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 already closed and associative.
It remains to be shown that it has an identity and that every element of $S$ has an inverse in $S$.
First we show that $\struct {S, \circ}$ has an identity.
Choose $a \in S$.
Let the mapping $\lambda_a: S \to S$ be the left regular representation of $\struct {S, \circ}$ with respect to $a$.
- all elements of $S$ are cancellable
- $S$ is finite.
By Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection, $\lambda_a$ is a bijection.
Hence $a \circ e = a$ for some $e \in S$.
Let $x \in S$.
Then because of cancellability:
\(\ds a \circ e \circ x\) | \(=\) | \(\ds a \circ x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e \circ x\) | \(=\) | \(\ds x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ e \circ x\) | \(=\) | \(\ds x \circ x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ e\) | \(=\) | \(\ds x\) |
Thus $e$ is the identity.
The existence of inverses comes from the surjectivity of $\lambda_a$.
As $\lambda_a$ is surjective:
- $\exists y \in S: \map {\lambda_a} y = e$
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, $\struct {S, \circ}$ 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 $3$