Finitely many subgroups iff finite

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

1. Every group is a union of cyclic subgroups

Proof

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.