# 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:

$\struct {S, \circ}$ has an identity
every element of $S$ has an inverse in $S$.

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 $a$.

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.

$\blacksquare$