The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups

Let $G$ be a group (Group and abelian group) and $S\subseteq G$ a subset. The set of subgroups of $G$ containing $S$ is nonempty, since $G$ itself is such a subgroup, so its intersection is a subgroup of $G$ by The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$. That intersection is the subgroup generated by $S$,

$⟨S⟩:=\bigcap \{H:H\le G\text{ and }S\subseteq H\}\mathrm{.}$

It contains $S$, being an intersection of sets each containing $S$, and it is contained in every subgroup of $G$ that contains $S$; so it is the smallest subgroup of $G$ containing $S$, and these two properties determine it uniquely. The elements of $S$ are called generators.

For a single element $g\in G$ we write $⟨g⟩:=⟨\{g\}⟩$ and call it the cyclic subgroup generated by $g$. A group $G$ is cyclic when $G=⟨g⟩$ for some $g\in G$.

By convention $⟨\mathrm{\varnothing }⟩=\{e\}$: the trivial subgroup is the smallest subgroup containing the empty set, and this is a consequence of the definition, not a stipulation, since every subgroup contains $e$ (Subgroup).

• Two descriptions, one object. The definition above is "from outside": cut down from all subgroups containing $S$. There is also a description "from inside", as the set of all finite products of generators and their inverses. For a single generator that inside description is $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, proved in $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian. The general case belongs to a later page; nothing here needs it.

• Cyclic does not mean finite. $(\mathbb{Z},+)$ is cyclic, generated by $1$, and infinite; the generator may also fail to be unique, since $-1$ generates it too.

• Every cyclic group is abelian ($⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian), so a non-abelian group is never cyclic; the converse fails, and the Klein four-group on the companion page is an abelian group that is not cyclic.