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

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.

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I guess $\,C(G)=Z(G)=$ the group's center...? –  DonAntonio May 1 '13 at 18:17
Here are some hints: By induction on $|H|$ you can assume that $H$ is characterically simple, ie elementary abelian. Thus, $G$ acts on $H$ via automorphisms but $H$ can be seen as a vector space over a field of $p$ elements, and a $p$-group acting on such a vector spaces will have a non-zero fixed point (show that such a fixed point will be in that intersection) –  Tobias Kildetoft May 1 '13 at 18:17
@DonAntonio yes the group center –  user42912 May 1 '13 at 18:18
@BabakS. it's the group center –  user42912 May 1 '13 at 18:18
@TobiasKildetoft what is "elementary abelian"? I didn't study vector spaces yet (I will study it only in the next chapter) –  user42912 May 1 '13 at 18:21

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.

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Why each conjugacy class has order a power of $p$? thank you for your answer. –  user42912 May 1 '13 at 18:44
yes, I think I know, because of the orbit-stabilizer theorem. –  user42912 May 1 '13 at 18:51
Indeed, @user42912...so after all we use here some actionish stuff. :) –  DonAntonio May 1 '13 at 18:52
I tried a lot and I couldn't prove that there is one conjugacy class with one single element :( –  user42912 May 1 '13 at 19:19
@user42912 How does $G$ acts on the identity element? –  Tobias Kildetoft May 1 '13 at 19:22

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 $H \cap Z(G)$.

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So much simpler than what I was doing. –  Tobias Kildetoft May 1 '13 at 18:30
Saying $G$ acts on H by conjugation you meant the following group action: $G\times H\to H$, $(g,h)\mapsto ghg^{-1}$? –  user42912 May 1 '13 at 18:39
Very nice: it's basically the same I proposed, but mine doesn't invite actions to play. –  DonAntonio May 1 '13 at 18:39
really? I didn't know that, why? aren't they equivalent definitions? –  user42912 May 1 '13 at 18:47
Are you sure? I've just proved that this conditions is satisfied also. –  user42912 May 1 '13 at 20:48