$G$ $p$-group. If $H\triangleleft G$, then $H\cap C(G)\ne \{e\}$

Question

I'm pretty new on this subject and I need a hint to begin to solve this question:

If $G$ is a finite p-group, $H\triangleleft G$ and $H\ne \{e\}$, then $H\cap C(G)\ne \{e\}$

Thanks for any help.

Answer 1

This much is true for any nilpotent group, and the proof there is very simple...alas, we're going to have to go the long way with hints:

1) $\,H\,$ is a union of conjugacy classes

2) Each conjugacy class has order a power of $\,p\,$

3) Since there's for sure one conjugacy class with one single element, then it must be at least another conjugacy class with one single element, say $\,w\,$

4) The element $\,w\,$ is central.

Answer 2

Since $H$ is normal in $G$, $G$ acts on $H$ by conjugation. The orbits having size $\gt 1$, must have size divisible by $p$ by orbit-stabilizer theorem. Then, the number of orbits having size $1$ must be divisible by $p$ since $H \neq 1$. But their union is $H \cap Z(G)$.