$N$ is a proper normal subgroup of group $G$ and $N \cap G'=<e>$ then prove that $N<C(G)$.
If $G$ were abelian and nilpotent then the result would hold trivially.
So, assuming that $G$ is a non-abelian nilpotent group, for any $g\in G$, $n\in N$, $ngn^{-1}g^{-1}\in N$ and $ngn^{-1}g^{-1}\in G'$. This implies that $ngn^{-1}g^{-1}=e$ and therefore, $n\in C(G)$ (since $ng=gn$ $\forall g \in G)$ and $N\subseteq C(G)$. This implies that $N\leqslant C(G)$. How do we infer that $N < C(G)?$
Can we say that if $N=C(G)$ then since $G'\leqslant C_2(G)$ and $C(G)=C_1(G) \trianglelefteq C_2(G)$ implies that $N \cap G'=G'$ or $N?$