Finite Monoid with Left Cancellable Operation is Group

Theorem

Let $\struct {S, \circ}$ be a finite monoid.

Let $\circ$ be a left cancellable operation.

Then $\struct {S, \circ}$ is a group.

Proof

Group Axiom $\text G 0$: Closure, Group Axiom $\text G 1$: Associativity and Group Axiom $\text G 2$: Existence of Identity Element are satisfied by dint of $\paren {S, \circ}$ being a monoid.

Recall the definition of left cancellable operation:

$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$

Let $\lambda_c: S \to S$ be the left regular representation of $\struct {S, \circ}$ with respect to $c$.

By Left Cancellable iff Left Regular Representation Injective, $\lambda_c$ is an injection.

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

Thus $a \circ b = e$ has a unique solution for all $a \in S$.

That is, Group Axiom $\text G 3$: Existence of Inverse Element holds on $S$.

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Proposition $2$