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

John Bamberg , 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 not incident with P, there is a unique line incident with P and concurrent with . 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,t2 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 xΩ under gG by xg. The orbit of xΩ under G is the set xG={xggG}, the subgroup Gx={gGxg=x} is the stabiliser of xΩ, and the Orbit–Stabiliser Theorem says that |G:Gx|=|xG|. The action of G is transitive if xG=Ω for some (and hence every) xΩ, and semiregular if Gx is trivial for all xΩ. 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 gG, define ρg, λg,ιgSym(Ω) by

ρg:xxg,λg:xg-1x,ιg:xg-1xg.

Set

GR={ρg:gG},GL={λg:gG},Inn(G)={ιg:gG}.

The holomorphHol(G) of G is the semidirect product GRAut(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:GH and a bijection β:ΩΔ such that β(ωg)=β(ω)θ(g) for all gG and ωΩ. If a group M acts regularly on Ω, then there is a permutational isomorphism θ:NSym(Ω)(M)Hol(M) with bijection β:ΩM, where β:αgg for some fixed αΩ, and θ:τβ-1τβ. 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 MInn(M)G, then G contains ML because MInn(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 MInn(M)GMAut(M) with MT 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, k2, 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 MTk. In either case, G contains MR and ML, as explained above.

If we write 𝒮=(𝒫,,I) for a partial linear space, then we mean that 𝒫 is the point set, 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 , 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 is then identified with the subset of M comprising all of the points incident with , and hence PI if and only if P. Moreover, the stabiliser M is the set of all elements of M that fix 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 be a line incident with the identity 1M=P, and suppose that its stabiliser M is non-trivial. Then:

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

2. if MInn(M)G, then M=.

## Proof.

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

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

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 MInn(M)GMAut(M). If the action of M on L is not semiregular, then the lines 1,,t+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=MG1=G1M. By assumption, GHol(M) and so G1Aut(M). By Lemma 2.1 (ii), the lines 1,,t+1 can be identified with subgroups of M. Each gG1, acting naturally as an element of Aut(M), fixes 1 and hence maps 1 to 1g=i for some i{1,,t+1}. Conversely, consider the map φ:GAut(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 i for some i>1. By line-transitivity, we have i=1g for some gG. On the other hand, since G=ker(φ)G1, we have g=zg1 for some zker(φ) and g1G1, so 1g=1g1. Therefore, 1,,t+1 are precisely the subgroups of the form 1g with gG1. Since the lines i and j intersect precisely in the point 1 for ij, the t+1 subgroups 1,,t+1 are distinct, and they form a single G1-conjugacy class of subgroups of M. In particular, G1 acts transitively on {1,,t+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)(st+1) points and (t+1)(st+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 {1,,t+1}. That is, for each gM, we have g-11g=i for some i{1,,t+1}. Suppose first that Inn(M) is intransitive on {1,,t+1}. Then, without loss of generality, 2 is in a different Inn(M)-orbit to 1, and so, for every gM, we have g-11g=i for some i2. Hence, every double coset 1g2, where gM, has size |1g2|=|g-11g2|=|i2|=(s+1)2. Here the final equality holds because |i2|=1 (because distinct concurrent lines intersect in a unique point, in this case the point 1). Since the double cosets of 1 and 2 partition M, it follows that (s+1)2 divides |M|=|𝒫|=(s+1)(st+1). Therefore, s+1 divides st+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 {1,,t+1}. Consider two lines incident with 1, say 1,2. Then a double coset D=1g2, where gM, has size (s+1)2 or s+1 according as g-11g2 or g-11g=2. Let us say that D is small in the latter case. There are exactly |M|/(t+1) elements gM for which g-11g=2, that is, for which D=1g2 is small. Moreover, each such D has s+1 representatives hM, because 1h2=D if and only if hD, and |D|=s+1. Hence, there are exactly |M|/((s+1)(t+1)) small double cosets of the form 1g2. Therefore, (s+1)(t+1) divides |M|=|𝒫|=(s+1)(st+1), and so t+1 divides st+1=(t+1)s-(s-1) and hence s-1. In particular, we have st+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 1 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 1g2 of 1 with g22 is a line of 𝒮. Similarly, since left multiplications are collineations, every left coset g12 of 2 with g11 is a line of 𝒮. Therefore,

={g12g11}{1g2g22}

is a subset of . Now, consider also the subset 𝒫=12 of 𝒫=M, and let I be the restriction of I to (𝒫×)(×𝒫). We claim that 𝒮=(𝒫,,I) is a subquadrangle of 𝒮 of order (s,1). Let us first check that 𝒮 satisfies the generalised quadrangle axiom. Let and take P𝒫 not incident with . Then, since 𝒮 satisfies the generalised quadrangle axiom, there is a unique point Q𝒫 incident with and collinear with P. Since 𝒫, we have Q𝒫, and so 𝒮 also satisfies the generalised quadrangle axiom. It remains to check that 𝒮 has order (s,1). Now, every line in is incident with s+1 points in 𝒫, being a coset of either 1 or 2, so it remains to show that every point in 𝒫 is incident with exactly two lines in . Given P=g1g2𝒫, where g11, g22, each line incident with P is either of the form h12 for some h11 or 1h2 for some h22, and since P, we must have h1=g1 or h2=g2, respectively. Therefore, P is incident with exactly two lines in , namely g12 and 1g2. ∎

We also check that, in the case of a thick generalised quadrangle, the assumption that M is not semiregular on 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 M is trivial for , then |M|=|M|=|𝒫|=(s+1)(st+1) divides ||=(t+1)(st+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 M is trivial. ∎

## 3 Proof of Theorem 1.1: HS type

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

TInn(T)GTAut(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)(st+1). By Higman’s inequality, ts2, and hence ts-1. Therefore,

|T|=(s+1)2(st+1)for some 1ts-1.

By Theorem 2.2, G1Aut(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+1gcd(t+1,|T|)|Out(T)|.

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

gcd(t+1,st+1)=gcd(t+1,t-t)=gcd(t+1,t+1),

and in particular gcd(t+1,|T|)22(t+1). Therefore,

t+14(t+1)|Out(T)|.

Together with (3.1), this implies t(s+1)+24(t+1)|Out(T)|, and because t1, it follows that

s8|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(st+1), where 1ts-1,

2. 2s8|Out(T)|-3,

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

## Proof.

Since (8x-2)((8x-3)3+1)(8x)4 for real x1, 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: TAltn or a sporadic simple group. If TAlt6, 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 TAlt5, Alt7 and M11, and for these cases one checks that there is no solution to (a) subject to (b).

Case 2: TA1(q). Suppose that TA1(q), and write q=pf with p prime and f1. 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(22f-1)(8f-2)((8f-3)3+1),

which holds only if f7. If f=1, then T is not simple; and if f=2, then TAlt5, which we have already ruled out. For 3f7, 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(p2f-1)2(16f-2)((16f-3)3+1).

If f6, then this inequality fails for all p3. 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 TAlt5, which we have ruled out; if q=7, then TA2(2), which is ruled out in Case 3 below; and if q=9, then TAlt6, which we have ruled out. For the remaining values of q, there is no solution to (a) subject to (b).

Case 3: TAn(q), n2. Suppose that TAn(q), with n2 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 n3. 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)/2i=1n(qi+1-1)216ln4(2)(q-1)5ln4(q).

This inequality fails for all q2 if n=4, and therefore fails for all q2 for every n4 (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 fln(q)/ln(2), (3.2) gives

q3(q2-1)(q3-1)21635ln4(2)ln4(q).

This implies that q15. 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: TAn2(q2). Suppose that TAn2(q2), where now q2=pf for some prime p and f1. We have n2,

|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 n4. Noting that

f=logp(q2)=ln(q2)ln(p)2ln(q)ln(2),

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

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

This inequality fails for all q2 for n=4, and hence fails for all q2 for every n4.

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 q4. 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)21635ln4(2)ln4(q),

which implies that q15. If q=2, then the group TA22(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) TBn(q) or Cn(q). First suppose that TCn(q), and write q=pf with p prime and f1. We have n3, |T|=qn2/gcd(2,q-1)i=1n(q2i-1), and |Out(T)|=gcd(2,q-1)f. Noting that fln(q)/ln(2) and that gcd(2,q-1) is at most 2, (3.2) implies that

qn2i=1n(q2i-1)217ln4(2)ln4(q).

However, this inequality fails for all q2 if n=3, and hence fails for all q2 for every n3.

Now suppose that TBn(q), writing q=pf as before. In this case we have n2, and again |T|=qn2/gcd(2,q-1)i=1n(q2i-1). If n3 and q is even, then Bn(q)Cn(q). If n3 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)24f(22f-1)(24f-1)216f4,

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 fln(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) TDn(q). Suppose that TDn(q), writing q=pf again. We have n4, |T|=qn(n-1)(qn-1)/gcd(4,qn-1)i=1n-1(q2i-1), and

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

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

qn(n-1)(qn-1)i=1n-1(q2i-1)22634ln4(3)ln4(q),

which fails for all q3 if n=4, and hence fails for all q3 for every n4. 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(q2i-1)21634ln4(2)ln4(q),

which fails for all q2 if n=4, and hence fails for all q2 for every n4.

(iii) TE6(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 q2. Hence

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

Since |Out(T)|2gcd(3,q-1)f6ln(q)/ln(2), (3.2) implies the following inequality, which fails for all q2:

q5222034ln4(2)ln4(q).

(iv) TG2(q). Suppose that TG2(q), with q=pf. Then

|T|=q6(q6-1)(q2-1).

If p=3, then |Out(T)|=2f, so (c) implies 36f(36f-1)(32f-1)216f4, which fails for all f1. If p3, then |Out(T)|=fln(q)/ln(2), and (3.2) implies the following inequality, which fails for all q2:

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) TDn2(q). Suppose that TDn2(q2), now writing q2=pf. Then n4,

|T|=qn(n-1)(qn+1)gcd(4,qn+1)i=1n-1(q2i-1),

and |Out(T)|=gcd(4,qn+1)f. Since f2ln(q)/ln(2) and gcd(4,qn+1)4, (3.2) implies that

qn(n-1)(qn+1)i=1n-1(q2i-1)226ln4(2)ln4(q).

This fails for all q2 if n=4, and hence fails for all q2 for every n4.

(vi) TE62(q2). Suppose that TE62(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 f2ln(q)/ln(2) and gcd(3,q+1)3, (3.2) implies the following inequality, which fails for all q2:

q36(q12-1)(q9+1)(q8-1)(q6-1)(q5+1)(q2-1)35216ln4(2)ln4(q).

(vii) TD43(q3). Suppose that TD43(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 q2:

q12(q8+q4+1)(q6-1)(q2-1)34212ln4(2)ln4(q).

(viii) TB22(q), G22(q), or F42(q). Finally, suppose that T is as in one of the lines of Table 1. Suppose first that n1. Then |Out(T)|=2n+1 in each case, and (3.2) therefore implies that |T|212(2n+1)4. This inequality holds only in the case TB22(22n+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)22n+1
G22(q)q3(q3+1)(q-1)32n+1
F42(q)q12(q6+1)(q4-1)(q3+1)(q-1)22n+1

## 4 Proof of Theorem 1.1: HC type

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

MInn(M)GMAut(M),

where M=T1××Tk, with k2 and T1TkT 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 k3, and then deal with the case k=2 separately.

### 4.1 The case k⩾3

Suppose, towards a contradiction, that k3. Denote by 1,,t+1 the lines incident with the identity 1M. By Lemma 2.1 (ii), we may identify i with the subgroup of M comprising all points incident with i. Let us write :=1 for brevity.

Claim 4.1

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

### Proof.

Suppose, towards a contradiction, that M=A×B with A{1} and B{1}. We may assume, without loss of generality, that (i) A contains T1, and (ii) 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 AA and BB). Take also yB with y1. For every aInn(A)Inn(M), we have ya=y and hence a=, because a also fixes the point 1. In particular, is fixed by every element of Inn(T1), regarded as a subgroup of Inn(A). Therefore, (z,x2,,xk) for all zx1T1, and hence contains the group 0:=(z,x2,,xk):zx1T1. Let π1 denote the projection onto T1. Then π1(0)=z:zx1T1=T1, and hence π1()=T1. Also, taking zx1, we see that T1 contains (z,x2,,xk)-1x=(z-1x1,1,,1)1. That is, T1 is non-trivial, and it is normal in the simple group π1()=T1, so T1=T1 and hence T1. Now, G1 acts transitively on both {T1,,Tk} (because G is transitive on {T1,,Tk} and G=MG1) and {1,,t+1} (because G is flag-transitive, by Theorem 2.2). Therefore, t+1 divides k, and, without loss of generality, =1 contains TU1:=T1××Tk/(t+1), 2 contains TU2:=Tk/(t+1)+1××T2k/(t+1), and so on.

Subclaim

=TU1.

### Proof of sub-claim.

It remains to show that TU1 contains . Suppose, towards a contradiction, that there exists wTU1. Then there exists i>k/(t+1) such that the ith component wi of w is non-trivial, and so there exists σInn(Ti) such that wiσwi. Regarding σ as an element of Inn(M)G1, we see that σ fixes , because it centralises T1. Hence, wσ, and so contains the element w-1wσTi{1}. However, TiTUjj for some j1, and hence intersects j 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(st+1), where t:=(t-1)/(s+1)s-1 as before. Since st+1s(s-1)+1<(s+1)2, this implies that (s+1)t-1<(s+1)2, so t=2, and hence s+1t-1=1, a contradiction. ∎

Claim 4.2

is isomorphic to a subgroup of T.

### Proof.

Let x{1} have minimal support U. Suppose, without loss of generality, that x1:=π1(x)1. Suppose further, towards a contradiction, that there exists y{1} with π1(y)=1. Then every aInn(T1) fixes y and hence fixes , so contains xa and therefore contains xax-1T1. Taking a not in CT(x1) makes xax-1 non-trivial, and the minimality of the support U of x implies that U={1}, so xT1. However, the existence of y now contradicts Claim 4.1, because taking A=T1 and B=T2××Tk gives xA and yB. Hence, if x has minimal support U containing 1, then every non-trivial element of must project non-trivially onto T1. Therefore, is isomorphic (under projection) to a subgroup of T1. ∎

We now use Claim 4.2 to derive a contradiction to the assumption that k3. By Claim 4.2, s+1=|| divides |T|, so in particular s+1|T|. Writing

|M|=(s+1)2(st+1)

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

(s+1)2>s(s-1)+1st+1=|M|(s+1)2|M||T|2=|T|k-2(s+1)k-2,

and hence 2>k-2, namely k3.

Now suppose, towards a contradiction, that k=3. Write |T|=n(s+1). Then st+1=|M|/(s+1)=|T|3/(s+1)=n3(s+1)2, and hence n31(mods). On the other hand, we have s3+1st+1=n3(s+1)2>n3s2+1, so n3<s. Therefore, n=1, so |T|=s+1 and t=s+2. Together with Claim 4.2, this implies that is isomorphic to T. Consider first the case where is a diagonal subgroup {(t,ta,tb):tT}M for some automorphisms a,bAut(T). As (c,d)Inn(T2)×Inn(T3)G1 runs over all possibilities, we obtain |T|2 distinct images (c,d)={(t,tac,tbd):tT} of . Indeed, if =(c,d), then ta=tac for all tT, or equivalently, u=uc for all uT; 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 is a diagonal subgroup {(t,ta,1):tT}T1×T2 for some aAut(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 cInn(T2)G1 runs over all possibilities, we obtain |T| distinct images c={(t,tac,1):tT} of . Hence, (s+3)/3=(t+1)/3s+1, a contradiction. This leaves only the possibility that T1, and hence =T1 because ||=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×T2T2, and

|M|=(s+1)(st+1)=(s+1)2(st+1),where 1ts-1.

Therefore,

|T|=(s+1)(st+1)1/2,where 1ts-1 and st+1 is a square.

Writing y2=st+1, this is equivalent to

|T|=(s+1)y,where 3y2s(s-1)+1 and sy2-1.

By Theorem 2.2, G1Aut(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(st+1)).

In particular, t+1gcd(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,st+1)=gcd(t+1,t+1).

Hence we have gcd(t+1,2|T|2)8(t+1), and so t+18(t+1)|Out(T)|2. Re-writing this as t(s+1)+28(t+1)|Out(T)|2, and noting that t1, we obtain

s16|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 3y2s(s-1)+1 and sy2-1,

2. 2s16|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. TAlt6, s=19, and y=18,

2. TA2(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 ys<y2.

Case 1: TAltn 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 TAlt5, then by (a), we have (s+1)y=60 and sy2-1, which is impossible. If TAlt7, then we again apply (a): (s+1)y=2520, sy2-1, and y2s(s-1)+1, which is again impossible. Finally, we examine the case TAlt6, where |Out(T)|=4. Applying (a), we have s=19, y=18 as the only valid solution.

Case 2: TA1(q). Suppose that TA1(q), and write q=pf with p prime and f1. 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

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

which holds only if f7. If f=1, then T is not simple; and if f=2, then TAlt5, which we have already ruled out. For 3f7, 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

p2f(p2f-1)28(32f2-1)((64f2-3)3+1),

which implies that either 11p19 and f=1; 5p7 and f2; or p=3 and f4. If q=3, then T is not simple; if q=5, then TAlt5, which we have ruled out; if q=7, then TA2(2), which is ruled out in Case 3 below; and if q=9, then TAlt6, 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: TAn(q), n2. 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 TA3(3), and TA2(q) for q13. The former is ruled out by (4.1), because

|A3(3)|=6,065,280>2844=65,536.

For TA2(q), (c) implies that

q6(q2-1)2(q3-1)29(576ln2(2)ln2(q)-2)((576ln2(2)ln2(q)-3)3+1).

Therefore, q10. 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: TAn2(q2). Since (4.1) implies (3.2), we only need to check that TA32(q2) for 2q4, and TA22(q2) for q13. 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)29(576ln2(2)ln2(q)-2)((576ln2(2)ln2(q)-3)3+1),

and hence q10. If q=2, then TA22(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 TB2(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 24f(22f-1)(24f-1)212f4 fails when f=2. If TB22(22n+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=st+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 TAlt6, and |Aut(M)|=225,792 if TA2(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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