Show that $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z}$ is not isomorphic to Aut($G$) for any abelian group $G$

Question

Here is a problem on which I'm stuck:

Show that $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z}$ is not isomorphic to Aut($G$) for any abelian group $G$.

On the Contrary, we assume that $ \phi$ : $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z} \to Aut(G)$ is an isomorphism,hence we get that $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z}$ acts on $G$.Also as $G$ is abelian therefore $G$ is a $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z}$-module.How to conclude from here?I'm pretty sure that we have to use the fact that $ \frac {\mathbb Z}{p \mathbb Z} \times \frac {\mathbb Z}{p \mathbb Z}$ is a $ \frac {\mathbb Z}{p \mathbb Z}$-module of rank 2.Any ideas?

Answer 1

Suppose the contrary. Since $G$ is abelian $\operatorname{Aut}G$ contains an element of order $2$ ($f:G\to G$, $f(x)=x^{-1}$), so $2\mid p^2$ and hence $p=2$. (However, if $x^{-1}=x$ for all $x\in G$, then $G$ is isomorphic to a direct sum of copies of $\mathbb Z/2\mathbb Z$ and unless $|G|=2$ its automorphism group is not abelian.) But in this case $G=\mathbb Z/8\mathbb Z$ has the property that $\operatorname{Aut}G\simeq\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$, so you may want to assume that $p>2$.