Prove if $G$ is a finite nonabelian $p$-group, then $p^2\mid |{\rm Aut}(G)|$.

Question

Prove if $G$ is a finite nonabelian $p$-group, then $p^2\mid|{\rm Aut}(G)|$.

Suppose $|G|=p^m, m\in \mathbb{N}$.

A fact I know about $p$-groups:

Since $G$ is a $p$-group, $\forall i\leq m \space \exists H\leq G$ such that $|H|=p^i.$

In addition, $Z(G)\neq \{{e}\}.$

I have no idea how to approach this problem, I would like for some help. Thanks.

Answer 1

Hint: For each $g\in G$ define $\sigma_g(a)=gag^{-1}$, then $g\mapsto\sigma_g$ is a morphism of groups ($\varphi:G\to\textrm{Aut}(G)$).

Look at the kernel of $\varphi$, and then apply the first isomorphism theorem.

You will see that $G/\ker\varphi$ is isomorphic to the group of interior automorphisms of $G$.