If $N$ is a normal subgroup of $G$, and $N \cap [G,G]=\{e\}$, then $N$ is contained in $Z(G)$.

Question

I found an answer to this here: If the intersection of a normal subgroup and the derived group is $\{e\}$, show that $N$ is a subset of $Z(G)$.. However I don't really understand some of the answers given and some other aspects such as, why does, $$N \cap [G,G]=\{e\},$$ imply that $N$ is abelian? I know it doesn't contain any commutators, but I'm not sure how the abelian property follows. Furthermore, knowing that $N$ is normal and abelian, I'm not sure how to go about showing that $N$ is contained in $Z(G)$.

Answer 1

Pick any two $m,n\in N$. Then $[m,n]\in [G,G]\cap N=\{e\}$ so $[m,n]=e$, i.e. $mn=nm$.

Pick any $g\in G,n\in N$. Then $[g,n]=(gng^{-1})n^{-1}$ is a product of two elements of $N$ (note $gng^{-1}\in N$ since we assume $N\trianglelefteq G$ is normal) so $[g,n]\in [G,G]\cap N=\{e\}$. Thus $[g,n]=e$...