## Intersection of Center of p-group and nontrivial normal subgroup is nontrivial

Let G be a p-group and H a nontrivial normal subgroup. Prove that $H\cap Z(G)$ is nontrivial.

Let G act on H by conjugation. Since H is a normal subgroup, this is a well-defined group action since $ghg^{-1}\in H$ for all $g\in G$.

Let $H_0=\{h\in H:ghg^{-1}=h\ \ \ \forall g\in G\}$.

$h\in H_0 \iff O(h)=\{h\}$.

Therefore we have $H=H_0+O(h_1)+O(h_2)+\dots+O(h_n)$ where $|O(h_i)|>1$.

By the Orbit-Stabilizer Theorem,

$|H|=|H_0|+\sum_{i=1}^k |O(h_i)|=|H_0|+\sum_{i=1}^k[G:G_{h_i}]$

Let $|G|=p^n$. By Lagrange’s Theorem, $p^n=|G|=[G:G_{h_i}]|G_{h_i}|$. Since $|G_{h_i}|\neq p^n$, therefore $p\mid [G:G_{h_i}]$.

Hence, $|H|\equiv |H_0|\pmod p$.

Note that

\begin{aligned} h\in H_0\subseteq H &\iff ghg^{-1}=h\ \ \ \forall g\in G\\ &\iff gh=hg\ \ \ \forall g\in G\\ &\iff h\in Z(G)\cap H \end{aligned}

Therefore $|H\cap Z(G)|\equiv |H|\equiv 0\pmod p$. Since $1\in H\cap Z(G)$, this implies that $p \mid |H\cap Z(G)| \neq 0$. Therefore $|H\cap Z(G)|\geq p$.

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