Automorphism group is transitive on non-identity elements implies characteristically simple

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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Statement

Verbal statement

If a group has the property that its automorphism group acts transitively on non-identity elements, then the group is characteristically simple.

Definitions used

Characteristically simple group

Further information: characteristically simple group

A characteristically simple group is a group whose automorphism group has no proper nontrivial characteristic subgroup.

Proof

Hands-on proof

Given: A group $G$, such that $\mathrm{A}\mathrm{u}\mathrm{t}(G)$ acts transitively on the set of non-identity elements of $G$. A characteristic subgroup $H$ of $G$

To prove: Either $H$ is trivial, or $H=G$

Proof: If $H$ is trivial, we are done. Otherwise, there exists a non-identity element $x\in H$. We want to show that $H=G$.

Pick any element $y\in G$. We want to argue that $y\in H$. Clearly, if $y$ is the identity element $y\in H$; otherwise, by assumption, there exists an automorphism $\sigma \in \mathrm{A}\mathrm{u}\mathrm{t}(G)$ such that $\sigma (x)=y$. But then, since $H$ is characteristic, and $x\in H$, we'd have $y=\sigma (x)\in H$. This completes the proof.

Conceptual proof

The automorphism group acting transitively, implies that if we start with any element of the group, its characteristic closure is the whole group -- this is precisely the condition of being characteristically simple.