# Statement(s) > [!NOTE] Theorem (Set of all words over $S\subset G$ is a subgroup of $G$) > Let $G$ be a [[Groups|group]]. Let $S$ be a [[Subsets|subset]] of $G$ and $\langle S\rangle$ denote the [[Generated Subgroup|subgroup generated]] by $S$ (the set of all words over $S$). Then $\langle S\rangle$ is indeed a [[Subgroup|subgroup]] of $G.$ # Proof(s) ###### Proof of Set of all words over $S\subset G$ is a subgroup of $G$ \[MA268\]: Given $S \subset G,$ we define the subgroup generated by $S$ by $\langle S \rangle = \{ s_{1}^{\varepsilon_{1}}s_{2}^{\varepsilon_{2}}\cdots s_{m}^{\varepsilon_{m}}\mid m\geq 0, s_{i} \in S, \varepsilon_{i}\in \{-1,1 \} \}.$ The identity of $G,$ $1_{G},$ is clearly a word in $S$ (when $m=0$). Any word is a finite product of the elements of $S$ or their inverses so the product of two words is again a word. Using [[Inverse of the Product of Group Elements]], $(s_{1}^{\varepsilon_{1}}s_{2}^{\varepsilon_{2}}\cdots s_{m}^{\varepsilon_{m}})^{-1}=s_{m}^{-\varepsilon_{m}} \cdots s_{2}^{-\varepsilon_{2}}s_{1}^{-\varepsilon_{1}}$which is also an element of $\langle S\rangle$ since for all $\varepsilon_{i}\in \{ -1,1 \}$ we have also $-\varepsilon_{i}\in \{-1,1\}.$ It follows from [[Two-Step Subgroup Test]] that $\langle S\rangle$ is indeed a subgroup of $G.$ $\blacksquare$