# Semidirect product with a $p$ group

let $P$ be a finite $p$-group that acts on a finite group $G$ and assume that $P$ is maximal subgroup of $G \rtimes P$. Show that $G$ is an abelian $q$-group for some prime $q$.

Hint: Show that $P$ stabilizes some sylow subgroup of $G$.

Thanks!

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What is the question? Hint for the answer: Frattini argument. – Geoff Robinson Jul 15 '12 at 17:13
I don't get it: AFAIK, $\,G\rtimes P\,$ means there's a homomorphism $\,G\longrightarrow Aut(P)\,$...what action do we have here? Perhaps it should be $\,P\rtimes G\,$? And still, we should have then an action by automorphisms of $\,P\,$ on $\,G\,$, something that isn't given...am I confusing something here? – DonAntonio Jul 15 '12 at 18:24
@DonAntonio: While there is some disagreement of what a semidirect product "of $H$ by $K$" means (for some, $H$ will be the normal subgroups, for others $K$), I am aware of no disagreement on the symbol. In $K\rtimes H$, you should read the triangle as indicating what is normal; in $G\rtimes P$, $G$ is the normal subgroup in the semidirect product, so that we have a homomorphism $P\to\mathrm{Aut}(G)$ (given by the fact that "$P$ acts on a finite group $G$"; that means that $P$ acts as automorphisms on $G$). Review your symbols... – Arturo Magidin Jul 15 '12 at 19:26

New edit: I am going to rewrite the proof in a more unified way to cover both the cases where $p$ divides $|G|$ and when $p$ does not. Set $K = GP$ (for ease of notation: it is to be understood that $G \lhd K$, and that $P$ is acting on $G$ by the action specified in the given semidirect product). Note that, as the hint suggests, it suffices to find a non-identity Sylow $q$-subgroup $Q$ of $G$ which is normalized by $P$ (for some prime $q$ which may or may not equal $p$). For then $P$ normalizes the non-identity characteristic subgroup $Z(Q)$ of $Q$, so $PZ(Q)$ is subgroup of $K$ which strictly contains $P$, so since $P$ is assumed to be maximal, we have $K = PZ(Q)$ and $G = Z(Q).$
We next note that if $p$ divides $|G|$, then $P$ permutes the Sylow $p$-subgroups of $G$ by conjugation. The number of these is congruent to $1$ (mod $p$) by Sylow's theorem, so one of them, $Q$ say, is fixed by $P$ in this action, that is, normalized by $P,$ and we have found the required Sylow subgroup of $G.$
Hence we suppose from now on that $p$ does not divide $|G|$. Let $q$ be a prime divior of $|G|$. Now $P$ permutes the Sylow $q$-subgroups of $G$ by conjugation. The number of such Sylow subgroups is a divisor of $|G|,$ so is not divisible by $p,$ as we are now assuming that $|G|$ is not divisible by $p.$ Hence $P$ must have an orbit of length $1$ in this action: in other words, $P$ normalizes some Sylow $q$-subgroup $Q$ of $G$ in this case too.
(It is also possible to produce a Sylow $q$-subgroup of $G$ normalized by $P$ by using the Frattini argument, which gives $K = GN_{K}(Q)$ for each Sylow $q$-subgroup $Q$ of $G$).
As was pointed out to me in my (incorrect) development, it is possible for $p$ to divide $|G|$.... but only if $|G|=p$, in which case the result sought holds. – Arturo Magidin Jul 16 '12 at 4:24
Yes thanks. I had thought for some reason that the question assumes that $|G|$ was coprime to $p.$ I'll re-edit. – Geoff Robinson Jul 16 '12 at 7:59