Every group is a union of cyclic subgroups

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

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

Related facts

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.