Reconstructing a V W Plane from Its Collineation Group

Reconstructing a V W Plane from Its Collineation Group

Reconstructing a VW plane from its collineation group Cafer Caliskan, Spyros S. Magliveras Department of Mathematical Sciences Florida Atlantic University, 777 Glades Road, Boca Raton, FL 33431, U.S.A. [email protected], [email protected] Abstract In this study we analyze the structure of the full collineation group of certain Veblen-Wedderburn(VW) planes of orders 52, 72 and 112. We also discuss a reconstruction method using their collineation groups. 1 Introduction In [1] some group{theoretical methods for constructing both the Hughes plane of order q2 and the Figueroa plane of order q3, q an odd prime power, are discussed. The method is using the well{known linear group GL(3; q). In this paper, we discuss a reconstruction method for certain non{ desarguesian VW planes of some particular orders from their collineation groups. In section 2 we introduce some notation, definitions and preliminaries. In section 3 we discuss a particular Planar Ternary Ring (R; T ) of order p2, p an odd prime, which gives rise to the non{desarguesian VW planes α, β and γ, of orders 52, 72 and 112, respectively. In section 4 we analyze the structure of the full collineation groups of α, β and γ. In Section 5 we discuss how to reconstruct these particular VW planes from their collineation groups. 1 2 Preliminaries We assume the reader is familiar with the basics of finite projective planes and group theory. If G is a group acting on the set X, we denote by GjX the group action of G on X. If π is a projective plane, we denote by Pπ and Lπ, the sets of points and lines of π, respectively. We denote by (a) the set of points incident with a 2 Lπ and by (A) the set of lines incident with A 2 Pπ. If A; B 2 Pπ and A 6= B, we denote by AB the line in Lπ incident with A and B. Symmetrically if a; b 2 Lπ, a 6= b, ab denotes the point in Pπ incident with a and b. By a quadrangle of a plane π we mean a set of four points no three of which are collinear. A collineation of a projective plane π of order n is a permutation of its points which maps lines onto lines [2]. The set of all collineations of π forms a group under composition, called the full collineation group Gπ of π. Veblen-Wedderburn (VW ) systems are algebraic systems used to coor- dinatize projective planes, and planes coordinatized by VW systems are called VW planes. A VW system (R; +; ·) of elements with operations + and · satisfies the following axioms: (i) (R; +) is commutative. (ii) (R n f0g; ·) is a loop. (iii) (a + b)c = ac + bc, a; b; c 2 R. (iv) If a 6= b, xa = xb + c has a unique solution x. See [4] for further information about VW systems. A Planar Ternary Ring (PTR) is a structure (R; T ), where R is a nonempty set containing distinct elements called 0 and 1, and T : R3 ! R satisfying the following: (i) T (a; 0; b) = T (0; a; b) = b, 8a; b 2 R. (ii) T (1; a; 0) = T (a; 1; 0) = a, 8a 2 R. (iii) For every a; b; c 2 R, T (a; b; x) = c has a unique solution x 2 R. (iv) For every a; b; c; d 2 R, where a 6= c, T (x; a; b) = T (x; c; d) has a unique solution x 2 R. (v) For every a; b; c; d 2 R, where a 6= c, each of T (a; x; y) = b and T (c; x; y) = d has a unique solution (x; y) 2 R2. Note that the fifth axiom is redundant if R is finite. For further infor- mation about PTR's see [3]. Given a certain PTR, the corresponding projective plane π, with points Pπ, lines Lπ and incidence I ⊂ Pπ × Lπ, is constructed as follows: (i) Pπ = f(x; y): x; y 2 Rg [ f(x): x 2 Rg [ f(1)g, (ii) Lπ = f[a; b]: a; b 2 Rg [ f[a]: a 2 Rg [ f[1]g, (iii) For all a; b; x; y 2 R,(x; y)I[a; b] if and only if T (a; x; y) = b, (iv) (x; y)I[a], (x)I[a; b] if and only if x = a, (v) (x)I[1], (1)I[a], (1)I[1], (vi) (x; y) 6 I[1], (x) 6 I[a], (1) 6 I[a; b]. 3 A VW plane Let F be a finite field of order p2, p an odd prime, and R the set of elements of F. Define T : R3 ! R as follows: T (a; b; c) = ab + c if b is a square in F, and T (a; b; c) = apb + c if b is not a square in F. Proposition 1 Let R and T be as described above. Then (R; T ) is a PTR. Proof: Let a; b; c 2 R and a be a square in R. Then T (a; 0; b) = a0 + b = b = 0a + b = T (0; a; b), T (a; 1; 0) = a1 + 0 = a = 1a + 0 = T (1; a; 0), and T (b; a; x) = ba + x = c has a unique solution x 2 R. If a is not a square in R, then T (a; 0; b) = a0 + b = b = 0pa + b = T (0; a; b), T (a; 1; 0) = a1 + 0 = a = 1pa + 0 = T (1; a; 0), and T (b; a; x) = bpa + x = c has also a unique solution x 2 R. Now, let a; b; c; d 2 R, where a 6= c and a; c 6= 0 . We have the following cases: (i) If a and c are both squares in R, then T (x; a; b) = T (x; c; d) , xa+b = xc + d and xa + b = xc + d has a unique solution x 2 R. (ii) If a is not a square and c is a square, then T (x; a; b) = T (x; c; d) , xpa + b = xc + d. This equation has a unique solution x 2 R. See [5] for a proof. (iii) If neither a nor c is a square in R, then T (x; a; b) = T (x; c; d) , xpa + b = xpc + d , xp = (v=u), where u = a − c 6= 0 and v = d − b. But there exists t0 2 R such that (t0)p = (v=u). Therefore, xp = (t0)p. Hence there is a unique solution for T (x; a; b) = T (x; c; d). Hence, (R; T ) is a PTR. 2 In this study we use this particular Planar Ternary Ring (R; T ) of order p2, p = 5; 7 or 11, to construct the non-desarguesian projective planes α, β, and γ of orders 52, 72 and 112, respectively. It follows easily from the definition that α, β, and γ are VW planes. We compute the full collineation groups Gα, Gβ and Gγ of the planes α, β, and γ, respectively. Then we ask the following question: \Is it possible to reconstruct the planes α, β, and γ by only using their collineation groups?". 4 Structure of Gπ Let π be one of the planes α, β, or γ. Since π is of order p2, p = 5; 7 or 11, we assume that Pπ = fA0;A1; :::; Ap4+p2 g and Lπ = fa0; a1; :::; ap4+p2 g throughout the article. We observe that Gπ is not transitive on points and lines. Furthermore, there are three orbits on points, namely Θ1,Θ2 and 2 4 2 Θ3, of lengths 1, 2p and p − p , and three orbits on lines, namely Γ1, 2 4 Γ2 and Γ3, of lengths 2, p − 1 and p , respectively. Let Γ1 = fa0; a1g, where (a0) = fA0;A1; :::; Ap2 g and (a1) = fA0;Ap2+1; :::; A2p2 g. Then we have that Γ2 = (A0) n Γ1 and Γ3 = Lπ n (A0). Moreover, Θ1 = fA0g, Θ2 = ((a0) [ (a1)) n fA0g and Θ3 = Pπ n ((a0) [ (a1)). Furthermore, the actions Gπ j Θ2 and Gπ j Θ3 are faithful. 2 2 There is a subgroup K ≤ Gπ, of order p (p − 1), and K is normal in a subgroup H < Gπ, where [H : K] = 2. See Figure 1. Furthermore, there is a cyclic subgroup C < K of order (p2 − 1)=2. If C = hxi, then 2 (p2−1)=4 there is an element y 2 K such that y x = 1Gπ if p ≡ 3 (4), 2 (p2−1)=8 and y x = 1Gπ if p ≡ 1 (4). Moreover, the Sylow p{Subgroup 2 Sylp < K is of order p and K is the split extension of Sylp by the subgroup hx; yi generated by x and y. See the appendix for the presentations of K in Gα, Gβ and Gγ . In addition, there is an involution m such that H = hK; mi. The generators of the subgroup H, namely x; y; a; b and m, are represented as permutations on the subset f1; :::; p2g. Further, there is 0 −1 0 an involution u 2 Gπ n H such that for H = u Hu, H \ H = hmi and 0 Gπ = hH; ui = hH; H ; ui. See the appendix for the size and generators of the full collineation groups Gα, Gβ and Gγ . GΠ H H ′ ′ K .m K Syl Syl ′ p .b p .a .b′ .y .a′ .y′ C ′ C .x . x′ .u Figure 1: The full collineation group Gπ 5 Reconstruction from Gπ Counting Principle.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    11 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us