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

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## 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

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

$\blacksquare$