Left Cancellable iff Left Regular Representation Injective

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure.

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

Proof
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:
 * $\lambda_a \left({x}\right) = \lambda_a \left({y}\right) \implies x = y$

and so the left regular representation is injective.

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:
 * $a \circ x = a \circ y \implies x = y$

and so $a$ is left cancellable.

Also see

 * Right Cancellable iff Right Regular Representation Injective
 * Cancellable iff Regular Representations Injective