# 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

\(\displaystyle H g_1\) | \(=\) | \(\displaystyle H g_2\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle g_1 {g_2}^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ Right Cosets are Equal iff Product with Inverse in Subgroup | $\quad$ | ||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren { {g_1}^{-1} }^{-1} {g_2}^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ Inverse of Group Inverse | $\quad$ | ||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle {g_1}^{-1} H\) | \(=\) | \(\displaystyle {g_2}^{-1} H\) | $\quad$ Left Cosets are Equal iff Product with Inverse in Subgroup | $\quad$ |

$\blacksquare$

## Sources

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