Let $H\leq G$. Prove $x^{-1}y^{-1}xy\in H\text{ }\forall x,y\in G \iff H\trianglelefteq G \text{ and } G/H \text{ is abelian}$. Question: Let $H\leq G$.  Prove $x^{-1}y^{-1}xy\in H\text{ }\forall x,y\in G \iff H\trianglelefteq G \text{ and } G/H \text{ is abelian}$.
my thoughts: In the forward direction, if $x^{-1}y^{-1}xy\in H$ for all $x,y\in G$, then $y^{-1}xy\in xH\subseteq G$, so, since $y\in G$, I just need to show that $x\in H$ to show that $H\trianglelefteq G$, right?
In the backwards direction, since $H\trianglelefteq G$, I can consider the cosets $xH$, $yH$, $x^{-1}H$, and $y^{-1}H$ where $x,y\in G$, thus so are their inverses.  Then, I consider $(xH)(x^{-1}H)(yH)(y^{-1}H)$, but I am not quite sure how to finish from here.  Any help is greatly appreciated!  Thank you.
 A: For the forward direction, you have that $y^{-1}xy \in xH$ for all $x$ and $y$ in $G$, as you say.  Therefore, if $y$ is an arbitrary element of $G$, and $x$ is an arbitrary element of $H$, then $xH = H$, and therefore $y^{-1}xy \in  H$.  This is exactly what it means for $H$ to be a normal subgroup of $G$.
To see that $G/H$ is abelian, you need only show that $xyH = yxH$ for any $x, y \in G$.  This is equivalent to saying that $x^{-1}y^{-1}xyH = H$ for all $x$ and $y$, which is of course true by your hypothesis.
It is quite easy to get the reverse direction from the previous paragraph.
A: For backward direction, you already started really well. Since $G/H$ is abelian, we have
$$H = (xH)(x^{-1}H)(yH)(y^{-1}H) =(x^{-1}H)(y^{-1}H)(xH)(yH) = (x^{-1}y^{-1}xy)H$$
since we can swap places of elements in $G/H$.
For forward direction, in order to show $H \unlhd G$, pick an element of the form $g^{-1}hg \in G$. Then,
$$h^{-1}g^{-1}hg \in H \implies g^{-1}hg \in hH = H \implies g^{-1}Hg \subseteq H \implies H \unlhd G$$
I leave as an exercise to you to show that $G/H$ is abelian.
