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

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.