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Publicly Available Published by De Gruyter September 14, 2016

Point-primitive, line-transitive generalised quadrangles of holomorph type

John Bamberg EMAIL logo , Tomasz Popiel and Cheryl E. Praeger
From the journal Journal of Group Theory

Abstract

Let G be a group of collineations of a finite thick generalised quadrangle Γ. Suppose that G acts primitively on the point set 𝒫 of Γ, and transitively on the lines of Γ. We show that the primitive action of G on 𝒫 cannot be of holomorph simple or holomorph compound type. In joint work with Glasby, we have previously classified the examples Γ for which the action of G on 𝒫 is of affine type. The problem of classifying generalised quadrangles with a point-primitive, line-transitive collineation group is therefore reduced to the case where there is a unique minimal normal subgroup M and M is non-Abelian.

1 Introduction

A partial linear space is a point–line incidence geometry in which any two distinct points are incident with at most one line. All partial linear spaces considered in this paper are assumed to be finite. A generalised quadrangle𝒬 is a partial linear space that satisfies the generalised quadrangle axiom: given a point P and line ℓ not incident with P, there is a unique line incident with P and concurrent with ℓ. This axiom implies, in particular, that 𝒬 contains no triangles. If each point of 𝒬 is incident with at least three lines, and each line is incident with at least three points, then 𝒬 is said to be thick. In this case, there exist constants s,t⩾2 such that each point (line) is incident with exactly t+1 lines (s+1 points), and (s,t) is called the order of 𝒬. Generalised quadrangles were introduced by Tits [9], together with the other generalised polygons, in an attempt to find a systematic geometric interpretation for the simple groups of Lie type. It is therefore very natural to ask which groups arise as collineation groups of generalised quadrangles.

A topic of particular interest is that of generalised quadrangles admitting collineation groups M that act regularly on points, where the point set is identified with M acting on itself by right multiplication. Ghinelli [6] showed that a Frobenius group or a group with non-trivial centre cannot act regularly on the points of a generalised quadrangle of order (s,t) if s is even and s=t, and Yoshiara [10] showed that a generalised quadrangle with s=t2 does not admit a point-regular collineation group. Regular groups arise, in particular, as subgroups of certain primitive groups. Bamberg, Giudici, Morris, Royle and Spiga [2] showed that a group G acting primitively on both the points and the lines of a generalised quadrangle must be almost simple. The present authors and Glasby [3, Corollary 1.5] sought to weaken this assumption to primitivity on points and transitivity on lines, and, using a result of De Winter and Thas [4], classified the generalised quadrangles admitting such a group in the case where the primitive action on points is of affine type. (There are only two examples, arising from hyperovals in PG⁡(2,4) and PG⁡(2,16).) In this case, the regular subgroup M of G is Abelian, and hence left multiplication by any element of M is also a collineation. We consider the situation where M is non-Abelian but G has a second minimal normal subgroup, which is necessarily the centraliser of M, so that all left multiplications are again collineations. In the context of the O’Nan–Scott Theorem [8, Section 5] for primitive permutation groups, this means that the action of G on points is of either holomorph simple (HS) or holomorph compound (HC) type (see Section 2 for definitions). We prove the following result.

Theorem 1.1

Let G be a collineation group of a finite thick generalised quadrangle with point set P and line set L. If G acts transitively on L and primitively on P, then G has a unique minimal normal subgroup; that is, the action of G on P does not have O’Nan–Scott type HS or HC.

The proof of Theorem 1.1 is given in Sections 3 and 4, using some preliminary results established in Section 2, and the Classification of Finite Simple Groups.

2 Preliminaries

We first recall some definitions and facts about permutation groups. Let G be a group acting on a set Ω, and denote the image of x∈Ω under g∈G by xg. The orbit of x∈Ω under G is the set xG={xg∣g∈G}, the subgroup Gx={g∈G∣xg=x} is the stabiliser of x∈Ω, and the Orbit–Stabiliser Theorem says that |G:Gx|=|xG|. The action of G is transitive if xG=Ω for some (and hence every) x∈Ω, and semiregular if Gx is trivial for all x∈Ω. It is regular if it is both transitive and semiregular. If G acts transitively on Ω and M is a normal subgroup of G, then all orbits of M on Ω have the same length, and in particular it makes sense to speak of M being semiregular.

Given g∈G, define ρg, λg,ιg∈Sym⁢(Ω) by

ρg:x↦x⁢g,λg:x↦g-1⁢x,ιg:x↦g-1⁢x⁢g.

Set

GR={ρg:g∈G},GL={λg:g∈G},Inn⁡(G)={ιg:g∈G}.

The holomorphHol⁡(G) of G is the semidirect product GR⋊Aut⁡(G) with respect to the natural action of Aut⁡(G) on GR (see [1, Section 2.6]). We have Hol⁡(G)=NSym⁢(G)⁢(GR), and GL=CSym⁢(G)⁢(GR). A group H acting on a set Δ is permutationally isomorphic to G acting on Ω if there is an isomorphism θ:G→H and a bijection β:Ω→Δ such that β⁢(ωg)=β⁢(ω)θ⁢(g) for all g∈G and ω∈Ω. If a group M acts regularly on Ω, then there is a permutational isomorphism θ:NSym⁢(Ω)⁢(M)→Hol⁡(M) with bijection β:Ω→M, where β:αg↦g for some fixed α∈Ω, and θ:τ↦β-1⁢τ⁢β. We have θ⁢(M)=MR, so the regular action of M on Ω is permutationally isomorphic to the action of M on itself by right multiplication, and hence we can identify Ω with M. Furthermore, θ⁢(CSym⁢(Ω)⁢(M)) equals ML. If M is a normal subgroup of G, then G is permutationally isomorphic to a subgroup of Hol⁡(M). If M⋊Inn⁡(M)⩽G, then G contains ML because M⋊Inn⁡(M)=〈MR,ML〉.

A transitive action of G on Ω is said to be primitive if it preserves no non-trivial partition of Ω. The structure of a primitive permutation group is described by the O’Nan–Scott Theorem [8, Section 5], which splits the primitive permutation groups into eight types. We are concerned with only two of these types. If M⋊Inn⁡(M)⩽G⩽M⋊Aut⁡(M) with M≅T for some non-Abelian finite simple group T, then G, being contained in the holomorph of a simple group, is said to have type HS. If instead M is isomorphic to a compound group Tk, k⩾2, then G has type HC. In this case, G induces a subgroup of Aut⁢(M)≅Aut⁢(T)≀Sk which acts transitively on the set of k simple direct factors of M≅Tk. In either case, G contains MR and ML, as explained above.

If we write 𝒮=(𝒫,ℒ,I) for a partial linear space, then we mean that 𝒫 is the point set, ℒ is the line set, and I is the incidence relation. An incident point–line pair is called a flag. A collineation of 𝒮 is a permutation of 𝒫, together with a permutation of ℒ, such that incidence is preserved. If 𝒮 admits a group of collineations M that acts regularly on 𝒫, then we identify 𝒫 with M acting on itself by right multiplication (as above). A line ℓ is then identified with the subset of M comprising all of the points incident with ℓ, and hence P⁢I⁢ℓ if and only if P∈ℓ. Moreover, the stabiliser Mℓ is the set of all elements of M that fix ℓ setwise by right multiplication.

Lemma 2.1

Let S=(P,L,I) be a partial linear space with no triangles, and let G be a group of collineations of S with a normal subgroup M that acts regularly on P. Let ℓ be a line incident with the identity 1∈M=P, and suppose that its stabiliser Mℓ is non-trivial. Then:

  1. ℓ is a union of left Mℓ-cosets, including the trivial coset,

  2. if M⋊Inn⁡(M)⩽G, then Mℓ=ℓ.

Proof.

(i) Let g∈Mℓ. Since 1⁢I⁢ℓ, namely 1∈ℓ, it follows that g=1g⁢I⁢ℓg=ℓ, namely g∈ℓ. Therefore, Mℓ⊆ℓ. Now, if h∉Mℓ\{1} is incident with ℓ, then every non-trivial element of Mℓ must map h to another point incident with ℓ, and hence the whole coset h⁢Mℓ is contained in ℓ.

(ii) By (i), we have Mℓ⊆ℓ, so it remains to show the reverse inclusion. Let m∈ℓ∖{1}. Since Mℓ is non-trivial, there exists a non-trivial element h∈Mℓ. Since M⋊Inn⁡(M)⩽G, left multiplication by h-1 is a collineation of 𝒮. Since 1 and m are both incident with ℓ, it follows that h-1 and h-1⁢m are collinear. On the other hand, h-1∈Mℓ⊆ℓ by (i), so h-1⁢m is collinear with m because right multiplication by m is a collineation. That is, h-1⁢m is collinear with two points h-1,m that are incident with ℓ, and so h-1⁢m is itself incident with ℓ because 𝒮 contains no triangles. Therefore, m maps two points 1,h-1 incident with ℓ to two points m,h-1⁢m incident with ℓ, and so m∈Mℓ. ∎

Theorem 2.2

Let S=(P,L,I) be a partial linear space with no triangles. Let G be a group of collineations of S that acts transitively on L, and suppose that G has a normal subgroup M that acts regularly on P and satisfies M⋊Inn⁡(M)⩽G⩽M⋊Aut⁡(M). If the action of M on L is not semiregular, then the lines ℓ1,…,ℓt+1 incident with 1 are a G1-conjugacy class of subgroups of M, and G acts transitively on the flags of S.

Proof.

Since M acts transitively on 𝒫, we have G=M⁢G1=G1⁢M. By assumption, G⩽Hol⁡(M) and so G1⩽Aut⁡(M). By Lemma 2.1 (ii), the lines ℓ1,…,ℓt+1 can be identified with subgroups of M. Each g∈G1, acting naturally as an element of Aut⁡(M), fixes 1 and hence maps ℓ1 to ℓ1g=ℓi for some i∈{1,…,t+1}. Conversely, consider the map φ:G→Aut⁢(M) defined by φ⁢(g)=ιg. The restriction of φ to G1 is the identity. Moreover, ker⁡(φ)=CG⁢(M), and hence θ⁢(ker⁡(φ))=ML, where θ is the permutational isomorphism defined above. In particular, ker⁡(φ) acts transitively (indeed, regularly) on 𝒫. Hence, we have ker⁡(φ)⁡G1=G, so Im⁡(φ)=φ⁢(G1)=G1. Now consider a line ℓi for some i>1. By line-transitivity, we have ℓi=ℓ1g for some g∈G. On the other hand, since G=ker⁡(φ)⁡G1, we have g=z⁢g1 for some z∈ker⁡(φ) and g1∈G1, so ℓ1g=ℓ1g1. Therefore, ℓ1,…,ℓt+1 are precisely the subgroups of the form ℓ1g with g∈G1. Since the lines ℓi and ℓj intersect precisely in the point 1 for i≠j, the t+1 subgroups ℓ1,…,ℓt+1 are distinct, and they form a single G1-conjugacy class of subgroups of M. In particular, G1 acts transitively on {ℓ1,…,ℓt+1}, so G acts transitively on the flags of 𝒮. ∎

Let us draw a corollary in the case where 𝒮 is a thick generalised quadrangle. In this case, 𝒮 has (s+1)⁢(s⁢t+1) points and (t+1)⁢(s⁢t+1) lines, where (s,t) is the order of 𝒮.

Corollary 2.3

If the partial linear space in Theorem 2.2 is a thick generalised quadrangle of order (s,t), then s+1 divides t-1.

Proof.

Begin by observing that Inn⁡(M) acts on {ℓ1,…,ℓt+1}. That is, for each g∈M, we have g-1⁢ℓ1⁢g=ℓi for some i∈{1,…,t+1}. Suppose first that Inn⁡(M) is intransitive on {ℓ1,…,ℓt+1}. Then, without loss of generality, ℓ2 is in a different Inn⁡(M)-orbit to ℓ1, and so, for every g∈M, we have g-1⁢ℓ1⁢g=ℓi for some i≠2. Hence, every double coset ℓ1⁢g⁢ℓ2, where g∈M, has size |ℓ1⁢g⁢ℓ2|=|g-1⁢ℓ1⁢g⁢ℓ2|=|ℓi⁢ℓ2|=(s+1)2. Here the final equality holds because |ℓi∩ℓ2|=1 (because distinct concurrent lines intersect in a unique point, in this case the point 1). Since the double cosets of ℓ1 and ℓ2 partition M, it follows that (s+1)2 divides |M|=|𝒫|=(s+1)⁢(s⁢t+1). Therefore, s+1 divides s⁢t+1=(s+1)⁢t-(t-1), and hence s+1 divides t-1, as claimed.

Now suppose, towards a contradiction, that the group Inn⁡(M) is transitive on {ℓ1,…,ℓt+1}. Consider two lines incident with 1, say ℓ1,ℓ2. Then a double coset D=ℓ1⁢g⁢ℓ2, where g∈M, has size (s+1)2 or s+1 according as g-1⁢ℓ1⁢g≠ℓ2 or g-1⁢ℓ1⁢g=ℓ2. Let us say that D is small in the latter case. There are exactly |M|/(t+1) elements g∈M for which g-1⁢ℓ1⁢g=ℓ2, that is, for which D=ℓ1⁢g⁢ℓ2 is small. Moreover, each such D has s+1 representatives h∈M, because ℓ1⁢h⁢ℓ2=D if and only if h∈D, and |D|=s+1. Hence, there are exactly |M|/((s+1)⁢(t+1)) small double cosets of the form ℓ1⁢g⁢ℓ2. Therefore, (s+1)⁢(t+1) divides |M|=|𝒫|=(s+1)⁢(s⁢t+1), and so t+1 divides s⁢t+1=(t+1)⁢s-(s-1) and hence s-1. In particular, we have s⩾t+2>t, and so [7, Result 2.2.2 (i)] implies that 𝒮 cannot contain a subquadrangle of order (s,1). For a contradiction, we now construct such a subquadrangle.

Since ℓ1 is a subgroup of M and right multiplication by any element of M is a collineation of 𝒮, we have in particular that every right coset ℓ1⁢g2 of ℓ1 with g2∈ℓ2 is a line of 𝒮. Similarly, since left multiplications are collineations, every left coset g1⁢ℓ2 of ℓ2 with g1∈ℓ1 is a line of 𝒮. Therefore,

ℒ′={g1⁢ℓ2∣g1∈ℓ1}∪{ℓ1⁢g2∣g2∈ℓ2}

is a subset of ℒ. Now, consider also the subset 𝒫′=ℓ1⁢ℓ2 of 𝒫=M, and let I′ be the restriction of I to (𝒫′×ℒ′)∪(ℒ′×𝒫′). We claim that 𝒮′=(𝒫′,ℒ′,I′) is a subquadrangle of 𝒮 of order (s,1). Let us first check that 𝒮′ satisfies the generalised quadrangle axiom. Let ℓ∈ℒ′ and take P∈𝒫′ not incident with ℓ. Then, since 𝒮 satisfies the generalised quadrangle axiom, there is a unique point Q∈𝒫 incident with ℓ and collinear with P. Since ℓ⊂𝒫′, we have Q∈𝒫′, and so 𝒮′ also satisfies the generalised quadrangle axiom. It remains to check that 𝒮′ has order (s,1). Now, every line in ℒ′ is incident with s+1 points in 𝒫′, being a coset of either ℓ1 or ℓ2, so it remains to show that every point in 𝒫′ is incident with exactly two lines in ℒ′. Given P=g1⁢g2∈𝒫′, where g1∈ℓ1, g2∈ℓ2, each line ℓ∈ℒ′ incident with P is either of the form h1⁢ℓ2 for some h1∈ℓ1 or ℓ1⁢h2 for some h2∈ℓ2, and since P∈ℓ, we must have h1=g1 or h2=g2, respectively. Therefore, P is incident with exactly two lines in ℒ′, namely g1⁢ℓ2 and ℓ1⁢g2. ∎

We also check that, in the case of a thick generalised quadrangle, the assumption that M is not semiregular on ℒ is satisfied when |M| is even.

Lemma 2.4

Let Q=(P,L,I) be a thick generalised quadrangle of order (s,t). Let G be a group of collineations of Q that acts transitively on L, and suppose that G has a normal subgroup M that acts regularly on P. If M has even order, then M does not act semiregularly on L.

Proof.

If Mℓ is trivial for ℓ∈ℒ, then |ℓM|=|M|=|𝒫|=(s+1)⁢(s⁢t+1) divides |ℒ|=(t+1)⁢(s⁢t+1), and hence s+1 divides t+1, so [2, Lemma 3.2] implies that gcd⁡(s,t)>1. However, |M| is even, so M contains an element of order 2, and because gcd⁡(s,t)>1, it follows from [2, Lemma 3.4] that every such element must fix some line, contradicting the assumption that Mℓ is trivial. ∎

3 Proof of Theorem 1.1: HS type

Suppose that 𝒬=(𝒫,ℒ,I) is a thick generalised quadrangle with a collineation group G that acts transitively on ℒ and primitively of O’Nan–Scott type HS on 𝒫. Then

T⋊Inn⁡(T)⩽G⩽T⋊Aut⁡(T)

for some non-Abelian finite simple group T, with T acting regularly on 𝒫. Since |T| is even by the Feit–Thompson Theorem [5], Lemma 2.4 tells us that 𝒬 satisfies the hypotheses of Theorem 2.2 and Corollary 2.3. In particular, s+1 divides t-1 (by Corollary 2.3), and we write

(3.1)t′:=t-1s+1.

Since T acts regularly on 𝒫, we have |T|=|𝒫|=(s+1)⁢(s⁢t+1). By Higman’s inequality, t⩽s2, and hence t′⩽s-1. Therefore,

|T|=(s+1)2⁢(s⁢t′+1) for some ⁢1⩽t′⩽s-1.

By Theorem 2.2, G1⩽Aut⁡(T) acts transitively on the t+1 lines incident with 1, and hence t+1 divides |Aut(T)|=|T|⋅|Out(T)|. Therefore, |Out(T)| is divisible by (t+1)/gcd⁡(t+1,|T|), so

t+1⩽gcd(t+1,|T|)|Out(T)|.

Since |T|=(s+1)⁢(s⁢t+1) is even, s must be odd; and since s+1 divides t-1, we have gcd⁡(t+1,s+1)=2. Moreover, s⁢t′+1=t-t′, so

gcd⁡(t+1,s⁢t′+1)=gcd⁡(t+1,t-t′)=gcd⁡(t+1,t′+1),

and in particular gcd⁡(t+1,|T|)⩽22⁢(t′+1). Therefore,

t+1⩽4(t′+1)|Out(T)|.

Together with (3.1), this implies t′(s+1)+2⩽4(t′+1)|Out(T)|, and because t′⩾1, it follows that

s⩽8|Out(T)|-3.

Since |T|⩽(s+1)⁢(s3+1) (by Higman’s inequality), we have

|T|⩽(8|Out(T)|-2)((8|Out(T)|-3)3+1).

The following lemma therefore completes the proof of Theorem 1.1 in the HS case.

Lemma 3.1

There is no finite non-Abelian simple group T satisfying

  1. |T|=(s+1)2⁢(s⁢t′+1), where 1⩽t′⩽s-1,

  2. 2⩽s⩽8|Out(T)|-3,

  3. |T|⩽(8|Out(T)|-2)((8|Out(T)|-3)3+1).

Proof.

Since (8⁢x-2)⁢((8⁢x-3)3+1)≤(8⁢x)4 for real x⩾1, condition (c) implies that

(3.2)|T|⩽212|Out(T)|4.

We use (3.2) instead of (c) to rule out certain possibilities for T.

Case 1: T≅Altn or a sporadic simple group. If T≅Alt6, then |Out(T)|=4 and there is no solution to (a) subject to (b). If T is an alternating group other than Alt6, or a sporadic simple group, then |Out(T)|⩽2, and so (c) implies that |T|⩽(13+1)⁢(133+1)=30,772. This rules out everything except T≅Alt5, Alt7 and M11, and for these cases one checks that there is no solution to (a) subject to (b).

Case 2: T≅A1⁢(q). Suppose that T≅A1⁢(q), and write q=pf with p prime and f⩾1. Then |T|=q⁢(q2-1)/gcd⁡(2,q-1), and |Out(T)|=gcd(2,q-1)f.

Suppose first that q is even, namely that p=2. Then gcd⁡(2,q-1)=1, and (c) implies that

2f⁢(22⁢f-1)⩽(8⁢f-2)⁢((8⁢f-3)3+1),

which holds only if f⩽7. If f=1, then T is not simple; and if f=2, then T≅Alt5, which we have already ruled out. For 3⩽f⩽7, there is no solution to (a) subject to (b).

Now suppose that q=pf is odd. Then gcd⁡(2,q-1)=2, and hence we have |Out(T)|=2f, so (c) reads

pf⁢(p2⁢f-1)⩽2⁢(16⁢f-2)⁢((16⁢f-3)3+1).

If f⩾6, then this inequality fails for all p⩾3. The inequality holds if and only if

q=pf∈{3,5,7,32,11,13,17,19,23,52,33,29,31,37,72,34,53,35}.

If q=3, then T is not simple; if q=5, then T≅Alt5, which we have ruled out; if q=7, then T≅A2⁢(2), which is ruled out in Case 3 below; and if q=9, then T≅Alt6, which we have ruled out. For the remaining values of q, there is no solution to (a) subject to (b).

Case 3: T≅An⁢(q), n⩾2. Suppose that T≅An⁢(q), with n⩾2 and q=pf. Then

|T|=qn⁢(n+1)/2gcd⁡(n+1,q-1)⁢∏i=1n(qi+1-1),

and |Out(T)|=2gcd(n+1,q-1)f.

First suppose that n⩾3. Noting that f=logp⁡(q)=ln⁡(q)/ln⁡(p)⩽ln⁡(q)/ln⁡(2) and gcd⁡(n+1,q-1)⩽q-1, and applying (3.2), we find

qn⁢(n+1)/2⁢∏i=1n(qi+1-1)⩽216ln4⁡(2)⁢(q-1)5⁢ln4⁡(q).

This inequality fails for all q⩾2 if n=4, and therefore fails for all q⩾2 for every n⩾4 (because the left-hand side is increasing in n while the right-hand side does not depend on n). It fails for n=3 unless q∈{2,3}, but A3⁢(2)≅Alt8 has already been ruled out, and (c) rules out A3⁢(3) because

|A3⁢(3)|=6,065,280>30⁢(293+1)=731,700.

Finally, suppose that n=2. Noting that gcd⁡(3,q-1)⩽3 and f⩽ln⁡(q)/ln⁡(2), (3.2) gives

q3⁢(q2-1)⁢(q3-1)⩽216⁢35ln4⁡(2)⁢ln4⁡(q).

This implies that q⩽15. For q∈{5,8,9,11,13}, the sharper inequality (c) fails. For q∈{2,3,4,7}, there are no solutions to (a) subject to (b).

Case 4: T≅An2⁢(q2). Suppose that T≅An2⁢(q2), where now q2=pf for some prime p and f≥1. We have n⩾2,

|T|=qn⁢(n+1)/2gcd⁡(n+1,q+1)⁢∏i=1n(qi+1-(-1)i+1),

and |Out(T)|=gcd(n+1,q+1)f.

First suppose that n⩾4. Noting that

f=logp⁡(q2)=ln⁡(q2)ln⁡(p)⩽2⁢ln⁡(q)ln⁡(2),

and that gcd⁡(n+1,q+1)⩽q+1, (3.2) gives

qn⁢(n+1)/2⁢∏i=1n(qi+1-(-1)i+1)⩽216ln4⁡(2)⁢(q+1)5⁢ln4⁡(q).

This inequality fails for all q⩾2 for n=4, and hence fails for all q⩾2 for every n⩾4.

Now suppose that n=3. Then we can replace the (q+1)5 on the right-hand side above by 45=210, because gcd⁡(n+1,q+1)=gcd⁡(4,q+1)⩽4. This yields

q6⁢(q2-1)⁢(q3+1)⁢(q4-1)⩽226ln4⁡(2)⁢ln4⁡(q),

which implies that q⩽4. If q∈{2,3}, then there are no solutions to (a) subject to (b). If q=4, then (c) fails.

Finally, suppose that n=2. Then gcd⁡(n+1,q+1)⩽3, and hence

q3⁢(q2-1)⁢(q3+1)⩽216⁢35ln4⁡(2)⁢ln4⁡(q),

which implies that q⩽15. If q=2, then the group T≅A22⁢(22) is not simple. If q∈{3,4,5,8}, then there are no solutions to (a) subject to (b). If q∈{7,9,11,13}, then (c) fails.

Case 5: Remaining possibilities for T. We now rule out the remaining possibilities for the finite simple group T.

(i) T≅Bn⁢(q) or Cn⁢(q). First suppose that T≅Cn⁢(q), and write q=pf with p prime and f⩾1. We have n⩾3, |T|=qn2/gcd⁡(2,q-1)⋅∏i=1n(q2⁢i-1), and |Out(T)|=gcd(2,q-1)f. Noting that f⩽ln⁡(q)/ln⁡(2) and that gcd⁡(2,q-1) is at most 2, (3.2) implies that

qn2⁢∏i=1n(q2⁢i-1)⩽217ln4⁡(2)⁢ln4⁡(q).

However, this inequality fails for all q⩾2 if n=3, and hence fails for all q⩾2 for every n⩾3.

Now suppose that T≅Bn⁢(q), writing q=pf as before. In this case we have n⩾2, and again |T|=qn2/gcd⁡(2,q-1)⋅∏i=1n(q2⁢i-1). If n⩾3 and q is even, then Bn⁢(q)≅Cn⁢(q). If n⩾3 and q is odd, then |Out(T)| is the same as for Cn⁢(q). We may therefore assume that n=2. First suppose that q=2f. Then |Out(T)|=2gcd(2,q-1)f=2f, so (3.2) implies that

(3.3)24⁢f⁢(22⁢f-1)⁢(24⁢f-1)⩽216⁢f4,

and hence f∈{1,2}. For f=1, B2⁢(2) is not simple but its derived subgroup B2⁢(2)′≅Alt6 is simple and has already been ruled out. For f=2, (c) fails. Now suppose that q is odd. Then |Out(T)|=gcd(2,q-1)f=2f and f⩽ln⁡(q)/ln⁡(3), so (3.2) implies that

q4⁢(q2-1)⁢(q4-1)⩽217ln4⁡(3)⁢ln4⁡(q),

and hence q=3. However, we have B2⁢(3)≅A32⁢(22), which has been dealt with in Case 4.

(ii) T≅Dn⁢(q). Suppose that T≅Dn⁢(q), writing q=pf again. We have n⩾4, |T|=qn⁢(n-1)⁢(qn-1)/gcd⁡(4,qn-1)⋅∏i=1n-1(q2⁢i-1), and

|Out(T)|={6gcd(2,q-1)2fif ⁢n=4,2gcd(2,q-1)2fif ⁢n<4⁢ and ⁢n⁢ is even,2⁢gcd⁡(4,qn-1)⁢fif ⁢n<4⁢ and ⁢n⁢ is odd.

If q is odd, then gcd⁡(4,qn-1)⩽4, |Out(T)|⩽24f, and f⩽ln⁡(q)/ln⁡(3), so (3.2) implies that

qn⁢(n-1)⁢(qn-1)⁢∏i=1n-1(q2⁢i-1)⩽226⁢34ln4⁡(3)⁢ln4⁡(q),

which fails for all q⩾3 if n=4, and hence fails for all q⩾3 for every n⩾4. If q is even, then gcd⁡(4,qn-1)=1, |Out(T)|⩽6f and f=ln⁡(q)/ln⁡(2), so (3.2) implies that

qn⁢(n-1)⁢(qn-1)⁢∏i=1n-1(q2⁢i-1)⩽216⁢34ln4⁡(2)⁢ln4⁡(q),

which fails for all q⩾2 if n=4, and hence fails for all q⩾2 for every n⩾4.

(iii) T≅E6⁢(q),E7⁢(q),E8⁢(q) or F4⁢(q). Suppose that T is one of E6⁢(q), E7⁢(q), E8⁢(q) or F4⁢(q), and write q=pf again. Observe that |Ei⁢(q)|⩾|F4⁢(q)| for every i∈{6,7,8}, for all q⩾2. Hence

|T|⩾|F4⁢(q)|=q24⁢(q12-1)⁢(q8-1)⁢(q6-1)⁢(q2-1)⩾q5224.

Since |Out(T)|⩽2gcd(3,q-1)f⩽6ln(q)/ln(2), (3.2) implies the following inequality, which fails for all q⩾2:

q52⩽220⁢34ln4⁡(2)⁢ln4⁡(q).

(iv) T≅G2⁢(q). Suppose that T≅G2⁢(q), with q=pf. Then

|T|=q6⁢(q6-1)⁢(q2-1).

If p=3, then |Out(T)|=2f, so (c) implies 36⁢f⁢(36⁢f-1)⁢(32⁢f-1)⩽216⁢f4, which fails for all f⩾1. If p≠3, then |Out(T)|=f⩽ln(q)/ln(2), and (3.2) implies the following inequality, which fails for all q⩾2:

q6⁢(q6-1)⁢(q2-1)⩽212ln4⁡(2)⁢ln4⁡(q).

Note that G2⁢(2) is not simple, but G2⁢(2)′≅A22⁢(32) is simple and has already been ruled out.

(v) T≅Dn2⁢(q). Suppose that T≅Dn2⁢(q2), now writing q2=pf. Then n⩾4,

|T|=qn⁢(n-1)⁢(qn+1)gcd⁡(4,qn+1)⁢∏i=1n-1(q2⁢i-1),

and |Out(T)|=gcd(4,qn+1)f. Since f⩽2⁢ln⁡(q)/ln⁡(2) and gcd⁡(4,qn+1)⩽4, (3.2) implies that

qn⁢(n-1)⁢(qn+1)⁢∏i=1n-1(q2⁢i-1)⩽226ln4⁡(2)⁢ln4⁡(q).

This fails for all q⩾2 if n=4, and hence fails for all q⩾2 for every n⩾4.

(vi) T≅E62⁢(q2). Suppose that T≅E62⁢(q2), with q2=pf. Then

|T|=1gcd⁡(3,q+1)⁢q36⁢(q12-1)⁢(q9+1)⁢(q8-1)⁢(q6-1)⁢(q5+1)⁢(q2-1),

and

|Out(T)|=gcd(3,q+1)f.

Noting that f⩽2⁢ln⁡(q)/ln⁡(2) and gcd⁡(3,q+1)⩽3, (3.2) implies the following inequality, which fails for all q⩾2:

q36⁢(q12-1)⁢(q9+1)⁢(q8-1)⁢(q6-1)⁢(q5+1)⁢(q2-1)⩽35⁢216ln4⁡(2)⁢ln4⁡(q).

(vii) T≅D43⁢(q3). Suppose that T≅D43⁢(q2), where now q3=pf. Then

|T|=q12⁢(q8+q4+1)⁢(q6-1)⁢(q2-1),

and |Out(T)|=f=3ln(q)/ln(p)⩽3ln(q)/ln(2), so (3.2) implies the following inequality, which fails for all q⩾2:

q12⁢(q8+q4+1)⁢(q6-1)⁢(q2-1)⩽34⁢212ln4⁡(2)⁢ln4⁡(q).

(viii) T≅B22⁢(q), G22⁢(q), or F42⁢(q). Finally, suppose that T is as in one of the lines of Table 1. Suppose first that n⩾1. Then |Out(T)|=2n+1 in each case, and (3.2) therefore implies that |T|⩽212⁢(2⁢n+1)4. This inequality holds only in the case T≅B22⁢(22⁢n+1) with n=1, but |2B2(23)|=29,120 cannot be written in the form (a) subject to (b). For n=0, we have that B22⁢(q) is not simple; G22⁢(3) is not simple, but G22⁢(3)′≅A1⁢(8) has been ruled out in Case 2 above; and F42⁢(2) is not simple, but F42⁢(2)′ is simple of order 17,971,200 and has outer automorphism group of order 2, so (3.2) fails.

This completes the proof of Lemma 3.1. ∎

Table 1

Orders of the Suzuki and Ree simple groups.

T|T|q
B22⁢(q)q2⁢(q2+1)⁢(q-1)22⁢n+1
G22⁢(q)q3⁢(q3+1)⁢(q-1)32⁢n+1
F42⁢(q)q12⁢(q6+1)⁢(q4-1)⁢(q3+1)⁢(q-1)22⁢n+1

4 Proof of Theorem 1.1: HC type

Suppose that 𝒬=(𝒫,ℒ,I) is a thick generalised quadrangle with a collineation group G that acts transitively on ℒ and primitively of O’Nan–Scott type HC on 𝒫. Then

M⋊Inn⁡(M)⩽G⩽M⋊Aut⁡(M),

where M=T1×⋯×Tk, with k⩾2 and T1≅⋯≅Tk≅T for some non-Abelian finite simple group T. Moreover, M acts regularly on 𝒫, and G induces a subgroup of Aut⁢(T)≀Sk which acts transitively on the set {T1,…,Tk} (see [8, Section 5]). Since |M|=|T|k is even by the Feit–Thompson Theorem [5], Lemma 2.4 tells us that 𝒬 satisfies the hypotheses of Theorem 2.2 and Corollary 2.3. In particular, s+1 divides t-1 (by Corollary 2.3), and we define t′ as in (3.1).

We first rule out the case k⩾3, and then deal with the case k=2 separately.

4.1 The case k⩾3

Suppose, towards a contradiction, that k⩾3. Denote by ℓ1,…,ℓt+1 the lines incident with the identity 1∈M. By Lemma 2.1 (ii), we may identify ℓi with the subgroup of M comprising all points incident with ℓi. Let us write ℓ:=ℓ1 for brevity.

Claim 4.1

The group M cannot be decomposed in the form M=A×B with ℓ∩A≠{1} and ℓ∩B≠{1}.

Proof.

Suppose, towards a contradiction, that M=A×B with ℓ∩A≠{1} and ℓ∩B≠{1}. We may assume, without loss of generality, that (i) A contains T1, and (ii) ℓ∩A contains an element x=(x1,…,xk) that projects non-trivially onto each simple direct factor of A (if not, then change the decomposition of M to A′×B′ with A′⩽A and B′⩾B). Take also y∈ℓ∩B with y≠1. For every a∈Inn⁢(A)⩽Inn⁢(M), we have ya=y and hence ℓa=ℓ, because a also fixes the point 1∈ℓ. In particular, ℓ is fixed by every element of Inn⁡(T1), regarded as a subgroup of Inn⁡(A). Therefore, (z,x2,…,xk)∈ℓ for all z∈x1T1, and hence ℓ contains the group ℓ0:=〈(z,x2,…,xk):z∈x1T1〉. Let π1 denote the projection onto T1. Then π1(ℓ0)=〈z:z∈x1T1〉=T1, and hence π1⁢(ℓ)=T1. Also, taking z≠x1, we see that ℓ∩T1 contains (z,x2,…,xk)-1⁢x=(z-1⁢x1,1,…,1)≠1. That is, ℓ∩T1 is non-trivial, and it is normal in the simple group π1⁢(ℓ)=T1, so ℓ∩T1=T1 and hence T1⩽ℓ. Now, G1 acts transitively on both {T1,…,Tk} (because G is transitive on {T1,…,Tk} and G=M⁢G1) and {ℓ1,…,ℓt+1} (because G is flag-transitive, by Theorem 2.2). Therefore, t+1 divides k, and, without loss of generality, ℓ=ℓ1 contains TU1:=T1×⋯×Tk/(t+1), ℓ2 contains TU2:=Tk/(t+1)+1×⋯×T2⁢k/(t+1), and so on.

Subclaim

ℓ=TU1.

Proof of sub-claim.

It remains to show that TU1 contains ℓ. Suppose, towards a contradiction, that there exists w∈ℓ∖TU1. Then there exists i>k/(t+1) such that the ith component wi of w is non-trivial, and so there exists σ∈Inn⁢(Ti) such that wiσ≠wi. Regarding σ as an element of Inn⁡(M)⩽G1, we see that σ fixes ℓ, because it centralises T1⩽ℓ. Hence, wσ∈ℓ, and so ℓ contains the element w-1⁢wσ∈ℓ∩Ti∖{1}. However, Ti⩽TUj⩽ℓj for some j≠1, and hence ℓ intersects ℓj in more than one point, a contradiction, proving the sub-claim. ∎

By the sub-claim, s+1=|T|u, where u=k/(t+1). Since |T|(t+1)⁢u=|M|, we have (s+1)t+1=(s+1)2⁢(s⁢t′+1), where t′:=(t-1)/(s+1)⩽s-1 as before. Since s⁢t′+1⩽s⁢(s-1)+1<(s+1)2, this implies that (s+1)t-1<(s+1)2, so t=2, and hence s+1∣t-1=1, a contradiction. ∎

Claim 4.2

ℓ is isomorphic to a subgroup of T.

Proof.

Let x∈ℓ∖{1} have minimal support U. Suppose, without loss of generality, that x1:=π1⁢(x)≠1. Suppose further, towards a contradiction, that there exists y∈ℓ∖{1} with π1⁢(y)=1. Then every a∈Inn⁡(T1) fixes y and hence fixes ℓ, so ℓ contains xa and therefore contains xa⁢x-1∈T1∩ℓ. Taking a not in CT⁢(x1) makes xa⁢x-1 non-trivial, and the minimality of the support U of x implies that U={1}, so x∈T1. However, the existence of y now contradicts Claim 4.1, because taking A=T1 and B=T2×⋯×Tk gives x∈ℓ∩A and y∈ℓ∩B. Hence, if x has minimal support U containing 1, then every non-trivial element of ℓ must project non-trivially onto T1. Therefore, ℓ is isomorphic (under projection) to a subgroup of T1. ∎

We now use Claim 4.2 to derive a contradiction to the assumption that k⩾3. By Claim 4.2, s+1=|ℓ| divides |T|, so in particular s+1⩽|T|. Writing

|M|=(s+1)2⁢(s⁢t′+1)

with t′:=(t-1)/(s+1)⩽s-1 as before, we have

(s+1)2>s⁢(s-1)+1⩾s⁢t′+1=|M|(s+1)2⩾|M||T|2=|T|k-2⩾(s+1)k-2,

and hence 2>k-2, namely k⩽3.

Now suppose, towards a contradiction, that k=3. Write |T|=n⁢(s+1). Then s⁢t+1=|M|/(s+1)=|T|3/(s+1)=n3⁢(s+1)2, and hence n3≡1⁢(mod⁡s). On the other hand, we have s3+1⩾s⁢t+1=n3⁢(s+1)2>n3⁢s2+1, so n3<s. Therefore, n=1, so |T|=s+1 and t=s+2. Together with Claim 4.2, this implies that ℓ is isomorphic to T. Consider first the case where ℓ is a diagonal subgroup {(t,ta,tb):t∈T}⩽M for some automorphisms a,b∈Aut⁢(T). As (c,d)∈Inn⁡(T2)×Inn⁡(T3)⩽G1 runs over all possibilities, we obtain |T|2 distinct images ℓ(c,d)={(t,ta⁢c,tb⁢d):t∈T} of ℓ. Indeed, if ℓ=ℓ(c,d), then ta=ta⁢c for all t∈T, or equivalently, u=uc for all u∈T; that is, c is the identity automorphism of T (and similarly, d is the identity). Hence, s+3=t+1≥(s+1)2, a contradiction. Now consider the case where ℓ is a diagonal subgroup {(t,ta,1):t∈T}⩽T1×T2 for some a∈Aut⁢(T). Then 3 divides t+1 because G1 is transitive on the Ti, and we have exactly (t+1)/3 lines incident with 1 that are diagonal subgroups of T1×T2. As c∈Inn⁢(T2)⩽G1 runs over all possibilities, we obtain |T| distinct images ℓc={(t,ta⁢c,1):t∈T} of ℓ. Hence, (s+3)/3=(t+1)/3⩾s+1, a contradiction. This leaves only the possibility that ℓ⩽T1, and hence ℓ=T1 because |ℓ|=s+1=|T1|. This implies that t+1=3, and hence s=0 because s+1 divides t-1, a contradiction.

4.2 The case k=2

Here we argue as in the case where the primitive action of G on 𝒫 has type HS. That is, we obtain an upper bound on |T| in terms of |Out(T)|, and consider the possibilities for T case by case using the Classification of Finite Simple Groups. We have M=T1×T2≅T2, and

|M|=(s+1)⁢(s⁢t+1)=(s+1)2⁢(s⁢t′+1),where ⁢1⩽t′⩽s-1.

Therefore,

|T|=(s+1)⁢(s⁢t′+1)1/2,where ⁢1⩽t′⩽s-1⁢ and ⁢s⁢t′+1⁢ is a square.

Writing y2=s⁢t′+1, this is equivalent to

|T|=(s+1)y,where 3⩽y2⩽s(s-1)+1 and s∣y2-1.

By Theorem 2.2, G1⩽Aut⁢(M)≅Aut⁢(T)≀S2 acts transitively on the lines incident with 1, and hence t+1 divides |Aut(M)|=2|T|2|Out(T)|2. Therefore, |Out(T)|2 is divisible by

t+1gcd⁡(t+1,2⁢|T|2)=t+1gcd⁡(t+1,2⁢(s+1)2⁢(s⁢t′+1)).

In particular, t+1⩽gcd(t+1,2|T|2)|Out(T)|2. We have

  1. gcd⁡(t+1,s+1)=2, so gcd⁡(t+1,2⁢(s+1)2)⩽8,

  2. gcd⁡(t+1,s⁢t′+1)=gcd⁡(t+1,t′+1).

Hence we have gcd⁡(t+1,2⁢|T|2)⩽8⁢(t′+1), and so t+1⩽8(t′+1)|Out(T)|2. Re-writing this as t′(s+1)+2⩽8(t′+1)|Out(T)|2, and noting that t′⩾1, we obtain

s⩽16|Out(T)|2-3.

Higman’s inequality then gives

|T|2=|M|⩽(16|Out(T)|2-2)((16|Out(T)|2-3)3+1).

The following lemma therefore rules out all but two possibilities for T.

Lemma 4.3

Let T be a finite non-Abelian simple group satisfying

  1. |T|=(s+1)⁢y, where 3⩽y2⩽s⁢(s-1)+1 and s∣y2-1,

  2. 2⩽s⩽16|Out(T)|2-3,

  3. |T|2⩽(16|Out(T)|2-2)((16|Out(T)|2-3)3+1).

Then one of the following holds:

  1. T≅Alt6, s=19, and y=18,

  2. T≅A2⁢(2), s=13, and y=12.

Proof.

The right-hand side of (c) is at most (16|Out(T)|2)4, so

(4.1)|T|⩽28|Out(T)|4.

Since (4.1) implies (3.2), any group T that was ruled out using (3.2) in the HS case (that is, in the proof of Lemma 3.1) is automatically ruled out here. To rule out the remaining possibilities for T, we use either (4.1) or (c), or check that (a) has no solution subject to (b). Note that (a) implies y⩽s<y2.

Case 1: T≅Altn or a sporadic simple group. If T is an alternating group other than Alt6, or a sporadic simple group, then |Out(T)|⩽2 and so (c) implies that |T|<3,752. Hence, T is one of Alt5, Alt6, or Alt7. If T≅Alt5, then by (a), we have (s+1)⁢y=60 and s∣y2-1, which is impossible. If T≅Alt7, then we again apply (a): (s+1)⁢y=2520, s∣y2-1, and y2⩽s⁢(s-1)+1, which is again impossible. Finally, we examine the case T≅Alt6, where |Out(T)|=4. Applying (a), we have s=19, y=18 as the only valid solution.

Case 2: T≅A1⁢(q). Suppose that T≅A1⁢(q), and write q=pf with p prime and f⩾1. Then |T|=q⁢(q2-1)/(2,q-1), and |Out(T)|=(2,q-1)f.

Suppose first that q is even, namely that p=2. Then gcd⁡(2,q-1)=1, and (c) implies that

22⁢f⁢(22⁢f-1)2⩽(16⁢f2-2)⁢((16⁢f2-3)3+1),

which holds only if f⩽7. If f=1, then T is not simple; and if f=2, then T≅Alt5, which we have already ruled out. For 3⩽f⩽7, there is no solution to (a) subject to (b).

Now suppose that q=pf is odd. Then gcd⁡(2,q-1)=2, and hence we have |Out(T)|=2f. By (c), we have

p2⁢f⁢(p2⁢f-1)2⩽8⁢(32⁢f2-1)⁢((64⁢f2-3)3+1),

which implies that either 11⩽p⩽19 and f=1; 5⩽p⩽7 and f⩽2; or p=3 and f⩽4. If q=3, then T is not simple; if q=5, then T≅Alt5, which we have ruled out; if q=7, then T≅A2⁢(2), which is ruled out in Case 3 below; and if q=9, then T≅Alt6, which we have already dealt with in Case 1. Hence, we only need to consider q∈{11,13,17,19,33,34,52,72}. For each of these values, there is no solution to (a) subject to (b).

Case 3: T≅An⁢(q), n⩾2. Since (4.1) implies (3.2), by comparing with the proof of Case 2 in Lemma 3.1, we see that we only need to check T≅A3⁢(3), and T≅A2⁢(q) for q⩽13. The former is ruled out by (4.1), because

|A3⁢(3)|=6,065,280>28⁢44=65,536.

For T≅A2⁢(q), (c) implies that

q6⁢(q2-1)2⁢(q3-1)2⩽9⁢(576ln2⁡(2)⁢ln2⁡(q)-2)⁢((576ln2⁡(2)⁢ln2⁡(q)-3)3+1).

Therefore, q⩽10. For q=2, there is a unique solution to (a) subject to (b), namely s=13, t′=11. For q∈{3,4,5,7,8,9}, there are no solutions to (a) subject to (b).

Case 4: T≅An2⁢(q2). Since (4.1) implies (3.2), we only need to check that T≅A32⁢(q2) for 2⩽q⩽4, and T≅A22⁢(q2) for q⩽13. If (n,q)=(3,3) or (3,4), then (4.1) fails; and for (n,q)=(3,2), there are no solutions to (a) subject to (b). For n=2, (c) gives

q6⁢(q2-1)2⁢(q3+1)2⩽9⁢(576ln2⁡(2)⁢ln2⁡(q)-2)⁢((576ln2⁡(2)⁢ln2⁡(q)-3)3+1),

and hence q⩽10. If q=2, then T≅A22⁢(q2) is not simple. If q∈{3,4,5,7,8,9}, then there are no solutions to (a) subject to (b).

Case 5: Remaining possibilities for T. We only need to check the groups from Case 5 of the proof of Lemma 3.1 that were not ruled out by (3.2) or by exceptional isomorphisms to groups that have already been handled. There are only two such cases. If T≅B2⁢(2f) with f=2, then, using (4.1) instead of (3.2), the 216 on the right-hand side of (3.3) becomes 212, and the resulting inequality 24⁢f⁢(22⁢f-1)⁢(24⁢f-1)⩽212⁢f4 fails when f=2. If T≅B22⁢(22⁢n+1) with n=1, then (4.1) fails (although (3.2) does not).

This completes the proof of Lemma 4.3. ∎

It remains to rule out cases (i) and (ii) from Lemma 4.3. Using y2=s⁢t′+1, we find that t=341 in case (i), and t=155 in case (ii). Both cases are then ruled out because the required divisibility condition t+1∣|Aut(M)|=2|T|2|Out(T)|2 fails. (Note that |Aut⁡(M)|=4,147,200 if T≅Alt6, and |Aut⁡(M)|=225,792 if T≅A2⁢(2).)


Communicated by Timothy C. Burness


Funding statement: The first author acknowledges the support of the Australian Research Council (ARC) Future Fellowship FT120100036. The second author acknowledges the support of the ARC Discovery Grant DP140100416. The research reported in the paper forms part of the ARC Discovery Grant DP140100416 of the third author.

Acknowledgements

We thank the referee for a careful reading of the paper, and in particular for bringing to our attention an error in a previous version of the proof of Corollary 2.3.

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Received: 2015-8-10
Revised: 2016-8-22
Published Online: 2016-9-14
Published in Print: 2017-3-1

© 2017 by De Gruyter

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