Suppose $G$ is a finite $p$-group (where $p$ is prime), so that $|G|=p^n$ for some positive integer $n\ge 2$.
How can we prove that $|\text{Aut}(G)|$ is divisible by $p$?
Here $\text{Aut}(G)$ is the group of all automorphisms of $G$. I know how to prove this when $G$ is non-abelian. In this case, we look at the action of group on itself by conjugation, i.e. we consider the map $\phi: G\to \text{Aut}(G)$ defined by $\phi(g)=\tau_g$ where $\tau_g: G\to G$ given by $\tau_g(x)=gxg^{-1}$ for each $x\in G$. The kernel of $\phi$ is then $Z(G)$, the center of the group. By First Isomorphism Theorem, we get that $G/Z(G)$ is isomorphically embedded in $\text{Aut}(G)$. Since $Z(G)$ is proper subgroup of $G$ (because $G$ is non-abelian), we see that $p$ divides $\left|G/Z(G)\right|$, so by Lagrange's Theorem, $\text{Aut}(G)$ is divisible by $p$.
But what happens when $G$ is abelian? The above homomorphism $\phi$ is no longer of use, since $\phi$ becomes the trivial map [i.e. $G=Z(G)=\text{ker}(\phi)$].
Thanks for your time 🙂
Best Answer
Use FTFGAG and consider two cases. (1) The group is elementary abelian and (2) it is not.
In the first case, the group is $\mathbb{Z}_p^n$. Its automorphism group clearly has order $(p^n-1)(p^n-p)\ldots(p^n-p^{n-1})$, which is divisible by $p$ if $n \geq 2$.
In the second case, the group is a direct sum of cyclic subgroups, and at least one of these subgroups has order greater than $p$. So $G \simeq \mathbb{Z}_{p^k} \oplus H$, where $k \geq 2$. $\mathrm{Aut} (G)$ then has a subgroup isomorphic to $\mathrm{Aut} (\mathbb{Z}_{p^k})$, and the latter has order $(p-1)p^{k-1}$, which is divisible by $p$.