Working on some problems after my first semester of abstract-algebra. I saw this question
$\mathbf{Theorem}$ Let $N$ be a normal subgroup of a group $G$. Then $N$ contains the commutator subgroup if and only if $G/N$ is abelian.
And saw this solution but wanted to try my hand at a "slick" proof using a fact that I overheard once (which tbh I might be misstating). Is there anything useful in this proof? Or should I stick to the traditional method?
$\mathbf{Proof}$ Every group $G$ is isomorphic to the quotient group of a free group. If a group is abelian it is isomorphic to the quotient group of a free group modded out by its commutator subgroup. Therefore $G/N$ is abelian if and only if $N$ contains an isomorphic copy of a commutator subgroup as in it contains the commutator subgroup of $G$.