# Regular Representation of Invertible Element is Permutation

## Theorem

Let $\struct {S, \circ}$ 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 $a \in \struct {S, \circ}$ 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.

### Injectivity

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

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

$\Box$

### Surjectivity

Let $y \in S$.

Then:

\((1):\quad\) | \(\displaystyle y\) | \(=\) | \(\displaystyle \paren {a \circ a^{-1} } \circ y\) | (as $a$ is invertible) | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \circ \paren {a^{-1} \circ y}\) | Associativity of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lambda_a \paren {a^{-1} \circ y}\) | Definition of $\lambda_a$ |

\((2):\quad\) | \(\displaystyle y\) | \(=\) | \(\displaystyle y \circ \paren {a^{-1} \circ a} \circ y\) | (as $a$ is invertible) | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {y \circ a^{-1} } \circ a\) | Associativity of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \rho_a \paren {y \circ a^{-1} }\) | Definition of $\rho_a$ |

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

$\Box$

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

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 7$: Theorem $7.1$