Def/Generate (group)

 =Construction= Suppose that $G$ is a group and $S$ is a subset of $G$. Let $M$ be the set of subgroups of $G$ which contain $S$ as a subset.

Then, $M$ is nonempty, since $G$ is a subgroup of $G$ which contains $S$, and hence $G \in M$.

Define $ \langle S \rangle = \bigcap M$ (the intersection of all subgroups containing $S$). Then, we find that:
 * $\langle S \rangle$ is a subgroup of $G$, since intersections of subgroups are subgroups
 * If $H$ is a subgroup of $G$, and $H$ contains $S$ (so that $H \in M$), then $H$ contains $\langle S \rangle$, using basic properties of intersections.

=Construction= Suppose that $G$ is a group, and $S$ is a subset of $G$. Define another subset of $G$ by: $$S^{-1} = \{ g \in G \mbox{ such that } \exists s \in S, g = s^{-1} \}.$$ Define $L = S \cup S^{-1}$. A defines::word, using letters from $L$, is a finite sequence $\ell_1, \ldots, \ell_k$ of elements of $L$. Each word can be viewed as an element of $G$, via composition: $$w = \ell_1 \circ \cdots \circ \ell_k.$$

In this case, the subgroup $\langle S \rangle$ can be described as the set of words, using letters from $L$. The elements of $\langle S \rangle$ are elements of $G$ which can be expressed as compositions of elements of $S$ and their inverses.

We have seen that $W$ is a subgroup of $G$, and clearly $S \subset W$, since $S \subset L \subset W$. Conversely,

The previous two proofs demonstrate that $W = \langle S \rangle$.

=Metadata=