Left Cancellable iff Left Regular Representation Injective

Theorem
Let $\struct {S, \circ}$ be an algebraic structure.

Then $a \in S$ is left cancellable the left regular representation $\map {\lambda_a} x$ 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:
 * $\map {\lambda_a} x = a \circ x$

Thus:
 * $\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$

and so the left regular representation is injective.

Suppose $\map {\lambda_a} x$ is injective.

Then:
 * $\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$

From the definition of the left regular representation:
 * $\map {\lambda_a} x = a \circ x$

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