Every proper normal subgroup of a $p$-group contains non-trivial central elements

Question

How do I prove the following statement?

Let $G$ be a $p$-group, $H$ - its normal subgroup, $H\neq \{e\}$. Then the center of $G$ $$ Z(G)=\{a\in G\ \ |\ \ \forall x\in G\ \ ax=xa\} $$ has non-trivial intersection with $H$, i.e. $H\cap Z(G) \neq \{e\}$.

As far as I understand, this is about finding central elements in a normal subgoup. Maybe there is some straightforward way to construct such an element (using inner automorphisms, for instance...)?

Are you missing some condition? The symmetric group $S_n$ for $n>2$ has trivial center, but non-trivial normal subgroup $A_n$. — moonlight, Oct 25 '15 at 08:49

score 2 · Accepted Answer · answered Oct 25 '15 at 11:08

2

Let $G$ act on $H$ by conjugation. As $H$ is a non-trivial p-group, then $\vert H\vert\equiv \mbox{fixed points} \mod p.$

This implies that $H\cap Z(G)$ has size divisible by p. Thus $H\cap Z(G)$ is nontrivial.

answered Oct 25 '15 at 11:08

JeSuis

697

I assume that you are familiar with the class equation, anyway it's a duplicate of here – JeSuis Oct 25 '15 at 11:10
Oh, I got it, thank you! I wish you told that it's a duplicate earlier, we could just delete the whole thing... – Glinka Oct 25 '15 at 14:57
@Glinka I saw it after I wrote the "answer"..sorry – JeSuis Oct 25 '15 at 18:13
How do you use the normality of H? – Ivan Gonzalez Apr 26 '20 at 00:19

Every proper normal subgroup of a $p$-group contains non-trivial central elements

1 Answers1