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Is this statement true that for a finite and non abelian $p$-group $G$; $p^2\big||Aut(G)|$? I just found $Q_8$ fulfillment of this claim.

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    @ChrisEagle: Thanks for noting that point.2012-05-18

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Yes, it's true.

Note that $G/Z(G)$ is isomorphic to a subgroup of $\mathrm{Aut}(G)$. Then we have this old chestnut:

Let $G$ be a group, and let $N\subseteq Z(G)$. If $G/N$ is cyclic, then $G$ is abelian.

In particular, $G/Z(G)$ cannot be nontrivial and cyclic. So if $G$ is a $p$-group, then either $Z(G)=G$ or else $[G:Z(G)]$ is at least $p^2$. Therefore, if $G$ is not abelian, then $p^2|[G:Z(G)]$, and $[G:Z(G)]$ is the order of a subgroup of $\mathrm{Aut}(G)$, establishing the desired claim.