Characterization of the natural numbers which are only the orders of cyclic groups, part 2

Prove the converse to the preceding exercise: if n \geq 2 is an integer such that every group of order n is cyclic, then n = p_1p_2 \cdots p_k is a product of distinct primes p_i such that p_i does not divide p_j - 1 for all i,j. [Hint: If n is not of this form, construct noncyclic groups using direct products and known noncyclic groups of order p^2 and pq where p|q-1.]


Let n = p_1p_2 \cdots p_k, where the p_i are prime.

  • Suppose (without loss of generality) that p_1 = p_2. Then Z_{p_1}^2 \times Z_{p_3 \cdots p_k} is noncyclic since (in particular) it has a noncyclic subgroup Z_{p_1}^2. Moreover, this group has order n. Thus not every group of order n is cyclic.
  • Suppose (without loss of generality) that p_1 divides p_2 - 1. By the example on page 143 in the text, there exists a nonabelian group Q of order p_1p_2. Now Q is not cyclic, since cyclic groups are abelian, hence Q \times Z_{p_3 \cdots p_k} is a noncyclic group of order n. Thus not every group of order n is cyclic.

Hence, if all groups of order n are cyclic, then the p_i are mutually distinct and p_i does not divide p_j-1 for all i,j.

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