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This is exercise 5.6 from Algebra, Isaacs.

Let $G$ be a finite group with $|G|>1$, and suppose $P\subseteq \operatorname{Aut}(G)\ $ is a $p$-subgroup. Show that there exists some nontrivial Sylow $q$-subgroup $Q$ of $G$ (for some prime $q$) such that $\sigma(Q)=Q$ for all $\sigma \in P \ $.

I can solve the case $ p \mid |G| \ $, but not the other.

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Hint: If $p$ does not divide $|G|$ and $q$ is a prime divisor of $|G|,$ what do you know about the number of Sylow $q$-subgroups of $G?$ –  Geoff Robinson Dec 25 '11 at 13:38
    
If Q is a q-sylow, the number $n_{q} \ $ of sylow q-subgroups of G is $|G:N_{G}(P)|\ $, and so $n_{q}| \ |G| \ $. Now we can consider the obvious action of P on the q-sylow subgroups of G, if there isn't a Q such that $\sigma(Q) = Q \ $ for all $\sigma \in P \ $ then each orbit has lenght divisible by p and so p | $n_{q}$ | |G|. Correct? –  WLOG Dec 25 '11 at 13:49
    
Looks right to me! –  Geoff Robinson Dec 25 '11 at 13:54
    
Well, slight typo: It should be $[G:N_G(Q)]$ in your first line. –  Geoff Robinson Dec 25 '11 at 15:19
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@stef: you can write your solution up as an answer so the question does not stay "unanswered". –  Jack Schmidt Dec 25 '11 at 18:05

1 Answer 1

up vote 4 down vote accepted

The solution in the case that p isn't a divisor of |G|:

If Q is a q-sylow, the number $n_{q}$ of sylow q-subgroups of G is $[G:N_{G}(Q)]$ , and so $n_{q}\mid |G|$ . Now we can consider the obvious action of P on the q-sylow subgroups of G, if there isn't a Q such that $\sigma(Q) = Q$ for all $\sigma \in P \ $ then each orbit has lenght divisible by p and so $p \mid n_{q} \mid |G|$ in contrast with the assumption.

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Thanks, I have edited. –  WLOG Dec 25 '11 at 23:14

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