Proving Normal subgroups and abelian My questions is
Let H be a subgroup of G which contains all products of type $x^{-1}y^{-1}xy\in H  \forall x,y\in G$.
Prove that $h\in H$that $ghg^{-1}\in H$. So H is normal.
Then consider the factor group $G/H$. Prove $G/H$ is abelian, so $(aH)(bH)=(bH)(aH)$. 
I know that I have to use $b^{-1}a^{-1}ba\in H$ to show that $ba\in (aH)(bH).$ 
Any tips? I dont know how to make this any clearer because this is all I have been given.
For the first part, I know that in order for $ghg^{-1}\in H$, $ghg^{-1}=gx^{-1}y^{-1}xyg^{-1}$=$(g^{-1})^{-1}x^{-1}y^{-1}xyg^{-1}.$ Im stuck after that
 A: First, we prove that $H$ is normal. $\forall h \in H,g\in G$, suppose $h=xyx^{-1}y^{-1}=:[x, y]$, we have
$$ghg^{-1}=g[x,y]g^{-1}=gxyx^{-1}y^{-1}g^{-1}=gx(g^{-1}g)y(g^{-1}g)x^{-1}(g^{-1}g)y^{-1}g^{-1}=(gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1}=[gxg^{-1},gyg^{-1}] \in H.$$
For arbitrary $h$, the proof follows from what have been proved.
Next, we prove that $G/H$ is abelian. $\forall x, y\in G$, we have
$$\bar x\bar y\bar x^{-1}\bar y^{-1}=\overline{xyx^{-1}y^{-1}}=\bar e,$$
since $xyx^{-1}y^{-1}\in H$. Hence, $\bar x\bar y=\bar y\bar x$, which completes the proof.
Note: $\bar x:=xH$.
A: Let $h \in H$ and $g \in G$. Then $ghg^{-1} = (ghg^{-1}h^{-1})h$ is the product of two elements of $H$, hence belongs to $H$. (Take $x = g^{-1}$ and $y = h^{-1}$.) Therefore $H$ is normal.
Now let $\overline{x},\overline{y}$ be arbitrary elements of $G/H$, represented respectively by $x,y \in G$. In $G/H$, we have
$$\left(\overline{x}\right)^{-1}\left(\overline{y}\right)^{-1}\overline{x}\overline{y} = \overline{x^{-1}y^{-1}xy} = 1,$$
since $x^{-1}y^{-1}xy \in H$. Multiplying on the left by $\overline{y}\overline{x}$, we find $\overline{x}\overline{y}= \overline{y}\overline{x}$, proving that $G/H$ is commutative.
NOTE. The stated condition is equivalent to the condition that $H$ should contain the commutator subgroup of $G$, which is the subgroup of $G$ generated by the commutators $xyx^{-1}y^{-1}$.
