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.