Right Cancellable iff Right Regular Representation Injective

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Theorem

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

Then $a \in S$ is right cancellable if and only if 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.

$\Box$


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.

$\blacksquare$


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