Regular Representation of Invertible Element is Permutation

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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 $S$.


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.


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.



Let $y \in S$.


\(\text {(1)}: \quad\) \(\ds y\) \(=\) \(\ds \paren {a \circ a^{-1} } \circ y\) 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.


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