Cyclic automorphism group implies abelian

This article gives a result about how information about the structure of the automorphism group of a group (abstractly, or in action) can control the structure of the group
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Statement

Suppose $G$ is a group that is an Aut-cyclic group (?): the Automorphism group (?) $\operatorname{Aut}(G)$ is a Cyclic group (?). Then, $G$ is an Abelian group (?).

Related facts

• Finite and cyclic automorphism group implies cyclic
• Classification of finite groups with cyclic automorphism group
• Cyclic automorphism group not implies cyclic
• Finite abelian and abelian automorphism group implies cyclic

Facts used

1. Group acts as automorphisms by conjugation
2. Cyclicity is subgroup-closed
3. Cyclic over central implies abelian

Proof

Given: A group $G$. The automorphism group $\operatorname{Aut}(G)$ is cyclic.

To prove: $G$ is abelian.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $Z$ be the center of $G$. Then, $G/Z$ is isomorphic to a subgroup of $\operatorname{Aut}(G)$, called the inner automorphism group $\operatorname{Inn}(G)$.	Fact (1)			[SHOW MORE] By Fact (1), $G$ acts as automorphisms on itself by conjugation, with the kernel being $Z$, so that $G/Z$ is isomorphic to a subgroup of $\operatorname{Aut}(G)$. This is the subgroup of inner automorphisms and is called $\operatorname{Inn}(G)$.
2	$G/Z$ is a cyclic group.	Fact (2)	$\operatorname{Aut}(G)$ is cyclic	Step (1)	Step-fact-given combination direct.
3	$G$ is abelian	Fact (3)		Steps (1), (2)	Fact-step combination direct.