Right Cancellable iff Right Regular Representation Injective

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

Then $a \in S$ is right cancellable the right regular representation $\map {\rho_a} x$ is injective.

Proof
Suppose $a \in S$ is right cancellable.

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

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

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

and so the right regular representation is injective.

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

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

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

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

and so $a$ is right cancellable.

Also see

 * Left Cancellable iff Left Regular Representation Injective
 * Cancellable iff Regular Representations Injective