Biplanes with flag-transitive automorphism groups of almost simple type, with classical socle

DOI 10.1007/s10801-007-0070-7

Biplanes with flag-transitive automorphism groups of almost simple type, with classical socle

Eugenia O’Reilly-Regueiro

Received: 31 October 2005 / Accepted: 9 March 2007 / Published online: 21 April 2007

© Springer Science+Business Media, LLC 2007

Abstract In this paper we prove that if a biplaneDadmits a flag-transitive automor- phism groupGof almost simple type with classical socle, thenDis either the unique (11,5,2) or the unique (7,4,2) biplane, andG≤P SL2(11)orP SL2(7), respectively.

Here ifXis the socle ofG(that is, the product of all its minimal normal subgroups), thenXG≤AutGandXis a simple classical group.

Keywords Automorphism group·Biplanes·Flag-transitive

1 Introduction

A biplane is a(v, k,2)-symmetric design, that is, an incidence structure ofv points andv blocks such that every point is incident with exactlyk blocks, and every pair of blocks is incident with exactly two points. Points and blocks are interchangeable in the previous definition, due to their dual role. A nontrivial biplane is one in which 2< k < v−1. A flag of a biplaneD is an ordered pair(p, B)wherep is a point ofD,Bis a block ofD, and they are incident. Hence ifGis an automorphism group ofD, thenGis flag-transitive if it acts transitively on the flags ofD.

The only values ofkfor which examples of biplanes are known arek=3, 4, 5, 6, 9, 11, and 13 [7, pp. 76]. Due to arithmetical restrictions on the parameters, there are no examples withk=7, 8, 10, or 12.

Fork=3, 4, and 5 the biplanes are unique up to isomorphism [6], fork=6 there are exactly three non-isomorphic biplanes [13], fork=9 there are exactly four non- isomorphic biplanes [26], fork=11 there are five known biplanes [3,10,11], and fork=13 there are two known biplanes [1], in this case, it is a biplane and its dual.

E. O’Reilly-Regueiro (

Instituto de Matemáticas, Universidad Nacional Autónoma de México, Mexico, DF 04510, Mexico

e-mail: eugenia@matem.unam.mx

In [24] it is shown that if a biplane admits an imprimitive, flag-transitive automor- phism group, then it has parameters (16,6,2). There are three non-isomorphic biplanes with these parameters [4], two of which admit flag-transitive automorphism groups which are imprimitive on points, (namely 2⁴S4and(Z2×Z8)S4[24]). Therefore, if any other biplane admits a flag-transitive automorphism group G, thenGmust be primitive. The O’Nan-Scott Theorem classifies primitive groups into five types [22].

It is shown in [24] that if a biplane admits a flag-transitive, primitive, automorphism group, it can only be of affine or almost simple type. The affine case was treated in [24]. The almost simple case when the socle ofGis an alternating or a sporadic group was treated in [25], in which it is shown that no such biplane exists. Here we treat the almost simple case when the socleXofGis a classical group. We now state the main result of this paper:

Theorem 1 (Main Theorem) If D is a nontrivial biplane with a primitive, flag- transitive automorphism groupGof almost simple type with classical socleX, then Dhas parameters either (7,4,2), or (11,5,2), and is unique up to isomorphism.

This, together with [24, Theorem 3] and [25, Theorem 1] yield the following:

Corollary 1 IfDis a nontrivial biplane with a flag-transitive automorphism group G, then one of the following holds:

(1) Dhas parameters (7,4,2), (2) Dhas parameters (11,5,2), (3) Dhas parameters (16,6,2),

(4) G≤AL₁(q), for some odd prime powerq, or

(5) Gis of almost simple type, and the socleXofGis an exceptional group of Lie type.

For the purpose of proving our Main Theorem, we will considerDto be a nontriv- ial biplane, with a primitive, flag-transitive, almost simple automorphism groupG, with simple socleX, such thatX=X_d(q)is a simple classical group, with a natural projective action on a vector spaceV of dimensiondover the fieldFq, whereq=p^e, (pprime).

For this we will proceed as in [27], in which the case for finite linear spaces with almost simple flag-transitive automorphism groups of Lie type is treated.

2 Preliminary results

In this section we state some preliminary results we will use throughout this paper.

Lemma 2 IfDis a(v, k,2)-biplane, then 8v−7 is a square.

Proof The result follows from [24, Lemma 3].

Corollary 3 If D is a flag-transitive (v, k,2)-biplane, then 2v < k², and hence 2|G|<|Gx|³.

Proof The equalityk(k−1)=2(v−1), impliesk²=2v−2+k, so clearly 2v < k². The result follows fromv= |G:G_x|andk≤ |G_x|.

From [9] we get the following two lemmas:

Lemma 4 If D is a biplane with a flag-transitive automorphism group G, then k divides 2di for every subdegreed_i ofG.

Lemma 5 IfGis a flag-transitive automorphism group of a biplaneD, thenkdivides 2·gcd(v−1,|Gx|).

Lemma 6 (Tits Lemma [28, 1.6]) IfXis a simple group of Lie type in characteristic p, then any proper subgroup of index prime topis contained in a parabolic subgroup ofX.

Lemma 7 IfXis a simple group of Lie type in characteristic 2, (XA₅orA₆), then any proper subgroupHsuch that[X:H]2≤2 is contained in a parabolic subgroup ofX.

Proof First assumeX=Cl_n(q)is classical (qa power of 2), and takeHmaximal in X. By Aschbacher’s Theorem [2],His contained in a member of the collectionCof subgroups ofL_n(q), or inS, that is,H^(∞) is quasisimple, absolutely irreducible, and not realisable over any proper subfield ofF(q).

We check for every family Ci that if H is contained inCi, then 2|H|2<|X|2, except whenHis parabolic.

Now we takeH∈S. Then by [18, Theorem 4.2],|H|< q²ⁿ⁺⁴, orH andXare as in [18, Table 4]. If|X|2≤2|H|2≤q²ⁿ⁺⁴, then eitherX=L_n(q)andn≤6, or X=Spn(q)orP _n(q)andn≤10. We check the list of maximal subgroups ofX for n≤10 in [15, Chapter 5], and we see that no group H satisfies 2|H|2≤ |X|2. We then check the list of groups in [18, Table 4], and again, none of them satisfy this bound.

Finally, assume X to be an exceptional group of Lie type in characteristic 2.

By [20], if 2|H| ≥ |X|2, thenH is either contained in a parabolic subgroup, orH andXare as in [20, Table 1]. Again, we check all the groups in [20, Table 1], and in

all cases 2|H|2<|X|2.

As a consequence, we have a strengthening of Corollary3:

Corollary 8 SupposeD is a biplane with a primitive, flag-transitive almost simple automorphism groupGwith simple socleXof Lie type in characteristicp, and the stabiliserGxis not a parabolic subgroup ofG. Ifpis odd thenpdoes not dividek;

and ifp=2 then 4 does not dividek. Hence|G|<2|G_x||G_x|²_p.

Proof We know from Corollary 3 that |G|<|G_x|³. Now, by Lemma 6, p di- vides v = [G:Gx]. Since k divides 2(v−1), if p is odd then (k, p)=1, and if p=2 then (k, p)≤2. Hence k divides 2|G_x|p, and since 2v < k², we have

|G|<2|G_x||G_x|²_p.

From the previous results we have the following lemma, which will be quite useful throughout this chapter:

Lemma 9 Supposepdividesv, andG_xcontains a normal subgroupHof Lie type in characteristicpwhich is quasisimple andp|Z(H )|; thenkis divisible by[H:P], for some parabolic subgroupP ofH.

Proof The assumption thatp dividesv and the fact thatk divides 2(v−1)imply (k, p)≤(2, p). Also, we knowk= [G_x:G_x,B](whereB is a block incident with x), so[H:HB]dividesk, and therefore([H:HB], p)≤(2, p). By Lemmas6and7 we conclude thatH_Bis contained in a parabolic subgroupP ofH, andP maximal inHimplies thatH_Bis contained inP, sokis divisible by[H:P].

Lemma 10 ([21, 3.9]) IfX is a group of Lie type in characteristicp, acting on the set of cosets of a maximal parabolic subgroup, and X is not P SL_d(q),P ⁺_2m(q) (withmodd), norE6(q), then there is a unique subdegree which is a power ofp.

3 Xis a linear group

In this case we consider the socle of G to be P SLn(q), and β= {v1, v2, . . . , vn} a basis for the naturaln-dimensional vector spaceV forX.

Lemma 11 If the groupXisP SL₂(q), then it is one of the following:

(1) P SL₂(7)acting on the (7,4,2) biplane with point stabiliserS₄, or (2) P SL2(11)acting on a (11,5,2) biplane with point stabiliserA5.

Proof Suppose X∼=P SL2(q), (q =p^m) is the socle of a flag-transitive automor- phism group of a biplaneD, soG≤P L2(q). AsGis primitive,Gxis a maximal subgroup ofG, and henceX_xis isomorphic to one of the following [12]: (Note that

|G_x|divides(2, q−1)m|X_x|):

(1) A solvable group of indexq+1.

(2) D_(2,q)(q−1). (3) D(2,q)(q+1).

(4) L₂(q₀)if(r >2), orP GL₂(q₀)if(r=2), whereq=q₀^r,rprime.

(5) S4ifq=p≡ ±1(mod 8).

(6) A4ifq=p≡3,5,13,27,37(mod 40).

(7) A₅ifq≡ ±1(mod 10).

(1)Herev=q+1, sok(k−1)=2(v−1)=2q, henceq=3, butP SL₂(3)is not simple.

(2)and(3)The degrees in these cases are a triangular number, but the number of points on a biplane is always one more than a triangular number.

(4) First assume r >2. Clearly, q₀ divides v =q₀^r−1q₀^2r−1 q₀²−1

, so k divides 2(v−1, mq0(q₀²−1)), hence k=^2m(q_n²⁰⁻¹⁾ for some n. Say q0=p^b, som=br and (except forp=2 and 2≤b≤4), we haveb <√

q0, (sinceb²< p^b=q0).

Now,k²>2vimplies

4m²(q₀²−1)²

n² >2q₀^r−1

q₀^2r−1 q₀²−1

,

so

n²<2m²(q₀²−1)³ (q₀^2r−1)q₀^r−1.

First considerr >3, so (r≥5). Hereq₀^r > b²r²=m². On the other hand, 2m²>

q^r₀⁻¹(q₀^2r−1)

(q₀²−1)³ , therefore

2q₀^r<q₀^r−1(q₀^2r−1) (q₀²−1)³ , which is a contradiction.

Next considerr=3. Fromk²>2v, we obtain 18b²(q₀²−1)³> n²q₀²(q₀⁶−1), this together withb²< q0, implyn²(q₀⁶−1) <18q₀⁵, thereforeq0≤17. We check for all possible values ofq₀that 8v−7 is not a square, contradicting Lemma2.

Now assumer=2. Thenv=^q_(2,q^0(q02₋⁺₁₎¹⁾. Asq=q₀²=2, we havem²< q, so 4b²<

q₀², which impliesq0=2.

First consider q even. From 2(v−1)=k(k−1), we have 2(q₀³+q0−1)=

2m(q₀²−1) n

2m(q₀²−1)

n −1

, however gcd(q₀³+q₀−1, q₀²−1)divides 3, which implies k=^6m_t , witht=1,3.

Ift=3 thenq₀³+q₀−1=2m²−m=m(2m−1) <2m², butm < q₀, so this is a contradiction.

Ift=1 thenq₀³+q0−1=18m²−6m, which impliesq0<18, that isq0=4,8, or 16. Howeverm=2bimpliesk=12b, sov−1 is divisible by 6, but this is not the case for any of these values ofq0.

Now consider q odd. The equality 2(v−1)=k(k−1) yields q₀³+q₀−2=

4m²

n² (q₀²−1)²− ^2m_n (q₀²−1), and the inequalityk²>2v implies ^4m_n2²(q₀²−1)²>

q0(q₀²+1). In this casem=2b, sok=^4b(p2b_n⁻¹⁾, andv= ^p3b₂^+pb >^b6+₂^b2, hence we have the following inequalities:

b⁶+b²< p^3b+p^b<4b(p^2b−1)

n <4b·p^2b

n .

This implies ^n(p3b+pb)

p^2b <4b, son(p^b+p^b2) <4b <4p^b2, thereforen(p^b2 +1) <4 which impliesn=1=b, andp=3, 5, or 7, but in all these casesk > v, which is a contradiction.

(5) In this caseq =p≡ ±1(mod 8), and m=1, soG₀∼=S₄. We haveq odd, v=^q(q₄₈²⁻¹⁾, andkdivides 2_q(q²₋₁₎₋₄₈

48 ,24

, sok|48. Nowk²>2vimpliesq≤37, henceq=7, 17, 23, or 31. The only one of these values for which 8v−7 is a square (Lemma2) isq =7, so v=7 and k=4, that is, we have the (7,4,2) biplane and G=X∼=P SL2(7).

(6) Here q=p≡3, 5, 13, 27, or 37 (mod 40), so m=1 andG_x∼=A₄. Here v= ^q(q₂₄²⁻¹⁾, and so k divides 2_q(q²₋₁₎₋₂₄

24 ,12

, sok|24. As 2v < k², we have q =3, 5, or 13. Forq =3 we havev=1, which is a contradiction. Forq=5 we havev=5, but there is no such biplane. Finally,q=13 impliesv=91, but then 8v−7 is not a square, contradicting Lemma2.

(7)Hereq=porp²≡ ±1(mod 10), andv=^q(q₁₂₀²⁻¹⁾, sokdivides 120m, with m=1 or 2. The inequality 2v < k²impliesq³−q <60k²<60(120)²m², soq=9, 11, 19, 29, 31, 41, 49, 59, 61, 71, 79, 81, 89, or 121. Of these, the only value for which 8v−7 is a square isq=11. In this case,v=11 andk=5, that is, we have a (11,5,2) biplane, withG=X∼=P SL2(11), andG_x∼=A5.

This completes the proof of Lemma11.

Lemma 12 The groupXis notP SLn(q), withn >2, and(n, q)=(3,2).

Proof Suppose X∼=P SL_n(q), with n >2 and (n, q)=(3,2)(since P SL₃(2)∼= P SL2(7)). We have q =p^m, and take {v1, . . . , v_n} to be a basis for the natural n-dimensional vector space V for X. Since G_x is maximal in G, then by Aschbacher’s Theorem [2], the stabiliserG_x lies in one of the families Ci of sub- groups ofL_n(q), or in the setS of almost simple subgroups not contained in any of these families. We will analyse each of these cases separately. In describing the Aschbacher subgroups, we denote byˆHthe pre-image of the groupH in the corre- sponding linear group.

C1) HereG_xis reducible. That is,G_x∼=P_i stabilises a subspace ofV of dimen- sioni.

SupposeG_x∼=P1. ThenGis 2-transitive, and this case has already been done by Kantor [14].

Now supposeG_x∼=P_i (1< i < n) fixesW, ani-subspace ofV. We will assume i≤ⁿ₂since our arguments are arithmetic, and foriandn−iwe have the same calcu- lations. Considering theG_x-orbits of thei-spaces intersectingWini−1-dimensional spaces, we seekdivides

2q(qⁱ−1)(qⁿ⁻ⁱ−1) (q−1)² . Also,

v=(qⁿ−1) . . . (qⁿ⁻ⁱ⁺¹−1)

(qⁱ−1) . . . (q−1) > qⁱ⁽ⁿ⁻ⁱ⁾, butk²>2v, so eitheri=3 andn <10, ori=2.

First assumei=3 andq=2.

Ifn=9 thenk=2²·3²·7², but the equationk(k−1)=2(v−1)does not hold.

Ifn=8 thenk=4·7·31 but again the equationk(k−1)=2(v−1)does not hold.

Forn=7k=420 or 210, but again,kdoes not divide 2(v−1).

Finally, ifn=6 thenk=196 or 98, but neither is a divisor of 2(v−1).

Now assumei=3 andq >2. Thenn=6 or 7.

Ifn=7 thenkdivides 2

q(q³−1)(q⁴−1)

(q−1)² ,(q⁷−1)(q⁶−1)(q⁵−1) (q³−1)(q²−1)(q−1) −1

,

but thenk²< v, which is a contradiction.

Ifn=6 thenkdivides 2

q(q³−1)²

(q−1)² ,(q⁶−1)(q⁵−1)(q⁴−1) (q³−1)(q²−1)(q−1) −1

,

But againk²<2v.

Hencei=2. Herev=^(q_(qⁿ⁻2^1)(q−1)(qⁿ⁻−¹⁻1)¹⁾, andGhas suborbits with sizes:

|{2-subspacesH : dim(H∩W )=1}| =^q(q+1)(q_q−1ⁿ⁻²⁻¹⁾ and

|{2-subspacesH :H∩W=0}| =^q4(q_(qⁿ⁻2²−⁻1)(q^1)(q−ⁿ⁻1)³⁻¹⁾.

Ifnis even thenkdivides ^q(q_(qⁿ₂⁻₋²⁻₁₎¹⁾, sinceq+1 is prime to^(qn_q⁻₋³⁻₁¹⁾, this implies k²< v, which is a contradiction.

Hencenis odd, andkdivides^2q(q_qⁿ⁻²₋₁⁻¹⁾(q+1,ⁿ⁻₂³).

First assume n=5. Thenv=(q²+1)(q⁴+q³+q²+q+1), andk divides 2q(q²+q+1). The fact thatk²>2vforcesk=2q(q²+q+1).

The conditionk(k−1)=2(v−1)implies

4q²(q²+q+1)²−2q(q²+q+1)=2(q⁶+q⁵+2q⁴+2q³+2q²+q), so

(q²+q+1)

2q(q²+q+1)−1

=(q⁵+q⁴+2q³+2q²+2q+1).

If we expand we get the following equality:

q⁵+3q⁴+4q³+q²−q−2=0, which is a contradiction. Thereforen≥7. Here

v=(qⁿ⁻¹+qⁿ⁻²+ · · · +q+1)(qⁿ⁻³+qⁿ⁻⁵+ · · · +q²+1),

andkdivides 2dc, whered=q(qⁿ⁻³+qⁿ⁻⁴+ · · · +q+1)andc=(q+1,ⁿ⁻₂³).

Sayk=^2dc_e , thenv < k²forcese≤2q. We have the following equality:

v−1

d =qⁿ⁻²+qⁿ⁻⁴+ · · · +q³+q+1, and also, sincek(k−1)=2(v−1), we have

k=2(v−1)

k +1=2e(v−1)

2dc =eqⁿ⁻²+eqⁿ⁻⁴+ · · · +eq³+eq+e+c

c .

Now,(kc, d)dividesd, and also (kc, q(eqⁿ⁻³+eqⁿ⁻⁵+ · · · +eq²+e))

=(eqⁿ⁻²+eqⁿ⁻⁴+ · · · +eq+e+c, q(eqⁿ⁻³+eqⁿ⁻⁵+ · · · +eq²+e))

=(eqⁿ⁻²+ · · · +eq+e+c, e+c),and (kc,^ed_q)

=(eqⁿ⁻²+ · · · +eq+e+c, eqⁿ⁻³+eqⁿ⁻⁴+ · · · +eq+e)

=(eqⁿ⁻²+ · · · +eq+e+c, (2e+c)q+e+c).

Therefore k divides c(e+c)((2e+c)q +e+c), and since e≤2q and c=

q+1,ⁿ⁻₂³

, the only possibilities fornandq aren=7 andq≤3, orn=9 and q=2. However in none of these possibilities is 8v−7 a square, again contradicting Lemma2.

C₁) HereGcontains a graph automorphism and Gx stabilises a pair{U, W}of subspaces of dimensioniandn−i, withi <ⁿ₂. WriteG⁰forG∩P L_n(q)of index 2 inG.

First assumeU⊂W. By Lemma10, there is a subdegree which is a power ofp.

On the other hand, ifpis odd then the highest power ofpdividingv−1 isq, it is 2q ifq >2 is even, and is at most 2ⁿ⁻¹ifq=2. Hencek²< v, which is a contradiction.

Now supposeV =U⊕W. Herepdividesv, so(k, p)≤2. First assumei=1.

Ifx= {v1,v2. . . vn}, then considery= {v1, . . . , vn−1,vn}, so[Gx:Gxy] =

qⁿ⁻²(qⁿ⁻¹−1)

q−1 andkdivides ^2(q_qⁿ⁻¹₋₁⁻¹⁾. Howeverv=^qn−1_q^(q₋ⁿ₁⁻¹⁾ > q²⁽ⁿ⁻¹⁾, which im- pliesk²< v, a contradiction.

Now assumei >1. Considerx= {v₁, . . . , v_i,v_i+1, . . . , v_n}andy= {v₁, . . . , vi−1, vi+vn,vi+1, . . . , vn}. Then[G⁰_x:G⁰_xy]pdivides 2(qⁱ−1)(qⁿ⁻ⁱ−1), which impliesk <2qⁿ, butv > q^2i(n−i), so againk²< v, a contradiction.

C2) HereGx preserves a partitionV =V1⊕ · · · ⊕Va, with eachVi of the same dimension, say,b, andn=ab.

First consider the case b=1 and n=a, and let x= {v1, . . . ,v_n} andy = {v1+v2,v2, . . . ,v_n}. Since n >2, we see k divides 4n(n−1)(q−1)= 2[G_x : G_xy]. Now v > ^qn(n−1)_n! and k² > v, so n=3 and q ≤4, that is v =

q³(q³−1)(q+1)

(3,q−1)6! . Ask|2(v−1), only forq =2 can k >2, so considerq =2. Then k|6 andv=28, but there is no such value ofksatisfyingk(k−1)=2(v−1).

Now let b >1, and consider x = {v₁, . . . , v_b,v_b+1, . . . , v_2b, . . .} and y = {v₁, . . . , v_b−1, v_b+1,v_b, v_b+2, . . . , v_2b, . . . ,v_n−b+1, . . . , v_n}. Then k divides

2a(a−1)(q^b−1)²

q−1 , sov >^qn(n−b)_a! , forcingn=4,q≥5, anda=2=b. In none of these cases can we obtaink >2.

C3) In this caseG_x is an extension field subgroup. Since 2|G_x||G_x|²_p>|G|by Corollary8, either:

(1) n=3 andX∩G_x=ˆ(q²+q+1)·3< P SL3(q)=X, or (2) nis even andG_x=N_G(ˆP SLⁿ

2(q²)).

First consider case(1). Herev=^q3(q2−₃^1)(q−1), sokdivides 6(q²+q+1)(log_pq), andk²> vimpliesq=3, 4, 5, 8, 9, 11, 13, or 16. In none of these cases is 8v−7 a square.

Now consider case(2)and writen=2m. Aspdividesv, we have(k, p)≤2. First supposen≥8, and letWbe a 2-subspace ofV considered as a vector space over the field ofq²elements, so thatW is a 4-subspace over a field ofqelements. If we con- sider the stabiliser ofWinG_xand inGthen inG_W\G_xW there is an elementgsuch thatGx∩G^gxcontains the pointwise stabiliser ofW inGxas a subgroup. Therefore kdivides 2(qⁿ−1)(qⁿ⁻²−1), contrary to 2v < k², which is a contradiction.

Now letn=6. Then since(k, p)≤2, Lemma9implieskis divisible by the index of a parabolic subgroup ofG_x, so it is divisible by the primitive prime divisorq₃of q³−1, but this divides the index ofGxinG, which isv, a contradiction.

Hencen=4. Thenv=^q4(q3−₂^1)(q−1), and sokis odd and prime toq−1. The fact that(v−1, q+1)=1 implieskis also prime toq+1, and hencek|(q²+1)log_pq, contrary tok²>2v, another contradiction.

C4) HereGx stabilises a tensor product of spaces of different dimensions, and n≥6. In all these casesv > k².

C5) In this case G_x is the stabiliser in G of a subfield space. So G_x = NG(P SLn(q0)), withq=q₀^mandmprime.

Ifm >2 then 2|G_x||G_x|²_p>|G|forcesn=2, a contradiction.

Hencem=2. Ifn=3 thenv=^(q03+_(q^1)(q_0+1,3)^02+1)q03.

Sincep dividesv, we have (k, p)≤2, so Lemma9impliesG_xB (whereB is a block incident withx) is contained in a parabolic subgroup ofG_x. Thereforeq₀²+ q0+1 dividesk, and(v−1, q₀²+q0+1)divides 2q0+(q0+1,3), forcingq0=2 andv=120, but then 8v−7 is not a square.

Ifn=4, then by Lemma 9we seeq₀²+1 dividesk, but q₀²+1 also divides v, which is a contradiction.

Hencen≥5. Considering the stabilisers of a 2-dimensional subspace ofV, we seekdivides 2(q₀ⁿ−1)(q₀ⁿ⁻¹−1), but thenk²< v, which is also a contradiction.

C6) HereG_xis an extraspecial normaliser. Since 2|G_x||G_x|²_p >|G|, we haven≤ 4. Now,n >2 implies thatGx∩Xis either 2⁴A6or 3²Q8, withXeitherP SL4(5)or P SL₃(7)respectively. Sincekdivides 2(v−1,|G_x|), we check thatk≤6, contrary tok²>2v.

Ifn=2 thenG_x∩X=A₄.a < L₂(p)=X, witha=2 precisely whenp≡ ±1 (mod 8), anda=1 otherwise, (and there areaconjugacy classes inX). From|G|<

|G_x|³we obtainp≤13. Ifp=7 then the action is 2-transitive. The remaining cases are ruled out by the fact thatkdivides 2(v−1,|G_x|), andk(k−1)=2(v−1).

C7) Here G_x stabilises the tensor product of a spaces of the same dimension, sayb, andn=b^a. Since|G_x|³>|G|, we haven=4 andG_x∩X=(P SL2(q)× P SL2(q))2^d< X=P SL4(q), withd=(2, q−1). Thenv=^q4(q2+1)(q_x ³⁻¹⁾ >^q_x⁹, withx=2 unlessq≡1(mod 4), in which casex=4. Hence 4k, and sokdivides 2(q²−1)log_pq, and ifqis odd thenkdivides^(q

2−1)log_pq

32 .

Ifqis odd, thenk²<^q₃₂⁹ <^q_x⁹ =v, a contradiction. Henceqis even, and so

k=2(q²−1)²log_pq

r ,

and sincek²>2vwe haver²<^4(q+1)

4log_pq

q⁵ , thereforeq≤32.

However, the five cases are dismissed by the fact thatkdivides 2(v−1).

C8) Now considerG_xto be a classical group.

(1) First assumeG_xis a symplectic group, sonis even. By Lemma6kis divisible by a parabolic index inG_x. Ifn=4 thenv=^q_(2,q^2(q3₋⁻₁₎¹⁾, and ^q_q⁴₋⁻₁¹ dividesk, however (v−1, q²+1)divides 2, which is a contradiction.

If n=6 then v= ^q6(q_(3,q^5−1)(q₋₁₎³⁻¹⁾ and q³+1 divides k, but q³+1 divides 2(v−1)only ifq=2, sok=9, too small.

Now supposen≥8. If we consider the stabilisers of a 4-dimensional subspace of G_x andG, we see thatk divides twice the odd part of(qⁿ−1)(qⁿ⁻²−1).

Also,(k, q−1)≤2, sokdivides 2^(qn−_(q^1)(q₋₁₎ⁿ⁻²2 ⁻¹⁾, and thereforek≤8q²ⁿ⁻⁴. The inequalityk²>2vforcesn=8. In this casev=^q12(q7−_(q^1)(q_−1,4)^{5−1)(q3−1)} which im- pliesq≤3, and in neither of these two cases is 8v−7 a square.

(2) Now letG_x be orthogonal. Thenq is odd, since that is the case with odd dimen- sion, and with even dimension it is a consequence of the maximality ofGx inG.

The case in whichn=4 andGx is of typeO₄⁺will be investigated later, in all other cases Lemma6implies thatkis divisible by a parabolic index inG_xand is therefore even, but it is not divisible by 4 sincevis also even and(k, v)≤2. This and the fact thatqdoes not dividekimpliesk < v, a contradiction.

(3) Finally letGx be a unitary group over the field of q0 elements, whereq=q₀². If n≥4 then considering the stabilisers of a nonsingular 2-subspace of V in G andG_x, we seek divides 2(q₀ⁿ−(−1)ⁿ)(q₀ⁿ⁻¹−(−1)ⁿ⁻¹). The inequality k²>2v forcesn=4, and in this casev= ^q06(q04+1)(q_(q0−⁰³_1,4)^+1)(q20+1). Sincek divides 2(q₀⁴−1)(q₀³+1)and(k, (q₀²+1)(q0−1)≤2, we seekdivides 2(q₀³+1)(q0+1), sok²≤2v, a contradiction. Thereforen=3, and by Lemma6q₀²−q₀+1 divides k, andkdivides 2(v−1)withv=^{q03(q03−1)(q}_x ⁰²⁺¹⁾withxeither 1 or 3. This implies q0=2, but thenv=280, and 8v−7 is not a square.

S) We finally consider the case whereG_xis an almost simple group, (modulo the scalars), not contained in the Aschbacher subgroups ofG. From [18, Theorem 4.2]

we have the possibilities|G_x|< q²ⁿ⁺⁴,G_x=A_n−1orA_n−2, orG_x∩X andXare as in [18, Table 4].

Also,|G|<|G_x|³by Corollary3and|G| ≤qⁿ²⁻ⁿ⁻¹, son≤7, and by the bound 2|G_x||G_x|²_p >|G|we need only to consider the following possibilities [15, Chapter 5]:

n=2, andGx∩X=A5, withq=11, 19, 29, 31, 41, 59, 61, or 121.

n=3, andG_x∩X=A6< P SL3(4)=X.

n=4, andG_x∩X=U₄(2) < P SL4(7)=X.

In the first case, with A5< L2(11)the action is 2-transitive. In the remaining cases, the fact thatkdivides 2|Gx|and 2(v−1)forcesk²< v, which is a contradic-

tion.

This completes the proof of Lemma12.

4 Xis a symplectic group

Here the socle ofGisX=P Sp2m(q), withm≥2 and(m, q)=(2,2). As a standard symplectic basis forV, we haveβ= {e1, f1, . . . , em, fm}.

Lemma 13 The groupXis notP Sp2m(q)withm≥2, and(m, q)=(2,2).

Proof We will considerG_x to be in each of the Aschbacher families of subgroups, and finally, an almost simple group not contained in any of the Aschbacher families ofG. In each case we will arrive at a contradiction.

When(p,2m)=(2,4)the groupSp₄(2^f)admits a graph automorphism, this case will be treated separately after the eight Aschbacher families of subgroups.

C1) IfG_x∈C1, then G_x is reducible, so either it is parabolic or it stabilises a nonsingular subspace ofV.

First assume thatG_x=P_i, the stabiliser of a totally singulari-subspace ofV, with i≤m. Then

v=(q^2m−1)(q^2m−2−1) . . . (q^2m−2i+2−1) (qⁱ−1)(qⁱ⁻¹−1) . . . (q−1) .

From this we seev≡q+1 (modpq), soq is the highest power ofp dividing v−1. By Lemma10there is a subdegree which is a power ofp, and sincekdivides twice every subdegree,kdivides 2q, contrary tov < k².

Now suppose thatG_x=N_2i, the stabiliser of a nonsingular 2i-subspaceU ofV, withm >2i. Thenpdividesv, so(k, p)≤2.

TakeU= e₁, f₁, . . . e_i, f_i, andW = e₁, f₁, . . . e_i−1, f_i−1, e_i+1, f_i+1. Thep- part of the size of theGx-orbit containingW is

(q²ⁱ−1)(q^2m−2i−1) (q²−1)² .

Since v < q^4i(m−i), we can only havev < k² if q=2 andm=i+1, which is a contradiction.

C2) IfG_x ∈C2 then it preserves a partitionV =V₁⊕ · · · ⊕V_a of isomorphic subspaces ofV.

First assume all theVj’s to be totally singular subspaces ofV of maximal dimen- sionm. ThenGx∩X=ˆGLm(q).2, andGxmaximal impliesqis odd [17]. Then

v=q^{m(m+2 1)}(q^m+1)(q^m−1+1) . . . (q+1)

2 >q^m(m+1)

2 ,

and(k, p)=1.

Let

x= {e₁, . . . , e_m,f₁, . . . , f_m}, and

y= {e₁, . . . , e_m−1, f_m,f₁, . . . , f_m−1, e_m}.

Then thep-part of theG_x-orbit ofy divides 2(q^m−1), and sokdivides 4(q^m−1), contrary tov < k².

Now assume that each of theV_j’s is nonsingular of dimension 2i, soG_x∩X= ˆSp2i(q)wrSt, withit=m. Let

x= {e₁, f₁, . . . , e_i, f_i,e_i+1, f_i+1, . . . , e_2i, f_2i, . . .}, and take

y= {e1, f1, . . . , e_i, f_i+e_i+1,e_i+1, f_i+1−e_i, e_i+2, . . . , e2i, f2i, . . .}. Considering the size of theG_x-orbit containingy, we seekdivides

t (t−1)(q²ⁱ−1)²

q−1 .

Now,

q^{2i2t (t−1)} t! < v,

sov < k²impliest!t⁴> q^{2i2t (t−1)+2−8i}, henceq^{2t (t−1)−6}< t^t+4and thereforet <4.

First assumet=3. Then by the above inequalitiesi=1 andq=2, but thenG_x is not maximal [8, p. 46], a contradiction.

Now lett=2. Thenk <2q⁴ⁱ⁻¹, soq^4i2−8i+2<8 and thereforei≤2.

Ifi=2 thenq=2 andv=45696=2⁷·3·7·17, but then 8v−7 is not a square, which is a contradiction.

Ifi=1 thenX=P Sp4(q),

v=q²(q²+1)

2 ,

andkdivides 2(q+1)²(q−1). Sincekdivides 2(v−1), we havekdivides(q²(q²+ 1)−2,2(q+1)²(q−1)), that is,kdivides

((q²+2)(q²−1),2(q+1)²(q−1))=(q²−1)(q²+2,2(q+1))≤6(q²−1).

Therefore

k=6(q²−1)

r ,

with 1≤r≤6. Now 2(v−1)=(q²+2)(q²−1), and also 2(v−1)=k(k−1), but we check that for all possible values ofrthis equality is not satisfied.

C3) IfGx∈C3, then it is an extension field subgroup, and there are two possibili- ties.

Assume first thatGx∩X=P Sp2i(q^t).t, withm=itandta prime number. From

|G|<|G_x|³, we obtaint=2 or 3.

Ift=3, thenv < k²impliesi=1, and so

Gx∩X=P Sp2(q³) < P Sp6(q)=X,

and

v=q⁶(q⁴−1)(q²−1)

3 .

This implies thatkis coprime toq+1, but applying Lemma9toP Sp₂(q³)yields q³+1 dividesk, which is a contradiction.

Ift=2, then

v=q²ⁱ²(q⁴ⁱ⁻²−1)(q⁴ⁱ⁻⁶−1) . . . (q⁶−1)(q²−1)

2 .

Consider the subgroupSp₂(q²)◦Sp_2i−2(q²)ofG_x∩X. This is contained inSp₄(q)◦ Sp4i−4(q)inX. Taking g∈Sp4(q)\Sp2(q²), we seeSp2i−2(q²)is contained in Gx∩G^gx, sokdivides 2(q⁴ⁱ−1)log_pq. The inequalityv < k²forcesi≤2.

First assumei=2. Then

v=q⁸(q⁶−1)(q²−1) 2

andk divides 2(q⁸−1)log_pq, but since(k, v)≤2 andq²−1 dividesv, we seek divides 2(q⁴+1)(q²+1)log_pq, forcingq=2. In this casev=2⁷·3³·7=24192, andk=2·5·17=170 (otherwisek²< v), but thenkdoes not divide 2(v−1), which is a contradiction.

Hencei=1, so

v=q²(q²−1)

2 ,

andGx∩X=P Sp2(q²).2< P Sp4(q)=X, Thereforek divides 4q²(q⁴−1), but since (k, v)≤2, then k divides 4(q²+1), so k= ^4(q2_r⁺¹⁾ for somer ≤8 (since v < k²). Now 2(v−1)=k(k−1), and also 2(v−1)=(q²−2)(q²+1), so we have

r²(q²−2)=16(q²+1)−4r, that is,

(r+4)(r−4)q²=2(8+r(r−2)).

This implies 4< r≤8, but solving the above equation for each of these possible values ofrgives non-integer values ofq, a contradiction.

Now assumeG_x∩X=ˆGU_m(q).2, withqodd. Sincevis even, 4 does not divide k. Also,kis prime top, so by the Lemma9, the stabiliser inG_x∩Xof a block is contained in a parabolic subgroup. But thenq+1 divides the indices of the parabolic subgroups in the unitary group, soq+1 dividesk, butq+1 also dividesv, which is a contradiction.

C4) IfG_x∈C4, thenG_xstabilises a decomposition ofV as a tensor product of two spaces of different dimensions, andG_xis too small to satisfy

|G|<2|Gx||Gx|²_p.