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...)?
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.