One of the most outstanding problems in combinatorial mathematics
and geometry is the problem of existence of finite projective planes
whose order is not a prime power

Hoxha, Razim

Bulgarian Digital Mathematics Library at IMI-BAS

Serdica Math. J. 27 (2001), 279-286ON PROJECTIVE PLANE OF ORDER 13 WITH AFROBENIUS GROUP OF ORDER 39 AS A COLLINEATIONGROUPRazim HoxhaCommunicated by S. L. TroyanskiAbstract. One of the most outstanding problems in combinatorial math-ematics and geometry is the problem of existence of finite projective planeswhose order is not a prime power.It is well known that a Desarguesian plane of order p, p prime number,has not a proper subplane, also every known non Desarguesian plane (of orderpr, p – odd prime number and r ≥ 2) has a subplane of order 2.It was shown by E. Ademaj [1] that: If P is a projective plane of order11 on which operates a group G = 〈ρ, τ/ρ7 = τ9 = 1, ρτ = ρ2〉 of order 63, thenG cannot fix a subplane of order 2.The next number is p = 13.In this paper we shall look projective plane of order 13 with a Frobeniusgroup of order F3.13 . Using the method of tactical decomposition, we shall con-struct the orbit structure of Frobenius group as a collineation group of projectiveplane of order 13.2000 Mathematics Subject Classification: 05B25.Key words: projective plane, collineation group, orbit structure.280 Razim HoxhaWe begin with two definitionsDefinition 1. A projective plane of order n ≥ 2 is a set P with n2+n+1elements called points with n2 + n+ 1 subsets of P called lines, such that:1. each line contains (is incident with) n+ 1 points,2. any two (distinct) lines intersect in exactly one point.Definition 2. A permutation α of the points P which maps lines of Ponto lines of P is called a collineation (automorphims).The purpose of this paper is to prove the following result:Theorem. It exists orbit structure of a Frobenius group of order 39 as acollineation group of projective plane of order 13.P r o o f. Let P be a projective plane of order 13 which possesses a Frobe-nius groupG = 〈ρ, µ/ρ13 = µ3 = 1, ρµ = ρ3〉of order 39 as a collineation group. Then P has 132 + 13 + 1 = 183 points andlines. We take a partition of the point set P.P = {P∞, P1, P2, . . . , P14},where Pi = {i0, i1, i2, . . . , i12}, i = 1, 2, . . . , 14 and bloch setB = {B∞, B1B2, . . . , B14} so that the partitionP = {P∞, P1, P2, . . . , P14, B∞, B1B2, . . . , B14},is a tactical decomposition of P.According to the definition of tactical decomposition, let r to a collineationof order 13 of AutP. The we can set:(∗) ρ = (∞)(10, 11, . . . , 112)(20, 21, . . . , 212) . . . (140, 141, . . . , 1412)where ∞ is a fixed point and 10, 11, . . . , 112, 20, 21, . . . , 212, . . . , 140, 141, . . . , 1412are all other points of P.Since the number of orbits of blocs and points is the same, for the fixedbloc we can setv = {∞, 10, 11, . . . , 112}Since through every point of P passes exactly 14 = 13 + 1 lines, for the otherlines through ∞ without loos of generality we can setP = {∞, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140}On projective plane of order 13. . . 281with the action of 〈ρ〉 on line p, we get the other lines through ∞:pρ = {∞, 21, 31, . . . , 141}pρ2= {∞, 22, 32, . . . , 142}. . . . . . . . . . . . . . . . . . . . . . . . . . . .pρ12= {∞, 212, 312, . . . , 1412}We have constructed one nontrivial orbit of lines of P (the lines through∞). There are exactly 13 futher nontrivial 〈ρ〉- orbits of lines of P.The collineation ρ regarded as the permutation of the points of P is givenby (∗). We can see that restriction of ρ at Z13 = {0, 1, . . . , 12} is ρ/Z13 =(0, 1, 2, . . . , 12) that is ρ/Z13 : x → x + 1(mod 13). The assumption is that on〈ρ〉 acts the collineation µ of order 3. That is µ normalizes 〈ρ〉 and replaces〈ρ〉-orbits (of points) again into the 〈ρ〉 orbits and also it permutes the ordinalnumbers (indeces) 0, 1, 2, . . . , 12. We can take of restriction ρ/Z13.Fromρ1 = ρ/Z13 = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)ρ13 = (0, 3, 6, 9, 12, 2, 5, 8, 11, 1, 4, 7, 10)it follovs thatµ1 = µ/Z13 = (0)(1, 3, 9)(2, 6, 5)(4, 12, 10)(7, 8, 11).One can see that 〈µ1〉 to the conjugation (also conjugation with help of ρi1, i ∈Z13) is the only subgroup of order 3 (in Σ13) normalizing 〈ρ1〉 = 〈(0, 1, 2, . . . , 12)〉.Besides we can easily verify thatµ1 : x→ kx (x ∈ Z13, compution mod 13), k ∈ {3, 9}.In the above presentation of µ1 we have chosen µ1 so that it fixes index0. In fact, µ1 which is of order 3, fixes one index from Z13, which we take to bezero. Other wise, it can be any other index by the conjugation µ1 → µρi11.The presentation of the permutation of the big (orbital) numbers 1, 2, . . . , 14— under µ, without loss of generality, can be deseribed byµ = (∞)(1)(2)(3, 4, 5)(6, 7, 8)(9, 10, 11)(12, 13, 14)282 Razim HoxhaThe collineation µ (as it normalizes ρ) fixes globaly the sets {∞}, {1}, {2},of the points and {v}, {p}, {l} of the lines of P. The other lines pρi, i =0, 1, 2, . . . , 12 leavs invariant, as m permutes the orbital numbers.Hence we have two possibilities of the action of G = 〈ρ, µ〉 on a projectiveplane P of order 13.Case I: G acts faithfully on a subplane of order 3.Case II: Does not fix a subplane of order 3.We must contruct the representatives other remaining 13 nontrivial orbits of linesof P.At first we difine the Haming number:H = (|ρ| − 1)λ = (|13| − 1)1 = 12We have the following presentation of number 12:(a) 12 = 4 · 3(b) 12 = 3 · 2 + 2 · 1 + 2 · 1 + 2 · 1(c) 12 = 3 · 2 + 3 · 2(d) 12 = 2 · 1 + 2 · 1 + 2 · 1 + 2 · 1 + 2 · 1 + 2 · 1.Thus, we shall look the case (a) for invariant line, but other lines we getof from type (b).Since the next line must be µ-invariant, we have the following presentationfor the µ-invariant line l:I: l = {10, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}II: l = {1, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 2}In case I, we construct other 12 representatives (of 〈ρ〉-orbits lines) whichpasses through 10.We define number (game product):(k1, l) = |r|λ = 13 · 1 = 13Since, Haming number H = 3 · 2 + 2 · 1 + 2 · 1 + 2 · 1, then for line k1 wecan set:k1 = {1, a, a, a, b, b, c, c, d, d, e, f, g, h}so that (k1, l) = 13. Let a = 2, then (k1, l) = 1+ 12 = 13. Since b, c, d, e, f, g, h ∈{12, 13, 14}, and on the one other and b, c, d, e, f, g, h /∈ {3, 4, 5, 6, 7, 8, 9, 10, 11},then we conclude a 6= 2.Suppose a = 3. Then we havek1 = {1, 3, 3, 3, b, b, c, c, d, d, e, f, g, h}On projective plane of order 13. . . 283and (k1, l) = 1 + 3 = 4. For b = 2 have (k1, l) = 4 + 8 = 12. Since c, d /∈{4, 5, 6, 7, 8, 9, 10, 11} and c, d ∈ {12, 13, 14}, then we must have c = 12, d = 13and have (k1, l) = 12 + 0 = 12. Since e ∈ {4, 5, 6, 7, 8, 9, 10, 11, 14}, then havee = 4, and have (k1, l) = 12 + 1 = 13. Since f, g, h /∈ {5, 6, 7, 8, 9, 10, 11} andf, g, h ∈ {14}, then we conclude b 6= 2. Permuting numbers 4, 5, 6, we concludethat b /∈ {4, 5, 6}.Let b = 7, thenk1 = {1, 3, 3, 3, 7, 7, c, c, d, d, e, f, g, h}and (k1, l) = 4 + 2 = 6. For c = 2 we have (k1, l) = 6 + 8 = 14 it means c 6= 2.Permuting numbers 4, 5, 6 we conclude that c /∈ {4, 5, 6}. If c = 8, thenk1 = {1, 3, 3, 3, 7, 7, 8, 8, d, d, e, f, g, h}and (k1, l) = 6 + 2 = 8. Since, and d 6= 2, d /∈ {4, 5, 6} then we take d = 9, andwe havek1 = {1, 3, 3, 3, 7, 7, 8, 8, 9, 9, e, f, g, h}where (k1, l) = 8 + 2 = 10. For e = 2 have (k1, l) = 10 + 4 = 14 it means e 6= 2.Since e ∈ {4, 5, 6, 10, 11, 12, 13, 14} then let e = 4, and we havek1 = {1, 3, 3, 3, 7, 7, 8, 8, 9, 9, 4, f, g, h}where (k1, l) = 10 + 1 = 11. Permuting numbers 5, 6, 10, 11, 12, 13, 14 we mayconclude f, g /∈ {5, 6} and f, g ∈ {10, 11, 12, 13, 14}. If f = 10 and g = 11, thenwe havek1 = {1, 3, 3, 3, 7, 7, 8, 8, 9, 9, 4, 10, 11, h}where (k1, l) = 11+2 = 13. Since, h /∈ {5, 6} and h ∈ {12, 13, 14} suppose h = 12.Then, we have the following solution for line k1k1 = {1, 3, 3, 3, 7, 7, 8, 8, 9, 9, 4, 10, 11, 12}.We act with 〈µ〉 on line k1, thenkµ1= {1, 4, 4, 4, 5, 8, 8, 6, 6, 10, 10, 11, 9, 13}kµ21= {1, 5, 5, 5, 3, 6, 6, 7, 7, 11, 11, 9, 10, 14}Since, the conditions are filled (k1, l) = (k1, kµ1) = (k1, kµ21) = 13 and (kµ1, kµ21) =13 then we conclude that these are three representatives lines (of 〈ρ〉 – nontrivialorbits) which passes through 10. Now, we shall construct the line k2:k2 = {1, a, a, a, b, b, c, c, d, d, e, f, g, h}284 Razim Hoxhaso that (kµi2, l) = 13, (kµi2, kµj2) = 13, where i 6= j and i, j ∈ {0, 1, 2}.Using the same method we may construct other representatives line k2:k2 = {1, 7, 7, 7, 4, 4, 6, 6, 12, 12, 9, 13, 14, 2}and the conditions are filled (k2, l) = (k2, kµi1) = (kµi1, kµj1) = 13, where i 6= j andi, j ∈ {0, 1, 2}.We act with 〈µ〉 on line k2 and we havekµ2= {1, 2, 5, 5, 8, 8, 8, 7, 7, 10, 13, 13, 14, 12}kµ22= {1, 2, 3, 3, 6, 6, 6, 8, 8, 11, 14, 14, 12, 13}Using the same method we may construct other the representatives linek3:k3 = {1, 2, 5, 6, 8, 9, 9, 9, 10, 10, 12, 12, 14, 14}which fills given conditions.We act with 〈µ〉 on line k3, then we havekµ3= {1, 2, 3, 7, 6, 10, 10, 10, 11, 11, 13, 13, 12, 12}kµ23= {1, 2, 4, 8, 7, 11, 11, 11, 9, 9, 14, 14, 13, 13}Finally, using the same method, we obtain the following solution for linek4:k4 = {1, 2, 3, 3, 4, 4, 5, 7, 10, 10, 13, 14, 14, 14}and with the action of 〈µ〉 on in this line we may set:kµ4= {1, 2, 4, 4, 5, 5, 3, 8, 11, 11, 14, 12, 12, 12}kµ24= {1, 2, 5, 5, 3, 3, 4, 6, 9, 9, 12, 13, 13, 13}On projective plane of order 13. . . 285Thus, we obtain the following orbital structure (in case I):l = 1 2 2 2 2 3 4 5 6 7 8 9 10 11k1 = k1 = 1 3 3 3 4 7 7 8 8 9 9 10 11 12k2 = kµ1= 1 4 4 4 5 8 8 6 6 10 10 11 9 13k3 = kµ21= 1 5 5 5 3 6 6 7 7 11 11 9 10 14k4 = k2 = 1 2 4 4 7 7 7 6 6 9 12 12 13 14k5 = kµ2= 1 2 5 5 8 8 8 7 7 10 13 13 14 12k6 = kµ2= 1 2 3 3 6 6 6 8 8 11 14 14 12 13k7 = k3 = 1 2 5 6 8 9 9 9 10 10 12 12 14 14k8 = kµ3= 1 2 3 7 6 10 10 10 11 11 13 13 12 12k9 = kµ23= 1 2 4 8 7 11 11 11 9 9 14 14 13 13k10 = k4 = 1 2 3 3 4 4 5 7 10 10 13 14 14 14k11 = kµ4= 1 2 4 4 5 5 3 8 11 11 14 12 12 12k12 = kµ24= 1 2 5 5 3 3 4 6 9 9 12 13 13 13Thus, we showed that, there exists orbit structure of the Frobenius groupof order 39 as a collineation group of projective plane of order 13. We can saythat in orbit structure of the lines the Haming number is 12 but the game productis 13 that in fact is a condition that lines will be compatible (it means that theyfulfill conditions of projective plane).The prof of the Theorem is complete.REFERENCES[1] E. Ademaj. On a projective plane of order 11 on which operates a groupof order 63 which fixes a subplane of order 2. Glasnik matematicˇki 19, 39(1984) 217–224.286 Razim Hoxha[2] A Beutelspacher. Einfu¨hrung die Geometrie I. B.I. – Wissenchaftsverlag,Mannheim – Wien – Zu¨rich, 1982.[3] D. R. Hughes, F. C. Piper. Projektive planes. Springer-Verlag, Berlin –Heidelberg – New York, 1973.University of Prishtina38 000 Prishtine¨Kosove¨Received October 26, 2000Revised July 11, 2001

On Projective Plane of Order 13 with a Frobenius Group of Order 39 as a Collineation Group

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On Projective Plane of Order 13 with a Frobenius Group of Order 39 as a Collineation Group

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