# 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##.

## 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?

Mentor
2021 Award
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)##.

Mentor
2021 Award
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.

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Mentor
2021 Award
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)##.

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Mr Davis 97