Prove that a group of order 34 with no more than 33 automorphisms is cyclic

[Dedrosnute]

Homework Statement

I need to show that for a group G of order 34 that if the order of the automorphism group is less than or equal to to 33, then G is cyclic.

Homework Equations

none

The Attempt at a Solution

I'm mainly trying to do a proof by contradiction. First I assumed that G had <= 33 automorphisms and then supposed that G was not cyclic. So then I tried to construct a group of that sort.

If G has order 34, but G is not cyclic, i.e. generated by an element of order 34, then G must be generated by an element x of order 2 and an element y of order 17. Is this correct? It seems like G would still be cyclic of order 34 and would be generated by the element xy.

Anyhow, I tried constructing the possible automorphisms. There would be an automorphism mapping y to each power of y between 1 and 17, so that is 17 automorphisms. Then I wasn't sure how to use the factor of 2. If I could use the factor of 2, I might be able to double the amount of automorphisms, but it seems like x could only be mapped to x, since x = x^-1 is the only element of order 2.

Could someone help me out please?

 

[Dick]

Yes, G is generated by an element x of order 2 and an element y of order 17. But that doesn't mean G is cyclic of order 34. That's only true if x and y commute. Can you show that if the inner automorphisms generated by two elements c and d are the same then cd^(-1) is in the center of G? Can you show if G has a nontrivial center, then it's abelian?

 

[Dedrosnute]

Ok, this seems to be coming together.

Let G be a group generated by x of order 2 and y of order 17.

Let Ic be the inner automorphism defined by Ic(a) = cac^-1, for some c in G, all a in G.
Let Id be the inner automorphism defined by Id(a) = dad^-1, for some d in G, all a in G.
Then cac^-1 = dad^-1 ==>
(d^-1 c)a = a(d^-1 c) ==>
(d^-1 c) is in Z(G).

Then let s be in Z(G) and let s not be the identity. s exists because the inner automorphism by x is equal to the inner automorphism by the identity, so xe^-1 = x is in Z(G).

So the center is nontrivial. How exactly does this imply that G is abelian? Is it because the inner automorphism of x by y^n is equal to x, for any integer n, and the inner automorphism of any power of y^m by y^n is equal to y^m, for any integers m,n?

 

[Dick]

I would say that the center is nontrivial because all 34 inner automorphisms can't be different. Therefore two of them must be the same. So pick s to be a nonidentity element of the center. What are the possibilities for the order of s? Think again about the structure of the group considering s as one of the generators.

 

[Dedrosnute]

Ok, I think I got it.

Let G be a group generated by x of order 2 and y of order 17.

Let Ic be the inner automorphism defined by Ic(a) = cac^-1, for some c in G, all a in G.
Let Id be the inner automorphism defined by Id(a) = dad^-1, for some d in G, all a in G.
Then cac^-1 = dad^-1 ==>
(d^-1 c)a = a(d^-1 c) ==>
(d^-1 c) is in Z(G).

The order of the automorphism group is less than or equal to 33, so there must be at least nontrivial automorphism which fixes all g in G. Suppose this is Is. Then Is(g) = sgs^-1 = g, for all g in G. Then s is in Z(G).
Since s is not the identity, |s| must be 2, 17, or 34. If |s| is 34, then G = <s> is cyclic.

If s in <x>, then it commutes with all elements of <y>, so since <x> and <y> are cyclic, G = <x, y> is cyclic.
Same argument if s is in <y>.


Are there any problems here?

Also, a small question: Are there automorphisms besides inner automorphisms?

 

[Dick]

It probably safer to say if |s|=2 then there is another element of order 17 and between them they generate the whole group rather than saying 'if s is <x>'. A priori there may be more than one element of order two. And yes, there can be noninner automorphisms. In the cyclic group {0,1,2} interchanging 1 and 2 is an automorphism but it's not inner, since the group is abelian.