1. ## abstract algebra

a, b are in G, a group. H is a subgroup. If aH = bH then Ha^(-1) = Hb^(-1). Prove/disprove.

I can't even tell if this is true or false so I really can't prove it!

2. Originally Posted by fulltwist8
a, b are in G, a group. H is a subgroup. If aH = bH then Ha^(-1) = Hb^(-1). Prove/disprove.

I can't even tell if this is true or false so I really can't prove it!
Hint: $\displaystyle xH = yH$ if and only if $\displaystyle x^{-1}y \in H$ and same situation with $\displaystyle Hx=Hy$.

3. Originally Posted by ThePerfectHacker
Hint: $\displaystyle xH = yH$ if and only if $\displaystyle xy^{-1} \in H$
$\displaystyle xH = yH\ \Leftrightarrow\ \color{red}y^{-1}x\color{black}\in H$

4. Originally Posted by JaneBennet
$\displaystyle xH = yH\ \Leftrightarrow\ \color{red}y^{-1}x\color{black}\in H$
It makes no difference. Because if $\displaystyle y^{-1} x\in H$ then $\displaystyle (y^{-1} x)^{-1} = x^{-1}y\in H$.

EDIT: Okay, I see I wrote it the other way around.

5. ok thanks, i figured that out!!... but what about if it's "If aH = bH, then a^2 H = b^2 H"?