# Regular Representations wrt Element are Permutations then Element is Invertible

## 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.

$\blacksquare$