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 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 $a$.

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

We have that:
 * $\struct {S, \circ}$ is non-empty
 * all elements of $S$ are cancellable
 * $S$ is finite.


 * Existence of Identity Element

Let $x \in S$ be arbitrary.

Then:

Thus $e$ is an identity element.

By Identity is Unique, there is only one such.


 * Existence of Inverse Elements

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.

Also see

 * Cancellable Infinite Semigroup is not necessarily Group