Finitely many subgroups iff finite

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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.