Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 34 (1993), No. 2, 277-286.

On Flag-Transitive Anomalous C_{3}-Geometries

Satoshi Yoshiara and Antonio Pasini

Division of Mathematical Sciences, Osaka Kyoiku University 4-698-1 Asahigaoka Kashiwara, 582 Osaka, Japan Dipartimento di Matematica, Università di Siena Via del Capitano, 15, 53100 Siena, Italy

Abstract. A finite C_{3}-geometry is called anomalous if it is neither a building nor the A_{7}-geometry. It is conjectured that no flag-transitive thick anomalous C_{3}-geometry exists. For a flag-transitive thick anomalous C_{3}-geometry, we prove that its 2-order y is odd and that its full automorphism group is non-solvable. As a corollary, there are no flag-transitive circular extensions of duals of anomalous C_{3}-geometries.

1. Introduction

A flag-transitive C_{3}-geometry G of finite order (x, y) is a residually connected incidence geometry on I = {0, 1, 2}, in which the residue G_{v} at an element v of type i is isomorphic to a generalized quadrangle of finite order (x, y), to a generalized digon, or to a projective plane of order x, respectively for i = 0, 1, 2.

If x, y ≥ 2, G is called thick. (See [2] for the fundamental terminology.)

It is conjectured that a flag-transitive thick C_{3}-geometry is either a finite building of type C_{3} or the sporadic A_{7}-geometry. A C_{3}-geometry is called anomalous if it is neither a building nor the sporadic A_{7}-geometry. M.Aschbacher [1] proved this conjecture assuming that the residues of the elements of type 0 and 2 are classical. In the general case, the second author derived several results, which are summarized in [8], but he did not yet succeed to prove the conjecture.

In particular, he showed that the conjecture can be proved if we weaken the hypotheses of Aschbacher [1], assuming only that residues of elements of type 2 are classical projective planes. This is sufficient to prove that anomalous C_{3}-geometries cannot appear

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1993 Heldermann Verlag, Berlin

278 S. Yoshiara, A. Pasini: On Flag-Transitive Anomalous C_{3}-Geometries

as rank 3 residues in certain geometries of rank 4. For instance, there are no finite flagtransitive geometries for any of the following two diagrams where C_{3}-residues are thick and anomalous.

Indeed, the projective planes at the middle of the diagram are forced to be classical here (see [3]). This non-existence lemma is one of the first step in the classification of finite flag-transitive geometries belonging to the above diagrams (see [8],[4] and [5] for the first diagram; [12] and [6] for the second one).

In this paper, we will prove the following.

Theorem 1. Let G be a flag-transitive thick anomalous C_{3}-geometry of order (x, y). Then y is odd and the full automorphism group Aut(G) is not solvable.

Although this does not yet solve the conjecture completely, it suffices to eliminate the existence of some extensions of anomalous C_{3}-geometries. A residually connected geometry G = (G_{0}, G_{1}, G_{2}, G_{3}; ∗) of rank 4 is called a C_{3}c^{∗}-geometry, if it belongs to the following diagram

That is, the residue G_{π} of each element π of G_{3} is a C_{3}-geometry, the residues G_{F} of all flags F of type {1, 2} and {0, 2} are generalized digons, and the residue G_{F} of every flag F of type {0, 1} is the dual of a circle geometry.

Theorem 2. There is no flag-transitive C_{3}c^{∗}-geometry, in which the residue at each element of type 3 is a thick anomalous C_{3}-geometry.

The first author has started a classification program for flag-transitive locally finite C_{3}c^{∗}- geometries in [14] and [15], but assuming that C_{3}-residues are buildings. The case with C_{3}-residues isomorphic to the A_{7}-geometry is settled in [10]. By Theorem 2, these are the only two cases to consider if C_{3}-residues are thick.

2. Notation and Preliminaries

Throughout §2–3, G = (G_{0}, G_{1}, G_{2}; ∗) is a thick anomalous C_{3}-geometry of order (x, y) with a fixed maximal flag (a, l, u) (a ∈ G_{0}, l ∈ G_{1}, u ∈ G_{2}), admitting a flag-transitive automorphism group A. Elements of G_{i} are called points, lines and planes respectively for i = 0, 1, 2. For a flag F and a type i not contained in the type of F, we use G_{i}(F) to denote the set of elements of type i incident to all elements of F. The stabilizer of a flag F in A is denoted by A_{F} and we use K_{F} to denote the kernel of the action of A_{F} on the residue G_{F}.

We summarize here several known facts about an anomalous C_{3}-geometry [8], [9]:

Lemma 1. The following hold.

(1) There is at most one line incident to two distinct planes u and v. Such a line will be denoted by u ∩ v, if it exists.

(2) For every point-plane flag (a, u), we denote by α(a, u) the number of planes v(≠ u) of G_{2}(a) such that u ∩ v exists but a is not incident to u ∩ v. Then α := α(a, u) is a constant, not depending on the particular choice of (a, u), which is called the Ott-Liebler number of G. For any point-plane antiflag (b, u) of G, α + 1 is the number of planes v of G_{2}(b) such that u ∩ v exists.

(3) The locally finite geometry G is a finite geometry with (x^{2} + x + 1)(x^{2}y + 1)/(α + 1) points, (x^{2} + x + 1)(x^{2}y + 1)(xy + 1)/(α + 1) lines and (x^{2}y + 1)(xy + 1)(y + 1)/(α + 1) planes.

(4) The number x is an even number such that x + 1 ≡ 0 (mod 3) and x^{2} + x + 1 is a prime. We have x < y ≤ x^{2} − x and x + y divides xy(x + 1)(y + 1).

(5) For d := g.c.d.(x^{2}, y), we have d^{2} < x and (x − 1)d^{2} + d ≤ y. The Ott-Liebler number α is a multiple of xd, and α + 1 divides (xy/d) − (α/xd). Furthermore, the number (x + y)(α + 1) divides (1 + xy)(xy − (α/x)), and the number (x^{2} + y)(α + 1) divides (1 + x^{2}y)(x^{3}y − (α/x)).

(6) For a plane u, the group A_{u}/K_{u} is isomorphic to a Frobenius group F_{x^{2}+x+1}^{x+1} with kernel of order x^{2} + x + 1 and a complement a cyclic group of order x + 1. It acts regularly on the set of point-line flags of the projective plane G_{u}, in which the Frobenius kernel acts regularly on both the sets of points and lines, while a complement fixes a point-line antiflag (b, m) and acts regularly both on the set of lines through b and the set of points on m.

(7) Assume that |A| is even, and let C be the subgeometry of G fixed by an involution i of A. If y is odd, then C does not contain any plane and consists of some lines and all points on them. If y is even, then C consists of a unique plane and all lines and points on it.

(8) For a prime p = x^{2}+x+1, a Sylow p-subgroup of A is of order p and acts semiregularly on G_{0}.

3. Proof of Theorem 1

Lemma 2. Let Ω be a system of non-trivial proper imprimitivity blocks of G_{0} under the action of A. Then |G_{0}(u) ∩ Δ| ≤ 1 for each block Δ of Ω.

Proof. Suppose |Δ ∩ G_{0}(u)| ≥ 2 for a block Δ. Then the sets Δ^{g} ∩ G_{0}(u) for g ∈ A_{u} form a system of non-trivial imprimitivity blocks of G_{0}(u) under the action of A_{u}. By Lemma 1(6), A_{u} acts primitively on G_{0}(u), which implies that G_{0}(u) ⊆ Δ.

For any plane v(≠ u) for whch u ∩ v exists and any g ∈ A with v = u^{g}, we have ∅ ≠ G_{0}(u ∩ v) ⊆ G_{0}(u) ∩ G_{0}(v) ⊆ Δ ∩ Δ^{g}, and so G_{0}(v) ⊆ Δ = Δ^{g}. Since G is residually connected, any plane w can be joined to u by a chain u = u_{0}, u_{1}, …, u_{m} = w such that u_{i−1} ∩ u_{i} exists for each i = 1, 2, …, m. Then the above argument shows that G_{0}(w) ⊆ Δ for any plane w, that is, Δ = G_{0}, which contradicts the assumption.

Lemma 3. For two distinct points b, c, we denote by L(b, c) the set of lines incident to both b and c, and set n(b, c) = |L(b, c)|. Then the following hold.

(1) We have α = ∑_{b∈G_{0}(l)−{a}}(n(a, b) − 1).

(2) If n(a, b) is minimum among n(a, b) for b ∈ G_{0}(l) − {a}, we have n(a, b) ≤ α/x.

Proof. (1) Let A(a, u) be the set of planes v (≠ u) incident to a such that the line u ∩ v is not incident a. By Lemma 1 (2), α = |A(a, u)|. We will count the cardinal number of the set

X := {(v, m, b) ∈ A(a, u) × (G_{1}(a) − {l}) × (G_{0}(l) − {a})|m ∗ v, b ∈ G_{0}(m)}

in two different ways. For each plane v ∈ A(a, u), the line u ∩ v on u is not incident to a. Then l and u ∩ v are distinct lines on the projective plane u, and they intersect at the unique point b (≠ a). Since a and b are distinct points on the projective plane v, there is a unique line m (≠ l) on v joining a and b. Thus α = |X |. On the other hand, for each point b on l distinct from a, there are n(a, b) − 1 lines through a, b distinct from l. For each such line m, (m, u) is an antiflag of the generalized quadrangle G_{b}. Then there is a unique plane v (≠ u) of G_{b} incident to m such that u ∩ v exists. Since v ∈ A(a, u), we have |X | = ∑_{b∈G_{0}(l)−{a}}(n(a, b) − 1).

(2) By the minimality of n(a, b) and the equation in (1) above, we have x(n(a, b)−1) ≤ α or n(a, b) ≤ 1 + (α/x). Note that x divides α by Lemma 1(5), hence 1 + (α/x) is an integer. Suppose n(a, b) = 1 + (α/x). Then the minimality of n(a, b) implies that n(a, b) is constant for b ranging over the set of points on l distinct from a. By the flag-transitivity of A on G, this implies that n(a, b) is constant for (a, b) ranging over the set of pairs of collinear points. Thus by [8] p.9, Proposition 7, G is either a building or flat. In the latter case, G is the sporadic A_{7}-geometry by [8] p.25, Theorem 8. Since G is anomalous by our assumption, the case n(a, b) = 1 + (α/x) is ruled out. As (α/x) is an integer, we have the result.

Lemma 4. Assume that A is solvable. Then the following hold.

(1) There is a system Ω of p = x^{2} + x + 1 imprimitivity blocks of G_{0} on which A induces a Frobenius group A/N with kernel of order p, where N = O_{p′}(A).

(2) The group A_{a}N/N is a cyclic Frobenius complement, which is a direct product of A_{a,u}N/N and A_{a,l}N/N.

(3) We have A_{u} ∩ N = K_{u} = A_{a,l,u}.

Proof. (1) Set p = x^{2} + x + 1. By Lemma 1(8), the stabilizer A_{u} contains a Sylow psubgroup P of A of order p. Since A is a solvable group with a cyclic Sylow p-subgroup P, it follows from Ito’s lemma ([13] Ch.4, Theorem 5.13 and (5.15)) that the largest normal p′-subgroup N := O_{p′}(A) of A and the full inverse image H := O_{p′,p}(A) of O_{p}(A/N) in A satisfy the following conditions: A H N, A/H and N are p′-groups, H/N ≅ P and C_{A/N}(H/N) = H/N. In particular, A/N is isomorphic to a subgroup of the Frobenius group F_{p}^{x^{2}+x} with kernel of order p and a complement a cyclic group of order p−1 = x^{2}+x.

By Lemma 1(8), the stabilizer A_{a} of a point a is a p′-group. Since A is solvable, there is a Hall S_{p′}-subgroup S of A containing A_{a} ([13] Ch.4, Theorem 5.6). Then A = P S. Let Δ be the S-orbit on G_{0} containing the point a. If Δ ∩ Δ^{g} ≠ ∅ for some g ∈ A, we have a^{h} = a^{hg^{−1}k} for some h, k ∈ S. Then hg^{−1}kh^{−1} ∈ A_{a} ⊆ S, hence g ∈ S and Δ^{g} = Δ. Thus Ω := {Δ^{g}|g ∈ A} forms a system of imprimitivity blocks under the action of A. Since A = SP, Ω consists of exactly p blocks.

The group A acts transitively and so primitively on Ω. Since the normal p′-subgroup N = O_{p′}(A) is contained in any S_{p′}-subgroups of A, N acts trivially on Ω. Since any non-trivial normal subgroup of the Frobenius group A/N contains the Frobenius kernel of order p, A/N acts faithfully on Ω. Thus A/N is a Frobenius permutation group on Ω, that is, each non-trivial element of A/N fixes at most one block of Ω.

(3) Since Ω consists of p-blocks, |G_{0}(u) ∩ B| = 1 for each block B of Ω by Lemma 2. Then A_{u} ∩ N fixes each point of G_{0}(u), hence fixes each line on the projective plane G_{u}. Thus A_{u} ∩ N ⊆ K_{u}. The group K_{u}N/N is a p′-subgroup of the Frobenius group A/N normalized by the kernel P N/N. Thus A_{u} ∩ N = K_{u} = A_{a,l,u} by Lemma 1(6).

(2) The group A is generated by A_{a,l}, A_{a,u} and A_{l,u} by the residual connectedness of G. Since A_{u}N/N ≅ A_{u}/K_{u} by the claim (3) above, it follows from Lemma 1(6) that A_{u}N/N is a split extension of A_{a,u}N/N by P N/N. Then A/N = ⟨A_{u}N/N, A_{a}N/N⟩ = (P N/N)(A_{a}N/N), as P N/N is the Frobenius kernel of A/N. Since A_{a}N/N fixes the block Δ containing a, it is a Frobenius complement. Thus AN/N is a cyclic group generated by A_{a,u}N/N and A_{a,l}N/N.

Let g = hn be an element of A_{a,u} ∩ A_{a,l}N with h ∈ A_{a,l}, n ∈ N. For each point b on l, Ω(b) means the block of Ω containing b. We have Ω(b) = Ω(b^{n}) as N is the kernel of the action of A on Ω. Note that b^{n} ∈ G_{0}(l)^{n} = G_{0}(l)^{g} ⊆ G_{0}(u)^{g} = G_{0}(u). Then both b^{n} and b are points on the plane u contained in the block Ω(b). Thus b = b^{n} by Lemma 2. Since n fixes all points on l, the element g = hn acting on the projective plane G_{u} fixes the point a and permutes the points on the line l. Hence g fixes a flag (a, l, u), and therefore A_{a,u}N ∩A_{a,l}N = (A_{a,u} ∩A_{a,l}N)N = A_{a,l,u}N = N by the claim (3) above. Then the cyclic subgroup A_{a}N/N is a direct product of A_{a,u}N/N and A_{a,l}N/N.

Proposition 5. There is no locally finite thick anomalous C_{3}-geometry, admitting a flagtransitive solvable automorphism group.

Proof. We use the same notation as in Lemma 4. Let b be a point on l distinct from a such that n(a, b) is minimum (see Lemma 3). We denote by O(b) the orbit through b under the action of N_{a} := N ∩ A_{a}. By Lemma 4, O(b) is a subset of the block Ω(b) containing b. In particular, |O(b)| ≤ |Ω(b)| = |G_{0}|/(x^{2} + x + 1) = (x^{2}y + 1)/(α + 1) by Lemma 1(3).

The line l is incident to two points a and b. Let N be the N_{a}-orbit on G_{1}(a) containig l. By Lemma 4(2), we have |A_{a}/N_{a}| = |A_{a,u}/N ∩A_{a,u}||A_{a,l}/N_{a} ∩A_{l}|. Rewriting this using Lemma 4(3), we have |N_{a} : N_{a} ∩ A_{l}| = |A_{a} : A_{a,u}|/|A_{a,l} : A_{a,l,u}| = (xy + 1)(y + 1)/(y + 1), since A_{a} and A_{a,l} act transitively on the set G_{1}(a) of lines and on the set G_{2}(a, l) of planes through a line l in the generalized quadrangle G_{a} of order (x, y), respectively. Thus there are xy + 1 lines in the orbit N.

Since b lies on l, each line of N is incident to a and to at least one point of O(b). By Lemma 2, the points in the block Ω(b) are mutually non-collinear. Thus each line of N is incident with a unique point of O(b). For each point c ∈ O(b), we denote by N (c) the set of lines of N incident to c. By the remark above, N is a disjoint union of sets N (c) (c ∈ O(b)). Since O(b) is an N_{a}-orbit, |N (c)| is constant, and therefore we have |O(b)| = |N |/|N (b)| = (xy + 1)/|N (b)|. Since |O(b)| ≤ (x^{2}y + 1)/(α + 1), we have |N (b)| ≥ (xy + 1)(α + 1)/(x^{2}y + 1). However, since N (b) consists of lines incident to both a and b, we have (α/x) ≥ n(a, b) ≥ |N (b)| by Lemma 3(2), and therefore (x^{2}y + 1)(α/x) ≥ (xy + 1)(α + 1). This implies that ((1/x) − 1)α ≥ xy + 1, a contradiction.

We will now complete the proof by establishing the following statement.

Proposition 6. There is no flag-transitive anomalous C_{3}-geometry of finite order (x, y) with y even.

Proof. Assume that G is an anomalous C_{3}-geometry of finite order (x, y) with y even, admitting a flag-transitive automorphism group A. For any element g ∈ A − A_{u}, the subgroup A_{u} ∩ A_{u}^{g} fixes two distinct planes u and u^{g}, hence it has odd order by Lemma 1(7). That is, the proper subgroup A_{u} is a strongly embedded subgroup of A. Then either a Sylow 2-subgroup of A is cyclic or generalized quarternion, or A acts doubly transitively on the cosets A/A_{u}, which can be identified with the set G_{2} of planes (see e.g. [13] Chap.6, Theorem 4.8).

If A acts doubly transitively on G_{2}, any two distinct planes are incident to a common line. Thus for a point-plane antiflag (b, u) and for any plane v of G_{2}(b), the line u∩v exists. It follows from Lemma 1(2) that |G_{2}(b)| = (y + 1)(xy + 1) = α + 1. Since α + 1 divides x^{2}y + 1 by Lemma 1(3), we have (y + 1)(xy + 1) ≤ x^{2}y + 1, or x^{2} − x − 1 ≥ xy. However, xy > x^{2} by Lemma 1(4), which is a contradiction.

If A has a cyclic Sylow 2-subgroup S, A is a split extension of S by the odd core O(A), which is solvable by the odd order theorem. Such a possibility is eliminated by Proposition 5. Assume that a Sylow 2-subgroup S of A is a generalized quaternion group. Since |G_{2}| is odd by Lemma 1(3), we may take S ⊆ A_{u}. Then S ⊆ K_{u}, as |A_{u}/K_{u}| = (x^{2} +x+1)(x+1) is odd. The unique involution i of S is central in A modulo O(A) (e.g. [13] Chap.6, §2, Example 3). In particular, M := O(A)⟨i⟩ is a normal subgroup of A for which A/M has dihedral Sylow 2-subgroups. Since A is not solvable by Proposition 5, A/M is not solvable. Thus A/M is isomorphic either to A_{7} or to a subgroup of Aut(L_{2}(q)) containing L_{2}(q) for some odd prime power q (e.g. [13] Chap.6, Theorem 8.6, (8.9)).

Let Ω be the set of O(A)-orbits on G_{0}. Each O(A)-orbit contains a point a^{g} for some g ∈ A. Since gi = ign for some n ∈ O(A),we have (a^{g})^{i} = (a^{i})^{gn} = a^{gn}. Thus the involution i fixes every O(A)-orbit. Since the length of an O(A)-orbit is odd, i has a fixed point on each O(A)-orbit. Since the set of points fixed by i coincides with G_{0}(u) by Lemma 1(8), it follows from Lemma 2 that Ω consists of exactly x^{2} + x + 1 orbits. The subgroup S fixes every point of G_{0}(u), and therefore it acts trivially on Ω. Hence the normal subgroup N of A generated by all Sylow 2-subgroups together with M acts trivially on Ω.

If A/M ≅ A_{7}, a non-abelian simple group, we have A = N, which contradicts the transitivity of A on G_{0}. Assume that A/M is a subgroup of Aut(L_{2}(q)) containing L_{2}(q). Then A/N corresponds to a subgroup of the field automorphism group of L_{2}(q). In particular, A/N is a cyclic group. On the other hand, since N acts trivially on Ω, it follows from Lemma 2 that A_{u} ∩ N ⊆ K_{u}. Thus a subgroup A_{u}N/N of A/N has a factor group isomorphic to A_{u}/K_{u}, which is a non-abelian Frobenius group of order (x^{2} + x + 1)(x + 1). This contradiction eliminates the final case.

4. Proof of Theorem 2

In the remainder of this paper, we assume that G = (G_{0}, G_{1}, G_{2}, G_{3}; ∗) is a C_{3}c^{∗}-geometry (see §1), admitting a flag-transitive automorphism group G, in which the residue at π ∈ G_{3} is an anomalous C_{3}-geometry. We may apply the results in §2,3 to G_{π}. We call elements of G_{i} points, lines, planes and quads, respectively for i = 0, 1, 2 and 3. We fix a maximal flag (P, l, u, π) of G with P ∈ G_{0}, l ∈ G_{1}, u ∈ G_{2} and π ∈ G_{3}, and follow the other convention in §2,3.

Lemma 7. The group G acts regularly on the set of maximal flags of G. In particular,

(1) For each flag F, the stabilizer G_{F} acts faithfully and regularly on the set of maximal flags of the residue G_{F}.

(2) G_{u} is isomorphic to F_{x^{2}+x+1}^{x+1} × 2 with the kernel O_{2}(G_{u}) = G_{P,l,u} on the projective plane (G_{0}(u), G_{1}(u); ∗). Furthermore, G_{P,u,π} and G_{l,u,π} are cyclic subgroups of order x+1.

Proof. We divide the proof into several steps.

Step 1. The stabilizer G_{P,l,u} coincides with the kernel of G_{u} on Π(u) := (G_{0}(u), G_{1}(u); ∗). Furthermore, G_{P,l,u,π} = K_{u} and G_{u}/K_{u} ≅ F_{x^{2}+x+1}^{x+1} × 2.

Proof. Since G_{P,l,u} induces an automorphism group of the projective plane Π(u) fixing a point-line flag (P, l) of Π(u), G_{P,l,u} acts trivially on Π(u) by Lemma 1 (6). Then G_{P,l,u} is contained in the kernel of the action of G_{u} on Π(u), and the converse inclusion clearly holds.

Since (G_{3}(l), G_{2}(l); ∗) is a circle geometry, G_{3}(u) consists of two quads including π, on which G_{P,l,u} acts transitively by the flag-transitivity of G. Thus G_{u} = G_{P,l,u}G_{u,π} and G_{P,l,u} ∩ G_{u,π} = G_{P,l,u,π} coincides with the kernel K_{u} of the action of G_{u} on the residue G_{u} by the paragraph above. Moreover, G_{P,l,u}/K_{u} ≅ 2 is the kernel of the action of G_{u}/K_{u} on Π(u). On the other hand, since G_{u}/K_{u} induces the unique flag-transitive automorphism group F := F_{x^{2}+x+1}^{x+1} of Π(u), we have G_{u}/G_{P,l,u} ≅ G_{u,π}/K_{u} ≅ F. Thus G_{u}/K_{u} = (G_{u,π}/K_{u}) × (G_{P,l,u}/K_{u}) ≅ F × 2.

Step 2. The kernel of G_{P,π} on the generalized quadrangle G_{(P,π)} coincides with K_{P}.

Proof. Let K = K(P, π) be the kernel of the action of G_{P,π} on the generalized quadrangle G_{(P,π)}. Clearly K_{P} ⊆ K. Since (G_{3}(P), G_{2}(P), G_{1}(P); ∗) is a residually connected c.C_{2}- geometry, any quad μ of G_{3}(P) can be joined to π by a chain π_{0} = π, π_{1}, …, π_{n} = μ of quads of G_{3}(P) with G_{2}(π_{i−1}) ∩ G_{2}(π_{i}) ≠ ∅ for every i = 1, …, n. Thus it suffices to show that K(P, π) ⊆ K(P, π′) for any quad π′ with G_{3}(v) = {π, π′} for some plane v ∈ G_{2}(P).

The group K fixes v and the unique quad π′ of G_{3}(v) distinct from π. Let m be any line of G_{1}(P, v). The group K fixes every plane of G_{2}(m, π)(⊆ G_{2}(P, π)). On the other hand, G_{2}(m, π) can be identified with the set G_{3}(m) − {π}, representing a plane w ∈ G_{2}(m, π) by the unique quad incident to it and distinct from π′. Hence K fixes all quads and planes incident to m. Thus K ⊆ G_{P,m,w,π′} = K_{w} for all planes w ∈ G_{2}(m, π′) by Step 1. That is, the group K acts on the generalized quadrangle G_{(P,π′)} fixing a plane v and all planes w intersecting v, together with all lines on them. Then K acts trivially on G_{(P,π′)}. Thus K ⊆ K(P, π′), and therefore K_{P} = K(P, π). Proof of Lemma 7. Both the kernels K_{l} and K_{P} act on the projective plane Π(u) fixing a flag (P, l). Thus K_{l} (resp. K_{P}) acts trivially on (G_{0}∪G_{1})(u) (resp. G_{0}(l)). As G_{3}(u) ⊆ G_{3}(l) and (G_{2} ∪ G_{3})(l) ⊆ (G_{2} ∪ G_{3})(P), we have K_{P} ⊆ K_{l} ⊆ K_{u}. By Step 2, we have K_{π} ⊆ K_{P}, and so K_{π} ⊆ K_{P} ⊆ K_{l} ⊆ K_{u}. Since K_{u} acts trivially on the projective plane Π(u), it follows from Lemma 1(6) that K_{u} ⊆ K_{π}. Then the residual connectivity of G implies that K_{π} = K_{P} = K_{l} = K_{u} is a normal subgroup of G = ⟨G_{P} , G_{l}, G_{u}, G_{π}⟩, which, since it fixes (P, l, u, π), fixes all maximal flags of G. Thus K_{π} = K_{P} = K_{l} = K_{u} = 1. All statements of the lemma now follow from Steps 1,2 and Lemma 1(6).

We now consider the action of G_{l} on the set G_{3}(l) of y + 2 quads.

Lemma 8. Let M_{l} and N_{l} be the kernels of the action of G_{l} on G_{3}(l) and G_{0}(l), respectively. Then the following hold.

(1) The group M_{l} is a cyclic group of order dividing x + 1.

(2) The group G_{l}/M_{l} is a doubly transitive permutation group on G_{3}(l), in which the stabilizer of two distinct quads is a cyclic group of odd order.

(3) The group N_{l}M_{l}/M_{l} ≅ N_{l} is a normal subgroup of G_{l}/M_{l} of odd index, in which the stabilizer of two distinct quads of G_{3}(l) is trivial.

Proof. Since M_{l} is a subgroup of the cyclic group G_{l,u,π} of order x + 1 by Lemma 7(2), the claim (1) follows. Since (G_{3}(l), G_{2}(l); ∗) is a circle geometry, the set G_{3}(l) − {π} can be identified with the set G_{2}(l, π) of y + 1 planes. Then it follows from the flag-transitivity of G that G_{l} acts doubly transitively on G_{3}(l). The stabilizer G_{l,u,π} in G_{l} of two quads π, π′ ∈ G_{3}(l) with {π, π′} = G_{3}(u) is a cyclic group of order x + 1 by Lemma 7(2). The claim (2) follows.

The image N_{l}M_{l}/M_{l} is a normal subgroup of G_{l}/M_{l} isomorphic to N_{l}, since N_{l} ∩M_{l} = K_{l} = 1 by Prop.2(1). The stabilizer N_{l} ∩ G_{u,π} in N_{l} of two quads π, π′ above is contained in G_{P,l,u,π}, and therefore trivial by Lemma 7.

To complete the proof, it only remains to show that |G_{l}/N_{l}| is odd. Let g be an involution of G_{l}, which has order (x+1)(y +2)(y +1) (see Lemma 7). Since (G_{3}(l), G_{2}(l); ∗) is a circle geometry, there is a plane v of G_{2}(l) incident to two quads fixed or interchanged by the involution g. Since |G_{0}(l)| = x + 1 is odd, g fixes a point Q. Thus g acts on the projective plane Π(v) fixing a flag (Q, l). Since Aut(Π(v)) is of odd order, g fixes all elements of (G_{0} ∪ G_{1})(v). In particular, g ∈ N_{l}. Thus the group G_{l}/N_{l} is of odd order.

Lemma 9. The permutation group G_{l}/M_{l} on the set G_{3}(l) is equivalent to a subgroup of AΓL(1, F_{q^{n}}) containing T (1, F_{q^{n}}) on the set F_{q^{n}} for some prime power q = p^{m} and some n with y + 2 = q^{n}, where AΓL(1, F_{q^{n}}) (resp.T (1, F_{q^{n}})) denotes the group of maps f(a, b; α) : x ↦ ax^{α} + b (resp. f(1, b; id)) on F_{q^{n}} for all a ∈ F_{q^{n}} − {0}, b ∈ F_{q^{n}} and α ∈ Gal(F_{q^{n}}/F_{q}).

Proof. By Lemma 8(2), there is a unique minimal non-trivial normal subgroup U_{l}/M_{l} of G_{l}/M_{l} which is either a non-abelian simple group acting doubly transitively on G_{3}(l) or an elementary abelian p-group acting regularly on G_{3}(l). In the latter case, G_{l}/M_{l} is a split extension of G_{l,π}/M_{l} by U_{l}/M_{l}, in which U_{l}/M_{l} can be identified with G_{3}(l) so that the action of G_{l,π}/M_{l} on the non-trivial elements of U_{l}/M_{l} is equivalent to the action of G_{l,π} on G_{3}(l) − {π}.

Assume that U_{l}/M_{l} is a non-abelian simple group. The unique minimal normal subgroup U_{l}/M_{l} is contained in a non-trivial normal subgroup N_{l}M_{l}/M_{l} of G_{l}/M_{l}. By Lemma 3(3), U_{l}/M_{l} is a sharply doubly transitive permutation group on G_{3}(l). However, this forces the group U_{l}/M_{l} to be a Frobenius group on G_{3}(l) of maximal possible order, and therefore it follows from [11] p.62 Prop. 8.4 that it has a regular normal subgroup, which contradicts the simplicity of U_{l}/M_{l}. Thus U_{l}/M_{l} is an elementary abelian group of order y+2 = q = p^{m} for some prime p.

Since G_{l,u,π} is a cyclic group of odd order x + 1, the stabilizer in G_{l}/M_{l} of two quads is of odd order. Thus by applying [7] p.458, Theorem 6.7 to G_{l}/M_{l}, either we have the conclusion of the lemma or one of the following.

(i) G_{l}/M_{l} on G_{3}(l) is equivalent to a subgroup of GL(2, F_{q^{n}}) on a 2-dimensional space over F_{q^{n}} containing SL(2, F_{q^{n}}) for some prime power q = p^{m} and some n with y +2 = q^{2n}.

(ii) y + 2 = 5^{2}, 7^{2}, 9^{2}, 11^{2}, 19^{2}, 23^{2}, 29^{2}, 59^{2} or 3^{6}. By applying the conditions in Lemma 1(4) to the values of y in Case (ii), we obtain (x, y) = (434, 23^{2} −2), (1562, 59^{2} −2) or (1886, 59^{2} −2). In the first two cases, d = g.c.d.(x^{2}, y) = 31 or 71, which does not satisfy the inequality d^{2} < x in Lemma 1(5). In the last case, we have d = 1 and we can verify that for each value t = 1, …, y − 1, the number α = xt does not satisfy the divisibility condition α + 1|xy − (α/x) in Lemma 1(5). Thus the case (ii) does not occur.

Assume that Case (i) occurs. By Lemma 1(4)(5), we have q^{n} > 3. Then a perfect subgroup (G_{l,π}/M_{l})′ ≅ SL_{2}(q^{n}) of G_{l}/M_{l} is contained in the normal subgroup N_{l}M_{l}/M_{l} of odd index (see Lemma 8(3)). The stabilizer in (G_{l,π}/M_{l})′ ≅ SL_{2}(q^{n}) of a non-zero vector of F_{q}^{2} is isomorphic to F_{q}. In particular, there is a non-trivial element of N_{l}M_{l}/M_{l} fixing two distinct quads of G_{3}(l), which contradicts Lemma 8(3). Thus the lemma holds.

Lemma 10. A Sylow 2-subgroup T of G_{P,l,π} is a Sylow 2-subgroup of G_{π}, which is cyclic. In particular, G_{π} = O(G_{π})T is a solvable group.

Proof. By Lemma 7 and Lemma 1(3), |G_{π}| = ((x^{2}y+1)/(α+1))(x^{2}+x+1)(xy+1)(y+1)(x+ 1). As x is even by Lemma 1(4), the group G_{P,l,π} of order y+1 contains a Sylow 2-subgroup T of G_{π}. Since G_{P,l} ∩ M_{l} = G_{P,l,u,π} = 1 by Lemma 7, G_{P,l,π} is isomorphic to a subgroup of the stabilizer G_{P,l,π}M_{l}/M_{l} of the quad π in G_{l}/M_{l}. This stabilizer corresponds to a subgroup of the group {f(a, 0; α)|0 ≠ a ∈ F_{q^{n}}, α ∈ Gal(F_{q^{n}}/F_{q})} by Lemma 9, in which the stabilizer G_{l,u,π}/M_{l} of two quads corresponds to a subgroup of the group {f(1, 0; α)} of the field automorphisms. Since G_{l,u,π} is of odd order, a Sylow 2-subgroup T of G_{P,l,π}/M_{l} is isomorphic to a subgroup of the normal cyclic group {f(a, 0; id)|0 ≠ a ∈ F_{q^{n}}}. Thus T is cyclic.

Then the lemma follows from a standard result in group theory and the odd order theorem.

It now follows from Proposition 5 in §3 and Lemma 10 above that there is no flagtransitive C_{3}c^{∗}-geometry with residues at elements of type 3 isomorphic to a locally finite thick anomalous C_{3}-geometry. Theorem 2 is proved.

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