Finite Monoid with Right Cancellable Operation is Group

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

Let $\circ$ be a right cancellable operation.

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

Proof
Group axioms $\text G 0$, $\text G 1$ and $\text G 2$ are satisfied by dint of $\struct {S, \circ}$ being a monoid.

Recall the definition of right cancellable operation:
 * $\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$

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

By Right Cancellable iff Right Regular Representation Injective, $\rho_c$ is an injection.

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

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

That is, group axiom $\text G 3$ holds on $S$.