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