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If $G$ is a finite $p$-group and $H\neq 1$ is normal in $G$, then $H\cap Z(G)\neq 1$

I know that there is an answer here but it uses composition series.

My teacher gave me the following hint:

$$G = Z(g)\cup (\cup G_{x_i})$$

and to see $H$ as

$$|H| = |H\cap Z(G)| + \cdots$$

I think it has something to do with the class equation. If we let $G$ act on $H$ by conjugation, I know that the class equation becomes:

$$|H| = |\{h | ghg^{-1}=h\}| + \sum_{i}|\{gh_ig^{-1}| g\in G\}| = \\ |\{h \ | \ gh = hg\}| + \sum_{i}|\{gh_ig^{-1}| g\in G\}| = Z(G) + \sum_{i}|G_{x_i}| = Z(G) + \sum_{i}(G:G_{x_i})$$

(the last equality I'm just guessing... Shouldn't it have something to do with $H$?)

I also feel that the conclusion should follow from $|H|$ being $>1$ and divisibility, but I'm not quite sure. Could somebody help me and also verify if my class equation is right?

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Guerlando OCs
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1 Answers1

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You have the right idea, but have made a slight mistake, possibly due to lazy notation! The idea is that a normal subgroup is a union of conjugacy classes. This is because if $h\in H$ then $ghg^{-1}\in H$ for all $g\in G$.

Let $h_1,\ldots, h_n$ be representatives for the conjugacy classes contained in $H$ ordered such that $h_1,\ldots,h_s\in Z(G)$ and $h_{s+1},\ldots,h_n\notin Z(G)$. Since $1\in Z(G)$, $s\ge 1$ and since $h_1,\ldots,h_s\in Z(G)$ we have $\{h_1,\ldots,h_s\}=H\cap Z(G)$.

Using your notation $|H|=|H\cap Z(G)|+\sum_{i=s+1}^n|G_{h_i}|$. As $h_{s+1},\ldots,h_n\notin Z(G)$ we have $p\mid |G_{h_i}|$ for $i=s+1,\ldots,n$. But also $p\mid |H|$, so $p\mid |H\cap Z(G)|$.

Robert Chamberlain
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