# $H\vartriangleleft G$ and $|H|\not\equiv 1 (\mathrm{mod} \ p)$ then $H\cap C_{G}(P)\neq1$

Let $G$, a finite group, has $H$ as a proper normal subgroup and let $P$ be an arbitrary $p$-subgroup of $G$ ($p$ is a prime). Then $$|H|\not\equiv 1 (\mathrm{mod} \ p)\Longrightarrow H\cap C_{G}(P)\neq1$$

What I have done:

I can see the subsequent well-known theorem is an especial case of the above problem:

Let $G$ is a finite non trivial $p$-group and $H\vartriangleleft G$. Then if $H\neq1$ so $H\cap Z(G)\neq1$.

So I assume that $G$ acts on $H$ by conjugation and therefore $$|H|=1+\sum_{x\in H-\{1\}}|\mathrm{Orbit}_G(x)|$$ $|H|\not\equiv 1 (\mathrm{mod} \ p)$ means to me that there is $x_0\in H$ such that $p\nmid|\mathrm{Orbit}_G(x_0)|$. Am I doing right? Thanks.

This problem can be applied nicely in the following fact:

Let $p$ is an odd prime and $q$ is a prime such that $q^2\leqslant p$. Then $\mathrm{Sym}(p)$ cannot have a normal subgroup of order $q^2$.

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LaTeX tip: x \pmod{p} produces $x \pmod p$. – user2468 Aug 26 '12 at 19:21
@JenniferDylan: Thanks for the code. I searched for it through net. – Babak S. Aug 26 '12 at 19:26

Instead of letting all of $G$ act on $H$, consider just the action of $P$ on $H$. Then everything you wrote above still holds, but since $P$ is a $p$-group, the size of the orbit of $x_0$ must be a power of $p$. Can you conclude from here?
Oh yes. I see the proof. In fact as you both noted; there must be a $x_0\in H-\{1\}$ such that $\mathrm{Orbit}_P(x_0)=\{x_0\}$. This ends the proof. Thanks Ted. Thanks Steve. – Babak S. Aug 26 '12 at 19:24
Let $P$ act on $H$; the number of fixed points is the number of elements in $C_G(P)\cap H$. Now use the easy fact that the number of fixed points is congruent to $|H|\pmod{p}$.