Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 37 (1996), No. 2, 231-258.

Flag Transitive c.Af^{∗} Geometries

Barbara Baumeister  Alberto Del Fra
Institut für Algebra und Geometrie  Facoltà di Ingegneria
Martin-Luther-Universität Halle-Wittenberg  Università de L’Aquila
D-06099 Halle/Saale, Germany  I-67040 Roio Poggio (AQ), Italy
Thomas Meixner  Antonio Pasini
Mathematisches Institut  Dipartimento di Matematica
Justus-Liebig-Universität Giessen  Università di Siena
D-35392 Giessen, Germany  I-53100 Siena, Italy

Abstract. In this paper we classify flag-transitive geometries belonging to the following diagram

•  ^{•}  ^{•} c  ^{Af^{∗}}^{q}  ^{q − 1}  ^{(with 1 < q < ∞)}

It turns out that there are only six simply connected examples, with automorphism groups of the following forms: 2^{3}.S_{4}, L_{3}(2), 2^{6}.3.S_{6}, L_{3}(4).2, 2^{4}.S_{6} and 2^{9}.L_{2}(17). Four more examples arise as quotients of some of these.

1. Introduction

In this paper we study c.Af^{∗} geometries of order q < ∞, i.e. geometries of rank 3 belonging to the following diagram

•^{points}  ^{•}  ^{•}^{lines} c  ^{Af^{∗}}^{q}  ^{q − 1}^{planes}

where the first stroke denotes circular geometries of order q, i.e. complete graphs with q + 2 vertices, and the second one denotes dual affine planes of order q. We shall prove that there

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1996 Heldermann Verlag, Berlin

232 B. Baumeister et al.: Flag Transitive c.Af^{∗} Geometries

are precisely ten flag-transitive c.Af^{∗} geometries (three of order 2, five of order 4 and two of order 16). Six of them are simply connected.

Some Terminology and Notation. The reader is referred to [24] for basic notions of diagram geometry. According to [24], all geometries are residually connected, by definition. We denote the incidence relation by ∗. We call the elements of a c.Af^{∗} geometry “points”, “lines” and “planes” according to their types (see the above picture) and we freely use phrases as “the point x is on the plane y”, “the line z passes through the point x” etc. to mean that x ∗ y, x ∗ z etc.

We say that two points (planes) are collinear (dually collinear) if there is a line incident with both of them. We denote the collinearity relation (the dual collinearity relation) by ⊥ (by ⊤ respectively). The collinearity graph (the dual collinearity graph) of a c.Af^{∗} geometry Γ is the graph defined by ⊥ (by ⊤) on the set of points (planes) of Γ.

The diameter of the collinearity (dual collinearity) graph of Γ will be called the (dual) collinearity diameter of Γ.

Some of the geometries we consider satisfy one or both of the following properties (LL) For any two distinct points, there is at most one line incident with both of them. (T) For any three pairwise collinear points, there is at least one plane incident with all of

them.

In the context of this paper, (LL) is equivalent to the Intersection Property (IP) ([24], Lemma 7.25). If (LL) holds in Γ, then the point-line system of Γ and the collinearity graph of Γ are the same.

A c.Af^{∗} geometry Γ is said to be flat if all its points are incident with all planes. Clearly, (T) holds in every flat geometry, but (LL) fails to hold in them. Indeed, any two points of a flat c.Af^{∗} geometry of order q belong to q^{2} planes, each plane contains just one line through them and every line is in q planes. Hence any two points of a flat c.Af^{∗} geometry of order q are in precisely q lines.

2. Examples

2.1. Bi-affine geometries over GF(2)

Given a plane π and a point p of P G(3, 2), let us remove π, p and their residues from P G(3, 2). Then we get a geometry belonging to the diagram Af.Af^{∗}, with orders 1, 2, 1:

• 1 points  •  • lines Af  Af^{∗} 2  1 planes

We call this geometry a bi-affine geometry of order 2 and rank 3, of flag or anti-flag type according to whether p ∈ π or p ∉ π (compare [9]). We denote this geometry by Γ^{ε}(2), with ε = + or − according to whether p ∈ π or p ∉ π. Since

•  ^{•} ^{=} ^{•}  ^{•} c  ^{Af}^{2}  ^{1}  ^{2}

Γ^{ε}(2) is also a c.Af^{∗} geometry. It satisfies (LL) and it is flag-transitive. Aut(Γ^{ε}(2)) is the stabilizer of p and π in the automorphism group L_{4}(2) of P G(3, 2).

Γ^{+}(2) satisfies (T) and Aut(Γ^{+}(2)) ≅ 2^{3}.S_{4} is flag-transitive. We can factorize Γ^{+}(2) by the (unique) elation of center p and axis π. Thus we obtain a flat geometry with the same diagram, on which 2^{2}.S_{4} acts flag-transitively. We denote this quotient by Γ^{+}(2)/2.

On the other hand, Aut(Γ^{−}(2)) ≅ L_{3}(2), flag-transitive. The collinearity graph of Γ^{−}(2) has diameter 1, (T) does not hold in Γ^{−}(2) and no quotient exists.

2.2. Laguerre structures

Given two sets X, Y with |X| = q+2 and |Y | = q, a set F of mappings from X to Y is said to be sharply 3-transitive (of degree (q +2, q)) if, given any triple (x_{1}, x_{2}, x_{3}) of distinct elements of X and any triple (y_{1}, y_{2}, y_{3}) of elements of Y, there is precisely one mapping f ∈ F such that f(x_{i}) = y_{i} for i = 1, 2, 3. Such a set F exists only if q is even [12].

Let Γ be the rank 3 geometry with X×Y as the point-set, F as the set of planes and where the lines are the pairs {(x_{1}, y_{1}), (x_{2}, y_{2})} with x_{i} ∈ X, y_{i} ∈ Y (i = 1, 2) and x_{1} ≠ x_{2}. The incidence relation is defined as follows. A point (x, y) ∈ X × Y (a line {(x_{1}, y_{1}), (x_{2}, y_{2})}) is declared incident with a plane f ∈ F when f(x) = y (respectively, f(x_{1}) = y_{1} and f(x_{2}) = y_{2}). A point (x, y) and a line {(x_{1}, y_{1}), (x_{2}, y_{2})} are declared to be incident when (x, y) ∈ {(x_{1}, y_{1}), (x_{2}, y_{2})}. This geometry is called a special Laguerre 1-structure or a special Laguerre plane ([27], [28]). We will use the word Laguerre structure, for short.

The only examples of Laguerre structures we know are obtained as follows. Let O be a hyperoval in the star of a given point p of P G(3, q), where q = 2^{h} for some integer h. Take as points the points of P G(3, q) different from p and belonging to the q + 2 lines of O, as lines the lines of P G(3, q) meeting precisely two lines of O and as planes those of P G(3, q) not passing through p. The incidence relation is the natural one, inherited from P G(3, q).

Let Γ_{O}(q) be the Laguerre structure obtained in this way. The geometry P G(3, q) can be recovered from Γ_{O}(q) ([14], Theorem 2). Hence Aut(Γ_{O}(q)) is the stabilizer of p and O in P ΓL_{4}(q). Namely, Aut(Γ_{O}(q)) = 2^{h} : Z_{q−1}A where A is the stabilizer of O in P ΓL_{2}(q).

Let π be a plane through p not containing any of the lines of O and let H_{p,π} be the group of elations with center p and axis π. If we factorize Γ_{O}(q) by a subgroup K ≠ 1 of H_{p,π} then we obtain a c.Af^{∗} geometry not satisfying (LL). In particular, when K = H_{p,π} we get a flat geometry.

For instance, Γ_{O}(2) = Γ^{+}(2) and Γ_{O}(2)/H_{p,π} = Γ^{+}(2)/2.

Remark. It is worth describing the above construction of Γ_{O}(q) in the dual way. We start from a dual hyperoval O in the plane at infinity of AG(3, q). The planes of Γ_{O}(q) are the points of AG(3, q) whereas the lines (points) of Γ_{O}(q) are the lines (planes) of AG(3, q) whose point (line) at infinity belongs to (is) a line of O. The incidence relation is inherited from AG(3, q).

2.3. The geometry Γ_{O}(4)

Clearly, Γ_{O}(q) is flag-transitive if and only if the stabilizer A of O in P ΓL_{2}(q) is 2-transitive on the set of points of O. It is well known that A is 2-transitive on O if and only if q = 2 or 4 and that all hyperovals of P G(2, q) (q = 2 or 4) are classical. Therefore, if q = 2 or 4 then Γ_{O}(q) is flag-transitive and the isomorphism type of Γ_{O}(q) does not depend on the particular choice of O. As the case of q = 2 is examined in 2.2, we focus on the case of q = 4 here.

We have Aut(Γ_{O}(4)) = 2^{6}.3.S_{6}. The stabilizer in Aut(Γ_{O}(4)) of a plane of Γ_{O}(4) is 3.S_{6} acting as S_{6} on the six points of that plane. The stabilizer of a point is 2^{4}.3.S_{5}. The stabilizer of a line can be described as ((2^{2}.3) × S_{4}).2 and the stabilizer of a chamber is 3 × S_{4}. The following subgroups of Aut(Γ_{O}(4)) are also flag-transitive on Γ_{O}(4):

2^{6}.3.A_{6}, 2^{6}.3.S_{5}, 2^{6}.(3 × A_{5}), 2^{6}.S_{5}, 2^{6}.A_{5}

Let H = H_{p,π} be as in 2.2. Then H is elementary abelian of order 4, normal in the stabilizer 2^{6}.3.S_{5} of π in Aut(Γ_{O}(4)). Hence Aut(Γ_{O}(4))/H = 2^{4}.3.S_{5} is flag-transitive on the (flat) quotient Γ_{O}(4)/H (but the subgroups 2^{4}.3.A_{5}, 2^{4}.S_{5} and 2^{4}.A_{5} of Aut(Γ_{O}(4))/H also act flagtransitively on Γ_{O}(4)/H). We denote Γ_{O}(4)/H by Γ_{O}(4)/2^{2}. As Γ_{O}(4) is simply connected (Section 4, Corollary 4), every automorphism of Γ_{O}(4)/2^{2} lifts to an automorphism of Γ_{O}(4). Hence Aut(Γ_{O}(4)/2^{2}) = Aut(Γ_{O}(4))/H.

The group H is central in the subgroup X = 2^{6}.A_{5} of the stabilizer Y = 2^{6}.3.S_{5} of π in Aut(Γ_{O}(4)). The factor group Y/X acts as S_{3} on the three 2-subgroups of H. If we factorize Γ_{O}(4) by any of the subgroups of H of order 2 we still get a flag-transitive c.Af^{∗} geometry of order 4. Let us denote this quotient by Γ_{O}(4)/2. As Γ_{O}(4) is simply connected, every automorphism of Γ_{O}(4)/2 lifts to an automorphism of Γ_{O}(4). Therefore Aut(Γ_{O}(4)/2) = 2^{5}.S_{5} (but 2^{5}.A_{5} is also flag-transitive on Γ_{O}(4)/2). Clearly, Γ_{O}(4)/2 neither satisfies (LL) nor is flat.

It is not difficult to check that Γ_{O}(4)/2 and Γ_{O}(4)/2^{2} are the only two flag-transitive proper quotients of Γ_{O}(4).

2.4. Gluings

A 1-factorization of a graph G is a mapping from the set of edges of G to a given set of colours such that every vertex belong to precisely one edge of each colour.

Let G be a complete graph with q + 2 vertices. It is well-known that G admits a 1factorization if and only if q is even. Let q be even. Given a 1-factorization φ of G and an affine plane Π of order q, let α be a bijection from the set of colours of φ to the line at infinity of Π.

Let us take the vertices of G as points and the points of Π as planes. As lines we take the pairs (e, l) where e is an edge of G and l is a line of Π with α(φ(e)) as its points at infinity. Every point is declared to be incident with every plane, whereas the incidence between a point or a plane and a line is defined in the natural way. The geometry obtained in this way is a flat c.Af^{∗} geometry of order q. We call it the gluing of (G, φ) with Π via α.

The flat quotient Γ_{O}(q)/H_{p,π} (notation as in 2.2) is a gluing of the affine plane AG(2, q) with the complete graph with q + 2 vertices. Indeed, by removing the star of p from π we get a model for the dual of AG(2, q). Thus the lines of π through p are the points at infinity of AG(2, q). On the other hand, they can also be taken as colours for the edges of the complete graph having the lines of O as vertices (every pair of such lines uniquely determines a plane through p, which intersects π in a line ...)

In particular, Γ_{O}(4)/2^{2} is a gluing of AG(2, 4) with the complete graph on six vertices, whereas Γ^{+}(2)/2 is a gluing of AG(2, 2) with the complete graph on four vertices.

If we want a flag-transitive gluing, we need to assume q = 2 or 4 (see [4] and recall that q must be a prime power in a flag-transitive affine plane). Since there is a unique way of gluing AG(2, 4) (resp. AG(2, 2)) with the complete graph on six (four) vertices ([2], 6.2), the quotients Γ_{O}(4)/2^{2} and Γ^{+}(2)/2 already described in 2.3 and 2.1 are the only flag-transitive examples that can be obtained by gluing.

2.5. An example for L_{3}(4)

The following example is due to Sprague [26]. Let H be one of the three orbits of L_{3}(4) on the set of hyperovals of P G(2, 4). We take H as the set of planes and the points of P G(2, 4) as points. As lines we take the pairs of distinct points of P G(2, 4). As incidence relation we take the natural one (symmetrized inclusion). Thus, we obtain a c.Af^{∗} geometry of order 4 satisfying (LL) but not (T). We denote it by Γ_{21}(4), where the subscript 21 should remind us of the number of points of this geometry. Γ_{21}(4) is flag-transitive and its full automorphism group is L_{3}(4).2 (but L_{3}(4) also acts flag-transitively on it).

(Clearly, the point-plane system of Γ_{21}(4) can also be described as the design obtained from S(3, 6, 22) by removing a point p and all hexads on it. Compare [26].)

2.6. An example for 2^{4}.S_{6}

The Steiner system S(3, 6, 22) gives rise to another flag-transitive c.Af^{∗} geometry of order 4. Given a hexad E of S(3, 6, 22), take the 6 points of E as points, the hexads intersecting E in two points as lines and the 16 points outside E as planes. All points are declared to be incident with all planes. The incidence relation between lines and points or planes is the natural one, inherited from S(3, 6, 22). We denote this geometry by Γ_{6}(4). It is flagtransitive and its full automorphism group is 2^{4}.S_{6} (but 2^{4}.A_{6}, 2^{4}.P GL_{2}(5) and 2^{4}.L_{2}(5) also act flag-transitively on it).

Clearly, Γ_{6}(4) is flat. However, Γ_{6}(4) ≅ Γ̸_{O}(4)/2^{2}. Indeed, the automorphism groups of these two geometries are different. Thus, Γ_{6}(4) is not a gluing.

2.7. The geometry Γ(16)

The permutation module U = V (18, 2) for the linear group L = L_{2}(17) contains two reducible 9-dimensional L-submodules, say V^{+} and V^{−}, interchanged by an outer automorphism of L_{2}(17). (See [23]; also [17] or [16] (1.4)). They can be described as follows. Regarding P G(1, 17) as the set GF(17) ∪ {∞}, let S be the set of non-zero squares in GF(17). The vectors of U can be viewed as characteristic functions of subsets of P G(1, 17). The submodule V^{+} is generated by the characteristic functions of the images of S under L, whereas V^{−} is generated by the characteristic functions of the images of S ∪ {0, ∞} under L. We have V^{+} ∩ V^{−} = C_{U}(L) = ⟨ζ⟩ (where ζ denotes the characteristic function of P G(1, 17)) and V^{+} + V^{−} is the hyperplane of U consisting of the characteristic functions of subsets of P G(1, 17) of even cardinality. For V = V^{+} or V^{−}, the L-module V/⟨ζ⟩ is irreducible.

Note that, if α is a non-square element of GF(17), then αS is the set of non-square elements of GF(17). Hence P G(1, 17)\αS = S ∪ {0, ∞}. Let g_{α} be the element of P GL_{2}(17) mapping i ∈ P G(1, 17) onto αi (with α∞ = ∞, by convention). Then g_{α} maps the characteristic function of S, which is a vector of V^{+}, onto a vector of V^{−}. Therefore, and since g_{α} normalizes L, it interchanges V^{+} and V^{−}. This makes it clear that the subgroups V^{+}L and V^{−}L of U.L are isomorphic. (However, V^{+} and V^{−} are not isomorphic L-modules, as elements of order 8 have different minimal polynomials on V^{+} and V^{−}.) Henceforth, we set G = V.L with V = V^{+} or V^{−}. (Which of V^{+} or V^{−} we choose as V makes no difference for the sequel.)

A flag-transitive c.Af^{∗} geometry of order 16 can constructed as follows. We take the vectors of V as planes, viewed as functions from P G(1, 17) to GF(2). As points we take the pairs (i, ε) with i ∈ P G(1, 17) and ε ∈ GF(2), declaring a plane u and a point p = (i, ε) to be incident precisely when u(i) = ε. That is, the set of planes incident with p = (i, ε) is the affine hyperplane of V of equation u(i) = ε. It is worth noticing that the hyperplane of V of equation u(i) = 0 is the commutator [V, S_{i}] of the (unique) Sylow 17-subgroup S_{i} of L that fixes i.

We still need to define lines. We shall define them as sets of planes, as follows. Let p_{∞} = (∞, 0) and p_{0} = (0, 0). There are 2^{7} planes incident with both p_{∞} and p_{0}. The stabilizer in L of the unordered pair {∞, 0} is a dihedral group of order 16. Let a be an element of order 8 in that group. Clearly, the centralizer of a^{4} in V_{i} = [V, S_{i}], which is ainvariant, is at least 4-dimensional. However, as a^{4} inverts S_{i} elementwise, a^{4}s is an involution conjugate to a^{4}, for every s ∈ S_{i}. Hence, for s ≠ 1, s = a^{4}(a^{4}s) fixes C_{V_{i}}(a^{4}) ∩ C_{V_{i}}(a^{4}s), which has to be null as s acts irreducibly on V_{i}. Therefore C_{V_{i}}(a^{4}) is 4-dimensional and hence [V, a^{4}] = [V_{i}, a^{4}] = C_{V_{i}}(a^{4}) is also 4-dimensional.

Take as lines of the geometry the images of [V, a^{4}] under the group V L. Clearly, they are just all cosets of certain 4-spaces in V, namely of the spaces [V, t] for t in the conjugacy class of a^{4}. We declare them to be incident to exactly the planes which they contain (as vectors of V). We declare a coset v + [V, t] to be incident with the point (i, ε) when t fixes i and v(i) = ε. Clearly, this definition is independent on the choice of the representative v of v + [V, t], as all vectors of [V, t] are contained in [V, S_{i}] for i fixed by t.

We denote this geometry by Γ(16). We have defined it as a shadow geometry on planes, embedded in AG(9, 2). Clearly, V L is flag-transitive on Γ(16). Let us check that Γ(16) is a c.Af^{∗} geometry of order 16.

By the discussion above, all planes (vectors) incident to (contained in) the line v + [V, t] are also incident to the two points (i, v(i)) and (j, v(j)), where i and j are the two fixed points of t on P G(1, 17). Hence the residue of the line v + [V, t] is a generalized digon. Turning to residues of planes (vectors of V), let us consider the residue of the null vector of V. This plane is incident to all points (i, 0) (i ∈ P G(1, 17)) and to all lines [V, t], t ∈ (a^{4})^{L}. As a 2-point stabilizer in L contains precisely one involution (which is in (a^{4})^{L}), for any 2-set {i, j} in P G(1, 17) there is precisely one line [V, t] incident to the points (i, 0) and (j, 0). Hence the residues of planes are circular spaces with 18 points.

Consider the residue of the point p_{∞} = (∞, 0). The planes (vectors) in its residue make up the 8-dimensional subspace [V, S_{∞}] of V. The lines (cosets of 4-spaces) incident to p_{∞} are the subspaces v + [V, t], with t ∈ (a^{4})^{S_{∞}} and v ∈ [V, S_{∞}]. However, we have proved above that [V, t] ∩ [V, t′] = 0 for t, t′ ∈ (a^{4})^{S_{∞}}. Hence the seventeen 4-spaces [V, t] (t ∈ (a^{4})^{S_{∞}}) form a spread of the 8-space [V, S_{i}]. This shows that the residue of p_{∞} is a dual affine plane of order 16. (In fact, being flag-transitive, it is a copy of AG(2, 16); see [8].)

As we shall see in 7.3, the geometry Γ(16) is simply connected. Furthermore, G = V.L is the full automorphism group of Γ(16) (compare 7.1, Corollary 44). The center Z = ⟨ζ⟩ of G defines a flag-transitive proper quotient of Γ. We denote it by Γ(16)/2. This is the unique flag-transitive proper quotient of Γ.

The points of Γ(16)/2 bijectively correspond to the elements of GF(17)∪{∞}, whereas the planes of Γ(16)/2 are the vectors of W = V/Z. The lines are the cosets of [W, g^{4}] = C_{W} (g^{4}), where g is any element of L of order 8 with two fixed points on GF(17) ∪ {∞}. A point p ∈ GF(17) ∪ {∞} (a plane u ∈ W) and a line v + [W, g^{4}] of Γ(16)/2 are incident if and only if g fixes p (respectively, (u + v)^{g^{4}} = u + v).

Remark. As there are two conjugacy classes of subgroups isomorphic to L_{2}(17) in G = V L (see [23]), the reader might wonder if the isomorphism type of Γ(16) depends on the particular conjugacy class from where the complement L of V has been picked up. But it doesn’t. Indeed the elements of Γ(16) are vectors of V and cosets of suitable subspaces of V of the form [V, X] with X ≤ L. These only depend on the projection XV/V of X onto G/V and not on the particular complement of V which X belongs to.

2.8. Another description of Γ^{+}(2), Γ_{O}(4) and Γ_{O}(4)/2

A description of the same type as the one given for Γ(16) in 2.7 can also be given for Γ^{+}(2), Γ_{O}(4) and Γ_{O}(4)/2.

Let L = P GL_{2}(3) or P GL_{2}(5) and p = 3 or 5, respectively. Then, the 2-point stabilizers of L on P G(1, p) are cyclic and contain a unique involution. Let I denote the class of these involutions.

Let U be the permutation module over GF(2) for L in the action of L on P G(1, p) and let V be the even part of U. The subspace Z spanned by the characteristic function of P G(1, p) is clearly contained in V and V/Z is irreducible. For p = 3 it is a natural module for L_{2}(2) induced by P GL_{2}(3) on V/Z, for p = 5 it is a “natural” module for ΓL_{2}(4) ≅ P GL_{2}(5). In any case, the L-module V is indecomposable.

Let S ∈ Syl_{p}(L) with fixed point i on P G(1, p) and ι ∈ I fixing the letters i and j. The element ι normalizes precisely two Sylow p-subgroups of L, namely S and the conjugate S^{g} of S fixing the letter j. The subspace [V, ι] = [Z, ι]⊕[[V, S], ι] is clearly contained in [V, S]. In fact, it is contained precisely in [V, S] and [V, S]^{g} = [V, S^{g}]. With the same argument as for a^{4} in 2.7, we conclude that dim[V, ι] = (dim[V, S])/2 and the spaces [V, ι], ι ∈ I, ι ∈ N_{L}(S), form a spread in [V, S].

Take the vectors of V as planes, the cosets in V of [V, ι] (ι ∈ I) as lines and the cosets of [V, S^{g}] (g ∈ L) as points and define incidence by symmetrized inclusion. Then we get a c.Af^{∗} geometry whose dual affine planes have order q − 1. Thus, we denote it by Γ(p − 1). The geometry Γ(p−1) admits G = V L as a flag-transitive automorphism group. It turns out that Γ(2) ≅ Γ^{+}(2) and Γ(4) ≅ Γ_{O}(4)/2. In both cases, G is the full automorphism group.

If we perform the same procedure with the group L = L_{2}(17) and its permutation module U over GF(2), there are two L-invariant subspaces V^{+} and V^{−} of dimension (17 + 1)/2 = 9 and we construct the geometry Γ(16) in V = V^{+} or V^{−} (as we have remarked in 2.7, the isomorphism type of Γ(16) does not depend on whether V = V^{+} or V = V^{−}). The same thing happens for the permutation module U over GF(4) for the group L_{2}(5). There are two L-invariant subspaces V^{+} and V^{−} of dimension (5 + 1)/2 = 3 (see [23]) and the construction in V = V^{+} or V = V^{−} gives a geometry isomorphic to Γ_{O}(4) with V L = 2^{6}.A_{5} as a flagtransitive group of automorphisms (but Aut(Γ_{O}(4)) is larger than V L).

2.9. Summary

Here is the list of the flag-transitive examples constructed in the previous paragraphs.

order name  automorphism group description
2 Γ^{+}(2)  2^{3}.S_{4}  2.1. Also 2.8.
2 Γ^{+}(2)/2  2^{2}.S_{4}  2.1
2 Γ^{−}(2)  L_{3}(2)  2.1.
4 Γ_{O}(4)  2^{6}.3.S_{6}  2.3. Also 2.8.
4 Γ_{O}(4)/2  2^{5}.S_{5}  2.3. Also 2.8.
4 Γ_{O}(4)/2^{2}  2^{4}.3.S_{5}  2.3
4 Γ_{21}(4)  L_{3}(4).2  2.5.
4 Γ_{6}(4)  2^{4}.S_{6}  2.6.
16 Γ(16)  2^{9}.L_{2}(17)  2.7. Also 2.8.
16 Γ(16)/2  2^{8}.L_{2}(17)  2.7

3. The main theorem

We shall prove the following:

Theorem 1 There are precisely ten flag-transitive c.Af^{∗} geometries of finite order. They are those listed in 2.9.

The rest of this paper is devoted to the proof of the above theorem. The case of order 2 is settled by the next proposition, which is a special case of the main theorem of [15]:

Proposition 2 There are precisely three c.Af^{∗} geometries of order 2, namely Γ^{+}(2), Γ^{+}(2)/2 and Γ^{−}(2).

Thus, we will only consider c.Af^{∗} geometries of order q > 2. In Section 4 we state some preliminary results, without using flag transitivity. In particular, we obtain that Γ_{O}(4) and Γ_{21}(4) are simply connected and we prove that, if Γ is a non-flat c.Af^{∗} geometry of order q > 2 with collinearity diameter δ_{⊥} = 1, then either Γ ≅ Γ_{21}(4) or q = 18 (Proposition 18). Thus, the flat case and the case with δ_{⊥} = 1 only remain to study. The case of q = 18 is of no interest in this paper. Indeed, in Section 5 we prove that a flag-transitive c.Af^{∗} geometry has order q = 2, 4 or 16.

The case of q = 4 is considered in Section 6. We show that Γ_{6}(4) and Γ_{O}(4)/2^{2} are the unique flat flag-transitive c.Af^{∗} geometries of order 4 and we prove that, if Γ is a flagtransitive c.Af^{∗} geometry of order 4 with collinearity diameter δ_{⊥} > 1, then Γ is a cover of Γ_{O}(4)/2^{2}. Hence Γ = Γ_{O}(4) or Γ_{O}(4)/2, since Γ_{O}(4) is simply connected.

The case of q = 16 is discussed in Section 7. In that section, given a flag-transitive c.Af^{∗} geometry Γ of order 16, we prove that either Γ = Γ(16)/2 or Γ is a double cover of Γ(16)/2. Therefore Γ(16) is simply connected and Γ = Γ(16) or Γ(16)/2.

4. Some preliminary results

In this section Γ is a c.Af^{∗} geometry of order q > 2. We do not assume flag transitivity. We assume q > 2 only because the case of q = 2 is covered by Proposition 2. However, this assumption is not really necessary for the sequel (but it will make some statements a bit shorter).

The following is easy to prove ([24], Chapter 7, Exercise 7.42):

Lemma 3 The geometry Γ is a Laguerre structure if and only if it satisfies both (LL) and (T).

The pair of properties (LL) and (T) is preserved under taking covers. On the other hand, the collinearity diameter of a Laguerre structure is 2. Hence (LL) fails to hold in any proper quotient of a Laguerre structure. Therefore:

Corollary 4 All Laguerre structures are simply connected.

(In particular, Γ_{O}(4) is simply connected.) The next result is due to Sprague [26], [27]:

Proposition 5 If (LL) holds in Γ, then either Γ is a Laguerre structure or Γ ≅ Γ_{21}(4).

(One more case was allowed in [26] and [27], with q = 10. However, that case is impossible since no projective planes of order 10 exist [20]). Laguerre structures of order q and optimal (q + 2, 3)-codes of order q are equivalent objects ([28], Theorem 4). It is well known that there is just one optimal (6, 3)-code of order 4. Therefore:

Proposition 6 If q = 4 and (LL) holds in Γ, then Γ ≅ Γ_{O}(4) or Γ_{21}(4).

Corollary 7 The geometry Γ_{21}(4) is simply connected.

(Obvious, by Proposition 6 and because (LL) is preserved when taking covers.) We need some definitions before to state the next lemmas. Given an element x of Γ, the set of points incident with x will be denoted by σ_{0}(x). We call it the 0-shadow of x. Similarly, the 1shadow σ_{1}(x) and the 2-shadow σ_{2}(x) of x are respectively the set of lines and the set of planes incident with x. We write σ_{i}(x, y) for σ_{i}(x) ∩ σ_{i}(y), σ_{i}(x, y, z) for σ_{i}(x) ∩ σ_{i}(y) ∩ σ_{i}(z) etc.

We denote the residue of an element x by Γ_{x}, as in [24]. When x is a point, Γ_{x} can be viewed both as an affine plane or as dual affine plane, according to whether we take the lines and the planes of Γ_{x} as lines and points or, dually, as points and lines, respectively. We choose the latter point of view. Thus, Γ_{x} is an affine plane. We denote its dual by Γ_{x}^{∗}.

We denote the collinearity graph (the dual collinearity graph) of Γ by G_{⊥}(Γ) (by G_{⊤}(Γ)) and the collinearity diameter (dual collinearity diameter) by δ_{⊥} (by δ_{⊤}). Given two points a and b, we denote their distance in G_{⊥}(Γ) by δ(a, b). Given a point a, the neighbourhood of a in G_{⊥}(Γ) is denoted by a^{⊥}.

Two lines (or a line and a point) are said to be coplanar if there is a plane incident with both of them.

Lemma 8 Given a point a of Γ, a line x on a and a plane u incident with a but not with x, the line x is coplanar with all but one lines of u on a.

(Obvious, since Γ_{a}^{∗} is an affine plane.)

Lemma 9 We have δ_{⊥} ≤ 2.

Proof. Let a and d points at distance 3, if possible, and let (a, b, c, d) be a path of G_{⊥}(Γ) from a to d. Let x, y, z be lines joining a with b, b with c and c with d, respectively. Let us take a plane u on y and a point e of u other than b or c and let y_{b}, y_{c} be the lines of u joining e with b and c, respectively. The lines x and y are not coplanar, as a and c are at distance 2. Hence x is coplanar with y_{b} by Lemma 7. Therefore a ⊥ e. Similarly, e ⊥ d. Hence δ(a, d) ≤ 2, a contradiction. q.e.d.

Lemma 10 We have |σ_{0}(u) \ a^{⊥}| ≤ 1 for every plane u and every point a.

Proof. The statement is trivial if a ∗ u. Let a ∉ σ_{0}(u) and assume a^{⊥} ⊉ σ_{0}(u). Let b ∈ σ_{0}(u) \ a^{⊥}. We have δ(a, b) = 2 by Lemma 9. Let c ∈ a^{⊥} ∩ b^{⊥} and let x be a line on c and b. The line x is coplanar with all but one lines of u on b, by Lemma 8. Hence there are a line y of u on b and a plane v incident with both x and y. Let z be a line joining c with a. By Lemma 8, z is coplanar with all lines of v on c but one. However, z and x are not coplanar, since a ⊥̸ b. Therefore z is coplanar with the line of v joining c with the point d of y other than b. Hence a ⊥ d. Let x′ be a line joining a with d. By Lemma 8, x′ is coplanar with all lines of u on d but one. Clearly, y is the only line of u on d non-coplanar with x′. Therefore a is collinear with all points of u but b. q.e.d.

Given two points a and b of Γ, let us write aΦb to mean that a ⊥̸b or a = b.

Lemma 11 Φ is an equivalence relation.

Proof. Let aΦbΦc and a⊥c, if possible. Let u be a plane on a and c. By Lemma 10, b is non-collinear with at most one point of u, thus it is collinear with at least one of a and c, contrary the hypothesis that aΦbΦc. q.e.d.

Given two distinct points a, b we denote by ν_{a,b} the number of lines joining a with b.

Lemma 12 The number ν_{a,b} does not depend on the choice of the collinear pair (a, b).

Proof. Let a, b, c be a triple of coplanar points, let u be a plane on a, b, c and let x_{a}, x_{b}, x_{c} be the lines of u joining b with c, c with a and a with b, respectively. Let z be any other line on a and b. By Lemma 8, z is coplanar with x_{a}, since it cannot be coplanar with x_{c}. Let v be the plane on z and x_{a} and let f(z) be the line of v joining a with c. We have f(z) ≠ x_{b}, otherwise u and v are two different planes on c with two distinct lines x_{a}, x_{b} in common, contrary to the fact that Γ_{c}^{∗} is an affine plane.

It is now clear that, if we also set f(x_{c}) = x_{b}, then we get an injective mapping from the set of lines on a and b to the set of lines on a and c. Therefore ν_{a,b} ≤ ν_{a,c}. Similarly, ν_{a,c} ≤ ν_{a,b}. Hence ν_{a,b} = ν_{a,c}. The conclusion follows from the connectedness of Γ. q.e.d.

We write ν for ν_{a,b} and we call ν the collinearity index of Γ. Clearly ν ≤ q, since pointresidues are dual affine planes of order q and two lines of Γ with the same points cannot be coplanar. Obviously we also have ν ≥ 1, and ν = 1 if and only (LL) holds.

Lemma 13 The geometry Γ is flat if and only if ν = q.

Proof. Let ν = q. Then two lines on a point a have the same points if and only if they are parallel in Γ_{a}^{∗}. Therefore |a^{⊥}| = q +1 and any two points of a^{⊥} are collinear since they belong to non-parallel lines of Γ_{a}^{∗}, i. e. to coplanar lines of Γ. Hence G_{⊥}(Γ) is a complete graph and Γ has q + 2 points, namely it is flat. Conversely, if Γ is flat, then |σ_{2}(a, b)| = q^{2} for any two distinct points a, b. Hence ν = q. q.e.d.

Given a point a in Γ, an easy computation shows that |a^{⊥}| = (q + q^{2})/ν. Denoted by φ the size of the equivalence class of a with respect to the relation Φ (see Lemma 11), the point-set of Γ has size φ+(q +q^{2})/ν. Hence φ does not depend on the choice of the point a. Therefore:

Lemma 14 The graph G_{⊥}(Γ) is a complete t-partite graph with classes of size φ and

t = 1 + \frac{q + q^{2}}{φν}

(Note that φ might be 1. In that case G_{⊥}(Γ) is a complete graph, that is δ_{⊥} = 1). It follows from the above that φν divides q + q^{2}. Moreover:

Lemma 15 The number νφ divides q.

Proof. Let a be a point of Γ. As G_{⊥}(Γ) is a complete t-partite graph, a^{⊥} is a join of classes of Φ. The relation “to join a with points of the same class of Φ” is an equivalence relation on the set of lines through a. It is in fact a refinement of the parallelism of Γ_{a}^{∗} and its equivalence classes have size νφ. Parallelism classes in Γ_{a}^{∗} have size q. Hence νφ divides q. q.e.d.

Lemma 16 One of the following holds:

(i) φν = q; (ii) ν = φ = 1 and q = 4; (iii) φ = 1, ν = 3 and q = 18.

Proof. Since Γ has φ + (q + q^{2})/ν points, the number of planes of Γ is

\frac{(φ + \frac{q+q^{2}}{ν})q^{2}}{q + 2}

Therefore q+2 divides q^{2}(φν+q+q^{2}). The greatest common divisor of q+2 and q^{2} is a divisor of 4. Consequently, q + 2 divides 4(φν + q + q^{2}). That is, q + 2 divides 4φν + 8. Moreover, νφ divides q by Lemma 15. Consequently, mφν + 2 divides 4φν + 8, with m = q/φν. It is straightforward to prove that the latter condition holds only if m and φν are as in one of the following cases:

(1) m = 1;

(2) m = 3 and νφ = 2 (hence q = 6);

(3) m = 4 and νφ = 1 (hence q = 4);

(4) m = 6 and νφ = 3 (hence q = 18);

(5) m = 7 and νφ = 2 (hence q = 14).

Cases (2) and (5) are impossible by the well known Bruck-Ryser condition for the existence of a projective plane of order q. In case (4) we might have either ν = 3 and φ = 1 or ν = 1 and φ = 3. However, if ν = 1, then Γ is a Laguerre structure by Proposition 5. The collinearity graph of a Laguerre structure of order q is a complete (q + 2)-partite graph with all classes of size q. Therefore the case of ν = 1, φ = 3 and q = 18 is impossible. The statement is proved. q.e.d.

Corollary 17 If δ_{⊥} = 2, then G_{⊥}(Γ) is a complete (q + 2)-partite graph and the 0-shadows of the planes are maximal cliques of G_{⊥}(Γ).

(Easy, by Lemmas 14 and 16.)

Proposition 18 If δ_{⊥} = 1, then one of the following holds:

(i) Γ is flat;

(ii) Γ ≅ Γ_{21}(4);

(iii) q = 18 and ν = 3.

Proof. We have δ_{⊥} = 1 if and only if φ = 1. Furthermore, (LL) holds if and only if ν = 1. The statement follows from Lemma 16 and Proposition 5. q.e.d.

Lemma 19 If Γ does not satisfy (LL) and q ≠ 18, then |σ_{2}(a, x)| = ν for any non-incident point-line pair (a, x) with σ_{0}(x) ⊆ a^{⊥}.

Proof. Let (LL) fail to hold (i.e. ν > 1) and q ≠ 18. Then q = νφ by Lemma 16. If δ_{⊥} = 1, then Γ is flat by Proposition 18. Hence ν = q by Lemma 13 and the statement follows.

Let δ_{⊥} = 2 (compare Lemma 9) and let (a, x) be a non-incident point-line pair with σ_{0}(x) ⊆ a^{⊥}. Let {b, c} = σ_{0}(x). Every plane incident with x and a uniquely determines a line on a and b. Conversely, given any line y on a and b, the lines x and y are non-parallel in Γ_{b}^{∗} since a ⊥ c and q = νφ. Hence they are coplanar in Γ. There are ν lines on a and b. The statement is now clear. q.e.d.

Lemma 20 We have |σ_{0}(u, v)| = 2|σ_{1}(u, v)| for any two distinct planes u, v.

Proof. Let x, y be distinct lines incident with both u, v. Then x and y have no points in common. Indeed, if a were a common point of x and y, then x, y would be distinct lines of Γ_{a}^{∗} meeting in two distinct points u, v, but this is impossible. On the other hand, any point a incident with both u and v is incident with precisely one line x ∈ σ_{1}(u, v) (namey the line of Γ_{a}^{∗} through the points u and v). Whence the statement.q.e.d.

Proposition 21 We have q ≡ 0 (mod 2).

Proof. When q = 18 there is nothing to prove. Thus, let q ≠ 18.

If (LL) holds (that is, ν = 1), then the statement follows from Proposition 5 (recall that Laguerre structures have even order [12]). If q = ν (that is, if Γ is flat), then q ≡ 0 (mod 2) by Lemma 20.

Let 1 < ν < q. Given a plane u and two points a, b of it, let a′, b′ be points distinct from a and b and such that a ⊥̸ a′ and b ⊥̸ b′ (they exist by Lemma 14). By Lemma 14, we have a′ ⊥ b′ and every point of u different from a and b is collinear with both a′ and b′. Let x be a line on a′ and b′. For every point c ∈ σ_{0}(u) \ {a, b}, there are ν planes of σ_{2}(x, c) by Lemma 19. Hence, νq is the number of point-plane flags (c, v) with c ∈ σ_{0}(u) \ {a, b} and v ∗ x. However, we can count these flags in another way too, by choosing a plane v on x and counting the points incident with both u and v. Therefore qν = ∑_{v∈σ_{2}(x)} |σ_{0}(u, v)|. However, the number |σ_{0}(u, v)| is even (possibly 0) for every v ∈ σ_{2}(x), by Lemma 20. Hence φν is even. On the other hand, q = φν by Lemma 16. Therefore q is even. q.e.d.

5. Feasible orders

From now G is a flag-transitive subgroup of Aut(Γ), with Γ a c.Af^{∗} geometry of order q. Given an element x of Γ, we denote by G_{x} the stabilizer of x in G and by K_{x} the elementwise stabilizer of Γ_{x}. We set \overline{G}_{x} = G_{x}/K_{x}. That is, \overline{G}_{x} is the action of G_{x} on Γ_{x}. Given a flag (x, y), we set G_{x,y} = G_{x} ∩ G_{y}, we denote by K_{x,y} the elementwise stabilizer of the residue of (x, y) and we set \overline{G}_{x,y} = G_{x,y}/K_{x,y}.

As in the previous section, given a point p of Γ we denote by Γ_{p}^{∗} the dual of Γ_{p}. We shall prove the following,

Proposition 22 We have q = 2, 4 or 16 and the affine plane Γ_{p}^{∗} is desarguesian. Furthermore, when q = 16, given a point p and a plane u of Γ we have L_{2}(17) ≤ \overline{G}_{u} ≤ P GL_{2}(17) and \overline{G}_{p} ≤ AΓL_{1}(2^{8}).

Proof. Since the order of a flag-transitive affine plane is a prime power [7] and q is even by Proposition 21, we have q = 2^{h} for some h ≥ 1. It is well known that for q = 2 or 4 there is a unique affine plane of order q, which is desarguesian. Thus, we may assume q > 4, that is h > 2.

Given a plane u of Γ, the flag transitivity of G on Γ implies the 2-transitivity of \overline{G}_{u} on the q + 2 points of Γ_{u}. By comparing the list of 2-transitive groups ([3]; also [21]) and recalling that q is a power of 2, it is not difficult to prove that the following are the only possibilities for \overline{G}_{u}:

(1.a)  A_{q+2} ≤ \overline{G}_{u} ≤ S_{q+2}

(1.b)  L_{n}(s) ≤ \overline{G}_{u} ≤ P ΓL_{n}(s)

with

q + 2 = \frac{s^{n} − 1}{s − 1}

in (1.b), for some prime power s and some integer n > 1. On the other hand, given a point p of u, Γ_{p,u} is the bundle of lines of Γ_{p}^{∗} on u, which is a point of the affine plane Γ_{p}^{∗}. Therefore,

(2)  |\overline{G}_{u,p}| ≤ |P ΓL_{2}(q)|

(see [1], for instance). Our proof will go on as follows. By comparing (2) and (1.a), (1.b) we will get some control over \overline{G}_{u,p}. Then we shall turn to \overline{G}_{p} and we will obtain the conclusion.

In case (1.a) we have \overline{G}_{u,p} ≥ A_{q+1}. Therefore (2^{h} + 1)! ≤ 2^{h+1}(2^{2h} − 1)h, by (2) and by the fact that q = 2^{h}. The latter inequality forces h ≤ 2, contrary to the assumption h > 2. Thus, (1.a) is ruled out and (1.b) holds.

Let s, n be as in (1.b). As 2^{h} + 2 = (s^{n} − 1)/(s − 1), s is odd, n is even and

(3)  2^{h} = \frac{s^{n} − 2s + 1}{s − 1}

The stabilizer of p in the subgroup L_{n}(s) of \overline{G}_{u} is E : GL_{n−1}(s)/Z, with Z the group of n^{th}-roots of unit in GF(s) and E elementary abelian of order s^{n−1}. Clearly, |Z| ≤ n. By (2) we get the following:

(4)  s^{n(n−1)/2} ∏_{i=1}^{n−1}(s^{i} − 1) ≤ n2^{h}(2^{2h} − 1)h

Let n ≥ 6. Then

∏_{i=1}^{n−1}(s^{i} − 1) = (s − 1) ∏_{i=2}^{n−1}(s^{i} − 1) > (s − 1) ∏_{i=2}^{n−1} s^{i−1} =

= (s − 1) ∏_{i=1}^{n−2} s^{i} = (s − 1)s^{(n−2)(n−1)/2}

Comparing this and (4) we get

(s − 1)s^{(n−1)^{2}} < n2^{h}(2^{2h} − 1)h

By substituting (3) in the above inequality we obtain the following

(s − 1)s^{(n−1)^{2}} < nh\frac{(s^{n} − 2s + 1)(s^{n} − 3s + 2)(s^{n} − s)}{(s − 1)^{3}}

Hence

(5)  (5)(s − 1)^{4}s^{(n−1)^{2}−3n} < nh

On the other hand, h < n log_{2} s, by (3). Since s is odd and > 1, we have log_{2} s < s − 1. Therefore h < n(s − 1). Comparing (5) we obtain

(s − 1)^{3}s^{(n−1)^{2}−3n} < n^{2}

However, this inequality does not hold for n ≥ 6 and s ≥ 3. Therefore n < 6. As n is even, n = 2 or 4. Assume n = 4. Then

2^{h} = s^{3} + s^{2} + s − 1

by (3). As s is odd, the right side of the above equality is congruent to 2 or 6 modulo 8. Hence h = 1 and s = 1, contradiction.

Therefore n = 2. Also, s = 2^{h} + 1 by (3) and q = 2^{h}. On the other hand, s is a prime power. Thus, by a classical result of number theory [11] either s is prime or s = 9. In the first case we have

(6.a)  Z_{s} : Z_{(s−1)/2} ≤ \overline{G}_{p,u} ≤ Z_{s} : Z_{s−1}

whereas

(6.b)  3^{2} : Z_{4} ≤ \overline{G}_{p,u} ≤ (3^{2} : Z_{8}).2

in the latter case.

Now we turn to Γ_{p}^{∗}. By [1] one of the following holds:

(7.a)  \overline{G}_{p,u} ≅ Suz(2^{h/2}).Z_{h/2} with Γ_{p}^{∗} a Lüneburg plane

(7.b)  \overline{G}_{p,u} ≤ ΓL_{1}(2^{2h})

(7.c)  \overline{G}_{p,u} ≤ P ΓL_{2}(2^{h}) with Γ_{p}^{∗} a desarguesian affine plane

(note that (7.b) and (7.c) are not incompatible).

Assume (7.a). Then h ≡ 2 (mod 4) and h ≥ 6. Hence (6.a) holds for \overline{G}_{u,p}. According to that, Suz(2^{h/2}) should contain a cyclic group of order 2^{h}. This is false. Indeed the largest cyclic subgroups of a Sylow 2-subgroup of a Suzuki group have order 4. Thus, (7.a) is ruled out.

Let (7.b) hold. By combining (7.b) with any of (6.a) or (6.b) we get that 2^{h−1}(2^{h} + 1) divides (2^{2h} − 1)2h. This forces h = 4 (recall that h > 2 by assumption). Therefore Γ_{p}^{∗} = AG(2, 16), since this is the unique flag-transitive affine plane of order 16 (see [8]). Furthermore, \overline{G}_{u} and \overline{G}_{p} are as in the second claim of the proposition, by (1.b) and (7.b).

In order to finish the proof we only need to show that (7.c) forces h = 4 (whence G_{p} ≤ AΓL_{1}(2^{8}) by (6.a)). Assume (7.c) and let s = 9, as in (6.b). Then \overline{G}_{p,u} ≥ Z_{4}. Hence P ΓL_{2}(8) contains Z_{4}. However this is false (the Sylow 2-subgroup of P ΓL_{2}(8) is an elementary abelian subgroup of order 8). Thus, (6.b) is ruled out. Therefore (6.a) holds and h ≠ 3.

By (6.a), the Sylow 2-subgroup of \overline{G}_{p,u} contains a cyclic group of order 2^{h−1}, whereas cyclic subgroups of the Sylow 2-subgroup of P ΓL_{2}(2^{h}) have order at most 2^{r+1}, where r is the exponent of the maximal power of 2 that divides h (possibly, r = 0). Consequently h − 1 ≤ r + 1, that is r ≥ h − 2. On the other hand, 2^{r} ≤ h. Therefore h ≤ 4. Hence h = 4, as h ≠ 3 by the above and h > 2 by assumption. q.e.d.

6. The case of q = 4

Let q = 4. The geometry Γ is assumed to be flag-transitive in this section, but we will exploit this hypothesis only in a very few places.

The case with (LL) is classified in Proposition 6. Thus, we assume that (LL) fails to hold in Γ, namely ν > 1. Therefore ν = 2 or 4, since ν divides q (Lemma 16). Note that Γ is flat when ν = 4 (Lemma 13). In 6.2 we prove that, if Γ is flat then Γ is isomorphic to either Γ_{O}(4)/2^{2} or Γ_{6}(4). We consider the case of ν = 2 in 6.3, proving that if ν = 2 then Γ is a cover of Γ_{O}(4)/2^{2}. On the other hand, Γ_{O}(4) is simply connected by Corollary 4. Hence Γ is a quotient of Γ_{O}(4). In fact, Γ ≅ Γ_{O}(4)/2 as ν = 2.

6.1. More definitions and notation

We say that two distinct lines overlap if they meet in two points (i.e., they have the same points). Clearly, “being overlapping or equal” is an equivalence relation on the set of lines of Γ. We denote it by ∥_{0}. All equivalence classes of ∥_{0} have size ν. Furthermore, overlapping lines have no planes in common. Thus, if x ∥_{0} y and a ∈ σ_{0}(x) = σ_{0}(y), then x and y are parallel lines of the affine plane Γ_{a}^{∗}. Given two lines x, y, we write x ∥_{2} y to mean that σ_{2}(x) = σ_{2}(y). This is also an equivalence relation. We shall denote the set of points of Γ by P.

6.2. The flat case

Let ν = 4. Then Γ is flat (Lemma 13). We can assume that P is the set {1, 2, 3, 4, 5, 6} of the first six positive integers. The number of planes of Γ is 16 and the number of lines is 60.

If u and v are distinct planes of Γ, then σ_{1}(u, v) consists of three lines (Lemma 20), which partition P (= σ_{0}(u) = σ_{0}(v)) in three pairs.

Lemma 23 Given a line x of Γ, let u_{1}, u_{2}, u_{3}, u_{4} be the four planes on x. If |σ_{1}(u_{1}, u_{2}, u_{3})| = 3 then σ_{1}(u_{1}, u_{2}, u_{3}) ⊆ σ_{1}(u_{4}).

Proof. Let σ_{1}(u_{1}, u_{2}, u_{3}) = {x, y, z}. By flag transitivity, some g ∈ G_{x} maps u_{3} onto u_{4}. We have |g({u_{1}, u_{2}, u_{3}}) ∩ {u_{1}, u_{2}, u_{3}}| = 2. Let

{u, u′} = g({u_{1}, u_{2}, u_{3}}) ∩ {u_{1}, u_{2}, u_{3}}

Then σ_{1}(u_{4}, u, u′) = σ_{1}(u, u′). However, σ_{1}(u, u′) = {x, y, z} because u and u′ are two of the planes u_{1}, u_{2}, u_{3} and σ_{1}(u_{i}, u_{j}) = {x, y, z} for i, j = 1, 2, 3, i ≠ j. Thus u_{4} is incident with all of x, y, z. q.e.d.

Lemma 24 Given a line x of Γ, let u_{1}, u_{2}, u_{3} be three of the four planes on x. If |σ_{1}(u_{1}, u_{2}, u_{3})| ≥ 2 then |σ_{1}(u_{1}, u_{2}, u_{3})| = 3.

Proof. Let σ_{1}(u_{1}, u_{2}, u_{3}) ⊇ {x, y} and let z_{i} be the third line of σ_{1}(u_{i}, u_{3}), for i = 1, 2. The condition σ_{0}(z_{i}) = P \ (σ_{0}(x) ∪ σ_{0}(y)) uniquely determines z_{i} in σ_{1}(u_{3}). Hence z_{1} = z_{2}. q.e.d.

Lemma 25 Given a line x of Γ, let u_{1}, u_{2}, u_{3}, u_{4} be the four planes on x. One of the following occurs

(i) |σ_{1}(u_{1}, u_{2}, u_{3}, u_{4})| = 3;

(ii) σ_{1}(u_{i}, u_{j}, u_{k}) = {x} for any 3-subset {i, j, k} of {1, 2, 3, 4}.

Moreover, if one of the above cases occurs for some line x, then the same holds for every line x.

(Obvious, by Lemmas 24 and 23 and flag transitivity). By this lemma, either all classes of ∥_{2} have size 3 (case (i) of the lemma) or ∥_{2} is the identity (case (ii)).

6.2.1. Case (i)

Assume that the classes of ∥_{2} have size 3.

We denote by ∥_{0} · ∥_{2} the product of the relations ∥_{0} and ∥_{2}, in this order (that is, x ∥_{0} · ∥_{2} y for two lines x, y when x ∥_{0} z ∥_{2} y for some line z). We say that ∥_{0} and ∥_{2} commute if ∥_{0} · ∥_{2}=∥_{2} · ∥_{0}.

Lemma 26 The relations ∥_{0} and ∥_{2} commute.

Proof. Given two overlapping lines x and x′, let u_{1}, u_{2}, u_{3}, u_{4} be the four planes incident with x and let u′_{1}, u′_{2}, u′_{3}, u′_{4} be those incident with x′. Let σ_{1}(u_{1}, u_{2}, u_{3}, u_{4}) = {x, y, z} and σ_{1}(u′_{1}, u′_{2}, u′_{3}, u′_{4}) = {x′, y′, z′}. We need to prove that y′ overlaps y or z.

Suppose that y′ does not overlap any of y or z. Then the same holds for z′. We can assume without loss that σ_{0}(x) = σ_{0}(x′) = {1, 2}, σ_{0}(y) = {3, 4} and σ_{0}(z) = {5, 6}. Therefore x′, y′, z′ determine one of the following two partitions:

{{1, 2}, {3, 5}, {4, 6}}, {{1, 2}, {3, 6}, {4, 5}}

We can assume without loss that they determine the first partition. Given i = 1, 2, 3 or 4, we denote by P_{i} the partition of P formed by the three lines of σ_{1}(u_{1}, u′_{i}). None of the pairs {1, 2}, {3, 4} and {5, 6} is a member of P_{i} because u_{1}, u_{2}, u_{3}, u_{4} are the four planes incident with each of x, y and z and none of them is equal to any of u′_{1}, u′_{2}, u′_{3}, u′_{4} (indeed these are the planes on x′, and x′ overlaps x). Similarly, neither {3, 5} nor {4, 6} is a class of P_{i}. Thus, the following partitions only remain to consider:

\begin{matrix}\{\{1, 3\}, \{2, 6\}, \{4, 5\}\}, & \{\{1, 6\}, \{2, 3\}, \{4, 5\}\}, \\ \{\{1, 4\}, \{2, 5\}, \{3, 6\}\}, & \{\{1, 5\}, \{2, 4\}, \{3, 6\}\}. \end{matrix}

Furthermore, P_{i} ≠ P_{j} if i ≠ j, otherwise σ_{1}(u_{1}, u′_{i}) = σ_{1}(u_{1}, u′_{j}) = σ_{1}(u′_{i}, u′_{j}) = {x′, y′, z′}, whereas x′ ∉ σ_{1}(u_{1}). Thus, we can assume that

\begin{matrix}P_{1} = \{\{1, 3\}, \{2, 6\}, \{4, 5\}\}, & P_{2} = \{\{1, 6\}, \{2, 3\}, \{4, 5\}\}, \\ P_{3} = \{\{1, 4\}, \{2, 5\}, \{3, 6\}\}, & P_{4} = \{\{1, 5\}, \{2, 4\}, \{3, 6\}\}. \end{matrix}

Consequently, the line of u_{1} incident with the points 4 and 5 is incident with both u′_{1} and u′_{2}. However, this is impossible. Indeed σ_{1}(u′_{1}, u′_{2}) = {x′, y′, z′}, but none of these lines has the pair {4, 5} as its 0-shadow. q.e.d.

By Lemma 26, the product ∥_{0} · ∥_{2} of the relations ∥_{0} and ∥_{2} is the equivalence relation generated by them. Let us denote it by ∥. The classes of ∥ have size 12. Hence ∥ has precisely five classes. Each class of ∥ uniquely determines a partition of P in three pairs. The five partitions of P obtained in this way form a 1-factorization of the complete graph that has P as the set of vertices. Furthermore, given a point a, two distinct lines x, y on a overlap if and only if they are parallel in the affine plane Γ_{a}^{∗} (indeed the classes of ∥_{0} have size 4).

It is now clear that Γ is a gluing of AG(2, 4) with the complete graph on six vertices (compare 2.4). As we have remarked at the end of 2.4, there is a unique way of gluing AG(2, 4) with the complete graph on six vertices and that gluing is precisely Γ_{O}(4)/2^{2}. Thus,

Proposition 27 If ∥_{2} has classes of size 3, then Γ ≅ Γ_{O}(4)/2^{2}.

6.2.2. Case (ii)

We now assume that ∥_{2} is the identity. Chosen a point a of Γ, we define a rank 2 geometry Π as follows. The points of Π are the five points of Γ other than a and the planes of Γ. The lines of Π are the 20 lines of Γ on a and a symbol ∞. The incidence relation of Π is the one inherited from Γ, with the additional clause that ∞ is declared to be incident with precisely the points of Γ other than a.

Clearly, if we drop ∞ and its points from Π then we get Γ_{a}^{∗}, which is AG(2, 4). Therefore Π = P G(2, 4).

For every line x of Γ not on a we set O_{x} = σ_{0}(x) ∪ σ_{2}(x). We denote by O the family of these sets.

Lemma 28 For every line x of Γ not on a, the set O_{x} is a hyperoval of the projective plane Π.

Proof. Clearly, |O_{x}| = 6. We have σ_{1}(u, v, w) = {x} for any three distinct elements u, v, w ∈ O_{x} because ∥_{2} is the identity relation. Therefore u, v, w cannot be collinear in Π. q.e.d.

Lemma 29 We have O_{x} ≠ O_{y} for any two distinct lines x, y of Γ not on a.

Proof. If O_{x} = O_{y}, then σ_{2}(x) = σ_{2}(y). Hence x = y because ∥_{2} is the identity relation. q.e.d.

It is well known that there are 168 hyperovals in Π = P G(2, 4) and that the relation “to intersect in an even number of points” (actually, 0, 2 or 6) is an equivalence relation on that set of hyperovals. The relation splits the set into three classes, say O_{1}, O_{2}, O_{3}, each of size 56.

Lemma 30 The family O is contained in one of the classes O_{1}, O_{2} or O_{3}.

Proof. Let x, y be two distinct lines of Γ not on a and let z ∈ O_{x} ∩ O_{y}. This element might be a point or a plane of Γ.

Let z be a point of Γ, that is z ∈ σ_{0}(x) ∩ σ_{0}(y). If x and y have the same points, then they cannot be incident with any common plane. Therefore O_{x} ∩ O_{y} = σ_{0}(x) = σ_{0}(y), and this is a pair of points. On the other hand, let z be the only point incident with both x and y. Then x and y are not parallel in the affine plane Γ_{z}^{∗}, since ν = 4. This forces σ_{2}(x) ∩ σ_{2}(y) to be nonempty. However, |σ_{2}(x) ∩ σ_{2}(y)| = 1 as x and y have a point in common. Therefore |O_{x} ∩ O_{y}| = 2.

Assume now that x and y have no points in common. Hence z is a plane of Γ. As ∥_{2} is the identity relation, the three planes on x determine on σ_{0}(z) three distinct partitions in pairs and the pair σ_{0}(x) belongs to each of these partitions. It is clear that only one of these partitions contains σ_{0}(y). Therefore only one of the three planes on x other than z is incident with y. This shows that |O_{x} ∩ O_{y}| = 2. q.e.d.

Assume O ⊆ O_{1}, to fix ideas. Since Γ contains precisely 40 lines not on a, we have |O| = 40. It is well known that every line of P G(2, 4) meets 40 hyperovals in each of the classes. On the other hand, every hyperoval of O meets ∞. Therefore, O is the family of the hyperovals of O_{1} intersecting ∞.

Let S be the rank 2 geometry that has a and the points of Π as points and the lines of Π and the hyperovals of Π as blocks, with the incidence relation inherited from Π and the additional clause a ∗ ∞. Then S is a model of S(3, 6, 22). It is now clear that Γ consists of the six points of the hexad ∞, the 16 points of Π not on ∞ (these are taken as planes) and the 40 hexads of S intersecting the hexad ∞ in two points. That is,

Proposition 31 We have Γ ≅ Γ_{6}(4).

By Propositions 27 and 31, Γ_{O}(4)/2^{2} and Γ_{6}(4) are the only flat flag-transitive c.Af^{∗} geometries of order 4.

6.3. The case of ν = 2

Let ν = 2. Hence φ = 2, by Lemma 16. That is, every point is non-collinear with precisely one point. Given a point a, we denote by π(a) the unique point non-collinear with it. We call π(a) the antipodal point of a.

Since φ = 2, we have δ_{⊥} = 2 and |P | = 12. Therefore Γ has 32 planes and 120 lines. Two lines incident with two distinct planes have no points in common. Hence two distinct

planes are incident with at most three common lines. We say that two planes u, v are i-secant if |σ_{1}(u, v)| = i. Given a line-plane flag (x, u), let n_{i} be the number of planes incident with x and i-secant with u. Note that the numbers n_{1}, n_{2}, n_{3} do not depend on the choice of the flag (x, u), by flag transitivity.

We denote by m_{i} the number of planes i-secant with u. By flag transitivity, the numbers m_{0}, m_{1}, m_{2}, m_{3} do not depend on the choice of the plane u. Clearly, m_{0}+m_{1}+m_{2}+m_{3} = 31, since 32 is the number of planes of Γ.

Lemma 32 We have m_{3} = n_{3} = 0.

Proof. Since any two planes on the same point a meet in a line on a, two planes are 3secant if and only if they have the same set of points. Thus, “being 3-secant or equal” is an equivalence relation. Let us denote it by ∥^{∗}. Given a point a, its stabilizer G_{a} in G preserves the equivalence relation induced by ∥^{∗} on the set of planes on a. However, G_{a} acts primitively on the set of planes of Γ_{a} ([7], 2.3.7). Therefore either all planes on a belong to the same class of ∥^{∗}, or no two of them correspond in ∥^{∗}. In the latter case m_{3} = 0 (hence n_{3} = 0).

Let the first case occur, if possible. By connectedness, any two planes of Γ correspond in ∥^{∗}. That is, all planes of Γ are incident with the same set of points. Namely, Γ is flat, contrary to the hypothesis that ν = 2. q.e.d.

Lemma 33 We have m_{1} = m_{2} = 15, n_{1} = 1, n_{2} = 2.

Proof. By counting the flags (v, y) with y a line of u and v a plane i-secant with u, we get m_{1} = 15n_{1} and 2m_{2} = 15n_{2}. Clearly, n_{1} + n_{2} + n_{3} = 3. Hence n_{1} + n_{2} = 3 because n_{3} = 0 (Lemma 32). Therefore m_{1} +2m_{2} = 45. Hence either m_{1} = m_{2} = 15 or m_{1} = 45 and m_{2} = 0. In the latter case, any two planes have at most one line in common. This forces (LL) to hold in Γ, by Lemma 7.25 of [24] in the dual of Γ. However, (LL) fails to hold in Γ, as we have assumed ν = 2. Therefore m_{1} = m_{2} = 15. q.e.d.

Lemma 34 We have δ_{⊤} = 2 and m_{0} = 1.

Proof. By Lemma 14, given any two planes u, v, some points of u are collinear with some points of v. Let x be a line joining a point a of u with a point b of v and let w be a plane on x. A line incident with u and w can be found in Γ_{a}. In Γ_{b} we find a line incident with w and v. Hence u and v have distance ≤ 2 in the dual collinearity graph of Γ. Therefore δ_{⊤} ≤ 2.

On the other hand, m_{0} + m_{1} + m_{2} = 31. Hence m_{0} = 1, by Lemma 33. Consequently, δ_{⊤} = 2. q.e.d.

As m_{0} = 1, for every plane u there is precisely one plane that is not dually collinear with u. We call it the antipodal plane of u and we denote it by π(u).

Lemma 35 Given a plane u, the planes that are 1-secant with u are 2-secant with π(u), whereas those that are 2-secant with u are 1-secant with π(u).

Proof. Let v be a plane 2-secant with both u and π(u). Then

|σ_{0}(u, v)| = |σ_{0}(π(u), v)| = 4

Hence σ_{0}(u, π(u), v) ≠ ∅, which is impossible because σ_{0}(u, π(u)) = ∅. Therefore no planes are 2-secant with both u and π(u). The conclusion follows from Lemma 33. q.e.d.

Lemma 36 The function π mapping every point and every plane onto its antipodal is an automorphism of the point-plane system of Γ and it commutes with every automorphism of Γ.

Proof. Clearly, π^{2} is the identity. Whence π is bijective. It is also clear that it commutes with every automorphism of Γ.

By contradiction, assume that π is not an automorphism of the point-plane system of Γ. Then π(a) ∉ σ_{0}(π(u)) for some point-plane flag (a, u). Then π(b) ∉ σ_{0}(π(u)) for every point b ∗ u, by flag transitivity. That is, every point of u is collinear with all points of π(u).

Let N be the number of 5-tuples (b, c, x, y, v) with b a point of u, c a point of π(u), v a plane incident with both b and c, x the line on b incident with both u and v, and y the line on c incident with both v and π(u). Since all points of u are collinear with all points of π(u), there are precisely 36 choices for the pair (b, c). Given one of these pairs, there are eight planes incident with both b and c. If v is one of them, the lines x and y of Γ_{b} and Γ_{c} joining v with u and π(u) respectively are uniquely determined. Therefore N = 36 · 8.

However, we can count N in the following way, too. There are 30 planes distinct from u and π(u). By Lemma 35, fifteen of these planes are 1-secant with u and 2-secant with π(u). The remaining ones are 2-secant with u and 1-secant with π(u). Hence there are 30 · 2 triples (x, y, v) with v a plane different from u and π(u) and x, y lines of v incident with u and π(u), respectively. Each of these lines has two points. Therefore N = 30 · 4. However, we have previously obtained N = 36 · 8. The contradiction is reached. q.e.d.

Lemma 37 Given a line-plane flag (x, u), let a, b be the two points of x and let v be another plane on x. Then π(v) is incident with the line of π(u) that is incident with π(a) and π(b).

Proof. Both π(a) and π(b) are incident with both π(u) and π(v), by Lemma 36. Let y be the line of π(u) incident with π(a) and π(b). By contradiction, assume that y ∉ σ_{1}(π(v)).

Then π(v) and π(u) are 2-secant and σ_{1}(π(u), π(v)) consists of two lines y_{a}, y_{b} containing π(a) and π(b), respectively. By Lemma 35, π(u) is 1-secant with v, as it is 2-secant with π(v). Therefore v is 2-secant with u, by Lemma 35. Let a′, b′, a′′, b′′ be the points of σ_{0}(u) \ σ_{0}(x), with a′, b′ ∈ σ_{0}(u, v). We have σ_{0}(π(u), π(v)) = {π(a), π(b), π(a′), π(b′)} by Lemma 36, and we can assume that {π(a), π(a′)} = σ_{0}(y_{a}) and {π(b), π(b′)} = σ_{0}(y_{b}).

As \overline{G}_{u} is 2-transitive on the six points of u, it contains L_{2}(5), acting on those points as on P G(1, 5). Thus, there is an element ḡ ∈ \overline{G}_{u} fixing both a and b and acting with two orbits of size 2 on the remaining four points of u.

Let u′, v′ be the two planes on x different from both u and v. By Lemma 33, one of them is 1-secant with u and 2-secant with v, the other one is 2-secant with u and 1-secant with v. We can assume that u′ and v′ are 1-secant with u and with v, respectively. Then σ_{0}(u, v′) = {a′′, b′′}.

Let g be an element of G_{u} inducing ḡ on σ_{0}(u). Clearly, g fixes u′ and either it fixes both v and v′ or it permutes them. In the latter case ḡ permutes the pairs {a′, b′} and {a′′, b′′}. Assume that this is the case. In the affine plane Γ_{a}^{∗} = AG(2, 4), u, u′, v, v′ are the four points of the line x. As g fixes two of them, it acts as a semilinear transformation on x. Therefore the action of \overline{G}_{a} on the line at infinity of the affine plane Γ_{a}^{∗} contains P ΓL_{2}(4) ≅ P GL_{2}(5). Hence \overline{G}_{u} ≥ P GL_{2}(5). Consequently, the stabilizer of a and b in \overline{G}_{u} contains some element f̄ of order 4 such that f̄^{2} = ḡ. Let f be an element of G_{u} inducing f̄ on σ_{0}(u). Clearly, f fixes u′ and permutes v and v′. Hence f̄ interchanges the pairs {a′, b′} and {a′′, b′′}. However, this forces ḡ to fix both those pairs, contrary to the hypothesis that g permutes v and v′.

Therefore g fixes both v and v′ and permutes a′ with b′ and a′′ with b′′. It also fixes π(a), π(b), π(u) and π(v) and permutes π(a′) with π(b′), since g commutes with π. However, this forces g to map the lines y_{a}, y_{b} onto lines different from both y_{a} and y_{b}. Thus, g cannot fix π(v). We have reached the contradiction. q.e.d.

Given a line x, let a, b be its two points and let u be any of the four planes on x. We denote by π(x) the line of π(u) joining π(a) with π(b). By Lemma 37, this definition is consistent. Thus we have extended π to an automorphism of Γ. Clearly, π^{2} = 1 and π belongs to the center of Aut(Γ). It is also clear that π defines a quotient of Γ. We denote this quotient by Γ/π.

Proposition 38 We have Γ/π ≅ Γ_{O}(4)/2^{2}.

Proof. As Γ is flag-transitive and π belongs to the center of Aut(Γ), the geometry Γ/π is also flag-transitive. Clearly, it is flat. We need to prove that Γ/π is as in 6.2.1. Then Proposition 27 will give us the conclusion.

Given a plane u of Γ and a line x of it, let v, v′ be the two planes on x that are 2secant with u. Let a, b, a′, b′, a′′, b′′ be the points of u. We can assume that {a, b} = σ_{0}(x), {a, b, a′, b′} = σ_{0}(u, v) and {a, b, a′′, b′′} = σ_{0}(u, v′). Let x′, x′′ be the lines of u joining a′ with b′ and a′′ with b′′, respectively.

The plane π(v) is 2-secant with π(u), whence it is 1-secant with u, by Lemma 35. Obviously, σ_{0}(u, π(v)) = σ_{0}(u) \ σ_{0}(u, v) = {a′′, b′′}. Therefore {x, x′} = σ_{1}(u, v) and {x′′} = σ_{1}(u, π(v)). Similarly, {x, x′′} = σ_{1}(u, v′) and {x′} = σ_{1}(u, π(v′)). Hence the images of u, v, v′ in Γ/π are incident with the images of x, x′ and x′′. Thus, Γ/π is as in 6.2.1. q.e.d.

Corollary 39 We have Γ ≅ Γ_{O}(4)/2.

Proof. The Laguerre structure Γ_{O}(4) is the universal cover of Γ_{O}(4)/2^{2}, by Corollary 4. Hence Γ is a quotient of Γ_{O}(4), by Proposition 38 and by the definition of universal covers. On the other hand, Γ_{O}(4)/2 is the only flag-transitive non-flat proper quotient of Γ_{O}(4). Therefore Γ ≅ Γ_{O}(4)/2. q.e.d.

7. The case of q = 16

Let q = 16. As in Section 5, G stands for a flag-transitive subgroup of Aut(Γ).

7.1. Preliminaries

Choose a chamber (p, l, u) of Γ, with p, l and u a point, a line and a plane, respectively. By Proposition 22,

(1)  L_{2}(17) ≤ \overline{G}_{u} ≤ L_{2}(17).2 (= P GL_{2}(17))

and \overline{G}_{p} ≤ AΓL_{1}(2^{8}). On the other hand, 2^{8} : Z_{17} ≤ \overline{G}_{p} by flag-transitivity. Therefore,

(2)  2^{8} : Z_{17} ≤ \overline{G}_{p} ≤ AΓL_{1}(2^{8}) (= (2^{8} : (Z_{17} × Z_{5} × Z_{3})).Z_{8})

Lemma 40 We have K_{p,u} = K_{u}.

Proof. Clearly, K_{u} ≤ K_{p,u}. On the other hand, the subgroup K_{p,u} fixes p, u and all lines of u on p. Hence it fixes all points of u. Therefore K_{p,u} ≤ K_{u}. q.e.d.

Let (\overline{G}_{u})_{p} be the stabilizer of p in \overline{G}_{u}. We have Z_{17}.Z_{8} ≤ (\overline{G}_{u})_{p} ≤ Z_{17}.Z_{16} by (1). Furthermore (\overline{G}_{u})_{p} = \overline{G}_{p,u} by Lemma 40. Recalling that \overline{G}_{p,u} is a factor of (\overline{G}_{p})_{u} and comparing (2), we obtain the following,

(3)  (\overline{G}_{u})_{p} = Z_{17}.Z_{8} (= \overline{G}_{u,p})

Hence

(4.1)  \overline{G}_{u} = L_{2}(17)
(4.2)  Z_{17}.Z_{8} ≤ (\overline{G}_{p})_{u} ≤ ΓL_{1}(2^{8}) (= (Z_{17} × Z_{5} × Z_{3}).Z_{8})

Lemma 41 We have K_{p} = 1.

Proof. Let g ∈ K_{p}. Then g fixes all points of p^{⊥}. Let q be one of them. Given a line m joining p with q, let v be a plane on m. Since g ∈ K_{p}, g fixes v and all lines of v on p. Hence it fixes all points of v. Consequently, it fixes all lines of v. However, v is any of the planes on m. Therefore g fixes m, all planes on m and all lines incident with any of those planes. It is now clear that g acts trivially on Γ_{q}. That is, g ∈ K_{q}. Hence K_{p} ≤ K_{q}. Conversely, K_{q} ≤ K_{p}. Therefore K_{p} = K_{q}. By connectedness, K_{x} = K_{p} for every point x of Γ. This forces K_{p} = 1. q.e.d.

Lemma 42 We have K_{u} = 1.

Proof. Since K_{u} = K_{u,p}, we have K_{u} ≤ Z_{5} × Z_{3} by (3), (4.2) and Lemma 41. Thus, K_{u} = 1, Z_{3}, Z_{5} or Z_{5} × Z_{3}. On the other hand, G_{u} = K_{u}.L_{2}(17) by (4.1). As L_{2}(17) is simple, this extension must be central. The Schur multiplier of L_{2}(17) is 2 (see [5]). Hence the extension K_{u}.L_{2}(17) is a direct product, G_{u} = K_{u} × L_{2}(17).

This forces K_{u} to be central in G_{p,u}. However G_{p,u} = K_{u}.Z_{17}.Z_{8} < ΓL_{1}(2^{8}) by (3) and Lemma 41 and this group does not contain any central subgroup of order 3, 5 or 15. Therefore K_{u} = 1. q.e.d.

Corollary 43 We have K_{p,u} = 1.

(Easy, by Lemmas 40 and 42.) By Lemmas 41 and 42, by Corollary 43 and by (3), (4.1) and (2) we get the following:

(5.1)  G_{u} = L_{2}(17)

(5.2)  G_{p} = 2^{8} : Z_{17}.Z_{8} ≤ AΓL_{1}(2^{8})

Furthermore G_{l} = G_{p,l}G_{l,u} = G_{l,u}G_{p,l}, with

(5.3)  G_{p,l} = 2^{4} : Z_{8}, G_{l,u} = Z_{8}.2, and

(5.4)  Z_{8} = G_{p,l,u} (= G_{p,l} ∩ G_{l,u}).

Therefore,

(5.5)  G_{l} = 2^{4} : Z_{8}.Z_{2}.

Corollary 44 We have G = Aut(Γ).

(Indeed (5.1) leaves just one possibility for G_{u}.)

7.2. The structure of G

We have 16 = νφ by Lemma 16 and the collinearity graph G_{⊥}(Γ) of Γ is a complete 18-partite graph with all classes of size φ (Lemma 14). Thus, φ = 2^{t} for some t ∈ {0, 1, 2, 3, 4} and Γ has 18φ points and (18φ · 2^{8})/18 = 2^{8}φ = 2^{8+t} planes.

Let X_{1}, X_{2},..., X_{18} be the eighteen classes of G_{⊥}(Γ) and let V be the kernel of the action of G on X = {X_{i}}_{i=1}^{18}.

Lemma 45 The group V acts regularly on the set of planes of Γ.

Proof. By Corollary 17, every plane of Γ is incident with one point in every class of X. This fact and Lemma 42 imply that V acts semi-regularly on the set of planes of Γ. Let O_{1}, O_{2},... O_{k} be the orbits of V on the set of planes of Γ.

Given a point p, assume that the set σ_{2}(p) of the planes of Γ on p intersects at least two of O_{1}, O_{2},..., O_{k} in at least two elements. Since V is normal in G, it is normalized by G_{p}. Hence the orbits of V intersecting σ_{2}(p) non-trivially are imprimitivity classes for G_{p} in its action on σ_{2}(p), which is the set of the points of the affine plane Γ_{p}^{∗}. However, this is impossible ([7], 2.3.7). Therefore, either σ_{2}(p) intersects every orbit of V in at most one point or it is contained in one orbit of V.

Assume the first case, that is |σ_{2}(p) ∩ O_{i}| ≤ 1 for i = 1, 2, ...k. Then G_{p} ∩ V = 1. Therefore, G_{p} acts faithfully on the eighteen classes of X. On the other hand, G_{p} contains a normal subgroup W of order 2^{8} and an element b of order 17. Clearly, W stabilizes the class of X containing p. On the other hand, b permutes the classes of X fixed by W, since it normalizes W. However, b has order 17, stabilizes the class of X containing p and cyclically permutes the remaining classes of X. Therefore the class containing p is the unique member of X that is stabilized by W. However, this is impossible, because W is a 2-group and X has even size.

Therefore σ_{2}(p) is contained in one orbit of V. For instance, σ_{2}(p) ⊆ O_{1}. Then, given a point p′ collinear with p, the planes incident with p and p′ belong to O_{1}. Hence σ_{2}(p′) ⊆ O_{1}. By connectedness, all planes of Γ belong to O_{1}. That is, V is transitive (whence regular) on the set of planes of Γ. q.e.d.

Corollary 46 We have G = V : G_{u} (split extension) and |V | ≤ 2^{8+t}.

(Easy, by Lemma 45 and because 2^{8+t} is the number of planes of Γ.) Henceforth we set V_{p} = G_{p} ∩ V, for short.

Corollary 47 We have V_{p} = O_{2}(G_{p}) = 2^{8} and V acts transitively on each of the classes of X.

Proof. Let X be the class of X containing the point p. We have |X| = φ and |V | = 2^{8+t} by Lemma 45. Therefore |V_{p}| ≥ 2^{8}, with |V_{p}| = 2^{8} if and only if V is transitive on X. On the other hand, V_{p} is normal in G_{p} and O_{2}(G_{p}) = 2^{8} by (5.2). Therefore V_{p} = O_{2}(G_{p}). q.e.d.

We have G_{p,l,u} = Z_{8} by (5.4). Let a be a generator of G_{p,l,u}. Then

Corollary 48 We have G_{l} = [V_{p}, a^{4}]⟨a⟩⟨c⟩ for some involution c ∈ G_{l,u}\G_{p,l,u}.

Proof. We have G_{p,l} = [O_{2}(G_{p}), a^{4}]⟨a⟩ by (5.3). On the other hand, |G_{l} : G_{p,l}| = 2 (see (5.5)) and G_{l,u} contains some involutions c permuting the two points of l. Corollary 47 gives us the conclusion. q.e.d.

Lemma 49 We have V = V_{p}V_{p′} for any two collinear points p and p′.

Proof. Clearly, V_{p} ∩ V_{p′} is in the center of V, as V_{p} and V_{p′} generate V. Let p, p′ be distinct collinear points. By Corollary 47, the groups V_{p} and V_{p′} act regularly on σ_{2}(p) and on σ_{2}(p′), respectively. Therefore, given two planes u, u′ ∈ σ_{2}(p) ∩ σ_{2}(p′), there are elements g ∈ V_{p} and g′ ∈ V_{p′} mapping u onto u′. However, g = g′ because V is regular on the set of planes of Γ. Therefore V_{p} ∩ V_{p′} is regular on σ_{2}(p) ∩ σ_{2}(p′). On the other hand, |σ_{2}(p) ∩ σ_{2}(p′)| = 2^{4}ν, since there are ν lines incident with both p and p′ and every line is incident with 2^{4} planes. Hence |V_{p} ∩ V_{p′}| = 2^{4}ν. Therefore,

|V_{p}V_{p′}| = \frac{2^{16}}{2^{4}ν} = 2^{8+t} = |V |

Hence V = V_{p}V_{p′}. q.e.d.

Lemma 50 The group V is elementary abelian.

Proof. As V is a 2-group, we have Z(V) ≠ 1. Assume Z(V) ≠ V and let O_{1}, O_{2},..., O_{k} be the orbits of Z(V) on the set of planes of Γ. As in the proof of Lemma 45, one can prove that σ_{2}(p) meets each of O_{1}, O_{2},..., O_{k} in at most one element. Therefore Z(V) ∩ G_{p} = 1.

Let p′ be a another point, collinear with p. We have V_{p} ∩V_{p′} ≤ Z(V). Indeed both V_{p} and V_{p′} are abelian by Corollary 47 and they generate V by Lemma 49. Hence Z(V) ∩ G_{p} ≠ 1, contrary to the conclusion that we have previously obtained from the hypothesis that Z(V) ≠ V. Therefore V is abelian. It is also elementary abelian. Indeed V_{p} and V_{p′} generate V and all their elements other than 1 are involutions. q.e.d.

By Lemmas 45 and 50 we obtain the following:

Corollary 51 Let m = 8 + t. Then V is an m-dimensional GF(2)-module for G_{u} = L_{2}(17).

Lemma 52 One of the following holds:

(i) We have t = 0 (that is, V is 8-dimensional ), Z(G) = 1, the G_{u}-module V is irreducible and Γ is flat.

(ii) We have t = 1 (hence V is 9-dimensional ), Z(G) is a 1-dimensional G_{u}-submodule of V and V/Z(G) is irreducible; furthermore, V = Z(G) × V_{p} for any point p of Γ (but V is an indecomposable module for G_{u}).

Proof. The module V is at most 12-dimensional, by Corollary 51. Therefore, and by [17], a G_{u}-invariant section of V is either trivial or isomorphic to one of the two irreducible 8dimensional modules called V^{+} and V^{−} in 2.7. By Lemma 49, V = V_{p}V_{p′} with p and p′ distinct collinear points. Hence V is indecomposable with central submodule Z = Z(G), we have V = V_{p} × Z and V/Z is an 8-dimensional irreducible module for G_{u}. (We keep the multiplicative notation for V here, thus writing V_{p} × Z instead of V_{p} ⊕ Z.)

By (5.4), given a chamber (p, l, u) of Γ, we have G_{p,l,u} = ⟨a⟩ with a^{8} = 1. Let p′ be the point of l other than p. As V = V_{p} × Z = V_{p′} × Z, we have [V, a] = [V_{p}, a] = [V_{p′}, a]. Hence [V, a] ≤ V_{p} ∩ V_{p′}. However, |C_{V_{p}}(a)| = 2 by (5.2). Hence [V_{p}, a] ≅ V_{p}/C_{V_{p}}(a) is 7dimensional. Therefore V_{p} ∩V_{p′} is at least 7-dimensional. Consequently, V = V_{p}V_{p′} is at most 9-dimensional. That is, either dim(V) = 8 and Z = 1 or dim(V) = 9 and Z = Z_{2}.

If dim(V) = 8 (that is, t = 0), then Γ has 2^{8} planes, whence it is flat. We are in case (i). When dim(V) = 9 we have case (ii). q.e.d.

Corollary 53 There are at most 2^{9} planes in Γ.

(Easy, by the previous lemma and Lemma 45.)

7.3. End of the proof

By Lemma 52, either Z(G) = 1 and Γ is flat or V = V_{p} × Z(G) with Z(G) = Z_{2}.

Proposition 54 If Z(G) = 1, then Γ = Γ(16)/2.

Proof. Let Z(G) = 1. By Lemma 52(i), V is an irreducible 8-dimensional module for G_{u} = L_{2}(17). This description uniquely determines G (see 2.7). The planes of Γ bijectively correspond to the vectors of V and G_{u} acts on the eighteen points of Γ as L = L_{2}(17) does on P G(1, 17). The module V is in fact a section of the permutation module of L. We have G_{p} = V : G_{p,u}, G_{p,u} = Z_{17}Z_{8} and G_{p,l,u} = Z_{8}. Let a be a generator of G_{p,l,u} and let c be an involution of G_{l,u} permuting the two points of l. Then G_{l} = [W, a^{4}]⟨a⟩⟨c⟩ by Corollary 48. By comparing 2.7, one can easily see that Γ ≅ Γ(16)/2. q.e.d.

Proposition 55 If Z(G) = Z_{2}, then Γ = Γ(16).

Proof. Let Z = Z(G) = Z_{2}. We have Z ∩ G_{u} = 1 as V is regular on the set of planes. Furthermore, Z ∩ G_{p} = 1 as V = V_{p} × Z. Also, Z stabilizes all classes of X, since V is just the elementwise stabilizer of X. This and the fact that Z ∩ G_{p} = 1 imply that Z ∩ G_{l} = 1 for any line l. Thus, Z defines a quotient of Γ. Clearly, Γ/Z is flat and G/Z is flag-transitive on Γ/Z. Therefore Γ/Z ≅ Γ(16)/2 by Proposition 57.

On the other hand, Γ(16) is the universal cover of Γ(16)/2. (Indeed, a proper cover of Γ(16) would have more than 2^{9} planes, contrary to Corollary 53.) Therefore Γ is a quotient of Γ(16). However, Γ has as many planes as Γ(16). Hence Γ = Γ(16). q.e.d.

Acknowledgments. The authors wish to thank G. Stroth, D. Pasechnik and C. Huybrechts. Stroth and Pasechnik have helped us in the investigation of the case of q = 16 when, at an earlier stage of our work, we wanted to settle that case by computer-aided computations. Huybrechts has given us many suggestions for sections 2, 4 and 5.

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Received March 17, 1995