Prove the converse to the preceding exercise: if is an integer such that every group of order is cyclic, then is a product of distinct primes such that does not divide for all . [Hint: If is not of this form, construct noncyclic groups using direct products and known noncyclic groups of order and where .]
Let , where the are prime.
- Suppose (without loss of generality) that . Then is noncyclic since (in particular) it has a noncyclic subgroup . Moreover, this group has order . Thus not every group of order is cyclic.
- Suppose (without loss of generality) that divides . By the example on page 143 in the text, there exists a nonabelian group of order . Now is not cyclic, since cyclic groups are abelian, hence is a noncyclic group of order . Thus not every group of order is cyclic.
Hence, if all groups of order are cyclic, then the are mutually distinct and does not divide for all .