TITLE: Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian
QUESTION [36 upvotes]: Trying to get my head around the commutator subgroup. This is an excercise from Artin's Algebra:
Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian.
Here is what I've done:
Let $xC,yC \in G/C$ then $xyx^{-1}y^{-1}C = C$ since $xyx^{-1}y^{-1}$ is a commutator hence belongs to $C$. But then $xyC = yxC$ so $xC$ and $yC$ commute in $G/C$. This can be done for any elements, so $G/C$ is abelian.
This seems somewhat surprisingly short. Is that all there is to it? 
Regards

REPLY [5 votes]: Recall that $gh = hg[g,h]$ and that $[G,G]$ is the subgroup of $G$ generated by all commutators. $[G,G] \unlhd G$ because $$[h,k]g = g[h,k][[h,k],g].$$
The map $$G \longrightarrow \frac{G}{[G,G]}$$ is called abelianization (precisely because of the theorem we are about to prove).
Every element of $G/[G,G]$ is of the form $g [G,G]$ and this group is abelian because $$(g [G,G]) (g' [G,G]) = g g' [G,G] =  g g' [g',g] [G,G] = g' g [G,G] = (g' [G,G]) (g [G,G]).$$