Set of Words Generates Group

Theorem
Let $S \subseteq G$ where $G$ is a group.

Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.

Then $\left \langle {S} \right \rangle = W \left({\hat S}\right)$, where $W \left({\hat S}\right)$ is the set of words of $\hat S$.

Proof
Let $H = \left \langle {S} \right \rangle$ where $S \subseteq G$.

$H$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$. Thus $\hat S \subseteq H$.

By the closure axiom, $H$ must also contain all products of any finite number of elements of $\hat S$. Thus $W \left({\hat S}\right) \subseteq H$.

Now we prove that $W \left({\hat S}\right) \le G$.

By the Two-Step Subgroup Test:

Let $x, y \in W \left({\hat S}\right)$. As $x$ and $y$ are both products of a finite number of elements of $\hat S$, it follows that so is their product $x y$, thus $x y \in W \left({\hat S}\right)$ and the closure axiom is satisfied.

Let $x = s_1 s_2 \ldots s_n \in W \left({\hat S}\right)$.

Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in W \left({\hat S}\right)$.

Thus the conditions of the Two-Step Subgroup Test are fulfilled, and $W \left({\hat S}\right) \le G$.

Thus $W \left({\hat S}\right)$ is the subgroup of $G$ generated by $S$.