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 ( n 2 + n + 1)( n + 1) and n 2 + 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 C G ( 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 = A n ?

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 PG 2 (3 , 2) and G = A 7 .

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 + 4 λ ( 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) k 2 > λv , and | G x | 3 > λ | G | , where x ∈ P ; (ii) (Davies, 1987) k | λd i , where d i is any subdegree of G . Furthermore, if ( k, λ ) = 1 then k | d i . � From this lemma we have | G x | > 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 ) x 2 + ( − 1) v − 1 2 λy 2 = z 2 has a solution in integers x, y, z not all zero.

Lemmas Lemma If G is A n or S n , acting on a set Ω of size n , and H is any maximal subgroup of G with H � = A n , then H satisfies one of the following: (i) H = ( S k × S ℓ ) ∩ G , with n = k + ℓ and k � = ℓ ( intransitive ) ; (ii) H = ( S k ≀ S ℓ ) ∩ G , with n = kℓ , k > 1 and ℓ > 1 ( imprimitive ) ; (iii) H = AGL k ( p ) ∩ G , with n = p k and p prime ( affine ) ; (iv) H = ( T k . ( OutT × S k )) ∩ G , with T a nonabelian simple group, k ≥ 2 and n = | T | k − 1 ( diagonal ) ; (v) H = ( S k ≀ S ℓ ) ∩ G , with n = k ℓ , k ≥ 5 and ℓ > 1 ( wreath ) ; (vi) T � H ≤ Aut ( T ) , with T a nonabelian simple group, T � = A n 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 = G x , x ∈ Ω . If X ∼ = A c , an alternating group, then one of the following holds: (i) H is intransitive, and H = ( S k × S c − k ) ∩ G where 1 ≤ k < 1 2 c ; (ii) H is transitive and imprimitive, and H = ( S a ≀ S b ) ∩ G where ab = c, a > 1 , b > 1 ; (iii) H is primitive, n = 15 and G ∼ = A 7 .

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 � > t 4 > s 2 t 2 . s

> t 4 holds. Since � s + t � Proof . It is necessary to prove that s t > s ≥ 7 then � s + t � � t + 1 �� t + 2 � t + s � � s = · · · t 4 t 4 2 s > ( t + 2)( t + 3)( t + 4)( t + 5)( t + 6)( t + 7) 7! t 3 = t 6 + 27 t 5 + 295 t 4 + 1665 t 3 + 5104 t 2 + 8028 t + 5040 7! t 3 > t 3 + 27 t 2 + 295 t + 1665 7! ≥ 8 3 + 27 × 8 2 + 295 × 8 + 1665 = 179 144 > 1 , 7! > t 4 > s 2 t 2 . � s + t � so that � s

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 ) = A n . Let x be a point of P and H = G x .

Proof of Main Theorem Since G acts primitively on P . So that H is a maximal subgroup of G and v = | G : H | . Furthermore, by the flag-transitivity of G , we have that k divides | H | , and k 2 > v . Suppose first that n = 6 and G ∼ = M 10 , PGL 2 (9) or P Γ L 2 (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 = A n or S n with n ≥ 5 . The point stabilizer H = G x acts both on P and on the set Ω n := { 1 , 2 , · · · , n } . The action of H on Ω n is primitive, imprimitive, or intransitive as a subgroup of S n . Our proof consists of three steps.

Proof of Main Theorem: Step 1 Step 1. H acts primitively on Ω n . Proposition Let D and G satisfy Hypothesis. Let the point stabilizers act primitively on Ω n . Then D is the projective space PG 2 (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 of G. Let g ∈ G be a p -cycle, then there is a conjugate of g belong to H. This implies that H acting on Ω n 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 p 2 ∤ | G | , and then p 2 ∤ 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 Ω n and H � A n [ n +1 2 ]! implies that v ≥ by [9, Theorem 14.2], combining this with 2 k 2 > v gives k 2 > [ n +1 2 ]! . 2 Therefore, n = 5 , 6 , 7 , 8 or 13 , and k = 3 , 5 , 7 , 11 , 13 , 15 , 35 or 143 . [ n +1 2 ]! According to our Lemmas and the fact v ≥ , we obtain 14 2 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 = A 7 , i.e., D is the projective space PG 2 (3 , 2) .

Proof of Main Theorem: Step 2 Step 2. H acts transitively and imprimitively on Ω n . Proposition Let D and G satisfy Hypothesis. Then the point stabilizers cannot be transitive and imprimitive on Ω n . Proof . Suppose on the contrary that Σ := {△ 0 , △ 1 , . . . , △ t − 1 } is a nontrivial partition of Ω n preserved by H , where |△ i | = s, 0 ≤ i ≤ t − 1 , s, t ≥ 2 and st = n . Then � ts �� ( t − 1) s � 3 s �� 2 s � � ... s s s s v = t ! � ts − 1 �� ( t − 1) s − 1 � � 3 s − 1 �� 2 s − 1 � = ... . (1) s − 1 s − 1 s − 1 s − 1