# Action of ${\rm Aut}(G)$ on $G$

Let $G$ be a finite group and consider the natural action of ${\rm Aut}(G)$ on $G$ and let there are two orbits under this action. How could we show that $G$ is an (elementary) abelian group? Is the converse true? (${\rm Aut}(G)$ denotes the automorphism group of $G$).

Thank you very much!

Let me add my motivation from this question. If we consider the action of ${\rm Inn}(G)$ on $G$ with two orbits, then we can show that $|G|=2$. Actually this question is a generalization of this property.

• What are your thoughts on the problem? What tools are you familiar with? Expanding on your attempts to solve the problem will help others give suitable answers. – Servaes Aug 21 '15 at 11:35
• Which two orbits should that be? – Arthur Aug 21 '15 at 11:39
• What is an "elementary" group? – Alex M. Aug 21 '15 at 12:03
• @Alex M. : "Elementary abelian group" is a standard group theory term, just google it. It means that the exponent of $G$ is a prime number, or if you prefer, that every non-trivial element of $G$ has the same order (which then has to be prime by Lagrange's theorem). – Patrick Da Silva Aug 22 '15 at 15:22
• @PatrickDaSilva: Thank you, I was not aware of this concept. – Alex M. Aug 22 '15 at 16:21

## 1 Answer

We can easily assume $G$ is not trivial. By Lagrange's theorem, if $p$ divides $|G|$, then there exists $g \in G \backslash \{e\}$ with $g^p = 1$ (denote the identity of $G$ by $e$). Since the action of $\mathrm{Aut}(G)$ is transitive on $G \backslash \{e\}$ and that automorphisms preserve order, every non-trivial element of $G$ has order $p$.

To show $G$ is abelian, note that since $G$ is a $p$-group, it has order $p^n$ for some $n$, hence its center is not trivial (this follows from the class equation : $$|G| = |Z(G)| + \sum_{i=1}^r |G:C_G(g_i)|$$ where $g_1,\cdots,g_r$ are representatives of the remaining non-trivial conjugacy classes of $G$). An automorphism must map the center to itself since if $g \in Z(G)$ and $h \in G$, then $$\varphi(g) \varphi(h) = \varphi(gh) = \varphi(hg) = \varphi(h) \varphi(g).$$ Since the action of $\mathrm{Aut}(G)$ on $G \backslash \{e\}$ is transitive, we conclude that $G$ is abelian.

Hope that helps,

• Dear Patricka Da Silva, thanks for your nice answer! – sebastian Aug 22 '15 at 5:32