# Homework Help: Basic group theory prove xH=yH

1. Oct 7, 2012

### gottfried

1. The problem statement, all variables and given/known data
Let H be a subgroup of G
Prove xH=yH ⇔ x-1.y$\in$H

2. Relevant equations

3. 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?

2. Oct 7, 2012

### 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.

3. Oct 11, 2012

### 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

4. Oct 11, 2012

### Robert1986

yes, that seems correct to me.

5. Oct 12, 2012

### gottfried

Thanks for the help