Normal subgroup in a $p$-group

Question

Let $G$ be a $p$-group, and $H$ is a normal subgroup of $G$ with $|H| = p$. Prove that $H \leq Z(G)$. More general, if $K$ is a normal subgroup of $G$, then $K\cap Z(G) \neq \{e\}$, with $e$ is the unity of $G$, where $Z(G)$ is the center of $G$, $Z(G) = \{x\in G: xa = ax,\forall\; a\in G\}$.

Answer 1

Equivalently to @user8268's answer, since $H=\cup_{x\in H}~ C_x$, so we see that $$p=|H|=|H\cap Z(G)|+\sum_{x\in H-Z(G)}|C_x|$$ but we know that if $x\in H-Z(G)$ happens then $p\big|~|C_x|=[G:C_G(H)]$ and therefore $p$ divides $|H\cap Z(G)|$. What this means? Indeed, it means that $|H\cap Z(G)|=|H|=p$.

Answer 2

The orbits of the action of $G$ by conjugation on $K$ have $p^n$ elements ($n$ depends on the orbit; the number must divide the order of $G$, so it is a power of $p$) and $p^n=1$ iff the (unique) element of $K$ in the orbit is in $Z(G)$. The sum of these numbers is the order of $K$, i.e. a power of $p$, so by divisibility by $p$, $p$ divides $|K\cap Z(G)|$.