ON GROUPS WITH PERFECT ORDER SUBSETS KEVIN FORD, SERGEI KONYAGIN, AND FLORIAN LUCA ABSTRACT. A finite group G is said to have Perfect Order Subsets if for every d, the number of elements of G of order d (if there are any) divides |G|. Answering a question of Finch and Jones from 2002, we prove that if G is Abelian, then such a group has order divisible by 3 except in the case G = Z/2Z. We also place additional restrictions on the order of such groups. 1 Introduction Consider the multiplicative function f (n)=  p a ‖n (p a − 1). A finite group G is said to have Perfect Order Subsets if for every d, the number of elements of G of order d (if there are any) divides |G|. This notion was introduced in the paper [1] by C. Finch and L. Jones. In the case of finite Abelian groups, the authors reduced the problem of which groups have this property to the case of groups of the form G =  k i=1 (Z/p i Z) a i , where p i are primes and a i  1. For these groups, it follows from results in [1] that G has Perfect Order Subsets if and only if f (n)|n. Only 11 examples of such n are known, given below, and only one of these is divisible by the square of an odd prime. 2 2 · 3 2 2 · 3 2 3 · 3 · 7 2 4 · 3 · 5 2 5 · 3 · 5 · 31 2 8 · 3 · 5 · 17 2 16 · 3 · 5 · 17 · 257 2 17 · 3 · 5 · 17 · 257 · 131071 2 32 · 3 · 5 · 17 · 257 · 65537 2 11 · 3 · 5 · 11 2 · 23 · 89 The authors of [1] asked several basic questions about such groups. One of which asks if |G| is not a power of 2, must 3 divide |G|? We prove that this is the case for Abelian groups. Theorem 1. If f (n)|n and n> 2, then 3|n. We also show that f (n)|n implies that n/f (n) is bounded. Note that the divergence of  p (1 − 1/p) −1 implies that n/f (n) is unbounded for general n. On the other hand, all of the known examples of n such that n> 6 and f (n)|n (given in [1]) satisfy n =2f (n). Theorem 2. For any n ∈ N, if f (n)|n, then n/f (n)  85. The most important property of numbers n with f (n)|n is given by the following easy proposition. Proposition 1. If f (n)|n, then for every prime p|n, every prime divisor of p − 1 also divides n. Date: September 20, 2012. 2000 Mathematics Subject Classification. 11N25 (20K01). 1