7: Structure of Groups 7.1: Isomorphism Theorems; Automorphisms 1

Isomorphism Theorems; Automorphisms

1. In G=ℤ32× find cyclic subgroups H of order 2 and K of order 8 with HK = G and H ∩K={e }.

Conclude that ℤ32× ≃ℤ2×ℤ8.

First, ℤ32× ={1,3,5,7,9,11,13,15,17,19,21, 23,25,27,29,31}. We see that o (3)=8 and o (15)=2, so

⟨3⟩={1,3,9,11,17,19,25,27}≃ℤ8 and ⟨15⟩={1, 15}≃ℤ2 We see that ⟨3⟩∩⟨15⟩={1} and⟨3⟩ ⟨15⟩={1,3,5,7,9,11,13,15,17,19,21,25,27,29,31}. Thus, by Theorem 7.1.3, ℤ32

× ≃ℤ2×ℤ8.

2. Prove that D6≃S 3×ℤ2 .

Consider the sets {e ,a b} and {e ,a2 ,a4,b , a2b ,a4 b}. Both are subgroups: the first is cyclic of order 2 and the second has the same structure as S3 . We also see that the intersection of the subgroups is the identity and that their product yields all of D6 . Thus, by Theorem 7.1.3, D6≃S 3×ℤ2 .

3. Determine Aut (ℤ2×ℤ2).

First, ℤ2×ℤ2={(0,0) ,(0,1),(1,0) ,(1,1)}. Now, since ∣ℤ2×ℤ2∣=4, the only possible proper, nontrivial subgroups are of order 2. (This is by Lagrange's Theorem. Since it is such a fundamental result, I will no longer cite it when it is used.) There are clearly 3 subgroups of this type: {(0,0) ,(0,1)},{(0,0) ,(1,0)}, and {(0,0) ,(1,1)}. Since an isomorphism must preserve group structure, these

subgroups must be mapped among each other. In other words, Aut (ℤ2×ℤ2)≃S3 .

4. Let G be a finite abelian group of order n, and let m be a positive integer with (n, m) = 1. Show thatϕ :G →G defined by ϕ (g )=g m for all g∈G belongs to Aut(G).

(1-1) We know that ker ϕ={g∈G : g m=e }. Thus, if g∈ker ϕ , we know g m=e , so o (g )∣m . As always, o (g )∣n as well. But since (n, m)=1, we must have g = e. Thus, ker ϕ={e}.(Onto) Any finite map that is 1-1 is also onto.(Preservation) ϕ (ab)=(ab)m=am bm=ϕ (a )ϕ(b) .Thus, the map is an automorphism.

5. Let ϕ :G →G be the function defined by ϕ (g )=g−1 for all g∈G. Find conditions on G such thatϕ is a homomorphism.

The map is 1-1 by uniqueness of inverses. It is also onto since inverses must exist in a group. However, ϕ (ab)=(ab)−1=b−1 a−1 . The only way ϕ (ab)=a−1b−1 is if the elements commute. Thus, G must be abelian for the map to be an automorphism.

7: Structure of Groups 7.1: Isomorphism Theorems; Automorphisms 2

6. Show that for G=S 3 , Inn (G)≃G .

Examine Z (G). Since S3≃D3, which is defined by two elements a and b which do not commute, we know Z (G)={e}. Thus, by Theorem 7.1.8, Inn (S 3)≃S 3.

7. Determine Aut (S 3) .

We have already shown (Q2) that S3 has four subgroups: one isomorphic to ℤ3 and three isomorphic to ℤ2 . Now, the subgroup of order 3 must be mapped to itself, leaving the three isomorphic copies ofℤ2 to be permuted with each other. Thus, Inn (S 3)≃S 3 .

8. For groups G1 and G 2 , determine the center of G1×G2 .

Let (a , b)∈Z (G1×G2) . Then, (a , b)( g1 , g2)=(g1 , g 2)(a , b) for all (g1 , g2)∈G1×G2 . This implies that (ag1 , bg2)=( g1 a , g2 b). Thus, (a , b)∈Z (G1)×Z (G 2) . Conversely, if (a , b)∈Z (G1)×Z (G 2) , then(ag1,bg2)=( g1 a , g2 b) implying that (a ,b)( g1, g2)=(g1, g 2)(a ,b). Thus, (a , b)∈Z (G1×G2) .

Therefor, Z (G1×G 2)=Z (G1)×Z (G2) .

9. Show that G /Z (G) cannot be a nontrivial cyclic group.

Suppose G /Z (G)={Z (G) ,a Z (G),a2 Z (G) ,...}. Let b ,c∈G. Without loss of generality, assumeb∈a j Z (G) and c∈ak Z (G) . In other words, b=a j z1 and c=ak z2 . Then,bc=a j z 1ak z 2=a j(ak z2)z 1=ak a j z2 z1=ak z2 a j z1=cb . Thus, c ,b∈Z (G) and Z (G)=G.

10. Describe the centers Z ( Dn) of the dihedral groups D n , for all integers n≥3.

If for some m, am=an−m , then am∈Z ( Dn). These elements and the identity can be the only elements that commute since the group is generated by ⟨a , b : am=b2 , am b=bam−n⟩ .

11. In the group GL2(ℂ) of all invertible 2×2 matrices with complex entries, let Q be the following

set of matrices: ±[1 00 1],±[ i 0

0 −i] ,±[ 0 1−1 0],±[0 i

i 0].(a) Show that Q is not isomorphic to D 4 .(b) Find the center Z (Q)of Q .

(a) In Q, the elements ±[1 00 1] (to name two) commute with everything, but in D 4 , only the

element a2 commutes with everything. Thus, Q≄D4.

(b) From p. 122, it is clear that only commutes with every member of Q.

7: Structure of Groups 7.1: Isomorphism Theorems; Automorphisms 3

12. Let F 20 be the subgroup of GL2(ℤ5) consisting of all matrices of the form [m n0 1] , such that

m ,n∈ℤ5 and m≠0, as defined in Exercise 23 of Section 3.8. This group will be called the Frobenius group of degree 5. Find the center of F 20 .

Suppose [m0 n0

0 1]∈Z (F 20) . Then [a b0 1][m0 n0

0 1]=[m0 n0

0 1][a b0 1]. Thus,

[a m0 a n0+b0 1]=[m0 a m0b+n0

0 1]. So, we need a n0+b=m0 b+n0 for all a and b in ℤ5 . In

particular, when a = b = 0, we must have n0=0. Thus, Z ( F20)=[m0 00 1] where the first entry is 1, 2,

3, or 4.

13. Show that the Frobenius group F 20 defined in Exercise 12 can be defined by generators and

relations as follows. Let a=[1 10 1] and b=[2 0

0 1].(a) Show that o (a)=5 , o(b)=4 , and b a=a2 b .(b) Show that each element of F 20 can be expressed in the form a i b j for 0≤i≤4 and 0≤ j≤3.

(a) ⟨a ⟩={[1 00 1],[1 1

0 1],[1 20 1],[1 3

0 1],[1 40 1]}. ⟨b⟩={[1 0

0 1] ,[2 00 1],[4 0

0 1],[3 00 1]}.

ba=[2 20 1]=a2 b.

(b) Let [m n0 1]∈F20 . Then, [m n

0 1]=[1 n0 1][m 0

0 1]=an bk .

14. Let G be the subgroup of GL2(ℝ ) consisting of all matrices [a11 a12

a21 a22] such that a21=0 and

a22=1.(a) Let N be the set of matrices in G with a11=1. Show that N is a normal subgroup of G.

(b) Let a=[1 10 1] and b=[2 0

0 1]. Show that if H =⟨a ⟩ , then b H b−1 is a proper subset of H.

Conclude that H is not normal in any group that contains b.

(a) Let A=[x y0 1] and B=[r s

0 1] . Suppose AB−1=[x / r y−sx /r0 1]∈N. This implies that x=r

since the first entry must be 1. Then, A−1 B=[1 1/ x (s− y)0 1]∈N . Conversely, if we evaluate

A−1 B=[r / x 1/ x (s− y)0 1] first, we again see r = x. Thus, N ⊴G .

7: Structure of Groups 7.1: Isomorphism Theorems; Automorphisms 4

(b) bam b−1=[1 2m0 1]=a2m∈H. Since we get only even multiples, we know b H b−1⊊ H . This

implies that b H≠H b so H ⋬G.

15. Give another proof of Theorem 7.1.1 by constructing an isomorphism from (HN )/ N ontoH /( H∩N ) .

Define f : HN →G /( H∩N ) via f (hn)=h( H∩N ). Clearly ker f =N and f (HN )=H /(H∩N ).Thus, HN ≃H /(H ∩N ) .