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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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    @Jenni$f$erDylan: Thanks $f$or the code. I searched for it through net.2012-08-26

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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?

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    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.2012-08-26
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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 $|H|\pmod{p}$.