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.

$\ds H g_1$

$=$

$\ds H g_2$

$\ds \leadstoandfrom \ \ $

$\ds g_1 {g_2}^{-1}$

$\in$

$\ds H$

$\ds \leadstoandfrom \ \ $

$\ds \paren { {g_1}^{-1} }^{-1} {g_2}^{-1}$

$\in$

$\ds H$

$\ds \leadstoandfrom \ \ $

$\ds {g_1}^{-1} H$

$=$

$\ds {g_2}^{-1} H$

$\blacksquare$

1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $10$