TITLE: If the intersection of a normal subgroup and the derived group is $\{e\}$, show that $N$ is a subset of $Z(G)$.
QUESTION [5 upvotes]: I think my reasoning is wrong, but if the intersection only contains the identity, doesn't that imply that the only commutator in $N$ is $\{e\}$, so doesn't that mean $N$ is automatically commutative? Why was it necessary to state that $N$ was a normal subgroup? Thanks!

REPLY [3 votes]: (1) First, note that $\;N\lhd G\iff [N,G]\le N\;$ 
(2) Also, note that $\;x\in G\;$ is a central element iff $\;[x,G]=1\;$
So since $\;N\,,\,G'\lhd G\; $, we get
$$[N,G]\le[N,G']\le N\cap G'=1\iff N\le Z(G)$$