Every group is a union of cyclic subgroups

Statement
Any group can be expressed as a union of cyclic subgroups.

Related facts in group theory

 * Union of two subgroups is not a subgroup unless they are comparable: No group can be expressed as the union of two proper subgroups. More generally, any subgroup contained in the union of two subgroups is contained in either of them.
 * Cyclic iff not a union of proper subgroups: A group is cyclic if and only if it cannot be expressed as a union of proper subgroups.
 * Every group is a union of maximal among abelian subgroups: Any group can be expressed as a union of Abelian subgroups, with each of those subgroups having the property that is is maximal among Abelian subgroups -- it is not properly contained in any other Abelian subgroup.
 * Union of all conjugates is proper: In a finite group, the union of all conjugates of a proper subgroup is a proper subset of the group.

Related ideas in other subjects
The idea behind this proof is that of a point-indexed cover: for every element of the group, we find a cyclic subgroup containing it, and these cyclic subgroups therefore cover the whole group. This idea is used in many other parts of mathematics, notably in point-set topology. This is used in proving that a subset of a topological space is open if and only if every point is contained in an open subset inside it. The idea is also used when proving or applying compactness.

Applications

 * Artin's induction theorem states that if $$X$$ is a collection of subgroups of $$G$$ such that the union of conjugates of elements of $$X$$ equals $$G$$, then class functions induced from class functions on elements of $$X$$ span the space of class functions of $$G$$. The fact that every group is a union of cyclic subgroups tells us that we can take $$X$$ to be the collection of cyclic subgroups of $$G$$.

Proof
Given: A group $$G$$

To prove: $$G$$ is a union of cyclic subgroups

Proof: Observe first that:

$$G = \bigcup_{g \in G} \{ g \}$$

i.e. $$G$$ is the union of the singleton sets for all its elements. Now, we have:

$$\{ g \} \subset \langle g \rangle \subset G$$

i.e. the singleton subset for $$g$$ is contained in the subgroup generated by $$G$$. Thus, we have:

$$G = \bigcup_{g \in G} \{ g \} \subseteq \bigcup_{g \in G}\langle g \rangle \subseteq G$$

Equality holds throughout, so:

$$G = \bigcup_{g \in G}\langle g \rangle$$

Further, each of the subgroups in the union is cyclic.