Show that if $N$ is a normal subgroup of $G$ which contains all commuters then $G/N$ is abelian. I am working on my proof for class and I was wondering if this look ok?
Let $N$ be a normal subgroup of $G$ we want to show that $G/N$ is abelian, or $(aN)(aN) = abN = baN = (bN)(aN)$.  Since $N$ contains all commutors,
then let $aba^{-1}b^{-1}N = N$ for some $a,b \in G$ then,
\begin{align*}
aba^{-1}b^{-1}N &= N && \text{ Given}\\
ab(ba)^{-1}N &= N && \text{ Definition of inverse} \\
abN(ba)^{-1} &= N && \text{ since N is normal}\\
abN &= N(ba) && \text{ right multiply by ba}\\
abN &= baN && \text{ since N is normal}\\
\end{align*}
Which is what we wanted to show.
We are using Abstract Algebra by Judson and I tried to mimic one of the proofs in the book plus add my reasons behind doing so.

Commutators. No, not "for some $a,b\in G$, rather, "for all $a,b\in G$, $[a,b]N=N$. This says precisely that $G/N$ is abelian.
