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

$\blacksquare$


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