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 $$\hat S$$.

Corollary
It follows directly that, if $$T \subseteq G$$ where $$T$$ is closed under taking inverses, then $$W \left({T}\right) \le G$$.

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:

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 $$x, y \in W \left({\hat S}\right)$$.


 * 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 $S$.

Proof of Corollary
This follows directly from the fact that $$T$$ has the same properties as $$\hat S$$.