The order of the ${\rm Aut}(G)$, where $G$ is a $p$-group is divisible by $p$.

Question

The order of the ${\rm Aut}(G)$, where $G$ is a $p$-group is divisible by $p$.

I saw that this claim is commonly used in many papers but was wondering what the proof for this comes from?

Answer 1

For a nonabelian $p$-group, note that the inner automorphism group is isomorphic to $G/Z(G)$ and hence has order divisible by $p$. Being a subgroup of $\mathrm{Aut}(G)$, this forces $\mathrm{Aut}(G)$ to have order divisible by $p$.

Answer 2

This is not true if $G=C_p$.

If $G$ is non-abelian then $p\mid |G/Z(G)|=|{\rm Inn}(G)|\mid |{\rm Aut}(G)|$.

If $G$ is abelian then $G\cong C_{p^{d_1}}\times\cdots\times C_{p^{d_k}}$ for some $k,d_i\ge 1$. Suppose $C_{p^{d_i}}$ is generated by $a_i$.

If $k=1$ then we require $d_1>1$. In this case $\phi(a_1)=a_1^{1+p^{d_1-1}}$ defines an automorphism of order $p$.

If $k>1$ then the automorphism $\phi:G\to G$ with $\phi(a_i)=a_i$ for $i<k$ and $\phi(a_k)=a_ka_1^{p^{d_1-1}}$ has order $p$.