Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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, 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 $H \cap Z(G)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.