4so they
are isomorphic.
45. C∗has only one element of order 2 whereas Z2⊕Z2has three
elements of order 2.
46. If exactly one niis even then xis the unique element of order 2.
Otherwise xis the identity.
47. Each cyclic subgroup of order 6 has two elements of order 6. So,
the 24 elements of order 6 yield 12 cyclic subgroups of order 6. In
general, if a group has 2nelements of order 6 it has ncyclic
subgroups of order 6. (Recall from the Corollary of Theorem 4.4 if
a group has a finite number of elements of order 6 the number is
divisible by φ(6) = 2).
48. Use the observation that (h, k)n= (hn, kn).
49. Aut(U(25)) ≈Aut(Z20)≈U(20) ≈U(4) ⊕U(5) ≈Z2⊕Z4.
50. S3
51. In each position we must have an element of order 1 or 2 except for
the case that every position has the identity. So, there are 2k−1
choices. For the second question, we must use the identity in every
position for which the order of the group is odd. So, there are
2t−1 elements of order 2 where tis the number of n1, n2, . . . , nk
that are even.
52. Z10 ⊕Z12 ⊕Z6≈Z2⊕Z5⊕Z12 ⊕Z6≈Z2⊕Z60 ⊕Z6≈Z60 ⊕Z6⊕Z2.
53. No. Z10 ⊕Z12 ⊕Z6has 7 elements of order 2 whereas
Z15 ⊕Z4⊕Z12 has only 3.
54. x→(xmod 4, x mod 3).
8/External Direct Products 71
55. Using the fact that an isomorphism from Z12 is determined by the
image of 1 and the fact that a generator must map to a generator,
we determine that there are 4 isomorphisms.
56. Since φ(2,3) = 2 we have 8φ(2,3) = 16 = 1. So,
1 = φ(16,24) = φ(1,4).
57. Since (2,0) has order 2, it must map to an element in Z12 of order
2. The only such element in Z12 is 6. The isomorphism defined by
(1,1)x→5xwith x= 6 takes (2,0) to 6. Since (1,0) has order 4, it
must map to an element in Z12 of order 4. The only such elements
in Z12 is 3 and 9. The first case occurs for the isomorphism defined
by (1,1)x→7xwith x= 9 (recall (1,1) is a generator of Z4⊕Z3);
the second case occurs for the isomorphism defined by (1,1)x→5x
with x= 9.
58. First observe that any non-identity element in Z5⊕Z5has order 5
and any two subgroups of order 5 overlap in only the identity. Thus
ha1iwhere a1is any non-identity element in Z5⊕Z5has order 5. A
second subgroup of order 5 is ha2iwhere a2is any element not in
ha1i. A third subgroup of order 5 is ha3iwhere a3is any element
not in ha1ior ha2i. Continuing this way until we reach ha6iwe have
used all 25 elements in Z5⊕Z5.
59. Since a∈Zmand b∈Zn, we know that |a|divides mand |b|
divides n. So, |(a, b)|= lcm(|a|,|b|) divides lcm(m, n).
60. Map ax2+bx +cto (a, b, c). In general,
{an−1xn−1+···+a0|an−1, . . . , a0∈Zm}under addition modulo m
is isomorphic to Zm⊕ ··· ⊕ Zm(ncopies).
61. Up to isomorphism, Zis the only infinite cyclic group and it has 1
and −1 as its only generators. The number of generators of Zmis
|U(m)|so we must determine those msuch that |U(m)|= 2. First
consider the case where m=pn, where pis a prime. Then the
number of generators is pn−1(p−1). So, if p > 3 we will have more
than 2 generators. When p= 3 we must have n= 1. Finally,
|U(2n)|= 2n−1= 2 only when n= 2. This gives us Z3and Z4.
When m=p1p2···pk, where the p’s are distinct primes we have
|U(m)|=|U(p1)||U(p2)|···|U(pk))|. As before no prime can be
greater than 3. So, the only case is m= 2 ·3 = 6.
8/External Direct Products 72
62. Identify A with (0,0), T with (1,1), G with (1,0) and C with (0,1).
Then a string of length nof the four bases is represented by a
string of 0s and 1s of length 2nand the complementary string of
a1a2. . . a2nis a1a2. . . a2n+ 11 . . . 1.
63. Each subgroup of order pconsists of the identity and p−1 elements
of order p. So, we count the number of elements of order pand
divide by p−1. In Zp⊕Zpevery nonidentity element has order p
so there are (p2−1)/(p−1) = p+ 1 subgroups of order p.
64. Z⊕D3.
65. In Z⊕Z2|(1,1)|=∞,|(−1,0)|=∞,|(1,1)(−1,0)|=|(0,1)|= 2.
66. U(165) ≈U(11) ⊕U(3) ⊕U(5) ≈Z10 ⊕Z2⊕Z4.
67. U(165) ≈U(15) ⊕U(11) ≈U(5) ⊕U(33) ≈U(3) ⊕U(55) ≈
U(3) ⊕U(5) ⊕U(11).
68. Use the fact that Aut(Z20)≈U(20) ≈U(4) ⊕U(5) ≈Z2⊕Z4.
69. We use the fact that Aut(Z720)≈U(720) ≈Z2⊕Z4⊕Z6⊕Z4. In
order for (a, b, c, d) to have order 6 we must have |c|= 6 and a, b,
and dhave orders 1 or 2. So we have 2 choices for each of a, b, c,
and d. This gives 16 in all.
70. Use the fact that U(27) ≈Z18.
71. U(900) ≈Z2⊕Z6⊕Z20 so the element of largest order is the
lcm(2, 6, 20) = 60.
72. U(pm)⊕U(qn) = Zpm−pm−1⊕Zqn−qn−1and both of these groups
have even order. Now use Theorem 8.2.
73. Observe that U(55) ≈U(5) ⊕U(11) ≈Z4⊕Z10 and U(75) ≈
U(3) ⊕U(25) ≈Z2⊕Z20 ≈Z2⊕Z5⊕Z4≈Z10 ⊕Z4≈Z4⊕Z10.
74. U(144) ≈U(16) ⊕U(9) ≈Z2⊕Z4⊕Z6;
U(140) ≈U(4) ⊕U(5) ⊕U(7) ≈Z2⊕Z4⊕Z6.
75. By the Finite Subgroup Test, the set is closed since a2b2= (ab)2.
The set is proper because 12= (n−1)2.
76. Since U(55) ≈Z4⊕Z10,x20 = 1 for all xin U(55). Thus, a3=b3
implies (a3)7= (b3)7or a=b. So, all cubes in U(55) are distinct.
8/External Direct Products 73
77. We need to find relatively prime mand nso that both |U(m)|and
|U(n)|are divisible by 5. Since for every odd prime pwe have
|U(pk)|=pk−1(p−1), we can choose m= 52and n= 11. Then
U(mn) = U(275) ≈U(25) ⊕U(11) ≈Z20 ⊕Z10, which contains
h4i⊕h2i ≈ Z5⊕Z5.
78. U100(700)
79. This is the same argument as in Exercise 77 with 5 replaced by 3.
So, m= 9 and n= 7 gives us U(63) ≈U(9) ⊕U(7) ≈Z6⊕Z6,
which contains h2i⊕h2i ≈ Z3⊕Z3.
80. Observe that
Z2⊕Z4⊕Z9≈Z4⊕Z2⊕Z9≈Z4⊕Z18 ≈U(5) ⊕U(27) ≈U(135).
81. Since U(pq)≈U(p)⊕U(q)≈Zp−1⊕Zq−1if follows that k=
lcm(p−1, q −1).
82. Um(n) is a subgroup of Uk(n).
83. In the first case there are 2k−1; in the second case there are
2k+2 −1.
84. Observe that neither pn−pn−1= 14 nor pn−pn−1= 7 is possible
for any prime p.
85. We need a prime-power pkso that pk−1(p−1) is divisible by 14.
Obviously, 72is one choice.
86. Suppose U(m)≈Z4⊕Z4and write m= 2tnwhere nis odd. Then
U(m)≈U(2t)⊕U(n). Since U(2t)≈Z2⊕Z2t−2for t≥3, we have
t≤4. Let pbe a prime such that pk|nbut pk+1 6| n. Then U(n)
contains an element of order pk−pk−1. Thus pk−pk−1divides 4.
This implies that pk= 3 or 5 and m= 2t·3, 2t·3·5 where t≤4.
But none of these yields Z4⊕Z4.
87. We need psand qt, where pand qare distinct primes and both
ps−1(p−1) and qt−1(q−1) to be divisible by 4. One such choice is
p= 5 and q= 13. The U(65) ≈U(5) ⊕U(13) ≈Z4⊕Z12, which
contains h1i⊕h3i ≈ Z4⊕Z4.
88. None. Step 3 of the Sender part of the algorithm fails.
89. Since 5 ·29 = 1 mod 36, we have that s= 29. So, we need to
compute 3429 mod 2701. The result is 1415, which converts to NO.
74
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8
1. Consider the finite and infinite cases separately. In the finite case,
note that |H|=|φ(H)|. Now use Theorem 4.3. For the infinite
case, use Exercise 2 of Chapter 6.
2. See Theorem 6.2, part 3.
3. Observe that φ(x−1y−1xy)=(φ(x))−1(φ(y))−1(φ(x))(φ(y)), so φ
carries the generators of Gto the generators of G0.
4. When Kis characteristic, and automorphism of Gis an
automorphism of K, which in turn, is an automorphism of Nwhen
Nis characteristic.
5. All nonidentity elements of Gand Hhave order 3. Since Gis
Abelian and His non-Abelian, G6≈ H.
6. Suppose HK is a group and x∈HK. Then x−1∈HK so we may
write x−1=hk where h∈Hand k∈K. Then
x= (x−1)−1= (hk)−1=k−1h−1∈KH.
Thus HK ⊆KH. Now suppose x∈KH, say x=kh. Then
x−1=h−1k−1∈HK. But then x∈HK also.
Next suppose HK =KH and let x, y ∈HK. Say x=hk and
y=h0k0. Then
xy =hkh0k0=h(kh0)k0=h(h00k00)k0=hh00k00k0∈HK.
(Remember that HK =KH does not mean kh0=h0k.) Also,
x−1= (hk)−1=k−1h−1=h1k1∈HK.
7. Let aand bbe distinct nonidentity elements in Gof order 2. Let
H1={e, a, b, ab}. Note that H1is a subgroup of Gand the product
of all its elements is e. If H1=G, we are done. If not, then let cbe
an element of Gnot in H1. Then H2=H1∪cH1is a subgroup of
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 75
Gand the product of all the elements in H2is c4=e. If H2=G,
we are done. If not, then let dbe an element of Gnot in H2. Then
H3=H2∪dH2is a subgroup of Gand the product of all the
elements in H3is d8=e. Continuing in this way finishes the proof.
8. D4⊕Z3has an element of order 12 but S4does not.
9. Let xand ybelong to H. Then x=aiz1and y=ajz2where
0≤i, j < 4 and z1and z2are in Z(G). So,
xy−1=aiz1(ajz2)−1=ai−jz1z−1
2where i−j∈ {0,1,2,3}and
z1z−1
2∈Z(G). In the general case
H=Z(G)∪aZ(G)∪a2Z(G)∪ ··· ∪ ak−1Z(G) is a subgroup of G.
10. Observe that the exponent of a finite group is the least common
multiple of the order of all the elements of the group.
11. For U(n) to have exponent 2 every nonidentity element must have
order 2. If a prime pdivides nand p > 3, then U(n) has an element
of order p−1 so the only primes that can divide nare 2 and 3.
Since U(3m) has an element of order 3m−1,32cannot divide n.
Since U(2m) has an element of order 2m−2when m > 3 and U(2) is
the identity we know the only possible powers of 2 that can divide
nare 2, 4 and 8. Thus, we have shown that nmust have the form
3s2twhere s= 0 or 1 and t= 0,1,2,or 3. This gives us
n= 1,2,4,8,3,6,12, or 24. But U(1) and U(2) have exponent 1 so
we are left with U(n), where n= 4,8,3,6,12,24.
12. H∩Kis a subgroup of both Hand Kso |H∩K|divides |H|and
|K|. Thus |H∩K|= 1.
13. Observe that
a+bi =pa2+b2a
√a2+b2+b
√a2+b2iand pa2+b2belongs
to R+and a
√a2+b2+b
√a2+b2i∈T. So every element can be
expressed in the desired form.
Now suppose that r1z1=r2z2, where r1and r2belong to R+and
z1and z2belong to T. Then r1r−1
2=z2z−1
1∈T. Since Tis the
unit circle in the complex plane, the only real numbers in Tare 1
and −1. Then, because r1r−1
2is real and r1and r2are positive, we
have r1r−1
2= 1 and therefore r1=r2and z1=z2.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 76
14. Suppose φ:R∗→Q∗is an isomorphism. Let r∈R∗be such that
φ(r) = 2. Then
2 = φ(r) = φ(3
√r3
√r3
√r)=(φ(3
√r))3.
But there is no rational number whose cube is 2.
15. Suppose φ:Q→R is an isomorphism. Let φ(1) = x0. By property
2 of Theorem 6.2, for any nonzero integer b, we have
x0=φ(b1
b) = bφ(1
b) so that φ(1
b) = 1
bx0. Also,
φ(a
b) = aφ(1
b) = a1
bx0=a
bx0. So, if x0is irrational, then 1 is not in
the image and if x0is rational the √2 is not in the image.
16. In Runder addition every nonzero element has infinite order. In
R∗under multiplication −1 has order 2.
17. In Q, the equation 2x=ahas a solution for all a. The
corresponding equation x2=bin Q+does not have a solution for
b= 2.
18. Since Gis non-Abelian, xy 6=yx. Cancellation shows xy 6=e,x,x2,
or y.
19. Suppose xp−2= 1. Since |U(p)|=p−1, we have that xp−1= 1 for
all x∈U(p). So, cancellation, x= 1.
20. (g1, ng1)(g2, ng2)−1= (g1, ng1)(−g2,−n(g2))
= (g1−g2, n(g1−g2)) ∈H.
His the line y=nx in the real plane.
21. h3i⊕h4i.
22. Fix (a1, . . . , an). Then
(x1, . . . , xn)(a1, . . . , an)=(a1, . . . , an)(x1, . . . , xn)
if and only if for i= 1, . . . , n we have xiai=aixifor all xiin Gi.
That is, if and only if ai∈Z(Gi) for i= 1, . . . , n.
23. Z18,Z2⊕Z3⊕Z3,D9,D3⊕Z3. To see that these are not
isomorphic note that Z18 is cyclic and the others are not,
Z2⊕Z3⊕Z3is Abelian and the last two are not, D9has an
element of order 9 and D3⊕Z3does not.
24. 12
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 77
25. Say α=a1a2···anand β=b1b2. . . bm, where the a’s and b’s are
cycles. Then αβ−1=a1a2···anb−1
m···b−1
1is expressed as a product
of a finite number of cycles.
26. Elements of Z(G)Hhave the form zh where z∈Z(G) and h∈H.
Observe that (z1h1)(z2h2) = z1z2h1h2∈Z(G)Hand
(zh)−1=h−1z−1=z−1h−1∈Z(G).
27. D11 ⊕Z3has 11 elements of order 2 whereas D3⊕Z11 has only 3.
28. Count elements of order 2.
29. D33 has 33 elements of order 2 whereas D3⊕Z11 has only 3.
30. See Exercises 25-27.
31. If |Inn(G)|= 1, then for all ain G, x =φa(x) = axa−1, so that
xa =ax. Conversely, if Gis Abelian, φais the identity.
32. Observe that U(1000) ≈U(8) ⊕U(125) ≈Z4⊕Z100 and use
Theorem 8.1.
33. U50(450) ≈U(9) ≈Z6
34. {(0,0,0),(10,0,0),(0,0,30),(10,0,30)}
35. (4,10)
36. Z4⊕Z15 has an element of order 4 whereas Z6⊕Z10 does not.
37. D12 has 13 elements of order 2 whereas Z3⊕D4has only 5.
38. A pyramid with a regular pentagonal base.
39. 20; (8,7,(3251))
40. His not closed.
41. Let H={x∈Zp2⊕Zp2|xp= (0,0)}. Then |H|=p2and every
nonidentity element of (Zp2⊕Zp2)/H has order p.
42. It suffices to show that σ(i1i2. . . ik) = (σ(i1)σ(i2). . . σ(ik))σ.
Obviously, we need only check that σ(i1i2. . . ik) and
(σ(i1)σ(i2). . . σ(ik))σagree on all ij. Observe that
(σ(i1i2. . . ik)(ij) = σ(ij+1) while
((σ(i1)σ(i2). . . σ(ik))σ)(ij) = σ(ij+1). (When j=k, we use 1
instead j+ 1).
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 78
43. (12)(34)(56789).
44. For the matrix form see the rows in the Cayley table for D8in
Chapter 1 headed by R90 and TH. In cycle form,
T90 = (R0, R90, R180, R270)(H, D0, V, D); TH=
(R0, H)(R90, D)(R180, V )(R270, D0).
45. There are 7!/6 elements that are 6-cycles. There are
(7 ·6·5/3!)(4 ·3)/2 disjoint cycles of the form (a1a2a3)(a4a5) and
(7 ·6·5/3!)(4 ·3)/2(2 ·1)/2 disjoint cycles of the form
(a1a2a3)(a4a5)(a6a7). This makes 1260 in all.
46. S3⊕S4has an element of order 12 but S6does not.
47. Since |β2|= 15, we know that |β|= 15 or 30. But A9has no
element of order 30, so |β|= 15. Then
β=β16 = (β2)8= (17395)(286)
48. (2,5) + H={(2 + x, 3x+ 5|x∈R}. Notice the second coordinate is
triple the first minus 1. Thus they satisfy the equation y= 3x−1.
49. Say the points in Hlie on the line y=mx. Then
(a, b) + H={(a+x, b +mx)|x∈R}. This set is the line
y−b=m(x−a).
50. Certainly every member of γstabG(1) sends 1 to k. If βsends 1 to
k, then γ−1β∈stabG(1), so that β∈stabG(1)γ.
51. aH =bH implies a−1b∈H. So (a−1b)−1=b−1a∈H. Thus,
Hb−1a=Hor Hb−1=Ha−1. These steps are reversible.
52. Let H=haibe a cyclic subgroup of Gof largest order. If H6=G,
let bbelong to Gbut not belong to H. If gcd(|a|,|b|) = 1, then
hai⊕hbiis cyclic subgroup of larger order than H. So, there is a
prime pthat divides both |a|and |b|. Let c∈ haihave order pand
let d∈ hbihave order p. Then Gcontains hci⊕hdi. This is a
contradiction.
53. Since |(a, b)|=pif and only if (a, b)p= (ap, bp) = (e, e) and not
both aand bare the identity, counting the identity, there are p
choices for aand pchoices for b. So, excluding the case the
a=e=bthere are p2−1 choices for (a, b).
54. Use the previous exercise.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 79
55. First observe that by Theorem 8.1 every element of Zp2⊕Zp2has
order 1, p, or p2. From Exercise 53, Zp2⊕Zp2has p2elements of
order 1 or p. So, there are p4−p2elements of order p2. Since a
cyclic group of order p2has φ(p2) = p2−pgenerators, the number
of cyclic groups of order p2is (p4−p2)/(p2−p) = p(p+ 1).
56. One possibility is A8⊕Z5·7·22.
57. By Theorem 8.3, U(pq)≈U(p)⊕U(q), so an element xnin U(pq)
corresponds to an element (xn
1, xn
2)∈U(p)⊕U(q). It follows from
Corollary 4 of Theorem 7.1 that (xn
1, xn
2) = (1,1), the identity of
U(p)⊕U(q).
58. D6has an element of order 6 but S4does not.
59. First observe that (n, n −1, . . . 2,1)(12)(123 . . . n) = (1n).Also,
(1n)(123 . . . n) = (123 . . . n −1). So, by induction, (12) and
(123 . . . n) generate Sn−1. This means that every 2-cycle not
involving ncan be generated. Now note that (1k)(1n)(1k)=(kn),
so all 2-cycles are generated.
60. Since σis an (n−1)-cycle in Snthere is some ibetween 1 and nso
that σ(i) = i. Say βhas order 2 and commutes with σ. First
consider the case that β(i)6=i. Then,
σ(β(i)) = (σβ)(i)=(βσ)(i) = β(σ(i)) = β(i). But σfixes only i
and i6=β(i). Next consider the case that β(i) = i. Since nis even
and βcan be written as a product of 2-cycles, βmust fix some
j6=i. Then σ(j) = σ(β(j)) = (σβ)(j)=(βσ)(j) = β(σ(j)). So, β
fixes σ(j). But because βcommutes with σit commutes with all
powers of σ. It follows that βfixes σ2(j), σ3(j), . . . , σn−1(j). This
implies that βis the identity.
61. Let βhave order 2. In disjoint cycle form, βis a product of
transpositions, and since nis odd there must be some imissing
from this product. Thus, β(i) = i. Pick jso that β(j)6=j. Since σ
is an n-cycle, some power of σ, say σt, takes ito j. If βcommutes
with σ, it commutes with σtas well. Then
(σtβ)(i) = σt(β(i)) = σt(i) = j, whereas
(βσt)(i) = β(σt(i)) = β(j)6=j. This proves that σtβ6=βσt.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 5-8 80
62. Let β∈C(12) and suppose that β(1) is not 1 or 2. Evaluating
(12)β(12) = βat 2 on the left we have that 2 goes to β(1).
Evaluating (12)β(12) = βat 2 on the right we have that 2 goes to
β(2). A similar contradiction occurs if β(2) is not 1 or 2. This
proves that C(12) ⊆H. Conversely, if β∈H, then in disjoint cycle
form βconsists of the two 1-cycles (1)(2) and cycles disjoint from
(12) or the 2-cycle (12) and cycles disjoint from (12). In either case
βcommutes with (12).
63. Write σas (a1a2···an). Then for each entry aiin the cycle σm
takes aito ai+mwhere the subscript i+mis taken mod n. So, the
cycle decomposition of σmhas the cycle
(a1, a1+m, a1+2m, . . . , a1+(r−1)m) where ris the smallest positive
integer such that 1 = (1 + rm) mod n. Thus ris the smallest
positive integer with the property that rm = 0 mod n. But
n/gcd(m, n) is the smallest such integer. The same argument
applies to all the other cycles in σm. Since the sum of the lengths
of the cycles in the decomposition of σmis nand each cycle in the
decomposition of σmhas length n/gcd(m, n) there must be
gcd(m, n) cycles.
81
CHAPTER 9
Normal Subgroups and Factor Groups
1. No, (13)(12)(13)−1= (23) is not in H.
2. For every αin Sn, αAnα−1is even.
3. HR90 =R270H;DR270 =R90D;R90V=V R270
4. Solving (12)(13)(14) = α(12) for αwe have α= (12)(13)(14)(12).
Solving (1234)(12)(23) = α(1234) for αwe have α= (234).
5. Say i<jand h∈Hi∩Hj. Then h∈H1H2···Hj−1∩Hj={e}.
6. No. Let A="1 0
0−1#and B="1 0
1 1 #. Then Ais in Hand B
is in GL(2,R) but BAB−1is not in H.
7. Hcontains the identity so His not empty. Let A, B ∈H. Then
det (AB−1) = (det A)(det B)−1∈K. This proves that His a
subgroup. Also, for A∈Hand B∈Gwe have det (BAB−1) =
(det B)(det A)(det B)−1= det A∈Kso BAB−1∈H.
8. If kdivides n, then hki/hniis a cyclic group of order n/k. So it is
isomorphic to Zn/k.
9. Let x∈G. If x∈H, then xH =H=Hx. If x6∈ H, then xH is the
set of elements in G, not in Hand Hx is also the elements in G,
not in H.
10. a α5H6=Hα5. b It proves that coset multiplication is not a
binary operation.
11. In G/H the element (1,1)Hhas order 4, so G/H ≈Z4. In G/K
the elements (1,1)K, (3,3)K, and (2,3)Khave order 2, so
G/K ≈Z2⊕Z2.
12. Let G=hai. Then G/H =haHi.
9/Normal Subgroups and Factor Groups 82
13. Observe that aHbH =abH =baH =bHaH.
14. 4
15. Since (4U5(105))2= 16U5(105) = U5(105),|4U5(105)|= 2.
16. 3
17. H={0 + h20i,4 + h20i,8 + h20i,12 + h20i,16 + h20i}.
G/H ={0 + h20i+H, 1 + h20i+H, 2 + h20i+H, 3 + h20i+H}
18. 60/4 = 15
19. 40/10 = 4
20. Let H=U5(20). Then U(20)/H ={H, 3H, 9H, 7H}is a cyclic
group generated by 3H. The Cayley table for U(20)/H is identical
to the Cayley table for the subgroup h3iof U(20) with ireplaced
with iH.
21. By Theorem 9.5 the group has an element aof order 3 and an
element bof order 11. Then |ab|= 33.
22. ∞; no, (1,1) + h(2,2)ihas order 2.
23. Since (1,1) + h(4,2)ihas infinite order, (Z⊕Z)/h(4,2)iis infinite.
It is not cyclic because (6,3) + h(4,2)ihas order 2 and any infinite
cyclic group is isomorphic to Z, which has no elements of finite
order other than the identity.
24. Z4⊕Z2.
25. Since the element |3H|of G/H has order 8, G/H ≈Z8.
26. Z2⊕Z2.
27. Since Hand Khave order 2 they are both isomorphic to Z2and
therefore isomorphic to each other. Since |G/H|= 4 and |3H|= 4
we know that G/H ≈Z4. On the other hand, direct calculations
show that each of the three nonidentity elements in G/K has order
2, so G/K ≈Z2⊕Z2.
28. Z2⊕Z2;Z4.
9/Normal Subgroups and Factor Groups 83
29. If His a subgroup of A4⊕Z3of order 18, then the factor group
(A4⊕Z3)/H exists and has order 2. Then, for every element
(β, k)∈A4⊕Z3we have ((β, k)H)2= (β2,2k)H=H. Thus, H
contains every element of A4⊕Z3of the form (β2,2k), where
β∈A4and k∈Z3. However, in A4there are nine distinct elements
of the form β2and in Z3there are three of the form 2k. This
means that Hhas at least 27 elements.
30. U(165) = U15(165) ×U11(165) = U33(165) ×U5(165) =
U55(165) ×U3(165).
31. Certainly, every nonzero real number is of the form ±r, where ris a
positive real number. Real numbers commute, and
R+∩ {1,−1}={1}.
32. First observe that every proper subgroup of D4is Abelian. Now
use Theorem 9.6 and Exercise 4 of Chapter 8.
33. In the general case that G=HK there is no relationship. If
G=H×K, then |g|= lcm(|h|,|k|) provided the |h|and |k|are
finite. If |h|or |k|is infinite, so is |g|.
34. Since 1 = 3 ·5+(−2)7, m= (3m)5 + (−2m)7. No, since
35 ∈H∩K. Suppose His any subgroup of index 2. Then, since
|R∗/H|= 2, we have a2∈Hfor all a∈R∗. This implies that
R+⊆H. If there is some a∈Hwith a < 0, then since −a∈Hwe
have a−1(−a) = −1∈Halso. But this implies that R∗⊆H.
35. For the first question, note that h3i∩h6i={1}and
h10i∩h3ih6i={1}. For second question, observe that 12 = 3−162
so h3i×h12i 6=∅.
36. Certainly, R+has index 2.
37. Say |g|=n. Then (gH)n=gnH=eH =H. Now use Corollary 2
to Theorem 4.1.
38. 3, 6, 15, 30.
39. Let x∈C(H) and g∈G. We must show that gxg−1∈C(H). That
is, for any hin H,gxg−1h=hgxg−1. Note that in the expression
(gxg−1)h(gxg−1)−1=gxg−1hgx−1g−1the terms xand x−1cancel
since g−1hg ∈Hand xcommutes with every element of H. Then
9/Normal Subgroups and Factor Groups 84
we have (gxg−1)h(gxg−1)−1=gxg−1hgx−1g−1=gg−1hgg−1=h.
So, gxg−1∈C(H).
40. Let φ(h)∈φ(H) and let g=φ(g) where g∈G. Then gφ(h)g−1=
φ(g)φ(h)φ(g)−1=φ(g)φ(h)φ(g−1) = φ(ghg−1)∈φ(gHg−1) = φ(H).
41. Suppose that His a proper subgroup of Qof index n. Then Q/H
is a finite group of order n. By Corollary 4 of Theorem 7.1 we know
that for every xin Qwe have nx is in H. Now observe that the
function f(x) = nx maps Qonto Q. So, Q⊆H.
42. Let g∈G. Then there is an element bin Gso that
gH = (bH)2=b2H. Thus g=b2hfor some h∈H. But there is a c
in Hsuch that h=c2So, g= (bc)2. The proof is valid for any
integer.
43. Take G=Z6,H={0,3},a= 1, and b= 4.
44. Because xHx−1and Hare both subgroups of the same order.
45. Gand the trivial subgroup are normal. By Lagrange’s Theorem all
other subgroups have order por 2. By Exercise 9 of this chapter
46. By Example 14 of Chapter 3, Z(D13) is the identity. Then by
Theorem 9.4, D13 is isomorphic to Inn(D13).
47. Since N⊆NH ⊆G, we have |G:N|=|G:NH||NH :N|. Thus,
|G:H|= 1 or |NH :N|= 1. It follows that G=NH or NH =N.
48. Use Example 3 in Chapter 8 and the G/Z Theorem (Theorem 9.3).
49. By Lagrange’s Theorem, |Z(G)|= 1, p, p2, or p3. By assumption,
|Z(G)| 6= 1 or p3(for then Gwould be Abelian). So, |Z(G)|=por
p2. However, the “G/Z” Theorem (Theorem 9.3) rules out the
latter case.
50. Use the G/Z Theorem.
51. If His normal in G, then xNhN(xN )−1=xhx−1N∈H/N, so
H/N is normal in G/N. Now assume H/N is normal in G/N . Then
xhx−1N=xNhN(xN )−1∈H/N. Thus xhx−1N=h0Nfor some
h0∈Hand therefore xhx−1∈H. So, xhx−1=h0nfor some n∈N.
9/Normal Subgroups and Factor Groups 85
52. Say |aH|has finite order n. Then H= (aH)n=anHso that anis
in H. But this implies that |an|and therefore |a|is finite. Thus
aH =H.
53. Say Hhas an index n. Then (R∗)n={xn|x∈R∗} ⊆ H. If nis
odd, then (R∗)n=R∗; if nis even, then (R∗)n=R+. So, H=R∗
or H=R+.
54. a. In Exercise 4 of Supplementary Exercises for Chapters 1-4,
replace eby 1, aby i,a2by −1, a3by −i,bby −k,ba by −j,
ba2by k, and ba3by j.
b. Use part a of this exercise and part a of Exercise 4 of
Supplementary Exercises for Chapters 1-4.
c. 1 i j k
1 1 i j k
i i 1k j
j j k 1i
k k j i 1
55. By Exercise 9, we know that Kis normal in Land Lis normal in
D4. But V K ={V, R270}whereas KV ={V, R90}. So, Kis not
normal in D4.
56. x(H∩N)x−1=xHx−1∩xNx−1=H∩N. The same argument
works for the intersection of any family of normal subgroups.
57. In S3, let H={(1),(12)}and K={(1),(13)}. Then the set
HK ={(1),(13),(12),(12)(13)}={(1),(13),(12),(132)}does not
contain (13)(12) = (123).
58. By Exercise 55, NM is a subgroup. Also
xNM = (xN)M= (N x)M=N(xM) = N(Mx) = NMx.
59. Let H=hakibe any subgroup of N=hai. Let x∈Gand let
(ak)m∈H. We must show that x(ak)mx−1∈H. Note that
x(ak)mx−1=x(akm)x−1= (xax−1)km = (ar)km = (ak)rm ∈ haki.
(Here we used the normality of Nto replace xax−1by ar.)
60. Use Theorem 9.4.
61. gcd(|x|,|G|/|H|) = 1 implies gcd(|xH|,|G/H|) = 1. But |xH|
divides G/H. Thus |xH|= 1 and therefore xH =H.
9/Normal Subgroups and Factor Groups 86
62. a. Observe that for any gin G,
g(x−1y−1xy)g−1= (gx−1g−1)(gy−1g−1)(gxg−1)(gyg−1)∈S.
b. Observe that xG0yG0=yG0xG0if and only if x−1y−1xy ∈G0.
c.) Observe that xNyN =yN xN implies x−1y−1xyN =N. Thus
x−1y−1xy ∈N.
d. Let h∈Hand g∈G. Then ghg−1h−1∈G0≤Hso that
ghg−1∈Hh =H.
63. By Corollary 4 of Theorem 7.1, xmN= (xN)m=N, so xm∈N.
64. Let Cthe collection of all subgroups of Gof order n. Then, since
|H|=nimplies |xHx−1|=n, we have
x(∩H∈CH)x−1=∩H∈CxHx−1=∩H∈CH.
65. Suppose that Aut(G) is cyclic. Then Inn(G) is also cyclic. So, by
Theorem 9.4, G/Z is cyclic and from Theorem 9.3 it follows that G
is Abelian. This is a contradiction.
66. It follows from Example 5 of this chapter and Theorem 7.2 that
|H|=|H∩K|.
67. Say |gH|=n. Then |g|=nt (by Exercise 37) and |gt|=n. For the
second part observe that every nonidentity element of Zhas infinite
order while 1 + h3ihas order 3 in Z/h3i.
68. The mapping φ:g→xgx−1for all gis an automorphism of Gso
N=φ(N) = xNx−1.
69. Note DKand VKare two of the four cosets but their product is
not one of the four. So, closure fails. This does not contradict
Theorem 9.2 because Kis not normal.
70. Suppose that His a subgroup of S4of order 12 distinct from A4.
Then Example 5 in this chapter and Theorem 7.2 HA4=S4and
|HA4|= 12 ·12/|H∩A4|. It follows that |H∩A4|= 6. But this
contradicts Example 5 of Chapter 7.
71. First note that |G/Z(G)|=|G|/|Z(G)|= 30/5 = 6. By Theorem
7.3, the only groups of order 6 up to isomorphism are Z6and D3.
But G/Z(G) can’t be cyclic for if so, then by Theorem 9.3, G
would be Abelian. In this case we would have Z(G) = G.
9/Normal Subgroups and Factor Groups 87
72. Let g∈Gand let H={e, h}. Then {g, gh}=gH =Hg ={hg, g}.
So, gh =hg.
73. If A5had a normal subgroup of order 2 then, by Exercise 72, the
subgroup has a nonidentity element that commutes with every
element of A5. An element of A5of order 2 has the form (ab)(cd).
But (ab)(cd) does not commute with (abc), which also belong to A5.
74. Since G/H has two elements, call them Hand aH. Let |H|=n
and let g1, g2, . . . , g2nbe some listing of the elements of G. To show
that g1g2···g2nis not in Hit is enough to show that
g1g2···g2nH6=H. Note that g1g2···g2nH=g1Hg2H···g2nHand
that exactly nterms of this product are Hand exactly nare aH,
since these are the only two distinct cosets of Hin G. So,
g1Hg2H···g2nH= (aH)n=aH since nis odd and aH has order 2
in G/H.
75. To prove that His normal, observe that xgpx−1= (xgx−1)p.
Moreover, for any g∈Gwe have (gH)p=gpH=H.
76. The set H∪gH ∪g2Hhas 12 elements and is closed.
77. Since Hhas index 2 in Gis it a normal subgroup of Gand
|G/H|= 2. It follows that for every ain Gwe have (aH)2=H. If
ais an element of Gof order 2n+ 1, then
H=a2n+1H= ((aH)2)naH =aH. Thus, ais in H.
78. Suppose that Z(G) is a maximal subgroup of a group Gand let
a∈Gbut a6∈ Z(G). Since haiZ(G) is a subgroup of G(see
Example 5) we have G=haiZ(G). Because C(a) contains haiand
Z(G) we also have, C(a) = G. But this means a∈Z(G).
79. Observe that for any xin Gthe inner automorphism φxof Gthat
maps Hto the subgroup x−1Hx and |x−1Hx|=|H|. So,
x−1Hx =H.
9/Normal Subgroups and Factor Groups 88
80. First observe that Gand exactly 4 elements of order 5 and elements
of order 25 come in multiples of φ(25) = 20 (see the Corollary of
Theorem 4.4). So, elements of order 5 or 25 account for at most 84
elements of G. By Lagrange’s Theorem every element other
element of Gmust have order 2, 4, 10, 20, 50 or 100. It now follows
from Theorem 4.3 that Ghas a subgroup Kof order 2. a group
with an element of order 10, 20, 50 or 100 also has a subgroup of
order 10 and a subgroup group of order 25 has a subgroup of 2. By
Exercise 79 of this chapter His normal and by Example 5 HK is a
subgroup. By Theorem 7.2 we have |HK|= 10.
89
CHAPTER 10
Group Homomorphisms
1. Note that det (AB) = (det A)(det B).
2. Observe that |ab|=|a||b|.
3. Note that (f+g)0=f0+g0.
4. Let Edenote any even permutation and Oany odd. Observe that
φ(EE) = φ(E) = 0 = 0 + 0 = φ(E) + φ(E). φ(EO) = φ(O) = 1 =
0 + 1 = φ(E) + φ(O). The other cases are similar.
5. Observe for every positive integer rwe have (xy)r=xryrso the
mapping is a homomorphism. When ris odd the kernel is {1}so
the mapping is one-to-one and an isomorphism. When nis even the
kernel is {±1}and the mapping is two-to-one.
6. Recall, R(f+g) = Rf+Rg. Kernel = {0}. No.
7. (σφ)(g1g2) = σ(φ(g1g2)) = σ(φ(g1)φ(g2)) = σ(φ(g1))σ(φ(g2)) =
(σφ)(g1)(σφ)(g2). It follows from Theorem 10.3 that
|G/Ker φ|=|H|and |G/Ker σφ|=|K|. Thus.
[Ker σφ : Ker ] = |Ker σφ/Ker φ|=|H|/|K|.
8. See Exercise 20 of Chapter 5. The kernel is the set of even
permutations in G. When Gis Snthe kernel is Anand from
Theorem 10.3 we have that Sn/Anis isomorphic to {+1,−1}. So,
Anhas index 2 in Snand is normal in Sn. The kernel is the
subgroup of even permutations in G. If the members of Gare not
all even then the coset other than the kernel is the set of odd
permutations in G. All cosets have the same size.
9. φ((g, h)(g0, h0)) = φ((gg0, hh0)) = gg0=φ((g, h))φ((g0, h0)). The
kernel is {(e, h)|h∈H}.
10/Group Homomorphisms 90
10. See Exercise 9 of Chapter 1. The kernel is the subgroup of
rotations in G. If the members of Gare not all rotations then the
coset other than the kernel is the set of reflections in G. All cosets
have the same size.
11. The mapping φ:Z⊕Z→Za⊕Zbgiven by
φ((x, y)) = (xmod a, y mod b) is operation preserving by
Exercise 9 in Chapter 0. If (x, y)∈Ker φ, then x∈ haiand y∈ hbi.
So, (x, y)∈ h(a, 0)i×h(0, b)i. Conversely, every element in
h(a, 0)i×h(0, b)iis in Ker φ. So, by Theorem 10.3,
Z⊕Z→Za⊕Zbis isomorphic to h(a, 0)i×h(0, b)i.
12. x→xmod kis a homomorphism with kernel hki.
13. (a, b)→bis a homomorphism from A⊕Bonto Bwith kernel
A⊕ {e}. So, by Theorem 10.3, (A⊕B)/(A⊕ {e})≈B. Chapter 5.
The kernel is the set of even permutations in G. When Gis Snthe
kernel is Anand from Theorem 10.3 we have that Sn/Anis
isomorphic to {+1,−1}. So, Anhas index 2 in Snand is normal in
Sn. The kernel is the subgroup of even permutations in G. If the
members of Gare not all even then the coset other than the kernel
is the set of odd permutations in G. All cosets have the same size.
14. Observe that φ(6 + 7) = φ(1) = 3 while φ(6) + φ(7) = 8 + 1 = 9.
15. By property 6 of Theorem 10.1, we know
φ−1(9) = 23 + Ker φ={23,3,13}.
16. Observe that such a mapping would be an isomorphism and
isomorphisms preserve order.
17. Suppose φis such a homomorphism. By Theorem 10.3,
Ker φ=h(8,1)i,h(0,1)ior h(0,1)i. In these cases, the element
(1,0) + Ker φin (Z16 ⊕Z2)/Ker φhas order either 16 or 8. So,
(Z16 ⊕Z2)/Ker φis not isomorphic to Z4⊕Z4.
18. No, because of part 3 of Theorem 10.1. No, because the
homomorphic image of a cycle group must be cyclic.
19. Since |Ker φ|is not 1 and divides 17, φis the trivial map.
20. 0 onto Z8; 4 to Z8.
10/Group Homomorphisms 91
21. By Theorem 10.3 we know that |Z30/Ker φ|= 5. So, |Ker φ|= 6.
The only subgroup of Z30 of order 6 is h5i.
22. Let |φ(g)|= 8. By Theorem 10.1 part 3, |g|= 8k. Then |gk|= 8.
To generalize replace 8 by n.
23. a. The possible images are isomorphic to Z1,Z2,Z3,Z4,Z6, and
Z12.
b. h1i ≈ Z36,h2i ≈ Z18,h3i ≈ Z12,h4i ≈ Z9,h6i ≈ Z6, and
h12i ≈ Z3.
24. a. Let φ(1) = k. Then φ(7) = 7kmod 15 = 6 so that k= 3 and
φ(x)=3x.
b. h3i.
c. h5i.
d. 1 + h5i.
25. To define a homomorphism from Z20 onto Z10 we must map 1 to a
generator of Z10. Since there are four generators of Z10 we have
four homomorphisms. (Once we specific that 1 maps to an element
a, the homomorphism is x→xa.) To define a homomorphism from
Z20 to Z10 we can map 1 to any element of Z10. (Be careful here,
these mappings are well defined only because 10 divides 20.)
26. There are four: x→(xmod 2,0); x→(0, x mod 2);
x→(xmod 2, x mod 2); x→(0,0).
27. If φis a homomorphism from Znto Znwith φ(1) = k, then by
property 2 of Theorem 10.1 φ(x) = kx. Moreover, for each kwith
0≤k≤n−1, the mapping φ(x) = kx is a homomorphism.
28. Ker φ=A4. The trivial homomorphism and the one given in
Example 11 are the only homomorphisms. To verify this use
Theorem 10.3 and Exercise 70 of Chapter 9.
29. Say the kernel of the homomorphism is K. By Theorem 10.3,
G/K ≈Z10. So, |G|= 10|K|. In Z10, let H=h2i. By properties 5,
7, and 8 of Theorem 10.2, φ−1(H) is a normal subgroup of Gof
order 2|K|. So, φ−1(H) has index 2. To show that there is a
subgroup of Gof index 5, use the same argument with H=h5i. If
there is a homomorphism from a finite group Gonto Zn, then the
10/Group Homomorphisms 92
same argument shows that Ghas a normal subgroup of index dfor
any divisor Dof n.
30. Use parts 5 and 8 of Theorem 10.2.
31. By property 6 of Theorem 10.1, φ−1(7) = 7Ker φ={7,17}.
32. Write U(30) = U3(30) ×U10(30) and use Exercise 9.
33. By property 6 of Theorem 10.1,
φ−1(11) = 11Ker φ={11,19,27,3}.
34. Write U(40) = U5(40) ×U8(40) and use Exercise 9.
35. φ((a, b)+(c, d)) = φ((a+c, b +d)) = (a+c)−(b+d) =
(a−b)+(c−d) = φ((a, b)) + φ((c, d)). Ker φ={(a, a)|a∈Z}.
φ−1(3) = {(a+ 3, a)|a∈Z}.
36. 4a−4b.
37. Consider the mapping φfrom C∗onto R+, given by φ(x) = |x|.
(Recall from Chapter 0 that |a+bi|=√a2+b2.) By straight
forward algebra we have |xy|=|x||y|. Thus φis a homomorphism
with Ker φ=H. So, by Theorem 10.3, C∗/H is isomorphic to R+.
38. Ker γ= Ker α⊕Ker β.
39. φ(xy)=(xy)6=x6y6=φ(x)φ(y). Ker φ=hcos 60◦+isin 60◦i.
40. h12i;h12i; in general, the kernel is hlcm(m, n)i.
41. Consider the mapping φfrom Kto KN/N given by φ(k) = kN.
Since φ(kk0) = kk0N=kNk0N=φ(k)φ(k0) and kN ∈KN/N,φis
a homomorphism. Moreover, Ker φ=K∩H. So, by Theorem 10.3,
K/(K∩N)≈KN/N .
42. Show that the mapping from G/N to G/M given by gN →gM is
an onto homomorphism with kernel M/N.
43. For each divisor dof kthere is a unique subgroup of Zkof order d,
and this subgroup is generated by φ(d) elements. A homomorphism
from Znto a subgroup of Zkmust carry 1 to a generator of the
subgroup. Furthermore, since the order of the image of 1 must
divide n, so we need consider only those divisors dof kthat also
divide n.
10/Group Homomorphisms 93
44. Uk(n) is the kernel.
45. Let Nbe a normal subgroup of D4. By Lagrange’s Theorem the
only possibilities for |N|are 1, 2, 4, and 8. By Theorem 10.4, the
homomorphic images of D4are the same as the factor groups D4/N
of D4. When |N|= 1, we know N={e}and D4/N ≈D4. When
|N|= 2, then N={R0, R180}, since this is the only normal
subgroup of D4of order 2, and D4/N ≈Z2⊕Z2because D4/N is a
group of order 4 with three elements of order 2. When
|N|= 4,|D4/N|= 2 so D4/N ≈Z2. When |N|= 8, we have
D4/N ≈ {e}.
46. Use Theorem 10.4 and part 3 of Theorem 10.1.
47. It is divisible by 10. In general, if Znis the homomorphic image of
G, then |G|is divisible by n.
48. It is divisible by 30. In general, the order of Gis divisible by the
least common multiple of the orders of all its homomorphic images.
49. It is infinite. Z
50. Let Abe the coefficient matrix of the system. If Ais an n×m
matrix, then matrix multiplication by Ais a homomorphism from
Rminto Rnwhose kernel is S.
51. Let γbe a natural homomorphism from Gonto G/N. Let Hbe a
subgroup of G/N and let γ−1(H) = H. Then His a subgroup of G
and H/N =γ(H) = γ(γ−1(H)) = H.
52. Use Theorem 10.1, part 2.
53. The mapping g→φgis a homomorphism with kernel Z(G).
54. The identity belongs to H. If a, b ∈H, then
α(ab−1) = α(a)α(b−1) = α(a)α(b)−1=β(a)β(b)−1=β(a)β(b−1) = β(ab−1).
So, His a subgroup.
55. Since (f+g)(3) = f(3) + g(3), the mapping is a homomorphism.
The kernel is the set of elements in Z[x] whose graphs pass through
the point (3,0).
56. For the first part use trig identities. The kernel is h2πi.
10/Group Homomorphisms 94
57. Let gbelong to G. Since φ(g) belongs to
Z2⊕Z2=h1,0i∪h0,1i∪h1,1i, it follows that
G=φ−1(h1,0i)∪φ−1(h0,1i)∪φ−1(h1,1i). Moreover, each of these
three subgroups is proper since φis onto and each is normal by
Part 8 of Theorem 10.2.
58. Try g→(gH, gK).
59. By Exercise 56 of Chapter 9, H∩Kis a normal subgroup of G. By
Exercise 41, |H|/|H∩K|=|HK|/|K|=|G|/|K|= 2. Thus,
|G/(H∩K)|= (|G|/|H|)(|H|/|H∩K|)=2·2 = 4. Since
H/(H∩K) and K/(H∩K) are distinct subgroups of G/(H∩K)
of order 2, G/(H∩K) is not cyclic.
60. ns;ns
61. Let Gbe a group of order 77. By Lagrange’s Theorem every
nonidentity of Ghas order 7, 11, or 77. If Ghas an element of
order 77, then Gis cyclic. So, we may assume that all nonidentity
elements of Ghave order 7 or 11. Not all nonidentity elements can
have order 11 because, by the Corollary of Theorem 4.4, the
number of such elements is a multiple of 10. Not all nonidentity
elements of Gcan have order 7 because the number of such
elements is a multiple of 6. So, Gmust have elements aand bsuch
that |a|= 11 and |b|= 7. Let H=hai. Then His the only
subgroup of Gof order 11 for if Kis another one then by Theorem
7.2 |HK|=|H||K|/|H∩K|= 11 ·11/1 = 121. But HK is a subset
of Gand Gonly has 77 elements. Because for every xin G,xHx−1
is also a subgroup of Gof order 11 (see Exercise 1 of the
Supplementary Exercises for Chapters 5-8), we must have
xHx−1=H. So, N(H) = G. Since Hhas prime order, His cyclic
and therefore Abelian. This implies that C(H) contains H. So, 11
divides |C(H)|and |C(H)|divides 77. This implies that C(H) = G
or C(H) = H. If C(H) = G, then |ab|= 77. If C(H) = H, then
|N(H)/C(H)|= 7. But by the “N/C” Theorem (Example 15)
N(H)/C(H) is isomorphic to a subgroup of Aut(H)
≈Aut(Z11)≈U(11) (see Theorem 6.5). Since U(11) = 10, we have
a contradiction.
10/Group Homomorphisms 95
62. There are no homomorphisms from Zonto S3since the image of a
cyclic group must be cyclic. For each element xin S3the mapping
from Zto S3given by φ(n) = xnis a homomorphism. It follows
from part 2 of Theorem 10.2 that there are no others.
63. Let φbe a homomorphism from S3to Zn. Since |φ(S3)|must divide
6 we have that |φ(S3)|= 1,2,3,or 6. In the first case φmaps every
element to 0. If |φ(S3)|= 2, then nis even and φmaps the even
permutations to 0, and the odd permutations to n/2. The case that
|φ(S3)|= 3 cannot occur because it implies that Ker φis a normal
subgroup of order 2 whereas S3has no normal subgroup of order 2.
The case that |φ(S3)|= 6 cannot occur because it implies that φis
an isomorphism from a non-Abelian group to an Abelian group.
64. Observe that
γ(a+b) = nφ(a+b) = φ(na +nb) = φ(na) + φ(nb) = γ(a) + γ(b).
65. φ(zw) = z2w2=φ(z)φ(w). Ker φ={1,−1}and, by Theorem 10.3,
C∗/{1,-1}is isomorphic to C∗.
66. p2. To verify this note that for any homomorphism φfrom Zp⊕Zp
into Zpwe have φ(a, b) = aφ(1,0) + bφ(0,1). Thus we need only
count the number of choices for φ(1,0) and φ(0,1). Since pis prime
we may let φ(1,0) be any element of Zp. The same is true for
φ(0,1).
67. Suppose that His a proper subgroup of Gthat is not properly
contained in a proper subgroup of G. Then G/H has no nontrivial,
proper subgroup. It follows from Exercise 26 of Chapter 7 that
G/H is isomorphic to Zpfor some prime p. But then for every
coset g+Hwe have p(g+H) = Hso that pg ∈Hfor all g∈G.
But then G=pG ⊆H. Both Qand Rsatisfy the hypothesis.
96
CHAPTER 11
Fundamental Theorem of Finite Abelian
Groups
1. n= 4
Z4,Z2⊕Z2
2. n= 8; Z8, Z4⊕Z2, Z2⊕Z2⊕Z2
3. n= 36
Z9⊕Z4,Z3⊕Z3⊕Z4,Z9⊕Z2⊕Z2,Z3⊕Z3⊕Z2⊕Z2
4. order 2: 1, 3, 3, 7; order 4: 2, 4, 12, 8
5. The only Abelian groups of order 45 are Z45 and Z3⊕Z3⊕Z5. In
the first group, |3|= 15; in the second one, |(1,1,1)|= 15.
Z3⊕Z3⊕Z5does not have an element of order 9.
6. Z27 ⊕Z4;Z27 ⊕Z2⊕Z2
7. In order to have exactly four subgroups of order 3, the group must
have exactly 8 elements of order 3. When counting elements of
order 3 we may ignore the components of the direct product that
represent the subgroup of order 4 since their contribution is only
the identity. Thus, we examine Abelian groups of order 27 to see
which have exactly 8 elements of order 3. By Theorem 4.4, Z27 has
exactly 2 elements of order 3; Z9⊕Z3has exactly 8 elements of
order 3 since for |(a, b)|= 3 we can choose |a|= 1 or 3 and |b|= 1
or 3, but not both |a|and |b|of order 1; In Z3⊕Z3⊕Z3every
element except the identity has order 3. So, the Abelian groups of
order 108 that have exactly four subgroups of order 3 are
Z9⊕Z3⊕Z4and Z9⊕Z3⊕Z2⊕Z2. The subgroups of
Z9⊕Z3⊕Z4of order 3 are h(3,0,0)i,h(0,1,0)i,h(3,1,0)iand
h(3,2,0)i. The subgroups of Z9⊕Z3⊕Z2⊕Z2of order 3 are
h(3,0,0,0)i,h(0,1,0,0)i,h(3,1,0,0)iand h(3,2,0,0)i.
8. Z3⊕Z3⊕Z3⊕Z4;Z3⊕Z3⊕Z3⊕Z2⊕Z2
11/Fundamental Theorem of Finite Abelian Groups 97
9. Elements of order 2 are determined by the factors in the direct
product that have order a power of 2. So, we need only look at
Z8, Z4⊕Z2and Z2⊕Z2⊕Z2. By Theorem 4.4, Z8has exactly one
element of order 2; Z4⊕Z2has exactly three elements of order 2;
Z2⊕Z2⊕Z2has exactly 7 elements of order 2. So,
G≈Z4⊕Z2⊕Z3⊕Z5.
10. Z8⊕Z9⊕Z5;Z4⊕Z2⊕Z9⊕Z5;Z2⊕Z2⊕Z2⊕Z9⊕Z5;Z8⊕Z3⊕
Z3⊕Z5;Z4⊕Z2⊕Z3⊕Z3⊕Z5;Z2⊕Z2⊕Z2⊕Z3⊕Z3⊕Z5.
11. By the Fundamental Theorem, any finite Abelian group Gis
isomorphic to some direct product of cyclic groups of prime-power
order. Now go across the direct product and, for each distinct
prime you have, pick off the largest factor of that prime-power.
Next, combine all of these into one factor (you can do this, since
their orders are relatively prime). Let us call the order of this new
factor n1. Now repeat this process with the remaining original
factors and call the order of the resulting factor n2. Then n2
divides n1, since each prime-power divisor of n2is also a
prime-power divisor of n1. Continue in this fashion. Example: If
G≈Z27 ⊕Z3⊕Z125 ⊕Z25 ⊕Z4⊕Z2⊕Z2,
then
G≈Z27·125·4⊕Z3·25·2⊕Z2.
Now note that 2 divides 3 ·25 ·2 and 3 ·25 ·2 divides 27 ·125 ·4.
12. By the corollary to the Fundamental Theorem of Finite Abelian
Groups the given group has a subgroup of order 10. But this group
must be isomorphic to Z2⊕Z5≈Z10.
13. Z2⊕Z2
14. If Gis an Abelian group of order nand mis a divisor of n, then G
has a cyclic subgroup of order mif mis squarefree (i.e., each prime
factor of moccurs to the 1st power only).
15. a. 1 b. 1 c. 1 d. 1 e. 1 f. There is a unique Abelian
group of order nif and only if nis not divisible by the square of
any prime.
16. a. same b. same c. same d. same
(e) twice as many of order mcompared with the number of order n
11/Fundamental Theorem of Finite Abelian Groups 98
17. This is equivalent to asking how many Abelian groups of order 16
have no element of order 8. From the Fundamental Theorem of
Finite Abelian Groups the only choices are Z4⊕Z4, Z4⊕Z2⊕Z2,
and Z2⊕Z2⊕Z2⊕Z2.
18. 5n
19. The symmetry group is {R0, R180, H, V }. Since this group is
Abelian and has no element of order 4, it is isomorphic to Z2⊕Z2.
20. Consider every possible isomorphism class one by one and show
each has the desired subgroup. For instance, in
Z2⊕Z2⊕Z2⊕Z27 ⊕Z5the subgroup is
{(a, b, 0, c, d)|a, b ∈Z2, c ∈ h3i, d ∈Z5}
21. Because the group is Abelian and has order 9, the only possibilities
are Z9and Z3⊕Z3. Since Z9has exactly 2 elements of order 3 and
9, 16, and 22 have order 3, the group must be isomorphic to
Z3⊕Z3.
22. Because of the Fundamental Theorem and Corollary 1 of Theorem
8.2, we may assume |G|is a prime-power. Let xbe an element of G
of maximum order. Then for any yin Gwe have |y|divides |hxi|.
Since hxihas a unique subgroup for each divisor of |hxi| it follows
that hyi ⊆ hxi.
23. By the Corollary of Theorem 8.2, nmust be square-free (no prime
factor of noccurs more than once).
24. n=p2
1p2
2or p2
1p2
2p3p4···pkwhere k≥3 and p1, p2, . . . , pkare
distinct primes.
25. Among the first 11 elements in the table, there are 9 elements of
order 4. None of the other isomorphism classes has this many.
26. Z4⊕Z2; One internal direct product is h7i×h17i.
27. First observe that Gis Abelian and has order 16. Now we check
the orders of the elements. Since the group has 8 elements of order
4 and 7 of order 2 it is isomorphic to Z4⊕Z2⊕Z2. One internal
direct product is h7i×h101i×h199i.
28. Z2⊕Z2⊕Z3; One internal direct product is h19i×h26i×h31i.
11/Fundamental Theorem of Finite Abelian Groups 99
29. Since Z9has exactly 2 elements of order 3 once we choose 3
nonidentity elements we will either have at least one element of
order 9 or 3 elements of order 3. In either case we have determined
the group. The Abelian groups of order 18 are Z9⊕Z2≈Z18 and
Z3⊕Z3⊕Z2. By Theorem 4.4, Z18 group has 6 elements of order
18, 6 elements of order 9, 2 of order 6, 2 of order 3, 1 of order 2,
and 1 of order 1. Z3⊕Z3⊕Z2has 8 elements of order 3, 8 of order
6, 1 of order 2, and 1 of order 1. The worst case scenario is that at
the end of 5 choices we have selected 2 of order 6, 2 of order 3, and
1 of order 2. In this case we still have not determined which group
we have. But the sixth element we select will give us either an
element of order 18 or 9, in which case we know the group Z18 or a
third element of order 6 or 3, in which case we know the group is
Z3⊕Z3⊕Z2.
30. The element of order 8 rules out all but Z16 and Z8⊕Z2and two
elements of order 2 precludes Z16.
31. If a26=b2, then a6=band a6=b3. It follows that hai∩hbi={e}.
Then G=hai×hbi ≈ Z4⊕Z4.
32. Observe that the elements of order 2 together with the identity
form a subgroup.
33. By Theorem 11.1, we can write the group in the form
Zp1n1⊕Zp2n2⊕ ··· ⊕ Zpknkwhere each piis an odd prime. By
Theorem 8.1 the order of any element
(a1, a2, . . . , ak) = lcm(|a1|,|a2|,...,|ak|). And from Theorem 4.3 we
know that |ai|divides pni
i, which is odd.
34. Z2⊕Z2⊕ ··· ⊕ Z2(nterms)
35. By Theorem 7.2 we have,
|haiK|=|a||K|/|hai ∩ K|=|a||K|=|a||K|p=|G|p=|G|.
36. If |G|=pn, use the Fundamental Theorem and Theorem 9.6. If
every element has order a power of puse the corollary to the
Fundamental Theorem.
37. By the Fundamental Theorem of Finite Abelian Groups, it suffices
to show that every group of the form Zpn1
1⊕Zpn2
2⊕ ··· ⊕ Zpnk
kis a
subgroup of a U-group. Consider first a group of the form
Zpn1
1⊕Zpn2
2(p1and p2need not be distinct). By Dirichlet’s
11/Fundamental Theorem of Finite Abelian Groups 100
Theorem, for some sand tthere are distinct primes qand rsuch
that q=tpn1
1+ 1 and r=spn2
2+ 1. Then
U(qr) = U(q)⊕U(r)≈Ztpn1
1⊕Zspn2
2, and this latter group
contains a subgroup isomorphic to Zpn1
1⊕Zpn2
2. The general case
follows in the same way.
38. Observe that Aut(Z2⊕Z3⊕Z5)≈Aut(Z30)≈U(30) ≈
U(2) ⊕U(3) ⊕U(5) ≈Z2⊕Z4.
39. It follows from Exercise 4 of Chapter 8 and Theorem 9.6 that if D4
could be written in the form hai × Kwhere |a|= 4 it would be
Abelian.
101
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11
1. Say aH =Hb. Then a=hb for some hin H. Then
Ha =Hhb =Hb =aH.
2. Let Gbe a finite Abelian group of order nand let pbe a prime
divisor of m. By Theorem 9.5, Ghas an element of order p, call it
a. Then, by induction, G/haihas a subgroup H/haiof order m/p.
This His a subgroup of order m.
3. Suppose diag(G) is normal. Then
(e, a)(b, b)(e, a)−1= (b, aba−1)∈diag(G). Thus b=aba−1and
therefore ba =ab. If Gis Abelian
(g, h)(b, b)(g, h)−1= (gbg−1, hbh−1)=(b, b). The index of diag(G)
is |G|.
4. Let x∈D4and let Kbe the cyclic subgroup of D4consisting of all
rotations in D4. Then xHx−1is a group of rotations and
|H|=|xHx−1|. Now use Theorem 4.3.
5. Let α∈Aut(G) and φa∈Inn(G). Then
(αφaα−1)(x)=(αφa)(α−1(x)) = α(aα−1(x)a−1) = α(a)xα(a−1) =
α(a)x(α(a))−1=φα(a)(x).
6. Assume that His normal and ab ∈H. Then
ba =a−1(ab)a∈a−1Ha =H. For the other half, let h∈H. Then
h=x−1(xh)∈H. So, (xh)x−1∈H.
7. R∗(See Example 2 of Chapter 10.)
8. Zk.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 102
9. a. To determine the center of Hsuppose that
1a b
0 1 c
0 0 1
is in the center of H. Then for all choices of a0, b0and c0we have
1a b
0 1 c
0 0 1
1a0b0
0 1 c0
0 0 1
=
1a0+a b0+ac0+b
0 1 c0+c
0 0 1
And,
1a0b0
0 1 c0
0 0 1
1a b
0 1 c
0 0 1
=
1a+a0b+a0c+b0
0 1 c+c0
0 0 1
.
So, b0+ac0+b=b+a0c+bfor all choices of a0, b0and c0. Thus,
a0c=ac0for all choices of a0and c0. Taking a0= 0 and c0= 1 gives
a= 0. Taking a0= 1 and c0= 0 gives c= 0. Finally, we note that
1 0 b
010
001
does commute with every element of Hso Z(H)
consists of matrices of this form. Thus,
Z(H) =
1 0 b
0 1 0
0 0 1
b∈Q
.
b. The mapping
φ
1 0 b
0 1 0
0 0 1
=b
is an isomorphism. φis 1-1 and onto by observation. To see that φ
is operation preserving note that
φ
1 0 b
0 1 0
0 0 1
1 0 b0
0 1 0
0 0 1
=φ
1 0 bb0
0 1 0
0 0 1
=bb0=
φ
1 0 b
0 1 0
0 0 1
φ
1 0 b0
0 1 0
0 0 1
.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 103
c. To prove part c we define the mapping
φ
1a b
0 1 c
0 0 1
= (a, c).
By observation, φis onto Q⊕Q. To see that φis operation
preserving note that
φ
1a b
0 1 c
0 0 1
1a0b0
0 1 c0
0 0 1
=
φ
1a+a0b0+ac0+b
0 1 c+c0
0 0 1
= (a+a0, c0+c)=(a, c)(a0, c0) =
φ
1a b
0 1 c
0 0 1
φ
1a0b0
0 1 c0
0 0 1
.
Obviously, the kernel of φis Z(H). So, by Theorem 10.3,
H/Z(H)≈Q⊕Q.
d. The proofs are valid for Rand Zp.
10. Note that the factor group has order 4 and that the square of any
element in the factor group is the identity. Now use Exercise 47.
11. Every element of Q/Z can be written in the form a/b +Zwhere b
is a positive integer. Thus b(a/b +Z) = a+Z=Zso that a/b +Z
has order at most |b|.
12. Let Cbe a collection of normal subgroups. Then
x(∩H∈CH)x−1=∩H∈CxHx−1=∩H∈CH.
13. For h∈Hand x∈Gnote that (xhx−1)n=xhnx−1. So,
|xhx−1|=nif and only if |h|=n. For Hto be a subgroup nmust
be prime. To verify this note that if nis not prime we may write
n=mp where pis a prime and m > 1. But then if h∈H, we have
that |hm|=pso that hmis a not in H. For the first example, take
Dpwhere pis any odd prime. For the second exmaple, use observe
that in D4,|D|= 2 and |V|= 2 but DV =R90 has order 4.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 104
14. First observe that h1/n +Zihas order n. Now suppose that H/Z is
any subgroup of Q/Z of order n. Let a/b ∈Hwhere aand bare
integers and gcd(a, b) = 1. Then n(a/b)∈Zand it follows that b
divides n. Say, n=bq. Then a/b =aq/n. Thus
a/b +Z=aq(1/n +Z)∈ h1/n +Zi. So, H/Z ⊆ h1/n +Zi.
15. Observe that hkh−1k−1= (hkh−1)k−1∈Kand
hkh−1k−1=h(kh−1k−1)∈H. So, hkh−1k−1=eand therefore
hk =kh.
16. Let |G|=nand suppose x2=y2. Write 1 = 2t+ns. Then
x=x2t+ns =x2t=y2t=y2t+ns =y.
17. By Theorem 7.3, stabG(5) has index 2 and by Exercise 9 of
Chapter 9, stabG(5) is normal.
18. Since every g∈Gcan be written in the form g=hk where h∈H
and k∈Kand kcommutes with the elements of Hwe have
gN =hkN =hNk =Nhk =N g.
19. Suppose that φis a homomorphism from Z8⊕Z2⊕Z2onto
Z4⊕Z4. It follows from Theorem 10.3 that
(Z8⊕Z2⊕Z2)/Ker φ≈Z4⊕Z4and |Ker φ|= 2. Now observe
that φ((4,0,0)) = φ(4(1,0,0)) = 4φ(1,0,0) = (0,0), so that
Ker φ={(0,0,0),(4,0,0)}. However, in (Z8⊕Z2⊕Z2)/Ker φeach
of the four distinct elements
(1,0,0) + Ker φ, (1,1,0) + Ker φ, (1,0,1) + Ker φ, and
(1,1,1) + Ker φhas order 4 whereas Z4⊕Z4has only three
elements of order 4.
20. Suppose there is a homomorphism φonto a group of order 2 or 4.
Then |Ker φ|= 6 or 3. But for any element βof order 3, |φ(β)|
must divide both 3 and 2 or 4. Thus |φ(β)|= 1. Since A4has 8
elements of order 3, we have a contradiction. For the case that the
image has order 6 observe that the kernel would have order 2 so
that the mapping is 2 to 1. Since the 8 elements of order 3 in A4
must map to elements of order 3 or 1 and the image has only 3 such
elements, we have a contradiction. For the last portion use
Theorem 10.4.
21. Since |S4/H|= 6, by Theorem 7.3, S4/H is either cyclic or
isomorphic to D3(which is isomorphic to S3). But S4has no
element of order 6 (see Exercise 7 in Chapter 5).
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 105
22. By part 5 of Theorem 10.1, we obtain 5, 5 ·13 = 29,5·25 = 17.
23. Recall that when nis even, Z(Dn) = {R0, R180}. For each of the n
reflections Fin Dnthe element F Z(Dn) in Dn/Z(Dn) has order 2.
The subgroup hR360/ni/Z(Dn) has order n/2 = m. Since mis odd
the only elements in Dmof order 2 are the mreflections.
24. By the N/C Theorem, |G/C(H)|divides |Aut(H)|= 4. Since |G|is
odd, it follows that C(H) = G.
25. The mapping g→gnis a homomorphism from Gonto Gnwith
kernel Gn. So, by Theorem 10.3, G/Gn≈Gn.
26. Let r=√a2+b2. Then a+bi =r(a
r+bi
r).
27. Let |H|=p. Theorem 7.2 shows that His the only subgroup of
order p. But xHx−1is also a subgroup of order p. So xHx−1=H.
28. Use the subgroup test to show that His a subgroup. |G/H|= 4.
Z2⊕Z2.
29. Say aand bare integers and a/b +Zhas order nin Q/Z. Then
na/b =mfor some integer m. Thus,
a/b +Z=m/n +Z=m(1/n +Z)∈ h1/n +Zi.
30. First observe that xex−1=e=exx−1implies
φ(x)φ(e)φ(x−1) = φ(e)φ(x)φ(x−1). By right cancellation it follows
that φ(e)∈Z(G). Next define ψ(x) = φ(e)−1φ(x). To show that ψ
is a homomorphism we must show that
φ(e)−1φ(x)φ(e)−1φ(y) = φ(e)−1φ(xy). Canceling and using the fact
that φ(e)−1∈Z(G) it is enough to show that
φ(x)φ(y) = φ(e)φ(xy). Note that xy(xy)−1=e=eee implies that
φ(x)φ(y)φ((xy)−1) = φ(e)φ(e)φ(e).Moreover,
(xy)(xy)−1e=e=eee implies that
φ(xy)φ((xy)−1)φ(e) = φ(e)φ(e)φ(e). So,
φ(xy)φ((xy)−1) = φ(e)φ(e). Then , by substitution, it follows that
φ(x)φ(y)φ((xy)−1) = φ(e)φ(xy)φ((xy)−1). Finally, cancel the term
φ((xy)−1) on each side.
31. Let φbe a homomorphism from Z⊕Zinto Zand let φ((1,0)) = a
and let φ((0,1)) = b. Then
φ((x, y)) = φ(x(1,0) + y(0,1)) = xφ((1,0)) + yφ((0,1)) = ax +by.
32. Mimic Exercise 31.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 106
33. First note that by Exercise 11 every element in Q/Z has finite
order. For each positive integer n, let Bndenote the set of elements
of order nand suppose that φis an isomorphism from Q/Z to
itself. Then, by property 5 of Theorem 6.2, φ(Bn)⊆Bn. By
Exercise 29 we know that Bnis finite, and since φpreserves orders
and is one-to-one, we must have φ(Bn) = Bn. Since it follows from
Exercise 11 and Exercise 29 of Supplementary Exercises for
Chapters 9–11 that Q/Z =SBn, where the union is taken over all
positive integers n, we have φ(Q/Z) = Q/Z.
34. Use Exercises 29 and Theorem 4.4.
35. If the group is not Abelian, for any element anot in the center, the
inner automorphism induced by ais not the identity. If the group is
Abelian and contains an element awith |a|>2, then x→x−1
works; if every nonidentity element has order 2, then Gis
isomorphic to a group of the form Z2⊕Z2⊕ ··· ⊕ Z2. In this case,
the mapping that takes (a1, a2, a3, . . . , ak) to (a2, a1, a3, . . . , ak) is
an automorphism that is not the identity.
36. Suppose Kis a maximal subgroup of Q. Then Q/K is an Abelian
group whose only subgroups are itself and the identity. It follows
that Q/K ≈Zpfor some prime p. But then for any element
a+K∈Q/K we have p(a+K) = pa +K=K. This means that
pQ ⊆K. But pQ =Q.
37. Since aH, bH and baH are distinct and have order 2, G/H is
isomorphic to Z2⊕Z2.G/H is not isomorphic to a subgroup of G
since Ghas only one element of order 2.
38. Since reflections are their own inverses that Nis a normal subgroup
follows from Exercise 44 of Chapter 2 and the observation that
rotations commute with rotations. Since both F1Nand F5Nhave
order 2, D8/N is not cyclic.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 9-11 107
39. Observe that "1x
0 1 #" 1y
0 1 #="1x+y
0 1 #, so His closed.
Also, "1x
0 1 #−1
="1−x
0 1 #, which is in H. Thus,
His a subgroup of G.
Since "1a
0b#" 1x
0 1 #" 1a
0b#−1
=
"1a
0b#" 1x
0 1 #" 1−ab−1
0b−1#="1b−1x
0 1 #belongs to
H, we have that His normal in G.
40. Let H={x∈Zp2⊕Zp2|xp= (0,0)}. Then |H|=p2and every
nonidentity element of (Zp2⊕Zp2)/H has order p.
41. Let gbelong to G. Since gKg−1=K, conjugation is an
automorphism of K. Thus gHg−1=H.
108
CHAPTER 12
Introduction to Rings
1. For any n > 1, the ring M2(Zn) of 2 ×2 matrices with entries from
Znis a finite noncommutative ring. The set M2(2Z) of 2 ×2
matrices with even integer entries is an infinite noncommutative
ring that does not have a unity.
2. 6
3. In R,{n√2|n∈Z}is a subgroup but not a subring.
4. In Z4, 2x= 2 has solutions 1 and 3. In a group, x=a−1b.
5. The proof given in Theorem 2.1 for the uniqueness of the identity
in a group applies to the unity in a ring as well. The proof in
Theorem 2.3 of the uniqueness of inverses in groups is the same for
uniqueness of inverses in rings except we multiply ab =ac on the
left by b.
6. Consider Znwhere nis not prime.
7. First observe that every nonzero element ain Zphas a
multiplicative inverse a−1. For part a, if a6= 0, then a2=aimplies
that a−1a2=a−1aand therefore a= 1. For part b, if a6= 0, then
ab = 0 implies that b=a−1(ab) = a−10 = 0. For part c, ab =ac
implies that a−1(ab) = a−1(ac). So b=c.
8. Consider aba =aba.
9. If aand bbelong to the intersection, then they belong to each
member of the intersection. Thus a−band ab belong to each
member of the intersection. So, a−band ab belong to the
intersection.
10. Observe that all the sets in the examples are closed under
subtraction and multiplication.
12/Introduction to Rings 109
11. Part 3:
0 = 0(−b)=(a+(−a))(−b) = a(−b)+(−a)(−b) = −(ab)+(−a)(−b).
So, ab = (−a)(−b).
Part 4:
a(b−c) = a(b+ (−c)) = ab +a(−c) = ab + (−(ac)) = ab −ac.
Part 5: By Part 2, (−1)a= 1(−a) = −a.
Part 6: By Part 3, (−1)(−1) = 1 ·1 = 1.
12. If c=db, then c=d(a−1)ab = (da−1)ab.
If c= (ab)d, then c= (ad)b.
13. Let Sbe any subring of Z. By definition of a ring, Sis a subgroup
under addition. By Theorem 4.3, S=hkifor some integer k.
14. Use induction.
15. If mor nis 0 the statement follows from part 1 of Theorem 12.1.
For simplicity, for any integer kand any ring element xwe will use
kx instead of k·x. Then for positive mand n, observe that
(ma)(nb) = (a+a+···+a)+(b+b+···+b)=(ab +ab +···+ab),
where the terms a+··· +a, b +b+··· +b, and the last term have
mn summands.
For the case that mis positive and nis negative, we first observe
that nb means (−b)+(−b) + ··· + (−b) = (−n)(−b). So,
nb + (−n)b= ((−b)+(−b) + ··· + (−b)) + (b+b+··· +b) = 0.
Thus, 0 = (ma)(nb + (−n)b)=(ma)(nb)+(ma)(−n)b=
(ma)(nb) + m(−n)ab = (ma)(nb)+(−(mn))ab. So, adding (mn)ab
to both ends of this string of equalities gives (mn)ab = (ma)(nb).
For the case when mis negative and nis positive just reverse the
roles of mand nis the preceding argument. If both mand nare
negative, note that
(ma)(nb) = ((−a)+(−a) + ··· + (−a))((−b)+(−b) + ··· + (−b)) =
((−m)(−a))((−n)(−b)) = (−m)(−n)((−a)(−b)) = (mn)(ab).
16. Observe that n·(−a) + n·a= 0.
17. From Exercise 15, we have
(n·a)(m·a) = (nm)·a2= (mn)·a2= (m·a)(n·a).
18. Let a, b ∈S. Then (a−b)x=ax −bx = 0 −0 = 0. Also
(ab)x=a(bx) = a·0 = 0.
12/Introduction to Rings 110
19. Let a, b belong to the center. Then
(a−b)x=ax −bx =xa −xb =x(a−b). Also,
(ab)x=a(bx) = a(xb)=(ax)b= (xa)b=x(ab).
20. (" a b
c d #|ad −bc =±1)
21. (x1, . . . , xn)(a1, . . . , an)=(x1, . . . , xn) for all xiin Riif and only if
xiai=xifor all xiin Riand i= 1, . . . , n and xiai=xifor all xiin
Riif and only if xiis a unity of Ri.
22. By the One-Step Test we must show ab−1is a unit wherever aand
bare. But ab−1ba−1= 1.
23. By observation ±1 and ±iare units. To see that there are no
others note that (a+bi)−1=1
a+bi =1
a+bi
a−bi
a−bi =a−bi
a2+b2. But a
a2+b2is
an integer only when a2+b2= 1 and this holds only when a=±1
and b= 0 or a= 0 and b=±1.
24. Say eiis the unity of Ri.Then
(a1, . . . , an)(b1, . . . , bn)=(e1, . . . , en)
if and only if aibi=eifor i= 1, . . . , n. That is, if and only if
ai∈U(Ri).
25. Note that the only f(x)∈Z[x] for which 1/f(x) is a polynomial
with integer coefficients are f(x) = 1 and f(x) = −1.
26. {f(x) = c|c∈R, c 6= 0}.
27. If ais a unit, then b=a(a−1b).
28. In Z6, 4 ·2 = 2; in Z8, 3 ·5 = 7, in Z15, 9 ·3 = 12.
29. Note that (a+b)(a−1−a−2b)=1−a−1b+ba−1−a−2b2= 1.
30. If m<n, then a=an=aman−m= 0an−m= 0. If m=n, then
a=an=am= 0. If m>n, observe that
a=an= (aa ···a)n= (anan···an)n=ann. By repeating this
calculation we see that there are arbitrary large kso that a=ak.
Thus we may reduce to the first case.
31. In M2(Z), let a="0 1
0 0 #and b="1 0
0 0 #.
12/Introduction to Rings 111
32. ba = (ba)n=b(ab)a(ba)n−2= 0.
33. Note that 2x= (2x)3= 8x3= 8x.
34. Use induction.
35. For Z6use n= 3. For Z10 use n= 5. Say m=p2twhere pis a
prime. Then (pt)n= 0 in Zmsince mdivides (pt)n.
36. Say k=ms and k=nt. Then ka = (ms)a=m(sa) and
ka = (nt)a=n(ta) and therefore ka ∈mZ ∩nZ. Now suppose
b∈mZ ∩nZ. Then bis a common multiple of mand n. So, by
Exercise 10 of Chapter 0, b∈kZ.
37. Every subgroup of Znis closed under multiplication.
38. No. The operations are different.
39. Since ara −asa =a(r−s)aand (ara)(asa) = ara2sa =arsa,Sis a
subring. Also, a1a=a2= 1, so 1 ∈S.
40. The set is not closed under multiplication.
41. Let "a a −b
a−b b #and "a0a0−b0
a0−b0b0#∈R. Then
"a a −b
a−b b #−"a0a0−b0
a0−b0b0#=
"a−a0(a−a0)−(b−b0)
(a−a0)−(b−b0)b−b0#∈R. Also,
"a a −b
a−b b #" a0a0−b0
a0−b0b0#=
"aa0+aa0−ab0−ba0+bb0aa0−bb0
aa0−bb0aa0−ab0−ba0+bb0+bb0#
belongs to R.
42. The subring test is satisfied.
43. Sis not a subring because (1,0,1) and (0,1,1) belong to Sbut
(1,0,1)(0,1,1) = (0,0,1) does not belong to S.
44. Say n= 2m. Then
−a= (−a)n= (−a)2m= ((−a)2)m= ((a)2)m=a2m=an=a.
12/Introduction to Rings 112
45. Observe that n·1−m·1=(n−m)·1. Also,
(n·1)(m·1) = (nm)·((1)(1)) = (nm)·1.
46. 2Z∪3Zcontains 2 and 3, but not 2 + 3.
47. S={m/2n|m∈Z, n ∈Z+}contains 1/2. Since
m/2n−m0/2n0= (m2n0−2nm0)/2n+n0∈Sand
(m/2n)(m0/2n0) = mm0/2n+n0∈S, the subring test is satisfied. If T
is any subring that contains 1/2 then by closure under
multiplication it contains 1/2nand by closure under addition (and
subtraction) it contains m/2n. So, Tcontains S.
48. {an(2/3)n+an−1(2/3)n−1+··· +a1(2/3) |a1, a2, . . . , an∈
Z, n a positive integer}.
49. (a+b)(a−b) = a2+ba −ab −b2=a2−b2if and only if ba −ab = 0.
50. First note that
a+b= (a+b)2=a2+ab +ba +b2=a+ab +ba +b
so that 0 = ab +ba or −ab =ba. Then observe that
−ab = (−ab)2= (ab)2=ab.
51. Z2⊕Z2;Z2⊕Z2⊕ ··· (infinitely many copies).
52. 2x= 1 has no solution in Z4; 2x= 0 has two solution in Z4;
x=a−1(c−b) is the unique when a−1exists.
113
CHAPTER 13
Integral Domains
1. For Example 1, observe that Zis a commutative ring with unity 1
and has no zero divisors. For Example 2, note that Z[i] is a
commutative ring with unity 1 and no zero divisors since it is a
subset of C, which has no zero divisors. For Example 3, note that
Z[x] is a commutative ring with unity h(x) = 1 and if
f(x) = anxn+···+a0and g(x) = bmxm+···+b0with an6= 0 and
bm6= 0, then f(x)g(x) = anbmxn+m+··· +a0b0and anbm6= 0. For
Example 4, elements of Z[√2] commute since they are real
numbers; 1 is the unity; (a+b√2) −(c+d√2) = (a−c) + (b−d)√2
and (a+b√2)(c+d√2) = (ac + 2bd)+(bc +ad)√2 so Z[√2] is a
ring; Z[√2] has no zero divisors because it is a subring of R, which
has no zero divisors. For Example 5, note that Zpis closed under
addition and multiplication and multiplication is commutative; 1 is
the unity; In Zp,ab = 0 implies that pdivides ab. So, by Euclid’s
Lemma (see Chapter 0), we know that pdivides aor pdivides b.
Thus, in Zp,a= 0 or b= 0. For Example 6, if nis not prime, then
n=ab where 1 < a < n and 1 < b < n. But then a6= 0 and b6= 0
while ab = 0. For Example 7, note that
"1 0
0 0 #" 0 0
0 1 #="0 0
0 0 #.
For Example 8, note that (1,0)(0,1) = (0,0).
2. Example 5
3. Let ab = 0 and a6= 0. Then ab =a·0, so b= 0.
4. 2,4,5,6,8,10,12,14,15,16,18. The zero-divisors and the units
constitute a partition of Z20.
5. Let k∈Zn. If gcd(k, n) = 1, then kis a unit. If gcd(k, n) = d > 1,
write k=sd. Then k(n/d) = sd(n/d) = sn = 0.
13/Integral Domains 114
6. xin Z[x].
7. Let s∈R,s6= 0. Consider the set S={sr|r∈R}. If S=R, then
sr = 1 (the unity) for some r. If S6=R, then there are distinct r1
and r2such that sr1=sr2. In this case, s(r1−r2) = 0. To see
what happens when the “finite” condition is dropped, note that in
the ring of integers 2 is neither a zero-divisor nor a unit.
8. Suppose that ais a zero-divisor and let ab = 0 for some b6= 0.
Then a2b=a(ab) = 0 and b6= 0. Next assume that a2b= 0 for
some b6= 0. If ab = 0, then ais a zero-divisor. If ab 6== 0, then
a2b=a(ab) = 0 and we are done.
9. Take a= (1,0,0), b = (0,1,0) and c= (0,0,1).
10. The set of zero-divisors is {(a, b, c)|exactly one or two entries are
0}; The set of units is {(a, b, c)|a, c ∈ {1,−1}, b 6= 0}.
11. (a1+b1√d)−(a2+b2√d) = (a1−a2)+(b1−b2)√d;
(a1+b1√d)(a2+b2√d)=(a1a2+b1b2d)+(a1b2+a2b1)√d. Thus
the set is a ring. Since Z[√d] is a subring of the ring of complex
numbers, it has no zero-divisors.
12. 1
2= 4,−2
3= 4,√−3 = 2; −1
6= 1.
13. The ring of even integers does not have a unity.
14. Look in Z6.
15. (1 −a)(1 + a+a2+··· +an−1) =
1 + a+a2+··· +an−1−a−a2− ··· − an= 1 −an= 1 −0 = 1.
16. If an= 0 and bm= 0, consider (a−b)n+m.
17. Suppose a6= 0 and an= 0, where we take nto be as small as
possible. Then a·0 = 0 = an=a·an−1, so by cancellation,
an−1= 0. This contradicts the assumption that nwas as small as
possible.
18. a2=aimplies a(a−1) = 0.
19. If a2=aand b2=b, then (ab)2=a2b2=ab. The other cases are
similar.
13/Integral Domains 115
20. Note that if ncan be written is the form p2mwhere pis a prime,
then (pm)2=p2m2=nm = 0 in Znand pm is not 0. On the other
hand, if nis of the form p1p2···ptwhere the piare distinct primes
and ak= 0 mod nit follows from Euclid’s Lemma that each pi
divides a. Thus a= 0 in Zn.
21. Let f(x) = xon [−1,0] and f(x) = 0 on (0,1] and g(x) = 0 on
[−1,0] and g(x) = xon (0,1]. Then f(x) and g(x) are in Rand
f(x)g(x) = 0 on [−1,1].
22. We proceed by induction. The n= 1 case is trivial. Assume that
an=a. Then an+1 =aan=aa =a.
23. Suppose that ais an idempotent and an= 0. By the previous
exercise, a= 0.
24. (2 + i)(2 −i) = 0; (3 + 4i)2= 3 + 4i.
25. (3 + 4i)2= 3 + 4i.
26. Units: (1,1),(1,5),(2,1),(2,5);
zero-divisors: {(a, b)|a∈ {0,1,2}, b ∈ {2,3,4}};
idempotents: {(a, b)|a= 0,1, b = 1,3,4};
nilpotents: (0,0).
27. a2=aimplies a(a−1) = 0. So if ais a unit, a−1 = 0 and a= 1.
28. a. fis a zero-divisor if fis not the zero function and f(x) = 0 for
some xb. f(x) = 0 c. If f(x) is never 0, then 1/f(x) is defined for
all x.
29. Since Fis commutative so is K. The assumptions about Ksatisfy
the conditions for the One-Step Subgroup Test for addition and for
multiplication (excluding the 0 element). So, Kis a subgroup
under addition and a subgroup under multiplication (excluding 0).
Thus Kis a subring in which every nonzero element in a unit.
30. The proof that Q[√d] is an integral domain is the same as in
Exercise 11. Moreover,
(a+b√d)−1=a/(a2−db2)−(b/(a2−db2)√d.
31. Note that ab = 1 implies aba =a. Thus 0 = aba −a=a(ba −1).
So, ba −1 = 0.
13/Integral Domains 116
32. 6 is the unity; 4 and 6 are their own inverses and 2 and 8 are
inverses of each other.
33. A subdomain of an integral domain Dis a subset of Dthat is an
integral domain under the operations of D. To show that Pis a
subdomain, note that n·1−m·1=(n−m)·1 and
(n·1)(m·1) = (mn)·1 so that Pis a subring of D. Moreover,
1∈P, P has no zero divisors since Dhas none, and Pis
commutative because Dis. Also, since every subdomain contains 1
and is closed under addition and subtraction, every subdomain
contains P. Finally, we note that |P|= char Dwhen char Dis
prime and |P|is infinite when char Dis 0.
34. An integral domain of order 6 would be an Abelian group of order 6
under addition. So, it would be cyclic under addition. Now use
Theorems 13.3 and 13.4. The argument can not be adapted since
there is an integral domain with 4 elements. The argument can be
adapted for 15 elements.
35. By Theorem 13.3, the characteristic is |1|. By Lagrange’s Theorem
(Theorem 7.1), |1|divides 2n. By Theorem 13.4, the characteristic
is prime. Thus, the characteristic is 2.
36. Solve the equation x2= 1.
37. By Exercise 36, 1 is the only element of an integral domain that is
its own inverse if and only if 1 = −1. This is true only for fields of
characteristic 2.
38. If nis a prime then Znis a field and therefore has no zero divisors.
If nis not a prime we may write n=ab where both aand bare less
than n. If a6=b, then (n−1)! includes both aand bamong its
factors so (n−1)! = 0. If a=band a > 2, then
(n−1)! = (a2−1)(a2−2) ···(a2−a)···(a2−2a)···2·1. Since
this product includes a2=n, it is 0. The only remaining case is
n= 4 and in this case 3! = 2 is a zero divisor.
39. a. First note that a3=b3implies that a6=b6. Then a=b
because we can cancel a5from both sides (since a5=b5).
13/Integral Domains 117
b. Since mand nare relatively prime, by the corollary of
Theorem 0.2, there are integers sand tsuch that 1 = sn +tm.
Since one of sand tis negative we may assume that sis
negative. Then
a(an)−s=a1−sn = (am)t= (bm)t=b1−sn =b(bn)−s=b(an)−s.
Now cancel (an)−s.
40. In Z, take a= 1, b=−1, m= 4, n= 2.
41. (1 −a)2= 1 −2a+a2= 1 −2a+a= 1 −a.
42. 0 1 i1 + i
0 0 0 0 0
1 0 1 i1 + i
i0i1 1 + i
1 + i0 1 + i1 + i0
No. No.
43. Observe that (1 + i)4=−1, so |1 + i|= 8 and therefore the group is
isomorphic to Z8.
44. In Zp[k] note that (a+b√k)−1=1
a+b√k
(a−b√k)
(a−b√k)=a−b√k
a2−b2kexists if
and only if a2−b2k6= 0 where a6= 0 and b6= 0.
45. Let S={a1, a2, . . . , an}be the nonzero elements of the ring. Then
a1a1, a1a2, . . . , a1anare distinct elements for if a1ai=a1ajthen
a1(ai−aj) = 0 and therefore ai=aj. If follows that
S={a1a1, a1a2, . . . , a1an}. Thus, a1=a1aifor some i. Then aiis
the unity, for if akis any element of S, we have a1ak=a1aiak, so
that a1(ak−aiak) = 0. Thus, ak=aiakfor all k.
46. Say (ab)c= 0 where c6= 0. If ac = 0, then ais a zero divisor. If
ac 6= 0, then (ac)b= 0 so that bis a zero divisor.
47. Suppose that xand yare nonzero and |x|=nand |y|=mwith
n<m. Then 0 = (nx)y=x(ny). Since x6= 0, we have ny = 0.
This is a contradiction to the fact that |y|=m.
48. Use Exercise 47 and the observation that if |a|=mn, then
|ma|=n.
49. a. By the Binomial Theorem, (x+y)p=xp+pxp−1+···+px + 1,
where the coefficient of every term between xpand 1 is
divisible by p. Thus, (x+y)p=xp+yp.
13/Integral Domains 118
b. We prove that (x+y)pn= (xpn+ypn) for every positive
integer nby induction. The n= 1 case is done in part a.
Assume that (x+y)pk=xpk+ypk. Then
(x+y)pk+1 = ((x+y)pk)p= (xpk+ypk)p=xpk+1 +ypk+1 .
c. Let S={0,3,6,9}in Z12. Then Sis a ring of characteristic 4
and (3 + 3)4= 64= 0. But 34+ 34= 9 + 9 = 6.
50. Use part b of Exercise 49.
51. By Theorems 13.3 and 13.4, |1|has prime order, say p. Then by
Exercise 47 every nonzero element has order p. If the order of the
field were divisible by a prime qother than p, Theorem 9.5 implies
that the field also has an element of order q. Thus, the order of the
field is pnfor some prime pand some positive integer n.
52. Z3[x]
53. na b
c d =0 0
0 0 for all members of M2(R) if and only if na = 0
for all ain R.
54. Observe char R= least common multiple {|x| | x∈R}(additive
order). Now use Corollary 2 to Theorem 7.1.
55. This follows directly from Exercise 54.
56. 2 + iand 2 + 2i
57. a. 2b. 2, 3 c. 2, 3, 6, 11 d. 2, 3, 9, 10
58. 0.
59. By Theorem 13.3, char Ris prime. From 20 ·1 = 0 and 12 ·1=0
and Corollary 2 of Theorem 4.1, we know that char Rdivides both
12 and 20. Since the only prime that divides both 20 and 12 is 2,
the characteristic is 2.
60. If a2=aand b2=b, then (a−b)2=a2+b2=a−band
(ab)2=a2b2=ab.
61. Note that K={a+b√2|a, b ∈Q}is a field that contains √2 (see
Example 10) and if Fis any subfield of the reals that contains √2
then Fcontains K.
62. Use Corollary 4 of Theorem 7.1.
13/Integral Domains 119
63. By Exercise 49, x, y ∈Kimplies that x−y∈K. Also, if x, y ∈K
and y6= 0, then (xy−1)p=xp(y−1)p=xp(yp)−1=xy−1. So, by
Exercise 29, Kis a subfield.
64. Since the characteristic of a field of order 2nis 2, it suffices to show
that a=bfor then 0 = a2+ab +b2= 3a2= 2a2+a2=a2. Note
that a3−b3= (a−b)(a2+ab +b2) = 0 so that a3=b3. In a field of
order 2n,x2n+1 =x2for all x(see Exercise 54). Also, for all odd n,
2n+ 1 is divisible by 3. Thus
a2=a2n+1 = (a3)(2n+1)/3= (b3)(2n+1)/3=b2n+1 =b2. Finally,
a3=b3and a2=b2imply a=b.
65. Let a∈F, where a6= 0 and a6= 1. Then
(1 + a)3= 13+ 3(12a) + 3(1a2) + a3= 1 + a+a2+a3. If
(1 + a)3= 13+a3, then a+a2= 0. But then a(1 + a) = 0 so that
a= 0 or a=−1 = 1. This contradicts our choice of a.
66. Let F∗=hai. Then −1 = anfor some n. Thus, 1 = a2nand |a|
divides 2n.
67. φ(x) = φ(x·1) = φ(x)·φ(1) so φ(1) = 1. Also,
1 = φ(1) = φ(xx−1) = φ(x)φ(x−1). So, φ(x) = 1.
68. Apply Lagrange’s Theorem to F∗.
69. Since a field of order 27 has characteristic 3, we have 3a= 0 for all
a. Thus, 6a= 0 and 5a=−a.
120
CHAPTER 14
Ideals and Factor Rings
1. Let r1aand r2abelong to hai. Then r1a−r2a= (r1−r2)a∈ hai. If
r∈Rand r1a∈ hai, then r(r1a) = (rr1)a∈ hai.
2. To prove that Ais an ideal note that f(x)∈Aif and only if
f(0) = 0. If f(x), g(x)∈A, then f(0) = 0 and g(0) = 0. So,
f(0) −g(0) = 0 −0 = 0 and h(0)f(0) = h(0)0 = 0 for all h(x) in
Z[x]. Finally, note that f(x)∈ hxiif and only if f(0) = 0.
3. Clearly, Iis not empty. Now observe that
(r1a1+··· +rnan)−(s1a1+··· +snan) =
(r1−s1)a1+··· + (rn−sn)an∈I. Also, if r∈R, then
r(r1a1+··· +rnan)=(rr1)a1+··· + (rrn)an∈I. That I⊆J
follows from closure under addition and multiplication by elements
from R.
4. {(a, a)|a∈Z}.
5. Let a+bi, c +di ∈S. Then (a+bi)−(c+di) = a−c+ (b−d)iand
b−dis even. Also, (a+bi)(c+di) = ac −bd + (ad +cb)iand
ad +cb is even. Finally, (1 + 2i)(1 + i) = −1+3i /∈S.
6. a. h2ib. h2iand h5ic. h2iand h3id. hpiwhere pis a prime
divisor of n.
7. Since ar1−ar2=a(r1−r2) and (ar1)r=a(r1r), aR is an ideal.
4R={. . . , −16,−8,0,8,16, . . .}.
8. Mimic Exercise 9 of Chapter 12.
9. If nis a prime and ab ∈Zthen by Euclid’s Lemma (Chapter 0), n
divides aor ndivides b. Thus, a∈nZ or b∈nZ. If nis not a
prime, say n=st where s<nand t<n, then st belongs to nZ but
sand tdo not.
10. (a1+b1)−(a2+b2)=(a1−a2)+(b1−b2)∈A+B;
r(a+b) = ra +rb ∈A+B; (a+b)r=ar +br ∈A+B
14/Ideals and Factor Rings 121
11. a. a= 1 b. a= 2 c. a= gcd(m, n)
12. Let a1b1+··· +anbnand a0
1b0
1+···a0
mb0
mthen
(a1b1+··· +anbb)−(a0
1b0
1+··· +a0
mb0
m)
=a1b1+··· +anbn+ (−a0
1)b0
1+··· + (−a0
m)b0
m∈AB.
Also
r(a1b1+··· +anbn) = (ra1)b1+··· + (ran)bn∈AB
and
(a1b1+··· +anbn)r=a1(b1r) + ··· +an(bnr)∈AB.
13. a. a= 12
b. a= 48. To see this, note that every element of h6ih8ihas the
form 6t18k1+ 6t28k2+··· + 6tn8kn= 48s∈ h48i. So,
h6ih8i⊆h48i. Also, since 48 ∈ h6ih8i, we have h48i ⊆ h6ih8i.
c. a=mn
14. Since Aand Bare ideals, ab ∈Aand ab ∈Bwhen a∈Aand
b∈B. Now AB is just the sum of such terms.
15. Let r∈R. Then r= 1r∈A.
16. By Exercise 14, we have AB ⊆A∩B. So, let x∈A∩B. To show
that x∈AB, start by writing 1 = a+bwhere a∈A, b ∈B.
17. Let u∈Ibe a unit and let r∈R. Then r=r(u−1u) = (ru−1)u∈I.
18. J=h5ior J=h7i;I=Z.
19. Observe that h2iand h3iare the only nontrivial ideals of Z6, so
both are maximal. More generally, Zpq, where pand qare distinct
primes, has exactly two maximal ideals.
20. Observe that as groups |R:I|= 3. So there is not a proper
subgroup of Rthat strictly contains I.
21. Iis closed under subtraction since the even integers are closed
under subtraction. Also, if b1, b2, b3,and b4are even, then every
entry of "a1a2
a3a4#" b1b2
b3b4#is even.
14/Ideals and Factor Rings 122
22. Observe that {f(x)∈Z[x]|f(0) is even}is a proper ideal that
properly contains I[x].
23. Use the observation that every member of Rcan be written in the
form "2q1+r12q2+r2
2q3+r32q4+r4#. Then note that
"2q1+r12q2+r2
2q3+r32q4+r4#+I="r1r2
r3r4#+I.
24. Look at the ring {0,2,4,6}under addition and multiplication mod
8.
25. (br1+a1)−(br2+a2) = b(r1−r2)+(a1−a2)∈B;
r0(br +a) = b(r0r) + r0a∈B.
26. (b+A)(c+A) = bc +A=cb +A= (c+A)(b+A). If 1 is the unity
of R, then 1 + Ais the unity of R/A.
27. Suppose that Iis an ideal of Fand I6={0}. Let abe a nonzero
element of I. Then by Exercise 17, I=F.
28. Use Example 15 and Theorem 14.4.
29. Since every element of hxihas the form xg(x), we have hxi ⊆ I. If
f(x)∈I, then f(x) = anxn+···+a1x=x(anxn−1+···+a1)∈ hxi.
30. Observe that Z⊕Z/A ={(0,0) + A, (1,0) + A, (2,0) + A} ≈ Z3
and use Theorem 14.4. In general, for
A={(nx, y)|x, y ∈Z}, Z ⊕Z/A =
{(0,0) + A, (1,0) + A, (2,0) + A, . . . , (n−1,0) + A} ≈ Zn. Thus, A
is a maximal ideal of Z⊕Zif and only if nis prime.
31. Suppose f(x) + A6=A. Then f(x) + A=f(0) + Aand f(0) 6= 0.
Thus,
(f(x) + A)−1=1
f(0) +A.
This shows that R/A is a field. Now use Theorem 14.4.
32. h1i⊕h2i,h2i⊕h1i,h1i⊕h3i,h1i⊕h5i; 2, 2, 3, 5.
33. Since (3 + i)(3 −i) = 10 we know 10 + h3 + ii= 0 + h3 + ii. Also,
i+h3 + ii=−3 + h3 + ii= 7 + h3 + ii. Thus, every element
a+bi +h3 + iican be written in the form k+h3 + iiwhere
14/Ideals and Factor Rings 123
k= 0,1,...,9. Finally, Z[i]/h3 + ii={k+h3 + ii | k= 0,1,...,9}
since 1 + h3 + iihas additive order 10.
34. Consider J={f∈Z[x]|f(0) is even}.
35. Note that in (Z⊕Z)/I, (a, b) + I= (a, 0) + (0, b) + I= (0, b) + I.
So, (Z⊕Z)/I is isomorphic to Z(map (0, b) + Ito b). Since Zis
an integral domain but not a field, we have by Theorems 14.3 and
14.4 that Iis a prime ideal but not a maximal ideal.
36. rs −sr ∈Iif and only if rs −sr +I=Ior rs +I=sr +I. This is
equivalent to (r+I)(s+I) = (s+I)(r+I).
37. Since every element in hx, 2ihas the form f(x) = xg(x)+2h(x), we
have f(0) = 2h(0), so that f(x)∈I. If f(x)∈I, then
f(x) = anxn+···+a1x+ 2k=x(anxn−1···+a1) + 2k∈ hx, 2i. By
Theorems 14.3 and 14.4 to prove that Iis prime and maximal it
suffices to show that Z[x]/I is a field. To this end note that every
element of Z[x]/I can be written in the form
anxn+··· +a1x+ 2k+I= 0 + Ior
anxn+··· +a1x+ (2k+ 1) + I= 1 + I. So, Z[x]/I ≈Z2.
38. 2 6∈ h2 + 2iiand 1 + i6∈ h2 + 2iibut 2(1 + i)∈ h2 + 2ii. Z[i]/I has 8
elements and has characteristic 4.
39. 3x+1+I
40. Let a, b ∈Ip.Say |a|=pnand |b|=pm.Then pn+m(a−b) = 0 so
|a−b|divides pn+m. Also, pn(ra) = r(pna) = 0 so |ra|divides pn.
41. Every ideal is a subgroup. Every subgroup of a cyclic group is
cyclic.
42. Since "a b
0d#" r s
0t#="ar as +bt
0dt #and
"r s
0t#" a b
0d#="ra rb +sd
0td #for all a, b, and d, we must
have that rand tare even.
43. Let Ibe any ideal of R⊕Sand let
IR={r∈R|(r, s)∈Ifor some s∈S}and
IS={s∈S|(r, s)∈Ifor some r∈R}. Then IRis an ideal of R
and ISis an ideal of S. Let IR=hriand IS=hsi. Since, for any
14/Ideals and Factor Rings 124
(a, b)∈Ithere are elements a0∈Rand b0∈Ssuch that
(a, b)=(a0r, b0s), we have that I=h(r, s)i.
44. Say ax =br and ax0=br0. Then
ax −ax0=br −br0=b(r−r0)∈bR.
Also,
(r1x)a=r1(ax) = r1(br) = b(r1r)∈bR.
45. Say b, c ∈Ann(A). Then (b−c)a=ba −ca = 0 −0 = 0. Also,
(rb)a=r(ba) = r·0 = 0.
46. Suppose b, c ∈N(A). Say, bn∈Aand cm∈A. Then the binomial
theorem shows that (b−c)n+m∈A. Also, (rb)n=rnbn∈A.
47. a. h3i
b. h3i
c. h3i
48. a. h6ib. h2ic. h6i
49. Suppose (x+N(h0i))n= 0 + N(h0i). We must show that
x∈N(h0i). We know that xn+N(h0i) = 0 + N(h0i), so that
xn∈N(h0i). Then, for some m, (xn)m= 0, and therefore
x∈N(h0i).
50. Clearly N(A)⊆N(N(A)). Suppose x∈N(N(A)). Then
xn∈N(A) for some n. Thus (xn)m∈Afor some m, so x∈N(A)).
51. Let I=hx2+x+ 1i. Then Z2[x]/I ={0 + I, 1 + I, x +I, x + 1 + I}.
1 + Iis its own multiplicative inverse and
(x+I)(x+1+I) = x2+x+I=x2+x+1+1+I= 1 + I. So,
every nonzero element of Z2[x]/I has a multiplicative inverse.
52. For simplicity, denote the coset
f(x) + hx2+x+ 1iby f(x).
The tables are
+ 0 1 x x + 1
0 0 1 x x + 1
1 1 0 x+ 1 x
x x x + 1 0 1
x+ 1 x+ 1 x1 0
14/Ideals and Factor Rings 125
·1x x + 1
1 1 x x + 1
x x x + 1 1
x+ 1 x+ 1 1 x
53. x+2+hx2+x+ 1iis not zero, but its square is.
54. {na +ba |n∈Z, b ∈R}
55. If fand g∈A, then (f−g)(0) = f(0) −g(0) is even and
(f·g)(0) = f(0) ·g(0) is even. f(x)=1/2∈Rand g(x)=2∈A,
but f(x)g(x)/∈A.
56. Observe that 1 + h1−ii=i+h1−iiso any coset can be written in
the form a+h1−iiwhere a∈Z. But
1 + h1−ii= (1 + h1−ii)2
= (i+h1−ii)2=−1 + h1−ii
so 2 + h1−ii= 0 + h1−ii. This means that there are only two
cosets: 0 + h1−iiand 1 + h1−ii.
57. Any ideal of R/I has the form A/I where Ais an ideal of R. So, if
A=hai, then A/I =ha+Ii/I.
58. 1. To see this note that in Z[x]/h1 + iiwe have
1 + h1 + ii=−i+h1 + ii. Thus, (1 + h1 + ii)2= (−i+h1 + ii)2and
therefore 2 + h1 + ii= 0 + h1 + ii. Multiplying both sides by 3 we
obtain 1 + h1 + ii= 0 + h1 + ii.
59. By Theorem 14.3, R/I is an integral domain. Since every element
in R/I is an idempotent and Exercise 16 in Chapter 13 says that
the only idempotents in an integral domain are 0 and 1 we have
that R/I ={0 + I, 1 + I}.
60. Say J6=Iis also a maximal ideal. Let xbe an element of Jthat is
not in I. Then 1 = x−1x∈Jso that J=R.
61. hxi⊂hx, 2ni ⊂ hx, 2n−1i ⊂ ··· ⊂ hx, 2i
62. Use the ideal test to show that Iis an ideal of R. To show that Iis
not generated by a single element observe that every element of I
has the form (b1, b2, b3, . . . , bk,0,0,0, . . .) where all terms beyond
the kth one are 0). Thus, h(b1, b2, b3, . . . , bk,0,0,0, . . .)idoes not
contain (b1, b2, b3, . . . , bk,1,0,0, . . .).
14/Ideals and Factor Rings 126
63. Taking r= 1 and s= 0 shows that a∈I. Taking r= 0 and s= 1
shows that b∈I. If Jis any ideal that contains aand b, then it
contains Ibecause of the closure conditions.
127
SUPPLEMENTARY EXERCISES FOR CHAPTERS 12-14
1. In Z10 they are 0, 1, 5, 6. In Z20, they are 0, 1, 5, 16. In Z30, they
are 0, 1, 6, 10, 15, 16, 21, 25.
2. Let sand tbe such that 1 = ms +nt. Then
ms =ms(ms +nt)=(ms)2+mnst = (ms)2in Zmn. Similarly,
nt = (nt)2.
3. Suppose that an= 0 for some positive integer n. Let 2kbe the
smallest power of 2 greater than n. Then a2k= (a2k−n)an= 0.
Since 0 = a2k= (a2k−1)2we conclude that a2k−1= 0. By iteration
(or induction) we obtain that a= 0.
4. Fix some ain R,a6= 0. Then there is a bin Rsuch that ab =a.
Now if x∈Rand x6= 0 then there is an element cin Rsuch that
ac =x. Then xb =acb =c(ab) = ca =x. Thus bis the unity. The
rest is easy.
5. Suppose A6⊆ Cand B6⊆ C. Pick a∈Aand b∈Bso that a, b /∈C.
But ab ∈Cand Cis prime. This contradicts the definition of a
prime ideal.
6. In Z,h2i∩h3i=h6iis not prime.
7. Suppose that (a, b) is a nonzero element of an ideal Iin R⊕R. If
a6= 0, then (r, 0) = (ra−1,0)(a, b)∈I. Thus, R⊕ {0} ⊆ I.
Similarly, if b6= 0, then {0} ⊕ R⊆I. So, the ideals of R⊕Rare
{0}⊕{0},R⊕R,R⊕ {0},{0} ⊕ R. The ideals of F⊕Fare
{0}⊕{0}, F ⊕F, F ⊕ {0},{0} ⊕ F.
8. Every factor ring has the form Z/hnifor some nonnegative integer
n.
9. Suppose that ammod n= 0. Since ndivides am, every prime p
divisor of ndivides am. By Euclid’s Lemma (Chapter 0), pdivides
aand since nis square-free it follows that ndivides a.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 12-14 128
10. Try the case a= 1 and b2= 0 first. Then generalize to a= 1 and
bn= 0. Finally, reduce the original problem to the previous case.
11. Suppose that A⊆B∪Cbut A6⊆ Band A6⊆ C. Let a∈Abut
a6∈ Band a0∈Abut a06∈ C. Since a∈A⊆B∪Cand a6∈ Bwe
have that a∈C. Similarly, a0∈B. Then since a+a0∈Awe know
that a+a0∈Bor a+a0∈C. If a+a0∈Bwe have that
a= (a+a0)−a0∈B, which contradicts the fact that a6∈ B. A
similar contradiction arises if a+a0∈C.
12. Say Iis an ideal that contains a. The aR ⊆I. Now note that aR is
an ideal that contains a. For the example consider the ring of even
integers.
13. Let A={s1at1+···+snatn}. We first prove that Ais a ideal of R.
Since (s1at1+··· +snatn)−(s0
1at0
1+··· +s0
nat0
n) =
s1at1+··· +snatn+ (−s0
1)at0
1+··· + (−s0
n)at0
n), A is closed under
subtraction. If r∈Rthen
r(s1at1+··· +snatn)=(rs1)at1+··· + (rsn)atn)∈Aand
(s1at1+··· +snatn)r= (s1a(t1r) + ··· +sna(tnr)∈A, so Ais an
ideal of Rthat contains 1a1 = a(here 1 is the unity of R) and
therefore hai ⊆ A. On the other hand, we know that hai ⊇ Asince
multiplying aon the left or right by elements of Rgives an element
in haiand haiis closed under addition, it contains all elements of
the form s1at1+··· +snatn.
14. Use Theorem 13.3.
15. Since Ais an ideal, ab ∈A. Since Bis an ideal, ab ∈B. So
ab ∈A∩B={0}.
16. Observe that (1,0) ·(0,1) = (0,0).
17. 6
18. lcm(m, n). In general, the characteristic of the direct product of
any finite number of rings with nonzero characteristics is the least
common multiple of the characteristics of the rings.
19. Since for every nonzero element ain R,aR is a nonzero ideal of R,
we have aR =R. Then, by Exercise 4, Ris a field.
20. Let ibe the unity of Iand r∈R. First show ri ∈I. Then use this
fact to show rx ∈Ifor all r∈Rand x∈I.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 12-14 129
21. Since 2x+1+h2x+ 1i= 0 + h2x+ 1i, we have
−2x+h2x+ 1i= 1 + h2x+ 1i. So,
(−2 + h2x+ 1i)(x+h2x+ 1i= 1 + h2x+ 1i.
22. Clearly, h0i={0};h1i=Z[x]; and h−1i=Z[x]. So assume that
a6= 0,1,or −1. Observe that ha, xiproperly contains hai.
Moreover, ha, xidoes not contain 1. For if so, then there are
f(x), g(x)∈Z[x] such that af(x) + xg(x) = 1. But then
af(0) + 0g(0) = af(0) = 1. This implies that a=±1.
23. x2+1+hx4+x2i.
24. Suppose Ais prime and Bis an ideal which properly contains A.
Say A=haiand B=hbi. It suffices to show bis a unit. Write
a=br. Then, since Ais prime, b∈Aor r∈A. If b∈A, then
B⊆Aso r∈A. Say r=ar0. Then a=br =bar0so that
a(1 −br0) = 0. Thus 1 = br0and bis a unit.
25. In Z8, 22= 4 = 62and 23= 0 = 63.
26. {a22/3+b21/3+c|a, b, c ∈Q}
27. Say char R=p(remember pmust be prime). Then
char R/A =|1 + A|, the additive order of 1 + A. But |1 + A|divides
|1|=p.
28. It is the external direct product of ncopies of Zp.
29. Let Abe a prime ideal of R. By Theorem 14.3, Ris an integral
domain. Then, by Theorem 13.2, R/A is a field and, by
Theorem 14.4, Ais maximal.
30. If the additive group of R/C(R) were cyclic, then there is some ain
Rsuch that every element has the form na +cwhere c∈C(R). For
this it follows that Ris commutative.
31. Observe that A=a b
0 0 a, b ∈Z2but
1 1
1 1 1 0
0 0 =1 0
1 0 is not in A.
32. No, Consider R=Zand A=h4i.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 12-14 130
33. We claim that Z[i]/A ={0 + A, 1 + A}. To verify this claim let
a+bi +Abe any element in Z[i]/A. If aand bare both even or
both odd, then a+bi +A= 0 + A. If ais odd and bis even, then
a+bi +A= 1 + 0i+A= 1 + A. If ais even and bis odd, then
a+bi +A= 0 + i+A= 1 + A. So, Z[i]/A ={0 + A, 1 + A}, which
is a field isomorphic to Z2. Thus, by Theorem 14.4, Ais maximal.
34. Let Ibe a prime ideal and a6∈ I. By Theorem 14.4 it suffices to
show that a+Iis a unit in R/I. Observe that
(a+I)n=an+I=a+Iimplies that an−a+I=a(an−1−1) ∈I
so that an−1−1∈I. Thus (a+I)n−1= 1 + I.
35. A finite subset of a field is a subfield if it contains a nonzero
element and is closed under addition and multiplication.
36. If char Fis odd, use the fact that for x6= 0, xand −xare distinct.
If char Fis 2, observe that the additive structure of Fhas the form
Z2⊕Z2⊕ ··· ⊕ Z2. Let H={h1, h2,···, hn}be a subgroup of
index 2 and let x6∈ H. Then
h1+h2+··· +hn+ (x+h1) + ··· + (x+hn) = nx = 0.
37. Observe that (a+bi)(a−bi) = a2+b2= 0, so a+bi is a
zero-divisor. In Z13[i],2+3iis a zero-divisor.
38. Let a∈R. Then 0 = ab2−ab = (ab −a)bso that ab −a= 0.
Similarly, ba −a= 0.
39. According to Theorem 13.3 we need only determine the additive
order of 1 + h2 + ii. Since
5(1 + h2 + ii) = 5 + h2 + ii= (2 + i)(2 −i) + h2 + ii= 0 + h2 + ii,
we know that 1 + h2 + iihas order 5.
40. Use the fact that a2+b2= (a+bi)(a−bi).
41. The inverse is 2x+ 3.
42. Observe that Z[x, y]/hx, yi ≈ Z, then use Theorem 14.3 and
Theorem 14.4.
43. Because elements of Z5[x, y]/hx, yihave the form f(x, y) + hx, yi
and hx, yiabsorbs all terms of f(x, y) that have an xor a ywe
know that Z5[x, y]/hx, yi=
{0 + hx, yi,1 + hx, yi,2 + hx, yi,3 + hx, yi,4 + hx, yi} ≈ Z5. Thus, by
Theorem 14.4, hx, yiis maximal.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 12-14 131
44. Observe that Z[x, y]/h2, x, yi ≈ Z2, then use Theorem 14.4.
45. Say (a, b)n= (0,0). Then an= 0 and bn= 0. If am= 0 and bn= 0,
then (a, b)mn = ((am)n,(bn)m) = (0,0).
46. If (a, b) is a zero-divisor in R⊕Sthen there is a (c, d)6= (0,0) such
that (a, b)(c, d) = (0,0). Thus ac = 0 and bd = 0. So, aor bis a
zero-divisor or exactly one of aor bis 0. Conversely, if ais a
zero-divisor in Rthen there is a c6= 0 in Rsuch that ac = 0. In
this case (a, b)(c, 0) = (0,0). A similar argument applies if bis a
zero-divisor. If a= 0 and b6= 0 then (a, b)(x, 0) = (0,0) where xis
any nonzero element in A. A similar argument applies if a6= 0 and
b= 0.
47. If a2=a, then pk|a(a−1). Since aand a−1 are relatively prime,
pk|aor pk|(a−1). So, a= 0 or a= 1.
48. If a2=a, then a(1 −a) = 0.Let b= 1 −a. If a+b= 1, then
1a= (a+b)a=a2+ba =a2.
49. First observe that in Z3, a2+b2= 1 or 2 except for the case that
a=0=b. So, for any nonzero element a+b√2 in
Z3[√2], a2+b2= 1 or −1. Next note that if a+b√26= 0, then
(a+b√2)−1=1
a+b√2=1
a+b√2
a−b√2
a−b√2=a−b√2
a2−2b2=
a−b√2
a2+b2=a−b√2 or −a+b√2. So, Z3[√2] is a field. In
Z7[√2],(1 + 2√2)(1 + 5√2) = 0.
50. If ais a zero divisor in Zpnthen pdivides a. Thus pndivides an.
51. If xn= 0, then(rx)n=rnxn= 0.
52. Let I=h3, x2+ 1i. Using the condition that 3 + I= 0 + Iwe see
that when when adding and multiplying in Z[x]/I we may treat the
coset representatives Z[x]/I as members of Z3[x]. Then we see that
Z[x]/I =
{0+I, 1+I, 2+I, x+I, x+1+I, x+2+I, 2x+I, 2x+1+I, 2x+2+I}
Moreover, 1 + Iand 2 + Iare their own inverses;
(x+I)(2x+I)=2x2+I= 1 + I; (x+1+I)(x+2+I) =
1 + I; (2x+1+I)(2x+2+I) = 1 + I.
132
CHAPTER 15
Ring Homomorphisms
1. Property 1: φ(nr) = nφ(r) holds because a ring is a group under
addition. To prove that φ(rn)=(φ(r))nwe note that by induction
φ(rn) = φ(rn−1r) = φ(rn−1)φ(r) = φ(r)n−1φ(r) = φ(r)n.
Property 2: If φ(a) and φ(b) belong to φ(A) then
φ(a)−φ(b) = φ(a−b) and φ(a)φ(b) = φ(ab) belong to φ(A).
Property 3: φ(A) is a subgroup because φis a group
homomorphism. Let s∈Sand φ(r) = s. Then
sφ(a) = φ(r)φ(a) = φ(ra) and φ(a)s=φ(a)φ(r) = φ(ar).
Property 4: Let aand bbelong to φ−1(B) and rbelong to R. Then
φ(a) and φ(b) are in B. So,
φ(a)−φ(b) = φ(a) + φ(−b) = φ(a−b)∈B. Thus, a−b∈φ−1(B).
Also, φ(ra) = φ(r)φ(a)∈Band φ(ar) = φ(a)φ(r)∈B. So, ra and
ar ∈φ−1(B).
Property 5: φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a).
Property 6: Because φis onto, every element of Shas the form
φ(a) for some ain R. Then φ(1)φ(a) = φ(1a) = φ(a) and
φ(a)φ(1) = φ(a1) = φ(a).
Property 7: If φis an isomorphism, by property 1 of Theorem 10.1
and the fact that φis one-to-one, we have Ker φ={0}. If Ker
φ={0}, by property 5 of Theorem 10.2, φis one-to-one.
Property 8: That φ−1is one-to-one and preserves addition comes
from property 3 of Theorem 6.3. To see that φ−1preserves
multiplication note that φ−1(ab) = φ−1(a)φ−1(b) if and only if
φ(φ−1(ab)) = φ(φ−1(a)φ−1(b)) = φ(φ−1(a))φ(φ−1(b)). But this
reduces to ab =ab.
2. Since φis a group homomorphism, Ker φis a subgroup. Let
a∈Ker φand r∈R. The φ(ar) = φ(a)φ(r) = 0φ(r) = 0. Similarly,
φ(ra) = 0.
3. We already know the mapping is an isomorphism of groups. Let
Φ(x+ Ker φ) = φ(x). Note that Φ((r+ Ker φ)(s+ Ker φ)) =
Φ(rs + Ker φ) = φ(rs) = φ(r)φ(s) = Φ(r+ Ker φ)Φ(s+ Ker φ).
15/Ring Homomorphisms 133
4. See the proof of Theorem 10.4.
5. φ(2 + 4) = φ(1) = 5, whereas φ(2) + φ(4) = 0 + 0 = 0.
6. xy →3xy 6= 3x3y.
7. Observe that (x+y)/1=(x/1) + (y/1) and (xy)/1=(x/1)(y/1).
8. If φis a ring-homomorphism from Znto itself then
φ(x) = φ(1x) = φ(1)x. Moreover, if φ(1) = a, then
a2= (φ(1))2=φ(12) = φ(1) = a.
9. a=φ(1) = φ(1 ·1) = φ(1)φ(1) = aa =a2. For the example note
that the identity function from Z6to itself is a ring homomorphism
but 32= 3.
10. a. No. Suppose 2 →aand consider 2 + 2 and 2 ·2.
b. No.
11. If aand b(b6= 0) belong to every member of the collection, then so
do a−band ab−1. Thus, by Exercise 29 of Chapter 13, the
intersection is a subfield.
12. Try a+bi →a+bx +hx2+ 1i.
13. By observation φis one-to-one and onto. Since
φ((a+bi) + (c+di)) = φ((a+c) + (b+d)i) = "a+c b +d
−(b+d)a+c#=
"a b
−b a #+"c d
−d c #=φ(a+bi) + φ(c+di)
addition is preserved. Also,
φ((a+bi)(c+di)) = φ((ac −bd)+(ad +bc)i) =
"ac −bd ad +bc
−(ad +bc)ac −bd #="a b
−b a #" c d
−d c #=
φ(a+bi)φ(c+di)
so multiplication is preserved.
14. Try a+b√2→"a2b
b a #.
15/Ring Homomorphisms 134
15. Since φ " a b
c d #" a0b0
c0d0#!=φ " aa0+bc0ab0+bd0
ca0+dc0cb0+dd0#!=
aa0+bc06=aa0=φ " a b
c d #!φ " a0b0
c0d0#!multiplication is
not preserved.
16. It is a ring homomorphism.
17. Yes. φ(x)=6xis well defined because a=bin Z5implies that 5
divides a−b. So, 30 divides 6a−6b. Moreover,
φ(a+b) = 6(a+b)=6a+ 6b=φ(a) + φ(b) and
φ(ab) = 6ab = 6 ·6ab = 6a6b=φ(a)φ(b).
18. No; multiplication is not preserved.
19. The set of all polynomials passing through the point (1,0).
20. a=a2implies that φ(a) = φ(a2) = φ(a)φ(a)=(φ(a))2.
21. For Z6to Z6, 1 →0, 1 →1, 1 →3, and 1 →4 each define a
homomorphism. For Z20 to Z30, 1 →0, 1 →6, 1 →15, and 1 →21
each define a homomorphism.
22. By Exercise 8, any isomorphism has the form φ(x) = ax where
a2=a. Then φ(1) = aand φ(a) = a2=aso that 1 = a. Thus, the
only ring-isomorphism of Znto itself is the identity.
23. Suppose that φis a ring homomorphism from Zto Zand φ(1) = a.
Then φ(2) = φ(1 + 1) = 2φ(1) = 2aand
φ(4) = φ(2 + 2) = 2φ(2) = 4a. Also,
φ(4) = φ(2 ·2) = φ(2)φ(2) = 2a·2a= 4a2. Thus, 4a2= 4aand it
follows that a= 0 or a= 1. So, φis the zero map or the identity
map.
24. (0,0),(1,0),(0,1),(1,1).
25. Suppose that φis a ring homomorphism from Z⊕Zto Z⊕Z. Let
φ((1,0)) = (a, b) and φ((0,1)) = (c, d). Then
φ((x, y)) = φ(x(1,0) + y(0,1)) = φ(x(1,0)) + φ(y(1,0)) =
xφ((1,0)) + yφ((0,1)) = x(a, b) + y(c, d) = (ax +cy, bx +dy). Since
φpreserves multiplication we know
φ((x, y)(x0, y0)) = φ((xx0, yy0)) = (axx0+cyy0, bxx0+dyy0) =
φ((x, y))φ(x0, y0)) = (ax +cy, bx +dy)(ax0+cy0, bx0+dy0) =
15/Ring Homomorphisms 135
((ax +cy)(ax0+cy0),(bx +dy)(bx0+dy0)). Now observe that
(ax +cy)(ax0+cy0) = axx0+cyy0for all x, x0, y, y0if and only if
a2xx0+acxy0+acyx0+c2yy0=axx0+cyy0for all x, x0, y, y0. This
implies that a= 0 or 1 and c= 0 or 1 and ac = 0. This gives
(a, c) = (0,0),(1,0), and (0,1). Likewise, we have
(b, d) = (0,0),(1,0), and (0,1). Thus, we have nine cases for
(a, b, c, d):
(0,0,0,0) corresponds to (x, y)→(0,0);
(0,1,0,0) corresponds to (x, y)→(0, x);
(0,0,0,1) corresponds to (x, y)→(0, y);
(1,0,0,0) corresponds to (x, y)→(x, 0);
(1,1,0,0) corresponds to (x, y)→(x, x);
(1,0,0,1) corresponds to (x, y)→(x, y);
(0,0,1,0) corresponds to (x, y)→(y, 0);
(0,1,1,0) corresponds to (x, y)→(y, x);
(0,0,1,1) corresponds to (x, y)→(y, y).
It is straight forward to show that each of these nine is a ring
homomorphism.
26. The group A/B is cyclic of order 4. The ring A/B has no unity.
27. Say 1 is the unity of R. Let s=φ(r) be any nonzero element of S.
Then φ(1)s=φ(1)φ(r) = φ(1r) = φ(r) = s. Similarly, sφ(1) = s.
28. Consider the mapping given by (x, y)→(xmod a, y mod b) and use
Theorem 15.3.
29. Suppose that φis a ring homomorphism from Z⊕Zto Z. Let
φ((1,0)) = aand φ((0,1)) = b. Then
φ((x, y)) = φ(x(1,0) + y(0,1)) = φ(x(1,0)) + φ(y(1,0)) =
xφ((1,0)) + yφ((0,1)) = ax +by. Since φpreserves multiplication
we know
φ((x, y)(x0, y0)) = φ((xx0, yy0)) = axx0+byy0=φ((x, y)φ(x0, y0)) =
(ax +by)(ax0+by0) = a2xx0+abxy0+abyx0+b2yy0. Now observe
that axx0+byy0=a2xx0+abxy0+abyx0+b2yy0for all x, x0y, y0if
and only if a2=a, b2=b, and ab = 0. This means that a= 0 or 1
and b= 0 or 1 but not both a= 1 and b= 1. This gives us three
cases for (a, b):
(0,0) corresponds to (x, y)→0;
(1,0) corresponds to (x, y)→x;
(0,1) corresponds to (x, y)→y.
Each of these is obviously a ring homomorphism.
15/Ring Homomorphisms 136
30. (n2+ (n+ 1)2+ (n+ 2)2) mod 3 = 2 while k2= 2 mod 3 has no
solution.
31. Say m=akak−1···a1a0and n=bkbk−1···b1b0. Then
m−n= (ak−bk)10k+(ak−1−bk−1)10k−1+···+(a1−b1)10+(a0−b0).
By the test for divisibility by 9 given in Example 8, m−nis
divisible by 9 provided that
ak−bk+ak−1−bk−1+··· +a1−b1+a0−b0=
(ak+ak−1+··· +a1+a0)−(bk+bk−1+··· +b1+b0) is divisible
by 9. But this difference is 0 since the second expression has the
same terms as the first expression in some other order.
32. (akak−1. . . a1a0) mod 11 = (a0+10a1+102a2+···+10kak) mod 11 =
(a0−a1+a2− ···(−1)kak) mod 11.
33. Since the sum of the digits of the number is divisible by 9 so is the
number (see Example 8); the test for divisibility by 11 given in
Exercise 32 is not satisfied.
34. Use Example 8 and Exercise 32.
35. Let αbe the homomorphism from Zto Z3given by
α(n) = nmod 3. Then noting that α(10i) = α(10)i= 1i= 1 we
have that n=akak−1···a1a0=ak10k+ak−110k−1+···+a110 + a0
is divisible by 3 if and only if, modulo 3, 0 = α(n) =
α(ak) + α(ak−1) + ···+α(a1) + α(a0) = α(ak+ak−1+···+a1+a0).
But α(ak+ak−1+··· +a1+a0) = 0 mod 3 is equivalent
ak+ak−1+··· +a1+a0being divisible by 3.
36. akak−1. . . a1a0mod 4 =
(akak−1. . . a1a0−a1a0) mod 4 + a1a0mod 4 = a1a0mod 4.
37. Use Exercise 35.
38. Using the divisibility by 9 test we have a+ 27 + bmust be divisible
by 9. Using the divisibility by 11 test we have b−a−1 must be
divisible by 11. Experimentations yield b= 5 and a= 4.
39. Look at both sides mod 2.
40. Use the base 5 representation for n.
15/Ring Homomorphisms 137
41. Observe that 10 mod 3 = 1. So,
(2 ·1075 + 2) mod 3 = (2 + 2) mod 3 = 1 and
(10100 + 1) mod 3 = (1 + 1) mod 3 = 2 = −1 mod 3. Thus,
(2 ·1075 + 2)100 mod 3 = 1100 mod 3 = 1 and
(10100 + 1)99 mod 3 = 299 mod 3 = (−1)99 mod 3 = −1 mod 3 = 2.
42. Since the only idempotents in Qare 0 and 1 we have from Exercise
20 that a ring homomorphism from Qto Qmust send 1 →0 or
1→1. In the first case the homomorphism is x→0 and in the
second case it is x→x.
43. By Theorem 13.3, the characteristic of Ris the additive order of 1
and by property 6 of Theorem 15.1, the characteristic of Sis the
additive order of φ(1). Thus, by property 3 of Theorem 10.1, the
characteristic of Sdivides the characteristic of R.
44. Use Exercise 45(a) of Chapter 13.
45. No. The kernel must be an ideal.
46. Use Exercise 27 of Chapter 14.
47. a. Suppose ab ∈φ−1(A). Then φ(ab) = φ(a)φ(b)∈A, so that
a∈φ−1(A) or b∈φ−1(A).
b. Let Φ be the homomorphism from Rto S/A given by
Φ(r) = φ(r) + A. Then φ−1(A) = Ker Φ and, by
Theorem 15.3, R/Ker Φ ≈S/A. So, φ−1(A) is maximal.
48. If φ−1(A) = hai, then A=hφ(a)i.
49. a. Since φ((a, b)+(a0, b0)) = φ((a+a0, b +b0)) = a+a0=
φ((a, b)) + φ((a0, b0)), φ preserves addition. Also,
φ((a, b)(a0, b0)) = φ((aa0, bb0)) = aa0=φ((a, b))φ((a0, b0)) so φ
preserves multiplication.
b. φ(a) = φ(b) implies that (a, 0) = (b, 0), which implies that
a=b.φ(a+b)=(a+b, 0) = (a, 0) + (b, 0) = φ(a) + φ(b).
Also, φ(ab) = (ab, 0) = (a, 0)(b, 0) = φ(a)φ(b).
c. Define φby φ(r, s)=(s, r). By Exercise 14 in Chapter 8, φis
one-to-one and preserves addition. Since φ((r, s)(r0, s0)) =
φ((rr0, ss0)) = (ss0, rr0) = (s, r)(s0, r0) = φ((r, s))φ((r0, s0))
multiplication is also preserved.
15/Ring Homomorphisms 138
50. Since a generator must map to a generator, the only possibilities
are n→ ±m. Consider n→m. Then
n+n+··· +n
| {z }
nterms →m+m+··· +m
| {z }
nterms
=nm
while nn →mm. But nm 6=mm.
51. Observe that x4= 1 has two solutions in Rbut four in C.
52. First show that any automorphism φof Racts as the identity map
on the rationals. Then show that if a < b then φ(a)< φ(b). Next
suppose that there is some asuch that φ(a)6=a. Say, a<φ(a).
Pick a rational number rsuch that a<r<φ(a). Then
φ(a)< φ(r) = r, a contradiction. A similar argument applies if
φ(a)< a.
53. By Exercise 46 every ring homomorphism from Rto Ris an
automorphism of R. And by Exercise 52 the only automorphism of
Ris the identity.
54. To check that multiplication is operation preserving, observe that
xy →a(xy) = a2xy =axay. For the second part take m= 4, n = 6
and a= 4. Then 0 = φ(0) = φ(2 ·2) but φ(2)φ(2) = 2 ·2 = 4.
55. If a/b =a0/b0and c/d =c0/d0, then ab0=ba0and cd0=dc0. So,
acb0d0= (ab0)(cd0)=(ba0)(dc0) = bda0c0. Thus, ac/bd =a0c0/b0d0and
therefore (a/b)(c/d) = (a0/b0)(c0/d0).
56. First observe that 1 →1 so that 2 →2. Suppose √2→a+b√5.
Then 2 = √2√2→(a+b√5)2=a2+ 2ab√5+5b2. This would
imply that √5 is rational.
57. Let Fbe the field of quotients of Z[i].By definition
F={(a+bi)/(c+di)|a, b, c, d ∈Z}. Since Fis a field that contains
Zand i, we know that Q[i]⊆F. But for any (a+bi)/(c+di) in F
we have a+bi
c+di =a+bi
c+di
c−di
c−di =(ac+bd)+(bc−ad)i
c2+d2=ac+bd
c2+d2+(bc−ad)i
c2+d2∈Q[i].
58. Map [a/b] to ab−1.
59. The subfield of Eis {ab−1|a, b ∈D, b 6= 0}. Define φby
φ(ab−1) = a/b. Then φ(ab−1+cd−1) = φ((ad +bc)(bd)−1)) =
(ad +bc)/bd =ad/bd +bc/bd =a/b +c/d =φ(ab−1) + φ(cd−1).
Also, φ((ab−1)(cd−1)) = φ(acb−1d−1) = φ((ac)(bd)−1) = ac/bd =
(a/b)(c/d) = φ(ab−1)φ(cd−1).
15/Ring Homomorphisms 139
60. Zero divisors do not have multiplicative inverses.
61. Reflexive and symmetric properties follow from the commutativity
of D. For transitivity, assume a/b ≡c/d and c/d ≡e/f. Then
adf = (bc)f=b(cf) = bde, and cancellation yields af =be.
62. The set of even integers is a subring of the rationals.
63. Let φbe the mapping from Tto Qgiven by φ(ab−1) = a/b. Now
see Exercise 59.
64. Say 1Ris the unity of Rand 1Sis the unity of S. Pick a∈Rsuch
that φ(a)6= 0.Then 1Sφ(a) = φ(1Ra) = φ(1R)φ(a).Now cancel.
For the example, consider the mapping from Z3to Z6that sends x
to 4x.
65. Let anxn+an−1xn−1+··· +a0∈R[x] and suppose that
f(a+bi) = 0. Then an(a+bi)n+an−1(a+bi)n−1+··· +a0= 0.
By Example 2, the mapping φfrom Cto itself given by
φ(a+bi) = a−bi is a ring isomorphism. So, by property 1 of
Theorem 10.1,
0 = φ(0) = φ(an(a+bi)n+an−1(a+bi)n−1+··· +a0) =
φ(an)φ((a+bi))n+φ(an−1)φ((a+bi))n−1+··· +φ(a0) =
an(a−bi)n+an−1(a−bi)n−1+··· +a0=f(a−bi).
66. a. Apply the definition.
b. Ker φ=(" a b
c d #|a∈Z).
c. Use Theorem 15.3.
d. Yes, by Theorem 14.3.
e. No, by Theorem 14.4.
67. Certainly the unity 1 is contained in every subfield. So, if a field
has characteristic p, the subfield {0,1, . . . , p −1}is contained in
every subfield. If a field has characteristic 0, then
{(m·1)(n·1)−1|m, n ∈Z, n 6= 0}is a subfield contained in every
subfield. This subfield is isomorphic to Q[map (m·1)(n·1)−1to
m/n].
68. By part 5 of Theorem 6.2, the only possible isomorphism is given
by 1 →n. If this mapping is an isomorphism then 1 = 12→n2. So
n2=nmod 2nand it follows that nis odd. Now suppose nis odd.
15/Ring Homomorphisms 140
Then n(n−1) is divisible by 2nand n2=nmod 2n. This
guarantees that 1 →nis an isomorphism.
69. The mapping φ(x)=(xmod m, x mod n) from Zmn to Zm⊕Znis
a ring isomorphism.
70. This follows directly from the divisibilty by 11 test give in Exercise
32.
141
CHAPTER 16
Polynomial Rings
1. f+g= 3x4+ 2x3+ 2x+ 2
f·g= 2x7+ 3x6+x5+ 2x4+ 3x2+ 2x+ 2
2. Let f(x) = x4+xand g(x) = x2+x. Then
f(0) = 0 = g(0); f(1) = 2 = g(1); f(2) = 0 = g(2).
3. The zeros are 1, 2, 4, 5
4. Since Ris isomorphic to the subring of constant polynomials,
charR≤char R[x]. On the other hand, char R=cimplies
c(anxn+··· +a0) = (can)xn+··· + (ca0) = 0.
5. Write f(x)=(x−a)q(x) + r(x). Since deg(x−a) = 1, deg r(x) = 0
or r(x) = 0. So r(x) is a constant. Also, f(a) = r(a).
6. x2,x2+ 1, x2+x,x2+x+ 1
7. x3+ 1 and x3+x2+x+ 1.
8. There are 2npolynomials over Z2There are 4 polynomial functions
from Z2to Z2.
9. If ais a zero of f(x) we know by Corollary 1 of Theorem 16.2 that
the remainder when f(x) is divided by x−ais 0. So, x−ais a
factor of f(x). If x−ais a factor of f(x), then the remainder when
f(x) is divided by x−ais 0. So, by Corollary 1 of Theorem 16.2,
f(a) = 0.
10. Map the element rto the constant polynomial f(x) = r.
11. Let f(x), g(x)∈R[x]. By inserting terms with the coefficient 0 we
may write
f(x) = anxn+··· +a0and g(x) = bnxn+··· +b0.
16/Polynomial Rings 142
Then
φ(f(x) + g(x)) = φ(an+bn)xn+··· +φ(a0+b0)
= (φ(an) + φ(bn))xn+··· +φ(a0) + φ(b0)
= (φ(an)xn+··· +φ(a0))+(φ(bn)xn+··· +φ(b0))
=φ(f(x)) + φ(g(x)).
Multiplication is done similarly.
12. Use Exercise 9 and observe that φ(an)xn+··· +φ(a0) = 0 if and
only if φ(an)=0, . . . , φ(a0) = 0. Since φis an isomorphism, this
holds if and only if an= 0, . . . , a0= 0.
13. 4x2+ 3x+ 6 is the quotient and 6x+ 2 is the remainder.
14. 4x2+ 3x+ 6 is the quotient and 6x+ 2 is the remainder.
15. Observe that (2x+ 1)(2x+ 1) = 4x2+ 4x+ 1 = 1. So, 2x+ 1 is its
own inverse.
16. No. (See Exercise 15.)
17. No, because if an, bm∈Zpand are not zero, then
(anxn+··· +a0)(bmxm+··· +b0) = anbmxn+m+··· +a0b0and
anbm6= 0.
18. Consider f(x) = ax where ais a zero-divisor.
19. If f(x) = anxn+··· +a0and g(x) = bmxm+··· +b0, then
f(x)·g(x) = anbmxm+n+···+a0b0and anbm6= 0 when an6= 0 and
bm6= 0.
20. Observe that Q[x]/hxiis isomorphic to Q. Now use Theorem 14.4.
21. Let mbe the multiplicity of bin q(x). Then we may write
f(x)=(x−a)n(x−b)mq0(x) where q0(x) is in F[x] and q0(b)6= 0.
This means that bis a zero of f(x) of multiplicity at least m. If bis
a zero of f(x) of multiplicity greater than mthen bis a zero of
g(x) = f(x)/(x−b)m= (x−a)nq0(x). But then
0 = g(b)=(b−a)nq0(b) and therefore q0(b) = 0, which is a
contradiction.
16/Polynomial Rings 143
22. If there were infinitely many elements of order at most nfor some
positive integer n, then for some k≤nthere would be infinitely
many elements of order k. But then there would be infinitely many
zeros in Fof xk−1. This contradicts Corollary 3 of Theorem 16.2.
23. Suppose that f(x)6= 0 and f(x) is a polynomial of degree n. Then,
by Corollary 3 of Theorem 16.2, f(x) has at most nzeros. This is a
contradiction to the assumption that f(x) has infinitely many zeros.
24. If f(x) = g(x) for infinitely many elements of Fthen
h(x) = f(x)−g(x) has infinitely many zeros. So, by Exercise 23,
h(x) = 0.
25. If f(x)6=g(x), then deg[f(x)−g(x)] <deg p(x). But the minimum
degree of any member of hp(x)iis deg p(x). So, f(x)−g(x) does
not have a degree. This means that f(x)−g(x) = 0.
26. Consider h2, xi={2f(x) + xg(x)|f(x), g(x)∈Z[x]}.
27. We start with (x−1/2)(x+ 1/3) and clear fractions to obtain
(6x−3)(6x+ 2) as one possible solution.
28. If ahad multiplicity greater than 1, then we could write
f(x)=(x−a)2g(x). Now use the product rule to calculate f0(x).
29. The proof given for Theorem 16.2 with g(x) = x−ais valid over
any commutative ring with unity. Moreover, the proofs for
Corollaries 1 and 2 of Theorem 16.2 are also valid over any
commutative ring with unity.
30. Notice that the proof of the division algorithm holds for integral
domains when g(x) has the form x−a. Likewise the proofs of the
Factor Theorem and Corollary 3 of Theorem 16.2 hold.
31. Observe that f(x)∈Iif and only if f(1) = 0. Then if fand g
belong to Iand hbelongs to F[x], we have
(f−g)(1) = f(1) −g(1) = 0 −0 and
(hf)(1) = h(1)f(1) = h(1) ·0 = 0. So, Iis an ideal. By
Theorem 16.4, I=hx−1i.
32. Use the Factor Theorem.
33. This follows directly from Corollary 2 of Theorem 15.5 and
Exercise 11 in this chapter.
16/Polynomial Rings 144
34. Consider the ideal hx3−xi.
35. For any ain U(p), ap−1= 1, so every member of U(p) is a zero of
xp−1−1. From the Factor Theorem (Corollary 2 of Theorem 16.2)
we obtain that g(x)=(x−1)(x−2) ···(x−(p−1)) is a factor of
xp−1−1. Since both g(x) and xp−1−1 have lead coefficient 1, the
same degree, and their difference has p−1 zeros, their difference
must be 0 (for otherwise their difference would be a polynomial of
degree less than p−1 that had p−1 zeros).
36. By Theorem 16.4 the only possibility for g(x) is ±(x−1). By
Theorem 15.3 Z[x]/Ker φis isomorphic Z. The only possibilities
for g(x) are a(x−1) where ais any nonzero rational number.
Q[x]/Ker φis isomorphic Q.xin Z. But then g(x) = f(x)−ahas
infinitely many zeros. This contradicts Corollary 3 of Theorem 16.2.
37. C(x) (field of quotients of C[x]). Since pdoes not divide (p−1) we
know that pdivides (p−2)! −1. Thus, (p−2)! mod p= 1.
38. When nis prime, use Exercise 37. When nis composite and
greater than 4, (n−1)! mod n= 0.
39. By Exercise 36, (p−1)! mod p=p−1. So, pdivides
(p−1)! −(p−1) = (p−1)((p−2)! −1).
40. By Exercise 39 we have 1 = 99! mod 101 = (−2)98! mod 101.
41. Observe that, modulo 101,
(50!)2= (50!)(−1)(−2) ···(−50) = (50!)(100)(99) ···(51) = 100!.
And by Exercise 38, 100! mod 101 = 100 = −1 mod 101.
42. This follows directly from the definitions.
43. Note that I=h2iis maximal in Zbut I[x] is not maximal in Z[x]
since I[x] is properly contained in the ideal
{f(x)∈Z[x]|f(0) is even }.
44. That I[x] is an ideal is straightforward. To prove that I[x] is prime
let f(x) = amxm+··· +a0and g(x) = bnxn+··· +b0and suppose
f(x)g(x)∈I[x]. By filling in with coefficients of 0 we may assume
that m=n. We must show that all ai∈Ior all bi∈I. Suppose
some bi6∈ Iand let kbe the least integer such that bk6∈ I. The
coefficient xkin f(x)g(x) is akb0+ak−1b1+··· +a0bkand belongs
16/Polynomial Rings 145
to I. Thus a0bk∈Iand therefore a0∈I. The coefficient of xk+1 in
f(x)g(x) is ak+1b0+akb1+··· +a1bk+a0bk+1 ∈I. Thus, a1bk∈I
and therefore a1∈I. Continuing in this fashion we obtain all
ai∈I.
45. Since F[x] is a PID, hf(x), g(x)i=ha(x)ifor some a(x)∈F[x].
Thus a(x) divides both f(x) and g(x). This means that a(x) is a
constant. So, by Exercise 17 in Chapter 14, hf(x), g(x)i=F[x].
Thus, 1 ∈ hf(x), g(x)i.
46. Mimic Example 3.
47. By the Factor Theorem (Corollary 2 of Theorem 16.2) we may
write f(x)=(x−a)g(x). Then f0(x)=(x−a)g0(x) + g(x). Thus,
g(a) = 0 and by the Factor Theorem x−ais a factor of g(x).
48. If fand g∈I, then (f−g)(a) = f(a)−g(a) = 0 for all a. If f∈I
and g∈F[x], then (gf)(a) = g(a)f(a) = 0 for all a. If |F|=n,
then xkn−k+1 −x∈Ifor all positive integers k. If Fis infinite use
the Factor Theorem. If F={a1, a2, . . . , an}then
g(x) = (x−a1)(x−a2)···(x−an).
49. Say deg g(x) = m, deg h(x) = n, and g(x) has leading coefficient a.
Let k(x) = g(x)−axm−nh(x). Then deg k(x)<deg g(x) and h(x)
divides k(x) in Z[x] by induction. So, h(x) divides
k(x) + axm−nh(x) = g(x) in Z[x].
50. The mapping φ(f(x)) = f(x2) is a ring-isomorphism from R[x] onto
R[x2].
51. If f(x) takes on only finitely many values then there is at least one
ain Zwith the property that f(x) = afor infinitely many xin Z.
But then g(x) = f(x)−ahas infinitely many zeros. This
contradicts Corollary 3 of Theorem 16.2.
52. Since f(x) takes on infinitely many values over Z, there is an ain
Zsuch that f(a)6=±1 and f(a)6= 0. Because every nonzero
element of hf(x)ihas degree at least 1 and f(a) has degree 0,
hf(x), f(a)iproperly contains hf(x)i. Moreover hf(x), f(a)idoes
not contain 1. For if so, then there are g(x) and h(x) in Z[x] such
that f(x)g(x) + f(a)h(x) = 1. Evaluating both sides at x=agives
f(a)g(a) + f(a)h(a) = 1, which is a contradiction since f(a) does
not divide 1.
16/Polynomial Rings 146
53. Observe that (1 + tmx)(1 −tmx) = 1.
54. Say f(x) = anxn+an−1xn−1+··· +a1x+a0. Then using the fact
that ap
i=aifor all iand Exercise 49 of Chapter 13, we have
f(bp) = ap
n(bp)n+ap
n−1(bp)n−1+··· +ap
1bp+ap
0=
ap
n(bn)p+ap
n−1(bn−1)p+··· +ap
1b+ap
0=
(anbn+an−1bn−1+··· +a1b+a0)p=f(b)p= 0.
55. Let f(x) = anxn+an−1xn−1+··· +a1x+a0and assume that p/q
is a zero of f(x) where pand qare integers and nis even. We may
assume that pand qare relatively prime. Substituting p/q for x
and clearing fractions we have
anpn+an−1pn−1q+··· +a1pqn−1=−a0qn.If pis even, then the
left side is even. If pis odd, then each summand on the left side is
odd and since there is an even number of summands, the left side is
still even. Because a0is odd we then have that qis even. It follows
that anpn=−(an−1pn−1q+··· +a1pqn−1+a0qn) is even since the
right side is divisible by q. This implies that pis even. This
contradicts the assumption that pand qare relatively prime.
56. Say (f(x)/g(x))2=x. We may assume that f(x) and g(x) have no
common factor for, if so, we can cancel them. Since
(f(x))2=x(g(x))2we see that f(0) = 0. Thus, f(x) has the form
xk(x). Then x2(k(x))2=x(g(x))2and therefore x(k(x))2= (g(x))2.
This implies that g(0) = 0. But then f(x) and g(x) have xas a
common factor.
57. By the Division Algorithm (Theorem 16.2) we may write
x43 = (x2+x+ 1)q(x) + r(x) where r(x) = 0 or deg r(x)<2.
Thus, r(x) has the form cx +d. Then x43 −cx −dis divisible by
x2+x+ 1. Finally, let a=−cand b=−d.
58. By way of contradiction suppose aibjis not an integer and iis the
least integer for which aibjis not an integer for some j. Then
aibj+ai−1bj+1 +···+a0bi+jis an integer (since it is the coefficient
of xi+j) and all terms after the first are integers. Thus aibjis an
integer.
59. Observe that every term of f(a) has the form ciaiand
ciaimod m=cibimod m. To prove the second statement assume
that there is some integer kso that f(k) = 0. If kis even, then
because kmod 2 = 0, we have by the first statement
16/Polynomial Rings 147
0 = f(k) mod 2 = f(0) mod 2 so that f(0) is even. This shows that
kis not even. If kis odd, then kmod 2 = 1, so by the first
statement f(k) = 0 is odd. This contradiction completes the proof.
60. Since x+ 4 = x−3 in Z7[x] we have by the Remainder Theorem
that the remainder is 351 mod 7. Since 3 is in U(7) we also know
that 36= 1 mod 7. Thus, 351 mod 7 = 34833mod 7 = 6.
61. A solution to x25 −1 = 0 in Z37 is a solution to x25 = 1 in U(37).
So, by Corollary 2 of Theorem 4.1, |x|divides 25. Moreover, we
must also have that |x|divides |U(37)|= 36. So, |x|= 1 and
therefore x= 1.
148
CHAPTER 17
Factorization of Polynomials
1. By Theorem 17.1, f(x) is irreducible over R. Over Cwe have
2x2+ 4 = 2(x2+ 2) = 2(x+√2i)(x−√2i).
2. f(x) factors over Das ah(x) where ais not a unit.
3. If f(x) is not primitive, then f(x) = ag(x), where ais an integer
greater than 1. Then ais not a unit in Z[x] and f(x) is reducible.
4. Say r=p/q where pand qare relatively prime. Viewing f(x) as an
element of Q[x] we have from the Factor Theorem (Corollary 2 of
Theorem 16.2) that f(p/q) = 0. Clearing fractions and collecting
all terms and isolating the pnterm on one side we see that qdivides
pn. Using the fact that pand qare relatively prime, we conclude
that q= 1.
5. a. If f(x) = g(x)h(x), then af(x) = ag(x)h(x).
b. If f(x) = g(x)h(x), then f(ax) = g(ax)h(ax).
c. If f(x) = g(x)h(x), then f(x+a) = g(x+a)h(x+a).
d. Let f(x)=8x3−6x+ 1. Then
f(x+ 1) = 8(x+ 1)3−6(x+ 1) + 1 =
8x3+ 24x2+ 24x+ 8 −6x−6 + 1 = 8x3+ 24x2−18x+ 3.
By Eisenstein’s Criterion (Theorem 17.4), f(x+ 1) is
irreducible over Qand by part c, f(x) is irreducible over Q.
6. Use Exercise 5(a).
7. Use Exercise 5a and clear fractions.
8. By Corollary 1 of Theorem 17.5 we know the set is a field. To see
that it has pnelements, note that any element
g(x) + hf(x)i=f(x)q(x) + r(x) + hf(x)i=r(x) + hf(x)i
where r(x) = 0 or deg r(x)< n. So, all cosets have the form
an−1xn−1+··· +a0+hf(x)i
17/Factorization of Polynomials 149
and they are all distinct.
9. It follows from Theorem 17.1 that p(x) = x2+x+ 1 is irreducible
over Z5. Then from Corollary 1 of Theorem 17.5 we know that
Z5[x]/hp(x)iis a field. To see that this field has order 25 note that
if f(x) + hp(x)iis any element of Z5[x]/hp(x)i, then by the Division
Algorithm (Theorem 16.2) we may write f(x) + hp(x)iin the form
p(x)q(x) + ax +b+hp(x)i=ax +b+hp(x)i. Moreover,
ax +b+hp(x)i=cx +d+hp(x)ionly if a=cand b=d, since
(a−c)x+b−dis divisible by hp(x)ionly when it is 0. So,
Z5[x]/hp(x)ihas order 25.
10. Find an irreducible cubic over Z3and mimic Exercise 9.
11. Note that −1 is a zero. No, since 4 is not a prime.
12. (a) Irreducible by Eisenstein
(b) Irreducible by the Mod 2 Test (but be sure to check for
quadratic factors as well as linear)
(c) Irreducible by Eisenstein
(d) Irreducible by the Mod 2 Test
(e) Irreducible by Eisenstein (after clearing fractions)
13. Let f(x) = x4+ 1 and g(x) = f(x+ 1) = x4+ 4x3+ 6x2+ 4x+ 2.
Then f(x) is irreducible over Qif g(x) is. Eisenstein’s Criterion
shows that g(x) is irreducible over Q. To see that x4+ 1 is
reducible over R, observe that
x8−1=(x4+ 1)(x4−1)
so any complex zero of x4+ 1 is a complex zero of x8−1. Also note
that the complex zeros of x4+ 1 must have order 8 (when
considered as an element of C). Let ω=√2/2 + i√2/2. Then
Example 2 in Chapter 16 tells us that the complex zeros of x4+ 1
are ω,ω3,ω5, and ω7, so
x4+ 1 = (x−ω)(x−ω3)(x−ω5)(x−ω7).
But we may pair these factors up as:
((x−ω)(x−ω7))((x−ω3)(x−ω5))
= (x2−√2x+ 1)(x2+√2x+ 1)
17/Factorization of Polynomials 150
to factor using reals (see DeMoivre’s Theorem, Example 8 in
Chapter 0).
14. Use Theorem 17.1.
15. (x+ 3)(x+ 5)(x+ 6)
16. (x+ 1)3
17. a. Since every reducible polynomial of the form x2+ax +bcan
be written in the form (x−c)(x−d) we need only count the
number of distinct such expressions over Zp. Note that there
are p(p−1) expressions of the form (x−c)(x−d) where c6=d.
However, since (x−c)(x−d) = (x−d)(x−c) there are only
p(p−1)/2 distinct such expressions. To these we must add the
pcases of the form (x−c)(x−c). This gives us
p(p−1)/2 + p=p(p+ 1)/2.
b. First note that for every reducible polynomial of the form
f(x) = x2+ax +bover Zpthe polynomial cf(x) (c6= 0) is
also reducible over Zp. By part a, this gives us at least
(p−1)p(p+ 1)/2 = p(p2−1)/2 reducible polynomials over Zp.
Conversely, every quadratic polynomial over Zpcan be written
in the form cf(x) where f(x) has lead coefficient 1. So, the
p(p2−1)/2 reducibles we have already counted includes all
cases.
18. Use Exercise 17.
19. By Exercise 18, for each prime pthere is an irreducible polynomial
p(x) of degree 2 over Zp. By Corollary 1 of Theorem 17.5,
Zp[x]/hp(x)iis a field. By the Division Algorithm (Theorem 16.2)
every element in Zp[x]/hp(x)ican be written in the form
ax +b+hp(x)i. Moreover, ax +b+hp(x)i=cx +d+h(p(x)ionly
when a=cand c=dsince (ax +b)−(cx +d) is divisible by p(x)
only when it is 0. Thus, Zp[x]/hp(x)ihas order p2.
20. By Eisenstein, xn+pwhere pis a prime is irreducible over Q.
21. Consider the mapping from Z3[x] onto Z3[i] given by
φ(f(x)) = f(i). Since φ(f(x) + g(x)) = φ((f+g)(x)) = (f+g)(i) =
f(i) + g(i) = φ(f(x)) + φ(g(x)) and
φ(f(x)g(x)) = φ((fg)(x) = (fg)(i) = f(i)g(i) = φ(f(x))φ(g(x)), φ
is a ring homomorphism. Because φ(x2+ 1) = i2+ 1 = −1 + 1 = 0
17/Factorization of Polynomials 151
we know that x2+ 1 ∈Ker φ. From Theorem 16.4 we have that
Ker φ=hx2+ 1i. Finally, Theorem 15.3 gives us that
Z3[x]/hx2+ 1i ≈ Z3[i].
22. If a2x2+a1x+a0∈Zp[x] is a factor of f(x) with a26= 0 then
a−1
2(a2x2+a1x+a0) is a factor of f(x).
23. x2+ 1, x2+x+ 2, x2+ 2x+ 2
24. If so, then πis a zero of x2−ax −b.
25. 1 has multiplicity 1, 3 has multiplicity 2.
26. For f(x), both methods yield 4 and 5. (Notice that
√−47 = √2 = ±3). Neither method yields a solution for g(x). The
quadratic formula applied to g(x) involves √−23 = √2 and there is
no element of Z5whose square is 2. ax2+bx +c(a6= 0) has a zero
in Zp[x] if and only if b2−4ac =d2for some din Zp.
27. We know that an(r/s)n+an−1(r/s)n−1+··· +a0= 0. So, clearing
fractions we obtain anrn+san−1rn−1+··· +sna0= 0. This shows
that s|anrnand r|sna0. By Euclid’s Lemma (Chapter 0), s
divides anor sdivides rn. Since sand rare relatively prime, s
must divide an. Similarly, rmust divide a0.
28. Since a1(x)a2(x)···ak(x) = a1(x)(a2(x)···ak(x)), we have by
Corollary 2 of Theorem 17.5 that p(x) divides a1(x) or p(x) divides
a2(x)···ak(x). In the latter case, the Second Principle of
Mathematical Induction implies that p(x) divides some ai(x) for
i= 2,3, . . . , k.
29. If there is an ain Zpsuch that a2=−1, then
x4+ 1 = (x2+a)(x2−a). If there is an ain Zpsuch that a2= 2,
then x4+ 1 = (x2+ax + 1)(x2−ax + 1). If there is an ain Zpsuch
that a2=−2, then x4+ 1 = (x2+ax −1)(x2−ax −1).
To show that one of these three cases must occur, consider the
group homomorphism from Z∗
pto itself given by x→x2. Since the
kernel is {1,−1}, the image Hhas index 2 (we may assume that
p6= 2). Suppose that neither −1 nor 2 belongs to H. Then, since
there is only 1 coset other than H, we have −1H= 2H. Thus,
H= (−1H)(−1H)=(−1H)(2H) = −2H, so that −2 is in H.
30. Use the corollary to Theorem 17.4 and Exercise 5b with a=−1.
17/Factorization of Polynomials 152
31. Since (f+g)(a) = f(a) + g(a) and (f·g)(a) = f(a)g(a), the
mapping is a homomorphism. Clearly, p(x) belongs to the kernel.
By Theorem 17.5, hp(x)iis a maximal ideal, so the kernel is hp(x)i.
32. Since x2+ 1 is irreducible over Zit follows from Theorem 17.6 that
hx2+ 1iis prime. Let A={f(x)∈Z[x]|f(1) is even }. Then Ais a
proper ideal of Z[x] that properly contains hx2+ 1i.
33. Consider the mapping φfrom Fto F[x]/hp(x)igiven by
φ(a) = a+hp(x)i. By observation, φis one-to-one and onto.
Moreover,
φ(a+b) = a+b+hp(x)i=a+hp(x)i+b+hp(x)i=φ(a) + φ(b)
and φ(ab) = ab +hp(x)i= (a+hp(x)i)(b+hp(x)i) = φ(a)φ(b) so φ
is a ring isomorphism.
34. Let f(x) = g(x)h(x) where deg g(x)<deg f(x) and deg h(x)<
deg f(x). Then g(x)h(x) belongs to hf(x)ibut neither g(x) nor
h(x) belongs to hf(x)i.
35. f(x) is primitive.
36. Suppose that r+ 1/r = 2k+ 1 where kis an integer. Then
r2−2kr −r+ 1 = 0. It follows from Exercise 4 of this chapter that
ris an integer. But the mod 2 irreducibility test shows that the
polynomial x2−(2k+ 1)x+ 1 is irreducible over Qand an
irreducible quadratic polynomial cannot have a zero in Q.
37. Although the probability of rolling any particular sum is the same
with either pair of dice, the probability of rolling doubles is
different (1/6 with ordinary dice, 1/9 with Sicherman dice). Thus,
the probability of going to jail is different. Other probabilities are
also affected. For example, if in jail one cannot land on Virginia by
rolling a pair of 2’s with Sicherman dice, but one is twice as likely
to land on St. James with a pair of 3’s with the Sicherman dice as
with ordinary dice.
38. The nonstandard pair of tetrahedron dice are labeled 1, 2, 2, 3 and
1, 3, 3, 5.
39. The analysis is identical except that 0 ≤q, r, t, u ≤n. Now just as
when n= 2, we have q=r=t= 1, but this time 0 ≤u≤n.
However, when u > 2, P (x) = x(x+ 1)(x2+x+ 1)(x2−x+ 1)uhas
(−u+ 2)x2u+3 as one of its terms. Since the coefficient of x2u+3
17/Factorization of Polynomials 153
represents the number of dice with the label 2u+ 3, the coefficient
cannot be negative. Thus, u≤2, as before.
40. Adding 1 and 4 to each label for the eighteen sided die yields the
same frequencies for the sums as does an ordinary pair of dice.
154
CHAPTER 18
Divisibility in Integral Domains
1. 1. |a2−db2|= 0 implies a2=db2. Thus a=0=b, since
otherwise d= 1 or dis divisible by the square of a prime.
2. N((a+b√d)(a0+b0√d)) = N(aa0+dbb0+ (ab0+a0b)√d) =
|(a2−db2)(a02−db02)|=|(aa0+dbb0)2−d(ab0+a0b)2|=
|a2a02+d2b2b02−da2b02−da02b2|=|a2−db2||a02−db02|=
N(a+b√d)N(a0+b0√d).
3. If xy = 1, then 1 = N(1) = N(xy) = N(x)N(y) and
N(x) = 1 = N(y). If N(a+b√d) = 1, then
±1 = a2−db2= (a+b√d)(a−b√d) and a+b√dis a unit.
4. This part follows directly from 2 and 3.
2. Say a=bu where uis a unit. The ra =rbu = (ru)b∈ hbiso that
hai ⊆ hbi. By symmetry, hbi ⊆ hai. If hai=hbi, then a=bu and
b=av. Thus, a=avu and uv = 1.
3. Let I=∪Ii. Let a, b ∈Iand r∈R. Then a∈Iifor some iand
b∈Ijfor some j. Thus a, b ∈Ik, where k= max{i, j}. So,
a−b∈Ik⊆Iand ra and ar ∈Ik⊆I.
4. Say ris irreducible and uis a unit. If ru =ab where aand bare
not units, then r=a(bu−1) where aand bu−1are not units.
5. Clearly, habi⊆hbi. If habi=hbi, then b=rab, so that 1 = ra and a
is a unit.
6. a∼asince a=a·1; if a∼b, say a=bu where uis a unit, then
b=au−1so b∼a; if a∼b, say a=bu where uis a unit and b∼c,
say b=cr where ris a unit, then a=bu =cru where ru is a unit.
7. Say x=a+bi and y=c+di. Then
xy = (ac −bd)+(bc +ad)i.
18/Divisibility in Integral Domains 155
So
d(xy)=(ac −bd)2+ (bc +ad)2= (ac)2+ (bd)2+ (bc)2+ (ad)2.
On the other hand,
d(x)d(y) = (a2+b2)(c2+d2) = a2c2+b2d2+b2c2+a2d2.
8. First observe that for any r∈D, d(1) ≤d(1 ·r) = d(r) so that d(1)
is the minimum value of d. Now if uis a unit, then
d(u)≤d(uu−1) = d(1) so that d(u) = d(1). If d(u) = d(1), then
1 = uq +rwhere r= 0 or d(r)< d(u) = d(1). So r= 0.
9. Suppose a=bu, where uis a unit. Then d(b)≤d(bu) = d(a). Also,
d(a)≤d(au−1) = d(b).
10. Mimic the proof of Theorem 17.5.
11. m= 0 and n=−1 give q=−i, r =−2−2i.
12. Use the Ascending Chain Condition.
13. First observe that 21 = 3 ·7 and that 21 = (1 + 2√−5)(1 −2√−5).
To prove that 3 is irreducible in Z[√−5] suppose that 3 = xy,
where x, y ∈Z[√−5] and xand yare not units. Then
9 = N(3) = N(x)N(y) and, therefore, N(x) = N(y) = 3. But there
are no integers aand bsuch that a2+ 5b2= 3. The same argument
shows that 7 is irreducible over Z[√−5]. To show that 1 + 2√−5 is
irreducible over Z[√−5] suppose that 1 + 2√−5 = xy, where
x, y ∈Z[√−5] and xand yare not units. Then
21 = N(1 + 2√−5) = N(x)N(y). Thus N(x) = 3 or N(x) = 7, both
of which are impossible.
14. Notice that d(1 −i) is a prime number.
15. First observe that 10 = 2 ·5 and that 10 = (2 −√−6)(2 + √−6).
To see that 2 is irreducible over Z[√−6] assume that 2 = xy, where
x, y ∈Z[√−6] and xand yare not units. Then
4 = N(2) = N(x)N(y) so that N(x) = 2. But 2 cannot be written
in the form a+ 6b2. A similar argument applies to 5. To see that
2−√−6 is irreducible suppose that 2 −√−6 = xy where
x, y ∈Z[√−6] and xand yare not units. Then
10 = N(2 −√−6) = N(x)N(y) and as before this is impossible. We
know that Z[√−6] is not a principle ideal domain because a PID is
a UFD (Theorem 18.3).
18/Divisibility in Integral Domains 156
16. C[x] is a UFD but contains Z[√−6].
17. Suppose 3 = αβ, where α, β ∈Z[i] and neither is a unit. Then
9 = d(3) = d(α)d(β), so that d(α) = 3. But there are no integers
such that a2+b2= 3. Observe that 2 = −i(1 + i)2and
5 = (1 + 2i)(1 −2i) and 1 + i, 1+2i, and 1 −2iare not units.
18. If 7 = (a+b√6)(c+d√6) and neither factor is a unit, then
|a2−6b2|= 7 and, modulo 7, a2= 6b2. However the only solutions
to this equation modulo 7 are a=b= 0. In this case c+d√6 is a
unit.
19. Use Exercise 1 with d=−1.5 and 1 + 2i; 13 and 3 + 2i; 17 and 4 + i.
20. Use Example 1 and Theorem 18.2.
21. Suppose that 1 + 3√−5 = xy, where x, y ∈Z[√−5] and xand yare
not units. Then 46 = N(1 + 3√−5) = N(x)N(y). Thus, N(x)=2
or N(x) = 23. But neither 2 nor 5 can be written in the form
a2+ 5b2so 1 + 3√−5 is irreducible over Z[√−5]. To see that
1+3√−5 is not prime, observe that
(1 + 3√−5)(1 −3√−5) = 1 + 45 = 46 so that 1 + 3√−5 divides
2·23. For 1 + 3√−5 to divide 2 we need 46 = N(1 + 3√−5) divides
N(2) = 4. Likewise, for 1 + 3√−5 to divide 23 we need that 46
divides 232. Since neither of these is true, 1 + 3√−5 is not prime.
22. To show the two elements are irreducible mimic the solution given
for Exercise 18 but use modulo 4. To show the two elements are
not prime use the observation that 2 ·2 = 4 = (1 + √5)(−1 + √5)
together with the fact that and 1 + √5 are irreducible.
23. First observe that (−1 + √5)(1 + √5) = 4 = 2 ·2 and by
Exercise 22, 1 + √5 and 2 are irreducible over Z[√5]. To see that
−1 + √5 is irreducible over Z[√5] suppose that −1 + √5 = xy
where x, y ∈Z[√5] and xand yare not units. Let x=a+b√5.
Then 4 = N(−1 + √5) = N(x)N(y) so that a2−5b2=±2.
Viewing this equation modulo 5 gives us a2= 2 or a2=−2 = 3.
However, every square in Z5is 0, 1, or 4.
24. Let Ibe a prime ideal in F[x] and let Mbe a proper ideal that
contains I. By Theorem 16.3 there are elements f(x) and g(x) in
F[x] such that M=hf(x)iand I=hg(x)i. By definition g(x) is a
prime and by Theorem 18.2 it is irreducible over F. Since M
18/Divisibility in Integral Domains 157
contains Ithere is an h(x) in F[x] such that g(x) = f(x)h(x) and
since g(x) is irreducible over F, h(x) is a unit. Thus by Exercise 2
in this chapter M=I.
25. Suppose that x=a+b√dis a unit in Z[√d]. Then
1 = N(x) = a2+ (−d)b2. But −d > 1 implies that b= 0 and
a=±1.
26. Observe that (3 + 2√2)(3 −2√2) = 1. So,
(3 + 2√2)n(3 −2√2)n= ((3 + 2√2)(3 −2√2))n= 1.
27. 1 = N(ab) = N(a)N(b) so N(a) = 1 = N(b).
28. To see that 2 is prime in Z12 note that 2 is not a unit in Z12 and
suppose that bc = 2twhere b, c and tbelong to Z12. Thus there is
an integer ksuch that bc −2t= 12kin the ring Z. This implies
that 2 divides bc in Z. Thus one of bor cis divisible by 2 in Z12.
The same argument applies when 2 is replaced by 3. To see that 2
is irreducible in Z12, suppose that 2 = bc in Z12. Then there is an
integer ksuch that 2 −bc = 12kin Z. By Euclid’s Lemma this
implies that 2 divides bor cin Z. Say, b= 2d. Then in Zwe have
2−2dc = 12kwhich reduces to 1 = dc + 6k. This implies that cis
relatively prime to 6 and therefore a unit in Z12. But the only
solutions to the equation 2 = 2xin Z12 are x= 1 and x= 7. If
dc = 1 then cis a unit and if dc = 7 then in Z12,7dc = 7 ·7 mod 12
= 1 and cis again a unit. To see that 3 is reducible in Z12 note
that 3 = 3 ·9 mod 12 and neither 3 nor 9 is a unit.
29. Suppose that bc =pt in Zn. Then there exists an integer ksuch
that bc =pt +kn. This implies that pdivides bc in Zand by
Euclid’s Lemma we know that pdivides bor pdivides c.
30. First suppose that that pis irreducible in Zn. If p2does not divide
nthen there are integers sand tsuch that 1 = ps + (n/p)t. Thus,
p=p(ps) mod n. Since both pand ps are zero divisors in Znthey
cannot be units. The contradicts our assumption that pis
irreducible in Zn. Now assume that p2divides nand
31. See Example 3.
32. If (a+bi) is a unit then a2+b2= 1. Thus, ±1,±i.
33. Note that p|(a1a2···an−1)animplies that p|a1a2···an−1or p|an.
Thus, by induction, pdivides some ai.
18/Divisibility in Integral Domains 158
34. 4x+ 1 and 2x+ 3 are associates of 3x+ 2 and x+ 4.
35. By Exercise 10, hpiis maximal and by Theorem 14.4, D/hpiis a
field.
36. Let abe a nonzero element of the domain. It suffices to show that
ais a unit. Consider the chain hai⊇ha2i⊇ha3i⊇ha4i ⊇ ···. By
hypothesis, we have hani=han+1ifor some n. Thus an=an+1bfor
some band 1 = ab.
37. Suppose Rsatisfies the ascending chain condition and there is an
ideal Iof Rthat is not finitely generated. Then pick a1∈I. Since
Iis not finitely generated, ha1iis a proper subset of I, so we may
choose a2∈Ibut a2/∈ ha1i. As before, ha1, a2iis proper, so we
may choose a3∈Ibut a3/∈ ha1, a2i. Continuing in this fashion, we
obtain a chain of infinite length ha1i ⊂ ha1, a2i⊂ha1, a2, a3i ⊂ . . ..
Now suppose every ideal of Ris finitely generated and there is a
chain I1⊂I2⊂I3⊂ ···. Let I=∪Ii. Then I=ha1, a2, . . . , anifor
some choice of a1, a2, . . . , an. Since I=∪Ii, each aibelongs to
some member of the union, say Ii0. Letting
k= max {i0|i= 1, . . . , n}, we see that all ai∈Ik. Thus, I⊆Ikand
the chain has length at most k.
38. The set of complex numbers is a Euclidean domain (take d(a)=0
for all a6= 0) containing Z[√−5] as a subdomain. Now use
Example 7 and the corollary to Theorem 18.4.
39. Say I=ha+bii. Then a2+b2+I= (a+bi)(a−bi) + I=Iand
therefore a2+b2∈I. For any c, d ∈Z, let c=q1(a2+b2) + r1and
d=q2(a2+b2) + r2, where 0 ≤r1, r2< a2+b2. Then
c+di +I=r1+r2i+I.
40. −1 + √2; infinite.
41. N(6 + 2√−7) = 64 = N(1 + 3√−7). The other part follows
directly from Exercise 25.
42. Let Ii={(a1, a2, . . . , ai,0,0,0, . . .)}, where the a’s are integers.
43. Theorem 18.1 shows that primes are irreducible. So, assume that a
is an irreducible in a UFD Rand that a|bc in R. We must show
that a|bor a|c. Since a|bc, there is an element din Rsuch that
bc =ad. Now replacing b, c, and dby their factorizations as a
18/Divisibility in Integral Domains 159
product of irreducibles, we have by the uniqueness property that a
(or an associate of a) is one of the irreducibles in the factorization
of bc. Thus, ais a factor of bor ais a factor of c.
44. Observe that both x2and x3are irreducible over Fbut
x3x3=x2x2x2.
45. See Exercise 21 in Chapter 0.
46. 0 is a solution.
160
SUPPLEMENTARY EXERCISES FOR CHAPTERS 15-18
1. Let φbe a ring homomorphism from Zonto a field and let Ker φ=
nZ. Then by Theorem 15.3 we have Z/nZ ≈Znis a field. From
Theorem 14.3 we have that nZ is a prime ideal of Z, and from
Exercise 8 in Chapter 14, we know that nis a prime.
2. The identity map and r+s√2→r−s√2.
3. Let x, y ∈A∩Band r∈A. The x−y∈A∩Bbecause A∩Bis a
subgroup of A. Also, rx and xr are in Asince Ais closed under
multiplication and rx and xr are in Bsince Bis an ideal. This
proves that A∩Bis an ideal of A.
To prove that second statement, define a mapping from Ato
(A+B)/B by φ(a) = a+B. Since for any b∈Bwe have
a+B=a+b+Bwe know that φis onto. φpreserves addition and
multiplication because
φ(a+a0) = a+a0+B= (a+B)+(a0+B) = φ(a) + φ(a0) and
φ(aa0) = aa0+B= (a+B)(a0+B) = φ(a)φ(a0). So, φis a ring
homomorphism. Now observe that φ(a) = a+B= 0 + Bif and
only if a∈A∩B. So, by the First Isomorphism Theorem
(Theorem 15.3), A/(A+B)≈(A+B)/B.
4. To show the isomorphism, use the First Isomorphism Theorem.
5. Define a mapping from F[x] onto F[x]/hf(x)i ⊕ F[x]/hg(x)iby
φ(h(x)) = (h(x) + hf(x)i, h(x) + hg(x)i). To verify that φis a ring
homomorphism note that
φ(h(x) + k(x)) = (h(x) + k(x) + hf(x)i, h(x) + k(x) + hg(x)i) =
(h(x) + hf(x)i+k(x) + hf(x)i,(h(x) + hg(x)i+k(x) + hg(x)i) =
(h(x) + hf(x)i, h(x) + hg(x)i)+(k(x) + hf(x)i, k(x) + hg(x)i) =
φ(h(x)) + φ(k(x)) and
φ((h(x)k(x)) = (h(x)k(x) + hf(x)i, h(x)k(x) + hg(x)i)=(h(x) +
hf(x)i, h(x)+hg(x)i)(k(x)+ hf(x)i, k(x)+hg(x)i) = φ(h(x))φ(k(x)).
Next observe that h(x) belongs to Ker φif and only if
h(x)∈ hf(x)iand h(x)∈ hg(x)i. This condition is equivalent to
f(x) divides h(x) and g(x) divides h(x). Since f(x) and g(x) are
SUPPLEMENTARY EXERCISES FOR CHAPTERS 15-18 161
irreducibles and not associates this is equivalent to f(x)g(x) divides
h(x). So, Ker φ=hf(x)g(x)iand the statement follows from the
First Isomorphism Theorem (Theorem 15.3).
6. Mimic Exercise 5.
7. The homomorphism from Z[x] onto Z2[x] whereby each coefficient
is reduced modulo 2 has kernel Kequal to the set of all
polynomials whose coefficients are even, so Kis an ideal. To show
that Kis prime it suffices to observe that, by Theorem 16.3, Z2[x]
is an integral domain and appeal to Theorem 14.3.
8. Observe that R/I has only two elements so it is a field. Then use
Theorem 14.4.
9. As in Example 7 of Chapter 6, the mapping is onto, is one-to-one,
and preserves multiplication. Also, a(x+y)a−1=axa−1+aya−1,
so that it preserves addition as well.
10. Use the norm Nfrom Example 8 of Chapter 18.
11. We can do both parts by showing that
Z[i]/h2+ii={0+h2+ii,1+h2 +ii,2+h2+ ii,3+h2+ ii,4+h2+ii},
which is obviously isomorphic to Z5. To this end note that
2 + i+h2 + ii= 0 + h2 + iiimplies that i+h2 + ii=−2 + h2 + ii.
Thus, any coset a+bi +h2 + iican be written in the form
c+h2 + ii. Moreover, because
5 + h2 + ii= (2 + i)(2 −i) + h2 + ii= 0 + h2 + ii, we may assume
that 0 ≤c < 5. Lastly, we must prove that these five cosets are
distinct. It is enough to show that if c= 1,2,3, or 4, then
c+h2 + ii 6= 0 + h2 + ii. If c+h2 + ii= 0 + h2 + iithen there is an
element a+bi such that c= (2 + i)(a+bi). But then
c2=N(c) = N(2 + i)N(a+bi) = 5(a2+b2), which implies that 5
divides c.
12. Every ideal of the homomorphic image would be generated by a
single element but the image need not be an integral domain.
13. Let anxn+··· +a1x+a0belong to Zp[x]. Then using the fact that
for all ain Zp, a =apand Exercise 49 in Chapter 13, we have
f(xp) = an(xp)n+··· +a1xp+a0=ap
n(xn)p+··· +ap
1xp+ap
0=
(anxn+··· +a1x+a0)p= (f(x))p.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 15-18 162
14. Suppose φis a ring homomorphism from Zmto Zpk. Let φ(1) = a,
where we may assume 0 ≤a<pk. Then
a=φ(1) = φ(1 ·1) = (φ(1))2=a2
so that a2=aand a(a−1) = 0 mod pk. Thus pk|aor pk|a−1. So
a= 0 or a= 1. The case a= 0 is always permissible. If a= 1, then
pk=|a|=|φ(1)|which divides |1|=m(see part 3 of Theorem
10.1).
15. In Znwe are given (k+ 1)2=k+ 1. So, k2+ 2k+ 1 = k+ 1 or
k2=−k=n−k. Also, (n−k)2=n2−2nk +k2=k2, so
(n−k)2=n−k.
16. By Exercise 15, idempotents come in pairs unless
1 + k=n−k=−kmod n. Suppose there is such a kwhere
0< k < n. Then 1 + 2k=nand k= (n−1)/2. Since
k2= (−k)2=−k, we have
n−1
22
=−n+ 1
2or n2−2n+ 1
4=−n+ 1
2.
This gives n2= 1.
17. If there is a solution x2+y2= 2003 using integers then by reducing
modulo 4, there is a solution to x2+y2= 3 in Z4. But the only
squares in Z4are 0 and 1.
18. Suppose 0 < k < n and, in Zn,k2=kand (k+ 1)2=k+ 1. Then
in Zn,
k+ 1 = k2+ 2k+ 1 = 3k+ 1
so that 2kmod n= 0. Thus 2k=n.
19. By the Mod 2 Irreducibility Test (Theorem 17.3 with p= 2) it is
enough to show that x4+x3+ 1 is irreducible over Z2. By
inspection, x4+x3+ 1 has no zeros in Z2and so it has no linear
factors over Z2. The only quadratic irreducibles in Z2[x] are
x2+x+ 1 and x2+ 1 and these are ruled out as factors by long
division.
20. Apply the mod 5 irreducibility test.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 15-18 163
21. By Theorem 14.4, we can do both parts by showing that
Z[√2]/h√2ihas only two elements. First note that the coset
a+b√2 + h√2i=a+√2. Moreover, since 2 + h√2i=
(√2 + h√2i)(√2 + h√2i) = (0 + h√2)i(0 + h√2i) = (0 + h√2i) we
can write every coset in Z[√2]/h√2iin the form
2k+h√2i= 0 + h√2ior 2k+1+h√2i= 1 + h√2i.
22. The argument given in Example 7 of Chapter 18 applies in both
cases.
23. Because Z[i]/h3i ≈ Z3[i] (map a+bi to amod 3 + (bmod 3)iand
use Theorem 15.3) is a field (see Example 9 in Chapter 13) we
know from Theorem 14.4 that h3iis maximal.
24. 13 = (2 + 3i)(2 −3i); 5 + i= (1 + i)(3 −2i).
25. Say a/b, c/d ∈R. Then a/b −c/d = (ad −bc)/(bd) and ac/(bd)∈R
by Euclid’s Lemma. By Corollary 3 Theorem 15.5, the field of
quotients is Q.
26. Z⊕Z3.
27. Since Z[i]/h3i ≈ Z3[i] it is a field (see Exercise 23). But
(1,0)(0,1) = (0,0) shows that Z3⊕Z3is not a field.
28. Try "a0
0b#→(a, b).
29. Define a mapping from R[x] to (R/I)[x] by
φ(anxn+··· +a0) = (an+I)xn+··· + (a0+I). Since
φ((anxn+··· +a0)+(bnxn+··· +b0)) =
φ((an+bn)xn+···+(a0+b0)) = (an+bn+I)xn+···+(a0+b0+I) =
(an+I)xn+··· + (a0+I)+(bn+I)xn+··· + (b0+I) =
φ(anxn+··· +a0) + φ(bnxn+··· +b0) and
φ((anxn+···+a0)(bnxn+···+b0)) = φ((anbn)xn+···+ (a0b0)) =
(anbn+I)xn+···+ (a0b0+I) = ((an+I)xn+···+ (a0+I))((bn+
I)xn+··· + (b0+I)) = φ(anxn+··· +a0)φ(bnxn+··· +b0), φis a
ring homomorphism. Next note that φ(anxn+··· +a0) = 0 + Iif
and only if an, . . . , a0∈I. So, by Theorem 15.3,
R[x]/I[x]≈(R/I)[x].
30. Let I=h2, xi={2f(x) + xg(x)|f(x), g(x)∈Z8[x]}. Then
Z8[x]/I is isomorphic to Z2.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 15-18 164
31. Let I=h2i. Then Z8[x]/I is isomorphic to Z2[x]. (The mapping
that takes anxn+···a0to (anmod 2)xn+··· +a0mod 2 is a ring
homomorphism with kernel I.) Since Z2[x] is an integral domain
(see Theorem 16.3) but not a field (f(x) = xdoes not an inverse),
the same is true for Z8[x]/I.
32. Let I=hx, 3i. Then any element anxn+··· +a0+Iof Z[x]/I
simplifies to a0mod 3 (since Iabsorbs all terms with an xand all
multiples of 3). So, Z[x]/I ={0 + I, 1 + I, 2 + I}.
165
CHAPTER 19
Vector Spaces
1. Each of the four sets is an Abelian group under addition. The
verification of the four conditions involving scalar multiplication is
straight forward. Rnhas basis
{(1,0,...,0),(0,1,0,...,0),...,(0,0,...,1)};
M2(Q) has basis 1 0
0 0 ,0 1
0 0 ,0 0
1 0 ,0 0
0 1 ;
Zp[x] has basis {1, x, x2, . . .};Chas basis {1, i}.
2. (−1)u=−uand u+u0∈Uguarantee that Uis an Abelian group.
Since the four numbered conditions in the definition of a vector
space are satisfied for all scalars and all elements of V, they are
satisfied for all scalars and all elements of U.
3. (a2x2+a1x+a0)+(a0
2x2+a0
1x+a0
0)=(a2+a0
2)x2+(a1+a0
1)x+(a0+a0
0)
and a(a2x2+a1x+a0) = aa2x2+aa1x+aa0. A basis is {1, x, x2}.
Yes, this set {x2+x+ 1, x + 5,3}is a basis because
a(x2+x+ 1) + b(x+ 5) = 3c= 0 implies that
ax2+ (a+b)x+a+ 5b+ 3c= 0. So, a= 0, a +b=oand
a+ 5b+ 3c= 0. But the two conditions a= 0 and a+b= 0 imply
b= 0 and the three conditions a= 0, b = 0, and a+ 5b+ 3c= 0
imply c= 0.
4. (a1v1+···+anvn)+(b1v1+···+bnvn)=(a1+b1)v1+···+(an+bn)vn
and a(a1v1+··· +anvn) = aa1v1+··· +aanvn.
5. They are linearly dependent, since
−3(2,−1,0) −(1,2,5) + (7,−1,5) = (0,0,0).
6. Linearly dependent since
2"2 1
1 0 #+ 2 "0 1
1 2 #+"1 1
1 1 #="0 0
0 0 #.
19/Vector Spaces 166
7. Suppose au +b(u+v) + c(u+v+w) = 0. Then
(a+b+c)u+ (b+c)v+cw = 0. Since {u, v, w}are linearly
independent, we obtain c= 0, b +c= 0, and a+b+c= 0. So,
a=b=c= 0.
8. Say a1v1+a2v2+··· +anvn= 0 where not all a’s are zero. If
ai6= 0 we can solve the above equation for vi.
9. If the set is linearly independent, it is a basis. If not, then delete
one of the vectors that is a linear combination of the others (see
Exercise 8). This new set still spans V. Repeat this process until
you obtain a linearly independent subset. This subset will still span
Vsince you deleted only vectors that are linear combinations of the
remaining ones.
10. Consider hv1, v2,...vni. If this is Vthere is no need to find w0s. If
not, pick w1outside of hv1, v2, . . . , vni. Then hv1, v2, . . . , vn, w1iis
linearly independent. If this set spans Vwe are done, otherwise
pick w2outside hv1, v2, . . . , vn, w1i. Then {v1, v2, . . . , vn, w1, w2}is
linearly independent. Since Vis finite dimensional this process will
eventually stop. When this happens we have a basis.
11. Let u1, u2, u3be a basis for Uand w1, w2, w3be a basis for W.
Since dim V= 5, there must be elements a1, a2, a3, a4, a5, a6in F,
not all 0, such that a1u1+a2u2+a3u3+a4w1+a5w2+a6w3= 0.
Then a1u1+a2u2+a3u3=−a4w1−a5w2−a6w3belongs to U∩W
and this element is not 0 because that would imply that
a1, a2, a3, a4, a5and a6are all 0.
In general, if dim U+ dim W > dim V, then U∩W6={0}.
12. If ai1x1+··· +ainxn= 0 and ai1y1+··· +ainyn= 0 then
ai1(x1+y1) + ··· +ain(xn+yn) =
ai1x1+··· +ainxn+ai1y1+··· +ainyn= 0 + 0 = 0 and
ai1(ax1) + ··· +ain(axn) = a(ai1x1+··· +ainxn=a·0=0.
13. No. x2and −x2+xbelong to Vbut their sum does not.
14. No. (1,1,√2) and (1,−1,√2) belong to Wbut their sum does not.
15. Yes, Wis a subspace. If (a, b, c) and (a0, b0, c0) belong to Wthen
a+b=cand a0+b0=c0. Thus,
a+a0+b+b0= (a+b)+(a0+b0) = c+c0so (a, b, c)+(a0+b0+c0)
19/Vector Spaces 167
belongs to Wand therefore Wis closed addition. Also, if (a, b, c)
belongs to Wand dis a real number then d(a, b, c)=(da, db, dc)
and ad +bd =cd so Wis closed under scalar multiplication.
16. "a b
b c #+"a0b0
b0c0#="a+a0b+b0
b+b0c+c0#∈Vand
d"a b
b c #="ad bd
ab cd #∈V. A basis is
(" 1 0
0 0 #,"0 1
1 0 #,"0 0
0 1 #).
17. a a +b
a+b b +a0a0+b0
a0+b0b0=
a+a0a+b+a0+b0
a+b+a0+b0b+b0and
ca a +b
a+b b =ac ac +bc
ac +bc bc .
18. Pis a plane passing through (0,0,0), (2,1,0) and (3,0,1).
19. Suppose Bis a basis. Then every member of Vis some linear
combination of elements of B. If
a1v1+··· +anvn=a0
1v1+··· +a0
nvn, where vi∈B, then
(a1−a0
1)v1+··· + (an−a0
n)vn= 0 and ai−a0
i= 0 for all i.
Conversely, if every member of Vis a unique linear combination of
elements of B, certainly Bspans V. Also, if a1v1+··· +anvn= 0,
then a1v1+···+anvn= 0v1+··· + 0vnand therefore ai= 0 for all
i.
20. Use Exercise 10.
21. Since w1=a1u1+a2u2+··· +anunand a16= 0, we have
u1=a−1
1(w1−a2u2− ··· − anun), and therefore
u1∈ hw1, u2, . . . , uni. Clearly, u2, . . . , un∈ hw1, u2, . . . , uni. Hence
every linear combination of u1, . . . , unis in hw1, u2, . . . , uni.
22. pn
23. Since (a, b, c, d)=(a, b, a, a +b) = a(1,0,1,1) + b(0,1,0,1) and
(1,0,1,1) and (0,1,0,1) are linearly independent, these two vectors
are a basis.
19/Vector Spaces 168
24. If vand v0∈U∩Wand ais a scalar, then
v+v0∈U, v +v0∈W, av ∈U, and av ∈W. So, U∩Wis a
subspace. (See Exercise 11.) If u1+w1and u2+w2∈U+W, then
(u1+w1)+(u2+w2)=(u1+u2)+(w1+w2)∈U+Wand
a(u1+w1) = au1+aw1∈U+W.
25. Suppose that B1={u1, u2, . . . , un}is a finite basis for Vand B2is
an infinite basis for V. Let w1, w2, . . . , wn+1 be distinct elements of
B2. Then, as in the proof of Theorem 19.1, the set {w1, w2, . . . , wn}
spans V. This means that wn+1 is a linear combination of
w1, w2, . . . , wn. But then B2is not a linearly independent set.
26. Yes, because Z7is a field and, therefore, 1/2,−2/3,and −1/6 exist
in Z7. Specifically, 1/2=4,−2/3=4,−1/6 = 1.
27. If Vand Ware vector spaces over F, then the mapping must
preserve addition and scalar multiplication. That is, T:V→W
must satisfy T(u+v) = T(u) + T(v) for all vectors uand vin V,
and T(au) = aT (u) for all vectors uin Vand all scalars ain F. A
vector space isomorphism from Vto Wis a one-to-one linear
transformation from Vonto W.
28. This follows directly from the definition of linear transformation
and the subspace test.
29. Suppose vand ubelong to the kernel and ais a scalar. Then
T(v+u) = T(v) + T(u) = 0 + 0 = 0 and T(av) = aT (u) = a·0 = 0.
30. Let w∈Wand suppose T(v) = w. Write v=a1v1+··· +anvn.
Then w=T(v) = a1T(v1) + ··· +anT(vn).
31. Let {v1, v2, . . . , vn}be a basis for V. The mapping given by
φ(a1v1+a2v2+··· +anvn)=(a1, a2, . . . , an) is a vector space
isomorphism. By observation, φis onto. φis one-to-one because
(a1, a2, . . . , an)=(b1, b2, . . . , bn) implies that
a1=b1, a2=b2, . . . , an=bn. Since
φ((a1v1+a2v2+··· +anvn)+(b1v1+b2v2+··· +bnvn)) =
φ((a1+b1)v1+ (a2+b2)v2+··· + (an+bn)vn) =
(a1+b1, a2+b2, . . . , an+bn)=(a1, a2, . . . , an)+(b1, b2, . . . , bn) =
φ(a1v1+a2v2+··· +anvn) + φ(b1v1+b2v2+··· +bnvn) we have
shown that φpreserves addition. Moreover, for any cin Fwe have
φ(c(a1v1+a2v2+··· +anvn)) = φ(ca1v1+ca2v2+··· +canvn) =
19/Vector Spaces 169
(ca1, ca2, . . . , can) = c(a1, a2, . . . , an) = cφ(a1v1+a2v2+··· +anvn)
so that φalso preserves scalar multiplication.
32. Suppose that V=Sn
i=1 Viwhere nis minimal and Fis the field.
Then no Viis the union of the other Vj’s for otherwise nis not
minimal. Pick v1∈V1so that v16∈ Vjfor all j6= 1. Pick v2∈V2so
that v26∈ Vjfor all j6= 2. Consider the infinite set
L={v1+av2|a∈F}. We claim that each member of Lis
contained in at most one Vi. To verify this suppose both
u=v1+av2and w=v1+bv2belong to some Vi. Then
u−w= (a−b)v2∈Vi∪V2.By the way that v2was chosen this
implies that i= 2.Also, bu −aw = (b−a)v1∈Vi∪V1, which
implies that i= 1. This contradiction establishes the claim.
Finally, since each member of Lbelongs to at most one Vi, the
union of the Vihas at most nelements of L. But the union of the
Viis Vand Vcontains L.
170
CHAPTER 20
Extension Fields
1. {a52/3+b51/3+c|a, b, c ∈Q}.
2. See Example 6 of Chapter 21.
3. Since x3−1=(x−1)(x2+x+ 1) the zeros of x3−1 are
1,(−1 + √−3)/2, and (−1−√−3)/2. So, the splitting field is
Q(√−3).
4. See the answer to Exercise 11 of Chapter 17.
5. Since the zeros of x2+x+ 1 are (−1±√−3)/2 and the zeros of
x2−x+ 1 are (1 ±√−3)/2, the splitting field is Q(√−3).
6. Certainly R(a+bi)⊆C. But
i=b−1(a+bi)−b−1a∈R(a+bi)
so C⊆R(a+bi).
7. Note that a=q1 + √5 implies that a4−2a2−4 = 0. Then
p(x) = x4−2x2−4 is irreducible over Q(to see this let y=x2and
apply the quadratic formula to y2−2y−4) and p(a) = 0. So, by
Theorem 20.3, Q(q1 + √5) is isomorphic to Q[x]/hp(x)i.
8. 8. Use Theorem 20.3. To construct the multiplication table observe
that a3=a+ 1.
9. Since a3+a+ 1 = 0 we have a3=a+ 1. Thus,
a4=a2+a;a5=a3+a2=a2+a+ 1. To compute a−2and a100,
we observe that a7= 1, since F(a)∗is a group of order 7. Thus,
a−2=a5=a2+a+ 1 and a100 = (a7)14a2=a2.
10. Use the fact that a3=a+ 1, a4=a2+a,a5=a2+a+ 1,
a6=a2+ 1 and a7= 1.
20/Extension Fields 171
11. Q(π) is the set of all expressions of the form
(anπn+an−1πn−1+··· +a0)/(bmπm+bm−1πm−1+··· +b0),
where bm6= 0.
12. {1, π, π2}
13. x7−x=x(x6−1) = x(x3+ 1)(x3−1) = x(x−1)3(x+ 1)3;x10 −x=
x(x9−1) = x(x−1)9(see Exercise 49 of Chapter 13).
14. The identity is the only one.
15. Let bbe a zero of f(x) in some extension of F. Then bp=aand
f(x) = xp−bp= (x−b)p(see Exercise 49 of Chapter 13). So, if
b∈Fthen f(x) splits in Fand if b6∈ Fthen f(x) is irreducible
over F.
16. (x+β)(x+β+ 1)(x+β2)(x+β2+ 1).
17. Solving 1 + 3
√4=(a+b3
√2 + c3
√4)(2 −23
√2) for a, b, and cyields
a= 4/3, b = 2/3, and c= 5/6.
18. a=−3/23, b= 4/23
19. Since 1 + i=−(4 −i) + 5, Q(1 + i)⊆Q(4 −i); conversely,
4−i= 5 −(1 + i) implies that Q(4 −i)⊆Q(1 + i).
20. Since ac +b∈F(c) we have F(ac +b)⊆F(c). But
c=a−1(ac +b)−a−1b, so F(c)⊆F(ac +b).
21. If the zeros of f(x) are a1, a2, . . . , anthen the zeros of f(x+a) are
a1−a, a2−a, . . . an−a. So, by Exercise 20, f(x) and f(x−a) have
the same splitting field.
22. Since f(x) and g(x) are relatively prime in F[x] there are
polynomials t(x) and s(x) in F[x] such that
1 = f(x)t(x) + g(x)s(x).
(See Exercise 41 for Chapter 16.) If f(x) and g(x) had a
nonconstant factor in common this factor would divide 1.
23. Clearly, Qand Q(√2) are subfields of Q(√2). Assume that there is
a subfield Fof Q(√2) that contains an element a+b√2 with b6= 0.
Then, since every subfield of Q(√2) must contain Q, we have by
Exercise 20 that Q(√2) = Q(a+b√2) ⊆F. So, F=Q(√2).
20/Extension Fields 172
24. F(a, b)⊆F(a)(b) since F(a)(b) is a field that contains F,aand b.
Also, F(a)(b)⊆F(a, b) since F(a, b) contains F(a) and b. So,
F(a, b) = F(a)(b) and, by symmetry, F(a, b) = F(b, a).
25. Let F=Z3[x]/hx3+ 2x+ 1iand denote the coset x+hx3+ 2x+ 1i
by βand the coset 2 + hx3+ 2x+ 1iby 2. Then βis a zero of
x3+ 2x+ 1 and therefore β3+ 2β+ 1 = 0. Using long division we
obtain x3+ 2x+ 1 = (x−β)(x2+βx + (2 −β2)). By trial and error
we discover that β+ 1 is a zero of x2+βx + (2 −β2) and by long
division we deduce that −2β−1 is the other zero of
x2+βx + (2 −β2). So, we have
x3+ 2x+ 1 = (x−β)((x−β−1)(x+ 2β+ 1).
26. x(x+ 1)(x3+x2+ 1)(x3+x+ 1)
27. Suppose that φ:Q(√−3) →Q(√3) is an isomorphism. Since
φ(1) = 1, we have φ(−3) = −3. Then
−3 = φ(−3) = φ(√−3√−3) = (φ(√−3))2. This is impossible, since
φ(√−3) is a real number.
28. The field of quotients of Zp[x] is not perfect.
29. By long division we obtain x2+x+ 1 = (x−β)(x+1+β) so the
other zero is −1−β.
30. Use Theorem 20.5.
31. Since f(x) = x21 + 2x8+ 1 and f0(x) = xhave no common factor of
positive degree we know by Theorem 20.5 that f(x) has no multiple
zeros in any extension of Z3.
32. Use Theorem 20.5.
33. Since f(x) = xpn−xand f0(x) = −1 have no common factor of
positive degree we know by Theorem 20.5 that f(x) has no multiple
zeros in any extension of Z3.
34. Z3[x]/hx2+x+ 2i. Let βbe a zero of x2+x+ 2. Then
f(x)=(x−β)2(x−2β−2)(x−β−1).
35. Since Lis a splitting field of f(x) over F, we may write
f(x)=(x−a1)(x−a2). . . (x−an), where the coefficients of f(x)
belong to F. But then these coefficients also belong to L.
20/Extension Fields 173
36. By Theorem 20.5, the zeros of xn−aare distinct, say
α1, α2, . . . , αn. Then βi=α1/αifor i= 1,2, . . . , n are all the nth
roots of unity.
37. Since |(Z2[x]/hf(x)i)∗|= 31 is prime and the order of every element
must divide it, every nonidentity is a generator.
38. Observe that x4−6x2−7(x2−7)(x2−1).
39. Proceeding as in Example 9 we suppose that h(t)/k(t) is a zero in
Zp(t) of f(x) where deg h(t) = mand deg k(t) = n. Then
(h(t)/k(t))p=t, and therefore (h(t))p=t(k(t))p. Then by Exercise
49 of Chapter 13 we have h(tp) = tk(tp). Since deg h(tp) = pm and
deg tk(tp) = 1 + pn we have pm = 1 + pn. But this implies that p
divides 1, which is false. So, our assumption that f(x) has a zero in
Zp(x) has lead to a contradiction. That f(x) has a multiple zero in
Kfollows as in Example 9.
40. By the corollary to Theorem 20.9 deg f(x) has the form nt where t
is the number of distinct zeros of f(x).
174
CHAPTER 21
Algebraic Extensions
1. It follows from Theorem 21.1 that if p(x) and q(x) are both monic
irreducible polynomials in F[x] with p(a) = q(a) = 0, then
deg p(x) = deg q(x). If p(x)6=q(x), then
(p−q)(a) = p(a)−q(a) = 0 and deg (p(x)−q(x)) <deg p(x),
contradicting Theorem 21.1.
To prove Theorem 21.3 we use the Division Algorithm
(Theorem 16.2) to write f(x) = p(x)q(x) + r(x), where r(x) = 0 or
deg r(x)<deg p(x). Since 0 = f(a) = p(a)q(a) + r(a) = r(a) and
p(x) is a polynomial of minimum degree for which ais a zero, we
may conclude that r(x) = 0.
2. If f(x)∈F[x] does not split in E, then it has a nonlinear factor
q(x) which is irreducible over E. But then E[x]/hq(x)iis a proper
algebra extension of E.
3. Let F=Q(√2,3
√2,4
√2, . . .). Since [F:Q]≥[Q(n
√2) : Q] = nfor
all n, [F:Q] is infinite. To prove that Fis an algebraic extension
of Q, let a∈F. There is some ksuch that
a∈Q(√2,3
√2,4
√2,..., k
√2). It follows from Theorem 21.5 that
[Q(√2,3
√2,4
√2,..., k
√2) : Q] is finite and from Theorem 21.4 that
Q(√2,3
√2,4
√2,..., k
√2) is algebraic.
4. Suppose E(a) is an algebraic extension of E. Then ais algebraic
over F(Theorem 21.7). So if f(x) is the minimal polynomial for a
over Fthen f(x) can be written in the for
(x−a1)(x−a2)···(x−an) where each ai∈E. But then f(a) = 0
implies a=aifor some iand a∈E.
5. Since every irreducible polynomial in F[x] is linear, every
irreducible polynomial in F[x] splits in F. So, by Exercise 4, Fis
algebraically closed.
21/Algebraic Extensions 175
6. Suppose g(x) = h(x)k(x) where h(x) is irreducible over F(a). Let b
be a zero of h(x) in some extension of F(a). Then
[F(a, b) : F(a)][F(a) : F] = [F(a, b) : F(b)][F(b) : F].
Also,
deg f(x)=[F(a) : F],deg g(x) = [F(b) : F]
and
deg h(x) = [F(a, b) : F(a)].
It follows that deg g(x) divides deg h(x). Thus g(x) and h(x) are
associates and g(x) is irreducible over F(a).
7. Suppose Q(pa) = Q(√b). If √b∈Q, then √a∈Qand we may
take c=√a/√b. If √b /∈Q, then √a /∈Q. Write √a=r+s√b
where rand sbelong to Q. Then r= 0 for, if not, then
a=r2+ 2rs√b+band therefore (a−r2−b)/2r=s√b. But
(a−r2−b)/2ris rational whereas s√bis irrational.
Conversely, if there is a element c∈Qsuch that a=bc2(we may
assume that cis positive) then, by Exercise 20 in Chapter 20,
Q(√a) = Q(√bc2) = Q(c√b) = Q(√b).
8. Since (√3 + √5)2∈Q(√15),[Q(√3 + √5) : Q(√15)] = 2. For the
second question, first note that Q(√2,3
√2,4
√2) = Q(3
√2,4
√2). Then
observe [Q(3
√2,4
√2) : Q] is divisible by [Q(3
√2) : Q] = 3 and
[Q(4
√2) : Q] = 4. Thus it follows that [Q(3
√2,4
√2) : Q] = 12.
9. Since [E:F] = [E:F(a)][F(a) : F] we have [F(a) : F]=[E:F], in
which case F(a) = E, or [F(a) : F] = 1, in which case F(a) = F.
10. If ais a zero of p(x)∈Q[x], then n
√ais a zero of p(xn)∈Q[x].
11. Note that [F(a, b) : F] is divisible by both m= [F(a) : F] and
n= [F(b) : F] and [F(a, b) : F]≤mn. So, [F(a, b) : F] = mn.
12. Take F=Q,a=4
√2, b=6
√2. Then [F(a, b) : F] = 12.
13. Since ais a zero of x3−a3over F(a3), we have [F(a) : F(a3)] ≤3.
For the second part, take F=Q, a = 1;
F=Q, a = (−1 + ip3)/2; F=Q, a =3
√2.
14. Note that x=√−3 + √2 implies that x4+ 2x2+ 25 = 0.
21/Algebraic Extensions 176
15. Suppose E1∩E26=F. Then [E1:E1∩E2][E1∩E2:F] = [E1:F]
implies [E1:E1∩E2] = 1, so that E1=E1∩E2. Similarly,
E2=E1∩E2.
16. Let x=3
√2 + 3
√4 = 3
√2(1 + 3
√2).Then x3= 6(1 + x). Thus
3
√2 + 3
√4 is a zero of x3−6x−6 and x3−6x−6 is irreducible by
Eisenstein.
17. Since Emust be an algebraic extension of R, we have E⊆Cand
so [C:E][E:R] = [C:R] = 2. If [C:E] = 2, then [E:R] = 1
and therefore E=R. If [C:E] = 1, then E=C.
18. Use the quadratic formula and Exercise 20 of Chapter 20.
19. Let abe a zero of p(x) in some extension of F. First note
[E(a) : E]≤[F(a) : F] = deg p(x). Then observe that
[E(a) : F(a)][F(a) : F] = [E(a) : F]=[E(a) : E][E:F]. This
implies that deg p(x) divides [E(a) : E], so that
deg p(x) = [E(a) : E]. It now follows from Theorem 20.3 that p(x)
is irreducible over E.
20. If Eis a finite extension of F, pick a1∈Ebut not in F. Then
[E:F]≥[F(a1) : F]. If F(a1)6=E, pick a2∈Enot in F(a1).
Then [E:F]≥[F(a1, a2) : F]. Since [E:F] is finite, we can
continue this process only a finite number of times. The converse
follows from Theorem 21.5.
21. Suppose that α+βand αβ are algebraic over Qand that α≥β.
Then p(α+β)2−4αβ =pα2−2αβ +β2=p(α−β)2=α−βis
also algebraic over Q. Also, α= ((α+β)−(α−β))/2 is algebraic
over Q, which is a contradiction.
22. If f(a) is a zero of g(x)∈F[x], then ais a zero of (g◦f)(x)∈F[x].
23. It follows from the Quadratic Formula that √b2−4ac is a primitive
element.
24. Since x4−x2−2=(x2−2)2,Z3[x]/hx2−2iis a splitting field.
Alternatively, observe that iand −iare zeros of multiplicity 2 of
x2−2 so that x4−x2−2 splits in Z3[i]. Since Z3[i] has 9 elements,
x4−x2−2 does not split in any smaller field.
21/Algebraic Extensions 177
25. By the Factor Theorem (Corollary 2 of Theorem 16.2), we have
f(x)=(x−a)(bx +c), where b, c ∈F(a). Thus,
f(x) = b(x−a)(x+b−1c).
26. It suffices to show that √a∈Q(a). Observe that
0 = a2+a+ 1 = (a+√a+ 1)(a−√a+ 1). So, a+√a+ 1 = 0 or
a−√a+ 1 = 0. Thus, √a=−a−1 or √a=a+ 1.
27. Say ais a generator of F∗. Then F=Zp(a), and it suffices to show
that ais algebraic over Zp. If a∈Zp, we are done. Otherwise,
1 + a=akfor some k6= 0. If k > 0, we are done. If k < 0, then
a−k+a1−k= 1 and we are done.
28. Let f(x) be the minimal polynomial for aover Q, and let r=m/n
where mand nare integers. Consider g(x) = f(xn).
29. If [K:F] = n, then there are elements v1, v2, . . . , vnin Kthat
constitute a basis for Kover F. The mapping
a1v1+···+anvn→(a1, . . . , an) is a vector space isomorphism from
Kto Fn. If Kis isomorphic to Fn, then the nelements in K
corresponding to (1,0,...,0), (0,1,...,0), . . . , (0,0,...,1) in Fn
constitute a basis for Kover F.
30. [Q(41/6) : Q] = 3.
31. Observe that
[F(a, b) : F(a)] ≤[F(a, b) : F(a)][F(a) : F]=[F(a, b) : F].
32. Let deg f(x) = m, deg g(x) = nand suppose that f(x) is
irreducible over F(b). Then m= [F(a, b) : F(b)] and
[F(a, b) : F]=[F(a, b) : F(b)][F(b) : F] = mn = [F(a, b) :
F(a)][F(a) : F] = [F(a, b) : F(a)]m. Thus,
n= deg g(x)=[F(a, b) : F(a)] and therefore g(x) is irreducible
over F(a). The other half follows by symmetry.
33. Note that if c∈Q(β) and c6∈ Q, then
5=[Q(β) : Q]=[Q(β) : Q(c)][Q(c) : Q] so that [Q(c) : Q] = 5. On
the other hand, [Q(√2) : Q] = 2,[Q(3
√2) : Q] = 3, and
[Q(4
√2) : Q] = 4.
34. Since √2 = ( 6
√2)3and 3
√2=(6
√2)2we know that
Q(√2,3
√2) ⊆Q(6
√2). Also, [Q(√2,3
√2) : Q] is divisible by
[Q(√2) : Q] = 2 and [Q(3
√2) : Q]=3.It follows that
[Q(√2,3
√2) : Q]=6=[Q(6
√2) : Q].
21/Algebraic Extensions 178
35. By closure, Q(√a+√b)⊆Q(√a, √b). Since
(√a+√b)−1=1
√a+√b
√a−√b
√a−√b=√a−√b
a−band a−b∈Q(√a+√b) we
have √a−√b∈Q(√a+√b). (The case that a−b= 0 is trivial.)
It follows that √a=1
2((√a+√b)+(√a−√b)) and
√b=1
2((√a+√b)−(√a−√b)) are in Q(√a, √b). So,
Q(√a, √b)⊆Q(√a+√b).
36. Since [L:F]=[L:K][K:F] we have [K:F] = 1. It follows that
F=K.
37. Observe that K=F(a1, a2, . . . , an), where a1, a2, . . . , anare the
zeros of the polynomial. Now use Theorem 21.5.
38. A splitting field of a polynomial in Q[x] is an algebraic extension of
Qwhereas Cis an transcendental extension of Q.
39. Elements of Q(π) have the form
(amπm+am−1πm−1+···+a0)/(bnπn+bn−1φn−1+···+b0), where
the a’s and b’s are rational numbers. So, if √2∈Q(π), we have an
expression of the form
2(bnπn+bn−1φn−1+··· +b0)2= (amπm+am−1πm−1+··· +a0)2.
Equating the lead terms of both sides, we have 2b2
nπ2n=a2
mπ2m.
But then we have m=n, and √2 is equal to the rational number
am/bn.
40. αis a zero of x7−1=(x−1)(x6+x5+x4+x3+x2+x+ 1) and
the second factor is irreducible over Q(see Corollary to Theorem
17.4.) Similarly, βis a zero of x5−1=(x−1)(x4+x3+x2+x+ 1)
and the second factor is irreducible over Q(see Corollary to
Theorem 17.4.). So, [Q(α) : Q] = 6 and [Q(β) : Q] = 4. If
β∈Q(α), then Q(β) is a subfield of Q(α) and therefore
[Q(β) : Q] = 4 would divide [Q(α) : Q] = 6.
41. Observe that F(a) = F(1 + a−1).
42. Let {x1, x2, . . . , xn}be a basis for Kover F. Then
K=F(x1, x2, . . . , xn) for some x1, x2, . . . , xnin K.
43. Let |K:F|=nand let {x1, x2, . . . , xn}be a basis for Kover F.
Let F0=F, F1=F0(x1), F2=F1(x2), . . . , Fn=Fn−1(xn) = K.
Then n=|K:F|=|Fn:Fn−1||Fn−1:Fn−2|···|F1:F0|and each
|Fi+1 :Fi|is 1 or p.
21/Algebraic Extensions 179
44. √3 + √5 + √−2).
180
CHAPTER 22
Finite Fields
1. Since 729 = 93, [GF (729) : GF (9)] = 3; since 64 = 82,
[GF (64) : GF (8)] = 2.
2. Use Theorem 21.5.
3. The lattice of subfields of GF(64) looks like Figure 21.3 with GF(2)
at the bottom, GF(64) at the top, and GF(4) and GF(8) on the
sides.
4. Since α3+α2+ 1 = 0 we have α2(α+ 1) = 1.
5. 2α+ 1.
6. From α3+α+ 1 = 0 we have α3+ 1 = α. Squaring both sides we
have α6+ 1 = α2. Thus, 0 = (α2)3+α2+ 1 = f(α2). To find the
third zero, long divide x3+x+ 1 by
(x−α)(x−α2)=(x+α)(x+α2) to obtain the quotient
x+α2+α. Thus α2+αis a zero.
7. By Theorem 22.2, there is an element ain Ksuch that K∗=hai.
Thus K=F(a).
8. Use Theorem 22.2 and Theorem 4.4.
9. The only possibilities for f(x) are x3+x+ 1 and x3+x2+ 1. If ais
a zero of x3+x+ 1, then |Z2(a)|=|Z2[x]/hx3+x+ 1i| = 8.
Moreover, testing each of a2, a3, a4shows that the other two zeros
of x3+x+ 1 are a2and a4. So, Z2(a) is the splitting field for
x3+x+ 1.
For the second case, let abe a zero of x3+x2+ 1. As in the first
case, |Z2(a)|= 8. Moreover, testing each of a2, a3, a4shows that the
other two zeros are a2and a4. So, Z2(a) is the splitting field for
x3+x2+ 1.
10. Use Theorem 22.1.
22/Finite Fields 181
11. By Exercise 44 in Chapter 15, φis a ring homomorphism. Since the
only ideals of a field are {0}and the field itself (Exercise 27 in
Chapter 14), Ker φ={0}. Thus, φis an automorphism. To show
that φnis the identity, we first observe by Corollary 4 of Lagrange’s
Theorem (Theorem 7.1) that apn−1= 1 for all ain GF(pn)∗. Thus,
φn(a) = apn=apn−1a=afor all ain GF(pn)∗. Obviously,
φn(0) = 0.
12. GF(210), GF(215), GF(225).
13. If g(x) is an irreducible factor of x8−xover Z2and deg g(x) = m,
then the field Z2[x]/hg(x)ihas order 2mand is isomorphic to a
subfield of GF(8). So, by Theorem 22.3, m= 1 or 3.
14. It follows from Theorem 27.3 that the desired field is GF(212).
15. Since GF(pn)∗is a cyclic group of order pn−1 we seek the smallest
nsuch that pn−1 is divisible by 5. By observation, n= 4 for p= 2
or 3.
16. Expanding we obtain x3+ (a4+a2+a)x2+ (a6+a5+a3)x+a7.
Then note that a4+a2+a= 1; a6+a5+a3= 0; and a7= 1.
17. Since |(Z3[x]/hx3+ 2x+ 1i)∗|= 26, we need only show that
|x| 6= 1,2 or 13. Obviously, x6= 1 and x26= 1. Using the fact that
x3+ 2x+ 1 = 0 and doing the calculations we obtain x13 = 2. (Or
use the computer software for Chapter 22 at
www.d.umn.edu/˜jgallian.)
18. Since |(Z2[x]/hf(x)i)∗|= 31,|x|must be 31.
19. Direct calculations show that x13 = 1, whereas (2x)26= 1 and
(2x)13 6= 1. Thus 2xis a generator.
20. Since 1, x and x2form a basis for Z3[x]/hf(x)iover Z3we have
that |x| 6= 2. By Lagrange’s theorem |x|is 13 or 26. If |x|= 13,
then |2x|= 26.
21. Note that if Kis any subfield of GF(pn) then K∗is a subgroup of
the cyclic group GF(pn)∗. So, by Theorem 4.3, K∗is the unique
subgroup of GF(pn)∗of its order.
22. Use Exercises 64 and 41 of Chapter 4.
22/Finite Fields 182
23. Since x2+ 1 has no zeros in Z3, it is irreducible over Z3(see
Corollary 2 of Theorem 16.2). By Corollary 1 of Theorem 17.5,
Z3[x]/hx2+ 1iis a field. Since every element of Z3[x]/hx2+ 1ihas
the form ax +b+hx2+ 1i, the field has order 9. To see the
conversion table use the computer software for Chapter 22 at
www.d.umn.edu/˜jgallian.)
24. Mimic the proof of the case where the field is GF(pn) and the group
is GF(pn)∗given in Theorem 22.2.
25. Let a, b ∈K. Then, by Exercise 49b in Chapter 13,
(a−b)pm=apm−bpm=a−b. Also, (ab)pm=apmbpm=ab. So, K
is a subfield.
26. If g(x) divides xpn−x, then g(x) has a zero in GF(pn) and so
GF(pn) contains a subfield isomorphic to GF(p)[x]/hg(x)i, which
has degree dover GF(p). By Theorem 22.3, ddivides n.
27. By Corollary 4 of Lagrange’s Theorem (Theorem 7.1), for every
element ain F∗we have apn−1= 1. So, every element in F∗is a
zero of xpn−x.
28. Theorem 22.3 reduces the problem to constructing the subgroup
lattices for Z18 and Z30.
29. They are identical.
30. Without loss of generality, we may assume that p(x) is monic. In
some splitting field Kof p(x) we can write
p(x) = (x−a1)(x−a2)···(x−am) where the aiare distinct and
K= GF(p)(a1, a2, . . . , am). Then K=GE(pn) for some nand
since every element of Kis a zero of xpn−xwe have that
p(x) = (x−a1)(x−a2)···(x−am) divides xpn−x.
31. The hypothesis implies that g(x) = x2−ais irreducible over
GF(p). Then ais a square in GF(pn) if and only if g(x) has a zero
in GF(pn). Since g(x) splits in GF(p)[x]/hg(x)i ≈ GF(p2), g(x) has
a zero in GF(pn) if and only if GF(p2) is a subfield of GF(pn). The
statement now follows from Theorem 22.3.
32. Let abe a zero of f(x) in E=Zp[x]/hf(x)i. Then |E|=p3. If f(x)
splits in Ewe are done. If not, f(x) = (x−a)g(x) where g(x) is
irreducible over E. Let bbe a zero of g(x) in K=E[x]/hf(x)i.
22/Finite Fields 183
Then |K|=|E|2= (p3)2and Kis the splitting field for f(x) over
Zp.
33. This is a direct consequence of Exercise 11.
34. GF(25)∗=hα33i; GF(22)∗=hα341i; GF(2)∗=hα1023i={1}.
35. Since both α62 and −1 have order 2 in the cyclic group F∗and a
cyclic group of even order has a unique element of order 2 (see
Theorem 4.4), we have α62 =−1.
36. Note that GF(p2n) is an algebraic extension of GF(pn).
37. See the solution to Exercise 25.
38. pkwhere k= gcd(s, t).
39. Consider the field of quotients of Zp[x]. The polynomial f(x) = xis
not the image of any element.
40. Since ddivides n, GF(pn) contains the subfield GF(pd), and by
Corollary 2 of Theorem 22.2, GF(pd) has the form GF(p)(a) for
some element ain GF(pd) of degree d. Moreover, by Theorem 21.1,
GF(p)(a) has the form GF(p)[x]/hp(x)iwhere p(x) is an irreducible
polynomial over GF(p), the degree of p(x) is d, and p(a) = 0. Since
p(x) is irreducible over GF(p) the only common divisor of p(x) and
xpn−xis 1 or p(x). If the common divisor is 1 then by Exercise 41
of Chapter 16 there are elements h(x) and k(x) of GF(p)[x] such
that p(x)h(x)+(xpn−x)k(x) = 1. But since x−adivides the left
side we have a contradiction.
41. Observe that p−1 = −1 has multiplicative order 2 and a(pn−1)/2is
the unique element in haiof order 2.
42. Since 5 mod 4 = 1, we have that 5n−1 is divisible by 4 for all n.
Now observe that 2 has multiplicative order 4 and a(5n−1)/4has
order 4. (The only other element of order 4 is a3(5n−1)/4.)
43. Since p= 1 (mod 4) we have pn= 1 mod 4 and GF(pn)∗is a cyclic
of order pn−1. So, by Theorem 4.4 there is exactly two elements of
order 4.
22/Finite Fields 184
44. When nis odd, pn= 3 mod 4 and therefore pn−1 is not divisible
by 4. By Theorem 4.3 haihas no element of order 4. When
n= 2m,pn= (p2)m= (32)m= 1 mod 4 and therefore pn−1 is
divisible by 4. Thus, by Theorem 4.4, GF(pn)∗has exactly two
elements of order 4.
185
CHAPTER 23
Geometric Constructions
1. To construct a+b, first construct a. Then use a straightedge and
compass to extent ato the right by marking off the length of b. To
construct a−b, use the compass to mark off a length of bfrom the
right end point of a line of length a. The remaining segment has
length a−b.
2. Let xdenote the length of the long side of the triangle. Then a
1=x
b.
3. Let ydenote the length of the hypotenuse of the right triangle with
base 1 and xdenote the length of the hypotenuse of the right
triangle with the base |c|. Then y2= 1 + d2,x2+y2= (1 + |c|)2
and |c|2+d2=x2. So, 1 + 2|c|+|c|2= 1 + d2+|c|2+d2, which
simplifies to |c|=d2.
4. Let xdenote the length of the side of along the base of the small
triangle in figure in the text. Then xis constructible and x
1=a
b.
5. Suppose that sin θis constructible. Then, by Exercises 1, 2, and 3,
p1−sin2θ= cos θis constructible. Similarly, if cos θis
constructible then so is sin θ.
6. Look at Figure 23.1.
7. From the identity cos 2θ= 2 cos2θ−1 we see that cos 2θis
constructible if and only if cos θis constructible.
8. Use Exercise 6.
9. By Exercises 5 and 7 to prove that a 45◦angle can be trisected it is
enough to show that sin 15◦is constructible. To this end note that
sin 45◦=√2/2 and sin 30◦= 1/2 are constructible and
sin 15◦= sin 45◦cos 30◦−cos 45◦sin 30◦. So, sin 15◦is constructible.
10. Use Exercises 5, 6 and 7.
23/Geometric Constructions 186
11. Note that solving two linear equations with coefficients in F
involves only operations that Fis closed under.
12. Say the line is ax +by +c= 0 and the circle is
x2+y2+dx +ey +f= 0. We seek the simultaneous solution of
these two equations. If a= 0, then y=−c/b and the equation of
the circle reduces to a quadratic in xwith coefficients from F.
Since the solution of a quadratic involves only the operations of F
and a square root of an element from F, the value for xlies in For
in F(√α) where α∈Fand α > 0. If a6= 0, the xterms in the
circle can be replaced by (−b/a)y−c/a to obtain a quadratic
equation. Then, as in the previous case, yand therefore xlie in F
or in F(√α).
For the second portion, consider x2+y2= 1 and x−y= 0.
13. From Theorem 17.1 and the Rational Root Theorem (Exercise 27
in Chapter 17) it is enough to verify that none of ±1,±1/2,±1/4,
and ±1/8 is a zero of 8x3−6x−1.
14. If the polygon is constructible, so is cos(2π/7). Thus, it suffices to
show that 8x3+ 4x2−4x−1 is irreducible over Q. This follows
from Theorem 17.1 and Exercise 27 of Chapter 17.
15. If a regular 9-gon is constructible then so is the angle 360◦/9 = 40◦.
But Exercise 10 shows that a 40◦angle is not constructible.
16. Use Exercises 5, 6, and 7.
17. This amounts to showing √πis not constructible. But if √πis
constructible, so is π. However, [Q(π) : Q] is infinite.
18. It suffices to show that 2π/5 is constructible. Clearly,
[Q(cos 2π/5) : Q] = 2 so that cos 2π/5 is constructible. Now use
Exercise 6.
19. “Tripling” the cube is equivalent to constructing an edge of length
3
√3. But [Q(3
√3) : Q] = 3, so this can’t be done.
20. No, since [Q(3
√4) : Q] = 3.
21. “Cubing” the circle is equivalent to constructing the length 3
√π.
But [Q(3
√π) : Q] is infinite.
22. Use Exercises 1-4.
187
SUPPLEMENTARY EXERCISES FOR CHAPTERS 19-23
1. Since f(x) = x50 −1 and f0(x) = 50x49 have no common factor of
positive degree in common we know by Theorem 20.5 that x50 −1
has no multiple zeros in any extension of Z3.
2. Use the Fundamental Theorem of Field Theory (Theorem 20.1) and
the Factor Theorem (Corollary 2 of Theorem 16.2).
3. Suppose bis one solution of xn=a. Since F∗is a cyclic group of
order q−1, it has a cyclic subgroup of order n, say hci. Then each
member of hciis a solution to the equation xn= 1. It follows that
bhciis the solution set of xn=a.
4. Pick ain Kbut not in F. Now use Theorem 21.5.
5. (5a2+ 2)/a = 5a+ 2a−1. Now observe that since a2+a+ 1 = 0, we
know a(−a−1) = 1, so that a−1=−a−1. Thus,
(5a2+ 2)/a =−2+3a.
6. They are of the form a+b4
√2 where a, b ∈Q(√2).
7. Since [F(a) : F] = 5, {1, a, a2, a3, a4}is a basis for F(a) over F.
Also, from 5 = [F(a) : F]=[F(a) : F(a3)][F(a3) : F] we know that
[F(a3) : F] = 1 or 5. However, [F(a3) : F] = 1 implies that a3∈F
and therefore the elements 1, a, a2, a3, a4are not linearly
independent over F. So, [F(a3) : F] = 5.
8. By the proof of the Fundamental Theorem of Field Theory p(x) has
a zero in some extension Eof Fand [E:F]≤n. Now write
p(x) = (x−a)g(x) where g(x)∈E[x]. By induction on the degree
of the polynomial, g(x) splits in some extension Kof Eand
[K:E]≤(n−1)!. Thus p(x) splits in Kand
[K:F] = [K:E][E:F]≤(n−1)!n=n!.
9. Since F(a) = F(a−1), we have that
degree of a= [F(a) : F] = [F(a−1) : F] = degree of a−1.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 19-23 188
10. Let x=π2−1. Then (x+ 1)3=π6. It follows that π2−1 is a zero
of x3+ 3x2+ 3x+ 1 −π6∈Q(π3).
11. If ab is a zero of cnxn+··· +c1x+c0∈F[x], then ais a zero of
cnbnxn+··· +c1bx +c0∈F(b)[x].
12. We need only show that if a∈R, then a−1∈R. But
a−1∈F(a)⊆R(see Theorem 20.3).
13. Every element of F(a) can be written in the form f(a)/g(a), where
f(x), g(x)∈F[x]. If f(a)/g(a) is algebraic and not a member of F,
then there is some h(x)∈F[x] such that h(f(a)/g(a)) = 0. By
clearing fractions and collecting like powers of a, we obtain a
polynomial in awith coefficients from Fequal to 0. But then a
would be algebraic over F.
14. 64
15. If Kis a finite extension of a finite field Fthen Kitself is a finite
field. So, K∗=haifor some a∈Kand therefore K=F(a).
16. Let rbe a nonzero element of R. We know that r−1exists in the
field of quotients of R. It suffices to show that r−1is in R. If Rhas
dimension nover F, then 1, r, r2, . . . , rnis a linearly dependent set
over F. Thus, ris the zero of some irreducible polynomial over F.
By Theorem 20.3, there is a basis for the field F(r) viewed as a
vector space over Fof the form 1, r, r2, . . . , rm. But then
r−1⊆F(r)⊆R.
17. If the basis elements commute, then so would any combination of
basis elements. However, the entire space is not commutative.
18. No, for 1 is not in the span of such a set.
19. Since every element in the set has the form ax3+bx2+cx, the
elements x, x2, x3form a basis.
20. Write a1f+a2f0+··· +anf(n)= 0 and take the derivative ntimes
to get a1= 0. Similarly, get all other a0
is = 0. So, the set is linearly
independent and has the same dimension as Pn.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 19-23 189
21. To prove that Lis a subfield we must show that a, b ∈Limplies
that a−b∈Land ab−1∈L. To this end, suppose that apm∈F
and bpn∈F. Then, by Exercise 49 in Chapter 13,
(a−b)pm+n=apm+n−bpm+n= (apm)pn−(bpn)pm∈F. Also,
(ab−1)pm+n= (apm)pn((bpn)pm)−1∈F. So, Lis a subfield of K.
Lastly, note that for any a∈Fwe have a=ap0so that a∈L.
22. Let f(x) = xn−x. Observe that if f(x) does have a multiple zero r
then f(x) = x(xn−1−1) and f0(x) = nxn−1−1 have x−ras a
common factor. Clearly, r6= 0. Thus, x−ris a common factor of
xn−1−1 and nxn−1−1. But then x−ris a factor of
n(xn−1−1) −(nxn−1−1) = −n+ 1. This implies that
−n+ 1 mod p= 0 and therefore nmod p= 1, where pis the
characteristic of the field.
190
CHAPTER 24
Sylow Theorems
1. a=eae−1;cac−1=bimplies a=c−1bc =c−1b(c−1)−1;a=xbx−1
and b=ycy−1imply a=xycy−1x−1=xyc(xy)−1.
2. {e},{a2},{a, a3},{b, ba2},{ba, ba3}
3. Observe that T(xC(a)) = xax−1=yay−1=T(yC(a)) ⇔y−1xa =
ay−1x⇔y−1x∈C(a)⇔yC(a) = xC(a). This proves that Tis
well defined and one-to-one. Onto is by definition.
4. cl(a) = {a}if and only if for all xin G, xax−1=a. This is
equivalent to a∈Z(G).
5. It suffices to show that the correspondence from the set of left
cosets of N(H) in Gto the set of conjugates of Hgiven by
T(xN(H)) = xHx−1is well defined, onto, and one-to-one. Observe
that xN(H) = yN(H)⇔y−1xN (H) = N(H)⇔y−1x∈N(H)⇔
y−1xH(y−1x)−1=y−1xHx−1y=H⇔xHx−1=yHy−1. This
shows that Tis well defined and one-to-one. By observation, Tis
onto.
6. Let rdenote the number of conjugates of H. By Exercise 5 we have
r=|G:N(H)|. Since each conjugate of Hhas the same order as H
and contains the identity and H⊆N(H), we know that the union
of all the conjugates of Hhas fewer than
|G:N(H)||H|≤|G:H||H|=|G|elements.
7. Say cl(x) = {x, g1xg−1
1, g2xg−1
2, . . . , gkxg−1
k}. If x−1=gixg−1
i, then
for each gjxg−1
jin cl(x) we have
(gjxg−1
j)−1=gjx−1g−1
j=gj(gixg−1
i)g−1
j∈cl(x). Because |G|has
odd order, gjxg−1
j6= (gjxg−1
j)−1. It follows that |cl(x)|is even. But
this contradicts the fact that |cl(x)|divides |G|.
8. By Theorem 9.3, we know that in each case the center of the group
is the identity. So, in both cases the first summand is 1. In the case
24/Sylow Theorems 191
of 39 all the summands after the first one must be 3 or 13. In the
case of 55 all the summands after the first one must be 5 or 11.
Thus the only possible class equations are
39=1+3+3+3+3+13+13; 55=1+5+5+11+11+11+11.
9. Part a is not possible by the Corollary of Theorem 24.2. Part b is
not possible because it implies that the center would have order 2
and 2 does not divide 21. Part c is the class equation for D5. Part
d is not possible because of Corollary 1 of Theorem 24.1
10. See Example 3 of Chapter 5.
11. Let Hand Kbe distinct Sylow 2-subgroups of G. By Theorem 7.2,
we have 48 ≥ |HK|=|H||K|/|H∩K|= 16 ·16/|H∩K|. This
simplifies to |H∩K|>5. Since Hand Kare distinct and |H∩K|
divides 16 we have |H∩K|= 8.
12. h123i,h234i,h124i,h134i
13. By Example 5 of Chapter 9, hxiKis a subgroup. By Theorem 7.2,
|hxiK|=|hxi||K|/|hxi ∩ K|. Since Kis a Sylow p-subgroup it
follows that hxi=hxi ∩ K. Thus hxi ⊆ K.
14. By Theorem 24.5 the number of Sylow 5-subgroups of Ghas the
form 1 + 5kand divides 77. Thus the number is 11.
15. By Theorem 24.5, np, the number of Sylow p-subgroups has the
form 1 + kp and npdivides |G|. But if k≥1,1 + kp is relatively
prime to pnand does not divide m. Thus k= 0. Now use the
corollary to Theorem 24.5.
16. Use Exercise 13.
17. By Theorem 24.5, there are 8 Sylow 7-subgroups.
18. By Sylow, n7= 1 or 8. If n7= 8, the Sylow 7-subgroups contain 48
elements of order 7. This means all the elements whose orders are a
power of 2 belong to single Sylow 2-subgroup. So n2= 1.
19. There are two Abelian groups of order 4 and two of order 9. There
are both cyclic and dihedral groups of orders 6, 8, 10, 12, and 14.
So, 15 is the first candidate. And, in fact, Theorem 24.5 shows that
there is only one group of order 15.
24/Sylow Theorems 192
20. n3= 7, otherwise the group is the internal direct product of
subgroups of orders 3 and 7 and such a group is cyclic.
21. By Exercise 20, Ghas seven subgroups of order 3.
22. 6; h(12345)i,h(21345)i
23. By Theorem 24.5 the only possibilities are 1, 4, and 10. So, once we
find 5 we know there are actually 10. Here are five:
h(123)i,h(234)i,h(134)i,h(345)i,h(245)i.
24. By Sylow’s Third Theorem there is only one Sylow 5-subgroup of
G. By Theorem 9.7 the Sylow 5-subgroup is isomorphic to Z25 or
Z5⊕Z5. Thus the number of elements of order 5 is 4 or 24.
25. A group of order 100 has 1, 5 or 25 subgroups of order 4; exactly
one subgroup of order 25 (which is normal); at least one subgroup
of order 5; and at least one subgroup of order 2.
26. Mimic Example 3.
27. By the corollary of Theorem 24.5 it suffices to show that there is
only one Sylow p-subgroup. By Sylow’s Third Theorem, np, the
number of Sylow p-subgroups of Ghas the form 1 + kp and divides
m. But if k > 0,1 + kp > p ≥m. So k= 0.
28. Since |D2m|= 4mthe Sylow 2-subgroups have order 4. Let F1be
any reflection in D2m. One Sylow 2-subgroup is
{R0, R180, F1, F1R180}. Next let F2be any reflection in D2mother
than F1and F1R180. The a second Sylow 2-subgroup is
{R0, R180, F2, F2R180}. Since there are 2mreflections, continuing in
this way we have mSylow 2-subgroups. By Sylow’s Third Theorem
there are no others.
29. By Sylow’s Third Theorem Theorem and Exercises 5 and 28 we
know |D2m:N(K)|=m. Also, |D2m:K|=m. Thus we have
|N(K)|=|K|and since K⊆N(K) we are done.
30. If pdoes not divide q−1, and qdoes not divide p2−1, then a
group of order p2qis Abelian.
31. Primes are ruled by Corollary 3 of Theorem 7.1; the corollary to
Theorem 24.2 rules out 9; Theorem 24.6 rules out 15. This leaves
21.
24/Sylow Theorems 193
32. If n3= 1, then |HK|= 15 where |H|= 3 and |K|= 5. (See
Example 5 of Chapter 9.) If n3= 25, then Theorem 24.5 and
Exercise 5 guarantee the existence of a subgroup of order 15.
33. Sylow’s Third Theorem (Theorem 24.5) implies that the Sylow 3-
and Sylow 5-subgroups are unique. Pick any xnot in the union of
these. Then |x|= 15.
34. By Sylow, n7= 1 or 15, and n5= 1 or 21. Counting elements
reveals that at least one of these must be 1. Then the product of
the Sylow 7-subgroup and the Sylow 5-subgroup is a subgroup of
order 35.
35. By Sylow’s Third Theorem, n17 = 1 or 35. Assume n17 = 35. Then
the union of the Sylow 17-subgroups has 561 elements. By Sylow’s
Third Theorem, n5= 1. Thus, we may form a cyclic subgroup of
order 85 (Exercise 57 of Chapter 9 and Theorem 24.6). But then
there are 64 elements of order 85. This gives too many elements for
the group.
36. By Sylow, n5= 1 or 6. A5has 24 elements of order 5.
37. If |G|= 60 and |Z(G)|= 4, then by Theorem 24.6, G/Z(G) is
cyclic. The “G/Z” Theorem (Theorem 9.3) then tells us that Gis
Abelian. But if Gis Abelian, then Z(G) = G.
38. a. Form a factor group G/N of order 30. The discussion
preceding Theorem 24.6 shows G/N has normal subgroups of
orders 3, 5, and 15. Now pullback.
b. Let Hbe a Sylow 2-subgroup containing N. The product of H
with the subgroups of orders 6 and 10 have orders 12 and 20.
c. The product of Nand the subgroup of order 15 has order 30
and is an internal direct product.
39. Let Hbe the Sylow 3-subgroup and suppose that the Sylow
5-subgroups are not normal. By Sylow’s Third Theorem, there
must be six Sylow 5-subgroups, call them K1, . . . , K6. These
subgroups have 24 elements of order 5. Also, the cyclic subgroups
HK1, . . . , HK6of order 15 each have eight generators. Thus, there
are 48 elements of order 15. This gives us more than 60 elements in
G.
24/Sylow Theorems 194
40. Let Nbe the normal subgroup of order 4. Then by Sylow, G/N
has a normal Sylow 7-subgroup whose pullback is normal.
41. We proceed by induction on |G|. By Theorem 24.2 and
Theorem 9.5, Z(G) has an element xof order p. By induction, the
group G/hxihas normal subgroups of order pkfor every kbetween
1 and n−1, inclusively. By Exercise 51 in Chapter 10, every
subgroup of G/hxihas the form H/hxi, where His a subgroup of
G. Moreover, if |H/hxi| =pkthen |H|has order pk+1. So, all that
remains is to show is that if H/hxiis normal in G/hxi, then His
normal in G. Let g∈Gand h∈H. Then
ghxihhxi((ghxi)−1=ghxihhxig−1hxi=ghg−1hxi=h0hxifor some
h0∈H. So, ghg−1∈h0hxi ⊆ H.
42. Let x∈Ghave maximum order, |x|=pt. Now let ybelong to G.
Then |y|=ps≤pt. Since hxihas a subgroup of order ps, we have
hyi⊆hxi.
43. Pick x∈Z(G) such that |x|=p. If x∈H, by induction,
N(H/hxi)> H/hxi, say yhxi ∈ N(H/hxi) but yhxi 6∈ H/hxi. Then
yis not in H, and by the argument given in Exercise 33, y∈N(H).
If x6∈ H, then x∈N(H), so that N(H)> H.
44. Since both Hand xHx−1have the same set of conjugates this
statement follows directly from Exercise 5.
45. Since 3 divides |N(K)|we know that N(K) has a subgroup H1of
order 3. Then, by Example 5 in Chapter 9, Theorem 7.2 and
Theorem 24.6, H1Kis a cyclic group of order 15. Thus,
K⊆N(H1) and therefore 5 divides |N(H1)|. And since Hand H1
are conjugates it follows from Exercise 27 in the Supplementary
Exercises for Chapters 1–4 that 5 divides |N(H)|.
46. Let Kbe a Sylow p-subgroup. Then by Example 5 of Chapter 9
HK is a subgroup of Gand by Theorem 7.2
|HK|=|H||K|/|H∩K|. Since this is a power of p, we must have
H∩K=K.
47. Sylow’s Third Theorem shows that all the Sylow subgroups are
normal. Then Theorem 7.2 and Example 5 of Chapter 9 ensure
that Gis the internal direct product of it Sylow subgroups. Gis
cyclic because of Theorems 9.6 and 8.2. Gis Abelian because of
Theorem 9.6 and Exercise 4 in Chapter 8.
24/Sylow Theorems 195
48. Let |G|=pimand |H|=pjnwhere mand nare relatively prime to
p. Then |G⊕H|=pi+jmn and |Gp⊕Hp|=pi+j. Since mn is
relatively prime to pthe Sylow p-subgroup of G⊕Hhas order pi+j.
49. Since automorphisms preserve order, we know |α(H)|=|H|. But
then the corollary of Theorem 24.5 shows that α(H) = H.
50. Clearly, N(H)⊆N(N(H)). Let xbelong to N(N(H)). Since
H⊆N(H), for any hin Hwe have that xhx−1belongs to N(H).
By Theorem 7.2, |hxhx−1iH|=|hxhx−1i||H|/|hxHx−1i ∩ H|. Since
|hxhx−1i| is a power of pand His a Sylow p-subgroup we must
have hxhx−1i=hxhx−1i ∩ H⊆H. Thus, xhx−1is in Hand xis in
N(H).
51. That |N(H)|=|N(K)|follows directly from the last part of
Sylow’s Third Theorem and Exercise 5.
52. Mimic Example 6. Three pairs are: 5, 7; 7, 11; 11, 13.
53. Normality of Himplies cl(h)⊆Hfor hin H. Thus the conjugacy
classes of Hobtained by conjugating by elements from Gare
subsets of H. Moreover, since every element hin His in cl(h) the
union of the conjugacy classes of His H. This is true only when H
is normal.
54. If not, let qbe a prime that does not divide |G|. Then Sylow’s First
Theorem guarentees the existence of an element of order q.
55. The mapping from Hto xHx−1given by h→xhx−1is an
isomorphism.
56. It suffices to prove that |K|/|K∩S|is relatively prime to p.
Observe that Theorem 7.2 , |KS|=|K||S|/|K∩S|so that
|K|/|K∩S|=|KS|/|S|. Moreover, |KS|/|S|divides |G|/|S|, which
is relatively prime to p.
57. Suppose that Gis a group of order 12 that has nine elements of
order 2. By the Sylow Theorems Ghas three Sylow 2-subgroups
whose union contains the identity and the nine elements of order 2.
If Hand Kare both Sylow 2-subgroups, by Theorem 7.2
|H∩K|= 2. Thus the union of the three Sylow 2-subgroups has at
most 7 elements of order 2 since there are 3 in H, 2 more in Kthat
are not in H, and at most 2 more that are in the third but not in
Hor K.
24/Sylow Theorems 196
58. By way of contradiction, assume that His the only Sylow
2-subgroup of Gand that Kis the only Sylow 3-subgroup of G.
Then Hand Kare normal and Abelian (corollary to Theorem 24.5
and corollary to Theorem 24.2). So, G=H×K≈H⊕Kand,
from Exercise 4 of Chapter 8, Gis Abelian.
59. First note that pdivides n+ 1 is equivalent to nmod p=−1. Let
Hbe the unique Sylow p-subgroup of Gand let |H|=pm. By
Theorem 24.4 all the elements of order a power of pare contained in
H. Since every element of Gwhose order is a power of plies in H,
we have that n=pm−(number of elements of order pm+ number
of elements of order pm−1+··· + number of elements of order p2+
number of elements of order p0). By the Corollary of Theorem 4.4
and the fact that φ(pk) is divisible by pfor each k≥2 each term on
the right is divisible by pexcept p0= 1. Thus, nmod p=−1.
60. Mimic Example 6.
61. Let Gbe a group of order 21. By Sylow’s Third Theorem
(Theorem 24.5) and its corollary there is a unique Sylow 7-subgroup
H, which is normal in G. Let H=hxiand let y∈Ghave order 3.
Since hxiis normal in G,yxy−1=xifor some i= 1,2,3,4,5, or 6.
Then x=y3xy−3=y2(yxy−1)y−2=y2xiy−2=y(yxiy−1)y−1=
y(yxy−1)iy−1=yxi2y−1= (yxy−1)i2= (xi)i2=xi3. So, 7 divides
i3−1 and we see that the only possibilities for iare i= 1, 2 or 4.
62. By the corollary to Theorem 24.2, G/H is Abelian. To prove that
G0⊆Hit is enough to show that x−1y−1xy ∈Hfor all xand yin
G. To this end observe that x−1y−1xyH =x−1Hy−1HxHyH =H.
So, x−1y−1xy ∈H.
63. Say cl(e) and cl(a) are the only two conjugacy classes of a group G
of order n. Then cl(a) has n−1 elements all of the same order, say
m. If m= 2, then it follows from Exercise 47 Chapter 2 that Gis
Abelian. But then cl(a) = {a}and so n= 2. If m > 2, then cl(a)
has at most n−2 elements since conjugation of aby e,a, and a2
each yield a.
64. By Sylow’s Third Theorem the number of Sylow 7 subgroups is 1 or
8. So the number of elements of order 7 is 6 or 48.
65. Note that any subgroup of order 4 in a group of order 4mwhere m
is odd is a Sylow 2-subgroup. By Sylow’s Third Theorem, the
24/Sylow Theorems 197
Sylow 2-subgroups are conjugate and therefore isomorphic. S4
contains both the subgroups h(1234)iand {(1),(12),(34),(12)(34)}.
66. By the “G/Z Theorem” (9.3), |N(H)/C(H)|divides |Aut(H)|.
Since His cyclic we know that C(H)⊇Hand therefore
|N(H)/C(H)|is relatively prime to p. Letting |H|=pk, we have
by Theorem 6.5 that Aut(H) is isomorphic to U(pk) and by the
formula given in Chapter 8 we have |U(pk)|=pk−1(p−1). Since
the smallest prime divisor of |N(H)/C(H)|is greater than p, we
must have |N(H)/C(H)|= 1.
67. By Sylow’s Third Theorem, the number of Sylow 13-subgroups is
equal to 1 mod 13 and divides 55. This means that there is only
one Sylow 13-subgroup so it is normal in G. Thus
|N(H)/C(H)|= 715/|C(H)|divides both 55 and 12. This forces
715/|C(H)|= 1 and therefore C(H) = G. This proves that His
contained in Z(G). Applying the same argument to Kwe get that
Kis normal in Gand |N(K)/C(K)|= 715/|C(K)|divides both 65
and 10. This forces 715/|C(K)|= 1 or 5. In the latter case Kis
not contained in Z(G).
68. Let Kdenote a subgroup of Gof order 5 and Ldenote a Sylow
11-subgroup of G. By Sylow’s Third Theorem and its corollary, we
know that His only Sylow 7-subgroup and Lis the only Sylow
11-subgroup and both are normal in G. Thus, by Example 5 of
Chapter 9, and Theorem 24.6, the subgroup HK of order 35 is
cyclic and the subgroup HL of order 77 is cyclic. It follows that
C(H) contains the groups H, K, and Land so |C(H)|is divisible
by 5 ·7·11 = 385. If the Sylow 5-subgroup of Gis not cyclic then
the Sylow 5-subgroup has the form K1K2where K1and K2are
distinct subgroups of order 5. Thus the previous argument show
that C(H) contains both K1and K2and therefore K1K2. So,
|C(H)|is divisible by 52·7·11 = 1925. This means that His
contained in Z(G).
69. Let Hbe a Sylow 5-subgroup. Since the number of Sylow
5-subgroups is 1 mod 5 and divides 7 ·17, the only possibility is 1.
So, His normal in G. Then by the N/C Theorem (Example 15 of
Chapter 10), |G/C(H)|divides both 4 and |G|. Thus C(H) = G.
70. .5
24/Sylow Theorems 198
71. This follows directly from Theorem 24.1
72. |C(a)|/|G|.
73. Since D4has 5 conjugate classes, Pr(D4) = 5/8. Since S3has 2
conjugacy classes, Pr(S3) = 1/3. Since A4has 4 conjugacy classes,
Pr(A4) = 1/3.
74. Suppose that nis odd. Then there are n2commuting pairs of
rotations; ncommuting pairs of the form (R0, F ) where Fis a
reflection; ncommuting pairs of the form (F, R0); and n
commuting pairs of the form (F, F ) where Fis a reflection. No
other pairs commute. This gives Pr(Dn)
=(n2+ 3n)/4n2= (n+ 3)/4n. Now suppose that nis even. Then
there are n2commuting pairs of rotations; ncommuting pairs of
the form (R0, F ) where Fis a reflection; ncommuting pairs of the
form (F, R0); ncommuting pairs of the form (R180, F ) where Fis a
reflection; ncommuting pairs of the form (F, R180); ncommuting
pairs of the form (F, F ) where Fis a reflection; and ncommuting
pairs of the form (F, F R180) where Fis a reflection. There are no
others. This gives Pr(Dn) =(n2+ 6n)/4n2= (n+ 6)/4n.
75. By the discussion in this chapter, Pr(G⊕H) = m/(|G||H|) where
mis the number of conjugacy classes in G⊕H. Next prove that
|C(g, h)|=|C(g)||C(h)|. Then use Theorem 24.1.
76. Exactly as in the case for a group, we have for a ring
R={x1, x2, . . . , xn}, Pr(R) = |K|/n2, where
K={(x, y)|xy =yx, x, y ∈R}. Also,
|K|=|C(x1)|+|C(x2)|+··· +|C(xn)|. From Exercise 28 in the
Supplemental Exercises for Chapters 12–14, we know that R/C(R)
is not cyclic. Thus, |R/C(R)| ≥ 4 and so |C(R)| ≤ |R|/4. So, for at
least 3/4 of the elements xof R, we have |C(x)|≤|R|/2. Then
starting with the elements in the center and proceeding to the
elements not in the center, we have
|K|≤|R|/4 + (1/2)(3/4)|R|= (5/8)|R|.
199
CHAPTER 25
Finite Simple Groups
1. This follows directly from the “2·odd” Theorem (Theorem 25.2).
2. We may assume that n5= 56 and n7= 8. Then counting elements
forces n2= 1.
3. By the Sylow Theorems if there were a simple group of order 216
the number of Sylow 3-subgroups would be 4. Then the normalizer
of a Sylow 3-subgroup would have index 4. The Index Theorem
(corollary of Theorem 25.3) then gives a contradiction.
4. Observe that n5= 6 and use the Index Theorem.
5. Suppose Gis a simple group of order 525. Let L7be a Sylow
7-subgroup of G. It follows from Sylow’s theorems that
|N(L7)|= 35. Let Lbe a subgroup of N(L7) of order 5. Since
N(L7) is cyclic (Theorem 24.6), N(L)≥N(L7), so that 35 divides
|N(L)|. But Lis contained in a Sylow 5-subgroup (Theorem 24.4),
which is Abelian (see the Corollary to Theorem 24.2). Thus, 25
divides |N(L)|as well. It follows that 175 divides |N(L)|. The
Index Theorem now yields a contradiction.
6. The n5= 6 case is easy. Let n5= 36, L5a Sylow 5-subgroup, L3a
Sylow 3-subgroup. Consider |N(L3)N(L5)|to show N(L3)∩N(L5)
is a subgroup of order 3, call it K. Then show |N(K)|is divisible
by 45. Note that a group of order 45 is Abelian. This contradicts
|N(L5)|= 15.
7. Suppose that there is a simple group Gof order 528 and L11 is a
Sylow 11-subgroup. Then n11 = 12,|N(L11)|= 44, and Gis
isomorphic to a subgroup of A12. Since |N(L11)/C(L11)|divides
|Aut(Z11)|= 10, |C(L11)|= 22 or 44. In either case, C(L11) has
elements of order 2 and 11 that commute. But then C(L11) has an
element of order 22 whereas A12 does not.
8. A7has no element of order 15.
25/Finite Simple Groups 200
9. Suppose that there is a simple group Gof order 396 and L11 is a
Sylow 11-subgroup. Then n11 = 12, |N(L11)|= 33, and Gis
isomorphic to a subgroup of A12. Since |N(L11)/C(L11)|divides
|Aut(Z11)|= 10, |C(L11)|= 33. Then C(L11) has elements of order
3 and 11 that commute. But then C(L11) has an element of order
33 whereas A12 does not.
10. 211, 223, 227, 229 and 233 are prime. The 2·odd test rules out 202,
206, 210, 214, 218, 222, 226, 230 and 234. The Index Theorem rules
out 216 and 224. The Sylow test rules out the remaining cases.
11. If we can find a pair of distinct Sylow 2-subgroups Aand Bsuch
that |A∩B|= 8, then N(A∩B)≥AB, so that N(A∩B) = G.
Now let Hand Kbe any pair of distinct Sylow 2-subgroups. Then
16 ·16/|H∩K|=|HK| ≤ 112 (Theorem 7.2), so that |H∩K|is at
least 4. If |H∩K|= 8, we are done. So, assume |H∩K|= 4. Then
N(H∩K) picks up at least 8 elements from Hand at least 8 from
K(see Exercise 43 of Chapter 24). Thus, |N(H∩K)| ≥ 16 and is
divisible by 8. So, |N(H∩K)|= 16, 56, or 112. Since the latter
two cases imply that Ghas a normal subgroup, we may assume
|N(H∩K)|= 16. If N(H∩K) = H, then |H∩K|= 8, since
N(H∩K) contains at least 8 elements from K. So, we may assume
that N(H∩K)6=H. Then, we may take A=N(H∩K) and
B=H.
12. We may assume n7= 15 and n3= 7 or 10. Since A7does not have
an element of order 15, n36= 7. Then let L7be a Sylow 7-subgroup
and L3a Sylow 3-subgroup. Then |N(L7)|= 14 and |N(L3)|= 21.
But N(L3) must contain a unique subgroup of order 7. It follows
that N(L3) is cyclic. This means that N(L7) must contain a
subgroup of order 3, which is impossible.
14. Theorem 24.2 handles the case where p=q=r. The Sylow Test
for nonsimplicity (Theorem 25.1) shows a group of order p2qwith
p>qmust have a normal Sylow p-subgroup. The same theorem
implies that a simple group of order p2qwith p<qwould have p2
subgroups of order qand more than one subgroup of order p2.
Counting elements then yields a contradiction.
Finally consider a simple group of order pqr where p<q<r. Then
there are pq subgroups of order r, at least rsubgroups of order q
25/Finite Simple Groups 201
and at least qsubgroups of order p. But
pq(r−1) + r(q−1) + q(p−1) > pqr.
15. If A5had a subgroup of order 30, 20, or 15, then there would be a
subgroup of index 2, 3 or 4. But then the Index Theorem gives us a
contradiction to the fact that Gis simple.
16. If S5had a subgroup Hof order 40 or 30, then H∩S5would be a
subgroup of A5of order 20, 30 or 15 since every element of His
even or half of them are even (40 is excluded by Lagrange’s
Theorem). By Exercise 15, this is impossible.
17. Suppose that there is a simple group of order 120 and let L3be a
Sylow 3-subgroup of G. By Sylow’s Third Theorem
(Theorem 24.5), n5= 6, n3= 10, or 40. By the Embedding
Theorem (Corollary 2 of Theorem 25.3), Gis isomorphic to a
subgroup of A6. If n3= 10, then |N(L3)|= 12 and |N(L3)/C(L3)|
divides |Aut(Z3)|= 2. So, C(L3) has an element of order 2 and an
element of order 3 that commute. But then Ghas an element of
order 6 whereas A6does not. If n3= 40, the union of the Sylow
3-subgroups and the Sylow 5-subgroups has 105 elements. We now
claim that any two distinct Sylow 2-subgroups L2and L0
2have at
most 2 elements in common. If this is not the case, then
|L2∩L0
2|= 4. It then follows from Exercise 43 in Chapter 24 that
N(L2∩L0
2) contains L2L0
2. So, by Theorem 7.2
N(L2∩L0
2)≥8·8/4 = 16. We also know that |N(L2∩L0
2)|is
divisible by 8 and divides 120. So, |N(L2∩L0
2)| ≥ 24. Then, by the
Embedding Theorem, Gis isomorphic to a subgroup of A5, a group
of order 60. So, L2and L0
2have at most 2 elements in common.
Thus, when we take the union of three distinct Sylow 2-subgroups
(which exist by Sylow’s Third Theorem) we produce at least
7 + 6 + 5 = 18 new elements. This gives us more than 120.
18. Use Exercise 51 of Chapter 9.
19. Let αbe as in the proof of the Generalized Cayley Theorem
(Theorem 25.3). Then if g∈Ker αwe have gH =Tg(H) = Hso
that Ker α⊆H. Since α(G) consists of a group of permutations of
the left cosets of Hin Gwe know by the First Isomorphism
Theorem (Theorem 10.3) that G/Ker αis isomorphic to a subgroup
of S|G:H|. Thus, |G/Ker α|divides |G:H|!. Since Ker α⊆H, we
have that |G:H||H: Ker α|=|G: Ker α|must divide
25/Finite Simple Groups 202
|G:H|! = |G:H|(|G:H| − 1)!. Thus, |H: Ker α|divides
(|G:H| − 1)!. Since |H|and (|G:H| − 1)! are relatively prime, we
have |H: Ker α|= 1 and therefore H= Ker α. So, by the
Corollary of Theorem 10.2, His normal. In the case that a
subgroup Hhas index 2, we conclude that His normal.
20. Use Exercise 19.
21. If His a proper normal subgroup of S5, then H∩A5=A5or {ε}
since A5is simple and H∩A5is normal. But H∩A5=A5implies
H=A5, whereas H∩A5={ε}implies H={ε}or |H|= 2. (See
Exercise 23 of Chapter 5.) Now use Exercise 72 of Chapter 9 and
Exercise 66 of Chapter 5.
22. The Sylow Test for Nonsimplicity yields n5= 6 and n3= 4 or 10.
The Index Theorem rules out n3= 4. Now appeal to Sylow’s Third
Theorem (24.5).
23. If P SL(2, Z7) had a nontrivial proper subgroup H, then
|H|= 2,3,4,6,7,8,12,14,21,24,28,42,56, or 84. Observing that
"1 4
1 5 #has order 3 and using conjugation we see that P SL(2, Z7)
has more than one Sylow 3-subgroup; observing that "5 5
1 4 #has
order 7 and using conjugation we see that P SL(2, Z7) has more
than one Sylow 7-subgroup; observing that "5 1
3 5 #has order 4
and using conjugation we see that P SL(2, Z7) has more than one
Sylow 2-subgroup. So, from Sylow’s Third Theorem, we have
n3= 7, n7= 8, and n2is at least 3. So, P SL(2, Z7) has 14 elements
of order 3, 48 elements of order 7, and at least 11 elements whose
orders are powers of 2. If |H|= 3,6,or 12, then |G/H|is relatively
prime to 3, and by Exercise 61 of Chapter 9, Hwould contain the
14 elements of order 3. If |H|= 24, then Hwould contain the 14
elements of order 3 and at least 11 elements whose orders are a
power of 2. If |H|= 7,14,21,28, or 42, then Hwould contain the
48 elements of order 7. If |H|= 56, then Hwould contain the 48
elements of order 7 and at least 11 elements whose orders are a
power of 2. If |H|= 84, then Hwould contain the 48 elements of
order 7, but by Sylow’s Third Theorem a group of order 84 has
only one Sylow 7-subgroup. If |H|= 2 or 4, the G/H has a normal
25/Finite Simple Groups 203
Sylow 7-subgroup. This implies that Gwould have a normal
subgroup of order 14 or 28, both of which have been ruled out. (To
see that Gwould have a normal subgroup of order 14 or 28, note
that the natural mapping from Gto G/H taking gto gH is a
homomorphism then use properties 8 and 5 of Theorem 10.2.) So,
every possibility for Hleads to a contradiction.
24. Let H=h(12),(12345)i. First note that
(12345)−1(12)(12345)(12) = (125), so Hcontains an element of
order 3. Moreover, since
(12345)−2(12)(12345)2= (42531)(12)(13524) = (45), H contains
the subgroup {(1),(12),(45),(12)(45)}. This means that |H|is
divisible by 3, 4, and 5 and therefore |H|= 60 or 120. But |H|
cannot be 60 for if so, then the subset of even permutations in H
would be a subgroup of order 30 (see Exercise 23 in Chapter 5).
This means that A5would have a subgroup of index 2, which would
a normal subgroup. This contradicts the simplicity of A5.
25. Suppose that S5has a subgroup Hthat contains a 5-cycle αand a
2-cycle β. Say β= (a1a2). Then there is some integer ksuch that
αk= (a1a2a3a4a5). Note that
(a1a2a3a4a5)−1(a1a2)(a1a2a3a4a5)(a1a2) =
(a5a4a3a2a1)(a1a2)(a1a2a3a4a5)(a1a2)=(a1a2a5), so Hcontains an
element of order 3. Moreover, since
α−2βα2= (a4a2a5a3a1)(a1a2)(a1a3a5a2a4)=(a4a5), H contains
the subgroup {(1),(a1a2),(a4a5),(a1a2)(a4a5)}. This means that
|H|is divisible by 60. But |H|cannot be 60 for if so, then the
subset of even permutations in Hwould be a subgroup of order 30
(see Exercise 23 in Chapter 5). This means that A5would have a
subgroup of index 2, which would a normal subgroup. This
contradicts the simplicity of A5.
26. Let p=|G:H|and q=|G:K|. It suffices to show that p=q. But
if p<q, say, then qdoes not divide p!. This contradicts the Index
Theorem.
27. Suppose there is a simple group of order 60 that is not isomorphic
to A5. The Index Theorem implies n26= 1 or 3, and the Embedding
Theorem implies n26= 5. Thus, n2= 15. If every pair of Sylow
2-subgroups has only the identity element in common then the
union of the 15 Sylow 2-subgroups has 46 elements. But n5= 6, so
25/Finite Simple Groups 204
there are also 24 elements of order 5. This gives more than 60. As
was the case in showing that there is no simple group of order 144
the normalizer of this intersection has index 5, 3, or 1. But the
Embedding Theorem and the Index Theorem rule these out.
28. Use the Embedding Theorem.
29. Suppose there is a simple group Gof order p2qwhere pand qare
odd primes and q > p. Since the number of Sylow q-subgroups is 1
mod qand divides p2, it must be p2. Thus there are p2(q−1)
elements of order qin G. These elements, together with the p2
elements in one Sylow p-subgroup, account for all p2qelements in
G. Thus there cannot be another Sylow p-subgroup. But then the
Sylow p-subgroup is normal in G.
30. Let Land Mbe distinct maximal subgroups of G. Since N(K)
contains both Land Mit properly contains them. Since Lis
maximal, N(K) = G. Because Gis simple, we have K={e}.
31. Consider the right regular representation of G. Let gbe a generator
of the Sylow 2-subgroup and suppose that |G|= 2knwhere nis
odd. Then by Exercise 62 in Chapter 6 every cycle of the
permutation Tgin the right regular representation of Ghas length
2k. This means that there are exactly nsuch cycles. Since each
cycle is odd and there is an odd number of them, Tgis odd. This
means that the set of even permutations in the regular
representation has index 2 and is therefore normal. (See Exercise
23 in Chapter 5 and Exercise 9 in Chapter 9.)
205
CHAPTER 26
Generators and Relations
1. u∼ubecause uis obtained from itself by no insertions; if vcan be
obtained from uby inserting or deleting words of the form xx−1or
x−1xthen ucan be obtained from vby reversing the procedure; if
ucan be obtained from vand vcan be obtained from wthen ucan
be obtained from wby first obtaining vfrom wthen ufrom v.
2. Let abe any reflection in Dnand let b=aR360/n. Then aZ(Dn)
and bZ(Dn) have order 2 and generate Dn/Z(Dn). Now use
Theorem 26.5 and the fact that |Dn/Z(Dn)|=n=|Dn/2|.
3. b(a2N) = b(aN)a= (ba)Na =a3bN a =a3b(aN) = a3(ba)N
=a3a3bN =a6bN =a6Nb =a2Nb =a2bN
b(a3N) = b(a2N)a=a2bNa =a2b(aN) = a2a3bN
=a5bN =a5Nb =aNb =abN
b(bN) = b2N=N
b(abN) = baNb =a3bN b =a3b2N=a3N
b(a2bN) = ba2Nb =a2bN b =a2b2N=a2N
b(a3bN) = ba3Nb =abN b =ab2N=aN
4. Since b=b−1, we have bab =a2.Then a=a6= (bab)3=ba3bso
that ba =a3b. Thus, a3b=a2band a=e. Finally, note that Z2
satisfies the relations with a= 0 and b= 1.
5. Let Fbe the free group on {a1, a2, . . . , an}. Let Nbe the smallest
normal group containing {w1, w2, . . . , wt}and let Mbe the smallest
normal subgroup containing {w1, w2, . . . , wt, wt+1, . . . , wt+k}. Then
F/N ≈Gand F/M ≈G. The homomorphism from F/N to F/M
given by aN →aM induces a homomorphism from Gonto ¯
G.
To prove the corollary, observe that the theorem shows that Kis a
homomorphic image of G, so that |K|≤|G|.
6. Use the Corollary to Dyck’s Theorem.
26/Generators and Relations 206
7. Clearly, aand ab belong to ha, bi, so ha, abi⊆ha, bi. Also, aand
a−1(ab) = bbelong to ha, abi.
8. Use Theorem 26.5.
9. By Exercise 7, hx, yi=hx, xyi. Also,
(xy)2= (xy)(xy)=(xyx)y=y−1y=e, so by Theorem 26.5, Gis
isomorphic to a dihedral group and from the proof of
Theorem 26.5, |x(xy)|=|y|=nimplies that G≈Dn.
10. 3. hx, y, z |x2=y2=z2=e, xy =yx, xz =zx, yz =zyi.
11. Since x2=y2=e, we have (xy)−1=y−1x−1=yx. Also,
xy =z−1yz, so that (xy)−1= (z−1yz)−1=z−1y−1z=z−1yz =xy.
12. a. b0a=ab. ba
13. First note that b2=abab implies that b=aba.
a. So, b2abab3=b2(aba)b3=b2bb3=b6.
b. Also, b3abab3a=b3(aba)b3a=b3bb3a=b7a.
14. Observe that Gis generated by 6 and 4 where |6|= 2, |4|= 2 and
|6·4|=|3|=n. Now use Theorem 26.5.
15. Note that yxyx3=eimplies that yxy−1=x5and therefore hxiis
normal. So, G=hxi ∪ yhxiand |G| ≤ 16. From y2=eand
yxyx3=e, we obtain yxy−1=x−3. So,
yx2y−1=yxy−1yxy−1=x−6=x2. Thus, x2∈Z(G). On the other
hand, Gis not Abelian for if so we would have e=yxyx3=x4and
then |G| ≤ 8. It now follows from the “G/Z” Theorem
(Theorem 9.3) that |Z(G)| 6= 8. Thus, Z(G) = hx2i. Finally,
(xy)2=xyxy =x(yxy) = xx−3=x−2, so that |xy|= 8.
16. Use Theorem 7.2, Theorem 26.4, the corollary of Theorem 24.2,
and Theorem 24.6.
17. Since the mapping from Gonto G/N given by x→xN is a
homomorphism, G/N satisfies the relations defining G.
18. If Gwere Abelian then the relation st =ts could be derived from
sts =tst. But this same derivation would hold when s= (23) and
t= (13). However, (23)(13) 6= (13)(23).
26/Generators and Relations 207
19. For Hto be a normal subgroup we must have
yxy−1∈H={e, y3, y6, y9, x, xy3, xy6, xy9}. But
yxy−1=yxy11 = (yxy)y10 =xy10.
20. Every element has the form xior xiywhere 0 ≤i < 2n. Let
0< i < 2n. Then xi∈Z(G) if and only if y−1xiy=xi. But
y−1xiy= (y−1xy)i= (x−1)i=x−1.
So x2i=e. This implies i=n. A similar argument shows
xiy∈Z(G) implies i=n. But xny∈Z(G) and xn∈Z(G) imply
y∈Z(G), which is false. So xny6∈ Z(G).
To prove the second portion, observe that G/Z(G) has order 2n
and is generated by a pair of elements of order 2.
21. First note that b−1a2b= (b−1ab)(b−1ab) = a3a3=a6=e. So,
a2=e. Also, b−1ab =a3=aimplies that aand bcommute. Thus,
Gis generated by an element of order 2 and an element of order 3
that commute. It follows that Gis Abelian and has order at most
6. But the defining relations for Gare satisfied by Z6with a= 3
and b= 2. So, G≈Z6.
22. Since yx =x3y, the set S={hyi, xhyi, x2hyi, x3hyi} is closed under
multiplication on the left by xand y. Thus every element of Ghas
the form xiyjwith 0 ≤i < 4 and 0 ≤j < 4.
To compute the center observe that xyx =yand xy =yx3. So
x2y=x(xy) = xyx3= (xyx)x2=yx2.
Thus x2∈Z(G). Also,
xy2= (xy)y=yx3y=yx2xy =yxyx2=y(xyx)x=y2x
so that y2∈Z(G). It follows that Z(G) = {e, x2, y2, x2y2}.
(Theorem 9.3 shows |Z(G)| 6= 8.)
Finally, observe that G/hy2ihas order 8 and is generated by yhy2i
and xyhy2ieach of which has order 2.
26/Generators and Relations 208
23. In the notation given in the proof of Theorem 26.5 we have that
|e|= 1, |a|=|b|= 2, |ab|=|ba|=∞. Next observe that since every
element of D∞can be expressed as a string of alternating a’s and
b’s or alternating b’s and a’s, every element can be expressed in one
of four forms: (ab)n,(ba)n,(ab)na, or (ba)nbfor some n. Since
|ab|=|ba|=∞, we have |(ab)n|=|(ba)n|=∞(excluding n= 0).
And, since
((ab)na)2= (ab)na(ab)na= (ab)(ab)···(ab)a(ab)(ab)···(ab)a, we
can start at the middle and successively cancel the adjacent a’s,
then adjacent b’s, then adjacent a’s, and so on to obtain the
identity. Thus, |(ab)na|= 2. Similarly, |(ba)nb|= 2.
24. Use Theorem 26.4.
25. First we show that d=b−1, a =b2and c=b3so that G=hbi. To
this end observe that ab =cand cd =atogether imply that cdc =c
and therefore d=b−1. Then da =band d=b−1together imply
that a=b2. Finally, cd =aand d=b−1together imply c=b3.
Thus G=hbi. Now observe that bc =d, c =b3, and d=b−1yield
b5=e. So |G|= 1 or 5. But Z5satisfies the defining relations with
a= 1, b = 3, c= 4, and d= 2.
26. From Theorem 26.5 and its proof the group is dihedral and has
order 2|ab|. To compute |ab|, note that
(ab)3= (aba)(bab) = (aba(aba) = aba2ba =e. So, the group is D3.
27. Since aba−1b−1=e,Gis an Abelian group of order at most 6.
Then because Z6satisfies the given relations, we have that Gis
isomorphic to Z6.
28. F⊕Z3where Fis the free group on two letters.
30. There are only five groups of order 8: Z8and the quaternions have
only one element of order 2; Z4⊕Z2has 3; Z2⊕Z2⊕Z2has 7; and
D4has 5.
209
CHAPTER 27
Symmetry Groups
1. If Tis a distance-preserving function and the distance between
points aand bis positive, then the distance between T(a) and T(b)
is positive.
2. For any fixed v0in Rndefine
Tv0:Rn→Rnby Tv0(v) = v+v0.
Then {Tv0|v0∈Rn}is the set of translations of Rn. Closure and
associativity follow from the observation
Tv0◦Tw0=Tw0+v0;T0is the identity; (Tv0)−1=T−v0.
3. See Figure 1.5.
4. Use Theorem 7.2.
5. There are rotations of 0◦,120◦and 240◦about an axis through the
centers of the triangles and a 180◦rotation through an axis
perpendicular to a rectangular base and passing through the center
of the rectangular base. This gives 6 rotations. Each of these can
be combined with the reflection plane perpendicular to the base
and bisecting the base. So, the order is 12.
6. 16
7. There are nrotations about an axis through the centers of the
n-gons and a 180◦rotation through an axis perpendicular to a
rectangular base and passing through the center of the rectangular
base. This gives 2nrotations. Each of these can be combined with
the reflection plane perpendicular to a rectangular base and
bisecting the base. So, the order is 4n.
8. A drawing or model reveals the group consists of the identity, three
180◦rotations and 4 reflections and is Abelian.
27/Symmetry Groups 210
9. In R1, there is the identity and an inversion through the center of
the segment. In R2, there are rotations of 0◦and 180◦, a reflection
across the horizontal line containing the segment, and a reflection
across the perpendicular bisector of the segment. In R3, the
symmetry group is G⊕Z2, where Gis the plane symmetry group
of a circle. (Think of a sphere with the line segment as a diameter.
Then Gincludes any rotation of that sphere about the diameter
and any plane containing the diameter of the sphere is a symmetry
in G. The Z2must be included because there is also an inversion.)
10. No symmetry; symmetry across a horizontal axis only; symmetry
across a vertical axis only; symmetry across a horizontal axis and a
vertical axis.
11. There are 6 elements of order 4 since for each of the three pairs of
opposite squares there are rotations of 90◦and 270◦.
12. It is the same as a 180◦rotation.
13. An inversion in R3leaves only a single point fixed, while a rotation
leaves a line fixed.
14. A rotation of 180◦about the line L.
15. In R4, a plane is fixed. In Rn, a hyperplane of dimension n−2 is
fixed.
16. Consider a triangle whose sides have lengths a,b,c. The image of
this triangle is also a triangle whose sides have lengths a,b,c. Thus
the two triangles are congruent (side-side-side).
17. Let Tbe an isometry, let p,q, and rbe the three noncollinear
points, and let sbe any other point in the plane. Then the
quadrilateral determined by T(p), T(q), T(r), and T(s) is
congruent to the one formed by p,q,r, and s. Thus, T(s) is
uniquely determined by T(p), T(q), and T(r).
18. Use Exercise 17.
19. The only isometry of a plane that fixes exactly one point is a
rotation.
20. A translation a distance twice that between aand balong the line
joining aand b.
211
CHAPTER 28
Frieze Groups and Crystallographic
Groups
1. The mapping φ(xmyn) = (m, n) is an isomorphism. Onto is by
observation. If φ(xmyn) = φ(xiyj), then (m, n) = (i, j) and
therefore, m=nand i=j. Also, φ((xmyn)(xiyj)) =
φ(xm+iyn+j)=(m+i, n +j)=(m, n)(i, j) = φ(xmyn)φ(xiyj).
2. 4
3. Using Figure 28.9 we obtain x2yzxz =xy.
4. x−4y
5. Use Figure 28.9.
6. Use Figure 28.8.
7. x2yzxz =x2yx−1=x2x−1y=xy
x−3zxzy =x−3x−1y=x−4y
8. It suffices to show y−1xy =xiand z−1xz =xjfor some iand j.
9. A subgroup of index 2 is normal.
11. a. V, b. I, c. II, d. VI, e. VII, and f. III.
12. a. V b. III c. VII d. IV e. V
13. cmm
14. Reading down the columns starting on the left we have:
pgg, pmm, p2, p1, cmm, pmg, pg, pm, p3, p4, p4m, p4g, cm, p6,
p3m1, p31m, p6m.
15. a. p4m, b. p3, c. p31m, and d. p6m
28/Frieze Groups and Crystallographic Groups 212
16. The top row
α−3β2, α−2β2, α−1β2, β2, αβ2.
The bottom row is
α−2β−1, α−1β−1, β−1, αβ−1, α2β−1, α3β−1.
17. The principle purpose of tire tread design is to carry water away
from the tire. Patterns I and III do not have horizontal reflective
symmetry. Thus these designs would not carry water away equally
on both halves of the tire.
18. Let us call the motif a heart. Focus on the heart located at the
bottom middle, where the tip just touches the border. Now draw a
vertical axis of symmetry through the hearts in the next column
(vertical axis). Exactly midway between these two axes is a
glide-reflection axis that is not a reflection axis.
19. a. VI, b. V, c. I, d. III, e. IV, f. VII, g. IV
20. Focus on the two triangles located at the bottom left. The vertical
line through the tip of the right side of the base of the triangle on
the right is a glide-reflection axis that is not a reflection axis.
213
CHAPTER 29
Symmetry and Counting
1. The symmetry group is D4. Since we have two choices for each
vertex, the identity fixes 16 colorings. For R90 and R270 to fix a
coloring, all four corners must have the same color so each of these
fixes 2 colorings. For R180 to fix a coloring, diagonally opposite
vertices must have the same color. So, we have 2 independent
choices for coloring the vertices and we can choose 2 colors for each.
This gives 4 fixed colorings for R180. For Hand V, we can color
each of the two vertices on one side of the axis of reflection in 2
ways, giving us 4 fixed points for each of these rotations. For Dand
D0, we can color each of the two fixed vertices with 2 colors and
then we are forced to color the remaining two the same. So, this
gives us 8 choices for each of these two reflections. Thus, the total
number of colorings is
1
8(16 + 2 ·2+4+2·4+2·8) = 6.
2. 21
3. The symmetry group is D3. There are 53−5 = 120 colorings
without regard to equivalence. The rotations of 120◦and 240◦can
fix a coloring only if all three vertices of the triangle are colored the
same so they each fix 0 colorings. A particular reflection will fix a
coloring provided that fixed vertex is any of the 5 colors and the
other two vertices have matching colors. This gives 5 ·4 = 20 for
each of the three reflections. So, the number of colorings is
1
6(120 + 0 + 0 + 3 ·20) = 30.
4. 92 (γg1φg2)(x) = γg1(g2xH) = g1(g2xH)
29/Symmetry and Counting 214
5. The symmetry group is D6. The identity fixes all 26= 64
arrangements. For R60 and R300, once we make a choice of a
radical for one vertex all others must use the same radical. So,
these two fix 2 arrangements each. For R120 and R240 to fix an
arrangement every other vertex must have the same radical. So,
once we select a radical for one vertex and a radical for an adjacent
vertex we then have no other choices. So we have 22choices for
each of 2 these rotations. For R180 to fix an arrangement, each
vertex must have the same radical as the vertex diagonally opposite
it. Thus, there are 23choices for this case. For the 3 reflections
whose axes of symmetry joins two vertices, we have 2 choices for
each fixed vertex and 2 choices for each of the two vertices on the
same side of the reflection axis. This gives us 16 choices for each of
these 3 reflections. For the 3 reflections whose axes of reflection
bisects opposite sides of the hexagon, we have 2 choices for each of
the 3 vertices on the same size of the reflection axis. This gives us 8
choices for each of these 3 reflections. So, the total number of
arrangements is
1
12(64 + 2 ·2+2·4+8+3·16 + 3 ·8) = 13.
6. 9099
7. The symmetry group is D4. The identity fixes 6 ·5·4·3 = 360
colorings. All other symmetries fix 0 colorings because of the
restriction that no color be used more than once. So, the number of
colorings is 360/8 = 45.
8. 231
9. The symmetry group is D11. The identity fixes 211 colorings. Each
of the other 10 rotations fixes only the two colorings in which the
beads are all the same color. (Here we use the fact that 11 is prime.
For example, if the rotation R2·360/11 fixes a coloring then once we
choose a color for one vertex the rotation forces all other vertices to
have that same color because the rotation moves 2 vertices at a
time and 2 is a generator of Z11.) For each reflection, we may color
the vertex containing the axis of reflection 2 ways and each vertex
on the same side of the axis of reflection 2 ways. This gives us 26
colorings for each reflection. So, the number of different colorings is
1
22(211 + 10 ·2 + 11 ·26) = 126.
29/Symmetry and Counting 215
10. 57
11. The symmetry group is Z6. The identity fixes all n6possible
colorings. Since the rotations of 60◦and 300◦fix only the cases
where each section is the same color, they each fix ncolorings.
Rotations of 120◦and 240◦each fix n2colorings since every other
section must have the same color. The 180◦rotation fixes n3
colorings since once we choose colors for three adjacent sections the
colors for the remaining three sections are determined. So, the
number is 1
6(n6+ 2 ·n+ 2 ·n2+n3).
12. 51
13. The first part is Exercise 11 in Chapter 6. For the second part,
observe that in D4we have φR0=φR180 .
14. γg1g2(x)=(g1g2)xH
15. R0, R180, H, V act as the identity and R90, R270, D, D0interchange
L1and L2. Then the mapping g→γgfrom D4to sym(S) is a
group homomorphism with kernel {R0, R180, H, V }.
216
CHAPTER 30
Cayley Digraphs of Groups
1. 4 ∗(b, a)
2. 3 ∗((a, 0),(b, 0)),(a, 0),(e, 1), 3 ∗(a, 0),(b, 0), 3 ∗(a, 0),(e, 1)
3. (m/2) ∗{3∗[(a, 0),(b, 0)],(a, 0),(e, 1),3∗(a, 0),(b, 0),3∗(a, 0),(e, 1)}
5. a3b
6. Say we proceed from xto yvia the generators a1, a2, . . . , amand
via the generators b1, b2, . . . , bn. Then
y=xa1a2···am=xb1b2···bnso that a1a2···am=b1b2···bn.
7. Both yield paths from eto a3b.
8. Cay{{(1,0),(0,1)}:Z4⊕Z2}.
10.
6-
?
6-
?
pppppppppppppppppppp
ppppppppppppppppppppj
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
Y
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
I
pppppppppp
pppppppppp
R
0
1
2
3
4
5
6
7
11. Say we start at x. Then we know the vertices
x, xs1, xs1s2, . . . , xs1s2···sn−1are distinct and x=xs1s2···sn. So
if we apply the same sequence beginning at y, then cancellation
shows that y, ys1, ys1s2, . . . , ys1s2···sn−1are distinct and
y=ys1s2···sn.
30/Cayley Digraphs of Groups 217
12. Trace the sequence b, b, b, a, b, b, b. The digraph could be called
undirected because whenever xis connected to y,yis connected to
x. Such a digraph (that is, one in which all arrows go both
directions) is called a graph.
13. If there were a Hamiltonian path from (0,0) to (2,0), there would
be a Hamiltonian circuit in the digraph, since (2,0) + (1,0) = (0,0).
14. Cay({2,3}:Z6) does not have a Hamiltonian circuit.
15. a. If s1, s2, . . . , sn−1traces a Hamiltonian path and
sisi+1 ···sj=e, then the vertex s1s2···si−1appears twice.
Conversely, if sisi+1 ···sj6=e, then the sequence
e, s1, s1s2, . . . , s1s2···sn−1yields the nvertices (otherwise,
cancellation gives a contradiction).
b. This is immediate from part a.
16. The digraph is the same as those shown in Example 3 except all
arrows go in both directions.
17. The sequence traces the digraph in a clockwise fashion.
18. A circuit is 4 ∗((3 ∗a), b).
19. Abbreviate (a, 0), (b, 0), and (e, 1) by a,b, and 1, respectively. A
circuit is 4 ∗(4 ∗1, a), 3 ∗a,b, 7 ∗a, 1, b, 3 ∗a,b, 6 ∗a, 1, a,b, 3 ∗a,
b, 5 ∗a, 1, a,a,b, 3 ∗a,b, 4 ∗a, 1, 3 ∗a,b, 3 ∗a,b, 3 ∗a,b.
20. Notice that the digraph has four triangles. Start somewhere and
call that triangle 1. Now once you enter any of the other three
triangles, you must cover all three points before leaving it. The
digraph does have a Hamiltonian path, starting at vertex (124) and
ending at vertex (1).
21. Abbreviate (R90,0), (H, 0), and (R0,1) by R, H, and 1, respectively.
A circuit is 3 ∗(R, 1,1), H, 2∗(1, R, R), R, 1, R, R, 1, H, 1,1.
22. Abbreviate (a, 0), (b, 0) and (e, 1) by a,band 1 respectively. A
circuit is m
2∗(3 ∗(a, b), a, 1,3∗a, b, 3∗a, 1).
23. Abbreviate (a, 0),(b, 0), and (e, 1) by a, b, and 1, respectively. A
circuit is 2 ∗(1,1, a), a, b, 3∗a, 1, b, b, a, b, b, 1,3∗a, b, a, a.
30/Cayley Digraphs of Groups 218
24. Abbreviate (a, 0), (b, 0) and (e, 1) by a,band 1 respectively. A
circuit is (m−1) ∗1, a, 2∗1,((m−3)/2) ∗[2 ∗a, b, 3∗a, 1, b, b, a, 3∗
b, 1],2∗a, b, 3∗a, 1, b, b, a, 2∗b, 1,3∗a, b, a, a.
25. Abbreviate (r, 0),(f, 0), and (e, 1) by r, f, and 1, respectively. Then
the sequence is r,r,f,r,r, 1, f,r,r,f,r, 1, r,f,r,r,f, 1, r,r,
f,r,r, 1, f,r,r,f,r, 1, r,f,r,r,f, 1.
26. Abbreviate (r, 0), (f, 0) and (e, 1) by r,fand 1 respectively. A
circuit in Dn⊕Zn+1 is
(n−1) ∗(n∗1, r), n ∗1, f, n ∗((n−1) ∗r, 1),(n−1) ∗r, f.
27. m∗((n−1) ∗(0,1),(1,1))
28. Adapt the argument given in the proof of Theorem 30.1.
29. Abbreviate (r, 0), (f, 0), and (e, 1) by r, f, and 1, respectively. A
circuit is 1, r, 1,1, f, r, 1, r, 1, r, f, 1.
30. Abbreviate (a, 0), (b, 0) and (e, 1) by a, b and 1 respectively. Then a
circuit is
1, a, 1,1, b, a, 1, a, 1, a, b, 1,1,
2∗(a, 1,1, b, a, 1, a, 1, a, b, 1,1), a, 1,1, b, a, 1, a, 1, a, b, 1.
31. 5 ∗[3 ∗(1,0),(0,1)],(0,1)]
32. 12 ∗((1,0),(0,1)).
33. 12 ∗((1,0),(0,1))
34. Let (s1, s2, . . . , sn) = |N| ∗ (a1, a2, . . . , ar) and use Exercise 15(b).
Suppose s1s2···si=s1s2···sj. Letting pand qbe the quotient
and remainder upon division of iby r, and uand vthe quotient
and remainder upon division of jby r, we have
(a1a2···ar)pa1a2···aq=s1s2···si=s1s2···sj
= (a1a2···ar)ua1a2···av.
But, then,
(a1···ar)pa1a2···aqN= (a1a2···ar)ua1a2···avN
so that
a1Na2N···aqN=a1Na2N···avN.
30/Cayley Digraphs of Groups 219
Since (a1N,...,arN) is a circuit for G/N, we must have q=v.
Thus, (a1a2···ar)p= (a1a2···ar)u. And because a1a2···ar
generates Nand pand uare less than |N|, this means p=u.
35. In the proof of Theorem 30.3, we used the hypothesis that Gis
Abelian in two places: We needed Hto satisfy the induction
hypothesis, and we needed to form the factor group G/H. Now, if
we assume only that Gis Hamiltonian, then Halso is Hamiltonian
and G/H exists.
220
CHAPTER 31
Introduction to Algebraic Coding Theory
1. wt(000000) = 0; wt(0001011) = 3; wt(0010111) = 4;
wt(0100101) = 3; wt(1000110) = 3; wt(1100011) = 4;
wt(1010001) = 3; wt(1001101) = 4; etc.
2. 2, 3, 3
3. 1000110; 1110100
4. a. Both d(u, v) and d(v, u) equal the number of positions in
which uand vdiffer.
b. Use Theorem 31.1.
c. Use Theorem 31.1.
5. 000000, 100011, 010101, 001110, 110110, 101101, 011011, 111000
6. Argue that wt(v)+ wt(u+v)≥wt(u).
7. Not all single errors can be detected.
8. C0can detect any 3 errors whereas Ccan only detect any 2 errors.
9. Observe that a vector has even weight if and only if it can be
written as a sum of an even number of vectors of weight 1. So, if u
can be written as the sum of 2mvectors of even weight and vcan
be written as the sum of 2nvectors of even weight, then u+vcan
be written as the sum of 2m+ 2nvectors of even weight and
therefore the set of code words of even weight is closed. (We need
not check that the inverse of a code word is a code word since every
binary code word is its own inverse.)
10. Since the minimum weight of any nonzero member of Cis 4, we see
by Theorem 31.2 that Cwill correct any single error and detect any
triple error. (To verify this, use t= 3/2 in the last paragraph of the
proof for Theorem 31.2.)
31/Introduction to Algebraic Coding Theory 221
11. No, by Theorem 31.3.
12. H=
2 2
1 2
1 0
0 1
The code is {0000,1011,2022,0121,0212,1102,2201,2110,1220}. It
will correct any single error and detect any double error. 2201.
13. 0000000, 1000111, 0100101, 0010110, 0001011, 1100010, 1010001,
1001100, 0110011, 0101110, 0011101, 1110100, 1101001, 1011010,
0111000, 1111111.
H=
111
101
110
011
100
010
001
Yes, the code will detect any single error because it has weight 3.
14. Observe that the subset of code words that end with 0 is a
subgroup H. If His a proper subgroup, note that it has index 2.
The same is true for every component.
15. Suppose uis decoded as vand xis the coset leader of the row
containing u. Coset decoding means vis at the head of the column
containing u. So, x+v=uand x=u−v. Now suppose u−vis a
coset leader and uis decoded as y. Then yis at the head of the
column containing u. Since vis a code word, u=u−v+vis in the
row containing u−v. Thus u−v+y=uand y=v.
16. For 11101 we get 11100 or 11001. For 01100 we get 11100. No,
because the code word could have been 11100 or 11001. Yes, only
the code word 11100 differs in one position from the received word.
17. 000000, 100110, 010011, 001101, 110101, 101011, 011110, 111000
H=
110
011
101
100
010
001
31/Introduction to Algebraic Coding Theory 222
001001 is decoded as 001101 by all four methods.
011000 is decoded as 111000 by all four methods.
000110 is decoded as 100110 by all four methods.
Since there are no code words whose distance from 100001 is 1 and
three whose distance is 2, the nearest-neighbor method will not
decode or will arbitrarily choose a code word; parity-check matrix
decoding does not decode 100001; the standard-array and syndrome
methods decode 100001 as 000000, 110101, or 101011, depending on
which of 100001, 010100, or 001010 is a coset leader.
18. Here 2t+s+ 1 = 6. For t= 0 and s= 5, we can detect any 5 or
fewer errors; for t= 1 and s= 3, we can correct any one error and
detect any 2, 3 or 4 errors; for t= 2 and s= 1, we can correct any
1 or 2 errors and detect any 3 errors.
19. For any received word w, there are only eight possibilities for wH.
But each of these eight possibilities satisfies condition 2 or the first
portion of condition 30of the decoding procedure, so decoding
assumes that no error was made or one error was made.
20. The last row is obtained by adding 10000 to each code word. So the
code words can be obtained by subtracting 10000 from each
member of the last row. (Since the code is binary, this is the same
as adding 10000 to each member of the last row.)
21. There are 34code words and 36possible received words.
22. Yes, because the rows are nonzero and distinct.
23. No; row 3 is twice row 1.
24. Suppose that we can use the nearest-neighbor method to correct
any tor fewer errors and the weight of the code is k < 2t+ 1. Let u
be a code word of weight k. Let u0be the vector obtained from u
by changing dk/2e ≤ tcomponents of uto 0. If kis even, we have
d(u, u0) = k
2=d(0, u0) so that the nearest neighbor of u0is not
unique. If kis odd, then d(0, u0)< d(u, u0) and u0is not decoded as
u.
Now suppose that the nearest-neighbor method will detect any 2tor
fewer errors and that the weight of the code is at most 2t. Let ube
a code word whose weight is the weight of the code. Then the error
made by changing all the components of uto 0 is not detected.
31/Introduction to Algebraic Coding Theory 223
25. No. For if so, nonzero code words would be all words with weight
at least 5. But this set is not closed under addition.
26. Say G="1 0 a1a2a3
0 1 b1b2b3#. To detect 3 errors the minimum
weight of nonzero code words must be 4. Thus any nonzero code
word has at most one zero component. Since (10)G= 10a1a2a3and
(01)G= 01b1b2b3we have ai6= 0 and bi6= 0 for i= 1,2,3. Because
(21)G= 2,1,2a1+b1,2a2+b2,2a3+b3we must have ai6=bi. Thus
the last three columns for Gare 1
2or 2
1. But then
(11)G= 11000, a contradiction.
27. By Exercise 24, for a linear code to correct every error the minimum
weight must be at least 3. Since a (4,2) binary linear code only has
three nonzero code words, if each must have weight at least 3 then
the only possibilities are (1,1,1,0), (1,1,0,1), (1,0,1,1),(0,1,1,1) and
(1,1,1,1). But each pair of these has at least two components that
agree. So, the sum of any distinct two of them is a nonzero word of
weight at most 2. This contradicts the closure property.
28. 000010 110110 011000 111011 101100 001111 100001 010101.
29. Abbreviate the coset a+hx2+x+ 1iwith a. The following
generating matrix will produce the desired code:
1 0 1 1 x
0 1 x x + 1 x+ 1 .
30. G="1 0 2 1
0 1 1 2 #;
{0000,1021,2012,0112,1100,2121,0221,1212,2200};
H=
1 2
2 1
1 0
0 1
. The code will not detect all single errors.
31. By Exercise 14 and the assumption, for each component exactly
n/2 of the code words have the entry 1. So, determining the sum of
the weights of all code words by summing over the contributions
made by each component we obtain n(n/2). Thus, the average
weight of a code word is n/2.
31/Introduction to Algebraic Coding Theory 224
32. Suppose every vector of weight t+ 1 is a coset leader. Let vbe a
code word of weight 2t+ 1 and wthe vector obtained from vby
changing the first t+ 1 nonzero component to 0. Then
wt(w−v) = t+ 1 so that w−vis a coset leader. But
w+C=w−v+Cand whas weight t. This contradicts the
definition of coset leader.
33. Let c, c0∈C. Then, c+ (v+c0) = v+c+c0∈v+Cand
(v+c)+(v+c0) = c+c0∈C, so the set C∪(v+C) is closed under
addition.
34. Let vbe any vector. If uis a vector of weight 1, then wt(v) and
wt(v+u) have opposite parity. Since any vector uof odd weight is
the sum of an odd number of vectors of weight 1, it follows that
wt(v) and wt(v+u) have opposite parity. Now, mimic the proof of
Exercise 23.
35. If the ith component of both uand vis 0, then so is the ith
component of u−vand au, where ais a scalar.
225
CHAPTER 32
Introduction to Galois Theory
1. Note that φ(1) = 1. Thus φ(n) = n. Also, for
n6= 0,1 = φ(1) = φ(nn−1) = φ(n)φ(n−1) = nφ(n−1), so that
1/n =φ(n−1). So, by properties of automorphisms,
φ(m/n) = φ(mn−1) = φ(m)φ(n−1) = φ(m)φ(n)−1=mn−1=m/n.
2. Z2
3. If αand βare automorphisms that fix F, then αβ is an
automorphism and, for any xin F, we have
(αβ)(x) = α(β(x)) = α(x) = x. Also, α(x) = ximplies, by
definition of an inverse function, that α−1(x) = x. So, by the
Two-Step Subgroup Test, the set is a group.
4. Instead observe that Z2⊕Z2⊕Z2has 7 subgroups of order 2.
5. Suppose that aand bare fixed by every element of H. By
Exercise 29 in Chapter 13, it suffices to show that a−band ab−1
are fixed by every element of H. By properties of automorphisms
we have for any element φof H,
φ(a−b) = φ(a) + φ(−b) = φ(a)−φ(b) = a−b. Also,
φ(ab−1) = φ(a)φ(b−1) = φ(a)φ(b)−1=ab−1.
6. By Exercise 11 of Chapter 17, the splitting field is Q(√2, i).Since
[Q(√2, i) : Q]=4,|Gal(E/Q)|= 4. It follows that
Gal(E/Q) = {, α, β, αβ}where α(√2) = −√2 and
α(i) = i, β(√2) = √2, and β(i) = −iand the proper subfields of E
are Q, Q(√2), Q(√−2), and Q(i).
βhas fixed field Q(√2), α has fixed field Q(i), and αβ has fixed
field Q(√−2).
No automorphism of Ehas fixed field Q.
7. It suffices to show that each member of Gal(K/F ) defines a
permutation on the ai’s. Let α∈Gal(K/F ) and write
f(x) = cnxn+cn−1xn−1+··· +c0. Then
32/Introduction to Galois Theory 226
0 = f(ai) = cnan
i+cn−1an−1
i+··· +c0. So,
0 = α(0) = α(cn)(α(ai)n+α(cn−1)α(ai)n−1+··· +α(c0) =
cn(α(ai)n+cn−1α(ai)n−1+··· +c0=f(α(ai)). So, α(ai) = ajfor
some j, and therefore αpermutes the ai’s.
8. Use Corollary 3 of Theorem 16.2 and Exercise 7 of this chapter.
9. φ6(ω) = ω729 =ω.
φ3(ω+ω−1) = ω27 +ω−27 =ω−1+ω.
φ2(ω3+ω5+ω6) = ω27 +ω45 +ω54 =ω6+ω3+ω5.
10. |Gal(E/Q)|= [E:Q] = 4; |Gal(Q(√10)/Q|= [Q(√10) : Q] = 2.
11. a. Z20 ⊕Z2has three subgroups of order 10. b. 25 does not divide
40 so there is none. c. Z20 ⊕Z2has one subgroup of order 5.
12. See Example 4 in this chapter.
13. The splitting field over Ris R(√−3). The Galois group is the
identity and the mapping a+b√−3→a−b√−3.
14. Observe that D6has exactly three subgroups of order 6.
15. Use Theorem 22.3.
16. Use the Corollary to Theorem 24.2 and Theorem 11.1.
17. If there were a subfield Kof Esuch that [K:F] = 2 then, by the
Fundamental Theorem of Galois Theory (Theorem 32.1), A4would
have a subgroup of index 2. But, by Example 14 in Chapter 9, A4
has no such subgroup.
18. Let ω= (−1 + i√3)/2. The splitting field of x3−1 over Qis Q(ω).
Since [Q(ω) : Q] = 2, the Galois group of x3−1 over Qis Z2. The
splitting field of x3−2 over Qis Q(3
√2, ω). Since
[Q(3
√2, ω) : Q] = 6, the Galois group has order 6 and is generated
by αand βwhere α(3
√2) = ω3
√2, α(ω) = ωand β(3
√2) = 3
√2,
β(ω) = ω2. Since αβ 6=βα, the group must be S3(see Theorem 7.2
and the remark at the end of the proof).
19. This follows directly from the Fundamental Theorem of Galois
Theory (Theorem 32.1) and Sylow’s First Theorem (Theorem 24.3).
20. Use the subgroup lattice for D5.
32/Introduction to Galois Theory 227
21. Let ωbe a primitive cube root of 1. Then Q⊂Q(3
√2) ⊂Q(ω, 3
√2)
and Q(3
√2) is not the splitting field of a polynomial in Q[x].
22. Use the Fundamental Theorem and the fact that Gal(E/F ) is finite.
23. By the Fundamental Theorem of Finite Abelian Groups
(Theorem 11.1), the only Abelian group of order 10 is Z10. By the
Fundamental Theorem of Cyclic Groups (Theorem 4.3), the only
proper, nontrivial subgroups of Z10 are one of index 2 and one of
index 5. So, the lattice of subgroups of Z10 is a diamond with Z10
at the top, {0}at the bottom, and the subgroups of indexes 2 and
5 in the middle layer. Then, by the Fundamental Theorem of
Galois Theory, the lattice of subfields between Eand Fis a
diamond with subfields of indexes 2 and 5 in the middle layer.
24. Q(ω+ω4)
25. By Example 7, the group is Z6.
26. Z3
27. This follows directly from Exercise 21 in Chapter 25.
28. Let Kbe the subgroup of rotations in Dn. The desired series is
{R0} ⊂ K⊂Dn.
29. This follows directly from Exercise 41 in Chapter 24.
30. Note that A4has a normal Sylow 2-subgroup.
31. This follows directly from Exercise 42 in Chapter 10.
32. Let {e}=H0⊂H1⊂ ··· ⊂ Hn=Gbe the series that shows that
Gis solvable. Then H0∩H⊂H1∩H⊂ ··· ⊂ Hn∩H=Hshows
that His solvable.
33. Since K/N /G/N, for any x∈Gand k∈K, there is a k0∈K
such that k0N= (xN)(kN)(xN )−1=xNkNx−1N=xkx−1N. So,
xkx−1=k0nfor some n∈N. And since N⊆K, we have k0n∈K.
34. Use parts 7, 6 and 1 of Theorem 15.1.
228
CHAPTER 33
Cyclotomic Extensions
1. Since ω= cos π
3+isin π
3= cos 2π
6+isin 2π
6,ωis a zero of x6−1 =
Φ1(x)Φ2(x)Φ3(x)Φ6(x) = (x−1)(x+ 1))(x2+x+ 1)(x2−x+ 1), it
follows that the minimal polynomial for ωover Qis x2−x+ 1.
2. Use Theorem 33.1
3. Over Z,x8−1=(x−1)(x+ 1)(x2+ 1)(x4+ 1). Over Z2,
x2+ 1 = (x+ 1)2and x4+ 1 = (x+ 1)4. So, over Z2,
x8−1=(x+ 1)8. Over Z3,x2+ 1 is irreducible, but x4+ 1 factors
into irreducibles as (x2+x+ 2)(x2−x−1). So,
x8−1=(x−1)(x+ 1)(x2+ 1)(x2+x+ 2)(x2−x−1). Over Z5,
x2+ 1 = (x−2)(x+ 2), x4+ 1 = (x2+ 2)(x2−2), and these last
two factors are irreducible. So,
x8−1=(x−1)(x+ 1)(x−2)(x+ 2)(x2+ 2)(x2−2).
4. Use xn−1=(x−1)(xn−1+xn−2+··· +x+ 1) and that fact that
nth roots of unity form a cyclic group of order n.
5. Let ωbe a primitive nth root of unity. We must prove
ωω2···ωn= (−1)n+1. Observe that ωω2···ωn=ωn(n+1)/2. When
nis odd, ωn(n+1)/2= (ωn)(n+1)/2= 1(n+1)/2= 1. When nis even,
(ωn/2)n+1 = (−1)n+1 =−1.
6. Φ3(x)
7. If [F:Q] = nand Fhas infinitely many roots of unity, then there
is no finite bound on their multiplicative orders. Let ωbe a
primitive mth root of unity in Fsuch that φ(m)> n. Then
[Q(ω) : Q] = φ(m). But F⊇Q(ω)⊇Qimplies [Q(ω) : Q]≤n.
8. Observe that xn−1=(x−1)(xn−1+xn−2+··· +x+ 1) and use
Theorem 33.1.
9. Let 2n+ 1 = q. Then 2 ∈U(q) and 2n=q−1 = −1 in U(q) implies
that |2|= 2n. So, by Lagrange’s Theorem, 2ndivides
|U(q)|=q−1=2n.
33/Cyclotomic Extensions 229
10. We know Φ2(0) = 1. Now observe that
xn−1 = (x−1) Y
d|n
1<d<n
Φd(x)Φn(x)
and use induction.
11. Let ωbe a primitive nth root of unity. Then 2nth roots of unity
are ±1,±ω, . . . , ±ωn−1. These are distinct, since −1=(−ωi)n,
whereas 1 = (ωi)n.
12. Let αbe a primitive mnth root of unity. Then αnis a primitive
mth root of unity and αmis a primitive nth root of unity. This
shows that (xm−1)(xn−1) splits in the splitting field of xmn −1.
Conversely, let βbe a primitive mth root of unity and γbe a
primitive nth root of unity. It suffices to show that |βγ|=mn. Let
H=hβγi. Since (βγ)mn = (βm)n(γn)m= 1 ·1 we know |H| ≤ mn.
Since (βγ)m=βmγn=γnand, by Theorem 4.2, |γn|=|γ|, we
know that ndivides |H|. By symmetry, mdivides |H|. Thus
|H| ≥ mn. This proves that the splitting field of (xm−1)(xn−1)
contains a primitive mnth root of unity.
13. First observe that deg Φ2n(x) = φ(2n) = φ(n) and
deg Φn(−x) = deg Φn(x) = φ(n). Thus, it suffices to show that
every zero of Φn(−x) is a zero of Φ2n(x). But ωis a zero of Φn(−x)
means that | − ω|=n, which in turn implies that |ω|= 2n. (Here
|ω|means the order of the group element ω.)
14. Since the two sides are monic and have the same degree it suffices
to prove that every zero of Φpk(x) is a zero of Φp(xpk−1). Let ωbe a
zero of Φpk(x) and note that |ω|=pkimplies that |ωpk−1|=p.
Φ8(x) = x4+ 1,Φ27(x)=(x9)2+x9+ 1 = x18 +x9+ 1.
15. Let G= Gal(Q(ω)/Q) and H1be the subgroup of Gof order 2 that
fixes cos(2π
n). Then, by induction, G/H1has a series of subgroups
H1/H1⊂H2/H1⊂ ··· ⊂ Ht/H1=G/H1, so that
|Hi+1/H1:Hi/H1|= 2. Now observe that
|Hi+1/H1:Hi/H1|=|Hi+1/Hi|.
16. Use Theorem 33.4.
33/Cyclotomic Extensions 230
17. Instead, we prove that Φn(x)Φpn(x)=Φn(xp). Since both sides are
monic and have degree pφ(n), it suffices to show that every zero of
Φn(x)Φpn(x) is a zero of Φn(xp). If ωis a zero of Φn(x), then
|ω|=n. By Theorem 4.2, |ωp|=nalso. Thus ωis a zero of Φn(xp).
If ωis a zero of Φnp(x), then |ω|=np and therefore |ωp|=n.
18. Use Theorem 33.4.
19. Let ωbe a primitive 5th root of unity. Then the splitting field for
x5−1 over Qis Q(ω). By Theorem 33.4, Gal(Q(ω)/Q)
≈U(5) ≈Z4. Since h2iis the unique subgroup strictly between {0}
and Z4, we know by Theorem 32.1 that there is a unique subfield
strictly between Qand E.
20. Use Theorem 33.4 and Theorem 32.1.
21. The three automorphisms that take ω→ω4, ω →ω−1, ω →ω−4
have order 2.
231
SUPPLEMENTARY EXERCISES FOR CHAPTERS 24-33
1. First observe that since
xy = (xy)3(xy)4= (xy)7= (xy)4(xy)3=yx,xand ycommute.
Also, since y= (xy)4= (xy)3xy =x(xy) = x2ywe know that
x2=e. Then y= (xy)4=x4y4=y4and therefore, y3=e. This
shows that |G| ≤ 6. But Z6satisfies the defining relations with
x= 3 and y= 2. So, G≈Z6.
2. In H,|xy|= 6.
3. Let |G|= 315 = 9 ·5·7 and let Hbe a Sylow 3-subgroup and Ka
Sylow 5-subgroup. If H/G, then HK = 45. If His not normal,
then by Sylow’s Third Theorem the number of Sylow 3-subgroups
is 7 and therefore |G/N(H)|= 7 and |N(H)|= 45.
4. Use Sylow’s Third Theorem and its corollary.
5. Observe that K⊆N(H) implies that HK is a group of order
245 = 45 ·5 (see Exercise 57 in Chapter 9 and Theorem 7.2). By
Sylow’s Third Theorem (Theorem 24.5) and its corollary, Kis
normal in HK. Thus, H⊆N(K).
6. Say the group has order nand the three conjugacy classes are cl(e),
cl(a), and cl(b). The class equation then gives
|G| − 1 = |G|/|C(a)|+|G|/|C(b)|. Obviously, it can not be the case
that both |G|/|C(a)|= 1 and |G|/|C(b)|= 1. Nor can exactly one
of |G|/|C(a)|and |G|/|C(b)|be 1 since then |G| − 2≤ |G|/2 and
therefore |G| ≤ 4. It is also not possible that |G|/|C(a)|=|G|/2
and |G|/|C(b)|=|G|/2 either. In the remaining cases we have
|G|/|C(a)|+|G|/|C(b)|≤|G|/2 + |G|/3 = (5/6)|G|<|G| − 1when
|G|>6.
7. By the corollary to Sylow’s Third Theorem Kis the only Sylow
p-subgroup of H. But for any g∈G, we have
gKg−1⊆gHg−1=H. So, for all g∈G,gKg−1=K.
8. Mimic the analysis carried out for 3x5−15x+ 5 at the end of
Chapter 32.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 24-33 232
9. Use the same proof as Exercise 57 in Chapter 9.
10. Hint: If x∈N(H), then both Pand xP x−1are Sylow p-subgroups
of H. So, they are conjugate in H.
11. By Sylow’s Third Theorem n7= 8 and by the Embedding Theorem
(Chapter 25) we know that Gis isomorphic to a subgroup of A8.
But A8does not have an element of order 21.
12. Mimic Example 8 of Chapter 24 (Theorem 24.7 is relevant).
13. Let Gbe a non-Abelian group of order 105. By Theorem 24.6 we
know that Z15 and Z35 are the only groups of orders 15 are 35 and
by Theorem 9.3, G/Z(G) is not cyclic. So |Z(G)| 6= 3, 7, 15, 21, or
35. This leaves only 1 or 5 for |Z(G)|. Let H,K, and Lbe Sylow
3-, Sylow 5-, and Sylow 7-subgroups of G, respectively. Now,
counting shows that K/Gor L/G. Thus, |KL|= 35 and KL is
a cyclic subgroup of G. But, KL has 24 elements of order 35 (since
|U(Z35)|= 24). Thus, a counting argument shows that K/Gand
L/G. Now, |HK|= 15 and HK is a cyclic subgroup of G. Thus,
HK ⊆C(K) and KL ⊆C(K). This means that 105 divides
|C(K)|. So K⊆Z(G).
14. The Sylow 2-subgroup has order 2; the odd order Sylow
p-subgroups are contained in the rotation subgroup.
15.
u
u
u
u
u
u
u
u
u
u
u
u
u
u
-
-
-
-
?
?
?
?
?
?
6
16. Letting Vdenote a vertical move and Ha horizontal move and
starting at (1,0) a circuit is V, V, H, 6∗(V, V, V, H).
17. It suffices to show that xtravels by aimplies xab−1travels by a
(for we may successively replace xby xab−1). If xab−1traveled by
b, then the vertex xa would appear twice in the circuit.
SUPPLEMENTARY EXERCISES FOR CHAPTERS 24-33 233
18. To show that C⊥is a subspace, recall that if v,u, and ware
vectors and ais a scalar, then
(v−w)·u=v·u−w·uand (av)·u=a(v·u).
To prove C⊥has dimension n−k, choose a basis v1, v2, . . . , vkfor
C. The matrix with v1, v2, . . . , vkas columns has rank k(because
the columns are independent) so the kernel of the linear
transformation this matrix represents has dimension n−k. This
kernel is C⊥.
19. a. {00,11}
b. {000,111}
c. {0000,1100,1010,1001,0101,0110,0011,1111}
d. {0000,1100,0011,1111}
20. Since C⊆C⊥, we know u·u= 0 (remember, arithmetic is done
modulo 2) for all u∈C. If u= (u1, . . . , un) then
u·u=u1+u2+··· +un,so there is an even number of
components equal to 1.
21. The mapping Tv:Fn→ {0,1}given by Tv(u) = u·vis an onto
homomorphism. So |Fn/Ker Tv|= 2.
22. If v∈Fn, then the dot product of vand 11 ···1 is 0 if and only if v
has an even number of components equal to 1.
23. It follows from Exercise 18 that if Cis an (n, k) linear code, then
C⊥is an (n, n −k) linear code. Thus, in this problem, k=n−k.
Since C=C⊥, to prove that (11 ···1) is a code word it is enough
to prove that(11 ···1) ·u= 0 for all u∈C. By Exercise 21, we
know that (11 ···1) ·u=n/2 = 0 mod 2.
24. Since Gis solvable there is a series
{e}=K0⊂K1⊂ ··· ⊂ Km=G
such that Ki+1/Kiis Abelian. Now there is a series
Ki
Ki
=L0
Ki⊂L1
Ki⊂ ··· ⊂ Lt
Ki
=Ki+1
Ki
,
SUPPLEMENTARY EXERCISES FOR CHAPTERS 24-33 234
where |(Lj+1/Ki)/(Lj/Ki)|is prime. Then
Ki=L0⊂L1⊂L2⊂ ··· ⊂ Lt=Ki+1
and each |Lj+1/Lj|is prime (see Exercise 42 of Chapter 10). We
may repeat this process for each i.
