A non-trivial normal subgroup N of a finite group G or order $p^n$ is such that $N\cap Z(G)\neq \{ 1\}$ [duplicate]

Since $N$ is normal in $G$, then $G$ acts on $N$ by conjugation. So,

$$N = \sum \operatorname{Orbit}_{G}(x),$$ ($x\in N $) $ =$ 1 + $\sum \textrm{Orbit}_{G}(x)$, $x\in N-\{1\}$. Now, since the order of $\textrm{Orbit}_{G}(x)$ is a non negative power of $p$ and $p$||N|, then we see one orbit has order of 1. So, there is an non trivial element $s\in N$ that $\operatorname{Orbit}_{G}(s)$=${s}$. Therefore for all $g\in G$, $s^g=s$ and this means that $N \cap Z(G)$ is not 1.

Since $N$ is normal, it is an union of disjoint conjugacy classes. It contains the conjugacy class $\{1\}$, and thus not all of the other conjugacy classes can have order divisible by $p$. This means that besides the conjugacy class $\{1\}$, the subgroup $N$ contains also some other conjugacy class $\{z\}$ of size $1$. Then $z$ is a nontrivial central element in $N$.