Is it true that any quotient group G/N is abelian if N contains the commutator subgroup?

Question

Is it true that any quotient group G/N is abelian if N contains the commutator subgroup(including the case when N is the commutator subgroup)? I think this is true but I just want to confirm.

Answer 1

Yes, $G'=[G,G]$ is the smallest normal subgroup of $G$ such that $G/N$ is abelian: $abN=baN$ iff $b^{-1}a^{-1}baN=N$ iff $[a,b]\in N$ for any $a,b\in G$.

Answer 2

Yes. Let $\varphi: G \to G/N$ be the projection. Let $a', b' \in G/N$ have $a' = \varphi(a), b' = \varphi(b)$. Then $$[a',b'] = [\varphi(a),\varphi(b)] = \varphi([a,b]) = e,$$ since the commutator is in the kernel. So the bracket of any two elements in $G/N$ is trivial and $G/N$ is abelian.