# Automorphisms: chapter 2

1. Oct 4, 2007

### Benzoate

1. The problem statement, all variables and given/known data

Find Aut(Z(6)) Z(6) reads Z subscript 6 and Z represents integers.

2. Relevant equations

3. The attempt at a solution
Aut(Z(6))={L(1),L(5)}

L(5)= 5x

U(6) <=> Aut(Z(6))

Now I have to determine whether or not U(6) is cyclic. I start off by saying U(6)=<5> . <5> generates 5, and 5^2 =1 . Therefore <5>=Z(2). Aut(U(6) is similar to Aut(Z(2))= U subscript 2 = {1,5}

Aut(Aut(Z(10)) =Aut(U(10)) = Aut(Z subscript 4) = U(4) =Z subscript 2.

2. Oct 4, 2007

### morphism

Sorry - I can't understand most of that. What are L(1) and L(5)?

I think you're on the right track, but what you posted is too mangled up.

Let's think about what kind of automorphisms we can have on Z(6). An automorphism is going to be completely determined by how it acts on the generators of the group. So what are the generators of Z(6)? I think you managed to see that they were 1 and 5. So we have two automorphisms, namely the automorphism that sends 1->1 (the identity map) and the automorphism that sends 1->5. (Why am I not counting the ones that send 5->5 and 5->1 separately?)

So Aut(Z(6)) is a group that contains precisely two elements. And how many two-element groups do you know?