A question regarding subgroup $H$ of a group $G$

Question

I am looking at this proof from Let $x, y \in G$ be two arbitrary elements. Let $H \leq G$. Then, there is a proposition that says $Hx^{-1}y = H \implies x^{-1}y \in H$, which I understand, but is this a double implication or just one way? If it is a double implication, how do I prove that $x^{-1}y \in H \implies Hx^{-1}y = H$?

Answer 1

Let $z=x^{-1} y.$ If $z \in H$ then it is easy to see that $zH \subseteq H.$ Now suppose $h \in H.$ Since $z \in H$ so $z^{-1} \in H.$ Then $z^{-1} h \in H.$ Therefore $h=z(z^{-1} h) \in zH.$ Therefore $H \subseteq zH.$ This shows that $zH=H,$ as claimed.

Answer 2

Because, since $H$ is a subgroup $(\forall h\in H):Hh=H$. Now, apply this to $h=x^{-1}y$.

Answer 3

It should be double implication. Notice that if we have $x^{-1}y \in H$, let $x^{-1}y = h' \in H$. Then of course, we will have $Hh' = H$.