Proving $xH=yH$ if $x^{-1}y \in H$

Question

Given $G$ is a group and $H <G$ and if for arbitrary $x,y \in G$ if $x^{-1}y \in H$ Prove that $xH=yH$

I started like this: Since $x^{-1}y \in H$ we have $x^{-1}y=h$ for some $h \in H$

Then $y=xh$ $\implies$ $y \in xH$

Also $y \in yH$

How to proceed from here?

Answer 1

We will show that $xH=yH$ by showing that both $xH\subseteq yH$ and $yH\subseteq xH$. Since you have already shown that $yH\subseteq xH$, it suffices to show that $xH\subseteq yH$.

Let $z$ be an element in $xH$. Then, we can write $z=xh$ for some $h$ in $H$. Now note the following equalities: $$xh=yy^{-1}xh=y(x^{-1}y)^{-1}h.$$ Since $x^{-1}y$ is in $H$, its inverse is in $H$. In particular, the product $(x^{-1}y)^{-1}h$ is in $H$, and hence $xh$ is in $yH$, because we can rewrite it as $yh'$ for some $h'$ in $H$.

Answer 2

Hint:

You've shown that $yH \subseteq xH$ with your argument above. So now you want to show that $xH \subseteq yH$ too.

But your argument only needed $x^{-1}y \in H$, so if you can show $y^{-1}x \in H$ then you'll be done!

But $y^{-1}x = (x^{-1}y)^{-1}$...

Do you see how to finish the proof?

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I hope this helps ^_^