Let be the Euler totient function. Before getting into the main problem we give a useful lemma.

**Lemma**. Let be a subgroup of a group Let and be the normalizer and the centralizer of in respectively. Then and is isomorphic to a subgroup of

*Proof*. Define by for all and See that is a well-defined group homomorphism and

**Problem**. Let be a group of order Prove that if then is cyclic.

**Solution.** The proof is by induction on The case is trivial. For since we must have for some distinct primes Let be a Sylow -subgroup of and let

By the lemma, is isomorphic to a subgroup of But since is a cyclic group of order we have and thus Hence Clearly and so because So and thus, by the Burnside’s normal complement theorem, there exists a normal subgroup of such that

Thus and hence, by the induction hypothesis, is cyclic. Let

Again, by the lemma, is isomorphic to a subgroup of . Also, since is cyclic,

Therefore But clearly and thus i.e. So is in the center of and therefore is abelian because is abelian, is in the center of and Hence and so is cyclic because both and are cyclic with coprime orders.

**Remark 1**. By the fundamental theorem of finite abelian groups, a group of square-free order is abelian if and only if it is cyclic. This result can also be used instead of the last line of the solution.

**Remark 2**. The converse of the problem is also true, i.e. if is a positive integer and the only group of order is then