Characterization of natural numbers which are only the orders of cyclic groups

Prove the following classification: if G is a finite group of order p_1p_2 \cdots p_k where the p_i are distinct primes such that p_i does not divide p_j - 1 for all i,j, then G is cyclic.


We proceed by induction on the width of G. (Recall that the width of a finite group is the number of prime divisors (including multiplicity) in its prime factorization.)

For the base case, suppose G has width 1. Then |G| = p, and clearly p does not divide p-1. Moreover, every group of order p (a prime) is cyclic.

For the inductive step, suppose that every finite group of width at most k and whose order’s prime divisors are distinct and satisfy the divisibility requirement is cyclic. Let G be a finite group of width k+1, say |G| = p_1 p_2 \cdots p_{k+1}, whose order’s prime divisors are distinct and satisfy the divisibility requirement.

Note that every proper subgroup of G has width at most k and satisfies the distinctness and divisibility criteria; by the induction hypothesis, every proper subgroup of G is cyclic. Using this previous exercise, G is not simple (whether or not it is abelian). Let N \leq G be a nontrivial, proper, normal subgroup. Now N is cyclic, and (after a relabeling) we have |N| = p_1p_2 \cdots p_m for some 1 < m < k+1. By the N/C Theorem, G/C_G(N) \leq \mathsf{Aut}(N). Since N is cyclic, hence abelian, N \leq C_G(N). So |G/C_G(N)| divides p_{m+1} \cdots p_{k+1}. By Proposition 16 and properties of the Euler totient function, |\mathsf{Aut}(N)| = (p_1-1)(p_2-1) \cdots (p_m-1). Now if p_i is a prime dividing |G/C_G(N)|, then p_i divides some p_j-1, a contradiction. Thus G/C_G(N) = 1, and we have N \leq Z(G). In particular Z(G) is not trivial. Since |G/Z(G)| properly divides |G|, by the induction hypothesis G/Z(G) is cyclic. Hence G is abelian.

By Cauchy's Theorem, for each p_i, G has an element g_i of order p_i. Since G is abelian, |g_1g_2 \cdots g_{k+1}| = \mathsf{lcm}(p_1,p_2,\ldots,p_{k+1}) = |G|. Hence G is cyclic.

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