If the intersection of a normal subgroup and the derived group is $\{e\}$, show that $N$ is a subset of $Z(G)$.

Question

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!

Answer 1

Your argument is correct to show that $N$ is abelian (whether or not $N$ is a normal subgroup). But to show that $N$ is in the centre of $G$ is stronger --- you have to show that every element of $N$ commutes with every element of $G$. This will require the assumption that $N$ is normal.

Answer 2

(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

$$[N,G]\le[N,G']\le N\cap G'=1\iff N\le Z(G)$$