Regular Representation of Invertible Element is Permutation

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Theorem

Let $\struct {S, \circ}$ be a monoid.

Let $a \in S$ be invertible.


Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $S$.


Proof

Suppose $a \in \struct {S, \circ}$ is invertible.

A permutations is a bijection from a set to itself.

As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they are bijections.

To do that we can show that they are both injective and surjective.


Injectivity

From Invertible Element of Monoid is Cancellable, as $a$ is invertible, it is also cancellable.

From Cancellable iff Regular Representations Injective, it follows that both $\lambda_a$ and $\rho_a$ are injective.

$\Box$


Surjectivity

Let $y \in S$.


Then:

\(\text {(1)}: \quad\) \(\ds y\) \(=\) \(\ds \paren {a \circ a^{-1} } \circ y\) by hypothesis: $a$ is invertible
\(\ds \) \(=\) \(\ds a \circ \paren {a^{-1} \circ y}\) Monoid Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds \lambda_a \paren {a^{-1} \circ y}\) Definition of Left Regular Representation


\(\text {(2)}: \quad\) \(\ds y\) \(=\) \(\ds y \circ \paren {a^{-1} \circ a} \circ y\) by hypothesis: $a$ is invertible
\(\ds \) \(=\) \(\ds \paren {y \circ a^{-1} } \circ a\) Monoid Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds \rho_a \paren {y \circ a^{-1} }\) Definition of Right Regular Representation


Thus both $\lambda_a$ and $\rho_a$ are surjective.

$\Box$


So $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections, and thus permutations of $S$.

$\blacksquare$


Sources

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