Flag-transitive ( v, k, ) -symmetric designs with ( k, ) = 1 and - - PowerPoint PPT Presentation

Flag-transitive (v, k, λ)-symmetric designs with (k, λ) = 1 and alternating socle

Shenglin Zhou South China University of Technology Vilanova University, 2-5 June, 2014

t − (v, k, λ) designs

A 2-(v, k, λ) symmetric design is a pair D = (P, B), such that (a) P is a v-set, the elements is called points; (b) B is a collection of b k-subsets of P, its element called blocks; (c) b = v; (d) each 2-subset of P is contained in exactly λ blocks.

Automorphism groups of designs

An automorphism of a design D = (P, B) is a permutation π of P such that B ∈ B implies π(B) ∈ B. All automorphisms of D form a group which acts on P, denoted by Aut(D). Since an automorphism takes blocks to blocks, the group also has a permutation representation on the set B. If G ≤ Aut(D) then G is a automorphism group of D. Then G is called

◮ point-transitive (primitive): if G is transitive (primitive) on P; ◮ block-transitive (primitive): if G is transitive (primitive) on B; ◮ flag-transitive: if G is transitive on the set of flags

F = {(α, B)|α ∈ B};

◮ · · ·

2 − (v, k, λ) symmetric designs

1) 2 − (v, k, 1) designs: It is also called a finite linear space. 2) Here we focus on flag-transitive symmetric 2 − (v, k, λ) designs.

Flag-transitive projective planes

Symmetric design with λ = 1 are projective planes. If G ≤ Aut(D) is flag-transitive, the classification has been done by Kantor [6] (1987, J. Alg. 106, 15-45).

Theorem

If D is a projective plane of order n admitting a flag-transitive automorphism group G, then either:

• 1. D is Desarguesian and G ⊲ PSL(3, n), or
• 2. G is a sharply flag-transitive Frobenius group of odd order

(n2 + n + 1)(n + 1) and n2 + n + 1 is a prime.

Flag-transitive SD with λ small and alternating socle

Recently, there are some work on flag-transitive SD with λ small and alternating socle:

◮ For λ = 2, classified by Regueiro [8]; ◮ For λ ≤ 10, classified by Dong and Zhou [4, 5, 10]; ◮ For λ ≤ 100, classified by Zhu, Tian and Zhou [11].

Flag-transitive 2 − (v, k, λ) Symmetric Designs

In 1988, Zieschang [12] proved that the following theorem:

Theorem

Let G is a flag-transitive automorphism group of a 2-design with (r, λ) = 1 and let T be a minimal normal subgroup of G. If T is not abelian, then T is simple and CG(T) = 1.

Based on this Theorem, one natural question is:

Question

Can we classify the flag-transitive 2-designs with (r, λ) = 1 and T is nonabelian simple group, especially T = An?

Flag-transitive Symmetric 2 − (v, k, λ) Designs

Our main result is the following.

Theorem

If D is a (v, k, λ) symmetric design with (r, λ) = 1, which admits a flag-transitive automorphism group G with alternating socle, then D is the projective space PG2(3, 2) and G = A7.

Lemmas

Here are some basic lemmas.

Lemma

(Demboski, 1968, 2.3.7(a)) Let D be a (v, k, λ) design with flag-transitive automorphism group G. If (r, λ) = 1 then G is point-primitive.

Lemmas

Lemma

Let D be a (v, k, λ) symmetric design. Then (i) k(k − 1) = λ(v − 1). In particular, if (k, λ) = 1 then k | v − 1 and (k, v) = 1. (ii) 1 + 4λ(v − 1) is a square.

Lemma

If D is a (v, k, λ) symmetric design and G ≤ Aut(D) is flag transitive, point primitive, then (i) k2 > λv, and |Gx|3 > λ|G|, where x ∈ P; (ii) (Davies, 1987) k | λdi, where di is any subdegree of G. Furthermore, if (k, λ) = 1 then k | di. From this lemma we have |Gx| >

3

• |G|, which is called the cube

root bound.

Lemmas

Lemma

(BRC Theorem) Let v, k and λ be integers with k(k − 1) = λ(v − 1) such that there exists a (v, k, λ) SD. (i) If v is even, then k − λ is a square; (ii) If v is odd, then the equation (k − λ)x2 + (−1)

v−1 2 λy2=z2

has a solution in integers x, y, z not all zero.

Lemmas

Lemma

If G is An or Sn, acting on a set Ω of size n, and H is any maximal subgroup of G with H = An, then H satisfies one of the following: (i) H = (Sk × Sℓ) ∩ G, with n = k + ℓ and k = ℓ (intransitive); (ii) H = (Sk ≀ Sℓ) ∩ G, with n = kℓ, k > 1 and ℓ > 1 (imprimitive); (iii) H = AGLk(p) ∩ G, with n = pk and p prime (affine); (iv) H = (T k.(OutT × Sk)) ∩ G, with T a nonabelian simple group, k ≥ 2 and n = |T|k−1 (diagonal); (v) H = (Sk ≀ Sℓ) ∩ G, with n = kℓ, k ≥ 5 and ℓ > 1 (wreath); (vi) T H ≤ Aut(T), with T a nonabelian simple group, T = An and H acting primitively on Ω (almost simple).

Lemmas

Lemma

Let G be a primitive permutation group of odd degree n on a set Ω with simple socle X := Soc(G), and let H = Gx, x ∈ Ω. If X ∼ = Ac, an alternating group, then one of the following holds: (i) H is intransitive, and H = (Sk × Sc−k) ∩ G where 1 ≤ k < 1

2c;

(ii) H is transitive and imprimitive, and H = (Sa ≀ Sb) ∩ G where ab = c, a > 1, b > 1; (iii) H is primitive, n = 15 and G ∼ = A7.

Lemmas

In order to prove main Theorem, the following inequality is useful:

Lemma

Let s, t be positive integers such that t > s ≥ 7, then s + t s

• > t4 > s2t2.

• Proof. It is necessary to prove that

s+t

s

• > t4 holds. Since

t > s ≥ 7 then s+t

s

• t4

= t + 1 t4 t + 2 2

• · · ·

t + s s

• > (t + 2)(t + 3)(t + 4)(t + 5)(t + 6)(t + 7)

7!t3 = t6 + 27t5 + 295t4 + 1665t3 + 5104t2 + 8028t + 5040 7!t3 > t3 + 27t2 + 295t + 1665 7! ≥ 83 + 27 × 82 + 295 × 8 + 1665 7! = 179 144 > 1, so that s+t

s

• > t4 > s2t2.

Proof of Main Theorem

Hypothesis: Let D be a (v, k, λ) symmetric design with (r, λ) = 1, G ≤ Aut(D) be a flag-transitive automorphism group G with Soc(G) = An. Let x be a point of P and H = Gx.

Proof of Main Theorem

Since G acts primitively on P. So that H is a maximal subgroup

• f G and v = |G : H|. Furthermore, by the flag-transitivity of G,

we have that k divides |H|, and k2 > v. Suppose first that n = 6 and G ∼ = M10, PGL2(9) or PΓL2(9). Each of these groups has exactly three maximal subgroups with index greater than 2, and their indices are precisely 45, 36 and 10, by using the computer algebra system GAP. The possible parameters (v, k, λ) such that 2 < k < v − 1 and k(k − 1) = λ(v − 1) are (36,15,6), (36,21,12), (45,12,3) and (45,33,24). These can be ruled out since k ∤ |H|.

Proof of Main Theorem: Step 1

Now we consider G = An or Sn with n ≥ 5. The point stabilizer H = Gx acts both on P and on the set Ωn := {1, 2, · · · , n}. The action of H on Ωn is primitive, imprimitive, or intransitive as a subgroup of Sn. Our proof consists of three steps.

Proof of Main Theorem: Step 1

Step 1. H acts primitively on Ωn.

Proposition

Let D and G satisfy Hypothesis. Let the point stabilizers act primitively on Ωn. Then D is the projective space PG2(3, 2).

Proof of Main Theorem: Step 1

Now assume that k is odd. Let p | k then (p, v) = 1. Thus H contains a Sylow p-subgroup R

• f G. Let g ∈ G be a p-cycle, then there is a conjugate of g belong

to H. This implies that H acting on Ωn contains an even permutation with exactly one cycle of length p and n − p fixed

• points. Thus n − p ≤ 2 by [9, Theorem 13.9]. Therefore

n − 2 ≤ p ≤ n. It follows that p2 ∤ |G|, and then p2 ∤ k. So k = n − 2, n − 1, n, or k = (n − 2)n and n is odd.

Proof of Main Theorem: Step 1

Moreover, since the primitivity of H acting on Ωn and H An implies that v ≥

[ n+1

2 ]!

2

by [9, Theorem 14.2], combining this with k2 > v gives k2 > [ n+1

2 ]!

2 . Therefore, n = 5, 6, 7, 8 or 13, and k = 3, 5, 7, 11, 13, 15, 35 or 143. According to our Lemmas and the fact v ≥

[ n+1

2 ]!

2

, we obtain 14 possible parameters (v, k, λ) which listed in the following: (7, 3, 1), (21, 5, 1), (11, 5, 2), (15, 7, 3), (22, 7, 2), (43, 7, 1), (211, 15, 1), (106, 15, 2), (31, 15, 7), (71, 35, 17), (596, 35, 2), (1191, 35, 1), (10154, 143, 2), (20307, 143, 1).

Proof of Main Theorem: Step 1

For 4 parameters with v is even can be ruled out by BRC Theorem. For remaining 10 parameters with v is odd, there is only one case can occur, i.e., (v, k, λ) = (15, 7, 3) and G = A7, i.e., D is the projective space PG2(3, 2).

Proof of Main Theorem: Step 2

Step 2. H acts transitively and imprimitively on Ωn.

Proposition

Let D and G satisfy Hypothesis. Then the point stabilizers cannot be transitive and imprimitive on Ωn.

• Proof. Suppose on the contrary that Σ := {△0, △1, . . . , △t−1} is

a nontrivial partition of Ωn preserved by H, where |△i| = s, 0 ≤ i ≤ t − 1, s, t ≥ 2 and st = n. Then v = ts

s

(t−1)s

s

• ...

3s

s

2s

s

• t!

= ts − 1 s − 1 (t − 1)s − 1 s − 1

• ...

3s − 1 s − 1 2s − 1 s − 1

• .

(1)

Proof of Main Theorem: Step 2

Case 1): If s = 2, then t < 7; derived contradiction. Case 2): s ≥ 3. Let Oj ([2]) be an orbit of H on P, then dj = |Oj| = t

j

s

1

j = sjt

j

• . In particular, d2 =

t

2

s

1

2 = s2t

2

• and k | d2. Moreover, from

is − 1 s − 1

• = is − 1

s − 1 · is − 2 s − 2 · · · is − (s − 1) 1 > is−1, for i = 2, 3, . . . , t, we have v > 2(s−1)(t−1). Then 2(s−1)(t−1) < v < k2 ≤ s4 t 2 2 , and so 2(s−1)(t−1) < s4 t 2 2 . (2)

Proof of Main Theorem: Step 2

Then we calculate all pairs (s, t) satisfying the inequality (2). For this purpose, we give the following two basic facts. Fact 1): If s ≥ 6, t ≥ 2 then 2(s−1)(t−1) > s4t

2

2 implies 2s(t−1) > (s + 1)4t

2

2. Fact 2): If t ≥ 6, s ≥ 2 then 2(s−1)(t−1) > s4t

2

2 implies 2(s−1)t > s4t+1

2

2. Since 2(s−1)(t−1) = 225 > s4t

2

2 = 243652, i.e. the pair (s, t) = (6, 6) does not satisfy the inequality (2) but satisfies the conditions of Facts 1) and 2). Thus, we must have s < 6 or t < 6. We proceed our proof in two cases.

Proof of Main Theorem: Step 2

Subcase (i). 3 ≤ s < 6 and t ≥ 6; Subcase (ii). 2 ≤ t < 6. Then, we get 32 pairs (s, t) satisfying the inequality (2).

Proof of Main Theorem: Step 3

Step 3. H acts intransitively on Ωn.

Proposition

Let D and G satisfy Hypothesis. Then the point stabilizers cannot be intransitive on Ωn.

Further work

For general flag-transitive 2 − (v, k, λ) symmetric designs with Soc(G) = An?

Thank You!

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63(1987), 91-93.

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alternating socle, Algebraic Combinatorics and Applications(G¨

• ßeinstein, 1999), Springer, Berlin, 2001, 79-88.
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1968.

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2-(v, k, 4) symmetric design, J. Combin. Des., 19 (2011), 575-483.

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and application to finite projective planes, J. Algebra, 106(1)(1987), 15-45.

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(v, k, λ) symmetric designs with λ at most 100 and alternating socle, Math. Slovaca, submitted, 2013.

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