$H\vartriangleleft G$ 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 $q^2$.


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 $|H|\pmod{p}$.
