# Cancellable Finite Semigroup is Group

## Contents

## Theorem

Let $\struct {S, \circ}$ be a non-empty finite semigroup in which all elements are cancellable.

Then $\struct {S, \circ}$ is a group.

## Proof

As $\struct {S, \circ}$ is a semigroup, it is already closed and associative.

It remains to be shown that it has an identity and that every element of $S$ has an inverse in $S$.

First we show that $\struct {S, \circ}$ has an identity.

Choose $a \in S$.

Let the mapping $\lambda_a: S \to S$ be the left regular representation of $\struct {S, \circ}$ with respect to $a$.

- all elements of $S$ are cancellable

- $S$ is finite.

By Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection, $\lambda_a$ is a bijection.

Hence $a \circ e = a$ for some $e \in S$.

Let $x \in S$.

Then because of cancellability:

\(\displaystyle a \circ e \circ x\) | \(=\) | \(\displaystyle a \circ x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle e \circ x\) | \(=\) | \(\displaystyle x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \circ e \circ x\) | \(=\) | \(\displaystyle x \circ x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x \circ e\) | \(=\) | \(\displaystyle x\) |

Thus $e$ is the identity.

The existence of inverses comes from the surjectivity of $\lambda_a$.

As $\lambda_a$ is surjective:

- $\exists y \in S: \map {\lambda_a} y = e$

That is:

- $a \circ y = e$

So we see that $y$ acts as a right inverse for $a$.

This is the case for any $a \in S$: all of them have some right inverse.

So, from Right Inverse for All is Left Inverse, each of these elements is also a left inverse, and therefore an inverse.

Thus $S$ is closed, associative, has an identity and every element has an inverse.

So, by definition, $\struct {S, \circ}$ is a group.

$\blacksquare$

## Also see

## Sources

- 1964: Walter Ledermann:
*Introduction to the Theory of Finite Groups*(5th ed.) ... (previous) ... (next): $\S 4$: Alternative Axioms for Finite Groups: Theorem $1$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \delta$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $3$