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 $\hat S$.
Corollary
If $T$ is closed under taking inverses, then $\map W T$ is a subgroup of $G$.
Proof
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 $\map W {\hat S} \le G$.
By the Two-Step Subgroup Test:
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 $\text G 0$: 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 $S$.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.3$. Subgroup generated by a subset
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $20$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{K}$
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$