Prove the following classification: if is a finite group of order where the are distinct primes such that does not divide for all , then is cyclic.
We proceed by induction on the width of . (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 has width 1. Then , and clearly does not divide . Moreover, every group of order (a prime) is cyclic.
For the inductive step, suppose that every finite group of width at most and whose order’s prime divisors are distinct and satisfy the divisibility requirement is cyclic. Let be a finite group of width , say , whose order’s prime divisors are distinct and satisfy the divisibility requirement.
Note that every proper subgroup of has width at most and satisfies the distinctness and divisibility criteria; by the induction hypothesis, every proper subgroup of is cyclic. Using this previous exercise, is not simple (whether or not it is abelian). Let be a nontrivial, proper, normal subgroup. Now is cyclic, and (after a relabeling) we have for some . By the N/C Theorem, . Since is cyclic, hence abelian, . So divides . By Proposition 16 and properties of the Euler totient function, . Now if is a prime dividing , then divides some , a contradiction. Thus , and we have . In particular is not trivial. Since properly divides , by the induction hypothesis is cyclic. Hence is abelian.
By Cauchy's Theorem, for each , has an element of order . Since is abelian, . Hence is cyclic.