Show that $G/C$ is abelian if $C$ is normal closure of subgroup generated by elements $aba^{-1}b^{-1}$ Let $G$ be a group and $C$ the normal closure of the subgroup of $G$ generated by elements of the form $$aba^{-1} b^{-1}$$ for $a,b \in G$.  
Show that $G/C$ is Abelian. $C$ is called the commutator subgroup of $G$.  
Thanks in advance!
 A: If $x, y \in G$, then $xyx^{-1}y^{-1} \in C$. Hence, 
$$Cxyx^{-1}y^{-1} = C,$$
or
$$Cxy = Cyx,$$
which means that $G/C$ is abelian. 
Edit: corrected a typo pointed out by Loki Clock. 
A: Perhaps it is instructive to see how you get to construct such a subgroup.
Given a group $G$ and a normal subgroup $N$, you may inquire when is the quotient group $G/N$ abelian.
You see that this happens if and only if
$$
aN \cdot bN = bN \cdot aN
$$
for all $a, b \in G$, so that, multiplying on the right first by $(aN)^{-1} = a^{-1}N$ and then by $(bN)^{-1}$
$$
a b a^{-1} b ^{-1} N = N,
$$
that is, if and only if
$$
[a, b] = a b a^{-1} b ^{-1} \in N
$$
for all $a, b \in G$.
Now it is natural to consider the group generated by all these commutators $[a,b]$: 
$$
G' = \{ a b a^{-1} b ^{-1} : a, b \in G  \}.
$$
First of all it is normal, because if you conjugate a commutator, you get another commutator $[a, b]^x = [a^x, b^x]$. Alternatively, if $b \in G'$, then 
$$
b^{a^{-1}} = a b a^{-1} = a b a^{-1} b ^{-1} b = [a, b] b \in G'.
$$
And then, because the steps above are reversible, $G/G'$ is abelian.
We have thus seen that $G'$ is the smallest normal subgroup $N$ such that $G/N$ is abelian.
A: The commutator subgroup is always normal in $G$ so you don't need to take the normal closure.  Now the quotient is saying you are forcing the cosets of $aba^{-1}b^{-1}$ to be the identity coset.  What does this say about $(aC) (bC)$ and $(bC) (aC)$?
See this link to see a proof for why $[G,G]$ is normal
http://planetmath.org/DerivedSubgroupIsNormal.html
