Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian 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
 A: 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 $$(g [G,G]) (g' [G,G]) = g g' [G,G] =  g g' [g',g] [G,G] = g' g [G,G] = (g' [G,G]) (g [G,G]).$$
A: For some reason the OP won't post, or can't post, an answer, so summarizing the comments:
$(1)\,\,\forall\,g,x,y\in G\,$ , and putting $\,a^b:=b^{-1}ab\,\,,\,a,b\in G\,$:
$$[x,y]^g:=g^{-1}[x,y]g:=g^{-1}x^{-1}y^{-1}xy g=\left(x^{-1}\right)^g\left(y^{-1}\right)^gx^gy^g=[x^g,y^g]\in G'\Longrightarrow G'\triangleleft G$$and thus the quotient $\,G/G'\,$ is a group.
$(2)\,\,$ Let now $\,N\,$ be any normal subgroup of $\,G\,$  s.t. $\,G/N\,$ is abelian, then:
$$\forall\,x,y\in G,\,\,xNyN=yNxN\Longleftrightarrow xyN=yxN \Longleftrightarrow (yx)^{-1}xy\in N \Longleftrightarrow [x,y] \in N$$
and since $\,G':=\langle\,[x,y]\;:\;x,y\in G\,\rangle\,$ , then $\,G'\leq N\,\Longrightarrow \,G'$ is the minimal (normal) subgroup of 
$\,G\,$ s.t. its quotient is abelian -- "minimal" wrt set inclusion --.
Exercise:  Explain the parentheses around "normal" above, i.e. show that any subgroup of G containing the commutator subgroup is normal.
A: It is easy to show that the commutator subgroup is a characteristic subgroup , hence it is a normal subgroup. Proving G/C abelian is straightforward recalling that C is the subgroup of G generated by all commutators.
