Subgroup Generated by Commuting Elements is Abelian

Theorem
Let $\struct {G, circ}$ be a group.

Let $S \subseteq G$ such that:
 * $\forall x, y \in S: x \circ y = y \circ x$

Then the subgroup generated by $S$ is abelian.

Proof
Let $H = \gen S$ denote the subgroup generated by $S$.

Let $a, b \in H$.

Then:
 * $a = s_1$
 * $b = s_2$

for some words $s_1, s_2$ of the set of words $\map W S$ of $S$.

Then:

Hence the result, by definition of abelian group.