Let $N$ be a nilpotent group normal in a group $G$. Suppose that $G$ acts transitively upon the points of a finite non-Desarguesian projective plane $\mathcal{P}$. We prove that, if $\mathcal{P}$ has square order, then $N$ must act semi-regularly on $\mathcal{P}$.

In addition we prove that if a finite non-Desarguesian projective plane $\mathcal{P}$ admits more than one nilpotent group which is regular on the points of $\mathcal{P}$ then $\mathcal{P}$ has non-square order and the automorphism group of $\mathcal{P}$ has odd order

Gill, Nick

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arXiv

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Open Research OnlineThe Open University’s repository of research publicationsand other research outputsNilpotent Singer groupsJournal ItemHow to cite:Gill, Nick (2006). Nilpotent Singer groups. Electronic Journal of Combinatorics, 13(1) R94.For guidance on citations see FAQs.c© 2006 The American Mathematical SocietyVersion: Version of RecordLink(s) to article on publisher’s website:http://www.combinatorics.org/Volume 13/PDF/v13i1r94.pdfCopyright and Moral Rights for the articles on this site are retained by the individual authors and/or other copyrightowners. For more information on Open Research Online’s data policy on reuse of materials please consult the policiespage.oro.open.ac.ukNilpotent Singer groupsNick Gill∗9 Leonard Road, Redfield, Bristol, BS5 9NS, UK.nickgill@cantab.netSubmitted: Jun 11, 2006; Accepted: Oct 10, 2006; Published: Oct 27, 2006Mathematics Subject Classification: 20B25, 51A35AbstractLet N be a nilpotent group normal in a group G. Suppose that G acts transitivelyupon the points of a finite non-Desarguesian projective plane P. We prove that, ifP has square order, then N must act semi-regularly on P.In addition we prove that if a finite non-Desarguesian projective plane P admitsmore than one nilpotent group which is regular on the points of P then P hasnon-square order and the automorphism group of P has odd order.1 IntroductionA Singer group S of a projective plane P of order x is a collineation group of P which actssharply transitively on the points of P. The existence of such a Singer group is equivalentto a (v, k, 1) difference set in S where v = x2 + x + 1, k = x + 1 and the associated2− (v, k, 1) design is isomorphic to P.Ho [Ho98, theorem 1] has proved the following theorem concerning abelian Singergroups:Theorem C. A finite projective plane which admits more than one abelian Singer groupis Desarguesian.We will present an alternative proof of this theorem (our proof, unlike Ho’s, will bedependent on the Classification of Finite Simple Groups) and then will present work aimedat extending the result to nilpotent Singer groups. In particular we prove the following:Theorem B. Suppose that a non-Desarguesian finite projective plane P of order x admitsmore than one nilpotent Singer group. Then the automorphism group of P has odd orderand x is not a square.∗I wish to thank the University of Western Australia and the University of Gent for their supportduring the writing of this paper.the electronic journal of combinatorics 13 (2006), #R94 1In the course of proving Theorem B we will need to prove the following:Theorem A. Let F be a nilpotent group which is normal in a transitive automorphismgroup G of P, a projective plane of order x = u2. Then F acts semi-regularly on P.2 Background resultsLet P be a projective plane of order x and suppose that P admits G a point-transitiveautomorphism group. It is well known that an involution g ∈ G must fix x + 1, x + 2 orx+√x+1 points of P; furthermore no automorphism of P may fix more than x+√x+1points of P ([Dem97, 4.1.7]). If either of the first two possibilities hold then g is known asa perspectivity; Wagner has proved that if P admits a perspectivity then P is Desarguesianand PSL(3, x) E G ≤ PΓL(3, x) ([Wag59]).In order to prove Theorems A,B and C, it is not necessary to consider the situationwhere P is Desarguesian and PSL(3, x) E G ≤ PΓL(3, x). Hence for the remainder ofthis paper we will operate under the following hypothesis.Hypothesis. Let P be a projective plane of order x > 4. Let G be an automorphismgroup of P which acts transitively upon the points of P. If G contains any involutionsthen x = u2, u > 2, and each involution fixes u2 + u + 1 points. If h ∈ G then h fixes atmost x +√x + 1 points.We require the following result which is dependent on the Classification of FiniteSimple Groups:Theorem 1. [Gil, Theorem A] Suppose that G acts transitively on the points of a projec-tive plane P of order x. Then one of the following cases holds:• P is Desarguesian, PSL(3, x) E G ≤ PΓL(3, x) and the action is 2-transitive onpoints;• the Fitting group and the generalized Fitting group of G coincide, i.e. F ∗(G) =F (G).Clearly, under our hypothesis, Theorem 1 implies that the Fitting group and thegeneralized Fitting group of G coincide, i.e. F ∗(G) = F (G).We will also need the following result of Ljunggren:Lemma 2. [Lju43, p.11] If u2 + u + 1 = pa1where p1 is a prime, then either pa1= p1 orpa1= 73.Write α for a point of P. For a collineation group H of P, write Hα for the stabilizerof α.the electronic journal of combinatorics 13 (2006), #R94 23 Nilpotent collineation groupsIn order to prove Theorem A we need a well known result of Camina and Praeger. Westate a weaker version which is sufficient for our purposes:Theorem 3. [CP93, Theorem 1] Let G act transitively on the points of a projective planeP. Let N be a normal subgroup of G. Then N acts faithfully on each of its point orbits.Note that, in particular, Theorem 3 implies that a minimal normal subgroup of G willact semi-regularly on the points of P. We are now in a position to prove Theorem A.Proof. Let N be a Sylow p-group of F for some prime p dividing the order of F . Supposethat N does not act semi-regularly. If |N : Nα| = p, a prime, then, since N acts faithfully,N < Sp, the Symmetric group on p letters. But then |N | = p and N is semi-regular whichis a contradiction. Thus |N : Nα| ≥ p2 and x > 25.Observe that the average number of fixed points for non-identity elements of N is(x2 + x + 1)|Nα| − 1|N | − 1 >x2 + x + 12|N : Nα| .Now |N : Nα| divides into u2 ± u + 1. If |N : Nα| < u2 ± u + 1 then a non-identityelement of N fixes, on average, more than 32(u2 − u + 1) > u2 + u + 1 fixed points whichcontradicts our hypothesis.If |N : Nα| = u2 ± u + 1 then, by Lemma 2, |N : Nα| = 73.If |N : Nα| = 73 = u2 − u + 1 and |Nα| > 7 then the average number of fixed pointsfor non-identity elements of N is(x2 + x + 1)|Nα| − 1|N | − 1 = 343× 381|Nα| − 173|Nα| − 1 >343× 381351> 370.Now x = 361 and an element of order 7 must fix a multiple of 7 points. This implies thatthe most number of points such an element may fix is 357 which is a contradiction.If |N : Nα| = 73 = u2 − u + 1 and |Nα| = 7 then |Aut N | is not divisible by 127.Thus F (G) must contain a non-trivial Sylow 127-group and G has a normal semi-regularsubgroup P of order 127. Furthermore P centralizes Nα and so|Fix(Nα)| ≥ 7× 127 > x +√x + 1.This is a contradiction.If |N : Nα| = 73 = u2 + u + 1 then x = 324. In fact N = F = F (G) since otherwiseF (G) > N and a point semi-regular group of order 307 must centralize N which isimpossible.By Theorem 3 the centre of N acts semi-regularly on the points of P. This impliesthat Nα fixes a subplane. Suppose that |Z(N)| > 7; then Nα must fix a subplane of order18. Thus, for g ∈ Nα, Fix(g) = Fix(Nα). Then take β a point of P not in Fix(Nα). Wemust have Nα ∩ Nβ = {1}. Thus |Nβ| ≤ 73 and so |N | ≤ 76. But in this case 307 doesnot divide into |Aut N | which is a contradiction.the electronic journal of combinatorics 13 (2006), #R94 3Suppose, alternatively, that |Z(N)| = 7. Let X = NN(Nα). If X > NαZ(N) thenthere are at most 7 N -conjugates of Nα and so Nα must fix 0 or a multiple of 49 pointsin each orbit of N . Hence Nα fixes a Baer subplane and our previous argument can beapplied.Thus we assume that X = NαZ(N). Since N is nilpotent NN (X) > X and so wechoose α and β points in the same N -orbit such that Nα and Nβ are distinct subgroupsof NαZ(N). Then |Nα : (Nα ∩Nβ)| = 7. Furthermore Y := NN (Nα ∩Nβ) ≥ 〈X, n〉 > X,where n ∈ N such that αn = β, hence Y acts on the fixed set of Nα ∩Nβ with orbits ofsize a multiple of 49. We conclude that Nα ∩Nβ fixes a Baer subplane.If Nα ∩ Nβ fixes an entire N -orbit then, since N acts faithfully on its point-orbits,Nα ∩ Nβ = {1}. Thus |N | = 74. Alternatively Nα ∩ Nβ fixes exactly 0 or 49 points inany N -orbit. In which case we can find two other points γ and δ in the same N -orbitas α such that Nγ ∩ Nδ fix a Baer subplane. Then (Nα ∩ Nβ) ∩ (Nγ ∩ Nδ) = {1} andso |N | ≤ 78. Thus in all cases |N | ≤ q8 and 307 does not divide into |Aut N | which is acontradiction.4 Abelian Singer groupsThroughout this section P is non-Desarguesian and S ≤ G is an abelian Singer groupof P.We record some results of Ho [Ho98]:Theorem 4. [Ho98, theorem 2] An abelian Singer group contained in a soluble collineationgroup of a finite projective plane is always normal.Lemma 5. [Ho98, lemma 4.3] Let Q < G be a collineation group normalized by S. ThenS centralizes Q if one of the following holds:1. Q is abelian;2. |Q| is prime to |S|;3. x = u2 and Q is nilpotent.We are now able to give an alternative proof to Theorem C.Proof. Suppose that S and T are abelian Singer groups lying in G. We need to provethat S = T .If G is soluble then both S and T are normal in G and so lie in F (G). By Lemma 5,S and T centralize each other. Thus 〈S, T 〉 is abelian and transitive on the points of P.By Theorem 3, |〈S, T 〉| ≤ v and so S = T .If G is not soluble then we may assume that G contains a Baer involution. Hencex = u2 and, by Lemma 5, S and T both centralize F (G). Since F (G) = F ∗(G) thismeans that F (G) contains both S and T. Now F (G) is soluble and so we can apply thesame argument as when G was soluble and conclude that S = T .the electronic journal of combinatorics 13 (2006), #R94 45 Nilpotent Singer groupsThroughout this section P is non-Desarguesian and S ≤ G is a nilpotent Singer groupof P.Lemma 6. S contains F (G).Proof. Suppose the result does not hold and S 6≥ F (G). Let P ∈ SylpF (G) with P 6≤ S.Let H := PS = SP .Let Q ∈ SylpH such that Q = PP1 where P1 ∈ SylpS. Then consider h−1Qh forh ∈ H. We can write h = sp where p ∈ P and s ∈ S. Thenh−1Qh = p−1s−1PP1sp = Pp−1s−1P1sp = PPp1= Q.Thus Q is normal in H. But this is a nilpotent group normal in a transitive grouphence, by Theorem A, Q is semi-regular. But |Q| does not divide into x2 + x + 1 and wehave a contradiction.Corollary 7. Any prime dividing into v divides into |F (G)|.Proof. Since CG(F (G)) ≤ F (G) we know that if p divides into |S| then p divides into|F (G)|. So if p divides into v then p divides into |F (G)|.We are now in a position to prove Theorem B:Proof. Assume, for the sake of contradiction, that P admits two distinct nilpotent Singergroups S and T and set G := AutP. Let v = pa11. . . parr . Then F (G) = N1 × · · · × Nr,1 6= Ni ∈ SylpiF (G). In a similar way write,S = P1 × · · · × Pr, T = Q1 × · · · ×Qr.We will assume, without loss of generality, that P1 6= Q1. Now G ≤ (N1 × · · · ×Nr).(A1×· · ·×Ar) where Ai is a subgroup of the outer automorphism group of Ni. Thenboth Pi and Qi lie in (1× · · ·× 1×Ni × 1× · · ·× 1).(1× · · · × 1×Ai × 1× · · · × 1). Thisimplies thatCG(〈P1, Q1〉) ≥ P2 × · · · × Pr.Now 〈P1, Q1〉 must contain an element 1 6= g which fixes a point (since P1Q1 musthave an orbit of size strictly less than |P1Q1|.) Then we know that|v||v|p1∣∣ |Fix g|.In fact, consider 〈Q1, P1〉 acting on P/(P2×· · ·×Pr). Either this is a Frobenius actionor there exists g ∈ 〈Q1, P1〉 that fixes more than one element. But a Frobenius action hasa normal Frobenius kernel which must be Q1 and P1. This is a contradiction. Thus|v||v|p1< |Fix g|.Now, since x = u2, v = (x +√x + 1)(x− √x + 1). The bracketed terms are coprimeand so |Fix g| ≥ 2(x−√x + 1) > x +√x + 1. This gives a contradiction.the electronic journal of combinatorics 13 (2006), #R94 5References[CP93] Alan R. Camina and Cheryl E. Praeger, Line-transitive automorphism groupsof linear spaces, Bull. London Math. Soc. 25 (1993), 309–315.[Dem97] P. Dembowski, Finite geometries, Springer-Verlag, 1997.[Gil] Nicholas Gill, Transitive projective planes, Submitted.[Ho98] Chat Yin Ho, Finite projective planes with abelian transitive collineation groups,J. Algebra 208 (1998), 533–550.[Lju43] W. Ljunggren, Einige bemerkungen u¨ber die Darstellung ganzer Zahlen durchbina¨re kubische Formen mit positiver Diskriminante, Acta. Math. 74 (1943),1–21.[Wag59] A. Wagner, On perspectivities of finite projective planes, Math. Z. 71 (1959),113–123.the electronic journal of combinatorics 13 (2006), #R94 6

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