Regular Representation of Invertible Element is Permutation

Theorem
Let $\left({S, \circ}\right)$ 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 $S$.

Proof
Suppose $a \in \left({S, \circ}\right)$ 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.

Proof for Injective
From Invertible Elements of Semigroup Also Cancellable, as $a$ is invertible, it is also cancellable.

From Cancellable iff Regular Representation Injective, it follows that both $\lambda_a$ and $\rho_a$ are injective.

Proof for Surjective
Let $y \in S$.

Then:

Thus both $\lambda_a$ and $\rho_a$ are surjective.

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