If $\operatorname{Aut}(G)$ is abelian order $p^5$, is $G$ non-abelian?

Question

I was wondering if somebody could explain why '$G$ is clearly non-Abelian' just from knowing that $\operatorname{Aut}(G)$ is abelian and of order $p^5$, with $p$ prime (not 2) in this proof.

Answer 1

If $G$ is abelian and some $g\in G$ has $g^2\not = 1$, then $\operatorname{Aut}(G)$ contains the element $x \to x^{-1}$ of order $2\!\!\not|\,p^5$. The remaining case $G = \mathbb{Z}_2^{\oplus n}$ and $\operatorname{Aut}(G) = GL_n(\mathbb{Z}_2)$ can be handled directly. (That's assuming $G$ is finite, given the tags and the proof of the theorem.)

Answer 2

Inn($G$) is $p$-group, so $p$ divides $|G|$ and $p$ is odd.

So, if $G$ is abelian, then it is direct product of Sylow-$p$ subgroup and $p'$-subgroup.

The $x\mapsto x^{-1}$ is an automorphism of order $2$ of Sylow-$p$ and extends to automorphism of order $2$ of $G$ (identity on $p$'-subgroup).

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Thanks to Dietrich Burde for pointing out error in earlier argument.