Left Cancellable iff Left Regular Representation Injective

Theorem

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

Then $a \in S$ is left cancellable if and only if the left regular representation $\map {\lambda_a} x$ is injective.


Proof

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.

$\blacksquare$


Also see


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
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