Take the 2-minute tour ×

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 $\alpha$.

share|improve this question
Chandru, you appear to be going through random mathematics books and posting random exercises from them here. Where does this one come from? –  TonyK Sep 2 '10 at 13:10
@TonyK: I don't know what made you think like that! I have posted this question about math.stackexchange.com/questions/2121/… and that was why i was curious about this question. Anyway you can see my link. Its very useful! Sorry if by any means you felt like that! –  anonymous Sep 2 '10 at 15:32
@Chandru: silent editing of refuted questions is discouraged, because it turns correct answers into "wrong" ones, leading to (1) downvotes of the initially correct answers, and (2) confusing inconsistency between answers, where some say the (revised) question is correct and others that the (original) question is wrong, without any answer being mistaken. Correct editing practice is to retain the original question content, displaying modifications and corrections in a way that makes the evolution of the question clear. –  T.. Sep 2 '10 at 19:35
add comment

2 Answers

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 $1$.


share|improve this answer
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
add comment



share|improve this answer
+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
add comment

Your Answer

