Sylow’s theorem says the number of Sylow p-subgroups is congruent to 1 mod p. Does the converse hold? If n is a number congruent to 1 mod p, then is there a finite group with n Sylow p-subgroups?

For p=2, it turns out Sylow’s theorem is sharp. Sylow’s theorem states that the number of Sylow 2-subgroups of a finite group is odd. The dihedral group of order 4n+2 has exactly 2n+1 Sylow 2-subgroups, so every odd number is the number of Sylow 2-subgroups of some finite group.

Already for p=3, things are more complicated. The first few numbers are fine: 1, 4, 7, 10, 13, 16, 19. In fact other than 10, they all come from “Singer cycles” or AGL(1,q), my favorite example of group. For n=10 we get the exotic example of the simple group of order 60; there is no solvable group with exactly 10 Sylow subgroups!

We thus have two examples of groups to consider AGL(1,q) for q a prime power congruent to 1 mod p, and finite non-abelian simple groups. Philip Hall (1928) showed that the number of Sylow p-subgroups of a solvable group is equal to a product of numbers of Sylow p-subgroups of various AGL(1,q) with q varying over prime power divisors of the group (in particular, the size of its chief factors). Marshall Hall, Jr. (1967) showed that in the non-solvable case, the only additional complication were the finite non-abelian simple groups: the number of Sylow p-subgroups of a finite group is a product of numbers of Sylow p-subgroups for various finite non-abelian simple groups (occurring as composition factors) and for AGL(1,q) for various q (occurring as sizes of chief factors of certain subnormal subgroups).

Returning to our case-by-case analysis, things go wrong for n=22. There is no group with exactly 22 Sylow subgroups! M. Hall gives a proof based on Brauer’s theory of blocks of defect one, but one can avoid some of the details by working with tables of primitive groups of small degree.

Let G be a finite group, p a prime, P a Sylow p-subgroup of G, N=N_{G}(P) its normalizer, and K = ∩{N^{g}:g in G} its normal core. Then PK/K is a Sylow p-subgroup of G/K and its normalizer M/K = N_{G/K}(PK/K) contains N/K. Hence the number of Sylow p-subgroups of G/K divides the number of Sylow p-subgroups of G. Why are they equal? Well gK in G/K normalizes PK/K iff g in G normalizes PK. Every g in G normalizes K, so if g normalizes P then it also normalizes PK, and so N ≤ M. If g in G normalizes PK, then it normalizes its characteristic Sylow p-subgroup P (since P is normal in PK), and so M ≤ N. Hence G and G/K have exactly the same number of Sylow p-subgroups. To determine if a number is a possible number of Sylow p-subgroups of a finite group, we can restrict to groups that act faithfully on their set of Sylow p-subgroups by conjugation. In other words, we are only looking at **transitive groups** now.

For numbers like 22 we can do even better than transitive. If N < H ≤ G, then P is a Sylow p-subgroup of H with normalizer N, and [H:N] ≡ 1 mod p and [H:N] is a divisor of [G:N]=[G:H][H:N]. However, the only divisors of 22 that are equal to 1 mod 3 (or mod 7) are 1 and 22, so [H:N] = 22 and H=G. In other words, N must be a maximal subgroup, and G must be a **primitive** permutation group. While 22 is a little exotic, having both the Mathieu group of degree 22 and its automorphism groups as proper primitive groups of that degree, it still turns out that none of the four primitive groups of degree 22 have exactly 22 Sylow p-subgroups for any prime p (p=3 and p=7 being the only ones worth considering).

If G has n Sylow p-subgroups, then so does O^{p′}(G), the subgroup of G generated by the Sylow p-subgroups. In particular, a group minimal amongst its (normal) subgroups with n Sylow p-subgroups is equal to its p′ residual and so has no non-identity abelian quotients of order coprime to p — it is p′-perfect. If G has n Sylow p-subgroups, then so does G/O_{p}(G), the quotient of G by its p-core. The p-core is the intersection of the Sylow p-subgroups. If K is the intersection of the normalizers of every Sylow p-subgroup, then [K,P] ≤ K∩P is a p-group contained in every Sylow p-subgroup, so [K,P] ≤ O_{p}(G) = 1, and K centralizes every Sylow p-subgroup. Since we may assume G = O^{p′}(G), we get that K ≤ Z(G). At any rate, by the argument from a few paragraphs back, we can quotient by K and not change the number of Sylow p-subgroups, so we get that if n is the number of Sylow p-subgroups of some group, then n is the number of Sylow p-subgroups for a p′-perfect, transitive group G with O_{p}(G) = 1.

Of course using Hall’s result, we can just work arithmetically and look at finite non-abelian simple groups. For p=3, which new numbers do we get? For n≤1000, the “generators” of the numbers are just 10, 55, 82, 136, 253, 406, and 946 (along with the primes congruent to 1 mod 3, and the squares of the primes congruent to 2 mod 3). A little over a third of the numbers congruent to 1 mod 3 (up to 1000) cannot be the number of Sylow 3-subgroups of a finite group.

The numbers occurring as the number of Sylow 3-subgroups of simple groups along with the simple group in question, as long as they are not already Sylow numbers of soluble groups:

- 10 Alt(5) and PSL(2, 9)
- 55 PSL(2, 11) and MathieuGroup(11)
- 82 PSL(2, 3^4)
- 136 PSL(2, 17) and PSL(2, 16)
- 190 PSL(2, 19), but also AGL(1,19)×A5
- 253 PSL(2, 23)
- 280 PSL(3, 4), but also AGL(1,4)×AGL(1,7)×A5
- 406 PSL(2, 29)
- 730 PSU(3, 9) and PSL(2,3^6), but also AGL(1,73)×A5
- 820 PSL(2, 41), but also PSL(2,3^4)×A5
- 946 PSL(2, 43)

I wonder what the asymptotic density of Sylow numbers amongst numbers congruent to 1 mod p is?