A130751 - OEIS

%I #16 Jun 03 2019 15:42:23

%S 2,22,34,46,56,58,86,88,92,94,106,116,118,134,142,146,154,162,166,178,

%T 184,188,202,204,206,210,214,218

%N Numbers n such that for all finite groups G and all primes p, the number of Sylow p-subgroups of G does not equal n.

%C (i) If n is an odd positive integer, then the number of Sylow 2-subgroups of the dihedral group of order 2n is n. Therefore all terms of this sequence are even.

%C (ii) Let p be a prime and let q be a power of p. The number of Sylow p-subgroups of PGL(2,q) is q+1.

%C (iii) Let q be a power of a prime and let r be a prime factor of q-1. Then the group of all affine transformations from the finite field F_q to itself has exactly q Sylow r-subgroups. This implies that q is a nonmember of the sequence, as long as q-1 has prime factors. Thus it works for all q except 2. (iv) The Third Sylow Theorem: If G is a finite group and p is a prime factor of |G|, then the number of Sylow p-subgroups of G is congruent to 1 mod p. 2 is in the sequence, since 2 is not congruent to 1 modulo any primes. It is never again this easy. Therefore we need an easier equivalent form of the membership criterion:

%C Easier Test. n > 2 is in this sequence iff the symmetric group S_n lacks a transitive subgroup with exactly n Sylow p-subgroups.

%C Proof. If S_n has a transitive subgroup with exactly n Sylow p-subgroups, then n trivially is a nonmember of this sequence.

%C For the other direction, suppose that G has exactly n Sylow p-subgroups. Then we will show that S_n has a transitive subgroup with exactly n Sylow p-subgroups. Let P be a Sylow p-subgroup of G and let N be its normalizer in G. Let G act on its Sylow p-subgroups by conjugation. This gives a homomorphism phi: G --> S_n. We want to show that phi(G) is also a group with exactly n Sylow p-subgroups. It is transitive by the Second Sylow Theorem.

%C First, phi(P) is a nontrivial subgroup of G: By properties of Sylow subgroups, P normalizes none of the other Sylow p-subgroups of G. Therefore phi(P) is a p-subgroup of G. Also, [phi(G):phi(P)] divides [G:P], which is a nonmultiple of p. Therefore phi(P) is a Sylow p-subgroup of phi(G). Since N normalizes P, phi(N) normalizes phi(P). Finally, nothing outside phi(N) can normalize phi(P), by the definition of phi.

%C Likewise phi(N) is a point-stabilizer in phi(G). Since phi(G) is a transitive subgroup of S_n, the index of phi(N) in phi(G) is n. Since phi(N) is the normalizer of phi(P) in phi(G), this means the number of Sylow p-subgroups of phi(G) is n and we are done.

%C We continue the notation used above with G, N and P in what follows. (For these comments, isolated examples are given that show different variations on the reasoning used to prove membership in this sequence.)

%C 22 is in the sequence: If G is a finite group with exactly 22 Sylow p-subgroups, then p= 3 or 7. (Since there is no need to divide into cases based on the value of p, in what follows p= 3 or 7.) 22 has no proper divisors congruent to 1 mod p, so N is a maximal subgroup of G. Then the Easier Test can be strengthened to say "primitive" where it says "transitive". There are only 4 nonisomorphic primitive subgroups of S_22: M_22, Aut(M_22), A_22 and S_22. All of these have more than 22 Sylow p-subgroups.

%C 34 is in the sequence: If G is a finite group with exactly 34 Sylow p-subgroups, then p= 3 or 11. 34 has no proper divisors congruent to 1 mod p and (proceeding as before) the only primitive subgroups of S_34 are S_34 and A_34. Both have more than 34 Sylow p-subgroups.

%C 88 is a member: the only primitive permutation groups of degree 88 are S_88 and A_88. Neither of these has exactly 88 Sylow 29-subgroups and 88 has no proper divisors congruent to 1 mod 29. The divisors of 88 congruent to 1 mod 3 are 1, 4, 22 and 88. Then a minimal extension K of a Sylow 3-subgroup normalizer N must have [K:N]=4 (22 was already ruled out, as was 88 by the fact that S_88 and A_88 have more than 88 Sylow 3-subgroups). K will be maximal in G and have [G:K]=22. All of M_22, Aut(M_22), A_22 or S_22 have more than 88 Sylow 3-subgroups and G itself will have at least as many.

%D B. Sambale, Pseudo-Sylow numbers, Amer. Math. Monthly 126 (2019), 60-65; DOI: 10.1080/00029890.2019.1528825.

%H C. M. Roney-Dougal, <a href="https://doi.org/10.1016/j.jalgebra.2005.04.017">The primitive permutation groups of degree less than 2500</a>, Journal of Algebra 292 (2005) 154-183.

%H B. Sambale, <a href="https://arxiv.org/abs/1812.08988">Pseudo-Sylow numbers</a>, arXiv:1812.08988 [math.GR], 2018.

%e 120 is not a term in this sequence because 120 is the number of Sylow 7-subgroups of the symmetric group S_7 (or the alternating group A_7).

%e 4 is not a term in this sequence because 4 is the number of Sylow 3-subgroups of the alternating group A_4 (or the symmetric group S_4).

%K nonn,more

%O 1,1

%A _David L. Harden_, Jul 13 2007, Aug 11 2007