Quotient group commutator Denote $K(G)$ the commutator of $G$.
Let $G$ be a group and $N$ be a normal subgroup of $G$, is it true that $K(G/N)=K(G)N/N$ ?
I think that this is true, I said that $(aN)(bN)(a^{-1}N)(b^{-1}N)=(aba^{-1}b^{-1}N)$ and this is in $K(G)N/N$.
 A: I will use the more common $[G,G]$ to denote the commutator subgroup of $G$. ($K(G)$ is sometimes used to denote the set of commutators if $G$, which may fail to be a subgroup).
As has been noted in the comments, the equation written makes no sense as written. While $[G/N,G/N]$ is a subgroup of $G/N$, $[G,G]N$ is a subgroup of $G$. It should really be
$$\left[\frac{G}{N},\frac{G}{N}\right] = \frac{[G,G]N}{N}.$$
First, note that $[G,G]N/N$ is a normal subgroup of $G/N$ (since $[G,G]N$ is a normal subgroup of $G$). Therefore, $(G/N)/([G,G]N)/N$ makes sense, and by the isomorphism theorems we know that
$$\frac{G/N}{[G,G]N/N} \cong \frac{G}{[G,G]N}.$$
This is an abelian group (since it is a quotient of $G/[G,G]$, also by the isomorphism theorems. Since $[G/N,G/N]$ is contained in any normal subgroup $M$ of $G/N$ such that $(G/N)/M$ is abelian, it follows that $[G/N,G/N]\subseteq ([G,G]N)/N$.
For the converse inclusion, let $[x,y]nN = [x,y]N$ in $[G,G]N/N$. Then $[xN,yN] =(xN)^{-1}(yN)^{-1}(xN)(yN) = x^{-1}y^{-1}xyN = [x,y]N$, hence $[x,y]nN\in [G/N,G/N]$.
This proves the equality.
