Cancellable Finite Semigroup is Group

Theorem
Let $$\left({S, \circ}\right)$$ be a finite semigroup in which all elements are cancellable.

Then $$\left({S, \circ}\right)$$ is a group.

Proof
As $$S$$ is a semigroup, it is already closed and associative.

We just need to show that it has an identity and that every element has an inverse.


 * First we show that $$S$$ has an identity.

Choose $$a \in S$$. Let the mapping $$\lambda_a$$ be the left regular representation of $$\left({S, \circ}\right)$$ with respect to $$a$$.

Because all elements are cancellable, in particular, so is $$a$$, so $\lambda_a$ is injective.

As $$S$$ is finite, $\lambda_a$ is also surjective.

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

Let $$x \in S$$. Then because of cancellability:

$$ $$ $$ $$

Thus $$e$$ is the identity.


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

As $$\lambda_a$$ is surjective, $$\exists y \in S: \lambda_a \left({y}\right) = e$$.

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 the right inverse version of Left Inverse for All is Right 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 it's a group.

Comment
Note that the same does not apply to infinite semigroups.

Consider the semigroup $$\left({\N, +}\right)$$.
 * It is closed: the sum of two natural numbers is another natural number.
 * Addition is associative, so it is definitely a semigroup
 * $$b + a = c + a \implies b = c$$, so all elements of $$\left({\N, +}\right)$$ are cancellable.

But $$\left({\N, +}\right)$$ is not a group, as (apart from $$0$$) no element has an inverse.