Finitely many subgroups iff finite

Statement
The following are equivalent for a group:


 * The group is a fact about::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

 * 1) uses::Every group is a union of cyclic subgroups

Finite implies finitely many subgroups
Given: A finite group $$G$$.

To prove: $$G$$ has only finitely many subgroups.

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

Finitely many subgroups implies finite
Given: A group $$G$$ with only finitely many subgroups.

To prove: $$G$$ is finite.

Proof: We consider two cases:


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