# 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:

\(\text {(1)}: \quad\) | \(\ds y\) | \(=\) | \(\ds \paren {a \circ a^{-1} } \circ y\) | by hypothesis: $a$ is invertible | ||||||||||

\(\ds \) | \(=\) | \(\ds a \circ \paren {a^{-1} \circ y}\) | Monoid Axiom $\text S 1$: Associativity | |||||||||||

\(\ds \) | \(=\) | \(\ds \lambda_a \paren {a^{-1} \circ y}\) | Definition of Left Regular Representation |

\(\text {(2)}: \quad\) | \(\ds y\) | \(=\) | \(\ds y \circ \paren {a^{-1} \circ a} \circ y\) | by hypothesis: $a$ is invertible | ||||||||||

\(\ds \) | \(=\) | \(\ds \paren {y \circ a^{-1} } \circ a\) | Monoid Axiom $\text S 1$: Associativity | |||||||||||

\(\ds \) | \(=\) | \(\ds \rho_a \paren {y \circ a^{-1} }\) | Definition of Right Regular Representation |

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 permutations of $S$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Theorem $7.1$