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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. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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.

REFERENCES

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

LINKS

Table of n, a(n) for n=1..28.

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.

EXAMPLE

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).

CROSSREFS

Sequence in context: A166726 A156441 A094625 * A227534 A126913 A019593

Adjacent sequences:  A130748 A130749 A130750 * A130752 A130753 A130754

KEYWORD

nonn,more

AUTHOR

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

STATUS

approved

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Last modified November 28 20:45 EST 2021. Contains 349415 sequences. (Running on oeis4.)