# 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 $K(G)N/N$.

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You need to be more accurate on what you're saying, because what you wrote says only that $K(G/N) \subseteq K(G)N$. But the exact same equality can show that $K(G/N) \supseteq K(G)N$. If you worry a little about the words you say your proof is right there. – Patrick Da Silva Mar 17 '12 at 23:11
thank you for the help – Belgi Mar 17 '12 at 23:15
Shouldn't the right side be $K(G)N/N$? You want both side of the equation to be subgroups of $G/N.$ – Geoff Robinson Mar 17 '12 at 23:16
@GeoffRobinson - do you see some mistake in what I said ? I think I proved equality – Belgi Mar 17 '12 at 23:20
@Belgi: If you did mean it in the second way, then Geoff is right; the equation $K(G/N)=K(G)N$ makes no sense. The left-hand side is a set of cosets and the right-hand side is a set of elements. Your alleged proof doesn't work because "and this is in $K(G)N$" is wrong; both sides of that equation are cosets and thus can't be in $K(G)N$ if you interpret that in the usual sense as a set of elements. – joriki Mar 17 '12 at 23:36

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 $[x,y]nN\in [G/N,G/N]$.