Obviously every finite cyclic group has at most one subgroup of any order. We are going to prove the converse of this fact, i.e. if a finite group has at most one subgroup of any order, then is cyclic.

**Lemma**. Let be a finite -group. If has at most one subgroup of any order, then is cyclic.

*Proof*. Let be such that is as large as possible and put Let Then for some Thus We know that a -group has a subgroup of any order which divides the order of the group. Thus has a subgroup of order But has also order and so because, by our hypothesis, and therefore can have at most one subgroup of any order. So and hence We have proved that every element of is in and so

**Problem**. Let be a finite group. If has at most one subgroup of any order, then is cyclic.

**Solution**. Let be the set of all Sylow subgroups of and let Clearly for all and so because has at most one subgroup of any order. So are normal in and hence

On the other hand, cannot have more than one subgroup of any order because every subgroup of is also a subgroup of Thus, by the above lemma, all are cyclic. The result now follows from and the fact that for all