Product is Left Identity Therefore Left Cancellable

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Theorem

Let $\left({S, \circ}\right)$ 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 $\left({S, \circ}\right)$.


Proof

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


Then:

\(\displaystyle a \circ x\) \(=\) \(\displaystyle a \circ y\)
\(\displaystyle \implies \ \ \) \(\displaystyle b \circ \left({a \circ x}\right)\) \(=\) \(\displaystyle b \circ \left({a \circ y}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({b \circ a}\right) \circ x\) \(=\) \(\displaystyle \left({b \circ a}\right) \circ y\) as $\left({S, \circ}\right)$ 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