8/2/2019 alg-hw-ans-14

1/5

14.01 Find the order of the given factor group:Z

6/3Solution: 3 = {0, 3}. So the order ofZ6/3 is 3.

14.06 Find the order of the given factor group: (Z12 Z18)/(4, 3)

Solution: As a subgroup ofZ12 Z18,

(4, 3) = {(0, 0), (4, 3), (8, 6), (0, 9), (4, 12), (8, 15)}.

So the order of (Z12 Z18)/(4, 3) is (12 18)/6 = 36.

14.08 Find the order of the given factor group: (Z11 Z15)/(1, 1)

Solution:

It is easy to see that (1, 1) =Z11

Z15

. So the order of(Z11 Z15)/(1, 1) is 1.

14.10 Give the order of the element in the factor group: 26+12 in Z60/12.

Solution: 26 + 12 = 2 + 12. The cyclic subgroup of 2 + 12 is

{12, 2 + 12, 4 + 12, 6 + 12, 8 + 12, 10 + 12}

So the order of 26 + 12 in Z60/12 is 6.

14.12 Give the order of the element in the factor group: (3, 1) + (1, 1) in(Z4 Z4)/(1, 1).

Solution: The cyclic subgroup of (Z4 Z4)/(1, 1) generated by(3, 1) + (1, 1) consists of the following 2 elements:

(0, 0) + (1, 1),

(3, 1) + (1, 1) = (2, 0) + (1, 1)

So the order of (3, 1) + (1, 1) in (Z4 Z4)/(1, 1) is 2.

14.21

a. The elements in G/H are the cosets ofH in the abelian groupG,+, which are of the form a + H for a G. Let a and b betwo elements ofG/H is a wrong expression.

b. We must show that G/H is abelian. Let a + H and b + H betwo elements ofG/H.

27

8/2/2019 alg-hw-ans-14

2/5

c. We must show that G/H is abelian. Let a+H and b+H be twoelements ofG/H. Then

(a + H) + (b + H) = a + (H+ b) + H

= a + (b + H) + H

= (a + b) + H

= (b + a) + H+ H

= b + (a + H) + H

= b + (H+ a) + H

= (b + H) + (a + H).

So G/H is abelian.

14.24 Show that An is a normal subgroup ofSn and compute Sn/An; thatis, find a known subgroup to which Sn/An is isomorphic.

Solution: Suppose n 2. The group An consists of all even per-mutations in Sn. If g Sn, then g can be expressed as a productof transpositions in Sn, say g = 12 k. Then g

1 = kk1 1.Then

gAng1 = 12 kAnkk1 1

consists of all even permutations in Sn. This shows that gAng1 = An.

Hence An is a normal subgroup ofSn. Define : Sn Z2 by

() =

0, if is an even permutation;

1, if is an odd permtuation.

Then is an homorphism and ker = An. So

Sn/An = Sn/ ker (Sn) = Z2.

14.26 Prove that the torsion subgroup T of an abelian group G is a normalsubgroup ofG, and that G/T is torsion free.

Solution: Every subgroup of an abelian group is a normal subgroup.So T is a normal subgroup ofG.

Suppose on the contrary that G/T is not torsion free. Then thereexists a non-identity element a + T G/T, such that a + T has finiteorder in G/T. So n(a + T) = na + T = T for some positive integer

28

8/2/2019 alg-hw-ans-14

3/5

n. So na T. However, all elements in the torsion subgroup T are offinite orders. So there exists a positive integer m such that m(na) = 0.So (mn)a = 0. This shows that a T and hence a + T = T is theidentity in G/T, a contradiction to our assumption.

14.27 A subgroup H is conjugate to a subgroup K of a group G if thereexists an inner automorphism ig ofG such that ig[H] = K. Show thatthe conjugacy is an equivalence relation on the collection of subgroupsofG.

Solution: We use to define this relationship.

1. (Reflexive) For a subgroup H ofG we have i1[H] = 1 H 1 = H.

So H H.

2. (Symmetric) IfH1 and H2 are subgroups ofG such that H1 H2,then there exists g G such that ig[H1] = H2; that is, gH1g

1 =H2. So (g

1)H2(g1)1 = g1H2g = H1, that is ig1 [H2] = H1.

This shows that H2 H1.

3. (Transitive) IfH1, H2,H3 are subgroups ofG such that H1 H2and H2 H3, then there exists g, k G such that ig[H1] = H2and ik[H2] = H3. So

H3 = kH2k1 = k(gH1g

1)k1 = (kg)H1(kg)1.

This shows that H1 H3.

Therefore, the conjugacy is an equivalence relation on the collectionof subgroups ofG.

14.29 Referring to Exercises 27, find all subgroups ofS3 that are conjugateto {0, 2}.

Solution: {0, 2} =

1 2 3

1 2 3

,

1 2 3

3 2 1

. To find all subgroups con-

jugate to

1 2 3

1 2 3

,

1 2 3

3 2 1

, it suffices to choose one candidate g from

each left coset of

1 2 3

1 2 3,

1 2 3

3 2 1and do the conjugation by g (exer-

cise). Let us choose the conjugations by g1 = 1 2 31 2 3

, g2 = 1 2 32 3 1

, and

29

8/2/2019 alg-hw-ans-14

4/5

g3

=1 2 3

3 1 2

. Then

g1

1 2 3

1 2 3

,

1 2 3

3 2 1

g1

1=

1 2 3

1 2 3

,

1 2 3

3 2 1

,

g2

1 2 3

1 2 3

,

1 2 3

3 2 1

g1

2=

1 2 3

1 2 3

,

1 2 3

2 1 3

,

g3

1 2 3

1 2 3

,

1 2 3

3 2 1

g1

3=

1 2 3

1 2 3

,

1 2 3

1 3 2

.

These are all subgroups ofS3 that are conjugate to {0, 2}.

14.31 Show that an intersection of normal subgroups of a group G is again

a normal subgroup ofG.Solution: Let Hi be normal subgroups of G (for i I where I isan index set). Let H :=

iIHi. Then for g G by left and right

cancelation laws and the normality ofHi we have

gH=iI

(gHi) =iI

(Hig) =

iI

Hi

g = Hg.

This shows that H is a normal subgroup ofG.

14.32 Given any subset S of a group G. Show that it makes sense to speakof the smallest normal subgroup that contains S. [Hint: Use Exercise

31]

Solution: Let {Hi | i I} be the collection of all normal subgroupsofG that contain S. Then by Exercise 31, the subset

H :=iI

Hi

is a normal subgroup of G. Obviously H contains S. So H is thesmallest normal subgroup that contains S.

14.35 Show that ifH and N are subgroups of a group G, and N is normalin G, then HN is normal in H. Show by an example that HN

need not be normal in G.Solution: Suppose H is a subgroup, and N is a normal subgroup, ofG respectively. Given h H, we have h(H N)h1 H. On theother hand, h(HN)h1 hNh1 = N since N is normal. Therefore,

h(HN)h1 HN. (1)

30

8/2/2019 alg-hw-ans-14

5/5

Since h H is arbitrary, replacing h by h

1

in (1) we haveh1(HN)(h1)1 HN.

Multiplying h on the left and h1 on the right in the previous equation,we have

HN h(HN)h1. (2)

By (1) and (2) we see that h(H N)h1 = H N. So H N is anormal subgroup ofH.

To show that HN needs not be a normal subgroup ofG, we considerG = N := S3 and H :=

1 2 3

1 2 3, 1 2 3

1 3 2. Then N is the normalsubgroup ofG. But H is not a normal subgroup ofG, since for :=

1 2 3

2 1 3

we have

H =

1 2 3

2 1 3

,

1 2 3

2 3 1

= H =

1 2 3

2 1 3

,

1 2 3

3 1 2

.

Now HN= H is not a normal subgroup ofG.

14.40 Use the properties det(AB) = det(A) det(B) and det(In) = 1 forn n matrices to show the following:

a. The nn matrices with determinant 1 form a subgroup of GL(n,R).

b. The n n matrices with determinant 1 form a subgroup of

GL(n,R).

Solution:

a. Let H1 denote the collection of the n n matrices with determi-nant 1.

For two matrix A,B H1, we have det(AB) = det(A) det(B) =1 1 = 1. So AB H1 and thus H1 is closed under multiplication.

The identity In of GL(n,R) is in H1 since det(In) = 1.

For A H1, we have det(A) det(A1) = det(AA1) = det(In) =

1. So det(A1) = 1/det(A) = 1 and thus A1 H1. So H1 is

closed under the inverse.Therefore, H1 is a subgroup of GL(n,R).

b. Let H2 denote the collection of the n n matrices with determi-nant 1. (Similar to the argument in a., we can show that H2 isa subgroup of GL(n,R). Please complete the details by yourself.)

31

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-