How Does xH Equal yH Imply x⁻¹y Belongs to H in Group Theory?

[gottfried]

Homework Statement

Let H be a subgroup of G
Prove xH=yH ⇔ x^-1.y$\in$H

Homework Equations

The Attempt at a Solution

If x.H = y.H then x,y$\in$H
since H is a subgroup x^-1,y^-1$\in$H
and the closure of H means x^-1.y$\in$H

Proving the reverse is my problem despite the fact that I'm sure is very easy but i just can't see it.

What I want to do is show that x^-1.y$\in$H implies x,y$\in$H
In which case x.H=H=y.H.

How is the best way to show this?

 

[Robert1986]

If $xH = yH$ then $x,y \in H$. This is not true. Consider the integers. And conside the sub group $5Z$. Now, $3+5Z = 8+5Z$ but neither 8 nor 3 is in $5Z$. So, if $xH=yH$ then for every $h\in H$ there is a $h' \in H$ with $xh=yh'$. Now, use that fact to prove this direction.

Try to do something similar for the other direction. You know that there is some $h \in H$ with $y = xh$, see what you can make of that.

 

[gottfried]

Thanks.

So we know that x=y.h for some h$\in$H
therefore e=x^-1.y.h
e.h^-1= x^-1.y
Therefore x^-1.y$\in$H since h^-1$\in$H

So the reverse could be

x^-1.y=h$\in$H
therefore y=x.h and this tells us xH=yH by the theorem that xH=yH ⇔ x=y.h for some h$\in$H

 

[Robert1986]

yes, that seems correct to me.

 

[gottfried]

Thanks for the help