- #1

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## Homework Statement

Let G be a group, and H a subgroup of G. Let a and b denote elements of G. Prove the following:

1. ##Ha = Hb## iff ##ab^{-1} \epsilon H##.

## Homework Equations

Let ##e_H## be the identity element of H.

## The Attempt at a Solution

Proof: <= Suppose ##ab^{-1} \epsilon H##. Then ##abb^{-1} = a \epsilon Hb##. Since ##H## is a subgroup, ##e_H \epsilon H##. So ##e_Ha = a \epsilon Ha##. Since ##Ha## and ##Hb## share a common element, we must have ##Ha = Hb##.

=> Suppose ##Ha = Hb##. Since ##e_ha = a \epsilon Ha##, we have ##a \epsilon Hb##. So there must be some solution to ##a = xb## where ##x \epsilon H##. Observe, ##a = (ab^{-1})b## so ##ab^{-1} \epsilon H## necessarily?

I'm not sure why ##ab^{-1}## would be the only solution. I think it has something to do with, if we wrote out the Cayley table for H, and there were two solutions for a = xb, then it wouldn't look right. Because in a Cayley table every element is in every row and every column. So we'd have an element missing in a row.

Also I didn't use the fact H is abelian so did I missing something there?