Finitely many subgroups iff finite
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.
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.