# Product is Left Identity Therefore Left Cancellable

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## Contents

## Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $e_L \in S$ be a left identity of $S$.

Let $a \in S$ such that:

- $\exists b \in S: b \circ a = e_L$

Then $a$ is left cancellable in $\struct {S, \circ}$.

## Proof

Let $x, y \in S$ be arbitrary.

Then:

\(\displaystyle a \circ x\) | \(=\) | \(\displaystyle a \circ y\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle b \circ \paren {a \circ x}\) | \(=\) | \(\displaystyle b \circ \paren {a \circ y}\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \paren {b \circ a} \circ x\) | \(=\) | \(\displaystyle \paren {b \circ a} \circ y\) | as $\struct {S, \circ}$ is a semigroup, $\circ$ is associative | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle e_L \circ x\) | \(=\) | \(\displaystyle e_L \circ y\) | by hypothesis | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | Definition of Left Identity |

The result follows by definition of left cancellable.

$\blacksquare$

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $7.11$