# The2-(v,11,1) Designs with Unsolvable Block Transitive Automorphism

Authors: Kun Zhao; Shaojun Dai
Journal: ISRN Combinatorics
 (2013)

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## Abstract

This paper is a contribution to the study of the automorphism groups of2-(v,k,1) designs. Let 𝒟 be 2-(v,11,1) design and G≤ Aut(D) a block transitive and a point primitive. If G is unsolvable, then Soc(G), the socle of G, is not F42(q2).


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## Body

## 1. Introduction

A2-(v,k,1) design 𝒟=(𝒫,ℬ) is a pair consisting of a finite set 𝒫 of v points and a collection ℬ of  k-subsets of 𝒫, called blocks, such that any 2-subsets of 𝒫 are contained in exactly one block. We will always assume that 2<k<v.LetG≤Aut(𝒟) be a group of automorphisms of a  2-(v,k,1) design 𝒟. Then G is said to be block transitive on 𝒟 if G is transitive on ℬ and is said to be point transitive (point primitive) on 𝒟 if G is transitive (primitive) on 𝒫. A flag of 𝒟 is a pair consisting of a point and a block through that point. Then G is flag transitive on 𝒟 if G is transitive on the set of flags.The classification of block transitive2-(v,3,1) designs was completed about thirty years ago (see [1]). In [2], Camina and Siemons classified 2-(v,4,1) designs with a block transitive, solvable group of automorphisms. Li classified 2-(v,4,1) designs admitting a block transitive, unsolvable group of automorphisms (see [3]). Tong and Li [4] classified 2-(v,5,1) designs with a block transitive, solvable group of automorphisms. Han and Li [5] classified 2-(v,5,1) designs with a block transitive, unsolvable group of automorphisms. Liu [6] classified 2-(v,k,1) (where k=6,7,8,9,10) designs with a block transitive, solvable group of automorphisms. In [7], Han and Ma classified 2-(v,11,1) designs with a block transitive classical simple group of automorphisms.This paper is a contribution to the study of the automorphism groups of2-(v,k,1) designs. Let 𝒟 be 2-(v,11,1) design and G≤Aut(D) a block transitive and a point primitive. We prove the following theorem.Main Theorem. Let𝒟be2-(v,11,1)design andG≤Aut(D)a block transitive and a point primitive. IfGis unsolvable, thenSoc(G)≇F42(q2).


## 2. Preliminary Results

Let𝒟 be a 2-(v,k,1) design defined on the point set 𝒫 and suppose that G is an automorphism group of 𝒟 that acts transitively on blocks. For a 2-(v,k,1) design, as usual, b denotes the number of blocks and r denotes the number of blocks through a given point. If B is a block, GB denotes the setwise stabilizer of B in G and G(B) is the pointwise stabilizer of B in G. Also, GB denotes the permutation group induced by the action of GB on the points of B, and so GB≅GB/G(B).Lemma 1 (see [8]).
LetG=2F4(q2), q2=22n+1, n≥1, then every maximal subgroup of G is conjugate to one of the following: (1)
P1=[q22]:(PSL(2,q2)×(q2-1));(2)
P2=[q20]:(2B2(q2)×(q2-1));(3)
SU(3,q2):Z2;(4)
(Zq2+1×Zq2+1):GL(2,3);(5)
(Zq2-2q+1×Zq2-2q+1):[96], if q2>8;(6)
(Zq2+2q+1×Zq2+2q+1):[96];(7)
(Zq4-2q3+q2-2q+1):Z12;(8)
(Zq4+2q3+q2+2q+1):Z12;(9)
PGU(3,q2):Z2;(10)
2B2(q2):Z2;(11)
B2(q2):Z2;(12)
2F4(q02), where q02=22m+1 and (2n+1)/(2m+1) be prime.Lemma 2 (see [9]).
LetT=2F4(q2) be an exceptional simple group of Lie type over GF(q), and let G be a group with T⊴G≤Aut(T). Suppose that M is a maximal subgroup of G not containing T, then one of the following holds: (1)
|M|<q24|G:T|;(2)
T∩M=L(3,3):Z2, or L(2,25), if q2=2;(3)
T∩M is a parabolic subgroup of T.Lemma 3 (see [7]).
LetG and 𝒟=(𝒫,ℬ) be a group and a design, and G≤Aut(𝒟) be block transitive and point primitive but not flag transitive. Let Soc(G)=T. Then
(1)|T|≤vλ·|Tα|2·|G:T|,
where α∈𝒫 and λ is the length of the longest suborbit of G on 𝒫.


## 3. Proof of the Main Theorem

Proposition 4.
Let𝒟 be 2-(v,11,1) design and let G be block transitive and point primitive but not flag transitive, then v=110b2+1.Proof.
Letb1=(b,v),b2=(b,v-1),k1=(k,v),k2=(k,v-1). Obviously,
(2)k=k1k2,b=b1b2,r=b2k2,v=b1k1.
Since k=11, we get k1=1. Otherwise, k∣v, by [9], G is flag transitive, a contradiction. Thus, v=k(k-1)b2+1=110b2+1.Proposition 5.
Let𝒟 be 2-(v,11,1) design and let G be block transitive and point primitive but not flag transitive and |T| be even. If G be unsolvable, then |T|≤56|Tα|2|G:T|.Proof.
LetB={1,2,…,11}∈ℬ. Since G is unsolvable, then the structure of GB and the rank and subdegree of G do not occur:
(3)GBRankSubdegree〈1〉1111,b2,…,b2︷110.
Otherwise, |GB|=1 is odd and |B|=11. We have |GB| and b2 are odd. Since v=110b2+1=b1 and b=b1b2, then b is odd and |G|=b|GB| is also odd, a contradiction with |T| be even. Thus, λ≥2b2. By Lemma 3,
(4)|T||Tα|2≤vλ·|G:T|≤v2b2·|G:T|.
By Proposition 4,
(5)|T||Tα|2≤110b2+12b2·|G:T|≤56|G:T|.Now we may prove our main theorem.Suppose thatSoc(G)=2F4(q2)=T, then T⊴G≤Aut(T). We have G=T:〈x〉, where x∈Out(T), the outer automorphisms group of T which may be generated by an automorphism of field. By [8], we may assume that x is an automorphism of field. Set ∘(x)=f, then f∣(2n+1). Obviously, |2F4(q2)|=q24(q2-1)(q6+1)(q8-1)(q12+1). By [10] and k=11, G is not flag transitive. Since G is point primitive, Gα (α∈𝒫) is the maximal subgroup of G and T is block transitive in 𝒟. Hence, M=Gα satisfies one of the three cases in Lemma 2. We will rule out these cases one by one.Case 1 (|M|<q24|G:T|).
By Proposition5, we have an upper bound of |T|,
(6)|T|<56|Tα|2|G:T|<56q48f.
We get
(7)q24(q2-1)(q6+1)(q8-1)×(q12+1)<56q48f<56q48q2,
that is,
(8)(q-1)2<56.
Thus, q2=25 or 23 and T=2F4(25) or 2F4(23). Since v=110b2+1 is odd, then |Tα| contains a Sylow 2-subgroup of T. Clearly Tα is contained in some maximal subgroups of T. By Lemma 1, Tα≅Ti, where Ti≤Pi (i=1,2). Then v∣(25-1)(215+1)(220-1)(230+1) or v∣(23-1)(29+1)(212-1)(218+1). By Proposition 4 and rather long and repetitive numerical calculations, we get a contradiction.Case 2 (T∩M=L(3,3):Z2, or L(2,25), if q2=2).
Obviously,|T|=212·33·52·13 and v=|T:Tα|=26·52 or 25·3. By Proposition 4, we have 110∣26·52 or 110∣25·3, a contradiction.Case 3 (T∩M is a parabolic subgroup of T).
By Lemma1, the parabolic subgroup of 2F4(q2) is conjugate to P1 or P2. Then the order of parabolic subgroup is q24(q2-1)2(q2+1) or q24(q2-1)2(q4+1). We get v=(q4+1)(q6+1)(q12+1) or (q2+1)(q6+1)(q12+1). By Proposition 4, we have 55∣(q18+q14+q12+q8+q6+q2+1) or 55∣(q18+q16+q12+q10+q6+q4+1). But
(9)q18+q14+q12+q8+q6+q2+1≡{2(mod5)ifq≡1(mod5)1(mod5)ifq≡2(mod5)4(mod5)ifq≡3(mod5)2(mod5)ifq≡4(mod5),q18+q16+q12+q10+q6+q4+1≡{2(mod5)ifq≡1(mod5)1(mod5)ifq≡2(mod5)1(mod5)ifq≡3(mod5)2(mod5)ifq≡4(mod5).
Obviously, 5∤(q18+q14+q12+q8+q6+q2+1) or 5∤(q18+q16+q12+q10+q6+q4+1), a contradiction.This completes the proof the Main Theorem.



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*Source: 436987-2013-01-30.xml*