# 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:

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

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

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

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

Thus $e$ is the identity.

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

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

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

$\blacksquare$

## 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.
- 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.

## Sources

- Allan Clark:
*Elements of Abstract Algebra*(1971)... (previous)... (next): $\S 26 \delta$ - Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*(1978)... (previous)... (next): Exercise $6.3$