# Right Cosets are Equal iff Left Cosets by Inverse are Equal

## Theorem

Let $G$ be a group whose identity is $e$.

Let $H$ be a subgroup of $G$.

Let $g_1, g_2 \in G$.

Then:

$H g_1 = H g_2 \iff {g_1}^{-1} H = {g_2}^{-1} H$

where:

${g_1}^{-1}$ and ${g_2}^{-1}$ denote the inverses of $g_1$ and $g_2$ in $G$
$H g_1$ and $H g_2$ denote the right cosets of $H$ by $g_1$ and $g_2$ respectively
${g_1}^{-1} H$ and ${g_2}^{-1} H$ denote the left cosets of $H$ by ${g_1}^{-1}$ and ${g_2}^{-1}$ respectively.

## Proof

 $\ds H g_1$ $=$ $\ds H g_2$ $\ds \leadstoandfrom \ \$ $\ds g_1 {g_2}^{-1}$ $\in$ $\ds H$ Right Cosets are Equal iff Product with Inverse in Subgroup $\ds \leadstoandfrom \ \$ $\ds \paren { {g_1}^{-1} }^{-1} {g_2}^{-1}$ $\in$ $\ds H$ Inverse of Group Inverse $\ds \leadstoandfrom \ \$ $\ds {g_1}^{-1} H$ $=$ $\ds {g_2}^{-1} H$ Left Cosets are Equal iff Product with Inverse in Subgroup

$\blacksquare$