Let G be a group of collineations of a finite thick generalised quadrangle Γ. Suppose that G acts primitively on the point set of Γ, and transitively on the lines of Γ. We show that the primitive action of G on 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 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.
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 quadrangle 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 contains no triangles. If each point of is incident with at least three lines, and each line is incident with at least three points, then is said to be thick. In this case, there exist constants such that each point (line) is incident with exactly lines ( points), and is called the order of . Generalised quadrangles were introduced by Tits , 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  showed that a Frobenius group or a group with non-trivial centre cannot act regularly on the points of a generalised quadrangle of order if s is even and , and Yoshiara  showed that a generalised quadrangle with 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  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 , 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 and .) 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.
Let G be a collineation group of a finite thick generalised quadrangle with point set and line set . If G acts transitively on and primitively on , then G has a unique minimal normal subgroup; that is, the action of G on does not have O’Nan–Scott type HS or HC.
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 under by . The orbit of under G is the set , the subgroup is the stabiliser of , and the Orbit–Stabiliser Theorem says that . The action of G is transitive if for some (and hence every) , and semiregular if is trivial for all . 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 , define , by
The holomorph of G is the semidirect product with respect to the natural action of on (see [1, Section 2.6]). We have , and . A group H acting on a set Δ is permutationally isomorphic to G acting on Ω if there is an isomorphism and a bijection such that for all and . If a group M acts regularly on Ω, then there is a permutational isomorphism with bijection , where for some fixed , and . We have , 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 . Furthermore, equals . If M is a normal subgroup of G, then G is permutationally isomorphic to a subgroup of . If , then G contains because .
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 with 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 , , then G has type HC. In this case, G induces a subgroup of which acts transitively on the set of k simple direct factors of . In either case, G contains and , as explained above.
If we write for a partial linear space, then we mean that is the point set, is the line set, and is the incidence relation. An incident point–line pair is called a flag. A collineation of is a permutation of , together with a permutation of , such that incidence is preserved. If admits a group of collineations M that acts regularly on , then we identify 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 if and only if . Moreover, the stabiliser is the set of all elements of M that fix setwise by right multiplication.
Let be a partial linear space with no triangles, and let G be a group of collineations of with a normal subgroup M that acts regularly on . Let be a line incident with the identity , and suppose that its stabiliser is non-trivial. Then:
is a union of left -cosets, including the trivial coset,
if , then .
(i) Let . Since , namely , it follows that , namely . Therefore, . Now, if is incident with , then every non-trivial element of must map h to another point incident with , and hence the whole coset is contained in .
(ii) By (i), we have , so it remains to show the reverse inclusion. Let . Since is non-trivial, there exists a non-trivial element . Since , left multiplication by is a collineation of . Since 1 and m are both incident with , it follows that and are collinear. On the other hand, by (i), so is collinear with m because right multiplication by m is a collineation. That is, is collinear with two points that are incident with , and so is itself incident with because contains no triangles. Therefore, m maps two points incident with to two points incident with , and so . ∎
Let be a partial linear space with no triangles. Let G be a group of collineations of that acts transitively on , and suppose that G has a normal subgroup M that acts regularly on and satisfies . If the action of M on is not semiregular, then the lines incident with 1 are a -conjugacy class of subgroups of M, and G acts transitively on the flags of .
Since M acts transitively on , we have . By assumption, and so . By Lemma 2.1 (ii), the lines can be identified with subgroups of M. Each , acting naturally as an element of , fixes 1 and hence maps to for some . Conversely, consider the map defined by . The restriction of φ to is the identity. Moreover, , and hence , where θ is the permutational isomorphism defined above. In particular, acts transitively (indeed, regularly) on . Hence, we have , so . Now consider a line for some . By line-transitivity, we have for some . On the other hand, since , we have for some and , so . Therefore, are precisely the subgroups of the form with . Since the lines and intersect precisely in the point 1 for , the subgroups are distinct, and they form a single -conjugacy class of subgroups of M. In particular, acts transitively on , so G acts transitively on the flags of . ∎
Let us draw a corollary in the case where is a thick generalised quadrangle. In this case, has points and lines, where is the order of .
If the partial linear space in Theorem 2.2 is a thick generalised quadrangle of order , then divides .
Begin by observing that acts on . That is, for each , we have for some . Suppose first that is intransitive on . Then, without loss of generality, is in a different -orbit to , and so, for every , we have for some . Hence, every double coset , where , has size . Here the final equality holds because (because distinct concurrent lines intersect in a unique point, in this case the point 1). Since the double cosets of and partition M, it follows that divides . Therefore, divides , and hence divides , as claimed.
Now suppose, towards a contradiction, that the group is transitive on . Consider two lines incident with 1, say . Then a double coset , where , has size or according as or . Let us say that D is small in the latter case. There are exactly elements for which , that is, for which is small. Moreover, each such D has representatives , because if and only if , and . Hence, there are exactly small double cosets of the form . Therefore, divides , and so divides and hence . In particular, we have , and so [7, Result 2.2.2 (i)] implies that cannot contain a subquadrangle of order . For a contradiction, we now construct such a subquadrangle.
Since is a subgroup of M and right multiplication by any element of M is a collineation of , we have in particular that every right coset of with is a line of . Similarly, since left multiplications are collineations, every left coset of with is a line of . Therefore,
is a subset of . Now, consider also the subset of , and let be the restriction of to . We claim that is a subquadrangle of of order . Let us first check that satisfies the generalised quadrangle axiom. Let and take not incident with . Then, since satisfies the generalised quadrangle axiom, there is a unique point incident with and collinear with P. Since , we have , and so also satisfies the generalised quadrangle axiom. It remains to check that has order . Now, every line in is incident with points in , being a coset of either or , so it remains to show that every point in is incident with exactly two lines in . Given , where , , each line incident with P is either of the form for some or for some , and since , we must have or , respectively. Therefore, P is incident with exactly two lines in , namely and . ∎
We also check that, in the case of a thick generalised quadrangle, the assumption that M is not semiregular on is satisfied when is even.
Let be a thick generalised quadrangle of order . Let G be a group of collineations of that acts transitively on , and suppose that G has a normal subgroup M that acts regularly on . If M has even order, then M does not act semiregularly on .
3 Proof of Theorem 1.1: HS type
Suppose that is a thick generalised quadrangle with a collineation group G that acts transitively on and primitively of O’Nan–Scott type HS on . Then
for some non-Abelian finite simple group T, with T acting regularly on . Since is even by the Feit–Thompson Theorem , Lemma 2.4 tells us that satisfies the hypotheses of Theorem 2.2 and Corollary 2.3. In particular, divides (by Corollary 2.3), and we write
Since T acts regularly on , we have . By Higman’s inequality, , and hence . Therefore,
By Theorem 2.2, acts transitively on the lines incident with 1, and hence divides . Therefore, is divisible by , so
Since is even, s must be odd; and since divides , we have . Moreover, , so
and in particular . Therefore,
Together with (3.1), this implies , and because , it follows that
Since (by Higman’s inequality), we have
The following lemma therefore completes the proof of Theorem 1.1 in the HS case.
There is no finite non-Abelian simple group T satisfying
, where ,
Since for real , condition (c) implies that
Case 1: or a sporadic simple group. If , then and there is no solution to (a) subject to (b). If T is an alternating group other than , or a sporadic simple group, then , and so (c) implies that . This rules out everything except , and , and for these cases one checks that there is no solution to (a) subject to (b).
Case 2: . Suppose that , and write with p prime and . Then , and .
Suppose first that q is even, namely that . Then , and (c) implies that
Now suppose that is odd. Then , and hence we have , so (c) reads
If , then this inequality fails for all . The inequality holds if and only if
If , then T is not simple; if , then , which we have ruled out; if , then , which is ruled out in Case 3 below; and if , then , which we have ruled out. For the remaining values of q, there is no solution to (a) subject to (b).
Case 3: , . Suppose that , with and . Then
First suppose that . Noting that and , and applying (3.2), we find
This inequality fails for all if , and therefore fails for all for every (because the left-hand side is increasing in n while the right-hand side does not depend on n). It fails for unless , but has already been ruled out, and (c) rules out because
Finally, suppose that . Noting that and , (3.2) gives
Case 4: . Suppose that , where now for some prime p and . We have ,
First suppose that . Noting that
and that , (3.2) gives
This inequality fails for all for , and hence fails for all for every .
Now suppose that . Then we can replace the on the right-hand side above by , because . This yields
Finally, suppose that . Then , and hence
Case 5: Remaining possibilities for T. We now rule out the remaining possibilities for the finite simple group T.
(i) or . First suppose that , and write with p prime and . We have , , and . Noting that and that is at most 2, (3.2) implies that
However, this inequality fails for all if , and hence fails for all for every .
Now suppose that , writing as before. In this case we have , and again . If and q is even, then . If and q is odd, then is the same as for . We may therefore assume that . First suppose that . Then , so (3.2) implies that
and hence . However, we have , which has been dealt with in Case 4.
(ii) . Suppose that , writing again. We have , , and
If q is odd, then , , and , so (3.2) implies that
which fails for all if , and hence fails for all for every . If q is even, then , and , so (3.2) implies that
which fails for all if , and hence fails for all for every .
(iii) or . Suppose that T is one of , , or , and write again. Observe that for every , for all . Hence
Since , (3.2) implies the following inequality, which fails for all :
(iv) . Suppose that , with . Then
Note that is not simple, but is simple and has already been ruled out.
(v) . Suppose that , now writing . Then ,
and . Since and , (3.2) implies that
This fails for all if , and hence fails for all for every .
(vi) . Suppose that , with . Then
Noting that and , (3.2) implies the following inequality, which fails for all :
(vii) . Suppose that , where now . Then
and , so (3.2) implies the following inequality, which fails for all :
(viii) , , or . Finally, suppose that T is as in one of the lines of Table 1. Suppose first that . Then in each case, and (3.2) therefore implies that . This inequality holds only in the case with , but cannot be written in the form (a) subject to (b). For , we have that is not simple; is not simple, but has been ruled out in Case 2 above; and is not simple, but is simple of order and has outer automorphism group of order 2, so (3.2) fails.
This completes the proof of Lemma 3.1. ∎
4 Proof of Theorem 1.1: HC type
Suppose that is a thick generalised quadrangle with a collineation group G that acts transitively on and primitively of O’Nan–Scott type HC on . Then
where , with and for some non-Abelian finite simple group T. Moreover, M acts regularly on , and G induces a subgroup of which acts transitively on the set (see [8, Section 5]). Since is even by the Feit–Thompson Theorem , Lemma 2.4 tells us that satisfies the hypotheses of Theorem 2.2 and Corollary 2.3. In particular, divides (by Corollary 2.3), and we define as in (3.1).
We first rule out the case , and then deal with the case separately.
4.1 The case
Suppose, towards a contradiction, that . Denote by the lines incident with the identity . By Lemma 2.1 (ii), we may identify with the subgroup of M comprising all points incident with . Let us write for brevity.
The group M cannot be decomposed in the form with and .
Suppose, towards a contradiction, that with and . We may assume, without loss of generality, that (i) A contains , and (ii) contains an element that projects non-trivially onto each simple direct factor of A (if not, then change the decomposition of M to with and ). Take also with . For every , we have and hence , because a also fixes the point . In particular, is fixed by every element of , regarded as a subgroup of . Therefore, for all , and hence contains the group . Let denote the projection onto . Then , and hence . Also, taking , we see that contains . That is, is non-trivial, and it is normal in the simple group , so and hence . Now, acts transitively on both (because G is transitive on and ) and (because G is flag-transitive, by Theorem 2.2). Therefore, divides k, and, without loss of generality, contains , contains , and so on.
Proof of sub-claim.
It remains to show that contains . Suppose, towards a contradiction, that there exists . Then there exists such that the ith component of w is non-trivial, and so there exists such that . Regarding σ as an element of , we see that σ fixes , because it centralises . Hence, , and so contains the element . However, for some , and hence intersects in more than one point, a contradiction, proving the sub-claim. ∎
By the sub-claim, , where . Since , we have , where as before. Since , this implies that , so , and hence , a contradiction. ∎
is isomorphic to a subgroup of T.
Let have minimal support U. Suppose, without loss of generality, that . Suppose further, towards a contradiction, that there exists with . Then every fixes y and hence fixes , so contains and therefore contains . Taking a not in makes non-trivial, and the minimality of the support U of x implies that , so . However, the existence of y now contradicts Claim 4.1, because taking and gives and . Hence, if x has minimal support U containing 1, then every non-trivial element of must project non-trivially onto . Therefore, is isomorphic (under projection) to a subgroup of . ∎
with as before, we have
and hence , namely .
Now suppose, towards a contradiction, that . Write . Then , and hence . On the other hand, we have , so . Therefore, , so and . Together with Claim 4.2, this implies that is isomorphic to T. Consider first the case where is a diagonal subgroup for some automorphisms . As runs over all possibilities, we obtain distinct images of . Indeed, if , then for all , or equivalently, for all ; that is, c is the identity automorphism of T (and similarly, d is the identity). Hence, , a contradiction. Now consider the case where is a diagonal subgroup for some . Then 3 divides because is transitive on the , and we have exactly lines incident with 1 that are diagonal subgroups of . As runs over all possibilities, we obtain distinct images of . Hence, , a contradiction. This leaves only the possibility that , and hence because . This implies that , and hence because divides , a contradiction.
4.2 The case
Here we argue as in the case where the primitive action of G on has type HS. That is, we obtain an upper bound on in terms of , and consider the possibilities for T case by case using the Classification of Finite Simple Groups. We have , and
Writing , this is equivalent to
By Theorem 2.2, acts transitively on the lines incident with 1, and hence divides . Therefore, is divisible by
In particular, . We have
, so ,
Hence we have , and so . Re-writing this as , and noting that , we obtain
Higman’s inequality then gives
The following lemma therefore rules out all but two possibilities for T.
Let T be a finite non-Abelian simple group satisfying
, where and ,
Then one of the following holds:
, , and ,
, , and .
The right-hand side of (c) is at most , so
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 .
Case 1: or a sporadic simple group. If T is an alternating group other than , or a sporadic simple group, then and so (c) implies that . Hence, T is one of , , or . If , then by (a), we have and , which is impossible. If , then we again apply (a): , , and , which is again impossible. Finally, we examine the case , where . Applying (a), we have , as the only valid solution.
Case 2: . Suppose that , and write with p prime and . Then , and .
Suppose first that q is even, namely that . Then , and (c) implies that
Now suppose that is odd. Then , and hence we have . By (c), we have
which implies that either and ; and ; or and . If , then T is not simple; if , then , which we have ruled out; if , then , which is ruled out in Case 3 below; and if , then , which we have already dealt with in Case 1. Hence, we only need to consider . For each of these values, there is no solution to (a) subject to (b).
For , (c) implies that
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 with , then, using (4.1) instead of (3.2), the on the right-hand side of (3.3) becomes , and the resulting inequality fails when . If with , 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 , we find that in case (i), and in case (ii). Both cases are then ruled out because the required divisibility condition fails. (Note that if , and if .)
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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