Every group is a union of cyclic subgroups
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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:
i.e. G is the union of the singleton sets for all its elements. Now, we have:
i.e. the singleton subset for g is contained in the subgroup generated by G. Thus, we have:
Equality holds throughout, so:
Further, each of the subgroups in the union is cyclic.
