# Order of automorphism group of a $p$-group is divisible by $p$.

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. $G=Z(G)=\text{ker}(\phi)$].

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I'm confused, the automorphism group of $\mathbb Z_p$ is not divisible by $p$ for any prime $p$. –  JSchlather Aug 21 '13 at 4:30
If $n=1$ this is not true. –  Mark Bennet Aug 21 '13 at 4:30
@JSchlather: Sorry about that! We need $n\ge 2$. –  Prism Aug 21 '13 at 4:32
The automorphism group of $\mathbb Z_2 \times \mathbb Z_2$ has order $3$. –  JSchlather Aug 21 '13 at 4:32
@Prism You're right, I recalled that it was $S_3$ and my group theory is rusty enough that I forgot about the factorial. At any rate, you just need to exhibit an automorphism of order $p$. I would argue using the fundamental theorem of finitely generated abelian groups. First demonstrate that $\mathbb Z_p \times \mathbb Z_p$ has an automorphism of order $p$. Show this reduces to the case that the group is cyclic and then show that $\varphi(p^n)$ is divisible by $p$ for $n>2$. –  JSchlather Aug 21 '13 at 4:37

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 $p$.
Dan, thanks for your answer! Is it obvious that $\text{Aut}(K\oplus H)$ must have subgroup isomorphic to $\text{Aut}(K)$? I have the following idea: We can take all automorphisms of $K\oplus H$ that leaves $H$ fixed, and this should form a subgroup isomorphic to $\text{Aut}(K)$. Is my reasoning correct? –  Prism Aug 21 '13 at 4:54
@Prism Almost. If $\varphi \in \mathrm{Aut}(K)$, then there's an automorphism of $K \oplus H$ that sends $(k, h)$ to $(\varphi(k), h)$ for any $k \in K, h \in H$. This gives an embedding of $\mathrm{Aut}(K)$ into $\mathrm{Aut}(K \oplus H)$, and the image leaves $H$ fixed. –  Dan Shved Aug 21 '13 at 5:06
@Prism Although, in the general setting, I see no reason why the image of $\mathrm{Aut}(K)$ must exhaust all the automorphisms of $K \oplus H$ that leave $H$ fixed. –  Dan Shved Aug 21 '13 at 5:10
Right. That's a subtle point. So it is possible that some automorphism of $K\oplus H$ leaves $H$ fixed, but when restricted to $K$, it is not an automorphism of $K$. But the other direction (given in your first comment) always works. –  Prism Aug 21 '13 at 5:16