A question about automorphism groups of finite groups

I’ve encountered the following question whilst helping a colleague study for comprehensive exams, and I’m stuck on it:

Let $G$ be a finite group such that the natural action of the automorphism group $\mathrm{Aut}(G)$ on the set $G\setminus\{e\}$ of nonidentity elements of $G$ is transitive. Show that $G$ is an elementary Abelian $p$-group for some prime number $p$.

My Thoughts to Date: Transitivity of the action of $\mathrm{Aut}(G)$ on $G\setminus\{e\}$ means that, given any two nonidentity elements $g,h\in G$, there is an automorphism $\phi\in\mathrm{Aut}(G)$ such that $\phi(g)=h$. Since automorphisms preserve orders of elements, this implies that all nonidentity elements of $G$ have the same order, say $n$. By Lagrange’s theorem, $n\mid|G|$, and hence every divisor $d$ of $n$ also divides $|G|$. If $n$ had two distinct prime divisors $p$ and $q$, then $G$ would contain elements of order $p$ and $q$ by Cauchy’s theorem $\ (\Rightarrow\Leftarrow)$. Therefore $\exists!$ prime $p$ such that $p\mid n$. If $n=p^\ell$ for some $\ell>1$, then $G$ would contain elements of order $p^m$ for each $1\leq m\leq\ell\quad(\Rightarrow\Leftarrow)$. Therefore $n=p$, whence $G$ is a $p$-group. I could use some suggestions on the “elementary Abelian” part. All would be appreciated.

• In retrospect, that $G$ is a $p$-group follows from the fact that $p$ is the only prime dividing the order of $G$, and so I could have concluded that two lines earlier than I did. – anonymous Jul 27 '16 at 20:48
• The centre $Z(G)$ of a $p$-group is nontrivial and fixed by $\phi$ and so, since ${\rm Aut}(G)$ acts transitively on $G \setminus \{1\}$, we must have $Z(G)=G$ and hence $G$ is abelian. Since all elements have the same order $p$, it is elementary abelian. – Derek Holt Jul 27 '16 at 21:48
• @DerekHolt Why do you post an answer as a comment? – Arthur Jul 27 '16 at 21:50
• It was intended to be a sketch rather than a complete answer! – Derek Holt Jul 27 '16 at 21:51