 Solution

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.