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:

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:

Then we have:

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:

Then we have:

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:

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

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.