= Solution
Define $(xN)(yN)=xyN$. If representatives are changed to $xn_1,yn_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.
Let $G'$ be the set of finite products of commutators. It is closed under products, and $[x,y]^{-1}=[y,x]$, so it is a <subgroup>. Moreover
$$
g^{-1}[x,y]g=[g^{-1}xg,g^{-1}yg],
$$
so conjugation preserves products of commutators and $G'\triangleleft G$. In $G/G'$, one has $xyG'=yxG'$ because $(yx)^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G'\leq N$. This is the universal property of the <commutator subgroup>.
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