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Let $G$ be a finite group, and let $p^{\alpha} \mid |G|$, where $p$ is a prime. Now does this imply $p \mid |Aut(G)|$?

Clearly if $|G| \leq 2$, then the Automorphism group is the trivial group, so one can see that this need not be true for $\alpha =1$. I am curious to know for higher powers of $\alpha$.

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up vote 7 down vote accepted

Let $P$ be a Sylow p-subgroup of $G$. Now $G/Z(G)$ embeds in $Aut(G)$, so we can reduce to the case where $P\le Z(G)$, so $G=P\times H$, where $(|H|,p)=1$, and $P$ is abelian. Now if $\alpha>1$, then there will always be an element of order $p$ in $Aut(P)\subset Aut(G)$. You can see this by looking at cyclic groups of order $p^m$ with $m>1$, and elementary abelian p-groups of dimension greater than $1$.


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Possibly dumb question: Why is Aut(P) contained in Aut(G)? It is not clear to me that every automorphism of P extends to G. – David Speyer Sep 2 '10 at 12:06
@David Speyer: because $P$ is a direct factor of $G$. – Matt E Sep 2 '10 at 14:25
Oh, I see. You're skipping a lot of steps, right? Central --> normal. H normal and GCD(|H|, |G/H|)=1 --> semidirect product (Schur--Zassenhaus). Semidirect + central --> direct. Or is there a shorter way I'm missing? – David Speyer Sep 2 '10 at 16:01
Because $P$ is central, one can apply Burnside Transfer theorem, for example, to get $G=P\times H$, but sure, Schuz-Zassenhaus works as well. – user641 Sep 2 '10 at 17:14



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+1 for this is a valid answer before OP edited his question to exclude $\alpha = 1$. – Soarer Sep 2 '10 at 9:16
Thanks. Yes, the question that I answered (and its title) was whether "p^a divides |G| implies p^a divides |Aut(G)|". A side effect of the editing is that downvotes occur for correct answers that become incorrect. – T.. Sep 2 '10 at 19:46

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