Cancellable iff Regular Representations Injective

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Theorem

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


Then $a \in S$ is cancellable if and only if:

the left regular representation $\map {\lambda_a} x$ is injective

and:

the right regular representation $\map {\rho_a} x$ is injective.


Proof

Left Cancellable

Suppose $a \in S$ is left cancellable.

Then:

$\forall x, y \in S: a \circ x = a \circ y \implies x = y$

From the definition of the left regular representation:

$\map {\lambda_a} x = a \circ x$

Thus:

$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$

and so the left regular representation is injective.

$\Box$


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

Then:

$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$

From the definition of the left regular representation:

$\map {\lambda_a} x = a \circ x$

Thus:

$\forall x, y \in S: a \circ x = a \circ y \implies x = y$

and so $a$ is left cancellable.

$\blacksquare$


Right Cancellable

Suppose $a \in S$ is right cancellable.

Then:

$\forall x, y \in S: x \circ a = y \circ a \implies x = y$

From the definition of the right regular representation:

$\map {\rho_a} x = x \circ a$

Thus:

$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$

and so the right regular representation is injective.

$\Box$


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

Then:

$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$

From the definition of the right regular representation:

$\map {\rho_a} x = x \circ a$

Thus:

$\forall x, y \in S: x \circ a = y \circ a \implies x = y$

and so $a$ is right cancellable.

$\blacksquare$


Sources

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