Regular Representations wrt Element are Permutations then Element is Invertible

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Theorem

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

Let $\lambda_a: S \to S$ and $\rho_a: S \to S$ be the left regular representation and right regular representation with respect to $a$ respectively:

\(\ds \forall x \in S: \, \) \(\ds \map {\lambda_a} x\) \(=\) \(\ds a \circ x\)
\(\ds \forall x \in S: \, \) \(\ds \map {\rho_a} x\) \(=\) \(\ds x \circ a\)

Let both $\lambda_a$ and $\rho_a$ be permutations on $S$.


Then there exists an identity element for $\circ$ and $a$ is invertible.


Proof

We have that $\rho_a$ is a permutation on $S$.

Hence:

\(\ds \exists g \in S: \, \) \(\ds a\) \(=\) \(\ds \map {\rho_a} g\)
\(\ds \) \(=\) \(\ds g \circ a\) Definition of Right Regular Representation

Then we have:

\(\ds \forall b \in S: \, \) \(\ds \paren {b \circ g} \circ a\) \(=\) \(\ds b \circ \paren {g \circ a}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds b \circ a\) from above
\(\ds \leadsto \ \ \) \(\ds b \circ g\) \(=\) \(\ds b\) Right Cancellable iff Right Regular Representation Injective

which demonstrates that $g$ is a right identity for $\circ$.


In the same way, we have that $\lambda_a$ is also a permutation on $S$.

Hence:

\(\ds \exists g \in S: \, \) \(\ds a\) \(=\) \(\ds \map {\lambda_a} g\)
\(\ds \) \(=\) \(\ds a \circ g\) Definition of Left Regular Representation

Then we have:

\(\ds \forall b \in S: \, \) \(\ds \paren {a \circ g} \circ b\) \(=\) \(\ds a \circ \paren {g \circ b}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds a \circ b\) from above
\(\ds \leadsto \ \ \) \(\ds g \circ b\) \(=\) \(\ds b\) Left Cancellable iff Left Regular Representation Injective

which demonstrates that $g$ is a left identity for $\circ$.


So, by definition, $g$ is an identity element for $\circ$.


Again, we have that $\rho_a$ is a permutation on $S$, and so:

\(\ds \exists h \in S: \, \) \(\ds g\) \(=\) \(\ds \map {\rho_a} h\)
\(\ds \) \(=\) \(\ds h \circ a\) Definition of Right Regular Representation

and that $\lambda_a$ is also a permutation on $S$, and so:

\(\ds \exists h \in S: \, \) \(\ds g\) \(=\) \(\ds \map {\lambda_a} h\)
\(\ds \) \(=\) \(\ds a \circ h\) Definition of Right Regular Representation

So $h$ is both a left inverse and a right inverse for $a$.

Hence by definition $h$ is an inverse for $a$.

Hence $a$ is invertible by definition.

$\blacksquare$


Sources

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