Since the current year commonly features in math competition problems, it seems useful to begin the year by noting that .

The past couple years around this time I have set myself the further puzzle of working out the groups of that order. Usually it is not difficult, but occasionally it will be essentially impossible (e.g., in ).

This year is not very interesting actually, group-theoretically. This is an abelian year in the sense that all groups of order are abelian, and the only possibilities are and . This is an easy consequence of Sylow’s theorem. Let be a group of order . The number of Sylow -subgroups must divide and be , so it must be , and similarly the number of Sylow -subgroups must also be , so where has order . A group of order must be abelian (standard exercise), so or .

To make it more challenging, can we actually characterize the positive integers such that all groups of order are abelian? Call such an integer *abelian*.

**Proposition 1** Let be a positive integer. Then is abelian if and only if

(1) is cube-free,

(2) if are distinct primes such that then does not divide ,

(3) if are distinct primes such that then does not divide .

Let us first show that the conditions are necessary. Suppose (1) fails. Then for some prime . There is a nonabelian group of order (see extraspecial groups), so is a nonabelian group of order . Similarly if (2) fails then there is a nonabelian semidirect product , so we can again form a nonabelian group of order . Finally if (3) fails then note that , and contains an element of order (since the finite field of order has a primitive root), so we have another nonabelian semidirect product .

Now we show that the conditions are sufficient. We use induction on . Let be a group of order , where satisfies conditions (1), (2), and (3). Assume first is not simple. Let be a maximal normal subgroup of . Then and are both integers smaller than satisfying (1), (2), and (3), so and are abelian by induction. Since all subgroups of are normal and was chosen to be maximal normal, must be cyclic of prime order, say . Let be a Sylow -subgroup and let be the product of the Sylow -subgroups for . Then it follows that is the semidirect product .

Next consider the conjugation action of on . Let be a Sylow -subgroup. Then is isomorphic to one of , and is respectively isomorphic to one of . Note that . By the given conditions, none of these has order divisible by , so acts trivially on . Hence acts trivially on and is abelian, as claimed.

To finish we must rule out the possibility that is simple. This seems to require some nontrivial group theory. One lazy method is to note that condition (2) implies that is odd (or a power of ), so we are done by the Feit–Thompson theorem. An easier argument is to use Burnside’s transfer theorem, which states that if is a Sylow -subgroup and then has a normal subgroup of order (in particular is not simple). In our case, let be the smallest prime dividing and let be a Sylow -subgroup. Let and . Then is isomorphic to a subgroup of . As before is one of , , or . The orders of these groups have no prime factor larger than . Since is the smallest prime dividing and , it follows that is trivial, so we are done.

It is also possible to characterize *nilpotent* numbers in general, and even *solvable* numbers. See Pakianathan, J., & Shankar, K. (2000). Nilpotent Numbers. The American Mathematical Monthly, 107(7), 631–634. https://doi.org/10.2307/2589118.