On generalized quadrangles with a point regular group of automorphisms
aa r X i v : . [ m a t h . C O ] D ec On generalized quadrangles with a point regular group ofautomorphisms
Eric Swartz A BSTRACT . A generalized quadrangle is a point-line incidence geometry such that any twopoints lie on at most one line and, given a line ℓ and a point P not incident with ℓ , there is aunique point of ℓ collinear with P . We study the structure of groups acting regularly on the pointset of a generalized quadrangle. In particular, we provide a characterization of the generalizedquadrangles with a group of automorphisms acting regularly on both the point set and the line setand show that such a thick generalized quadrangle does not admit a polarity. Moreover, we provethat a group G acting regularly on the point set of a generalized quadrangle of order ( u , u ) or ( s, s ) , where s is odd and s + 1 is coprime to , cannot have any nonabelian minimal normalsubgroups.
1. Introduction A finite generalized quadrangle is an incidence geometry consisting of a finite set P ofpoints and a finite set L of lines such that, if P ∈ P and ℓ ∈ L such that P is not incidentwith ℓ , then there is a unique point on ℓ collinear with P . If every point is incident with at leastthree lines and every line is incident with at least three points, then the generalized quadrangleis said to be thick . From this definition, one can show that in a thick generalized quadrangle anypoint is incident with the same number of lines and any line is incident with the same numberof points. A generalized quadrangle where every point is incident with exactly t + 1 lines andany line is incident with exactly s + 1 points, where s and t are fixed positive integers, is saidto have order ( s, t ) . For further basic properties of generalized quadrangles, see [ ].A group G is said to act regularly on a set Ω if G is transitive on Ω and, for all ω ∈ Ω , thestabilizer G ω of ω in G is trivial. When a group G acts regularly on a set Ω , there is a naturalidentification of elements of G with elements of Ω : if we choose a distinguished element ω ∈ Ω ,then we may identify ω with the identity ∈ G , and, if α ∈ Ω , we may identify α with theunique element g ∈ G such that ω g = α .There has been substantial interest in generalized quadrangles with a group of automor-phisms G acting regularly on the point set P , and we briefly summarize here the current stateof knowledge. Such generalized quadrangles were first studied in [ ], where techniques fromthe study of difference sets of groups are used to show that, if Q is a generalized quadrangleof order ( s, t ) , where s = t and s is even, and G acts regularly on the point set of Q , then G has trivial center, G is not a Frobenius group, and s + 1 is not squarefree. These techniqueswere pushed even further in [ ], where it is shown that a generalized quadrangle with s = t cannot have a group of automorphisms that acts regularly on points. In [ ], it is shown thatif Q is a thick finite generalized quadrangle of order ( s, t ) admitting an abelian automorphismgroup that acts regularly on points, then Q is isomorphic to a generalized quadrangle T ∗ ( O ) Mathematics Subject Classification.
Primary 51E12, 05B25.
Key words and phrases. generalized quadrangle, regular action. arising from a generalized hyperoval, and the group acting on Q is elementary abelian of order n for some natural number n . (For further background information and descriptions of thegeneralized quadrangles referred to in this paragraph, the curious reader is again directed to[ ].) In [ ], it is shown that if a Heisenberg group of order q , where q is odd, acts regularlyon the point set of Q , then Q is a Payne-derived generalized quadrangle of a thick elationgeneralized quadrangle having a regular point . (A regular point of a generalized quadrangleis defined combinatorially and has no relation to a regular group action.) It was shown in [ ]that the only classical generalized quadrangles with a point-regular group of automorphisms are Q (5 , and Q (5 , , and the authors showed that the Payne-derived generalized quadrangles oforder ( q − , q + 1) obtained from the symplectic generalized quadrangle W ( q ) , where q is aprime power but not prime, can actually have several distinct groups of automorphisms actingregularly on the point set, showing that the class of such groups is much more varied than pre-viously thought. Finally, the study of groups acting regularly on the point set of generalizedquadrangles figures prominently in the study of skew-translation generalized quadrangles ; see[ ] and the subsequent work in [ ].The purpose of this paper is to study further the structure of such generalized quadranglesand the groups that act regularly on their point sets. A first step in the study of a generalizedquadrangle Q with a point regular group G of automorphisms is to consider the action of G onthe lines of Q . One possibility of interest is that G also acts regularly on lines, since, otherwise,some elements in the group acting regularly on the point set must fix a line, and one can proceedwith an analysis of line stabilizers. In the case when G also acts regularly on lines, we have s = t , in which case we say that Q has order s (see Lemma 2.2 (i)). The only known exampleof a generalized quadrangle with a group of automorphisms acting regularly on both points andlines is the unique thin generalized quadrangle of order (1 , , and it is not difficult to show that s would have to be even in any other examples (see Section 3).We are able to characterize further the groups that act regularly on both the point set and theline set of a generalized quadrangle. To that end, we have the following result, which followsimmediately from Propositions 3.1 and 3.2.T HEOREM
The group G acts regularly on both the point set and the line set of a gen-eralized quadrangle Q of order s if and only if G contains a subset Σ = { g = 1 , g , g , ..., g s } satisfying the following two conditions: (AX1) If g ∈ G \ Σ , then there exist unique integers i, j, k (with i = j and k = j ) such that g = g i g − j g k . (AX2) If g i g − j g k ∈ Σ , then either i = j or j = k . Since there is a natural identification of G with both the point set and the line set in thiscase, it is natural to think that Q should have a polarity θ ; that is, an order two isomorphismfrom Q to the dual generalized quadrangle Q ′ , which has point set L and line set P . In fact, thefollowing result shows that such a thick generalized quadrangle cannot have a polarity.T HEOREM
A generalized quadrangle Q of order s > that has a group of automor-phisms acting regularly on both the point set and line set cannot admit a polarity. On the other hand, the cyclic group of order satisfies the conditions listed in Theorem 1.1(see Observation 3.3). This begs the question of whether there exist any groups of order greaterthan satisfying the conditions in Theorem 1.1. Toward that end, we prove the followingtheorem, which places severe restrictions on the order of such a generalized quadrangle andthe group acting regularly on its point set and line set. (For information on the group theoreticnotation used in the statement of Theorem 1.3, see Subsection 2.1.)T HEOREM
Let the group G act regularly on both the point set and the line set of athick generalized quadrangle of order s . Then, G is solvable, s + 1 is coprime to both and , OINT REGULAR GENERALIZED QUADRANGLES 3 and s + 1 is not squarefree. Moreover, there exists a unique prime p dividing s + 1 such that F ( G ) = O p ( G ) and the following conditions are satisfied: (1) F ( G ) is not cyclic; (2) | F ( G ) | > s + 3 ; (3) If p n is the highest power of p dividing | G | , then | G | divides p n · Q n − k =0 ( p n − k − . We are able to use Theorem 1.3 to show that, if G acts regularly on the point set and the lineset of a thick generalized quadrangle of order s , then s > and | G | > + 10 + 10 + 1 (see Proposition 3.8). Ghinelli [ ] conjectures that no generalized quadrangle of order s , when s is even, has a group of automorphisms acting regularly on points. Based on Theorem 1.2 andthe numerical evidence, we make a more modest conjecture.C ONJECTURE
A generalized quadrangle Q of order s > cannot have a group ofautomorphisms acting regularly on both the point set and the line set. In this paper, we also study the structure of groups that could act regularly on the point setof a generalized quadrangle, leading to the following result.T
HEOREM
Let Q be a generalized quadrangle of order ( u , u ) or ( s, s ) , where u > and s is odd with s + 1 coprime to . If a group G acts regularly on the point set of Q , then G does not have any nonabelian minimal normal subgroups. It should be noted that Theorem 1.5 refines part of a result of Yoshiara (see Lemma 2.4(iv)) and is an immediate consequence of Theorems 4.8 and 5.4, which consider, respectively,the generalized quadrangles of order s and the generalized quadrangles of order ( u , u ) . It isnatural to ask about these orders in particular, since there exist infinite families of these orders;see [ , Chapter 3]This paper is organized as follows. Section 2 contains background information and sets thestage for the results proved later in this paper. In Section 3, we prove Theorems 1.1, 1.2 and1.3. In Section 4 we study the properties of a group G that acts regularly on the point set of ageneralized quadrangle of order s , where s is odd and s + 1 is coprime to . Finally, in Section5 we study the properties of a group G that acts regularly on the point set of a generalizedquadrangle of order ( u , u ) , where u > .
2. Background2.1. Group theory.
We begin with group theoretic terminology and notation. In a finitegroup G , the Fitting subgroup is the largest normal nilpotent subgroup of G and is denoted F ( G ) . Given a prime p , the p-core O p ( G ) of G is the largest normal p -subgroup of G . TheFitting subgroup is also the product of the p -cores of G for all primes p dividing | G | . A qua-sisimple group is a perfect central extension of a simple group, and a component of a group isa subnormal quasisimple group. The layer E ( G ) of a group G is the subgroup generated by allcomponents. The generalized Fitting subgroup F ∗ ( G ) is the subgroup generated by the Fittingsubgroup and the layer. In a solvable group, the generalized Fitting subgroup is the same as theFitting subgroup. Each of the Fitting subgroup and the generalized Fitting subgroup containsits own centralizer. For more information regarding these concepts, see [ ].We introduce now basic information about the Suzuki groups , an infinite family of finitesimple groups figuring prominently in later sections of this paper, since they are the only finitesimple groups whose orders are coprime to [ , Chapter II, Corollary 7.3]. The followingomnibus lemma contains results about the Suzuki groups Sz( q ) , where q = 2 m +1 for somenatural number m .L EMMA
Let m be a natural number, q = 2 m +1 , and G = Sz( q ) . ERIC SWARTZ (i) [ , Theorem 7] | G | = q ( q − q + 1) . (ii) [ , Theorem 3.6] does not divide | G | . (iii) [ , Theorem 3.9] If p is an odd prime, then the Sylow p -subgroups of G are cyclic. (iv) [ , Theorem 4.1] The maximal subgroups of G are isomorphic to one of the following: (a) N G ( Q ) , where Q is a Sylow -subgroup of G and | N G ( Q ) | = q ( q − ; (b) D q − , a dihedral group of order q − ; (c) C q + √ q +1 : C ; (d) C q −√ q +1 : C ; (e) Sz( q ) , where q = q r for some natural number r and q > . (v) The order of the smallest centralizer of an element in G is q − √ q + 1 , and the sizeof the largest conjugacy class is q ( q − q + √ q + 1) . (vi) [ , 2.2.5] The order of the centralizer of an element of order in G is q . (vii) [ , Section 4.2.4] G has trivial Schur multiplier unless m = 1 , in which case theSchur multiplier is elementary abelian of order . (viii) [ , Theorem 11] The outer automorphism group of G is isomorphic to the Galoisgroup of GF( q ) and is a cyclic group of order m + 1 . Note that Lemma 2.1 (v) follows from examination of the subgroups listed in Lemma 2.1(iv) and noting that a Sylow -subgroup Q of G has a center of size q . For more about thesegroups, the reader is referred to the original work of Suzuki [ ] or the more recent collectionof results in [ , Section 3]. We now discuss some results about generalized quadran-gles. Let Q be a finite generalized quadrangle with point set P , line set L , and order ( s, t ) .Given P, Q ∈ P , the notation P ∼ Q indicates that P and Q are distinct collinear points. Thefollowing lemma concerns the parameters s and t .L EMMA , 1.2.1, 1.2.2, 1.2.3] The following hold: (i) |P| = ( s + 1)( st + 1) and |L| = ( t + 1)( st + 1) ; (ii) s + t divides st ( s + 1)( t + 1) ; (iii) t s and s t . The first paper to consider a group G acting regularly on the point set P of a generalizedquadrangle Q was [ ]. Note that, if a group G acts regularly on the point set of a generalizedquadrangle of order s , where s is even, then | G | = ( s + 1)( s + 1) is odd, and so G is solvable.The following lemma collects various results proved by Ghinelli that will be useful later.L EMMA , Theorem 3.5, Theorem 4.2, Theorem 4.3, Theorem 4.6] Let G be an auto-morphism group acting regularly on the point set of a generalized quadrangle of order s . Thenthe following hold: (i) If N is an elementary abelian normal p -subgroup of G , p = 2 , then | N | divides s + 1 .In particular, if p > and O p ( G ) > , then p divides s + 1 . (ii) If s is even, then G is solvable, F ( G ) is a p -group with | F ( G ) | > s + 1 , and F ( G ) isnot cyclic. (iii) If s is even, then Z ( G ) is trivial and s + 1 is not squarefree. Yoshiara was able to generalize many of the results of Ghinelli in [ ]. The followinglemma collects various results proved by Yoshiara that will be useful later in this paper.L EMMA , Lemma 4, Lemma 6, Lemma 7, Theorem 8, Lemma 9, Lemma 10] Let G be a group that acts regularly on the point set of a generalized quadrangle of order ( s, t ) .Assume that gcd( s, t ) > , and let P be a distinguished point of the generalized quadrangle.Define ∆ := { g ∈ G : P g ∼ P } ∪ { } , and let ∆ c denote the complement of ∆ in G . OINT REGULAR GENERALIZED QUADRANGLES 5 (i) If H G and H is contained entirely in ∆ , then there is a unique line ℓ through P such that H G ℓ . (ii) If x is a nontrivial element of G and x G denotes the conjugacy class of x in G , then x G ∩ ∆ = ∅ and | x G ∩ ∆ c | is a multiple of gcd( s, t ) (possibly equal to ). (iii) If p = 2 , , then the following are equivalent: (a) | G | is divisible by p . (b) There is an element of order p fixing a line through P . (c) s + 1 is divisible by p . (iv) If s + 1 is coprime to and G has a nonabelian minimal normal subgroup N , then N is the internal direct product S S . . . S m and each S i ∼ = Sz( q ) for a fixed q = 2 e +1 . (v) If N is a nontrivial normal subgroup of G such that N is entirely contained in ∆ , then | N | divides gcd( s + 1 , t − t ) . (vi) If the conjugacy class x G of a nontrivial element x ∈ G is contained entirely in ∆ ,then | x | divides s + 1 . The following lemma is extremely useful and is used repeatedly in later sections.L
EMMA
Let G be a group that acts regularly on the point set of a generalized quad-rangle of order ( s, t ) . Assume that gcd( s, t ) > , and let P be a distinguished point of thegeneralized quadrangle. Define ∆ := { g ∈ G : P g ∼ P } ∪ { } , and let ∆ c denote thecomplement of ∆ in G . For any x ∈ G , if x G ∩ ∆ c = ∅ , then | x G | > gcd( s, t ) + 1 . P ROOF . Assume that G and x are as in the statement of the lemma. By Lemma 2.4 (ii), | x G ∩ ∆ | > and | x G ∩ ∆ c | ≡ s, t )) . Since | x G ∩ ∆ c | > by assumption, theresult follows. (cid:3)
3. Generalized quadrangles with a group of automorphisms acting regularly on bothpoints and lines
Throughout this section, we will assume that Q is a generalized quadrangle of order ( s, t ) with a group of automorphisms G that acts regularly on both the point set P and the line set L .This immediately implies that s = t and | G | = ( s + 1)( s + 1) .We now prove two propositions that characterize the generalized quadrangles with a groupthat acts regularly on both points and lines.P ROPOSITION
Let
Σ = { g = 1 , g , g , ..., g s } be a subset of a group G satisfying thefollowing: (AX1) If g ∈ G \ Σ , then there exist unique integers i, j, k (with i = j and k = j ) such that g = g i g − j g k . (AX2) If g i g − j g k ∈ Σ , then either i = j or j = k .Then there exists a generalized quadrangle Q such that Q has order s and G acts regularly onboth the point set P and the line set L of Q . P ROOF . First, there are precisely s + 1 elements in Σ , and g i g − j g k ∈ Σ only when i = j or j = k by (AX2). Moreover, the manner in which g ∈ G \ Σ can be written as a product g i g − j g k is unique by (AX1). This implies that there are s ( s +1) elements in G \ Σ , since there are ( s +1) choices for j and s choices for each of i and k , and so | G | = ( s +1)+ s ( s +1) = ( s +1)( s +1) .We define the point set P to be the elements of the group G (where the point associatedwith the group element g will be denoted P g ), and we define the line set L to be the sets Σ h ,where h ∈ G (we denote the line associated with the set Σ h by ℓ h ). Finally, the point P x willbe incident with the line ℓ y if and only if xy − ∈ Σ . ERIC SWARTZ
For any g ∈ G , note that the point P x is incident with the line ℓ y if and only if ( xg )( yg ) − = xy − ∈ Σ if and only if P xg is incident with ℓ yg . Hence each g ∈ G acts as a collineation of theputative generalized quadrangle Q , and this action is regular on both points and lines.Since | Σ | = s + 1 , each line contains s + 1 points. Moreover, if a point P x is on a line ℓ y ,then xy − ∈ Σ . For a fixed x ∈ G , there are exactly s + 1 elements y of G such that xy − ∈ Σ ,so each point is on exactly s + 1 lines.Let P x and P y be distinct points of Q , and assume that both points are on the lines ℓ g and ℓ h . This means that there exist integers i, j, m, n such that x = g i h = g m g and y = g j h = g n g .Thus xy − = g i g − j = g m g − n , which implies that g i = g m g − n g j ∈ Σ . By (AX2), this meansthat either g n = g m or g n = g j . If g n = g m , then y = g n g = g m g = x , a contradiction todistinctness. Thus g n = g j , which implies that g = h . Therefore, distinct points are incidentwith at most one line.Let ℓ x and ℓ y be distinct lines of Q and assume that they intersect in at least one point. Thismeans that g i x = g j y for some i, j s, and furthermore xy − = g − i g j . Assume first that xy − / ∈ Σ . This means that xy − = g g − i g j , and by (AX1), i and j are unique. In this case,the lines ℓ x and ℓ y meet in exactly one point. Assume now that xy − ∈ Σ . The lines ℓ x and ℓ y are distinct, so xy − = 1 = g . Thus xy − = g j for some j > , and x = g j y . Note that thesize of the intersection of Σ g j y and Σ y is the same as the size of the intersection of Σ g j and Σ .Suppose that g i and g k are such that g i g j = g k . Then g j = g − i g k = g g − i g k . By (AX2), either g i = 1 or g i = g k . If g i = 1 , then g k = g j , and the two lines intersect in a unique point. If g i = g k , then g j = 1 , a contradiction to the distinctness of the lines. In any case, two distinctlines are mutually incident with at most one point.Let P x be a point that is not incident with the line ℓ y . This is equivalent to xy − / ∈ Σ . By (AX1), there exist unique i, j, k such that xy − = g i g − j g k , i.e., there exists a unique i such that Σ y ∩ Σ g − i x is nonempty. Note that the lines through the point P are precisely thelines ℓ g − i ; hence the lines through P x are precisely the lines ℓ g − i x . Hence there is a uniqueline ℓ g − i x through P x such that ℓ g − i x and ℓ y are incident with a common point. Therefore, Q is a generalized quadrangle such that G acts regularly on both the points and lines of Q , asdesired. (cid:3) P ROPOSITION
Let G be a group that acts regularly on both the point set P and the lineset L of a finite generalized quadrangle Q of order s . Then there exists a subset Σ of elementsof G satisfying (AX1) and (AX2) of Proposition 3.1. P ROOF . Assume that such a generalized quadrangle Q exists. Fix a point P ∈ P andidentify it with the identity element ∈ G . Since G acts regularly on P , for each x ∈ G , we maydefine P x := P x . Let ℓ be a line incident with P , define Σ := { g ∈ G : P g is incident with ℓ } ,and label the points of Σ such that Σ = { g = 1 , g , ..., g s } . Note that G acts regularly on thelines of Q , so we may identify the lines of Q with the sets of elements of the form Σ h , where h ∈ G . Without a loss of generality we may denote the line Σ h by ℓ h , and so the line ℓ isidentified as ℓ .Assume first that g / ∈ Σ . This means that P g is not incident with ℓ . Since Q is a generalizedquadrangle, there exists a unique point P g k on ℓ such that P g is collinear with P g k . Since G actsregularly on both points and lines, P is incident with precisely the lines ℓ g − j , j s , whichimplies that the lines incident with P g k are of the form ℓ g − j g k . The point P g is on a unique lineof that form, and so g may be written uniquely as g = g i g − j g k , and Σ satisfies (AX1).Now suppose that the product g i g − j g k ∈ Σ , and assume that i = j . Since P g i g − j g k isincident with ℓ , we have that P g i g − j is incident with line ℓ g − k . Since i = j , g i g − j = 1 , so P g i g − j is distinct from P but collinear with P . On the other hand, P g i g − j is also on line ℓ g − j , which OINT REGULAR GENERALIZED QUADRANGLES 7 is also incident with P . Since two points are incident with at most one line, we conclude that ℓ g − j = ℓ g − k . Therefore, j = k , and Σ satisfies (AX2), as desired. (cid:3) We can now prove Theorem 1.1.P
ROOF OF T HEOREM (cid:3) O BSERVATION
The cyclic group C satisfies the conditions of Proposition 3.1. Indeed, if C = h x i , then the set Σ = { , x } satisfies all three conditions, and hencethe unique thin generalized quadrangle of order (1 , has a group of automorphisms that actsregularly on both its point set and its line set.Henceforth in this section, we fix a point P ∈ P , and we define ∆ := { g ∈ G : P g ∼ P } ∪ { } . L EMMA If s > , then | G | is coprime to both and . P ROOF . This follows directly from Lemma 2.4 (iii). (cid:3) L EMMA If s > , then G is solvable and F ( G ) = O p ( G ) , where p is an odd primedividing s + 1 . P ROOF . This follows immediately from Lemma 3.4 and Lemma 2.3 (iii). (cid:3) L EMMA If s > and F ( G ) = O p ( G ) , then | O p ( G ) | > s + 3 . P ROOF . Let = x ∈ O p ( G ) . Suppose that x is conjugate in G to x − . Then there exists g ∈ G such that x g = x − , and so x g = x . Since | G | has odd order by Lemma 3.4, g has oddorder, and there is some n ∈ N such that g = ( g ) n , which means that g centralizes x . However,this means that x = x − , and x must have order dividing , a contradiction. Hence x and x − cannot be conjugate in G , and O p ( G ) has at least three conjugacy classes: G , x G , and ( x − ) G .By Lemma 2.4 (vi) we know that neither x G nor ( x − ) G is contained entirely in ∆ . There-fore, by Lemma 2.5, | O p ( G ) | > | G | + | x G | + | ( x − ) G | > s + 1) + ( s + 1) = 2 s + 3 , as desired. (cid:3) We are now ready to prove Theorem 1.2.P
ROOF OF T HEOREM Q has a polarity. By [ , 1.8.2], s is a square,which means that s + 1 factors into ( s + √ s + 1)( s − √ s + 1) . By Lemmas 3.5 and 3.6,there exists a normal subgroup O p ( G ) of order p d > s + 3 , where p divides s + 1 . We alreadyknow that p is odd by Lemma 3.4, and, since p divides s + 1 , p is coprime to s , which impliesthat p d divides either s + √ s + 1 or s − √ s + 1 , but not both. However, for any s > , s − √ s + 1 < s + √ s + 1 < s + 3 p d , a contradiction. Therefore, Q does not admit apolarity. (cid:3) Note that Theorem 1.2 implies that the regular action of G on the set of points is necessarilydifferent than the action of G on the set of lines, in the sense that the set ∆ defined above mustbe distinct from ∆ L , where for some fixed line ℓ incident with the fixed point P , ∆ L := { g ∈ G : ℓ g is distinct from and concurrent with ℓ } ∪ { } . This begs the question of whether a group can possibly act regularly on both the pointset or the line set of a thick generalized quadrangle. Toward that end, we have the followingproposition, which is the last piece before we can prove Theorem 1.3.
ERIC SWARTZ P ROPOSITION
Let G act regularly on both the point set and the line set of a generalizedquadrangle of order s . If p is the unique odd prime dividing s + 1 such that F ∗ ( G ) = F ( G ) = O p ( G ) and p n is the highest power of p dividing | G | , then | G | divides p n · Q n − k =0 ( p n − k − . P ROOF . We note that G is solvable, and F ( G ) = O p ( G ) for some prime p dividing s +1 byLemma 3.5. Let p n be the order of a Sylow p -subgroup of G , and let H be a Hall p ′ -subgroupof G , which exists by Hall’s Theorem [ , Theorem 6.4.1]. Since F ( G ) = O p ( G ) , F ( G ) isself-centralizing, and, since H is coprime to p , by [ , Theorem 5.3.5] we have that H actsfaithfully on the Frattini quotient F ( G ) / Φ( F ( G )) , which is elementary abelian of rank at most n . Thus H . GL( n, p ) with order coprime to p , and, since | G | = p n | H | , the result follows. (cid:3) P ROOF OF T HEOREM | G | is not divisible by or , and by Lemma2.4 (iii), this means s + 1 is coprime to both and . Moreover, s + 1 is not squarefree byLemma 2.3 (iii). Of the remaining possible values of s , there must be a prime p dividing s + 1 such that ( s + 1) p > s + 3 by Lemma 3.6, where ( s + 1) p denotes the highest power of p dividing s + 1 . Moreover, ( s + 1) p cannot be prime by Lemma 2.3 (ii), and ( s + 1)( s + 1) must divide p n · Q n − k =0 ( p n − k − by Proposition 3.7. (cid:3) We next use Theorem 1.3 to rule out many possible “small” groups.P
ROPOSITION If G is a group that acts regularly on both the point set and a line set ofa thick generalized quadrangle of order s , then then s > and | G | > + 10 + 10 + 1 . P ROOF . The following computation was completed in GAP [ ]. Such a generalized quad-rangle of order s must satisfy all the conditions imposed on s listed in Theorem 1.3: namely, s + 1 is coprime to both and ; s + 1 is not squarefree; there is a prime p dividing s + 1 such that, if p n is the largest power of p dividing s + 1 , then n > and p n > s + 3 ; and ( s + 1)( s + 1) divides p n · Q n − k =0 ( p n − k − . Since no value of s less than or equal to satisfies all of these conditions, the result follows. (cid:3) It would be interesting to know whether there exist any even values of s that satisfy all ofthe restrictions mentioned in the statement of Theorem 1.3.
4. Generalized quadrangles of order s , s odd and s + 1 coprime to , with a group ofautomorphisms acting regularly on points Let Q be a finite generalized quadrangle of order ( s, s ) , where s > is odd and s + 1 iscoprime to , with point set P and line set L . Let G be a group of automorphisms of Q thatacts regularly on P , and suppose that N is a nonabelian minimal normal subgroup of G . For adistinguished point of P of Q , we define ∆ := { g ∈ G : P g ∼ P } ∪ { } . First, by Lemma 2.4 (iv), N ∼ = Sz( q ) m , where m ∈ N and q = 2 e +1 for some e ∈ N . Let N = S S ...S m , where each S i ∼ = Sz( q ) , each S i is normal in N , all S i and S j are conjugate in G , and S i ∩ S j = { } for all i = j .Since | G | = |P| = ( s + 1)( s + 1) and s + 1 ≡ , we have that the highest powerof dividing | G | , which we will denote by | G | , is s + 1) , i.e., two times the highest powerof dividing s + 1 . Since | Sz( q ) | = q by Lemma 2.1 (i), | N | = q m , and we immediatelyobtain the following inequality:(1) q m s + 1) . Furthermore, by Lemma 2.4 (ii), since gcd( s, s ) = s > , for all a ∈ G we have a G ∩ ∆ = ∅ .L EMMA
Let G be a group acting regularly on the set of points of a generalized quad-rangle of order ( s, t ) , where gcd( s, t ) > . Let N = T T ...T k be a nonabelian minimal normalsubgroup of G , where each T i is normal in N , has trivial intersection with the other T j , and is OINT REGULAR GENERALIZED QUADRANGLES 9 isomorphic to the same nonabelian simple group T . If each T i ⊆ ∆ , then there exists a line ℓ such that N G ℓ . P ROOF . Assume that T i ⊆ ∆ for each i k. By Lemma 2.4 (i) for each i there is a(necessarily unique) line ℓ i through the distinguished point P such that T i G ℓ i . Let x ∈ T and x ∈ T . Since x and x commute, both P x and P x are collinear with P x x . Thisimplies that either ℓ = ℓ and T T G ℓ or that no element x x , where neither x nor x is trivial, is in ∆ . Let x = x x ∈ T T for some x , x such that neither x nor x is trivial.This means that x G consists of elements of the form x i x j ∈ T i T j , i, j k. However, byLemma 2.4 (ii), x G ∩ ∆ = ∅ , so there exist i i , i k such that T i T i G ℓ i for some line ℓ i through P. We now proceed by induction, and assume that T j T j ...T j n G ℓ j for distinctindices j , ..., j n . Arguing as above, since ∆ meets every conjugacy class, there must be a setof indices l , ..., l n +1 such that T l T l ...T l n +1 G l . Hence N = T ...T k G ℓ for some line ℓ through the point P , as desired. (cid:3) L EMMA If G and N = S ...S m are as above, then m = 1 and N ∼ = Sz( q ) for some q. P ROOF . Applying Lemma 4.1 to our situation, if each S i ⊆ ∆ , then N G ℓ for some line ℓ. This implies that N ⊆ ∆ , and by Lemma 2.4 (v) implies that | N | divides gcd( s + 1 , s − s ) = 2 ,a contradiction. Hence we may assume without a loss of generality that S ∩ ∆ c = ∅ . Let a ∈ S ∩ ∆ c . Clearly a G ∩ ∆ c = ∅ , so(2) s + 1 | a G | by Lemma 2.5. On the other hand, by Lemma 2.1 (v) we have:(3) | a G | mq ( q − q + p q + 1) . Combining (2) with (1) and (3), we find that q m mq ( q − q + √ q + 1) < mq , andso q m − < m. Hence, m > q m − > (1 + 7) m − > m − , and, simplifying, we see that m < < , and so m = 1 or .Suppose that m = 2 , so N ∼ = Sz( q ) × Sz( q ) . The group G must act transitively on thesimple direct factors of N , so q divides | G | and hence q divides ( s + 1) . This improves (1)to q s + 1 . Going back to (3) and (2), we find now that:(4) q s + 1 q ( q − q + p q + 1) < q . Let p be any odd prime dividing s + 1 . Then pq divides s + 1 , and, combined with (4), thisyields q pq s + 1 < q , a contradiction.Hence s + 1 must be a power of , and s + 1 = | G | / . Moreover, since q s + 1 < q by (1), s + 1 must be either q or q . Since | Sz( q ) | divides | G | , q − divides | G | by Lemma 2.1(i). Since s + 1 is a power of , this means that q − must divide s + 1 . However, if s + 1 = q ,then s + 1 = ( q − + 1 = ( q −
1) ( q − q − , and, if s + 1 = 2 q , then s + 1 = (2 q − + 1 = ( q −
1) 4 q ( q − q − , and so q − does not divide s + 1 in either case, a contradiction. Therefore, m = 1 , and N ∼ = Sz( q ) for some q . (cid:3) L EMMA If G is as above, then it has a unique nonabelian minimal normal subgroup N . P ROOF . Suppose that G has two minimal normal subgroups, M and N . By Lemma 4.2, N ∼ = Sz( q ) and M ∼ = Sz( q ′ ) , where without a loss of generality q q ′ . Distinct minimal normalsubgroups must commute, and so we may proceed as in the proof of Lemma 4.2: q ( q ′ ) / divides s + 1 , and so q / q ( q ′ ) / s + 1 . Also, N ∩ ∆ c = ∅ by Lemma 2.4 (v), so forsome x ∈ N , x G ∩ ∆ c = ∅ . Thus, by Lemma 2.5 and Lemma 2.1 (v), s + 1 | x G | q ( q − q + p q + 1) < q . If an odd prime p divides s + 1 , then p > , and s + 1 > p ( s + 1) > q > q , a contradiction, and so no odd prime can divide s + 1 . However, if s + 1 is a power of ,then s + 1 = | G | / , and, since q divides | G | , q / divides s + 1 . We know from above that s + 1 < q , so either s + 1 = q / or s + 1 = q , since s + 1 is a power of . Moreover, if s + 1 is a power of , then, since q − divides | N | , q − must divide | G | and hence s + 1 . If s + 1 = q , then s + 1 = ( q − + 1 = ( q −
1) ( q − q − . If s + 1 = q / , then q − divides s + 1) , and so s + 1) = q − q + 8 = ( q − q + q + q + q − q − q − q −
3) + 5 , and hence q − cannot divide s + 1 in either case, a contradiction. Therefore, if G has anonabelian minimal normal subgroup, it must be unique. (cid:3) L EMMA If G is as above, then s + 1 cannot be a power of . P ROOF . We know that N ∼ = Sz( q ) is the unique minimal normal subgroup of G , and so q divides | G | = 2( s + 1) , i.e., q divides s + 1 . Let s + 1 = 2 n − q for some n ∈ N . Hence s + 1 = 2 n − q − n q + 2 . Now, since ( q − q + 1) divides | Sz( q ) | , ( q − q + 1) must divide s + 1 = 2 n − q − n q + 2 . Now, n − q − n q + 2 = ( q + 1)(2 n − q − (2 n − + 2 n )) + (2 n − + 1) + 1 , which implies that q + 1 must divide (2 n − + 1) + 1 , which in turn implies that q n − + 1 .Since q is a power of , we have q n − .Note also that gcd( q + 1 , s + 1) = 1 . Let p be an odd prime dividing q + √ q + 1 , and let a ∈ N be an element of order p (Sylow subgroups of odd order are cyclic in Sz( q ) by Lemma2.1 (iii)). Since p does not divide s + 1 , by Lemma 2.4 (vi) we have a G ∩ ∆ c = ∅ , and so, byLemma 2.5,(5) s + 1 | a G | = q ( q − q − p q + 1) < q . Furthermore, since ( q − q + 1) = q − q +1 and s + 1 = 2 n − q − n q + 2 = ( q − n − + (2 n − − n q + 2) , ( q − q + 1) must divide n − − n q + 2 . We divide into two cases, depending on whether n − − n q + 2 is nonnegative or negative.Suppose first that n − − n q + 2 > . Thus n q n − + 2 , and so q n − + n − ,and, since q is a power of , q n − . But then s + 1 = 2 n − q > q , contradicting (5). OINT REGULAR GENERALIZED QUADRANGLES 11
Now suppose that n − − n q + 2 < . Thus ( q − q + 1) divides n q − n − − ,which is a positive integer. We already know that q n − , so let n − = 2 m q for some m > . Substituting, we have ( q − q + 1) divides m +1 q − m q − . Since m +1 q − m q − q − q + 1)2 m +1 + (2 m +1 − m ) q − m +1 q + (2 m +1 − , ( q − q + 1) must divide (2 m +1 − m ) q − m +1 q + (2 m +1 − , and since (2 m +1 − m ) q − m +1 q + (2 m +1 −
2) = (2 m +1 − m )( q + 1) + (2 m − m +1 q − ,q + 1 must divide m − m +1 q − . We once more split into two cases, depending on the signof m − m +1 q − .If m − m +1 q − > , then m +1 q m − , and so q < m − . On the other hand, if m − m +1 q − < , then q + 1 m +1 q − m + 2 , which implies that q m +1 q andthat q m +1 . Moreover, if q = 2 m +1 in this last case, then m = 0 , which is impossible, since q > . Thus, in either case, q m . Recalling that s + 1 = 2 n − q and n − = 2 m q , we have ( s + 1) = 2 n − q = 2 m q > q , which contradicts (5). Therefore, s + 1 cannot be a power of . (cid:3) In order to further determine the structure of G , we will now examine the generalized Fittingsubgroup of the centralizer in G of N , F ∗ ( C G ( N )) = E ( C G ( N )) F ( C G ( N )) . Since N is theunique nonabelian minimal normal subgroup and the Schur multiplier of Sz( q ) is either trivialor, if q = 8 , an elementary abelian group of order by Lemma 2.1 (vii), E ( C G ( N )) E ( G ) = N or E ( C G ( N )) E ( G ) · N , and so E ( C G ( N )) = 1 (as in [ , Step 5]). This meansthat F ∗ ( C G ( N )) = F ( C G ( N )) . Note further that since F ( C G ( N )) is characteristic in C G ( N ) ,which is itself normal in G , F ( C G ( N )) ⊳ G , which implies that F ( C G ( N )) F ( G ) . L EMMA
There is at most one odd prime p such that O p ( G ) = 1 . Moreover, if such anodd prime p exists, then the following hold: (i) p divides s + 1 ; (ii) for any nonidentity element x ∈ G , | x G | > s + 1 ; (iii) | O p ( G ) | > s + 2 . P ROOF . The proof of this lemma is analagous to the proof of [ , Theorem 4.2]. By Lemma2.3 (i), O p ( G ) = 1 for all odd primes dividing s +1 . Let p , p be two odd primes dividing s +1 ,and let N = O p ( G ) and N = O p ( G ) . Without a loss of generality, | N | = p α > p α = | N | . Note that p α < s + 1 , since otherwise ( s + 1) | N || N | s + 1 , a contradiction. On theother hand, let = x ∈ N . Since | x G | | N | − < s , | x G ∩ ∆ c | < s. Since s divides | x G ∩ ∆ c | by Lemma 2.4 (ii), this implies that x G ∩ ∆ c = ∅ and x G ⊆ ∆ . But then by Lemma 2.4 (vi) p divides s + 1 , a contradiction. Hence there is at most one odd prime p such that p divides | F ( G ) | .Let | O p ( G ) | = p α and = x ∈ O p ( G ) . Since p does not divide s + 1 , x G ∩ ∆ c = ∅ byLemma 2.4 (vi), and so | x G | > s + 1 by Lemma 2.5. This proves (ii). Finally, ∈ O p ( G ) bydefinition and so | O p ( G ) | > | G | + | x G | > s + 2 . (cid:3) L EMMA If | O ( G ) | s + 1 , then O ( G ) . C , the cyclic group of order . P ROOF . By assumption, | O ( G ) | s + 1 . In particular, this implies that for each x ∈ O ( G ) , | x G | < s . This means that x G ⊆ ∆ by Lemma 2.4 (ii), which in turn implies that O ( G ) ⊆ ∆ . However, by Lemma 2.4 (v), this means that | O ( G ) | divides gcd( s +1 , s − s ) = 2 . Therefore, O ( G ) . C , as desired. (cid:3) L EMMA
The Fitting subgroup F ( C G ( N )) . C . P ROOF . Let p be an odd prime that divides | F ( C G ( N )) | , and let O p ( C G ( N )) = Q . Since Q is characteristic in F ( C G ( N )) , Q ⊳ G. Let x ∈ Q , x = 1 . By Lemma 4.5 (i), we have p | ( s + 1) , and so x G ∩ ∆ c = ∅ by Lemma 2.4 (vi). Thus, by Lemma 2.5, | Q | > | G | + | x G | > s + 2 . On the other hand, since gcd( s + 1 , s − s ) = 2 , by Lemma 2.4 (v) we know that N is notcontained entirely in ∆ , and so for some a ∈ N , a G ∩ ∆ c = ∅ , and hence, by Lemma 2.5, | N | > | G | + | a G | > s + 2 . Let = x ∈ Z ( Q ) . Then | C G ( x ) | > | Q || N | > ( s + 2)( s + 2) . Hence, we have: | x G | = | G || C G ( x ) | ( s + 1)( s + 1)( s + 2)( s + 2) < s. However, this implies that x G ⊆ ∆ by Lemma 2.4 (ii), and by Lemma 2.4 (vi) we have that p divides s + 1 , a contradiction. Therefore, F ( C G ( N )) = O ( C G ( N )) O ( G ) .Finally, we know that | O ( G ) | s + 1) . By Lemma 4.4, s + 1 cannot be a power of ,so there is an odd prime r that divides s + 1 . This implies that | O ( G ) | s + 1) < r ( s + 1) s + 1 . By Lemma 4.6, we have O ( G ) . C , and so F ( C G ( N )) . C , as desired. (cid:3) T HEOREM
Let s > be an odd integer such that s + 1 is coprime to . If G actsregularly on the point set of a generalized quadrangle of order s , then G has no nonabelianminimal normal subgroups. P ROOF . By Lemma 4.7, F ∗ ( C G ( N )) = F ( C G ( N )) . C . The generalized Fitting sub-group is normal in C G ( N ) and self-centralizing; however, this is only possible if C G ( N ) . C itself.First, suppose that C G ( N ) ∼ = C . Note that G/ ( N C G ( N )) . Out(Sz( q )) ∼ = C e +1 , where q = 2 e +1 (see Lemma 2.1 (viii)). As in the proof of [ , Step 5], this means that: | G | e + 1) | Sz( q ) | < q . On the other hand, q divides | G | = 2( s + 1) , so q divides ( s + 1) . By Lemma 4.4, s + 1 is not a power of , so q s + 1 . But then: | G | = ( s + 1)( s + 1) > q ((5 q − + 1) > q , a contradiction.Now suppose that C G ( N ) = 1 . Then
G/N . Out(Sz( q )) ∼ = C e +1 . Arguing as above, thisimplies that | G | (2 e + 1) | Sz( q ) | < q . On the other hand, q divides | G | = 2( s + 1) , and so, again proceeding as above, since s + 1 is not a power of and ∤ ( s + 1) , this implies that q s + 1 , which would in turn imply that | G | > q , a contradiction. Therefore, if G acts regularly on the points of such a generalizedquadrangle of order ( s, s ) , then G does not have a nonabelian minimal normal subgroup. (cid:3) We will now assume, like above, that s is odd, does not divide s + 1 , and G acts regularlyon the set of points of a generalized quadrangle of order ( s, s ) . We proved above that such a G cannot have a nonabelian minimal normal subgroup, and the following lemma summarizeswhat can be said in the case that G is not solvable.L EMMA If G is not solvable, then F ∗ ( G ) . O p ( G ) × C , where p is an odd primedividing s + 1 and for any = x ∈ O p ( G ) , | x G | > s + 1 . OINT REGULAR GENERALIZED QUADRANGLES 13 P ROOF . If such a group G exists, then every minimal normal subgroup of G is abelian byTheorem 4.8, and so F ∗ ( G ) = F ( G ) . By Lemma 4.5, there is at most one odd prime p forwhich O p ( G ) = 1 , and, if such a p exists, then p divides s + 1 and | x G | > s + 1 for allnonidentity elements x ∈ O p ( G ) . Thus F ∗ ( G ) . O p ( G ) × O ( G ) , and it only remains to showthat O ( G ) ∼ = C .Suppose that | O ( G ) | > s + 1 . Since | O ( G ) | | G | = 2( s + 1) , this would require that s + 1 is a power of , in which case | O ( G ) | = 2( s + 1) , and then O ( G ) is a Sylow 2-subgroupof G . Thus G/O ( G ) has odd order and is solvable. Since both O ( G ) and G/O ( G ) aresolvable, so is G , contrary to our hypotheses. Therefore, if G is not solvable, | O ( G ) | s + 1 ,which means O ( G ) . C by Lemma 4.6, completing the proof. (cid:3) We remark that, since gcd( s + 1 , s − s ) = 2 , by Lemma 2.4 (v), if F ∗ ( G ) has even order,then Z ( G ) ∼ = C . Otherwise, F ∗ ( G ) ∼ = O p ( G ) and Z ( G ) = 1 . It would be interesting to rule out entirely groups acting regularly on the point set of ageneralized quadrangle of order s , where s is odd and s + 1 is coprime to , in the same waythat Yoshiara ruled out such groups when s = t . A somewhat minor difference is that wecannot rule out the possibility that O ( G ) = 1 . A more substantial difference here is that, since |P| = |L| , we cannot guarantee that any lines are fixed by the elements of odd order of G ,which is crucial step in Yoshiara’s proof (see [ , Step 7, Final step]).
5. Generalized quadrangles of order ( u , u ) with a group of automorphisms actingregularly on points Throughout this section, let Q be a generalized quadrangle of order ( u , u ) , u > , withpoint set P and line set L . Let G be a group of automorphisms of Q that acts regularly on P . Note that in this case | G | = |P| = ( u + 1)( u + 1) .L EMMA If G is as above, then does not divide u + 1 . P ROOF . Suppose that u ≡ − . This means that u + 1 ≡ , and so divides | G | . However, by Lemma 2.4 (iii), this implies that divides u + 1 . However, neverdivides u + 1 for any integer u , a contradiction. Hence no group of automorphisms can actregularly on P in this instance. (cid:3) Assume henceforth that does not divide u +1 . Note that if u is even, then | G | is odd, and so G would be solvable. We thus assume that u is odd. Suppose that G has a nonabelian minimalnormal subgroup N . Since does not divide | G | , by Lemma 2.4 (iv), we have N ∼ = Sz( q ) m forsome m > and q = 2 e +1 > . Let N = S S ...S m , where each S i (i) is normal in N , (ii) isconjugate in G to and has trivial intersection with all other S j , and (iii) is isomorphic to Sz( q ) .First, u + 1 ≡ , so | G | = 2( u + 1) . Moreover, u + 1 = ( u + 1)( u − u + u − u + 1) , and u − u + u − u + 1 is odd, so | G | = 2( u + 1) . Hence, by Lemma 2.1 (i),(6) | N | = q m u + 1) u + 1) . Let x be an element of S of order . If x G ⊆ ∆ , then by Lemma 2.4 (vi), would divide u + 1 ,a contradiction. Hence x G ∩ ∆ c = ∅ , and by Lemma 2.5, this means | x G | > u + 1 . Since theorder of the centralizer of an element of order in Sz( q ) is q by Lemma 2.1 (vi), we have:(7) u + 1 | x G | = 12 mq ( q − q + 1) < mq . L EMMA
If such a G and N exist, then m = 1 and N ∼ = Sz( q ) . P ROOF . From (6) and (7), we have (cid:18) q m − (cid:19) + 1 u + 1 < mq . Simplifying, this yields q m < mq + 4 q m . If m > , this means that q m < mq + 4 q m mq m + 4 q m (2 m + 4) q m , and so: m + 4 > q m > (1 + 7) m > m ) , which simplifies to m < , a contradiction. Therefore, m = 1 , and N ∼ = Sz( q ) for some q . (cid:3) L EMMA
If such a G and such an N as above exist, then u + 1 cannot be a power of . P ROOF . Suppose that u + 1 is a power of . First, note that (6) becomes q u + 1) when m = 1 , which implies that q u + 1) . Note also that (7) becomes u + 1) < q when m = 1 . Putting these together, we see that: ( u + 1) < u + 1) < q u + 1) , which implies that ( u + 1) < q u + 1) . Since u + 1 is a power of by assumption, thismeans that q = 2( u + 1) . Hence q + 1 = 2 u + 3 . Since q + 1 divides | N | , q + 1 divides | G | .Since q + 1 divides | G | and q + 1 = 2 u + 3 and is odd, u + 3 must divide | G | u + 1 = ( u + 1)( u + 1) u + 1 = ( u +1)( u − u + u − u +1) = u − u +2 u − u +2 u − u +1 . On the other hand, u − u +2 u − u +2 u − u +1) = (2 u +3)(32 u − u +184 u − u +574 u − , so u + 3 must divide · . This means that u + 3 is one of , , , whichmeans that u +1 is one of , , , none of which is a power of , a contradiction. Therefore, u + 1 cannot be a power of . (cid:3) T HEOREM If G acts regularly on the set of points of a generalized quadrangle of order ( u , u ) , then G cannot have a nonabelian minimal normal subgroup. P ROOF . By Lemmas 5.1, 5.2, and 5.3, does not divide u + 1 , u + 1 is not a power of ,and N ∼ = Sz( q ) for some q = 2 e +1 . Let r be an odd prime dividing u + 1 . Then, noting that m = 1 , from (6) we have q = | N | | G | = 2( u + 1) < r ( u + 1) u + 1 , which implies that q u + 2 u + 1 . On the other hand, again noting that m = 1 , from (7) wehave: u + 1 < q . Putting these together, we see that u + 2 < u + 2 u + 1 , which simplifies to ( u − < , acontradiction. Therefore, no such N can exist, and such a G cannot have a nonabelian minimalnormal subgroup. (cid:3) Note that Theorem 1.5 follows immediately from Theorems 4.8 and 5.4.We will now assume, like above, that u is odd, does not divide u + 1 , and G acts regularlyon the set of points of a generalized quadrangle of order ( u , u ) . We proved above that such a G cannot have a nonabelian minimal normal subgroup, and the following lemma summarizeswhat can be said in the case that G is not solvable.L EMMA If G is not solvable, then F ∗ ( G ) . O p ( G ) × C , where p is an odd primedividing u − u + u − u + 1 and for any = x ∈ O p ( G ) , | x G | > u + 1 . P ROOF . The proof proceeds similarly to those in Section 4. Assume that G is not solvable. OINT REGULAR GENERALIZED QUADRANGLES 15 (Step 1)
If there is a normal subgroup N of G such that N ⊆ ∆ , then | N | divides gcd( u +1 , ( u ) − u ) = 2 . This follows from Lemma 2.4 (v).(Step 2) If p is an odd prime and O p ( G ) is nontrivial, then p divides u − u + u − u + 1 and | O p ( G ) | > u + 2 . Let p be an odd prime such that O p ( G ) is nontrivial. By (Step 1), O p ( G ) cannot beentirely contained in ∆ . This means that there is x ∈ O p ( G ) such that x ∈ ∆ c , and soby Lemma 2.5, | x G | > u + 1 . Thus | O p ( G ) | > | G | + | x G | > u + 2 . Since gcd( u + 1 , u + 1) = 2 and | O p ( G ) | > u + 1 , we have that p divides u + 1 .Moreover, since gcd( u + 1 , u − u + u − u + 1) = 1 , we have that p divides u − u + u − u + 1 .(Step 3) There is at most one odd prime p such that O p ( G ) is nontrivial. Assume that O p ( G ) and O p ( G ) are nontrivial for distinct odd primes p and p . By(Step 2), both p and p divide u − u + u − u + 1 and both | O p ( G ) | , | O p ( G ) | areat least u + 2 . However, since p and p are distinct, u − u + u − u + 1 > | O p ( G ) || O p ( G ) | > ( u + 2) , which simplifies to ( u + 3)( u + 1) , a contradiction.(Step 4) O ( G ) . C Since | O ( G ) | | G | = 2( u + 1) , | O ( G ) | u + 1) < u + 1 , and so | x G | < u for all x ∈ O ( G ) . By Lemma 2.4 (ii), this means x G ⊆ ∆ , and so O ( G ) ⊆ ∆ . Theresult follows from (Step 1).(Step 5) F ∗ ( G ) . O p ( G ) × C , where p is an odd prime dividing u − u + u − u + 1 By Theorem 5.4, G has no nonabelian minimal normal subgroups, and hence F ∗ ( G ) = F ( G ) . This now follows from (Step 3) and (Step 4).(Step 6) For all nonidentity elements of O p ( G ) , | x G | > u + 1 . Let x ∈ O p ( G ) , x = 1 . Since p ∤ u + 1 , by Lemma 2.4 (vi) we know that x G is notcontained in ∆ . The result follows from Lemma 2.5. (cid:3) We remark that, much as in the case when s = t in Section 4, it would be nice to rule outthe case when ( s, t ) = ( u , u ) entirely. While we can guarantee here that for every odd prime r dividing u + 1 , there are at least two lines stabilized by a nontrivial r -subgroup R , we cannotguarantee that exactly two lines are stabilized by R , which was the conclusion of [ , Step 7].(Yoshiara used [ , 1.2.4], which is a result specific to generalized quadrangles with s = t .)Moreover, even if we know that exactly two lines are stabilized by R (which would force R toact coprimely on O p ( G ) as in [ , Final step]), we have that u − u + u − u + 1 ≡ u + 1) , and so such an r -subgroup could in theory act coprimely on O p ( G ) .A CKNOWLEDGEMENTS . The author would like to thank John Bamberg and Cai-Heng Li formany interesting conversations on this topic as well as feedback on drafts of this manuscript,and the author would also like to thank the anonymous referees for many excellent suggestionsthat greatly improved the readability of this paper. This work was initially started when theauthor was employed at the University of Western Australia, and the author acknowledges thesupport of the Australian Research Council Discovery Grant DP120101336 during his timethere.
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