Cancellable iff Regular Representations Injective

Theorem
Let $\left({S, \circ}\right)$ be a semigroup.

Then:


 * $a \in S$ is left cancellable iff the left regular representation $\lambda_a \left({x}\right)$ is injective.


 * $a \in S$ is right cancellable iff the right regular representation $\rho_a \left({x}\right)$ is injective.

Proof

 * Left cancellable implies injective:

Suppose $a \in S$ is left cancellable. Then $a \circ x = a \circ y \implies x = y$.

From the definition of the left regular representation, $\lambda_a \left({x}\right) = a \circ x$.

Thus $\lambda_a \left({x}\right) = \lambda_a \left({y}\right) \implies x = y$


 * Injective implies left cancellable:

Suppose $\lambda_a \left({x}\right)$ is injective. Then $\lambda_a \left({x}\right) = \lambda_a \left({y}\right) \implies x = y$.

From the definition of the left regular representation, $\lambda_a \left({x}\right) = a \circ x$.

Thus $a \circ x = a \circ y \implies x = y$.

The proof for right cancellability is the same as for left.