A130751 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.

2, 22, 34, 46, 56, 58, 86, 88, 92, 94, 106, 116, 118, 134, 142, 146, 154, 162, 166, 178, 184, 188, 202, 204, 206, 210, 214, 218

Offset: 1

Author

David L. Harden, Jul 13 2007, Aug 11 2007

Keywords

• nonn
• more

Comments

(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.

(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.

(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:

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

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

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.

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.

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.

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.)

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.

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.

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.

Examples

			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).
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).
		

References

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

Links

• C. M. Roney-Dougal, The primitive permutation groups of degree less than 2500, Journal of Algebra 292 (2005) 154-183.
• B. Sambale, Pseudo-Sylow numbers, arXiv:1812.08988 [math.GR], 2018.