Proving that Aut(K), where K is cyclic, is abelian

[Mr Davis 97]

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?

 

[fresh_42]

But I am having a hard time proving that $\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times.$

If $\sigma \in \operatorname{Aut}(\mathbb{Z}_n)$ and $a$ generates $\mathbb{Z}_n$, then $\sigma(a)=a^m$ for some $m$. But $\sigma(a)$ has to generate the entire group, which is only possible if $(m,n)=1$ by Bezout.

 

[Mr Davis 97]

If $\sigma \in \operatorname{Aut}(\mathbb{Z}_n)$ and $a$ generates $\mathbb{Z}_n$, then $\sigma(a)=a^m$ for some $m$. But $\sigma(a)$ has to generate the entire group, which is only possible if $(m,n)=1$ by Bezout.

For part b), do you have any ideas? I was thinking that maybe since $C_G (K)$ is normal, we could show that $G/C_G(K)$ is abelian which would imply that $G' \subseteq C_G(K)$.

 

[fresh_42]

For part b), do you have any ideas? I was thinking that maybe since $C_G (K)$ is normal, we could show that $G/C_G(K)$ is abelian which would imply that $G' \subseteq C_G(K)$.

No, not yet. I was hoping you had one. Your plan looks nice, but we have a possibly pretty wild $G$ and $G/C_G(K)$ can still be quite large. But it being Abelian is equivalent to the stated assertion. As it is part of the same question, maybe we can establish an isomorphism $\operatorname{Aut}(K)\cong G/C_G(K)$ or at least an embedding $G/C_G(K) \hookrightarrow \operatorname{Aut}(K)$.

 

[Mr Davis 97]

No, not yet. I was hoping you had one. Your plan looks nice, but we have a possibly pretty wild $G$ and $G/C_G(K)$ can still be quite large. But it being Abelian is equivalent to the stated assertion. As it is part of the same question, maybe we can establish an isomorphism $\operatorname{Aut}(K)\cong G/C_G(K)$ or at least an embedding $G/C_G(K) \hookrightarrow \operatorname{Aut}(K)$.

So, we know that $G$ can act on $K$ by conjugation since $K$ is normal. Consider the permutation representation of this action, $\varphi : G \to S_K$. Note that $$\ker (\varphi) = \{g\in G \mid \varphi (g) = \operatorname{id}\} = \{g\in G \mid \varphi (g)(k) = k,\forall k\in K\} = \{g\in G \mid gkg^{-1} = k,\forall k\in K\} = C_G(K).$$ If I can establish that the image of $\varphi$ is $\operatorname{Inn}(K) \le \operatorname{Aut}(K)$, then by the first isomorphism theorem $G/C_G(K) \cong \operatorname{Inn}(K) \le \operatorname{Aut}(K)$. But $\operatorname{Aut}(K)$ is abelian, so $G/C_G(K) \cong \operatorname{Inn}(K)$ is also abelian, which means we're done I think.

I'm just having a bit of trouble seeing that the image of $\varphi$ is $\operatorname{Inn}(K)$ for some reason. I see that $\varphi (G) = \{\varphi (g) \mid g\in G\}$, but why is this set $\operatorname{Inn}(K)$? It looks like this set is $\operatorname{Inn}(G)$.

EDIT: Actually, maybe the image of $\varphi$ doesn't have to be $\operatorname{Inn}(K)$. Maybe it's just an unnamed subgroup of $\operatorname{Aut}(K)$, in which case I would still be done.

 

[fresh_42]

The inner automorphisms of any group, e.g. $K$, are conjugates by elements of $K$. What does this mean for an Abelian group? I don't see your isomorphism onto $\operatorname{Inn}(K)$ either, as $G/Z(G)\cong \operatorname{Inn}(G)$ resp. $K/Z(K) \cong \operatorname{Inn}(K)\cong 1$.

But we don't need this. $k \longmapsto gkg^{-1}$ are obviously automorphisms of $K$, so $\varphi(G) \subseteq \operatorname{Aut}(K)$. No inner automorphisms can be expected here, as $G$ is (possibly) so much larger than $K$. However, we have $\varphi\, : \,G \twoheadrightarrow \varphi(G) \leq \operatorname{Aut}(K)$ and kernel $C_G(K)$.