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 $$S$$.