# Finitely many subgroups iff finite

From Groupprops

## Contents

## Statement

The following are equivalent for a group:

- The group is a Finite group (?), i.e., its order (the number of elements in its underlying set) is finite.
- The group has only finitely many subgroups.

## Facts used

## Proof

### Finite implies finitely many subgroups

**Given**: A finite group .

**To prove**: has only finitely many subgroups.

**Proof**: Since is finite, the set of subsets of is finite. Since subgroups are subsets satisfying additional conditions, the set of subgroups of is also finite.

### Finitely many subgroups implies finite

**Given**: A group with only finitely many subgroups.

**To prove**: is finite.

**Proof**: We consider two cases:

- has an element of infinite order: In this case, the cyclic subgroup generated by is isomorphic to . has infinitely many subgroups (the subgroups are distinct for all natural numbers ). Thus, has infinitely many subgroups, contradicting the assumption.
- Every element in has finite order: In this case, by fact (1), is a union of cyclic subgroups,
*each*of which is finite. Since has only finitely many subgroups, is a finite union of finite subgroups, and thus, is finite.