# Cancellable Finite Semigroup is Group

## 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 *a fortiori* closed and associative.

It remains to be shown that:

Let $a \in S$ be arbitrary.

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

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

We have by hypothesis that:

- $\struct {S, \circ}$ is non-empty
- all elements of $S$ are cancellable
- $S$ is finite.

- Existence of Identity Element

\(\ds \exists e \in S: \, \) | \(\ds \map {\lambda_a} e\) | \(=\) | \(\ds a\) | as $\lambda_a$ is a surjection | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \circ e\) | \(=\) | \(\ds a\) | Definition of Left Regular Representation |

Let $x \in S$ be arbitrary.

Then:

\(\ds a \circ \paren {e \circ x}\) | \(=\) | \(\ds a \circ x\) | Semigroup Axiom $\text S 0$: Closure | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds e \circ x\) | \(=\) | \(\ds x\) | Definition of Cancellable Element | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x \circ \paren {e \circ x}\) | \(=\) | \(\ds x \circ x\) | Semigroup Axiom $\text S 0$: Closure | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ e} \circ x\) | \(=\) | \(\ds x \circ x\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x \circ e\) | \(=\) | \(\ds x\) | Definition of Cancellable Element |

Thus $e$ is an identity element.

By Identity is Unique, there is only one such.

- Existence of Inverse Elements

\(\ds \exists y \in S: \, \) | \(\ds \map {\lambda_a} y\) | \(=\) | \(\ds e\) | as $\lambda_a$ is a surjection | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds a \circ y\) | \(=\) | \(\ds e\) | Definition of Left Regular Representation |

Hence $y$ acts as a right inverse for $a$.

As $a$ is arbitrary, it follows that all $a \in S$ have a right inverse.

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): Chapter $\text {I}$: The Group Concept: $\S 4$: 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 $3$