by a p-sylow of PSL(2,q)

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1 International Mathematical Forum, Vol. 8, 201, no. 15, HIKARI Ltd, (2q +1)-Arcs in PG(,q ) Stabilized by a p-sylow of PSL(2,q) Hans-Joachim Kroll Zentrum Mathematik, Technische Universität München München, Germany kroll@ma.tum.de Rita Vincenti Dipartimento di Matematica e Informatica Università degli Studi di Perugia Via Vanvitelli 1, 0612 Perugia, Italy alice@unipg.it Copyright c 201 Hans-Joachim Kroll and Rita Vincenti. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract We construct arcs K of cardinality 2q + 1 in the projective space PG(,q ), q = p h, p> prime, from a cubic curve C. By construction, K is stabilized by a p-sylow subgroup of the projectivities preserving C and it is contained in no twisted cubic of PG(,q ). Mathematics Subject Classification: 51A0, 51E20, 51E21 Keywords: Twisted cubic, arc, finite projective geometry, group of projectivities, Sylow subgroup 1 Introduction A k-arc K in a d-dimensional projective space PG(d, F ) over a finite field F is a set of k points, no d + 1 of which belong to a hyperplane. A k-arc K is

2 728 H.-J. Kroll and R. Vincenti complete if there is no (k + 1)-arc containing K. It is well known that arcs are interesting both from a geometrical point of view and to get good linear codes. A rational normal curve C in PG(d, F ) is a complete ( F + 1)-arc. For char F 2 each ( F +1)-arc in PG(,F) is a twisted cubic (cf. [1], Theorems , 21.2., ). In [4] G. Korchmaros and N. Pace construct for an odd prime power q and odd n>1 a family of large complete arcs in PG(2,q n ) which are not contained in a conic, by using the group PSL(2,q) in its action on an irreducible conic in PG(2,q). They consider the union K of a conic in PG(2,q) and an orbit of a point of PG(2,q n ) outside PG(2,q). It turns out that K is an arc if the point fulfills some conditions. The group PGL(2,F) acts as an automorphism group on every rational normed curve of P G(d, F ). Therefore the question arises whether a construction similar to that of Korchmaros and Pace is possible for higher dimensions. Our first attempts to use PSL(2,q) like in [4] went wrong. Indeed one can show using a programm written by M. Giulietti that for q 1 prime and for q =25, 49, 121 the union K of a cubic C 0 in PG(,q) and an orbit of a point in PG(,q ) outside PG(,q) is an arc if and only if the point lies on the cubic C in PG(,q ) containing C 0. Therefore it is impossible to take over the construction with the group PSL(2,q). So we came to the conclusion to use a proper subgroup of PSL(2,q) accepting that the construction would yield a smaller arc. Therefore we tried to use a p-sylow group of order q, looking for points which can be added to a cubic curve of PG(,q) in order to obtain a (2q + 1)-arc of PG(,q ). The aim of this paper is to show the existence of arcs of small cardinality 2q +1 in PG(,q ), q 5 odd, which are stabilized by a group of projectivities isomorphic to a p-sylow subgroup of the group PSL(2,q) and which are not contained in cubic curves. We wish to thank M.Giulietti for his help. 2 Notations and preliminary results Denote F a finite field with F 5 and p = char F. Let PG(,F)= (P, L) be a -dimensional projective space over F, where P is the set of the points, L is the set of the lines. Denote Π the set of the planes. Let C Pbe a cubic curve of PG(,F).

3 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 729 The canonical form of C is C = {P (t) = t,t 2,t,1 t F } { } where x 0,x 1,x 2,x denotes the point determined by (x 0,x 1,x 2,x ) F 4 \{0} and = 1, 0, 0, 0. A plane π Πis-secant or 2-secant plane if C π =or C π =2, π is 1-secant plane if C π = 1 and if in some complexification PG(,F ) of PG(,F) 1 the corresponding extension of π is a - or a 2-secant plane, π is osculating plane if C π = 1 and π is not a 1-secant plane (cf.[1], p.24, Corollary 4). A -secant plane π with C π = {P (t 1 ),P(t 2 ),P(t )} can be represented by the equation (1) x 0 (t 1 + t 2 + t )x 1 +(t 1 t 2 + t 1 t + t 2 t )x 2 t 1 t 2 t x =0. Every 2-secant plane can be represented by an equation of the form (2) x 0 (2t 1 + t 2 )x 1 +(t t 1 t 2 )x 2 t 2 1t 2 x =0. Through each point P = P (t) Cthere is exactly one osculating plane π P represented by () x 0 tx 1 +t 2 x 2 t x =0. The osculating plane π at the point has the equation x =0. The points of A = P\π represented by (x 0,x 1,x 2, 1). are the affine points of PG(,F) and can be Denote G the group of projectivities of PG(,F) fixing C. As F 5 then G = PGL(2,F). The mapping g : t at+b, a, b, c, d F with ad bc 0 in PGL(2,F) ct+d corresponds to the projectivity represented by the matrix a a 2 c ac 2 c (4) M = a 2 b a 2 d +2abc bc 2 +2acd c 2 d ab 2 b 2 c +2abd ad 2 +2bcd cd 2 b b 2 d bd 2 d so that the action of g on C is described by 1 i.e. F is a quadratic extension of F. (t,t 2,t,1) (t,t 2,t,1)M

4 70 H.-J. Kroll and R. Vincenti (cf.[1], p.2, Lemma 21.1., (i)). The group G acts triply transitively on C (cf. [1], p.24, Corollary 1). Let S be the p-sylow subgroup of G isomorphic to the p-sylow subgroup of PGL(2,F) whose elements are the mappings σ b : t t + b, for all b F.By (4) the element of S corresponding to σ b is represented by the matrix (5) M b = b b 2 2b 1 0 b b 2 b 1. From the definition it follows that S is isomorphic to the additive group of F. The elements of S have a common fixed point, namely and every element of S\{id} has only this fixed point (cf. [], p.191, Satz 8.2, a)). For a point P and a group of projectivities H denote H(P )={hp h H} the orbit of P under the action of H. Proposition 1 Let P be an affine point of PG(,F). Then S(P ) { } is a twisted cubic in PG(,F). Proof. We have P 0 = 0, 0, 0, 1 C. By (5) the orbit of P 0 is S(P 0 ) = { b,b 2,b,1 b F }, hence C = S(P 0 ) { }. The translations with respect to the plane π are represented by matrices T a = for a =(a 0,a 1,a 2 ) F. a 0 a 1 a 2 1 An easy computation shows that for every matrix M b = b b 2 2b 1 0 b b 2 b 1

5 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 71 representing an element of S we have T a M b = M b T a. Since the group of all translations acts regularly on the set of affine points there is a translation τ with τ(p 0 )=P. Thus by the above consideration S(P )=Sτ(P 0 )=τs(p 0 )=τ(c \ { }) =τ(c) \ { }. Hence S(P ) { } = τ(c) is a twisted cubic. Note that the orbit S(P ) of a point P = x 0,x 1,x 2,x lies on a conic through precisely when P π and x 2 0, such a conic belonging to the plane x =0. S(P ) lies on a line precisely when P π and x 2 = 0, such a line being just the line represented by x 2 =0=x. Let F 0 be a subfield of F, P 0 = { x 0,x 1,x 2,x (x 0,x 1,x 2,x ) F0 4 \ {(0, 0, 0, 0)}} and L 0 = {L P 0 L L, L P 0 2}. Then (P 0, L 0 ) is a subgeometry PG(,F 0 )ofpg(,f). We call the points of P 0 the rational points of PG(,F). The set C 0 = C P 0 = {P (t) C t F 0 } { } is a twisted cubic in PG(,F 0 ). Let us name small cubic each twisted cubic in PG(,F 0 ). Set G 0 = {g G g(p 0 )=P 0 }. Since every element of G fixes C, then each element of G 0 fixes C 0 and G 0 = PGL(2,F0 ). Let S 0 = {σ S σ(p 0 )=P 0 }. Clearly S 0 < G 0 and every element of S 0 is represented by a matrix B b with b F 0. From now on we assume that F is a cubic extension of F 0 = GF (q) so that F = GF (q ). Lemma 2 G 0 acts transitively on C\C 0. Proof. Because of PGL(2,q) = q(q 2 1) = C \ C 0 it is enough to prove that for each point P C\C 0 the stabilizer of P in G 0 is trivial. Let P = P (t) C\C 0.Ifg G 0 fixes P then v =(t,t 2,t,1) is an eigenvector of the matrix A representing g. As g G 0 we may assume that the entries of A are in GF (q). Therefore v q and v q2 are eigenvectors of A as well. As P P\P 0, the points P, P q,p q2 are all different. Therefore g has three fixed points and g is the identity of G 0. Note that S 0 divides C\C 0 into a set Ω = {O i i =1,..., q 2 1} of q 2 1 orbits of cardinality q.

6 72 H.-J. Kroll and R. Vincenti Points of P 0 in planes intersecting C\C 0 The arc we show the existence is constructed by adding to a fixed orbit O Ω the points of a suitable small cubic curve C 0 C 0 in PG(,q). Hence we need to show the existence of a cubic C 0 C 0 in PG(,q) such that K = O C 0 is an arc in PG(,q ). For i =1, 2, define Π is = {π Π O π = i}. The curve C 0 must satisfy the following properties: 1) no point of C 0 belongs to a plane of Π s, 2) no 2 points of C 0 belong to a plane of Π 2s, ) no points of C 0 belong to a plane of Π 1s. For our construction it is therefore important to know how many points of PG(,q) lie on every plane which has a non-empty intersection with O. In this section we put together some results concerning this problem. Let x rx 2 sx u GF (q)[x] be an irreducible polynomial over GF (q) and let w GF (q ) \ GF (q) be a root of it so that w = rw 2 + sw + u and GF (q )=GF (q)(w). Fix the orbit O = S 0 (P (w)) = {P (w + b) b GF (q)}. Denote O 1 = S 0 (P ( 2 1 w)) and O 2 = S 0 (P ( 2w)) the two orbits defined by the points P ( 2 1 w),p( 2w) C\C 0. Let A 0 = P 0 \ π denote the set of the affine points of PG(,q). Lemma Let P 1,P 2 Oand P C with P 1 P 2, P P 1,P 2.Letπ Π be the plane containing the points P 1,P 2,P. Then we have: (i) If P / O 1 O 2, P then π A 0 =1. (ii) If P O 1 O 2 then π A 0 = q or π A 0 =0. (iii) If P = then π P 0 = { }. Proof. There are b i GF (q) with P i = P (w + b i ) for i =1, 2 and there are c, d GF (q) with P = P (cw + d) ifp. Firstly we consider the case P.

7 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 7 According to (1) the plane π is described by the equation x 0 ((2 + c)w + b 1 + b 2 + d)x 1 +((w + b 1 )(w + b 2 )+(2w + b 1 + b 2 )(cw + d))x 2 = =(w + b 1 )(w + b 2 )(cw + d)x. As it needs to determine the solutions (x 0,x 1,x 2, 1) with x i GF (q), we split the equation into the following system of three linear equations x 0 (b 1 + b 2 + d)x 1 +(b 1 b 2 +(b 1 + b 2 )d)x 2 = cu + b 1 b 2 d (6) (2 + c)x 1 +((b 1 + b 2 )(1 + c)+2d)x 2 = cs +(b 1 + b 2 )d + b 1 b 2 c (1 + 2c)x 2 = cr +(b 1 + b 2 )c + d. The rank of the matrix of the coefficients is precisely when c/ { 2 1, 2}, i.e. to P / O 1 O 2. Hence we obtain (i) and (ii). For P = the plane π is described by the equation x 1 (2w + b 1 + b 2 )x 2 +(w 2 +(b 1 + b 2 )w + b 1 b 2 )x =0. InGF (q) the only solution is x =0,x 2 =0,x 1 = 0. This shows (iii). If in equation (1) we replace t by t 1 we obtain equation (2), that is, the equation of a 2-secant plane intersecting the cubic C in the points P (t 1 ) and P (t 2 ). Therefore from the proof of Lemma we obtain Lemma 4 For each 2-secant plane π with π O =2follows π A 0 =1. By this lemma it is clear that the 2-secant planes belonging to Π 2s select points of A 0 that could be chosen for our construction. From Lemma (i) follows π A 0 = 1 for all π Π s. Lemma 5 The mapping φ :Π s A 0,π π A 0 is injective and φ(π s ) = Π s = ( q ). Proof. (a) For π Π s there are three different elements b i GF (q) with P i (w + b i ) π, i =1, 2,. By setting a 2 = b 1 + b 2 + b,a 1 = b 1 b 2 + b 1 b + b 2 b,a 0 = b 1 b 2 b, we have for the cubic polynomial f π (x) =x a 2 x 2 + a 1 x a 0 GF (q)[x] the factorization f π (x) =(x b 1 )(x b 2 )(x b ), i.e. each plane π Π s corresponds with exactly one cubic polynomial f(x) GF (q)[x] with three different roots.

8 74 H.-J. Kroll and R. Vincenti By solving the system (6) of linear equations for c =1,d= b, for the image φ(π) =(x 0,x 1,x 2, 1) we get x 2 = r + a 2, x 1 = 2a 2(r + a 2 ) (s + a 1 ), x 0 = a 2 x 1 a 1 x 2 + a 0 + u. 9 (b) Now let π Π s a further plane and b i GF (q) with P (w + b i ) π for i =1, 2, and b 1 b 2 b b 1. According to (a) set a 2 = b 1 + b 2 + b,a 1 = b 1 b 2 + b 1 b + b 2 b,a 0 = b 1 b 2b. If φ(π) =φ(π ) then (a) implies r + a 2 = x 2 = r + a 2, hence a 2 = a 2 and 2a 2 (r + a 2 ) (s + a 1 ) = x 1 = 2a 2(r + a 2 ) (s + a 1),thusa 1 = a and a 2 x 1 a 1 x 2 + u + a 0 = x 0 = a 2x 1 a 1 x 2 + u + a 0, hence a 0 = a 0. Therefore the polynomial f π (x) =f π (x), hence π = π by (a). Denote B = A 0 \ (C 0 φ(π s ) the set of affine points in PG(,q) lying neither on a plane in Π s nor in the small cubic C 0. Proposition 6 (i) The cardinality of B is B = 1 6 q(5q2 +q 8). (ii) For each g S 0 we have g(b) =B. (iii) B is the union of 1 6 (5q2 +q 8) orbits with respect to the group S 0. Proof. (i) From Lemma 5 follows φ(π s ) = Π s = ( q ). As C 0 φ(π s )=, then B = q q q(q 1)(q 2) = q(5q2 +q 8).! 6 (ii) holds as O is an orbit, hence g(π s )=Π s and thus g(φ(π s )) = φ(π s ) for all g S 0. (iii) follows from (i), (ii) and the fact that each orbit has cardinality q. Let Π 0 be the set of the planes in Π that are extension of planes of PG(,q) meeting O.

9 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 75 Lemma 7 (i) For each α Π 0 it holds α O =1, i.e., Π 0 Π 1s. (ii) For each point P Othere exists a unique plane α Π 0 with P α. This plane α does not contain the point. (iii) For α Π 0 we have S 0 (α) ={gα g S 0 } =Π 0. Proof. (i) follows from Lemmas, 4. (ii) Let P = P (w + f) = ((w + f), (w + f) 2,w + f,1) be a point of O. Computations show that (w+f) =(rw 2 +sw+u)+fw 2 +f 2 w+f =(r+f)w 2 +(s+f 2 )w+(u+f ) (w + f) 2 = w 2 +2fw + f 2. Denote ax 0 + bx 1 + cx 2 + d = 0 the equation of a plane with a, b, c, d GF (q). The point P belongs to it precisely when that is a((w + f) + b(w + f) 2 + cw + f + d =0 a((r +f)w 2 +(s +f 2 )w +(u + f )) + b(w 2 +2fw+ f 2 )+c(w + f)+d =0. Such equation splits into three linear homogeneous equations over GF (q) a(r +f)+b =0, a(s +f 2 )+2fb+ c =0, a(u + f )+bf 2 + cf + d =0in the unknown (a, b, c, d). It is a 0, otherwise b = c = d = 0. Therefore we may assume a = 1 so that it needs to solve the following non-homogeneous system in (b, c, d) from which we get the unique solution r +f + b =0 s +f 2 +2fb+ c =0 u + f + bf 2 + cf + d =0 (b, c, d) =( (r +f), f 2 +2rf s, (f + rf 2 sf + u)). Hence there is a unique plane α Π 1s through P whose equation is x 0 (r +f)x 1 +(f 2 +2rf s)x 2 (f + rf 2 sf + u) =0. Since a 0 it holds / α. (iii) follows from (i) and (ii).

10 76 H.-J. Kroll and R. Vincenti By Proposition 1 and Lemma we know that the elements of C B = {S 0 (P ) { } P B}are small twisted cubics partitioning the set B. By Proposition 6 follows C B = 1 6 (5q2 +q 8). Denote C F = {C 0 C B π Π 1s : C 0 π 2}. Proposition 8 C F 1 6 (q2 +q 8). Proof. For C b C B and π Π 1s it is π C b. (a) For α, β Π 0 it holds α C b = β C b by Lemma 7 (iii). (b) Let π Π 1s and let C s = {C b C B π C b =}. Then π Π 0, hence / π by Lemma 7(ii). Thus C s q2 as two different cubics in C s meet only in. Because of (a) we have C F = C B \C s.thus C F = C B C s 1 6 (5q2 +q 8) 1 q2 = 1 6 (q2 +q 8). Let P g P 0 be the set of the points incident with the curves of C F. It is P g = 1 6 (q +q 2 8q). The points of P g are called good points as each curve of C F satisfies the properties 1) and ). Now it needs to find a subset P P g so that at least one of the cubic curves partitioning P satisfies 2). 4 Points of P 0 in planes of Π 2s By Lemma, (ii) and in order to find the subset P one must know first how many planes of Π 2s meet PG(,q) in lines. To this end and to prove Lemma 10 we need Lemma 9 If is a non-square then r 2 +s 0. Proof. Let be a non-square. Then from [2, Lemma 4.5], follows that each element of GF (q) is a cube. Let us consider the function F (x) =x rx 2 sx. Assume r 2 +s = 0. Then F (x + r )=(x + r ) r(x + r )2 r2 (x + r)= x + rx 2 + r2 x + r rx 2 2 r2 x r + r2 x + r = x + r. Hence the function x F (x + r ) is bijective since each element of GF (q) is a cube, and therefore F is bijective as well. Consequently, the polynomial x rx 2 sx u is

11 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 77 reducible over GF (q), a contradiction to our assumption that x rx 2 sx u is irreducible. Denote Π l = {π π Π 2s,π P 0 L 0 }. Let π Π l. From Lemmas, 4 follows that π is described by the first equation of the proof of Lemma with c(π) :=c { 2 1, 2}. Let Π l1 := {π Π l c(π) = 2 1 } and Π l2 := Π l \ Π l1. Lemma 10 (1) Π l1 = q(q 1) 2 (2) 2 Π l2 = q(q +1) for 0 r 2 +s/, / q(q 1) for r 2 +s, / q(q 1) for 0 r 2 +s/, q(q ) for r 2 +s, 2q(q 1) for r 2 +s =0 Proof. (1) Here c = 2 1. Then the system of the three linear equations (6) has a solution precisely when 2 1 (r+b 1 +b 2 )+d =0,orford = 1(b 2 1 +b 2 +r), that is for q(q 1) choices for b 2 1,b 2,d GF (q) as(b 1,b 2 ) determines the same plane of Π l with c = 2 1 as (b 2,b 1 ). (2) Here c = 2. Then the system (6) has a solution precisely when ( (b 1 + b 2 )+2d)( 2(r + b 1 + b 2 )+d) + ((b 1 + b 2 )d 2(s + b 1 b 2 )=0 or d 2 ((b 1 + b 2 )+2r)d +(b 1 + b 2 ) 2 + r(b 1 + b 2 ) b 1 b 2 s =0 that has solutions d = ((b 1 + b 2 )+2r) ± (b 1 b 2 ) 2 +4(r 2 +s) 2 precisely when Δ= (b 1 b 2 ) 2 +4(r 2 +s) is zero or it is a square of GF (q). By setting x = b 1 b 2 and Δ = y 2 we get the equation (7) x 2 +4(r 2 +s) =y 2. The equation (7) represents a conic C Δ in the affine plane over GF (q).

12 78 H.-J. Kroll and R. Vincenti Exactly the points (x, y) C Δ with x 0 do correspond to sets of planes in Π l2. Let CΔ 0 := {(x, y) C Δ x =0}. Assume first r 2 +s 0, that is C Δ irreducible. Then C Δ is an ellipsis when is a non-square of GF (q), and it is a hyperbola otherwise. Thus C Δ = q +1if / and C Δ = q 1if. We have { 0 for r CΔ 0 = 2 + / 2 for r 2 + Assume now r 2 +s = 0. Then by Lemma 9. Hence the conic C Δ consists of two lines through (0, 0) with the equations y = ϱx and y = ϱx where ϱ 2 =. Hence C Δ =2(q 1) + 1 and C 0 Δ =1. Thus we get C Δ \C 0 Δ = q +1 for 0 r 2 +s/, / q 1 for r 2 +s, / q 1 for 0 r 2 +s/, q for r 2 +s, 2(q 1) for r 2 +s =0 Since each point (x, y) C Δ with x 0 corresponds to exactly q planes of Π l2 as b 2 runs through GF (q) and b 1 = x + b 2, d = ((b 1+b 2 )+2r)+y, and since ( x, y) 2 corresponds to the same set of planes as (x, y) we obtain the assertion. The following remark might be useful in looking for applications of Lemma 10. Remark 1. Let p be a prime number, h an odd positive integer, q = p h and r, s, u GF (q) such that x rx 2 sx u GF (q)[x] is an irreducible polynomial. Then by [2, Lemma 4.5] and Lemma 9 follows / if and only if q 2 mod if and only if q 1 mod. If r 2 +s = 0 then. Denote L l1 and L l2 the set of the lines of PG(,q) arising respectively from the planes (1) Π l1 and (2) Π l2 of Lemma 10. By construction it is L l1 L l2 =. From the equations (6) of Lemma follows that every line of L l1 L l2 may be identified by the pairs (b 1,b 2 ) GF (q 2 ) with b 1 b 2 so that we may denote it

13 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 79 by l(b 1,b 2 ) as an element of L l1, l (b 1,b 2 ) as an element of L l2. The lines L l1 Let GF (q) + GF (q) be a partition of GF (q) = GF (q) \{0} so that if a GF (q) + then a GF (q). For a GF (q) + let us consider the line l(0,a) L l1 and the orbit S 0 (l(0,a)). Set L (0,a) = S 0 (l(0,a)). Lemma 11 - The set L = {L (0,a) a GF (q) + } is a partition of L l1 so that L l1 = L (0,a). a GF (q) + Proof. The set L = {L (0,a) a GF (q) + } consists of orbits of lines of L l1, each orbit containing q lines. As GF (q) + = q 1 = L 2 (0,a) and L l1 = q q 1 2 from Lemma 10 we get L l1 = L (0,a). a GF (q) + For b 1 + b 2 = a, b 1 b 2 =0,d= a+r, the line l(0,a) is represented (comparing 2 with the system (6)) by the system { x0 (6 1 ) a+r x a(a+r) x 2 2 = 1u 2 x 1 +(a +2r)x 2 = s + a(a + r). By the second equation we get x 1 = (a +2r) a(a + r) s x 2 so that by substituting in the first equation we can express the line l1, (0,a) by the system (6 1 ) { x0 = Ax 2 + B x 1 = Cx 2 + D where A = 6a2 +6ar +2r 2 2a(a + r) 2s +u,b= 6 6 C = (a +2r) a(a + r) s,d=.

14 740 H.-J. Kroll and R. Vincenti Lemma 12 - For every a GF (q) + the orbit L (0,a) = S 0 (l(0,a)) consists of q mutually skew lines precisely when 2A +C 2 +6D = a2 +2r 2 +6s is a nonsquare of GF (q), i.e. q =5or, q>5,r 2 +s =0and 1 is a non-square. Proof. Each line of l a,b L (0,a) may be represented by l(0,a)m b =(Ax 2 +B, Cx 2 +D, x 2, 1)M b =(Ax 2 +B +b(cx 2 +D)+b 2 x 2 + b,cx 2 + D +2bx 2 + b 2,x 2 + b, 1), that is by ((A +bc +b 2 )x 2 + B +bd + b, (C +2b)x 2 + D + b 2,x 2 + b, 1) while b runs in GF (q). Therefore the intersection of two lines l a,b1,l a,b2 L (0,a) with b 1 b 2 is represented by the following system in 4 unknown x 21,x 22,b 1,b 2 in equations and 4 parameters A, B, C, D (A +b 1 C +b 2 1 )x 21 + B +b 1 D + b 1 = A +b 2C +b 2 2 )x 22 + B +b 2 D + b 2 (C +2b 1 )x 21 + D + b 2 1 =(C +2b 2 )x 22 + D + b 2 2 x 21 + b 1 = x 22 + b 2. By eliminating x 21 and x 22 we get Ab 1 Ab 2 2 C2 b C2 b 2 Db 1 +Db b 1 2 b2 1b b 1b b 2 =0. By dividing by b 2 b 1 follows 2A C 2 6D + b 2 1 2b 1 b 2 + b 2 2 =0 that is (b 1 b 2 ) 2 = 2A +C 2 +6D. Hence, the two lines are skew precisely when 2A +C 2 +6D = a2 +2r 2 +6s is a non-square of GF (q). Setting a = x, this is equivalent to say that the equation x 2 +2( 1 r 2 +s) =y 2 has no solution other than the pairs (x, 0) (in case 2( 1 r 2 + s) is a square in GF (q)). Moreover, as x GF (q) +, roots of kind (0,y) have to be excluded by hypothesis. Hence, the lines L (0,x) are mutually skew precisely when the conic C 2 : x 2 + y 2 2( 1 r 2 + s) = 0 has no points in AG(2,q) out the lines y =0,x=0. If q = 5 and the irreducible polynomial is x rx 2 sx u = x x +1, then C 2 has no points in AG(2, 5) out the lines y =0,x=0.

15 (2q +1)-arcs in PG(,q ) stabilized by a p-sylow of PSL(2,q) 741 If q>5and r 2 +s 0, there exist points (x, y) with x, y 0 fulfilling the equation of C 2, while C 2 contains only the point (0, 0) precisely when r 2 +s =0 and 1 is a non-square. The lines L l2 Let a GF (q) and let d be one of the two possible values arising in Lemma 10, (2) for which lines L l2 exist. Let us consider the line l (0,a) L l2. For b 1 + b 2 = a, b 1 b 2 = 0 from (6) follows that the line l (0,a) is represented by the system equivalently by (6 2 ) { x0 =(a + d)x 1 ax 2 2u x 2 = 2 (a + r)+ d. (6 2) { x0 = Ex 1 + F x 2 = G where E = a + d, F = a (2(a + r)+d) 2u, G = 2 (a + r)+d. Consider the orbit S 0 (l (0,a)). Denote L (0,a) = S 0(l 2(0,a) ) The set L = {L (0,a) a GF (q) } is a partition of L l2 in orbits, each orbit containing q lines. Lemma 1 - For every a GF (q ) the orbit L (0,a) = S 0(l (0,a)) consists of q mutually skew lines. Proof. Each line l a,b L (0,a) may be represented by l a,b = l (0,a)M b =(Ex 1 + F, x 1,G,1)M b =((E +b)x 1 + F +b 2 G + b,x 1 + 2bG + b 2,G+ b, 1) while b runs in GF (q). An affine common point of two lines l a,b 1,l a,b 2 L (0,a) with b 1 b 2 is represented by the following system in 4 unknown x 1,x 1,b 1,b 2 in equations and parameters E, F, G (E +b 1 )x 1 + F +b 2 1G + b 1 =(E +b 2 )x 1 + F +b 2 2G + b 2 x 1 +2b 1 G + b 2 1 = x 1 +2b 2G + b 2 2 G + b 1 = G + b 2.

16 742 H.-J. Kroll and R. Vincenti From the third equation we get immediately b 1 = b 2, a contradiction. Moreover the infinite points of l a,b 1 and of l a,b 2 are different as they are (E + b 1, 1, 0, 0) and (E + b 2, 1, 0, 0), respectively. Hence l a,b 1 l a,b 2 = for b 1 b 2. As the lines of L l1 in an orbit are not necessarily mutually skew, it needs to delete from P g the affine points incident the lines of L l1. Denote P 1 = q q2 l. As P 1 = and P 2 g \P 1 = q2 4q > 0, the set l L l1 P = P g \P 1 is not empty. Let P P and denote C 0 = S 0(P ) { }. Define K = O C 0. We show that K is an arc. By construction it needs only to prove that no two points of C 0 belong to a line of L l2. Assume there are two points Q, R C 0 such that the line l = QR is a line of L l2. The orbit S 0 (l) consists of q lines that divide in pairs the points of C 0. Then, as C 0 = q + 1 and S 0(l) = q, the orbit S 0 (l) should contain some pair of incident lines, a contradiction. Thus we have Theorem 14 In PG(,q ) there exist (2q +1)-arcs stabilized by S 0 which are not contained in a cubic curve. References [1] J.W.P. Hirschfeld, Finite Projective Spaces of Three Dimensions, Clarendon Press, Oxford, [2] M. Giulietti, On plane arcs contained in cubic curves, Finite Fields and Their Applications 8, (2002) pp [] B. Huppert, Endliche Gruppen I, Springer-Verlag Berlin Heidelberg New York, [4] G. Korchmaros and N. Pace, Infinite Family of Large Complete Arcs in PG(2,q n ), with q odd and n>1 odd, Designs, Codes and Cryptography, Volume 55, Numbers 2-, May, 2010, pp Received: February 15, 201