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.
$\blacksquare$
Also see
- Left Cancellable iff Left Regular Representation Injective
- Cancellable iff Regular Representations Injective
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$