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 $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$.

If $T$ is closed under taking inverses, then $W \left({T}\right)$ is a subgroup of $G$.

Let $H = \gen S$ 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 a finite number of elements of $\hat S$.

Thus $\map W {\hat S} \subseteq H$.

Now we prove that $\map W {\hat S} \le G$.

Let $x, y \in \map W {\hat S}$.

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 $x y \in \map W {\hat S}$ and Group Axiom $G \, 0$: Closure is satisfied.

Let $x = s_1 s_2 \ldots s_n \in \map W {\hat S}$.

Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in \map W {\hat S}$.

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

Thus $\map W {\hat S}$ is the subgroup of $G$ generated by $S$.

$\blacksquare$