# Regular Representations in Semigroup are Permutations then Structure is Group

## Theorem

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

For $a \in S$, let $\lambda_a: S \to S$ and $\rho_a: S \to S$ denote the left regular representation and right regular representation with respect to $a$ respectively:

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

For all $a$ in $S$, let $\lambda_a$ be a permutation on $S$.

Let there exist $b$ in $S$ such that $\rho_b$ is a permutation on $S$.

Then $\struct {S, \circ}$ is a group.

## Proof

We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.

In particular this applies to $b$.

So:

$\lambda_b$ is a permutation on $S$
$\rho_b$ is a permutation on $S$
$\struct {S, \circ}$ has an identity element
$b$ is invertible in $\struct {S, \circ}$.

It remains to be shown that $a$ is invertible in $S$ for all $a \in S$.

Let the identity element of $\struct {S, \circ}$ be $e$.

We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.

Hence:

 $\ds \forall a \in S: \exists a' \in S: \,$ $\ds e$ $=$ $\ds \map {\lambda_a} {a'}$ $\ds$ $=$ $\ds a \circ a'$

Thus for all $a$ in $S$, $a$ has a left inverse $a'$.

Now as $e$ is an identity element of $S$, it is by definition a left identity.

Hence from Left Inverse for All is Right Inverse, $a'$ is also a right inverse for $a$.

That is, every $a \in S$ has an element which is both a left inverse and a right inverse.

That is, every $a \in S$ has an inverse.

So we have that $\struct {S, \circ}$ is a semigroup such that:

$\struct {S, \circ}$ has an identity element
every element of $S$ has an inverse.

Hence, by definition, $\struct {S, \circ}$ is a group.

$\blacksquare$