# Every group is a union of cyclic subgroups

## Contents

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 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.