Cancellable iff Regular Representation Injective

Theorem

Let $\left({S, \circ}\right)$ be a semigroup.

Then $a \in S$ is cancellable iff:

the left regular representation $\lambda_a \left({x}\right)$ is injective

and

the right regular representation $\rho_a \left({x}\right)$ is injective.

Proof

Left Cancellable

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, $\lambda_a \left({x}\right) = a \circ x$.

Thus:

$\lambda_a \left({x}\right) = \lambda_a \left({y}\right) \implies x = y$

and so the left regular representation is injective.

$\Box$

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

Then:

$\lambda_a \left({x}\right) = \lambda_a \left({y}\right) \implies x = y$

From the definition of the left regular representation:

$\lambda_a \left({x}\right) = a \circ x$

Thus:

$a \circ x = a \circ y \implies x = y$

and so $a$ is left cancellable.

$\blacksquare$

Right Cancellable

Suppose $a \in S$ is right cancellable.

Then $x \circ a = y \circ a \implies x = y$.

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.

$\Box$

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:

$\rho_a \left({x}\right) = a \circ x$

Thus:

$a \circ x = a \circ y \implies x = y$

and so $a$ is right cancellable.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 7$