# Modern Algebra: Basic problem dealing with Cosets

## Homework Statement

If H is a subgroup of G and Ha = bH for elements a and b in G, show that aH = Hb.

None needed

## The Attempt at a Solution

I've basically just been fiddling around by right and left side multiplication of inverses and what not and can't seem to get it in the right form.. anyone want to guide me in the right direction?

My attempt:
Ha = bH --> (b^-1)Ha(a^-1) = (b^-1)bH(a^-1) = (b^-1)H = H(a^-1)

pasmith
Homework Helper
Involving the inverses of $a$ and $b$ seems unproductive, since you know nothing about them beyond that they exist.

Let $h \in H$. If $Ha = bH$, what can you say about the element $ha \in bH$?

HallsofIvy
I wouldn't try to do it that way. I would, instead, use individual elements of H. And then use inverse of the elements of H, not a and b. "Ha= bH" means that, for any x in H, there exist y in H such that xa= by. Multiply on both sides, on the right, by $y^{-1}$ to get $xay^{-1}= b$. Now multiply both sides, on the left, by $x^{-1}$: $ay^{-1}= x^{-1}b$.