# 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 is a collection of subgroups of such that the union of conjugates of elements of equals , then class functions induced from class functions on elements of span the space of class functions of . The fact that every group is a union of cyclic subgroups tells us that we can take to be the collection of cyclic subgroups of .

## Proof

**Given**: A group

**To prove**: is a union of cyclic subgroups

**Proof**: Observe first that:

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

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

Equality holds throughout, so:

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