Non-existence of Point-transitive 2-(106, 6, 1) Designs

Let S be a linear space with 106 points, with lines of size 6, and let G be an automorphism group of S. We prove that G cannot be point-transitive. In other words, there exists no point-transitive 2-(106, 6, 1) designs.


Introduction
For positive integers v and b satisfying b v 2, a finite linear space S is an incidence structure (P, L) consisting of a set P of v points and a collection L of b distinguished subsets of P called lines, with sizes 2 such that any two points are incident with exactly one line.Let α be a point of P, and k be a positive integer.Then r k α denotes the number of lines having size k through α, b k the number of lines of size k, and r α the number of all lines through α, called the degree of α.If all lines have a constant size k, then we say that S is regular, so it is a 2-(v, k, 1) design.
Let ∆ be a subset of P with |∆| 2, L ∆ = {λ ∩ ∆ : |λ ∩ ∆| 2 for λ ∈ L}.Then (∆, L ∆ ) forms an incidence structure, and the induced structure is a linear space.We are interested in the case that ∆ is Fix(g) (or Fix(H)), the set of fixed points of g ∈ G (or H G ) on P.An automorphism of S is a permutation of P which leaves L invariant.The full automorphism group of S is denoted by Aut(S) and any subgroup of Aut(S) is called an automorphism group of S. We say that the automorphism group G of S is point-transitive if G acts transitively on the set of points.Similarly, G is said to be line-transitive if G acts transitively on the set of lines.
Four 2-(91, 6, 1) designs have been found by Mills,McCalla and Colbourn ([6,7,12]).All of them have a cyclic automorphism group of order 91, and so they have pointtransitive automorphism groups.In 1989, Camina and Di Martino ( [3]) proved that any automorphism group of a point-transitive 2-(91, 6, 1) design is the natural split extension of a cyclic group of order 91 by a cyclic group of order d, where d | 12. Later, Janko and Tonchev ( [10]) showed that any cyclic 2-(91, 6, 1) design (i.e. one having an automorphism group of order 91) admitting an automorphism group whose order is larger than 91 is one of the four known designs.Here we are going to discuss the 2-(106, 6, 1) designs, where the number of points is also a product of two distinct primes, and 6 is the largest line-size for a non-trivial regular linear space with 106 points ( [8]).The only known 2-(106, 6, 1) design found by Mills ([11]) is not point-transitive, which is one of Miscellaneous Constructions ( [8]) and has a cyclic automorphism groups of order 53.It is a question whether there exist point-transitive 2-(106, 6, 1) designs, just like the 2-(91, 6, 1) designs.In this paper, we prove that there is no 2-(106, 6, 1) designs admitting a point-transitive automorphism group.
Theorem 1.Let S be a 2-(106, 6, 1) design, and G be an automorphism group of S. Then G cannot be transitive on points of S.
Our paper is organized as follows.Section 2 presents some preliminary results and notation.In Section 3, by considering the number of fixed points of an involutive automorphism, we bound the size of the 2-part of | Aut(S)|.In Section 4, we get a bound on | Aut(S)| and prove Theorem 1.

Preliminary results and notation
Let S be a linear space with v points, K be a set of positive integers such that v k for every k ∈ K and the set of line-sizes of S is contained in K.Note that it is not required that there is a line of size k for any k ∈ K. Let α be a point of P, then and for each k ∈ K, In particular, if S is a non-trivial finite regular linear space, then the following result is well-known.Lemma 2. [5, Lemma 2.1] Let S be a non-trivial finite regular linear space.Then the electronic journal of combinatorics 21(1) (2014), #P1.58 and where k is the line size of S, and r is the number of lines through a point.
The following results are very useful for the proof of Theorem 1. Lemma 4. [3, Lemma 1] Let S be a linear space, α a point of S, and r α be the degree of α.Then all lines of size > r α contain α, and for any point β of P, β = α, the number of lines of size > r α containing β is at most one.
Throughout this paper, we assume that the following hypothesis holds: Hypothesis: Let S = (P, L) be a 2-(106, 6, 1) design, and G be a point-transitive subgroup of Aut(S).Let N : Q be the semidirect product of groups N by Q, N × Q the direct product of groups N and Q, |G| p the p-part of |G|, and |G| p the p -part of |G|.

The 2-part of | Aut(S)|
In this section, our aim is to obtain the maximal size of the 2-part of | Aut(S)|.We begin this section with some information given in [3] about the linear spaces.Assume that 2 | | Aut(S)| and t is an involution of Aut(S).Let D = (Fix(t), L Fix(t) ) be the linear space induced by Fix(t) and K = {2, 4, 6} containing the set of its line sizes.In view of (1), we get for each α ∈ Fix(t).Since a non-fixed point of t cannot be on two fixed lines of it, all the non-fixed points t on its fixed lines of S are distinct.Thus Combing ( 2) and ( 4), we obtain Now for each point α ∈ Fix(t), define the weight So that (5) can be written as If r 2 α = x, r 4 α = y and r 6 α = z, then we say that α is of type (x, y, z).
Lemma 5. Assume that the Hypothesis holds and let t be an involution of Aut S. Then |Fix(t)| = 18, 20 or 22.
Proof.We prove this Lemma by dealing with the three cases separately.
If α 0 is of type (0, 3, 2), then α 0 lies on all lines of size 6 and r 6 α 1 for any other point α ∈ Fix(t) by Lemma 4. Exactly, the 10 points on the lines of size 6 through α 0 are of type (m, n, 1) having weight at least 6, and the 9 points on the lines of size 4 through α 0 are of type (x, y, 0) having weight at least 5, where m, n, x and y are non-negative integers.Then and we have a contradiction.Similarly, we can prove that α 0 is not of type (1, 1, 3).Therefore, |Fix(t)| = 20.
If α 0 is of type (0, 4, 1), then α 0 lies on all lines of size 6 and r 6 α 1 for any other point α ∈ Fix(t) by Lemma 4. Or rather, the 5 points on the line of size 6 through α 0 are of type (m, n, 1) with weight 2, and the 12 points on the lines of size 4 through α 0 are of type (x, y, 0) with weight 13  2 , where m, n, x and y are non-negative integers.Then a contradiction.Thus there is no point of type (0, 4, 1) and ω(α) 15 2 for any α ∈ Fix(t) with r 6 α 1.If α 0 is of type (1, 2, 2), then α 0 lies on all lines of size 6 and r 6 α 1 for any other point α ∈ Fix(t) by Lemma 4. In particular, the 10 points on the lines of size 6 through α 0 are of type (m, n, 1) with weight 15 2 , and the 7 points which do not lie on the lines of size 6 through α 0 are of type (x, y, 0) with weight 13  2 , where m, n, x and y are non-negative integers.This implies which is impossible.Similarly, α 0 cannot be of type (2, 0, 3).Thus |Fix(t)| = 18.
Lemma 6. Assume that the Hypothesis holds and let t be an involution of Aut (S).If there is a line λ of S contained in Fix(t), then |Fix(t)| = 12, 14 or 16.
the electronic journal of combinatorics 21(1) (2014), #P1.58 Proof.Since there is a line λ ∈ S such that λ ⊆ Fix(t), the linear space D induced by Fix(t) has at least one line of size 6.
Since D has at least one line of size 6, there is no point of type (2, 3, 0) by Lemma 4. If α 0 is of type (0, 2, 1), then α 0 lies on all lines of size 6, and r 6 α 1 for any other point α ∈ Fix(t).Precisely, the 5 points on the line of size 6 through α 0 are of type (6, 0, 1) having weight 12, and the 6 points on the lines of size 4 through α 0 are of type (8, 1, 0) having weight 33  2 .Then which is impossible.If α 0 is of type (1, 0, 2), then α 0 lies on all lines of size 6 and r 6 α 1 for any other point α ∈ Fix(t).The 10 points which lie on the lines of size 6 through α 0 are of type (6, 0, 1) having weight 12, and the point α 1 such that {α 0 , α 1 } is the line of size 2 is of type (11, 0, 0) having weight 22. Then If α 0 is of type (3, 1, 1), then α 0 lies on all lines of size 6 and r 6 α 1 for any other point α ∈ Fix(t).Let α 1 , α 2 and α 3 be the points that joined to α 0 form the there lines of size 2. Suppose that there is a line of size 4 not containing α 0 .Then this line must contain two points of {α 1 , α 2 , α 3 }.Thus r 4 Let λ 1 ∈ L be the line containing α j that has non-empty intersection with λ.Suppose that 1 = S 0 is the kernel of the action of S α j on the points of λ

Lemma 3 . [ 4 ,
Lemma 1]  Let S be a finite regular linear space, G an automorphism group of S, and H = 1 a subgroup of G. Then |Fix(H)| r unless every point lies on a fixed line and then |Fix(H)| r + k − 3.