Right Cancellable iff Right Regular Representation Injective

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

Then $a \in S$ is right cancellable the right regular representation $\rho_a \left({x}\right)$ 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:
 * $\rho_a \left({x}\right) = x \circ a$

Thus:
 * $\rho_a \left({x}\right) = \rho_a \left({y}\right) \implies x = y$

and so the right regular representation is injective.

Suppose $\rho_a \left({x}\right)$ is injective.

Then:
 * $\rho_a \left({x}\right) = \rho_a \left({y}\right) \implies x = y$

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

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