- #1

Mr Davis 97

- 1,462

- 44

## Homework Statement

Let ##G## be a group and ##K## a finite cyclic normal subgroup of ##G##.

a) Prove that ##\operatorname{Aut}(K)## is an abelian group

b) Prove that ##G' \subseteq C_G (K)##, where ##G'## is the commutator subgroup of ##G##.

## Homework Equations

## The Attempt at a Solution

I'm focusing on the first part. My plan is this. I remember the fact that ##\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times##, where the latter is an abelian group. So if claim that ##|K| = n##, then clearly by the uniqueness of cyclic groups of each order, ##K \cong \mathbb{Z}/n\mathbb{Z}##, so if I can show that ##\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times##, then ##\operatorname{Aut} (K) \cong \operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times##, and hence ##\operatorname{Aut} (K)## is abelian.

But I am having a hard time proving that ##\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times##. Here is what I might do. Characterize the elements of ##\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})## as the maps ##\phi_a## that map ##1## to ##a##, where ##\gcd(a,n) = 1##. And then show that the map ##\Phi : (\mathbb{Z}/n\mathbb{Z})^\times \to \operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) ## where ##a \mapsto \phi_a## is an isomorphism. This seems like a roundabout way of showing that ##\operatorname{Aut} (K)## is cylic, so would there be a better way or should I go with this?