Monoid of Self-Inverse Elements is Abelian Group

Theorem
Let $\left({S, \circ}\right)$ be a monoid such that:
 * $\forall x \in S: x \circ x = e$

where $e$ is the identity element of $\left({S, \circ}\right)$.

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

Proof
From Equivalence of Definitions of Self-Inverse, $x \circ x = e \implies x = x^{-1}$.

From Inverse in Monoid is Unique, it follows that every element of $\left({S, \circ}\right)$ has a unique inverse.

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

From All Elements Self-Inverse then Abelian, it follows that $\left({S, \circ}\right)$ is an abelian group.