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 Left Regular Representation

 * First we show that $\lambda_a$ is injective.

From Invertible also Cancellable, as $a$ is invertible, it is also cancellable.

As $a$ is cancellable, it is also left cancellable.

From Cancellable iff Regular Representation Injective, it follows that $\lambda_a$ is injective.


 * Now we show it is surjective.

Let $y \in S$. Then:

Thus $\lambda_a$ is surjective.


 * So $\lambda_a$ is injective and surjective, and therefore a bijection, and thus a permutations of $S$.

Proof for Right Regular Representation
A similar argument shows that $\rho_a$ has the same properties.


 * First we show that $\rho_a$ is injective.

From Invertible also Cancellable, as $a$ is invertible, it is also cancellable.

As $a$ is cancellable, it is also right cancellable.

From Cancellable iff Regular Representation Injective, it follows that $\rho_a$ is injective.


 * Now we show it is surjective.

Let $y \in S$. Then:

Thus $\rho_a$ is surjective.


 * So $\rho_a$ is injective and surjective, and therefore a bijection, and thus a permutations of $S$.