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 $q^2$.