I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:
Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in $Z(G)$ also commutes with the elements of $N$, $Z(G) \subseteq Z(N)$, but we know that $Z(N) \leq N$, so $ \{e\} \not = Z(G) \subseteq N.$
If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3\times \mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.