A computer free construction of J_{4}

A.A. Ivanov

U. Meierfrankenfeld

Abstract Define a finite simple group J to be of J_{4}-type (or simply J_{4}) provided that J contains an involution z with

C_{J}(z) ∼ 2_{+}^{1+12}3Aut Mat_{22}.

The purpose of this paper is to give the first computer free construction of a group of J_{4}- type. In addition we achieve yet another uniqueness proof for groups of J_{4}-type via the simple connectedness of the 2-local geometry of such a group.

1 Introduction

Initial evidence for the existence of groups of J_{4}-type was given by Z. Janko in [10]. He has shown that the order o(J_{4}) of such a group is

86, 775, 571, 046, 077, 562, 880 = 2^{21} · 3^{3} · 5 · 7 · 11^{3} · 23 · 29 · 31 · 37 · 43,

determined its conjugacy classes and much of the p-local structure. J. Conway, S. Norton, J. Thompson and D. Hunt used this information to determine the character table of J_{4} and, in particular, proved the existence of an irreducible (irrational) complex character of degree 1333. Looking at the 2-modular reduction of this character J. Thompson conjectured the existence of an irreducible 112-dimensional representation of J_{4} over GF (2). Based on this conjecture S. Norton with the help of D. Benson, J. Conway, R. Parker and J. Thackray constructed J_{4} as a subgroup of GL_{112}(2). Their construction is outlined in [13], discussed in more detail in [3] and depends on the use of a computer. In [12], W. Lempken gave explicit generators for J_{4} as a subgroup of GL_{1333}(11). The proof that the group generated is in fact J_{4} relies on its existence.

Definition 1.1 Let I be a set of size n. A (finite) amalgam of rank n ( over I) is a tuple (A; M_{i}, i ∈ I; ∗_{i}, i ∈ I) where A is a finite set, M_{i} is subset of A and ∗_{i} is binary operation defined on M_{i} so that the following conditions hold:

(i) (M_{i}, ∗_{i}) is a group for every i ∈ I;

(ii) A = ∪_{i∈I}M_{i};

(iii) ∩_{i∈I}M_{i} ≠Ø;

(iv) if x, y ∈ M_{i} ∩ M_{j} for i, j ∈ I then x ∗_{i} y = x ∗_{j} y.

We will write (M_{i} | i ∈ I) for the amalgam A as above (since A = ∪_{i∈I}M_{i}, there is no need to refer to A explicitly). Whenever x and y are in the same M_{i} there product x ∗_{i} y is defined and it is independent of the choice of i. We will normally write this product simply by xy. Since B := ∩_{i∈I}M_{i} is non-empty, one can easily see that B contains the identity element of (M_{i}, ∗_{i}) for every i ∈ I. Moreover, these identity elements must be equal. The reader may notice that a more common definition of amalgams in terms of morphisms is essentially equivalent to the above one. For J ⊆ I we put M_{J} = ∩_{i∈J}M_{i}. We will write, for instance M_{ijk} instead of M_{{i,j,k}} and consider M_{ij} as a subgroup in M_{i} and M_{ji} as a subgroup of M_{j}. An amalgam of rank 3 will also be called a triangle of groups. The isomorphism of amalgams is defined in the obvious way. Let (M, ∗) be a group, {M_{i} | i ∈ I} be a family of subgroups in M and ∗_{i} be the restriction of ∗ to M_{i}. Then (M_{i} | i ∈ I) is an amalgam. This is the most important example of amalgam but it is not difficult to construct an amalgam which is not isomorphic to a family of subgroups in a group.

Definition 1.2 A group M is said to be a completion of an amalgam (M_{i} | i ∈ I) if there is a mapping φ of ∪_{i∈I}M_{i} into M such that

(i) M is generated by the image of φ;

(ii) for every i ∈ I the restriction of φ to M_{i} is a group homomorphism with respect to ∗_{i} and the group operation in M.

If φ is injective then the completion M is said to be faithful.

Definition 1.3 A triangle (M_{1}, M_{2}, M_{3}) of groups is called a J_{4}-triangle provided that

(i) M_{1} is the semidirect product of the Mathieu group Mat_{24} of degree 24 and the 11-dimensional Todd module;

(ii) M_{2} is the semidirect product of L_{5}(2) and the exterior square of a natural module of L_{5}(2);

(iii) |O_{2}(M_{3})| = 2^{15} and M_{3}/O_{2}(M_{3}) ≅ Sym(5) × L_{3}(2);

(iv) |M_{2}/M_{21}| = 31, |M_{3}/M_{31}| = 5, |M_{3}/M_{32}| = 10 and |M_{23}/B| = 3.

It was shown in [10] (cf. Theorem A (4), (6), (9)) that every group of J_{4}-type is a faithful completion of a J_{4}-triangle of groups. This and the existence of the complex character of degree 1333 serve as motivation for our construction of J_{4}. The principal steps are as follows:

Step 1: Show that there exists a J_{4}-triangle of groups.

Step 2: Show that GL_{1333}(C) contains a faithful completion of a J_{4}-triangle of groups.

Step 3: Show that any faithful completion of a J_{4}-triangle of groups is a group of J_{4}-type.

Steps 1 and 2 are realized in Lemma 5.9 and Theorem 7.1, respectively. Step 3 was done in [2] (as the main step in the uniqueness proof for J_{4}) and independently in [8]. Both these proofs were achieved by establishing the simple connectedness of the 2-local geometry of J_{4}; hence rely on the existence of J_{4} and do not suit our purposes. In order to establish the existence of J_{4} we need to carry out Step 3 without assuming that J_{4} exists. In this form Step 3 is realized in Section 8 (cf. Theorem 8.26) and the proof is necessarily more complicated than the proofs in [2] and [8]. In particular our proof uses extremely detailed information about the 2-local geometries of Mat_{24} and Mat_{22}.

Although we do not need the uniqueness of J_{4} to establish its existence, we include a uniqueness proof since it can be achieved with only little extra effort. Namely, we prove in Lemma 5.7 that any two J_{4}-triangles are isomorphic and within the realization of Step 3 (cf. Theorem 8.26) we show that every faithful completion of a J_{4}-triangle is finite and that its order is equal to o(J_{4}). Since every completion of an amalgam is a quotient of the universal completion, this immediately implies that the unique J_{4}-triangle of groups has a unique faithful completion, namely the universal one.

2 Preliminaries

Our notation concerning groups is mostly standard. The symmetric, alternating and Mathieu group of degree n are denoted by Sym(n), Alt(n) and Mat_{n}, respectively. By writing G ∼ A_{1}A_{2}...A_{n} we mean that G has a normal series

1 ⊲ G_{1} ⊲ ... ⊲ G_{n} = G

such that G_{i}/G_{i−1} ≅ A_{i}. We write p^{a} for a p-group of order p^{a}; p^{a_{1}+a_{2}+...+a_{n}} for p^{a_{1}}p^{a_{2}}...p^{a_{n}} and 2_{ε}^{1+2n} for the extraspecial group of order 2^{1+2n} and of type ε ∈ {+, −}. Throughout the paper 3 · Alt(6) denotes the non-split extension of Alt(6) by a centre of order 3 and 3 · Sym(6) stands for the extension of 3 · Alt(6) by an outer automorphism which induces a transposition on the Alt(6)- quotient. Given a subgroup H in a group G we denote by G/H the set of right cosets of H in G.

Definition 2.1 Let G be a group, K ⊴ G, Ḡ = G/K and V a GF (2)G-module of dimension n with kernel K. Then

(i) if Ḡ ≅ L_{n}(2) then V is called a natural L_{n}(2)-module for G;

(ii) if Ḡ ≅ Ω_{n}(2) and G fixes a non-degenerate quadratic form of plus type on V then V is called a natural Ω_{n}^{+}(2)-module for G;

(iii) if Ḡ ≅ Sym(5), n = 4 and G′ preserves a GF (4)-structure on V, then V is called a natural ΓL_{2}(4)-module for G;

(iv) if Ḡ′ ≅ 3 · Alt(6) and n = 6, then V is called a hexacode module for G.

The module dual to V will be denoted by V^{∗}.

Notation 2.2 Let (M_{i} | i ∈ I) be an amalgam. Then for i, j ∈ I we put Q_{i} = O_{2}(M_{i}), Q_{i}^{∗} = O_{2,3}(M_{i}), Z_{i} = Z(Q_{i}) and T_{ij} = O_{2}(M_{ij}).

Definition 2.3 Let I be a set of size n and let Γ be an undirected n-partite graph without loops, whose parts are Γ_{i}, i ∈ I. This means that if a ∈ Γ_{i} is adjacent to b ∈ Γ_{j} then i ≠ j. Let d the usual distance function on Γ. Then

(i) if a ∈ Γ_{i}, then a is said to be of type i;

(ii) a path of type n_{1} − n_{2} − … − n_{k} is a tuple (a_{1}, a_{2}, …, a_{k}) of vertices in Γ such that a_{i} is of type n_{i} and a_{i} is adjacent to a_{i+1}; we denote such a path by

n_{1}^{a_{1}} − n_{2}^{a_{2}} − … − n_{k}^{a_{k}};

(iii) a non-degenerate path (or nd-path) is a path (a_{1}, a_{2}, …, a_{k}) such that a_{i−1} is neither equal nor adjacent to a_{i+1};

(iv) let Λ ⊆ Γ and a_{1}, a_{2}, …, a_{n} ∈ Γ, then

Λ(a_{1}a_{2} … a_{n}) = {b ∈ Λ | b is adjacent to a_{i} for all 1 ≤ i ≤ n}.

Definition 2.4 Let M be a group and (M_{i} | i ∈ I) a tuple of subgroups of M.

(i) The coset graph Γ = Γ(M; M_{i} | i ∈ I) is the graph with vertex set the disjoint union of the sets Γ_{i} = M/M_{i}, i ∈ I and where two vertices are adjacent if they are distinct and have non-empty intersection. Note that the Γ_{i} are parts of Γ and that M acts on Γ by right multiplication.

(ii) A flag in Γ is a set of pairwise adjacent vertices. The type of a flag is the set of types of its elements.

(iii) Let a ∈ Γ. Then Γ(a) is the graph whose vertices are the neighbours of a in Γ and two vertices are adjacent in Γ(a) if and only if they are adjacent in Γ.

(iv) Let a ∈ Γ. Then the graph Γ^{∗}(a) on the neighbours of a in Γ is defined as follows. Assume without loss that a = M_{i}. Let b, c be adjacent to a, where b = M_{j}r and c = M_{k}s with r, s ∈ M_{i}. Then b is adjacent to c in Γ^{∗}(a) if and only if j ≠ k and M_{ij}r ∩ M_{ik}s ≠Ø.

(v) Γ is called geometric if for all a ∈ Γ the graphs Γ(a) and Γ^{∗}(a) are equal.

(vi) If a, b, c, ... are vertices of Γ, then M_{abc...} denotes their elementwise stabilizer in M. If a ∈ Γ then Q_{a} = O_{2}(M_{a}), Q_{a}^{∗} = O_{2,3}(M_{a}) and Z_{a} = Z(Q_{a}).

(vii) Let a, b, c ∈ Γ. Then ^{}a̸bc = |c^{M_{ab}}|.

We remark that if b, c are adjacent in Γ^{∗}(a) then they are also adjacent in Γ(a). Furthermore, Γ^{∗}(M_{i}) is isomorphic to Γ(M_{i}; M_{ij} | j ∈ I \ {i}).

Lemma 2.5 Let Γ be as in 2.4.

(i) Let {a_{i}, a_{j}, a_{k}} be a flag in Γ where a_{l} is of type l. Then the following statements are equivalent:

(a) a_{x} and a_{y} are adjacent in Γ^{∗}(a_{z}) for every z ∈ {i, j, k} with {x, y, z} = {i, j, k};

(b) a_{x} and a_{y} are adjacent in Γ^{∗}(a_{z}) for some z ∈ {i, j, k} with {x, y, z} = {i, j, k};

(c) a_{i} ∩ a_{j} ∩ a_{k} ≠Ø;

(d) (a_{i}, a_{j}, a_{k}) is conjugate to (M_{i}, M_{j}, M_{k}).

(ii) Γ is geometric if and only if M acts transitively on each set of flags of size three of a given type, that is if and only if all flags {a_{i}, a_{j}, a_{k}} in Γ fulfill the equivalent conditions in (i).

Proof. (i) Clearly (a) implies (b). To show that (b) implies (c) we assume without loss that a_{k} = M_{k}, a_{i} = M_{i}r and a_{j} = M_{j}s for some r, s ∈ M_{k}. Since a_{i} and a_{j} are adjacent in Γ^{∗}(a_{k}), Ø≠ M_{ij}r ∩ M_{jk}s ⊆ M_{i} ∩ M_{j}r ∩ M_{k}s = a_{i} ∩ a_{j} ∩ a_{k} and so (c) holds. Clearly (c) implies (d), and (d) implies (a).

(ii) By the remark preceding this lemma, Γ is geometric if and only if (a) holds. □

Throughout the paper we will refer to the following easy principle concerning a set of size five.

Lemma 2.6 Let Γ_{1} be a 5-element set and Γ_{2} be the set of 2-element subsets in Γ_{1}. Let Γ be the bipartite graph on Γ_{1} ∪ Γ_{2} where a ∈ Γ_{1} is adjacent to b ∈ Γ_{2} if and only if a is not contained in b. Suppose that a ∈ Γ_{1} and b ∈ Γ_{2} are adjacent. Put R_{a}(b) = Γ_{1} \ (b ∪ {a}). Then every c ∈ Γ_{1} \ {a} is adjacent to exactly one of b and R_{a}(b).

Proof. This is clear since b, R_{a}(b) is a partition of Γ_{1} \ {a}. □

The next lemma contains some information on cohomologies of some small modules.

Lemma 2.7 (i) Let E ≅ L_{4}(2) and V be a natural Ω_{6}^{+}(2)-module for E. Then |H^{1}(V)| = 2, H^{2}(V) = 0.

(ii) Let E ≅ L_{4}(2) and V be a natural L_{4}(2)-module for E. Then H^{1}(V) = 0.

(iii) Let E ≅ L_{3}(2) and V be a natural L_{3}(2)-module for E. Then |H^{1}(V)| = 2.

Proof. [4] and [11]. □

The following lemma describes the actions of various subgroups of L_{5}(2) on the vectors of the exterior square of a natural L_{5}(2)-module.

Lemma 2.8 Let V be a 5-dimensional GF (2)-space and M = GL(V), so that M ≅ L_{5}(2). Let V_{1} < V_{3} < V where dim V_{1} = 1, dim V_{3} = 2. Let M_{i} = N_{M}(V_{i}), i = 1 and 3, B = M_{1} ∩ M_{3} and as usual let V^{∗} denote the dual of V. Then

(i) M has precisely two orbits H(2) and H(s) on the set of hyperplanes in ⋀^{2} V^{∗}; the hyperplanes in H(2) are indexed by the 2-spaces of V and the hyperplanes in H(s) are indexed by the pairs (W, s), where W is a hyperplane in V and s is a non-degenerate symplectic form on W;

(ii) if H ∈ H(s) then N_{M}(H) ∼ 2^{4}Sym(6);

(iii) the orbits of M_{1}; M_{3}; B on H(s) are of lengths 420 and 448; 84, 112 and 672; 84, 112, 224 and 448, respectively;

(iv) let H be a hyperplane from the orbit of length 84 of B on H(s); if N, N_{1}, N_{3}, and N_{0} are the normalizers of H in M, M_{1}, M_{3} and B, respectively, then

N = N′N_{0} and N′ ∩ N_{0} = (N′_{1} ∩ N_{0})(N′_{3} ∩ N_{0});

(v) the orbits of M_{3} on H(2) are of lengths 1, 42 and 112; the action of B on the M_{3}-orbit of length 42 is intransitive.

Proof. (i) By the definition of the exterior square ⋀^{2} V^{∗}, its hyperplanes are in one-to-one correspondence with the non-zero symplectic forms on V^{∗}. Hence (i) follows from the following well known facts: (a) any two non-degenerate symplectic forms on a finite dimensional vector space are isomorphic, (b) any vector space with a non-degenerate symplectic form is even dimensional and (c) there is exactly one non-degenerate symplectic form on a vector space with 4 elements.

(ii) Let W be a hyperplane in V, s a non-degenerate symplectic form on W and R = N_{M}(W, s). Then clearly C_{R}(W) = C_{M}(W) is elementary abelian of order 2^{4} and R/C_{R}(W) ≅ Sp_{4}(2) ≅ Sym(6).

(iii) Note that R acts transitively on the set of 1-spaces in W and C_{R}(W) acts regularly on the set of 1-spaces in V \ W. Thus R ∩ M_{1} ∼ 2^{4}(C_{2} × Sym(4)) if V_{1} ≤ W and R ∩ M_{1} ≅ Sym(6) otherwise. Moreover, R has three orbits on the set of 2-spaces in V, distinguished by V_{3} ≤ W and V_{3} is singular with respect to s; V_{3} ≤ W and V_{3} is non-degenerate with respect to s; and V_{3} ≰ W. The corresponding shapes of R ∩ M_{3} are 2^{4}(C_{2} × Sym(4)), 2^{4}(Sym(3) × Sym(3)) and 2(C_{2} × Sym(4)). Finally R has four orbits on the set of pairs of incident 1- and 2-spaces corresponding to the following four cases: V_{3} ≤ W and V_{3} is singular with respect to s; V_{3} ≤ W and is non-degenerate with respect to s; V_{1} ≤ W and V_{3} ≰ W; and V_{1} ≰ W. The corresponding shapes of R ∩ B are 2^{4}(C_{2} × D_{8}), 2^{4}(C_{2} × Sym(3)), 2(C_{2} × Sym(4)) and C_{2} × Sym(4). Thus we have described the orbits of R on 1- and 2-spaces in V and also on the incident pairs of such subspaces. This immediately gives us all the orbits of M_{1}, M_{3} and B on H(s) and the corresponding stabilizers. That the lengths of the orbits are as given in (iii) is now a trivial computation.

(iv) Let H correspond to (W, s). Then V_{3} ≤ W and V_{3} is singular with respect to s. Notice that C_{N}(W) ≅ W as N-module and in particular, C_{N}(W) = [C_{N}(W), N_{3}] ≤ N′_{3} ≤ N′. Thus (iv) holds if and only if it holds modulo C_{N}(W). But N_{i}/C_{N}(W) ≅ C_{2} × Sym(4) for i = 1, 3 and N_{0}/C_{N}(W) is a Sylow 2-subgroup of N/O_{2}(N) ≅ Sym(6). Now (iv) is readily verified.

(v) There is one 2-space equal to V_{3}, 42 = 3 · 14 2-spaces intersecting V_{3} in a 1-space and 112 = 7 · 16 2-spaces which intersect V_{3} trivially. Since some 2-spaces in the orbit of length 42 contain V_{1} while others do not, B does not act transitively on the M_{3}-orbit of length 42. □

3 Mat_{24}

We assume that the reader is familiar with the basic properties of the unique Steiner system S of type (5,8,24) (see for instance [1] or [9]). Let Ω be the set of size 24 underlying S and let Γ_{2} denote the block set of S. This means that Γ_{2} is a collection of 8-element subsets of Ω called octads such that every 5-element subset of Ω is in a unique octad. In particular |Γ_{2}| = (_{5}^{24})/(_{5}^{8}) = 759. A triple of pairwise disjoint octads is called a trio. Every 4-element subset T of Ω is contained in a unique sextet, which is a partition of Ω into six 4-element subsets T_{1} = T, T_{2}, ..., T_{6} called tetrads such that T_{i} ∪ T_{j} ∈ Γ_{2} for all 1 ≤ i < j ≤ 6.

Let Γ_{3} denote the set of trios, let Γ_{4} denote the set of sextets and let Γ = Γ_{2} ∪ Γ_{3} ∪ Γ_{4}. Define a graph on Γ as follows: a trio is adjacent to an octad if it contains the octad; a sextet is adjacent to an octad if the octad is the union of two of the tetrads in the sextet; and a sextet is adjacent to a trio if it is adjacent to all of the three octads in the trio.

Throughout this section M will stand for the automorphism group of S which is the Mathieu group Mat_{24} of degree 24. Let α, β and γ be pairwise adjacent octad, trio and sextet respectively, i.e. a maximal flag in Γ. If γ = {T_{1}, T_{2}, ..., T_{6}} we can put α = T_{1} ∪T_{2} and β = {T_{1} ∪T_{2}, T_{3} ∪T_{4}, T_{5} ∪T_{6}}. Let M_{2} = M_{α}, M_{3} = M_{β} and M_{4} = M_{γ} (the stabilizers in M of α, β and γ, respectively). Then (M_{2}, M_{3}, M_{4}) is a triangle of groups and M is a faithful completion of this triangle. Since M is flag transitive on Γ, Γ ≅ Γ(M; M_{2}, M_{3}, M_{4}). We have chosen the index set {2, 3, 4} rather then {1, 2, 3} since M_{2}, M_{3}, M_{4} will correspond to M_{12}, M_{13} and M_{14} in later sections.

We will need the following information on classes of elements in M of order 2 and 3 which can be deduced either from Section 21 in [1] together with the permutational characters of M on Γ_{2}, Γ_{3} and Γ_{4} given in [5] or from Sections 2.12 - 2.14 in [9].

Lemma 3.1 (i) M has two classes, 2a and 2b of involutions and two classes, 3a and 3b of elements of order 3.

(ii) If t ∈ 2a then t is 2-central, C_{M}(t) ∼ 2_{+}^{1+6}L_{3}(2), t fixes: 8 elements of Ω forming an octad, 71 octads, 99 trios and 91 sextets.

(iii) If s ∈ 2b then s is non-2-central, C_{M}(s) ∼ 2^{1+1+4}Sym(5), C_{M}(s) fixes a unique sextet, s fixes: 15 octads, 75 trios and 51 sextets.

(iv) If x ∈ 3a then C_{M}(x) ≅ 3 · Alt(6), x does not commute with a 2b-involution and fixes: 21 octads, 15 trios and 16 sextets.

(v) If y ∈ 3b then C_{M}(y) ≅ C_{3} × L_{3}(2), y commutes with a 2b-involution, acts fixed-points freely on the set of octads and fixes 15 trios and 7 sextets. □

The basic properties of the triangle (M_{2}, M_{3}, M_{4}) and of its completion M are given in the following lemma (cf. Section 19 in [1] or Section 2.10 in [9]).

Lemma 3.2 (i) |M/M_{2}| = |Γ_{2}| = 759, M_{2} ∼ 2^{4}L_{4}(2) and Q_{2} is a natural L_{4}(2)-module for M_{2}, Q_{2} is 2a-pure, M_{2} acts as Alt(8) ≅ L_{4}(2) on the elements in α and as the doubly transitive affine group AGL_{4}(2) on the elements outside α, in particular M_{2} splits over Q_{2}.

(ii) |M/M_{3}| = |Γ_{3}| = 3795; M_{3} ∼ 2^{6}(Sym(3) × L_{3}(2)), Q_{3} ≅ D_{1} ⊗ D_{2}, where D_{1} and D_{2} are natural L_{2}(2)- and L_{3}(2)-modules for M_{3}, respectively; M_{3} has two orbits on Q_{3}^{#} with lengths 21 and 42, consisting of involutions of type 2a and 2b, respectively and if y ∈ Q_{3}^{∗} \ Q_{3} then y is of type 3b and acts fixed-point freely on Q_{3}.

(iii) |M/M_{4}| = |Γ_{4}| = 1771, M_{4} ∼ 2^{6}3 · Sym(6), Q_{4} is a hexacode module for M_{4}, M_{4} has two orbits on Q_{4}^{#} with lengths 45 and 18, consisting of involutions of type 2a and 2b, respectively, if x ∈ Q_{4}^{∗} \ Q_{4} then x is of type 3a and acts fixed-point freely on Q_{4}, M_{4} induces Sym(6) on the tetrads constituting γ and the kernel induces Alt(4) on the elements in each tetrad.

(iv) |M_{3}/M_{34}| = 7, |M_{3}/M_{23}| = 3, |M_{4}/M_{24}| = 15 and |M_{34}/B| = 3.

(v) M_{24}/Q_{2} ∼ 2^{4}(Sym(3) × Sym(3)) and M_{23}/Q_{2} ∼ 2^{3}L_{3}(2).

(vi) M_{34}/Q_{3} ≅ Sym(3) × Sym(4) and M_{23}/Q_{3} ≅ Sym(2) × L_{3}(2).

(vii) M_{24}/Q_{4}^{∗} ≅ M_{34}/Q_{4}^{∗} ≅ Sym(4) × Sym(2).

(viii) |Q_{2} ∩ Q_{3}| = 8, |Q_{2} ∩ Q_{4}| = 4 and |Q_{3} ∩ Q_{4}| = 16. □

Comparing 3.1 and 3.2 one can observe the following. If t is an involution in Q_{2}, s is an involution in the orbit of length 18 of M_{4} on Q_{4}^{#}, x is an element of order 3 in Q_{4}^{∗} and y is an element of order 3 in Q_{3}^{∗}, then

C_{M}(t) = C_{M_{2}}(t), C_{M}(s) = C_{M_{4}}(s), C_{M}(x) = C_{M_{4}}(s), C_{M}(y) = C_{M_{3}}(y).

As a direct corollary of 3.2 we have the following.

Lemma 3.3 (i) M_{2} acts on Γ_{3}(α) and Γ_{4}(α) of size 15 and 35 as it acts on the 3- and 2-spaces in Q_{2}, respectively;

(ii) M_{3} acts on Γ_{2}(β) and Γ_{4}(β) of size 3 and 7 as it acts on the 1-spaces in D_{1} and on the 2-spaces in D_{2}, respectively;

(iii) M_{4} acts on Γ_{2}(γ) and Γ_{4}(γ) of size 15 each as it acts on the 2-element subsets of γ (considered as the set of six tetrads) and on the partitions of γ into three pairs. □

Lemma 3.4 Let B = M_{2} ∩ M_{3} ∩ M_{4} be the stabilizer in M of the flag F = {α, β, γ}. Let S be a Sylow 2-subgroup of B which is also a Sylow 2-subgroup of M.

(i) α is the unique octad, β is the unique tetrad and γ is the unique sextet stabilized by S.

(ii) Let H be a maximal subgroups of M containing S. Then H = M_{i} for i = 2, 3 or 4.

(iii) Let P be any subgroup of M containing S and let z denote the unique non-trivial element in Z(S). Then one of the following holds:

(a) P is the normalizer of a subflag in F.

(b) |P | = 2^{10} · 3^{2} and P = Q_{3}^{∗}C_{M_{3}}(z).

(c) |P | = 2^{10} · 3 · 7 and P = C_{M}(z).

(d) |P | = 2^{10} · 3 and P = Q_{3}^{∗}S, C_{M_{3}}(z) or C_{M_{4}}(z).

(e) P = S.

Proof. (ii) follows from [6], (i) follows from (ii) while (iii) follows from the structure of M_{2}, M_{3} and M_{4} as given in 3.2 (compare [15]). □

In subsequent sections we will need detailed information about the graph Γ and the action of M on this graph. For this purpose for every i ∈ {2, 3, 4} we describe the orbits of M_{i} on the vertex set of Γ and for any two such orbits A and B we calculate the number n_{i}(A, B) of vertices in B adjacent in Γ to a given vertex a ∈ A and finally determine how these vertices split into orbits under the stabilizer of a in M_{i}. It is clear that n_{i}(A, B) is zero unless A ⊆ Γ_{j}, B ⊆ Γ_{k} for j ≠ k and that |A| · n_{i}(A, B) = |B| · n_{i}(B, A). Finally for i ≠ j there is a natural correspondence between the orbits of M_{i} on Γ_{j} and the orbits of M_{j} on Γ_{i}.

Let Γ_{j}(m, i) denote an orbit of length m of M_{i} on Γ_{j}. It turns out that for every i, j ∈ {2, 3, 4} the orbits of M_{i} on Γ_{j} all have different lengths so the orbit Γ_{j}(m, i) is well defined. The information on the orbits of M_{i} on Γ is presented in the diagram D_{i}(Mat_{24}). In this diagram the orbit Γ_{j}(m, i) is denoted by m_{j} and the numbers n_{i}(A, B) and n_{i}(B, A) are attached to the edge joining A with B. When such a number is presented as a sum this indicates that there is more than one orbit of M_{i} ∩ M_{a} (where a ∈ A) on the vertices in B adjacent to a. Moreover the summands give the lengths of these orbits. The complete proof of the diagrams (originally given in the early version of the present work) can be found in Section 3.7 of [9].

[figure: 15_{3}; 420_{3}; 3360_{3}; 1; 280_{2}; 448_{2}; 1_{2}; 35_{4}; 840_{4}; 896_{4}; 15; 35; 1; 3; 7; 1; 8; 12; 18; 6; 3; 12; 8; 15; 6; 1; 2; 3; 12; 1; 3; 5; 16; 15; 4; 15; 2; 30_{2}; 2; 1; 14; 1; 20; 10; 28; 1; 7; 6; D_{2}(Mat_{24})] [figure: 3_{2}; 84_{2}; 672_{2}; 1; 56_{3}; 2688_{3}; 1_{3}; 7_{4}; 1344_{4}; 3; 7; 1; 3; 7; 1; 8; 12; 1; 2; 3; 1; 6+8; 12+16; 42_{3}; 1008_{3}; 84_{4}; 336_{4}; 14; 28; 1; 6; 12; 12; 16; 3; 6; 12; 1; 2; 1; 2; 3; 1; 6; 4; 2; 1; 6; 6; 1; 6; 3; 1; 6; 12; 8; 1; 3; 6; 12; 8; 3; 1; 12; D_{3}(Mat_{24})]

9

[figure: 15_{2}; 360_{2}; 384_{2}; 1; 1440_{4}; 1_{4}; 15_{3}; 2880_{3}; 15; 15; 1; 3; 3; 90_{4}; 18; 3; 6; 240_{4}; 180_{3}; 720_{3}; 12; 16; 1; 4; 6; 8; 4+24; 5; 30; 15; 3; 12; 1; 6; 8; 1+6; 8; 1; 6; 8; 3; 12; 3; 12; 3; 1; 2; 4; 1; 3; 6; 1; 1; 6; 2; D_{4}(Mat_{24})]

We will need the following refinement of the information given on the diagram D_{3}(Mat_{24}).

Lemma 3.5 Let b ∈ Γ_{3}(2688, 3). Then M_{βb} ∩ Q_{b} = 1 and M_{βb} ≅ Sym(4).

Proof. |M_{βb}| = |M_{3}|/2688 = 24 by direct calculation. By D_{3}(Mat_{24}) the subgroup M_{βb} acts transitively on Γ_{2}(b) and has two orbits in Γ_{4}(b) with lengths 1 and 6. Since the action of M_{βb} on Γ_{4}(b) is a subgroup of L_{3}(2), we conclude that the action is isomorphic either to Sym(4) or to Alt(4). Let K be the kernel of the action of M_{βb} on Γ_{4}(b). Then either K = 1 and M_{βb} ≅ Sym(4) or |K| = 2 and M_{βb}/K ≅ Alt(4). Assume the latter. Then K = Z(M_{βb}) and as M_{βb} acts transitively on Γ_{2}(b), K ≤ Q_{b}. By symmetry we get K = Q_{β} ∩ Q_{b} and so C_{M}(K) contains two elementary abelian groups of order 2^{6} (namely Q_{β} and Q_{b}) intersecting in a group of order 2 (namely K). But this contradicts to the structure of C_{M}(K) given by 3.1 (i) - (iii). □

Let P be the GF (2)-permutation module of M on Ω, that is the space of all the subsets of Ω with addition performed by the symmetric difference operator. The octads from Γ_{2} generate in P a 12-dimensional subspace Y_{0} known as the Golay code. The Golay code consists of: the empty set, the set Ω itself, 759 octads, 759 complements of octads and 2576 dodecads. The latter are 12-element subsets of Ω transitively permuted by M. The stabilizer of a dodecad is the Mathieu group Mat_{12} of degree 12 and it induces two non-equivalent 5-fold transitive actions on the dodecad and on its complement, which is also a dodecad. The setwise stabilizer of a pair of complementary dodecads is isomorphic to Aut Mat_{12}. The empty set together with the whole set Ω constitute the unique proper M-submodule in Y_{0}. The quotient Y of Y_{0} over this submodule is called the irreducible Golay code module (of dimension 11). Let P_{+} denote the subspace in P of even subsets of Ω. Then Y_{0} ≤ P_{+} and X = P_{+}/Y_{0} is the module dual to Y which is called the irreducible Todd module (of dimension 11). The following information can be found for instance in [1, 19.8].

Lemma 3.6 (i) The orbits of M on the non-zero vectors of Y (on the hyperplanes of X) are of length 759 and 1288. The vectors in these orbits are indexed by the octads and the complementary pairs of dodecads, respectively.

(ii) The orbits of M on the non-zero vectors of X (on the hyperplanes of Y) are of length 276 and 1771. The vectors in these orbits are indexed by the 2-element subsets of Ω and by the sextets, respectively. □

Lemma 3.7 (i) For i = 2, 3 and 4,

1 < C_{X}(Q_{i}) < [X, Q_{i}] < X

is the unique composition series for M_{i} on X;

(ii) C_{X}(Q_{2}) is isomorphic the exterior square of Q_{2}, [X, Q_{2}]/C_{X}(Q_{2}) is isomorphic to Q_{2} and |X/[X, Q_{2}]| = 2.

(iii) Let D_{1} and D_{2} be as in 3.2 (ii). Then C_{X}(Q_{3}) is dual to D_{2}, [X, Q_{3}]/C_{X}(Q_{3}) is isomorphic to D_{1} ⊗ D_{2} and X/[X, Q_{3}] is isomorphic to D_{1}.

(iv) |C_{X}(Q_{4})| = 2, [X, Q_{4}]/C_{X}(Q_{4}) is isomorphic to the dual of Q_{4} and |X/[X, Q_{4}]| = 2^{4}.

Proof. The irreducible Todd module is dual to the irreducible Golay code module. Hence the result can be obtained by dualizing some of the information found in Sections 19 and 20 of [1]. □

Let D be the set of dodecads and H be the set of complementary pairs of dodecads. Recall that if N(h) is the stabilizer of h ∈ H in M then N(h) ≅ Aut Mat_{12} and N(h)′ ≅ Mat_{12} is the subgroup of index 2 in N(h) which preserves each of the dodecads constituting h. We are interested in the orbits on H of M_{2}, M_{3} and M_{23}.

Lemma 3.8 M_{2} acting on the set D of dodecads has three orbits D_{2}, D_{4} and D_{6} with lengths 448, 1680 and 448, respectively. If d_{i} ∈ D_{i} and K_{i} is the stabilizer of d_{i} in M_{2} then |d_{i} ∩ α| = i, K_{2} ≅ K_{6} ≅ Sym(6) and K_{4} ∼ 2^{5}Sym(3).

Proof. [1, 19.6] □

Lemma 3.9 Let N ≅ Aut Mat_{12} and N_{2}, N_{3} and N_{23} subgroups of N such that |N_{2}| = |N_{3}| = 2^{7}·3, N_{23} = 2^{7} and N_{23} = N_{2} ∩ N_{3}. Then N_{23} ∩ N′ = (N_{23} ∩ N′_{2})(N_{23} ∩ N′_{3}).

Proof. For Z ≤ N let Z^{∗} = Z ∩ N′. Then N^{∗} ≅ Mat_{12}, |N/N^{∗}| = 2 and |N^{∗}|_{2} = 2^{6}. Thus N_{23}^{∗} is a Sylow 2-subgroup of N^{∗} and |N_{2}^{∗}| = |N_{3}^{∗}| = 2^{6} · 3. It follows ( see for example [15]) that N_{2}^{∗} and N_{3}^{∗} are two maximal subgroups of N^{∗} containing N_{23}^{∗}. Choose notation such that Z(N_{2}^{∗}) ≠ 1. Thus by the structure of N_{2}^{∗} and N_{3}^{∗}, O_{2}(N_{2}^{∗}) ≤ N_{2}^{∗′}, O_{2}(N_{3}^{∗}) ∩ N_{3}^{∗′} ≰ O_{2}(N_{2}^{∗}) and |N_{23}^{∗}/O_{2}(N_{2}^{∗})| = 2. Thus N_{23}^{∗} = O_{2}(N_{2}^{∗})(O_{2}(N_{3}^{∗}) ∩ N_{3}^{∗′}) = (N_{23} ∩ N_{2}^{∗′})(N_{23} ∩ N_{3}^{∗′}). □

Lemma 3.10 Let H be the set of complementary pairs of dodecads and for h ∈ H let N(h), N_{2}(h), N_{3}(h) and N_{23}(h) denote the stabilizers of h in M, M_{2}, M_{3} and M_{23}, respectively.

(i) M_{2} has precisely two orbits H_{1}(2) and H_{2}(2) on H, where |H_{1}(2)| = 840 and |H_{2}(2)| = 448. M_{3} has precisely three orbits H_{1}(3), H_{2}(3), and H_{3}(3) on H, where |H_{1}(3)| = 168, |H_{2}(3)| = 672 and |H_{3}(3)| = 448. M_{23} has precisely four orbits H_{1}, H_{2}, H_{3} and H_{4} on H, where |H_{1}| = 168, |H_{2}| = 224, |H_{3}| = 448 and |H_{4}| = 448. Moreover, H_{1}(2) = H_{1} ∪ H_{2} ∪ H_{3}, H_{2}(2) = H_{4}, H_{1}(3) = H_{1}, H_{2}(3) = H_{2} ∪ H_{4} and H_{3}(3) = H_{3}.

(ii) If h ∈ H_{1}, then N(h) = N_{23}(h)N(h)′ and N_{23}(h)∩N(h)′ = (N_{23}(h)∩N_{2}(h)′)(N_{23}(h)∩N_{3}(h)′).

(iii) If h ∈ H_{3}, then N_{3}(h) = N_{23}(h)N_{3}(h)′.

Proof. (i) The lengths of the orbits of M_{2} on H follow directly from 3.8. Observe also that N_{2}(h) ≤ N(h)′ (that is N_{2}(h) fixes the two dodecads forming h) if and only if h ∈ H_{2}(2).

Let Δ be the octad graph, that is a graph on Γ_{2} in which two octads are adjacent if they are disjoint. For a vertex x of Δ let Δ^{i}(x) denote the set of vertices which are at distance i from x in Δ. It is well known and also easily seen from the diagram D_{2}(Mat_{24}) that

Δ^{1}(α) = Γ_{2}(30, 2), Δ^{2}(α) = Γ_{2}(280, 2), Δ^{3}(α) = Γ_{2}(448, 2)

and that

| α ∩ δ |= 0, 4, 2 if δ ∈ Δ^{i}(α) for i = 1, 2, 3.

By 3.6 we can and will identify Δ∪H with the set of non-zero vectors in the irreducible Golay code module Y. Let e ∈ Δ^{3}(α). Then as α and e intersect in 2 elements, the symmetric difference of α and e is a dodecad intersecting α in 6 elements. Thus α+e ∈ H_{2}(2) and since |Δ^{3}(α)| = |H_{2}(2)| = 448, we have a one-to-one correspondence between H_{2}(2) and Δ^{3}(α). By D_{2}(Mat_{24}), M_{αe} acts transitively on Γ_{3}(α). Thus M_{23} acts transitively on Δ^{3}(α) and hence also on H_{4} =^{def} H_{2}(2).

Let h ∈ H_{1}(2). Then N_{2}(h) has order 2^{7} · 3 and the intersections of α with the dodecads in h form a partition of α into two sets of sizes 4. Thus N_{2}(h)Q_{α}/Q_{α}is contained in a subgroup 2^{4}(Sym(3) × Sym(3)) of M_{α}/Q_{α} and so normalizes a 2-subspace U_{2} in Q_{α}. Note that Q_{α} fixes 4 points in each of the two dodecads, Q_{α} ≤ Mat_{8} ≅ Q_{8}. As Q_{α} is elementary abelian, Q_{α} ∩ N(h) has order at most two. It follows that N_{2}(h)Q_{α}/Q_{α} has order 2^{6} · 3 and U_{1} =^{def} Q_{α} ∩ N(h) has order 2, U_{1} ≤ U_{2} and Q_{α}N_{2}(h) = N_{M_{α}}(U_{1}) ∩ N_{M_{α}}(U_{2}). Thus the orbits of Q_{α} on H_{1}(2) are in one-to-one correspondence with the pairs (U_{1}, U_{2}), where U_{i} is a i-space in Q_{α} and U_{1} ≤ U_{2}. This immediately implies that M_{αβ} has three orbits H_{1}, H_{2} and H_{3} on H_{1}(2) corresponding to the following three possibilities: (1) U_{2} ≤ Q_{α} ∩ Q_{β}, (2) U_{2} ≰ Q_{α} ∩ Q_{β} and U_{1} ≤ Q_{α} ∩ Q_{β} and (3) U_{1} ≰ Q_{α} ∩ Q_{β}. Now it is straightforward to calculate that |H_{1}| = 7 · 3 · 8 = 168, |H_{2}| = 28 · 1 · 8 = 224 and |H_{3}| = 28 · 2 · 8 = 448. (Notice that | Q_{α} ∩ Q_{β} |= 8 by 3.2 (i).)

Let L be the elementwise stabilizer in M of the octads in Δ(β). Then L is of index 2 in M_{23} and normal of index 6 in M_{3}. Hence each of the following holds ( for the last statement note that M = ⟨M_{2}, M_{3}⟩ acts transitively on H):

• For every i, L either acts transitively on H_{i} or has two orbits of the same length.

• Every M_{3}-orbit in H is the union of l of the orbits of equal lengths for L in H where l ∈ {1, 2, 3, 6}.

• There exists an M_{3}-orbit on H which has non-empty intersecting with both H_{1}(2) and H_{2}(2).

It is easy to check that these three conditions uniquely determine the fusion of the M_{23}-orbits into M_{3}-orbits.

(ii) and (iii): Let h ∈ H_{1}. As N(h) ≅ Aut Mat_{12}, |N(h)/N(h)′| = 2. Moreover, |N(h)/N_{23}(h)| is odd and so the first statement in (ii) holds. (iii) follows from a similar argument. By (i) we can apply 3.9 and so also the second part of (ii) holds. □

By [7] M has a 45-dimensional irreducible module V over the field C of complex numbers. Let χ be the corresponding character. Define V_{1}(3) = C_{V} (Q_{3}) and V_{2}(3) = [V, Q_{3}].

Lemma 3.11 (i) Let z be a 2-central involution in M. Then χ(z) = −3, C_{V} (z) is 21-dimensional and [V, z] is 24-dimensional.

(ii) C_{V} (Q_{2}) = 0 and C_{V} (H) is 3-dimensional for each hyperplane H of Q_{2}.

(iii) V = V_{1}(3) ⊕ V_{2}(3), V_{1}(3) is 3-dimensional and V_{2}(3) is 42-dimensional.

(iv) C_{V_{2}(3)}(H) = 0 for any hyperplane of Q_{3} containing Q_{2} ∩ Q_{3}, while C_{V_{2}(3)}(H) is 1-dimensional for any hyperplane of Q_{3} not containing any of the three conjugates of Q_{2} ∩ Q_{3} under M_{3}.

Proof. (i) The value for χ(z) is taken directly from the character table of Mat_{24} in [7]. Since dim C_{V} (z) + dim[V, z] = 45 and dim C_{V} (z) − dim[V, z] = χ(z) (i) holds.

(ii) Let d = dim[V, Q_{2}] and e = dim C_{[V,Q_{2}]}(H), where H is any hyperplane in Q_{2}. Since M_{2} acts transitively on the fifteen hyperplanes in Q_{2}, e is well defined and d = 15e. Let 1 ≠ z ∈ Q_{2}. Then exactly eight of the hyperplanes in Q_{2} do not contain z and so 24 = dim[V, z] = 8e. Thus e = 3, d = 45, V = [V, Q_{2}] and (ii) holds.

(iii) and (iv) By 3.2 (iii) the orbits of M_{3} on Q_{3}^{#} are of length 21 and 42 and by a dual argument the orbits of M_{3} on the hyperplanes of Q_{3} are of length 21 and 42. In particular, dim[V, Q_{3}] is divisible by 21. Moreover, by (ii) C_{V} (Q_{2} ∩ Q_{3}) has dimension 3 and so dim C_{V} (Q_{3}) ≤ 3. Thus dim V_{1}(3) = 3 and dim V_{2}(3) = 42. Let H be a hyperplane in Q_{3} with f =^{def} dim C_{V_{2}(3)}(H) ≠ 0. Then either |H^{M_{3}}| = 42 and f = 1; or |H^{M_{3}}| = 21 and f = 2. In particular H is unique up to conjugation. Suppose that |H^{M_{3}}| = 21. Let 1 ≠ z ∈ Q_{2} ∩ Q_{3}. Then it is easy to see that z lies in exactly 7 + 3 + 3 = 13 of the elements of H^{M_{3}} and so dim[V, z] = f · (21 − 13) = 16, a contradiction to (i). Thus |H^{M_{3}}| = 42 and the lemma is proved. □

Lemma 3.12 (i) M_{2} acts irreducibly on V and as M_{2}-module V ≅ V_{1}(3) ⊗_{CM_{23}} CM_{2}.

(ii) V_{1}(3) and V_{2}(3) are irreducible M_{3}-modules of dimension 3 and 42, respectively, and stay irreducible when restricted to M_{23} or O^{2}(M_{23}).

(iii) C_{M_{3}}(V_{1}(3)) = O_{2,3}(M_{3}) and M_{3} acts faithfully on V_{2}(3).

Proof. By 3.11 (ii), V_{1}(3) = C_{V} (Q_{2} ∩ Q_{3}) is a Wedderburn component for Q_{2} on V. Moreover, since M_{23} is maximal in M_{2}, M_{23} = N_{M_{2}}(Q_{2} ∩ Q_{3}) = N_{M_{2}}(V_{1}(3)) and so the second statement in (a) holds. Moreover, M_{2} is irreducible on V if and only if M_{23} is irreducible on V_{1}(3). Since L =^{def} O^{2}(M_{23}) acts transitively on the 42 hyperplanes in Q_{3} which have fixed-points in V_{2}(3), L acts irreducibly on V_{2}(3). Suppose that L does not act irreducibly on V_{1}(3). Since V_{1}(3) has odd dimension and |M_{23}/L| = 2 we conclude that M_{23} does not act irreducibly on V_{1}(3). Thus M_{23} has a 1- or 2-dimension submodule in V_{1}(3) and M_{2} has a 15- or 30-dimensional submodule in V. But this contradicts the fact that V_{2}(3) is a 42-dimensional irreducible L-module. Hence L is irreducible on V_{1}(3) and (i) and (ii) are proved.

To prove (iii) recall that M_{3} ∼ 2^{6}(Sym(3)×L_{3}(2)). Note that Q_{2} ∩Q_{3} is a hyperplane in Q_{2} and centralizes V_{1}(3). Since Q_{2} acts fixed-point freely on V we conclude that Q_{2}Q_{3}/Q_{3} inverts V_{1}(3). Furthermore, O_{2,3}(M_{3}) = [M_{3}, Q_{2}] and so O_{2,3}(M_{3}) centralizes V_{1}(3). Hence either C_{M_{3}}(V_{1}(3)) = O_{2,3}(M_{3}) or M′_{3} centralizes V_{1}(3). But in the later case M_{3} is not irreducible on V_{1}(3). The second statement in (iii) holds since Q_{3} is the unique minimal normal subgroup of M_{3} and does not centralize V_{2}(3). □

4 Mat_{22}

Definition 4.1 (i) A Mat_{22}-triangle is a triangle of groups (M_{1}, M_{2}, M_{3}) such that (a) M_{1} ∼ 2^{4}Alt(6), M_{2} ∼ 2^{3}L_{3}(2) and M_{3} ∼ 2^{4}Sym(5).

(b) |M_{2}/M_{23}| = |M_{2}/M_{12}| = 7, |M_{3}/M_{13}| = 5 and |M_{23}/B| = 3.

(ii) An Aut Mat_{22}-triangle is a triangle of groups (M̂_{1}, M̂_{2}, M̂_{3}) such that

(a) M̂_{1} ∼ 2^{4}Sym(6), M̂_{2} ∼ 2^{4}L_{3}(2) and M̂_{3} ∼ 2^{5}Sym(5).

(b) |M̂_{2}/M̂_{23}| = |M̂_{2}/M̂_{12}| = 7, |M̂_{3}/M̂_{13}| = 5 and |M̂_{23}/B| = 3.

Lemma 4.2 Let (M_{1}, M_{2}, M_{3}) be a Mat_{22}-triangle. Then the following assertions hold.

(i) B is a Sylow 2-subgroup of M_{2} and B = (Q_{1} ∩ M_{2})Q_{2}Q_{3}.

(ii) M_{13}/Q_{3} ≅ Sym(4) and M_{23}/Q_{3} ≅ Sym(3) × Sym(2).

(iii) M_{12}/Q_{2} ≅ M_{23}/Q_{2} ≅ Sym(4).

(iv) Q_{1} ≰ M_{2} and M_{13}/Q_{1} ≅ M_{12}Q_{1}/Q_{1} ≅ Sym(4).

(v) |Q_{1} ∩ Q_{2}| = 2, |Q_{1} ∩ Q_{3}| = 4, |Q_{2} ∩ Q_{3}| = 4 and Q_{1} ∩ Q_{2} ≤ Q_{1} ∩ Q_{3}.

(vi) T_{12} = (Q_{1} ∩ M_{2})Q_{2}, T_{13} = Q_{1}Q_{3}, T_{23} = Q_{2}Q_{3}.

Proof. Since M_{3} has a unique class of subgroups of index 5, M_{13}/Q_{3} ≅ Sym(4). Similarly Sym(4) ≅ M_{12}/Q_{2} ≅ M_{13}/Q_{1} ≅ M_{23}/Q_{2} and B is a Sylow 2-subgroup of M_{2}. Since |M_{23}/B| = 3, M_{23} has an orbit of length 3 on the cosets of M_{13} in M_{3}. Thus M_{23}Q_{3}/Q_{3} is contained a subgroup Sym(3) × Sym(2) of M_{3}/Q_{3}. As M_{23} has index 10 in M_{3} and Sym(3) × Sym(2) has index 10 in Sym(5) we conclude that Q_{3} ≤ M_{23} and M_{23}/Q_{3} ≅ Sym(3)×Sym(2). As |Q_{3}| > |Q_{2}|, Q_{3} ≰ Q_{2} and since M_{23} acts irreducibly on T_{23}/Q_{2}, we conclude that T_{23} = Q_{2}Q_{3} and |Q_{2}∩Q_{3}| = 4. Suppose that Q_{3} = Q_{1}. Then T_{23} = Q_{2}Q_{3} = Q_{2}Q_{1} is normalized both by M_{12} and M_{23}. Since M_{2i} = N_{M_{2}}(T_{2i}) for i = 1 and 3, this means that M_{12} = M_{23}, a contradiction to |M_{23}/B| = 3. Thus Q_{3} ≠ Q_{1}, T_{13} = Q_{1}Q_{3} and |Q_{1} ∩ Q_{3}| = 4. So Q_{1} ≤ O_{2}(M_{13}) ≤ M′_{3}Q_{3}, Q_{1} ≰ M_{12} and as no element of Q_{1} acts as a 2-cycle on M_{3}/M_{13}, Q_{1} ∩ M_{12} ≰ Q_{2}. Hence T_{12} = Q_{2}(Q_{1} ∩ M_{12}) and |Q_{2} ∩ Q_{1}| = 2. Since B = T_{12}T_{23}, the last statement in (i) holds and the proof is complete. □

We remark that a similar lemma holds for Aut Mat_{22}-triangles. Indeed the only changes necessary are that in part (iv), Sym(4) has to be replaced by Sym(2) × Sym(4) and in part (v), Q̂_{2} ∩ Q̂_{3} has order 8 and not 4.

As in the previous section let S be the Steiner system of type (5, 8, 24) and let p, q be a pair of elements from the basic set Ω. In this section M and M̂ will denote the elementwise and the setwise stabilizers of {p, q} in the automorphism group Mat_{24} of S, respectively. This means that M is the Mathieu group Mat_{22} of degree 22 with |M| = 2^{7} · 3^{2} · 5 · 7 · 11 and M̂ is the automorphism group of M.

Let γ be a sextet T_{1}, T_{2}, ..., T_{6} in S such that p and q are in the same tetrad (say in T_{1}). Let α and β be disjoint octads adjacent to γ such that {p, q} ⊆ α (say α = T_{1} ∪ T_{2} and β = T_{3} ∪ T_{4}). Let M_{α}, M_{β} and M_{γ} be the stabilizers in M of α, β and γ, respectively. Similarly define M̂_{α}, M̂_{β} and M̂_{γ}. The following lemma can be deduced directly from 3.2 (cf. Section 3.4 in [9]).

Lemma 4.3 (i) (M_{α}, M_{β}, M_{γ}) is a Mat_{22}-triangle.

(ii) (M̂_{α}, M̂_{β}, M̂_{γ}) is an Aut Mat_{22}-triangle. □

It is easy to deduce from the main result in [16] that every Aut Mat_{22}-triangle with a faithful completion is isomorphic to (M̂_{α}, M̂_{β}, M̂_{γ}) and that M̂ is the unique completion of the triangle. In order to explain the deduction we need some definitions.

Recall that the Petersen graph has 2-element subsets of a fixed 5-set as vertices and two subsets are adjacent if they are disjoint. The Petersen graph has 10 vertices, 15 edges and Sym(5) is its automorphism group.

Definition 4.4 Let Ξ = Ξ_{1} ∪ Ξ_{2} ∪ Ξ_{3} be a 3-partite graph and suppose that for a_{i} ∈ Ξ_{i}, 1 ≤ i ≤ 3 the following conditions hold.

(i) |Ξ_{2}(a_{1})| = 2, |Ξ_{3}(a_{1})| = 3 and every vertex from Ξ_{2}(a_{1}) is adjacent to every vertex from Ξ_{3}(a_{1}).

(ii) Ξ_{1}(a_{2}) are the points and Ξ_{3}(a_{2}) are the lines of a projective plane of order 2 with the natural adjacency relation.

(iii) Ξ_{1}(a_{3}) are the edges and Ξ_{2}(a_{3}) are the vertices of the Petersen graph with the natural adjacency relation.

Then Ξ is called a rank 3 Petersen type geometry.

Theorem 4.5 Up to isomorphism there are exactly two rank 3 flag-transitive Petersen type geometries: Ξ(Mat_{22}) and Ξ(3Mat_{22}). A flag-transitive automorphism group is isomorphic to M or M̂ for Ξ(Mat_{22}) and to 3M or 3M̂ (non-split extensions) for Ξ(3Mat_{22}). The stabilizer of a vertex from Ξ_{3} in M and 3M is 2^{4}Sym(5) while in M̂ and 3M̂ it is 2^{5}Sym(5).

Proof. [16]. □

Lemma 4.6 (i) Every Mat_{22}-triangle with a faithful completion is isomorphic to (M_{α}, M_{β}, M_{γ}) and M is the unique faithful completion of this triangle.

(ii) Every Aut Mat_{22}-triangle with a faithful completion is isomorphic to (M̂_{α}, M̂_{β}, M̂_{γ}) and M̂ is the unique faithful completion of this triangle.

Proof. (i) Let (M_{1}, M_{2}, M_{3}) be a Mat_{22}-triangle with a faithful completion N. Define a triangle (N_{1}, N_{2}, N_{3}) by N_{1} = C_{M_{1}}(Q_{1} ∩ Q_{2}), N_{2} = M_{2} and N_{3} = M_{3}. Then N_{1} ∼ 2^{4}Sym(4) by 4.2 and since M_{1} = ⟨N_{1}, M_{13}⟩, N is also a faithful completion of (N_{1}, N_{2}, N_{3}). Let Ξ = Γ(N; N_{1}, N_{2}, N_{3}). Then it is easy to check using the information in 4.2 that Γ is a rank 3 Petersen type geometry on which N acts flag-transitively. By 4.5 and since N_{3} = M_{3} ∼ 2^{4}Sym(5) we have N ≅ M or N ≅ 3M, but in the latter case M_{1} ∼ 2^{4}3 · Alt(6). Hence N ≅ M and (M_{1}, M_{2}, M_{3}) is isomorphic to (M_{α}, M_{β}, M_{γ}).

(ii) is proved similarly. □

The coset graph Γ = Γ(M; M_{1}, M_{2}, M_{3}) (which coincides with Γ(M̂; M̂_{1}, M̂_{2}, M̂_{3})) possesses a natural description in terms of the Steiner system S and a pair p, q of distinguished elements from the basic set Ω. Namely, Γ_{1} are the hexads which are octads containing {p, q} with p and q removed; Γ_{2} are the octets which are the octads disjoint from {p, q} and Γ_{3} are the pairs which are 2-element subsets of Ω \ {p, q}. A hexad and an octet are adjacent if they are disjoint; the adjacency between the hexads and pairs is via inclusion, finally an octet is adjacent to a pair {r, s} if it is the union of two tetrads from the sextet containing {p, q, r, s}. Below we present the diagrams D_{i}(Mat_{22}) describing the orbits of M_{i} on Γ and the adjacencies between the vertices in these orbits. These diagrams are analogous to the diagrams D_{i}(Mat_{24}). The proofs of the diagrams can be found in Section 3.9 in [9].

[figure: 30_{2}; 60_{2}; 240_{2}; 1; 60_{1}; 16_{1}; 1_{1}; 15_{3}; 120_{3}; 96_{3}; 30; 15; 1; 6; 3; 6; 4; 1; 4; 4; 1; 6; 3; 3; 6; 2; 15; 15; 4; 6; 1; 3; 3; 24; 6; 3; 5; 8; 10; 4; 15; 1; 3; D_{1}(Mat_{22})]

[figure: 7_{1}; 14_{1}; 56_{1}; 1; 42_{2}; 112_{2}; 1_{2}; 7_{3}; 84_{3}; 112_{3}; 7; 7; 1; 3; 3; 12; 12; 6; 2; 1; 3; 7_{2}; 168_{2}; 28_{3}; 6; 3; 6; 1; 4; 3; 3; 1; 8; 24; 16; 3; 1; 6; 12; 8; 3+12; 2; 7; 4; 1; 2; 4; 1; 6; 2; 1+4; 1; 2; 4; 3; 3; 2; 6; 1; 2; 4; 4; 4; 3; 4; 1; 4; 4; 6; 1; D_{2}(Mat_{22})]

[figure: 5_{1}; 40_{1}; 32_{1}; 1; 160_{3}; 1_{3}; 10_{2}; 160_{2}; 5; 10; 1; 3; 6; 4; 4; 3; 15; 2; 10; 30_{3}; 6; 1; 3; 6; 6; 2; 6; 40_{3}; 120_{2}; 40_{2}; 24; 8; 1; 6; 1+6; 16; 12; 5; 15; 4; 4; 4; 1; 4; 4; 6; 3; 3; 1; 1; 1; 2; 2; 4; 4; 3; 1+6; 4; 1; D_{3}(Mat_{22})]

We need some further refinement of the information given on the above diagrams.

Lemma 4.7 (i) Let a ∈ Γ_{1}(32, 3). View Γ_{3}(a) as the set of points in a 4-dimensional symplectic space S over GF (2) with M̂_{a}/Q_{a} acting as the full group of automorphisms. Then M̂_{γa} fixes a non-degenerate quadratic form Q of minus type on S and Γ_{3}(40, 3) is the set of singular points of Q. In particular, for each b ∈ Γ_{2}(a), there is a unique c ∈ Γ_{3}(ab) ∩ Γ_{3}(40, 3).

(ii) Q̂_{γ} act regularly on Γ_{1}(32, 3).

Proof. Note that any subgroup of index 32 in M̂_{γ} is isomorphic Sym(5) and so in particular M̂_{γa} ≅ Sym(5) and Q̂_{γ} acts regularly on Γ_{1}(32, 3). Thus the lemma follows directly from the diagram D_{3}(Mat_{22}) and elementary properties of the 4-dimensional symplectic GF (2)-geometry. □

Lemma 4.8 Let c ∈ Γ_{3}(160, 3). Then M̂_{γc}Q̂_{c}/Q̂_{c} ≅ Sym(3) × C_{2}, Q_{c} ∩ M_{γ} = 1 and Q̂_{c} ∩ M̂_{γ} ≅ C_{2}.

Proof. By D_{3}(Mat_{22}), γ ∪ c is not contained in a hexad. In particular γ and c are disjoint. Thus there exists exactly two hexads a_{1} and a_{2} such that c ⊂ a_{i} and γ ∩ a_{i} ≠Ø. Thus M̂_{γc} normalizes a subset of size two of the five hexads adjacent to c. Thus M̂_{γc}Q̂_{c}/Q̂_{c} is contained in a Sym(3) × C_{2} subgroup of M_{c}/Q_{c}. Let t ∈ Q_{c} ∩ M_{γ}. Then t normalizes a_{i} and fixes γ ∩ a_{i} for i = 1, 2 and also fixes the two elements in c. Thus t fixes three elements in a_{1}. Since M_{a_{1}}/Q_{a_{1}} ≅ Alt(6), t does not induce a 2-cycle on a_{1} and thus fixes a_{1} elementwise. Since t also fixes the point a_{2} ∩ γ outside of a_{1} we conclude t = 1 and Q_{c} ∩ M_{γ} = 1. Thus |Q̂_{c} ∩ M̂_{γ}| ≤ 2. Since |M̂_{γc}Q̂_{c}/Q̂_{c}| ≤ 12 and |M̂_{γc}| = 24, the lemma is established. □

5 J_{4}-triangles

In this section we establish the existence and uniqueness of a J_{4}-triangle of groups. We follow notation from Section 1.

Lemma 5.1 Let (M_{1}, M_{2}, M_{3}) be a J_{4}-triangle. Let K_{1} ≅ Mat_{24} be a complement to Q_{1} in M_{1}; K_{2} ≅ L_{5}(2) be a complement to Q_{2} in M_{2} and let L be the unique normal subgroup in M_{3} such that M_{3}/L ≅ Sym_{5}. Let S be the Steiner system of type (5, 8, 24) such that Q_{1} is the irreducible Todd module associated with the action of K_{1} on S and Ω(3) be an M_{3}-set of size 5 such that C_{M_{3}}(Ω(3)) = L. Then

(i) there are subsets Ω_{1}(3) and Ω_{2}(3) in Ω(3) of size 1 and 2 respectively with Ω_{1}(3) ⊈ Ω_{2}(3) such that M_{31} = N_{M_{3}}(Ω_{1}(3)) and M_{32} = N_{M_{3}}(Ω_{2}(3)); in particular,

M_{31}/Q_{3} ≅ Sym(4) × L_{3}(2), M_{32}/Q_{3} ≅ C_{2} × Sym(3) × L_{3}(2) and B/Q_{3} ≅ C_{2} × C_{2} × L_{3}(2),

moreover, Q_{1} ∩ Q_{2} ≤ Q_{3} and T_{13} ≰ M_{2};

(ii) there is a natural L_{5}(2)-module V (2) of M_{2}, a 1-space V_{1}(2) and a 2-space V_{3}(2) in V (2) with V_{1}(2) ≤ V_{3}(2) such that M_{23} = N_{M_{2}}(V_{3}(2)) and M_{21} = N_{M_{2}}(V_{1}(2)); in particular,

M_{23}/Q_{2} ∼ 2^{6}(Sym(3) × L_{3}(2)), M_{21}/Q_{2} ∼ 2^{4}L_{4}(2) and B/Q_{2} ∼ 2^{3+3+1}L_{3}(2);

(iii) there is an octad α and a trio β containing α such that M_{12}Q_{1} = N_{M_{1}}(α) and M_{13} = N_{M_{1}}(β); in particular,

M_{13}/Q_{1} ∼ 2^{6}(Sym(3) × L_{3}(2)), M_{12}Q_{1}/Q_{1} ∼ 2^{4}L_{4}(2) and BQ_{1}/Q_{1} ∼ 2^{3+3+1}L_{3}(2),

moreover, |Q_{1}/Q_{1} ∩ M_{2}| = 2;

(iv) For all i ≠ j, M_{ij} acts irreducibly on T_{ij}/Q_{j}.

(v) T_{13} = Q_{1}Q_{3}, T_{23} = Q_{2}Q_{3} and T_{12} = (Q_{1} ∩ M_{2})Q_{2}.

(vi) |Q_{2}/Q_{2} ∩ Q_{3}| = 2, |Q_{2}/Q_{1} ∩ Q_{2}| = 2^{4} and Q_{2} is isomorphic to ⋀^{2} V (2)^{∗} where V (2) is as in (ii).

(vii) Φ(Q_{3}) = Z(Q_{3}) is a natural L_{3}(2)-module for M_{3} and Q_{3}/Φ(Q_{3}) ≅ D_{1} ⊗ D_{2}, where D_{1} is a natural ΓL_{2}(4)-module for M_{3} and D_{2} is dual to Z(Q_{3}).

(viii) L = O^{2}(B).

(ix) N_{M_{i}}(Q_{i} ∩ Q_{j}) = M_{ij} if (i, j) ≠ (1, 2) and N_{M_{1}}(Q_{1} ∩ Q_{2}) = Q_{1}M_{12}.

Proof. Since |M_{3}/M_{13}| = 5 and M_{3} has a unique class of subgroups of index 5, we can put Ω(3) = M_{3}/M_{31} so that L = ∩_{g∈M_{3}}M_{31}^{g}. Since |M_{32}/B| = 3, M_{32} has on orbit of length 3 on Ω(3). Thus M_{32}L/L is contained in a Sym(3) × C_{2}-subgroup of M_{3}/L. Since the index of M_{23} in M_{3} and the index of Sym(3)×C_{2} in Sym_{5} are both 10, we conclude that L ≤ M_{32} and M_{32}/L ≅ Sym(3)×C_{2}. In particular, L ≤ B and since |M_{32}/B| = 3 we have B/L ≅ C_{2}×C_{2}. This implies that B/L contains 2-cycles and so B/L ≠ O_{2}(M_{31}/L). As O_{2}(M_{31}/L) = T_{31}L/L we get T_{31} ≰ B and T_{13} ≰ M_{2} which gives (i).

For (ii) let i ∈ {1, 3} and let V (2) be some natural L_{5}(2)-module for M_{2}. Since |M_{2}/M_{23}| = 155 and |M_{2}/M_{21}| = 31, M_{2i} contains a Sylow 2-subgroup of M_{2}. In particular, Q_{2} ≤ M_{2i} and M_{2i} is the normalizer of some flag in V (2). Since |M_{2}/M_{21}| = 31, M_{21} = N_{M_{2}}(V_{1}(2)) for some 1- or 4-space V_{1}(2) in V (2). Replacing V (2) by its dual if necessary we may assume that V_{1}(2) is a 1-space. Since |M_{2}/M_{23}| = 155, M_{23} = N_{M_{2}}(V_{3}(2)) for some 2- or 3-space V_{3}(2) in V (2). Since |M_{23}/M_{23} ∩M_{12}| = |M_{23}/B| = 3 which is odd, V_{1}(2) ≤ V_{3}(2) and V_{3}(2) is a 2-space. Thus (ii) holds.

(iii) Since |M_{1}/M_{13}| = 3795, M_{13} contains a Sylow 2-subgroup of M_{1} and so by 3.4 M_{13} = N_{M_{1}}(β) for some trio β in S. Suppose that Q_{1} ≤ Q_{3}. Since |Q_{1}| > |Q_{2}|, Q_{1} ≰ Q_{2}. By (ii) M_{12} acts irreducibly on T_{12}/Q_{2} and so T_{12} = Q_{1}Q_{2}. Hence T_{12} = Q_{1}Q_{2} ≤ Q_{3}Q_{2} ≤ T_{23} a contradiction since by (ii) T_{23} centralizes V_{3}(2) but T_{12} does not. Thus Q_{1} ≰ Q_{3} and so by (i), Q_{1}Q_{3} = T_{13} and Q_{1} ≰ M_{2}. Hence Q_{1} ≰ M_{12}, |Q_{1}/Q_{1} ∩ M_{2}| = 2, |M_{1}/M_{12}Q_{1}| = 759 and by 3.4, M_{12}Q_{1} = N_{M_{1}}(α) for some octad α. Also |M_{13}/Q_{1}B| = 3. By 3.2, M_{13}/Q_{1} has a unique class of subgroups of index 3 and so BQ_{1} = N_{M_{13}}(α^{∗}) for some octad α^{∗} in β. By 3.4 (ii) BQ_{1} fixes a unique octad, so α = α^{∗} and (iii) holds.

(iv) follows from (i), (ii) and (iii).

We already proved that Q_{1}Q_{3} = T_{13}. Now (T_{13}L/L)^{#} contains no 2-cycles and by (i) Q_{1} ∩ M_{2} ≰ T_{23}. Thus Q_{1} ∩ M_{2} ≰ Q_{2} and by (iv), (Q_{1} ∩ M_{2})Q_{2} = T_{12}. Since |Q_{3}| > |Q_{2}|, Q_{3} ≰ Q_{2} and by (iv) Q_{2}Q_{3} = T_{23} and (v) holds.

By (v) and (i) |Q_{2}/(Q_{2} ∩ Q_{3})| = |Q_{2}Q_{3}/Q_{3}| = |T_{23}/Q_{3}| = 2, by (v) and (iii) |Q_{2}/(Q_{2} ∩ Q_{1})| = |T_{12}/Q_{1}| = 2^{4} and by (i) Q_{2} ∩ Q_{1} ≤ Q_{3}. By the definition of J_{4}-triangle Q_{2} is isomorphic either to ⋀^{2} V (2) or to ⋀^{2} V (2)^{∗}. Since M_{21} = N_{M_{2}}(V_{1}(2)) for a 1-space V_{1}(2) in V (2), the only proper subspace in Q_{2} normalized by M_{12} has dimension 4 in the former case and dimension 6 in the latter case. Since Q_{1} ∩ Q_{2} is a 6-space (vi) follows.

Let Z_{3} = C_{Q_{1}}(Q_{3}). Since T_{13} = Q_{1}Q_{3}, we have Z_{3} = C_{Q_{1}}(T_{13}). Since Q_{1} is the irreducible Todd module, by 3.7 Z_{3} has order 2^{3} and Z_{3} ≤ Q_{1} ∩ Q_{3}. By 3.7 (iii) Z(Q_{3}) ≤ Z_{3} and hence Z_{3} = Z(Q_{3}). By (iv) and (v) Φ(Q_{3}) ≤ Q_{1} ∩ Q_{2}. Since Z_{3} < Q_{1} ∩ Q_{2} < Q_{1} ∩ Q_{3}, since M_{13} acts irreducibly on Q_{1} ∩ Q_{3}/Z_{3} and since M_{31} normalizes Φ(Q_{3}) we conclude that Φ(Q_{3}) ≤ Z_{3}. On the other hand by 3.7 [Q_{1} ∩ Q_{3}, Q_{3}] = Z_{3} and so Φ(Q_{3}) = Z_{3}. By 3.7, (Q_{1} ∩ Q_{3})/Z_{3} is the unique proper M_{31}-submodule in Q_{3}/Z_{3}. Moreover, all composition factors for L on Q_{3}/Z_{3} are dual to Z_{3} and the elements of order three in C_{M_{31}}(Z_{3}) act fixed-point freely on Q_{3}/Z_{3}. By (ii) Q_{2} ∩ Q_{3}/Z_{3} is the unique proper M_{23}-submodule in Q_{3}/Z_{3} and since Q_{1} ∩ Q_{2} < Q_{2} ∩ Q_{3}, Q_{1} ∩ Q_{3} ≠ Q_{2} ∩ Q_{3}. Thus M_{3} acts irreducible on Q_{3}/Z_{3} and (vii) holds.

(viii) By (i) |B/L| = 4 and by (vii) O^{2}(L) = L. Thus (viii) holds.

(ix) Clearly Q_{i} ∩ Q_{j} is normal in Q_{i}M_{ij} and the latter is equal to M_{ij} unless (i, j) = (1, 2). On the other hand in each case Q_{i}M_{ij}/Q_{i} is maximal in M_{i}/Q_{i} and hence the result follows. □

Our next result will be used as a characterization of M_{12}.

Lemma 5.2 For i = 1 and 2 let X_{i} be a group generated by subgroups Z_{i}, A_{i}, B_{i} and R_{i} such that

(i) R_{i} is isomorphic to L_{4}(2);

(ii) Z_{i}, A_{i} and B_{i} are elementary abelian 2-groups of order 2^{6}, 2^{4} and 2^{4}, respectively;

(iii) R_{i} normalizes Z_{i}, A_{i}, and B_{i}, A_{i} and B_{i} are isomorphic natural L_{4}(2)-modules for R_{i} and Z_{i} is isomorphic to the exterior square of A_{i}, that is Z_{i} is a natural Ω_{6}^{+}(2)-module for R_{i};

(iv) Z_{i} centralizes A_{i} and B_{i};

(v) [A_{i}, B_{i}] = Z_{i}.

Then

(a) there exists an isomorphism from X_{1} onto X_{2} mapping Y_{1} to Y_{2} for Y = Z, A, B and R; (b) Out X_{i} is elementary abelian of order 2^{2}.

Proof. Fix i ∈ {1, 2} and put Y = Y_{i} for Y ∈ {X, R, A, B, Z}. Pick 1 ≠ a ∈ A and put P = C_{R}(a). Note that A, B and Z are absolutely irreducible GF (2)R-modules and so there exist unique GF (2)R-isomorphisms ϕ : A → B and ψ : ⋀^{2} A → Z. Define ξ : A × A → Z by ξ(v, w) = [v, ϕ(w)]. Since A is irreducible and [A, B] ≠ 1, [a, B] ≠ 1. Note that [a, B] and B/C_{B}(a) are isomorphic as GF (2)P -modules. Moreover, P fixes no non-zero vector in Z and so [a, ϕ(a)] = 1. Thus ξ(a, a) = 1 and so ξ extends to a GF (2)R-homomorphism Ξ : ⋀^{2} A → Z. Thus Ξ = ψ and so [v, ϕ(w)] = ψ(v ∧ w) for all v, w ∈ A. It is now clear that (a) holds.

Put Q = ABZ. By 2.7 all complements to Q/Z in X/Z are conjugate in X/Z and ZR has two classes of complements to Z. Thus X has two classes of complements to Q and it follows easily from (a) that there exists an automorphism of X interchanging these two classes. Let α be an automorphism of X normalizing R. Since the module for R dual to A is not involved in Q, α induces an inner automorphism on R. So we may assume that α centralizes R.

Let C/Z be the unique irreducible R-submodule in Q/Z different from AZ/Z and BZ/Z. We claim that R does not normalize a complement to Z in C. First notice that if a complement in C exists, it should consist of the elements bϕ(b), b ∈ A, since aϕ(a) is the only element in C invariant under the maximal parabolic P in R. However these elements are not closed under multiplication, since

aϕ(a)bϕ(b) = abϕ(ab)[ϕ(a), b]

and the factor [ϕ(a), b] is non-trivial when a ≠ b. Thus the claim follows.

By the claim A^{α} ≰ C and {A, B} = {A^{α}, B^{α}}. Again by (a) there exists an automorphism of X normalizing R and interchanging A and B. So we may assume that α normalizes A and B. Since α centralizes R and since A, B and Z are absolutely irreducible GF (2)R-modules, α centralizes A, B and Z and α is the identity automorphism. □

Let V (2) be a 5-dimensional GF (2)-space, K_{2}^{∘} = GL(V (2)) ≅ L_{5}(2) and M_{2}^{∘} be the semidirect product of Q_{2}^{∘} := ⋀^{2} V (2)^{∗} and K_{2}^{∘} with respect to the natural action. Let V_{1}(2) be a 1-space in V (2), K_{21}^{∘} be the stabilizer of V_{1}(2) in K_{2}^{∘} and M_{21}^{∘} be the subgroup in M_{2}^{∘} which is the semidirect product of Q_{2}^{∘} and K_{21}^{∘}. Let S be a Steiner system of type (5,8,24), K_{1}^{∘} = Aut S ≅ Mat_{24}, Q_{1}^{∘} be the 11-dimensional Todd module associated with the action of K_{1}^{∘} on S and M_{1}^{∘} be the semidirect product of Q_{1}^{∘} and M_{1}^{∘}. Let α be an octad in S, K_{12}^{∘} be the stabilizer of α in K_{1}^{∘}, H_{1} be the unique hyperplane in Q_{1}^{∘} stabilized by K_{12}^{∘} (compare 3.2 and 3.7) and M_{12}^{∘} be the subgroup in M_{1}^{∘} which is the semidirect product of H_{1} and K_{12}^{∘}.

Lemma 5.3 Let X_{i} = Z_{i}A_{i}B_{i}R_{i} the group introduced in 5.2, then

(i) there is an isomorphism of M_{21}^{∘} onto X_{i} which sends K_{21}^{∘} onto A_{i}R_{i};

(ii) there is an isomorphism of M_{12}^{∘} onto X_{i} which sends K_{12}^{∘} onto A_{i}R_{i}.

Proof. By 2.8 and the obvious duality there is an orbit H(2)^{∗} of L_{5}(2) on the set of vectors in Q_{2}^{∘} indexed by the 3-spaces in V (2). Let A_{2} = O_{2}(K_{21}^{∘}), Q_{21}^{∘} = O_{2}(M_{21}^{∘}) and R_{2} a complement to A_{2} in K_{21}^{∘} normalizing a complement U to V_{1}(2) in V (2). Then R_{2} is isomorphic to L_{4}(2) and A_{2} is the kernel of the action of K_{21}^{∘} on the set of subspaces in V (2) containing V_{2}(1). This means that A_{2} is dual to U and the latter is canonically isomorphic to V (2)/V_{1}(2). The elements from H(2)^{∗} corresponding to 3-spaces containing V_{1}(2) are centralized by A_{2} and by a standard property of exterior squares they generate an R_{2}-submodule Z_{21} in Q_{2}^{∘} isomorphic to ⋀^{2} A_{2}. The elements from H(2)^{∗} corresponding to 3-spaces taken from U generate a complement B_{2} to Z_{21} in Q_{2}^{∘} normalized by R_{2} and isomorphic to A_{2}. In particular Z_{21} = C_{Q_{2}^{∘}}(A_{2}) = Z(Q_{21}^{∘}). Moreover, M_{21}^{∘} = Z_{21}A_{2}B_{2}R_{2} and (i) follows.

Next, let A_{1} = O_{2}(K_{12}^{∘}) and Z_{12} = C_{Q_{1}^{∘}}(A_{1}). Let t ∈ Q_{1}^{∘} \ H_{1}. By 3.2 and 3.7 (i) A_{1} acts regularly on the elements in tH_{1}/Z_{12}. Put R_{1} = N_{K_{12}^{∘}}(tZ_{12}). Then by the Frattini argument R_{1} is a complement to A_{1} in K_{12}^{∘}. In particular R_{1} ≅ L_{4}(2). Put B_{1} = A_{1}^{t}. Since t normalizes Z_{12}R_{1} and Z_{12}R_{1} normalizes A_{1} we conclude that Z_{12}R_{1} normalizes B_{1}. Thus B_{1} is R_{1}-invariant. Clearly A_{1} and B_{1} are isomorphic as R_{1}-modules and by 3.7 Z_{12} is isomorphic to the exterior square of A_{1}. Moreover, M_{12}^{∘} = Z_{12}A_{1}B_{1}R_{1} and so by 5.2 we obtain (ii). □

Lemma 5.4 With M_{1}^{∘}, M_{2}^{∘}, M_{12}^{∘} and M_{21}^{∘} as above there exists a unique amalgam (M_{1}^{∘}, M_{2}^{∘}) such that M_{1}^{∘} ∩ M_{2}^{∘} = M_{12}^{∘} = M_{21}^{∘} and K_{12}^{∘} = K_{21}^{∘}.

Proof. By 5.3 there is an isomorphism of M_{12}^{∘} onto M_{21}^{∘} which sends K_{12}^{∘} onto K_{21}^{∘} and hence the existence follows. In order to prove the uniqueness it is sufficient to show that for every automorphism σ of M_{12}^{∘} there is an automorphism δ of M_{1}^{∘} which normalizes M_{12}^{∘} such that the restriction of δ to M_{12}^{∘} coincides with σ. This is certainly true if σ is an inner automorphism and by 5.2 (b) and 5.3 the outer automorphism group of M_{12}^{∘} is of order 2^{2}. Thus it is sufficient to present a subgroup M̂_{1} in the automorphism group of M_{1}^{∘}, containing the inner automorphisms such that M_{12}^{∘} (identified with a subgroup of inner automorphisms of M_{1}^{∘}) has trivial centralizer in M̂_{1} and NM̂_{1}(M_{12}^{∘})/M_{12}^{∘} ≅ 2^{2}. Let Q̂_{1} be the 12-dimensional GF (2)M_{1}^{∘}-module obtained from the 24-dimensional permutational module on the element set Ω of the Steiner system S modulo the 12-dimensional Golay code. Let M̂_{1} be the semidirect product of Q̂_{1} and K_{1}^{∘}. Then M̂_{1} contains M_{1}^{∘} as a subgroup of index 2. It is well known (cf. [1] or [9]) that K_{1}^{∘} has four orbits on Q̂_{1}^{#} with lengths 24, 276, 2024 and 1771 indexed by 1-, 2-, 3-element subsets of Ω and by the sextets, respectively. This shows that CQ̂_{1}(K_{12}^{∘}) = 1 and hence CM̂_{1}(M_{12}^{∘}) = 1. On the other hand it is clear that M_{12}^{∘} is a normal subgroup of index 2^{2} in the subgroup in M̂_{1} which is the semidirect product of Q̂_{1} and K_{12}^{∘} and the result follows. □

In view of the preceding lemma we may and do identify M_{12}^{∘} with M_{21}^{∘} and K_{12}^{∘} with K_{21}^{∘}.

Lemma 5.5 Let (M_{1}, M_{2}, M_{3}) be a J_{4}-triangle of groups. There exists an isomorphism κ of the amalgam (M_{1}^{∘}, M_{2}^{∘}) as in 5.4 onto the subamalgam (M_{1}, M_{2}).

Proof. By 1.3 (i), (ii) there are isomorphisms κ_{1} : M_{1}^{∘} → M_{1} and κ_{2} : M_{2}^{∘} → M_{2}. By 5.1 (ii) and (iii) these isomorphisms can be chosen in such a way that that κ(M_{12}^{∘}) = M_{12} and κ(M_{21}^{∘}) = M_{21}. Now the uniqueness statement in 5.4 ensures existence of the isomorphism κ of amalgams. □

Notice that at this stage we do not know whether or not a J_{4}-triangle of groups exists but we do know that the rank two amalgam (M_{1}^{∘}, M_{2}^{∘}) exists.

Let β be a trio containing the octad α. Put M_{13}^{∘} = N_{M_{1}^{∘}}(β), B^{∘} = M_{12}^{∘} ∩ M_{13}^{∘}, L^{∘} = O^{2}(B^{∘}), M_{23}^{∘} = N_{M_{2}^{∘}}(L^{∘}), Q_{13}^{∘} = O_{2}(M_{13}^{∘}), Q_{3}^{∘} = O_{2}(L^{∘}) and Z_{3}^{∘} = Z(Q_{3}^{∘}).

Lemma 5.6 (i) L^{∘} = O^{2,3}(M_{13}^{∘}), M_{13}^{∘} = N_{M_{1}^{∘}}(L^{∘}), L^{∘}/Q_{3}^{∘} ≅ L_{3}(2), L^{∘} splits over Q_{3}^{∘}, Q_{1}^{∘}∩Q_{3}^{∘} ≤ H_{1} ≰ L^{∘}, Q_{1}^{∘}∩Q_{3}^{∘} = [Q_{1}^{∘}, Q_{13}^{∘}], Q_{13}^{∘} = Q_{1}^{∘}Q_{3}^{∘}, Z_{3}^{∘} = Φ(Q_{3}^{∘}) = (Q_{3}^{∘})′ = C_{Q_{3}^{∘}}(Q_{13}^{∘}), Z_{3}^{∘} is a natural L_{3}(2)-module for L^{∘} and Q_{3}^{∘}/Z_{3}^{∘} is the direct sum of four natural L_{3}(2)-modules dual to Z_{3}^{∘}.

(ii) M_{23}^{∘} = N_{M_{2}^{∘}}(V_{3}(2)) where V_{3}(2) is some 2-space in V (2) containing V_{1}(2).

(iii) M_{13}^{∘}/L^{∘} ≅ Sym(4) and M_{23}^{∘}/L^{∘} ≅ Sym(3) × C_{2}.

(iv) B^{∘} = M_{21}^{∘} ∩ M_{23}^{∘}, B^{∘}/L^{∘} ≅ C_{2} × C_{2} and B^{∘} is not normal in M_{13}^{∘}.

(v) C_{M_{1}^{∘}}(L^{∘}) = 1 = C_{M_{2}^{∘}}(L^{∘}).

(vi) the isomorphism κ in 5.5 can be chosen in such a way that κ(Y^{∘}) = Y for Y = B, L, M_{13}, M_{23}, Q_{3} and Z_{3}.

Proof. Let L′ be the kernel of the action of M_{13}^{∘} on the three octads in β. Since O^{2}(B^{∘}) fixes the two octads in β different from α, L^{∘} ≤ L′ ≤ B^{∘} and so L^{∘} = O^{2}(L′) ⊴ M_{13}^{∘}. Since M_{13}^{∘} is maximal in M_{1}^{∘}, M_{13}^{∘} = N_{M_{1}^{∘}}(L^{∘}). The remaining statements in (i) now follow from 3.2 and 3.7 (i) - (iii).

Recall that we identified M_{12}^{∘} and M_{21}^{∘}. Since α is contained in 15 trios, B^{∘} has index 15 in M_{12}^{∘}. Thus B^{∘} = N_{M_{12}^{∘}}(V_{3}(2)) where V_{3}(2) is 2- or 4-space in V (2) containing V_{1}(2). If V_{3}(2) is a 4-space then B^{∘}/Q_{2}^{∘} is an extraspecial group of order 2^{7} extended by L_{3}(2). Since L^{∘} = O^{2}(L^{∘}) = O^{2}(B^{∘}) we conclude that L^{∘} has a chief factor isomorphic to C_{2}, a contradiction to (i). Thus V_{3}(2) is a 2-space and L^{∘} = O^{2}(B^{∘}) = O^{2}(C_{M_{2}^{∘}}(V_{3}(2))). This means that L^{∘} is normal in N_{M_{2}^{∘}}(V_{3}(2)) and since the latter is maximal in M_{2}^{∘} it must be equal to M_{23}^{∘} = N_{M_{2}^{∘}}(L^{∘}) and (ii) follows.

(iii) By 3.2, M_{13}^{∘}/L^{∘}Q_{1}^{∘} ≅ Sym(3) and by 3.7 (iii) Q_{1}^{∘}/Q_{1}^{∘} ∩ Q_{3}^{∘} is isomorphic to the natural L_{2}(2)-module for M_{13}^{∘}. Thus M_{13}^{∘}/L^{∘} ≅ Sym(4). In M_{2}^{∘} we compute that M_{23}^{∘}/L^{∘}Q_{2}^{∘} ≅ Sym(3), M_{23}^{∘} splits over L^{∘}Q_{2}^{∘} and |Q_{2}^{∘}/Q_{2}^{∘} ∩ Q_{3}^{∘}| = 2. Thus M_{23}^{∘}/L^{∘} ≅ Sym(3) × C_{2}.

(iv) Clearly B^{∘} = N_{M_{12}^{∘}}(L^{∘}) = M_{12}^{∘} ∩M_{13}^{∘} = M_{21}^{∘} ∩M_{23}^{∘} = N_{M_{2}^{∘}}(V_{1}(2), V_{3}(2)). Hence we compute in M_{2}^{∘} that B^{∘}/L^{∘} ≅ C_{2}×C_{2}. Since Q_{1}^{∘} ≰ M_{12}^{∘}, we have B^{∘} ≠ Q_{1}^{∘}L^{∘}. Since Q_{1}^{∘}L^{∘}/L^{∘} = O_{2}(M_{13}^{∘}/L^{∘}), B^{∘} is not normal in M_{13}^{∘}.

(v) is readily verified in M_{1}^{∘} (see 3.2) and M_{2}^{∘}.

Finally (vi) follows from (i) - (v) and 5.1. □

Let M_{3}^{∘} be the universal completion of the amalgam (M_{13}^{∘}, M_{23}^{∘}) (which is the free amalgamated product of M_{13}^{∘} and M_{23}^{∘} over B^{∘}) and let (M_{1}^{∘}, M_{2}^{∘}, M_{3}^{∘}) be a triangle of groups where M_{i}^{∘} ∩ M_{j}^{∘} = M_{ij}^{∘} for 1 ≤ i < j ≤ 3. We are ready to prove the uniqueness statement for J_{4}-triangles.

Lemma 5.7 Every J_{4}-triangle of groups is isomorphic to the triangle (M_{1}^{∘}, M_{2}^{∘}, M_{3}^{∘}/N), where N = C_{M_{3}^{∘}}(L^{∘}).

Proof. Let (M_{1}, M_{2}, M_{3}) be a J_{4}-triangle of groups, κ be an isomorphism of (M_{1}^{∘}, M_{2}^{∘}) onto (M_{1}, M_{2}) as in 5.5, satisfying the condition in 5.6 (vi). Since M_{3} is generated by the subgroups M_{31}, M_{32} there is a mapping of (M_{1}^{∘}, M_{2}^{∘}, M_{3}^{∘}) onto (M_{1}, M_{2}, M_{3}) whose restriction to M_{1}^{∘} ∪ M_{2}^{∘} coincides with κ and whose restriction to M_{3}^{∘} is a homomorphism χ onto M_{3}. Thus the isomorphism type of (M_{1}, M_{2}, M_{3}) is uniquely determined by the kernel N of χ. We claim that N = C_{M_{3}^{∘}}(L^{∘}). On the one hand, N and L^{∘} are normal subgroups in M_{3}^{∘} and N ∩ L^{∘} = 1 since the restriction of κ to L^{∘} is an isomorphism onto L, hence N ≤ C_{M_{3}^{∘}}(L^{∘}). On the other hand by 5.6 (v) and since M_{3}/L ≅ Sym(5) we have C_{M_{3}}(L) = 1 and hence N ≥ C_{M_{3}^{∘}}(L^{∘}). Thus the claim follows and implies the result. □

For the remainder of the section we identify (M_{1}, M_{2}, M_{3}) with (M_{1}^{∘}, M_{2}^{∘}, M_{3}^{∘}/N) where N = C_{M_{3}^{∘}}(L^{∘}). In order to prove the existence we have to show that this is in fact a J_{4}-triangle of groups. For this we have to show that M_{3}/L ≅ Sym(5). By the definition M_{3} is the subgroup in Aut L generated by M_{13} and M_{23} (identified with their isomorphic images in Aut L). We need the following preliminary result.

Lemma 5.8 Let S be the symmetric group Sym(6) of degree 6. Let H_{1} and H_{2} be subgroups in S with H_{1} ≅ Sym(4), H_{2} ≅ Sym(3) × C_{2} and H_{1} ∩ H_{2} ≅ C_{2} × C_{2}. Then ⟨H_{1}, H_{2}⟩ ≅ Sym(5).

Proof. Let A_{1} and A_{2} be representatives of the conjugacy classes of Sym(5) subgroups in S. Put Ω_{i} = S/A_{i}, i = 1, 2. We choose representatives k_{1}, k_{2} and k_{3} of the conjugacy classes of involutions in S so that k_{i} acts as a transposition on Ω_{i} for i = 1, 2 and k_{3} ∈ S′ ≅ Alt(6). Then C_{S}(k_{1}) ≅ C_{S}(k_{2}) ≅ Sym(4)×C_{2} and C_{S}(k_{3}) ≅ D_{8} ×C_{2}. There are two conjugacy classes of Sym(4) subgroups in S not contained in S′. We choose representatives B_{1} and B_{2} of these classes so that B_{i} is the elementwise stabilizer in S of a pair of cosets from Ω_{i}, or equivalently, that B_{i} contains a conjugate of k_{i}, i = 1, 2. Applying the symmetry with respect to the full automorphism group of S, we assume that the central involution in H_{2} is k_{1}. Then H_{2} acting on Ω_{1} fixes a coset, say α and H_{1} ∩ H_{2} fixes two such cosets, say α and β. Since H_{1} contains k_{1}, it is a conjugate of B_{1} and hence fixes two cosets from Ω_{1}. Clearly these cosets must be α and β. This means that ⟨H_{1}, H_{2}⟩ fixes α and obviously it is the whole stabilizer of α in S, isomorphic to Sym(5). □

Lemma 5.9 Let M_{3} be the subgroup of Aut L generated by M_{13} and M_{23}. Then M_{3}/L ≅ Sym(5). In particular, (M_{1}, M_{2}, M_{3}) is J_{4}-triangle of groups.

Proof. We will use the information about M_{13} and M_{23} obtained in 5.6 without further reference. Let U be a complement to Q_{3} in L and S a Sylow 7-subgroup of U. Since Z_{3} is a natural module for L, M_{3} induces only inner automorphism on L/Q_{3} and so M_{3} = C_{M_{3}}(S)L. Put C = C_{M_{3}}(Q_{3}/Z_{3}) ∩ C_{M_{3}}(Z_{3}) and E = C_{C}(S). Then C = Q_{3}E. Let e ∈ E. Then the map

ξ : Q_{3}/Z_{3} → Z_{3}, xZ_{3} ↦ [x, e]

is a GF (2)S-homomorphism. Since Q_{3}/Z_{3} is the direct sum of four L_{3}(2)-modules dual to Z_{3}, none of the S-composition factors in Q_{3}/Z_{3} are isomorphic to Z_{3}. Hence the image of ξ is the identity and E centralizes Q_{3}. In particular, [U, E] ≤ C_{L}(Q_{3}) = Z_{3}, E normalizes Z_{3}U and E acts faithfully on Z_{3}U. By 2.7 Z_{3}U has two classes of complements and so |E| ≤ 2 and |C/Q_{3}| ≤ 2.

Put D = C_{M_{3}}(Z_{3}) and M̄_{3} = M_{3}/C. Then D centralizes L/Q_{3} and D̄ acts faithfully on Q_{3}/Z_{3}. Thus there exists a faithful four dimensional GF (2)D̄-module R so that as D-module Q_{3}/Z_{3} is isomorphic to the direct sum of three copies of R. Let D_{i} = C_{M_{i3}}(Z_{3}) = M_{i3} ∩ D. Notice that M̄_{3} = D̄ × L̄, M̄_{i3} = D̄_{i} × L̄ and M_{3} = ⟨M_{13}, M_{23}⟩. Thus D = ⟨D_{1}, D_{2}⟩, D̄_{1} ≅ Sym(4), D̄_{2} ≅ Sym(3) × C_{2} and D̄_{1} ∩ D̄_{2} ≅ C_{2} × C_{2}. By 3.7 (iii) Q_{1} ∩ Q_{3} is the only M_{13}-invariant subgroup between Z_{3} and Q_{3}. Similarly, Q_{2} is uniserial as GF (2)M_{23}-module and Q_{2} ∩ Q_{3} is the only M_{23}- invariant subgroup between Z_{3} and Q_{3}. In addition Q_{2} ∩ Q_{3} has index 2 in Q_{2} and Q_{1} ∩ Q_{2} has index 2^{4} in Q_{2}. Thus Q_{1} ∩ Q_{2} ∩ Q_{3} ≠ Q_{2} ∩ Q_{3} and Q_{1} ∩ Q_{3} ≠ Q_{2} ∩ Q_{3}. Hence D acts irreducibly on R.

We claim that D̄ preserves on R a non-degenerate symplectic form. Notice that Q_{3} is non-abelian and D centralizes Q′_{3} = Z_{3}. Let X ≤ Z_{3} with |X| = 4 and Q′_{3} ≰ X. Let Y be maximal in Q_{3} with respect to the condition [Q_{3}, Y] ≤ X. Let W/Y be an irreducible D-submodule of Q_{3}/Y and let K be maximal in Q_{3} with [W, K] ≤ X. Then we obtain a non-degenerate D-invariant bilinear map

ϕ : W/Y × Q_{3}/K → Z_{3}/X ≅ GF (2)
(wY, qK) ↦ [w, q]X.

Hence by linear algebra, Q_{3}/K is isomorphic to the dual of W/Y and so irreducible. On the other hand all composition factors of D in Q_{3}/Z_{3} are isomorphic to R. Hence ϕ induces a D-invariant non-degenerate bilinear map

ψ : R × R → GF (2).

It remains to show that we can choose ψ to be a symplectic form. Define ψ^{∗}(x, y) = ψ(x, y) + ψ(y, x). Then clearly ψ^{∗}(x, y) is symmetric and ψ^{∗}(x, x) = 0. As D acts irreducibly on R, either ψ^{∗}(x, y) = 0 for every x, y ∈ R or ψ^{∗} is non-degenerate D-invariant symplectic form. Suppose that ψ^{∗} is trivial. In this case ψ is symmetric. In particular, {r ∈ R | ψ(r, r) = 0} forms a D-invariant subspace of index at most 2 in R. As R is irreducible we conclude that ψ(r, r) = 0 for all r ∈ R and so ψ is a symplectic form and the claim follows. Thus D̄ is a subgroup in Sp_{4}(2) ≅ Sym(6) generated by D̄_{1} ≅ Sym(4) and D̄_{2} ≅ Sym(3) × C_{2} with D̄_{1} ∩ D̄_{2} ≅ C_{2} × C_{2}. Hence D̄ ≅ Sym(5) by 5.8.

Notice that M_{3}/C = DL/C, and |M_{3}/M_{13}C| = 5. Since M_{13}/L ≅ Sym(4) does not contain normal subgroups of order 2, M_{13} ∩ C ≤ Q_{3} and hence M_{13}/Q_{3} is a complement to C/Q_{3} in M_{13}C/Q_{3}. Thus by Gaschütz’s theorem, M_{3}/Q_{3} splits over C/Q_{3}. Since D_{i} = D′_{i}(D_{1} ∩ D_{2})Q_{3} and |D_{1} ∩ D_{2}/(D_{1} ∩ D_{2} ∩ D′_{1})Q_{3}| = 2, |M_{3}/M′_{3}Q_{3}| ≤ 2. But |Sym(5)/Sym(5)′| = 2, hence C/Q_{3} = 1 and the lemma is proved. □

Thus up to isomorphism there exists a unique J_{4}-triangle of groups.

6 Amalgams of Modules

In this section we prove a number of results to be used in the next section where a J_{4}-triangle of groups will be constructed inside GL_{1333}(C). The following lemma is of crucial importance.

Lemma 6.1 Let (M_{1}, M_{2}, M_{3}) be a triangle of groups, H be a group and A be a subgroup of Aut H. Suppose that for all 1 ≤ i ≤ 3, there exist homomorphisms α_{i} : M_{i} → H and elements a_{i} ∈ A such that

α_{1}|_{M_{13}}a_{2} = α_{3}|_{M_{13}}, α_{2}|_{M_{12}}a_{3} = α_{1}|_{M_{12}} and α_{3}|_{M_{23}}a_{1} = α_{2}|_{M_{23}}.

Put M_{23}^{∗} = M_{23}^{α_{3}a_{2}^{−1}}, M_{13}^{∗} = M_{13}^{α_{1}}, M_{12}^{∗} = M_{12}^{α_{1}} and B^{∗} = B^{α_{1}}. Then

(i) The following two statements are equivalent:

(a1) There exist b_{i} ∈ A, 1 ≤ i ≤ 3, such that

α_{i}b_{i}|_{M_{ij}} = α_{j}b_{j}|_{M_{ij}}, for all i ≠ j.

(a2) a_{2}a_{1}a_{3} ∈ C_{A}(M_{23}^{∗})C_{A}(M_{13}^{∗})C_{A}(M_{12}^{∗}).

(ii) B^{∗} ≤ M_{12}^{∗} ∩ M_{13}^{∗} ∩ M_{23}^{∗} and a_{2}a_{1}a_{3} ∈ C_{A}(B^{∗}). In particular, (a2) and (a1) hold if

(∗) C_{A}(B^{∗}) = C_{A}(M_{23}^{∗})C_{A}(M_{13}^{∗})C_{A}(M_{12}^{∗}).

(iii) Assume that (a1) holds and that each α_{i}, 1 ≤ i ≤ 3, is one to one. Put M_{i}^{∗} = M_{i}^{α_{i}b_{i}}. If M_{i}^{∗} ∩M_{j}^{∗} = M_{ij}^{α_{i}b_{i}} for all 1 ≤ i < j ≤ 3, then (M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗}) is a triangle of groups isomorphic to (M_{1}, M_{2}, M_{3})

Proof. Replacing α_{2} by α_{2}a_{3}, α_{3} by α_{3}a_{2}^{−1} and a_{1} by a_{2}a_{1}a_{3} we may assume that a_{2} = a_{3} = 1.

(i) Replacing b_{i} by b_{i}b_{1}^{−1}, for all i, we see that (a1) is equivalent to :

(1) α_{1}|_{M_{13}} = α_{3}b_{3}|_{M_{13}}, α_{1}|_{M_{12}} = α_{2}b_{2}|_{M_{12}} and α_{2}b_{2}|_{M_{23}} = α_{3}b_{3}|_{M_{23}} for some b_{2}, b_{3} ∈ A.

Since α_{1}|_{M_{13}} = α_{3}|_{M_{13}}, α_{1}|_{M_{12}} = α_{2}|_{M_{12}} and α_{2}|_{M_{23}} = α_{3}a_{1}|_{M_{23}}, (1) is equivalent to

(2) b_{3} ∈ C_{A}(M_{13}^{∗}), b_{2} ∈ C_{A}(M_{12}^{∗}) and a_{1}b_{2}b_{3}^{−1} ∈ C_{A}(M_{23}^{∗}) for some b_{2}, b_{3} ∈ A.

Now (2) is obviously equivalent to (a2).

(ii) Since a_{2} = a_{3} = 1, α_{2}|_{B} = α_{1}|_{B} = α_{3}|_{B} and so B^{∗} ≤ M_{12}^{∗} ∩ M_{13}^{∗} ∩ M_{23}^{∗}. Moreover, since α_{2}|_{M_{23}} = α_{3}|_{M_{23}}a_{1}, we get α_{1}|_{B} = α_{1}a_{1}|_{B} and a_{1} ∈ C_{A}(B^{∗}).

(iii) is obvious. □

Lemma 6.2 Let K be a field, G be a group, H be a subgroup of finite index m in G, W be a finite dimensional KG-module and U be a non-zero finite dimensional KH-module. Suppose that each of the following statements holds:

(i) U is isomorphic to a KH-submodule of W;

(ii) dim_{K}W = m · dim_{K}U;

(iii) At least one of W and U ⨂_{KH} KG is irreducible as a KG-module.

Then W ≅ U ⨂_{KH} KG as KG-modules.

Proof. By (i) and the universality property of induced modules, there exists a non-zero KHhomomorphism Φ : U ⨂_{KH} KG → W. By (iii) Φ is onto (in the first case) or one-to-one (in the second case). By (ii) dim_{K} W = m · dim_{K} U = dim_{K} U ⨂_{KH} KG and so Φ is an isomorphism. □

Fundamental to our construction of a J_{4}-triangle inside GL_{1333}(C) is the concept of ”amalgam of modules”. Amalgams of modules are a special case of sheaves ( see for example [14]) and can be discussed in broad generality, but we will restrict ourselves to what is needed in this paper.

Definition 6.3 Let H be a group and H_{1} and H_{2} subgroups of H with H = ⟨H_{1}, H_{2}⟩. Put H_{0} = H_{1} ∩ H_{2} and let K be a field.

(i) An amalgam of K-modules for H_{1} ← H_{0} → H_{2} is a tuple (W_{0}, W_{1}, W_{2}, ϕ_{1}, ϕ_{2}), where W_{i} is a KH_{i}-module, 0 ≤ i ≤ 2 and ϕ_{i} : W_{0} → W_{i} is a KH_{0}-monomorphism, 1 ≤ i ≤ 2. Such an amalgam of modules is denoted by

W_{1} ←^{ϕ_{1}} W_{0} →^{ϕ_{2}} W_{2}.

(ii) A faithful KH-completion for W_{1} ←^{ϕ_{1}} W_{0} →^{ϕ_{2}} W_{2} is a tuple (W, ψ_{1}, ψ_{2}), where W is a KHmodule and, for 1 ≤ i ≤ 2, ψ_{i} : W_{i} → W are KH_{i}-monomorphisms with ϕ_{1}ψ_{1} = ϕ_{2}ψ_{2}. Such a completion is denoted by

W_{1} →^{ψ_{1}} W ←^{ψ_{2}} W_{2}.

Let W be as in part (ii) of the above definition. In abuse of notation, we will refer to W itself as a completion of the amalgam of modules.

We following elementary lemma is at the heart of the construction of J_{4}.

Lemma 6.4 Let W_{1} ←^{ϕ_{1}} W_{0} →^{ϕ_{2}} W_{2} be an amalgam of K-modules for H_{1} ← H_{0} → H_{2}. Assume that each of the following three statements holds:

(1) W_{i} is irreducible for 0 ≤ i ≤ 2.

(2) There exists a normal elementary abelian subgroup Q of H contained in H_{0} with C_{W_{0}}(Q) = 0 and a hyperplane A in Q such that C_{W_{i}}(A) is one dimensional for 0 ≤ i ≤ 2.

(3) Put N_{i} = N_{H_{i}}(A) for 0 ≤ i ≤ 2 and N = N_{H}(A). Then N_{0} ∩ N′ = (N_{0} ∩ N′_{1})(N_{0} ∩ N′_{2}) and N = N_{0}N′.

Then W_{1} ← W_{0} → W_{2} has a faithful and irreducible KH-completion W of dimension |H/N|. Moreover, the Wedderburn components for Q on W are 1-dimensional and the action of H on these Wedderburn components is isomorphic to the action on A^{H}.

Proof. Let 0 ≤ i ≤ 2 and put X_{i} = C_{W_{i}}(A). Then from (1) and (2), X_{i} is a Wedderburn component for Q on W_{i} and so W_{i} ≅ X_{i} ⊗_{KN_{i}} KH_{i}. Since X_{i} is one dimensional, N′_{i} centralizes X_{i}. Let 1 ≤ j ≤ 2. Clearly X_{0}^{ϕ_{j}} = X_{j} and N_{0} ∩ N′_{j} centralizes X_{j} and X_{0}. By (3), N_{0} ∩ N′ centralizes X_{0}. Define the KN-module X by X = X_{0} as K-vector space and x^{g} = x^{h} whenever x ∈ X, g ∈ N and h ∈ N_{0} with N′g = N′h. Since N = N_{0}N′ such h always exists and since N′ ∩ N_{0} centralizes X_{0} this is well defined. Put W = X ⊗_{KN} KH . As W_{i} ≅ X_{i} ⊗_{KN_{i}} KH_{i} we conclude that W is a faithful KH-completion of W_{1} ← W_{0} → W_{2}. Clearly X is a Wedderburn component for Q on W, N_{H}(A) = N, W is irreducible and dim W = |H/N|. □

7 A J_{4}-triangle in GL_{1333}(C)

In this section (M_{1}, M_{2}, M_{3}) is an arbitrary J_{4}-triangle of groups and C is the field of complex numbers. Our goal is to define a J_{4}-triangle inside GL_{1333}(C).

The following notations will be used throughout this section. Let 1 ≤ i, j ≤ 3 with i ≠ j. If X is an CM_{i}-module, then R_{ij}(X) is the restriction of X to M_{ij}; if Y is an CM_{ij}-module then I^{i}(Y) = Y ⊗_{CM_{ij}} CM_{i} ( the module for CM_{i} induced from Y) and R_{0}(Y) is the restriction of Y to B; and if Z is an CB-module, then I^{ij}(Z) = Z ⊗_{CB} CM_{ij}.

In what follows X_{t}(i) will always denote an CM_{i}-module, Y_{t}(ij) an CM_{ij}-module and Z_{t} an CB-module. If G is a group, H ≤ G, U is an CH-module and W is an CG-module we write U → W or W ← U provided that U is isomorphic to a CH-submodule of W. (We remark that in all cases below the CH-submodule of W isomorphic to U will be unique).

Put L = O^{2}(B).

Let X_{1}(1) be an irreducible 45-dimensional CM_{1}/Q_{1}-module given by [7] regarded as an CM_{1}- module. Then clearly

(1) X_{1}(1) is irreducible of dimension 45 and C_{M_{1}}(X_{1}(1)) = Q_{1}.

The next three statements follow from 3.12.

(2) Put Y_{1}(12) = R_{12}(X_{1}(1)). Then Y_{1}(12) is irreducible of dimension 45 and C_{M_{12}}(Y_{1}(12)) = Q_{1} ∩ M_{12}.

(3) Restricted to M_{13}, X_{1}(1) is the direct sum of irreducible CM_{13}-modules Y_{1}(13) and Y_{2}(13), of dimension 3 and 42, respectively. Moreover, C_{M_{13}}(Y_{1}(13)) = O_{2,3}(M_{13}) and C_{M_{13}}(Y_{2}(13)) = Q_{1}.

(4) For i = 1, 2 put Z_{i} = R_{0}(Y_{i}(13)). Then Z_{1} and Z_{2} are irreducible of dimension 3 and 42, respectively. Moreover, restricted to B, Y_{1}(12) is isomorphic to Z_{1} ⊕ Z_{2}.

Let A/Q_{3} be the subgroup isomorphic to Alt(5) in M_{3}/Q_{3}. By 5.1 (A∩M_{13})/Q_{3} ≅ Alt(4). Thus A ∩ M_{13} ≤ O_{2,3}(M_{13}) and so by (3), A ∩ M_{13} centralizes Y_{1}(13). Moreover, M_{3} = AM_{13} and thus there exists an CM_{3}-module X_{1}(3) such that

(5) X_{1}(3) is irreducible of dimension 3, C_{M_{3}}(X_{1}(3))/Q_{3} ≅ Alt(5) and X_{1}(3) is isomorphic to Y_{1}(13) as an CM_{13}-module.

By (4) and (5)

(6) Put Y_{1}(23) = R_{23}(X_{1}(3)). Then Y_{1}(23) is irreducible of dimension 3 and restricted to B isomorphic to Z_{1}.

Put X_{1}(2) = I^{2}(Y_{1}(23)). There are 15 2-spaces of V (2) containing V_{1}(2) and 140 = 155 − 15 2-spaces of V (2) which do not contain V_{1}(2). Hence the orbits of M_{12} on M_{2}/M_{23} have length 15 and 140. Moreover, 15 = |M_{12}/B| , Y_{1}(12) is irreducible of dimension 45 = 15 · 3 = |M_{12}/B| · dim Z_{1} and so by 6.2, Y_{1}(12) ≅ I^{12}(Z_{1}). Since Z_{1} = R_{0}(Y_{1}(23)) the definition of X_{1}(2) now implies

(7) Y_{1}(23) → X_{1}(2), X_{1}(2) is 465-dimensional and is as an CM_{12}-module isomorphic to the direct sum of Y_{1}(12) and a 420 dimensional CM_{12}-module Y_{2}(12).

We remark that X_{1}(2) and Y_{2}(12) are irreducible. With some effort this could be proved directly at this stage, but we prefer to prove this later on ( see (17) and (29)) in shorter but indirect way.

Put X_{2}(3) = I^{3}(Y_{2}(13)). By 3.12 the restriction of Y_{2}(13) to L is an irreducible module U. Hence X_{2}(3) restricted to L is the sum of five irreducible CL-modules U_{1} = U, U_{2}, ..., U_{5}. By (3) C_{L}(U_{1}) = Q_{1} ∩ L. Since Z(Q_{3}) < Q_{1} ∩ L < Q_{3} and Z(Q_{3}) is the only proper M_{3}-invariant subgroup properly contained in Q_{3}, we have C_{L}(X_{2}(3)) ≠ C_{L}(U_{1}). Also M_{3} acts primitively on {U_{1}, ..., U_{5}} and hence C_{L}(U_{i}) ≠ C_{L}(U_{j}) for i ≠ j and we conclude:

(8) X_{2}(3) is the direct sum of five pairwise non-isomorphic 42-dimensional CL-modules naturally permuted by M_{3}/L ≅ Sym(5).

By 5.1 (i) the orbits of M_{13}, M_{23} and B on M_{3}/M_{13} have lengths 1 and 4; 3 and 2; and 1, 2 and 2, respectively. Thus (8) and Clifford theory implies the following four statements:

(9) X_{2}(3) is irreducible of dimension 210.

(10) Restricted to M_{13}, X_{2}(3) is isomorphic to the direct sum of Y_{2}(13) and Y_{3}(13), where Y_{3}(13) is an irreducible CM_{23}-module of dimension 168.

(11) Restricted to M_{23}, X_{2}(3) is the direct sum of irreducible CM_{23}-modules Y_{2}(23) and Y_{3}(23) of dimension 126 and 84, respectively.

(12) Restricted to B, Y_{2}(23) is isomorphic to the direct sum of Z_{2} and an irreducible 84-dimensional CB-module Z_{3}. Put Z_{4} = R_{0}(Y_{3}(23)). Then Z_{4} is an irreducible 84-dimensional CB-module and Z_{4} ≅ Z̸_{3}. Moreover, restricted to B, Y_{3}(13) is isomorphic to the direct sum of Z_{3} and Z_{4}.

Note that by definition (see (4)), Z_{2} is isomorphic to Y_{2}(13) as an CB-module, by (3) Y_{2}(13) → X_{1}(1), and by definition (see (2)) Y_{1}(12) is isomorphic to X_{1}(1) as an CM_{12}-module. Moreover, by (7) Y_{1}(12) → X_{1}(2). Hence as CB-modules

Z_{2} ≅ Y_{2}(13) → X_{1}(1) ≅ Y_{1}(12) → X_{1}(2).

Hence by (12), Y_{2}(23) ← Z_{2} → X_{1}(2). By (11) Y_{2}(23) is irreducible of dimension 126 = 3 · 42 = |M_{23}/B| · dim Z_{2} and we conclude from 6.2 that Y_{2}(23) ≅ I^{23}(Z_{2}). As Z_{2} → X_{1}(2), the universal property of induced representations implies that there exists a non-zero CM_{23}-homomorphism from Y_{2}(23) (≅ I^{23}(Z_{2})) to X_{1}(2). As Y_{2}(23) is irreducible, this homomorphism is one-to-one. So Y_{2}(23) → X_{1}(2). Then by (12) Z_{3} → Y_{2}(23) and so Z_{3} → X_{1}(2). Since dim Z_{3} > dim Y_{1}(12) and Z_{3} is irreducible, by (7) we get Z_{3} → Y_{2}(12). We record:

(13) Y_{2}(23) → X_{1}(2) and Z_{3} → Y_{2}(12).

By 5.1 we can pick t ∈ Q_{1} \ M_{2}. Then clearly t normalizes B and M_{12}. So if T is one of B and M_{12} and W is an CT -module, then T acts on W by w → w^{(g^{t})} for all w ∈ W, g ∈ T and we obtain a new CT -module denoted by W^{t}. Put B̂ = B⟨t⟩ and M̂_{12} = M_{12}⟨t⟩. Since B̂/L normalizes B/L ≅ C_{2} × C_{2} in M_{3}/L ≅ Sym(5), clearly B̂/L ≅ D_{8} and has orbits of length 1 and 4 on M_{3}/M_{13}. In particular, B̂ interchanges the two orbits of length 2 for B on M_{3}/M_{13}. Thus (8) - (12) imply

(14) Z_{4} ≅ Z_{3}^{t} and B̂ acts irreducibly on Y_{3}(13).

Let X = Y_{2}(12) ⨂_{CM_{12}} CM̂_{12} and Y = Z_{3} ⨂_{CB} CB̂. By (13) Z_{3} → Y_{2}(12) and by (12) Z_{3} → Y_{3}(13). Hence the universal property of induced modules implies the first part of the following statement (the second part is still to be proved):

(15) X ← Y → Y_{3}(13) and C_{Y_{2}(12)}(Q_{1} ∩ Q_{2}) = 0 .

Our nearest goal is to invoke 6.4 to find a faithful M_{1}-completion for the amalgam X ← Y → Y_{3}(13) of C-modules for M̂_{12} ← B̂ → M_{13}. We start by proving the second part of (15) which is equivalent to the claim that Q_{1} ∩ Q_{2} acts fixed-point freely on Y_{2}(12) and immediately implies that Q_{1} = ⟨Q_{1} ∩ M_{2}, t⟩ acts fixed-point freely on Y. For this notice that by 5.1 (vi) Q_{3} ∩ Q_{2} is a hyperplane in Q_{2}. Furthermore, by definition (see (6)), Y_{1}(23) = R_{23}(X_{1}(3)) and so by (5) Q_{3} and so also Q_{3} ∩ Q_{2} centralize Y_{1}(23). Since X_{1}(2) = I^{2}(Y_{1}(23)) and N_{M_{2}}(Q_{2} ∩ Q_{3}) = M_{23} by 5.1, every hyperplane of Q_{2} which centralizes a non-zero vector in X_{1}(2) is (Q_{2} ∩ Q_{3})^{m} for some m ∈ M_{2} and the vectors centralized by such a hyperplane form a 3-space in X_{1}(2). By 5.1 (vi) Q_{1} ∩ Q_{2} lies in 15 hyperplanes of Q_{2} and by (1), (2) Q_{1} ∩ Q_{2} centralizes the 45-dimensional space Y_{1}(12) in X_{1}(2). Since 45 = 15 · 3, this and (7) imply that Q_{1} ∩ Q_{2} acts fixed-point freely on Y_{2}(12) and the claim follows.

Recall that Y_{3}(13) restricted to B̂ is isomorphic to Y and Y restricted to B is the direct sum of two irreducible non-isomorphic CB-modules Z_{3} and Z_{4}. Hence both Y and Y_{3}(13) are irreducible 168-dimensional modules. Let A be a hyperplane in Q_{1} with d =^{def} dim C_{Y} (A) ≠ 0. Since Q_{1} centralizes neither Y nor Y_{3}(13), C_{Y} (A) is a Wedderburn component for Q_{1} on Y and Y_{3}(13). Hence d · |AB̂| = 168 = d · |A^{M_{13}}|. By 3.6 there are two M_{1}-orbits on the set of hyperplanes in Q_{1}, one is indexed by the octads and the other one by complementary pairs of dodecads in S. Suppose that A is from the former of the orbits. Recall that M_{13} is the stabilizer in M_{1} of a trio β and B̂ is the stabilizer of T and an octad α contained in T . By D_{3}(Mat_{24}) there are exactly two orbits S_{1} and S_{2} of M_{13} on the octads with length less than or equal to 168. Here S_{1} is the three octads in T and S_{2} contains the octads which are disjoint from exactly one octad in T . Since B̂ acts transitively neither on S_{1} nor on S_{2}, this is a contradiction. Thus A corresponds to a complementary pair of dodecads and by 3.10 (i), |A^{M_{13}}| = |AB̂| = 168, d = 1 and |AM̂_{12}| = 840. In particular, X is irreducible and C_{X}(A) is 1-dimensional. By 3.10 (ii) we can apply 6.4 and obtain a CM_{1}-module X_{2}(1) such that

(16) X → X_{2}(1) ← Y_{3}(13), X_{2}(1) is irreducible of dimension 1288, the Wedderburn components for Q_{1} on X_{2}(1) are 1-dimensional and the action of M_{1}/Q_{1} on these Wedderburn components is isomorphic to the action of M_{1}/Q_{1} on pairs of complementary dodecads.

Put Y_{3}(12) = Y_{2}(12)^{t}. Then by definition, X ≅ Y_{2}(12) ⨁ Y_{3}(12) as CM_{12}-module. Moreover, by (12) and (14) Z_{3} ≅ Z̸_{3}^{t} and since X is irreducible we get

(17) Y_{3}(12) and Y_{2}(12) are irreducible of dimension 420, Z_{4} → Y_{3}(12), Y_{3}(12) ≅ Y̸_{2}(12), Y_{2}(12) → X_{2}(1) and Y_{3}(12) → X_{2}(1).

(16) and 3.10 (i) imply the following two statements:

(18) Restricted to M_{13}, X_{2}(1) is isomorphic to the direct sum of Y_{3}(13), Y_{4}(13) and Y_{5}(13), where Y_{4}(13) and Y_{5}(13) are irreducible CM_{13}-modules of dimension 672 and 448, respectively.

(19) B̂ acts irreducibly on Y_{5}(13).

By (12) and (17) Y_{3}(23) ← Z_{4} → Y_{3}(12) and we will use 6.4 to find a faithful M_{2}-completion for this amalgam of C-modules for M_{23} ← B → M_{12}. By (15), C_{Y_{2}(12)}(Q_{1} ∩ Q_{2}) = 0 and as t normalizes Q_{1} ∩ Q_{2}, C_{Y_{3}(12)}(Q_{1} ∩ Q_{2}) = 0 = C_{Y_{3}(12)}(Q_{2}) = C_{Z_{4}}(Q_{2}). Let A be a hyperplane in Q_{2} with C_{Z_{4}}(A) ≠ 0.

The hyperplanes in Q_{2} are described in 2.8. Suppose that A corresponds to a 2-space in V (2). Then by 2.8e the orbits of M_{23} on A^{M_{2}} have lengths 1, 42 and 112. If A is normal in M_{23}, then since Y_{3}(23) is irreducible, Q_{2} inverts Y_{3}(23). This is a contradiction, since by (8) and (11) Q_{2} interchanges two of the three irreducible L-submodules in Y_{3}(23) . Moreover, 112 > dim Y_{3}(23) and hence the only possibility to consider is that |A^{M_{23}}| = 42. In this case by 2.8 B does not act transitively on A^{M_{23}}, contradicting the irreducibility of Z_{4}.

So A ∈ H(s). By 2.8 (iii) the orbits of M_{23} on H(s) have lengths 84, 112 and 672. It follows that |A^{M_{23}}| = 84, C_{Y_{3}(23)}(A) is 1-dimensional, |A^{B}| = 84, |A^{M_{12}}| = 420 and C_{Y_{3}(12)}(A) is 1-dimensional. By 2.8 (iv) we can apply 6.4 and so there exists an CM_{2}-module X_{2}(2) such that

(20) Y_{3}(23) → X_{2}(2) ← Y_{3}(12), X_{2}(2) is irreducible of dimension 868, the Wedderburn components for Q_{2} on X_{2}(2) are 1-dimensional and the action of M_{2}/Q_{2} on these Wedderburn components is isomorphic to the action of M_{2}/Q_{2} on H(s).

In particular, 2.8 (iii) yields the following three statements:

(21) Restricted to M_{23}, X_{2}(2) is isomorphic to the direct sum of Y_{3}(23), Y_{4}(23) and Y_{5}(23), where Y_{4}(23) and Y_{5}(23) are irreducible CM_{23}-modules of dimension 112 and 672, respectively.

(22) Restricted to M_{12}, X_{2}(2) is isomorphic to the direct sum of Y_{3}(12) and Y_{4}(12), where Y_{4}(12) is an irreducible CM_{12}-module of dimension 448.

(23) Put Z_{5} = R_{0}(Y_{4}(23)) and Z_{6} = R_{0}(Y_{4}(12)). Then Z_{5} and Z_{6} are irreducible of dimension 112 and 448, respectively. Moreover, restricted to B; X_{2}(2) is isomorphic to the direct sum of Z_{4}, Z_{5}, Z_{6} and Z_{7}; Y_{5}(23) is isomorphic to the direct sum of Z_{6} and Z_{7}; and Y_{3}(12) is isomorphic to the direct sum of Z_{4}, Z_{5} and Z_{7}. Here Z_{7} is an irreducible CB-module of dimension 224.

Put Z_{8} = Z_{5}^{t} and Z_{9} = Z_{7}^{t}. By (14), Z_{4} ≅ Z_{3}^{t} and by definition ( see after (17)) Y_{3}(12) = Y_{2}(12)^{t}. By (23) Y_{3}(12) ≅ Z_{4} ⨁ Z_{5} ⨁ Z_{7} as an CB-module and since t^{2} = 1 we conclude that

(24) Restricted to B, Y_{2}(12) is isomorphic to the direct sum of Z_{3}, Z_{8} and Z_{9}.

Put X_{3}(3) = I^{3}(Y_{4}(23)). Note that by (23) and (17) Z_{5} → Y_{3}(12) → X_{2}(1) and that by (22) and (23) dim Y_{5}(13) = 448 > 2 · 112 = 2 · dim Z_{5}. Thus by (19) and since |B̂/B| = 2, Z_{5} ↛ Y_{5}(13). Further by (12) Z_{5} ↛ Y_{3}(13) and so by (18), Z_{5} → Y_{4}(13). Since Y_{4}(13) is irreducible of dimension 672 = 6 · 112 = |M_{13}/B| · dim Z_{5}, 6.2 implies Y_{4}(13) ≅ I^{13}(Z_{5}). Thus

(25) Y_{4}(13) → X_{3}(3).

We claim that L acts irreducibly on Y_{4}(23). For this let A be a hyperplane in Q_{2} with C_{Y_{4}(23)}(A) ≠ 0. By (20) and (21), |A^{M_{23}}| = 112 and A corresponds to a pair (W, s), where W is a 4-space in V (2) and s is a non-degenerate symplectic form on W. Let U be the 2-space in V (2) normalized by M_{23}. Since |A^{M_{23}}| = 112 the proof of 2.8 (iii) implies U ≤ W and s|_{U} is nondegenerate. Let U = ⟨u_{1}, u_{2}⟩ and W = U + ⟨v_{1}, v_{2}⟩ with s(u_{i}, v_{j}) = 0. Note that each hyperplane of Q_{2} corresponds to a vector in V (2) ∧ V (2) and, in particular, Q_{2} ∩ L corresponds to u_{1} ∧ u_{2} and A corresponds to u_{1} ∧ u_{2} + v_{1} ∧ v_{2}. Thus the third hyperplane of Q_{2} containing A ∩ L corresponds to v_{1} ∧ v_{2} and A is the unique element of H(s) containing A ∩ L. Thus N_{L}(A) = N_{L}(A ∩ L) and C_{Y_{4}(23)}(A ∩ L) = C_{Y_{4}(23)}(A) is 1-dimensional. Since N_{M_{23}}(A) = N_{M_{2}}(U, W, s) acts as GL(U) on U and C_{M_{23}}(U) = LQ_{2}, M_{23} = N_{M_{23}}(A)L. Thus |(A ∩ L)^{L}| = |A^{L}| = 112 and L acts irreducibly on Y_{4}(23).

Since L is normal in M_{3} we conclude from the definition of X_{3}(3) that X_{3}(3) is the direct sum of ten irreducible CL-modules of dimension 112. Suppose these ten irreducibles are pairwise isomorphic. As Y_{4}(13) has dimension 6 · 112, we conclude from (25) that Y_{4}(13) is the direct sum of six isomorphic irreducible CL-submodules. Let H be a hyperplane in Q_{1} ∩ Q_{3} with C_{Y_{4}(13)}(H) ≠ 1. Since Q_{1} ∩ Q_{3} ≤ L, C_{Y_{4}(13)}(H) is at least 6-dimensional and H lies in at least six hyperplanes of Q_{1} corresponding to complementary pairs of dodecads. On the other hand all three hyperplanes of Q_{1} containing Q_{1} ∩ Q_{3} correspond to octads. Hence H is contained in at least nine hyperplanes of Q_{1}, a contradiction to |Q_{1}/H| = 8.

Thus X_{3}(3) is not the direct sum of isomorphic CL-modules. Since M_{23} is maximal in M_{2}, M_{3} acts primitively on the cosets of M_{23} in M_{3} and we conclude

(26) X_{3}(3) is irreducible of dimension 1120, Y_{4}(23) is an irreducible Wedderburn component for L on X_{3}(3) and N_{M_{3}}(Y_{4}(23)) = M_{23}.

Put Y_{6}(23) = I^{23}(Z_{8}). By definition, Z_{5} → Y_{4}(23) → X_{3}(3) and Z_{8} = Z_{5}^{t}. Thus 6.2 and (26) imply:

(27) Y_{6}(23) → X_{3}(3), Y_{6}(23) is irreducible of dimension 336 and Z_{8} ≅ Z̸_{5}.

By (24) and (7), Z_{8} → X_{1}(2). So by (27) and 6.2, Y_{6}(23) → X_{1}(2). By (13) Y_{2}(23) → X_{1}(2) and by (7) Y_{1}(23) → X_{1}(2). Since dim X_{1}(2) = 465 = 3+126+336 = dim Y_{1}(23)+dim Y_{2}(23)+dim Y_{6}(23) we conclude:

(28) Restricted to M_{23}, X_{1}(2) is isomorphic to the direct sum of Y_{1}(23), Y_{2}(23) and Y_{6}(23).

Since Y_{1}(23), Y_{2}(23), Y_{6}(23), Y_{1}(12) and Y_{2}(12) are all irreducible, by (7) and (28) we get

(29) X_{1}(2) is irreducible.

By (2), (3) and (4), Y_{1}(12) ≅ Z_{1} ⨁ Z_{2}, by (24) Y_{2}(12) ≅ Z_{3} ⨁ Z_{8} ⨁ Z_{9} and by (12) Y_{2}(23) ≅ Z_{2} ⨁ Z_{3} as CB-modules. Hence using (7) and (28) we get

(30) Restricted to B, X_{1}(2) is isomorphic to the direct sum of Z_{1}, Z_{2}, Z_{3}, Z_{8} and Z_{9}. Restricted to B, Y_{6}(23) is isomorphic to the direct sum of Z_{8} and Z_{9}.

In particular, Z_{9} → Y_{6}(23) → X_{3}(3) and since Z_{9} = Z_{7}^{t} we get Z_{7} → X_{3}(3) and thus by (26), Z_{7} ≅ Z̸_{9}. Moreover, by (23), Z_{7} → Y_{5}(23) and by 6.2, I^{23}(Z_{7}) ≅ Y_{5}(23). Hence Y_{5}(23) → X_{3}(3). By definition of X_{3}(3), Y_{4}(23) → X_{3}(3) and by (27), Y_{6}(23) → X_{3}(3). Therefore:

(31) Z_{7} ≅ Z̸_{9} and restricted to M_{23}, X_{3}(3) is isomorphic to the direct sum of Y_{4}(23), Y_{5}(23) and Y_{6}(23).

Put X(1) = X_{1}(1) ⨁ X_{2}(1), X(2) = X_{1}(2) ⨁ X_{2}(2) and X(3) = X_{1}(3) ⨁ X_{2}(3) ⨁ X_{3}(3). Then by the definition of Y_{1}(23), (11), (21), (28) and (31):

(32) X(2), X(3) and ⨁_{i=1}^{6} Y_{i}(23) are isomorphic as CM_{23}-modules.

By (6), (12), (23) and (30) each of the Y_{i}(23)′s can as CB-module be decomposed into a direct sum of some of the Z′_{j}s. Hence by (32):

(33) X(2), X(3) and ⨁_{i=1}^{9} Z_{i} are isomorphic as CB-modules.

Note that M_{13} has orbits of length 6 and 4 on M_{3}/M_{23}. Hence by (25) and (26), X_{3}(3) ≅ Y_{4}(13) ⨁ Y_{5}^{a}(13) as CM_{13}-modules, where Y_{5}^{a}(13) is an irreducible 448-dimensional CM_{13}-module. Let Z be the restriction of Y_{5}^{a}(13) to B̂. Then by (26) and (33), Z is irreducible and restricted to B isomorphic to Z_{7} ⨁ Z_{9}. Hence 6.2 implies that Z ≅ Z_{9} ⊗_{CB} CB̂. By (24) and (17), Z_{9} → Y_{2}(12) → X_{2}(1) and hence Z is isomorphic to a CB̂-submodule of X_{2}(1). From 3.10 (i) and (18) we conclude Y_{5}(13), Y_{5}^{a}(13) and Z are isomorphic as CB̂-modules. Let H be a hyperplane in Q_{1} with C_{Z}(H) ≠ 0 and N and N_{0} the normalizers of H in M_{13} and B̂, respectively. By 3.10 (iii), N = N_{0}N′. Let D and D^{a} be the centralizers of H in Y_{5}(13) and Y_{5}^{a}(13), respectively. Then D and D^{a} are 1-dimensional and so N′ centralizes D and D^{a}. Since D and D^{a} are isomorphic as CN_{0}-modules, we conclude that D and D^{a} are isomorphic as CN-modules. Thus Y_{5}(13) ≅ D⊗_{CN} CM_{13} ≅ D^{a}⊗_{CN} CM_{13} ≅ Y_{5}^{a}(13). We have proved:

(34) Restricted to M_{13}, X_{3}(3) is isomorphic to the direct sum of Y_{4}(13) and Y_{5}(13). Restricted to B, Y_{5}(13) is isomorphic to the direct sum of Z_{7} and Z_{9}.

From (3), (5), (10), (18) and (34) we conclude that:

(35) X(1), X(3) and ⨁_{i=1}^{5} Y_{i}(13) are isomorphic as CM_{13}-modules.

From (33) and (35)

(36) X(1), X(2), X(3) and ⨁_{i=1}^{9} Z_{i} are isomorphic as CB-modules.

By (17) and 3.10 (i), X_{2}(1) restricted to M_{12} is isomorphic to the direct sum of Y_{2}(12), Y_{3}(12) and Y_{4}^{a}(12), where Y_{4}^{a}(12) is a 448-dimensional CM_{12}-module. It follows from (36) that both Y_{4}(12) and Y_{4}^{a}(12) are isomorphic to Z_{6} as CB-modules. In particular, Y_{4}^{a}(12) is irreducible. Let H be a hyperplane in Q_{2} with C_{Z_{6}}(H) ≠ 0. Let N and N_{0} be the normalizers of H in M_{12} and B respectively. Then N/Q_{2} ≅ Sym(6), |N/N_{0}| = |M_{12}/B| = 15, N = N_{0}N′ and as in the proof of (34) we get Y_{4}(12) ≅ Y_{4}^{a}(12). Thus

(37) X_{2}(1) restricted to M_{12} is isomorphic to the direct sum of Y_{2}(12), Y_{3}(12) and Y_{4}(12).

Now (2), (7), (22) and (37) imply:

(38) X(1), X(2) and ⨁_{i=1}^{4} Y_{i}(12) are isomorphic as CM_{12}-modules.

We are now able to construct a completion of the J_{4}-triangle in GL_{1333}(C). Let {i, j, k} = {1, 2, 3}. Then by (32), (35) and (38) X(i) and X(j) are isomorphic as CM_{ij}-modules. Let X be a 1333-dimensional vector space over C. Then there exist monomorphisms α_{i} : M_{i} → GL(X), 1 ≤ i ≤ 3, and inner automorphisms a_{i} of GL(X) such that

α_{1}a_{2}|_{M_{13}} = α_{3}|_{M_{13}}, α_{2}a_{3}|_{M_{12}} = α_{1}|_{M_{12}} and α_{3}a_{1}|_{M_{12}} = α_{2}|_{M_{23}}.

Thus the assumptions of 6.1 are fulfilled with H = GL(X) and A = Inn(GL(X)). Note that if Y is one of B^{∗}, M_{23}^{∗}, M_{13}^{∗} and M_{12}^{∗}, then X is the direct sum of pairwise non-isomorphic, absolutely irreducible CY -modules and so C_{GL(X)}(Y) consists of exactly those linear transformations which act as non-zero scalars on each of the irreducible CY -submodules. By 6.1b (ii), B^{∗} ≤ Y and so C_{GL(X)}(Y) ≤ C_{GL(X)}(B^{∗}). It is now easy to verify that C_{A}(B^{∗}) = C_{A}(M_{23}^{∗})C_{A}(M_{13}^{∗})C_{A}(M_{12}^{∗}). Thus by 6.1 (i), (a1) in 6.1 holds. Put M_{i}^{∗} = M_{i}^{α_{i}b_{i}}. Then by 6.1 (iii)

Theorem 7.1 There exist subgroups M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗} of GL_{1333}(C) such that (M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗}) is a J_{4}- triangle isomorphic to (M_{1}, M_{2}, M_{3}).

8 Faithful Completions of J_{4}-triangles

This section is devoted to study completions of J_{4}-triangles. Let (M_{1}, M_{2}, M_{3}) be a J_{4}-triangle of groups with a faithful completion M. Let S be a Sylow 2-subgroup of B, Z_{4} = Z(S), M_{i4} = C_{M_{i}}(Z_{4}) and M_{4} the subgroup of M generated by M_{14}, M_{24} and M_{34}.

We will use the definitions introduced in 2.4 and 2.3 with respect to I = {1, 2, 3, 4} and Γ_{i} = M/M_{i}.

Let R = C_{B}(Z_{4}), Q_{4}^{∗} = O^{2}(R), Q_{4} = O_{2}(Q_{4}^{∗}) and Z_{3} = Z(Q_{3}). Let V (2), V_{1}(2) and V_{3}(2) be as in 5.1 (ii).

Lemma 8.1 (i) Q_{4} ≅ 2_{+}^{1+12}, Z(Q_{4}) = Z_{4}, Q_{4}^{∗}/Q_{4} ≅ C_{3} and Q_{4}^{∗}/Q_{4} acts fixed-point freely on Q_{4}/Z_{4}.

(ii) Q_{4}^{∗} is normal in M_{4} and M_{4}/Q_{4}^{∗} ≅ Aut Mat_{22}.

(iii) M_{i4} = M_{i} ∩ M_{4} for all 1 ≤ i ≤ 3.

(iv) M_{14} ∼ 2^{1+6+4}2^{6}3·Sym(6), M_{24} ∼ 2^{1+6+3}(2^{6}(Sym(3)×L_{3}(2))) and M_{34} ∼ 2^{1+2+8+4}(Sym(5)× Sym(4)).

Proof. First notice that Z_{4} is non-trivial since it is the centre of S which is a 2-group. Since Z_{4} ≤ Z(M_{4}), M_{i4} ≤ M_{i} ∩ M_{4} ≤ C_{M_{i}}(Z_{4}) = M_{i4} and so (iii) holds. Put Y_{i} = O_{2,3}(M_{i4}) for 1 ≤ i ≤ 3. Let us locate Z_{4} in M_{1} and determine M_{14}. Consider M_{1} as the semidirect product of Q_{1} and K_{1} where K_{1} ≅ Mat_{24} and Q_{1} is the irreducible Todd module for K_{1}. One can see, for instance from 3.7 (i) that a Sylow 2-subgroup of K_{1} acts faithfully on Q_{1} ∩ M_{2} and hence Z_{4} ≤ Q_{1}. Let K_{14} be the stabilizer in K_{1} of a sextet H and R_{14} = O_{2}(K_{14}). Since R_{14} stabilizes all octads and trios incident to H, R_{14} is contained in S. On the other hand by 3.7 (iv) R_{14} centralizes a unique non-zero vector in Q_{1} and clearly this is the vector which corresponds to H in the sense of 3.6 (ii). Thus |Z_{4}| = 2, Z_{4} = C_{Q_{1}}(R_{14}) and M_{14} = N_{M_{1}}(H). By 3.2 (iii) and 3.7 (iv) we have that M_{14} ∼ 2^{1+6+4}2^{6}3 · Sym(6). Now Y_{1} normalizes all the trios and octads adjacent to H and so Y_{1} ≤ RQ_{1}. Also, RY_{1}/Y_{1} is a Sylow 2-subgroup of M_{14}/Y_{1} and so Q_{4}^{∗} = O^{2}(R) = O_{2}(O^{2}(Y_{1}))). One can see from 3.7 (iv) that O_{2}(O^{2}(Y_{1}))) = ⟨[Q_{1}, R_{14}], R_{14}⟩ is extraspecial of order 2^{13}, a Sylow 3-subgroup of Y_{1} acts fixed-point freely on O_{2}(O^{2}(Y_{1})))/Z_{4} and so (i) follows.

Since Q_{2} is isomorphic to ⋀^{2}(V (2)^{∗}), and S is a Sylow 2-subgroup of M_{2}, by 2.8 Z_{4} corresponds to a 2-subspace in V (2)^{∗} and dually to a 3-space V_{4}(2) in V (2) with V_{3}(2) ≤ V_{4}(2). Thus M_{24} = N_{M_{2}}(V_{4}(2)) and from 3.2 (ii) together with standard properties of Q_{2} we have M_{24} ∼ 2^{1+6+3}(2^{6}(Sym(3) × L_{3}(2))). Note that Y_{2} ≤ C_{M_{2}}(V_{4}(2)) ≤ M_{123} = B. Thus similar to the above, Q_{4}^{∗} = O^{2}(Y_{2}) and M_{24}/Q_{4}^{∗} ∼ C_{2} × 2^{3}L_{3}(2). Since Q_{3} ≤ S, Z_{4} ≤ Z_{3} and so by 5.1 (vii), M_{34} ∼ 2^{1+2+8+4}(Sym(5) × Sym(4)). Thus Y_{3} ≤ L ≤ B (recall that L = O^{2}(B)). As above Q_{4}^{∗} = O^{2}(Y_{3}) and M_{34}/Q_{4}^{∗} ∼ 2^{4+1}Sym(5). Since Q_{4}^{∗} = O^{2}(Y_{i}) for 1 ≤ i ≤ 3, we conclude that Q_{4}^{∗} is normal in M_{i4} for all 1 ≤ i ≤ 3 and so Q_{4}^{∗} is normal in M_{4}. In particular all statements but the last one in (ii) are proved. It is now straightforward to verify that (M_{14}/Q_{4}^{∗}, M_{24}/Q_{4}^{∗}, M_{34}/Q_{4}^{∗}) is an Aut Mat_{22}-triangle of groups (compare 4.1) and thus by 4.6 M_{4}/Q_{4}^{∗} ≅ Aut Mat_{22}, completing the proof of the lemma. □

Lemma 8.2 (i) M_{12}Q_{1}, M_{13} and M_{14} are normalizers in M_{1} of an octad, a trio and a sextet, which are pairwise adjacent.

(ii) M_{12}, M_{23} and M_{24} are the normalizers in M_{2} of 1-space, 2-space and 3-space in a flag in V (2).

(iii) M_{13} and M_{23} are the normalizers in M_{3} of 1- and 2-subsets in the 5-set Ω(3), which are disjoint; M_{34} is the normalizer in M_{3} of a 1-space in Z_{3}.

(iv) M_{14}, M_{24} and M_{34} are the normalizers in M_{4} of a hexad, an octet and a pair, which form a flag.

(v) Γ is geometric.

Proof. (i) - (iv) follow from 8.1 and 5.1.

(v) We will appeal to 2.5. Let {a_{1}, a_{2}, a_{3}} be a flag of type {i, j, k}. If i = 3 and j = 4, then by (iii) M_{43}M_{k3} = M_{3} and so any two vertices of type 4 and k in Γ^{∗}(M_{3}) are adjacent. Hence a_{2} and a_{3} are adjacent. So we may assume that a_{1} = M_{1} and a_{2} = M_{2}. Since V_{1}(2) = C_{V (2)}(Q_{1} ∩ M_{2}) is 1-dimensional, any Q_{1} ∩M_{2}-invariant proper subspace of V (2) contains V_{1}(2). Now a_{3} is adjacent to M_{1} and its type is different from 2. So Q_{1} ≤ M_{a_{3}}, V_{a_{3}}(2) is Q_{1} ∩ M_{2}-invariant and V_{1}(2) ≤ V_{a_{3}}(2). Hence by (ii) a_{1} and a_{3} are adjacent in Γ^{∗}(a_{2}) and (v) follows from 2.5. □

Recall that ^{}a̸bc = |c^{M_{ba}}|. We remark that for all nd-paths (a, b, c), c^{M_{ba}} is completely determined by a, b, the type of c and ^{}a̸bc. Furthermore, if a and d are both adjacent to b and c, put

a̸ c^{b} d = |d^{M_{abc}}|.

Recall also that n_{1}^{a_{1}} − n_{2}^{a_{2}} − … − n_{k}^{a_{k}} stands for a path (a_{1}, a_{2}, …, a_{k}) of type n_{1} − n_{2} − … − n_{k}.

Given 1^{a} − 2^{b}, define R_{a}(b) by {b, R_{a}(b)} = b^{Q_{a}}.

For b of type 2, let V (b) be a natural 5-dimensional GF (2)M_{b}/Q_{b}-module such that Q_{b} is isomorphic to ⋀^{2} V (b)^{∗}, the exterior square of the dual of V (b). For c ∈ Γ(b) let V_{c}(b) = C_{V (b)}(Q_{c}) and note that V_{c}(b) is a 1-space in V (b), if c is of type 1, a 2-space if c has type 3 and 3-space if c is of type 4. Similarly, put V_{c}^{∗}(b) = C_{V (b)^{∗}}(Q_{c}) = {ϕ ∈ V (b)^{∗} | ϕ(V_{c}(b)) = 0} and note that V_{c}^{∗}(b) is a 4-, 3- and 2-space, respectively.

For c of type 3 let Ω(c) (see 5.1) be the M_{c}-set of size 5 with M_{c}/C_{M_{c}}(Ω(c)) ≅ Sym(5). For a ∈ Γ_{1}(c) let Ω_{a}(c) be the element of Ω(c) fixed by M_{ac} and for b ∈ Γ_{2}(c) let Ω_{b}(c) be the subset of size 2 in Ω(c) fixed by M_{bc}.

If a and b are adjacent, D_{a}(b) denotes the orbit-diagram for the orbits of M_{ab} on Γ(b). We will use these diagrams only for b of type 1 or 4, in which case they can be found in before 3.1 and 4.7.

Let a be a fixed vertex of type 1. For b, c of type 1, b ≠ c, write b ∼ c if b and c are adjacent to a common vertex of type 3, and let d(b, c) be the distance between b and c in (Γ_{1}, ∼). Let X_{0}(a) = {a} and X_{1}(a) = {b ∈ Γ_{1} | a ∼ b}.

Throughout this section we will often use 5.1 and 8.2 without further reference.

Lemma 8.3 Let a ∈ Γ_{1} and a_{i} ∈ Γ_{i}(a), 2 ≤ i ≤ 4. Then

(i) For k = 3, 4, a_{2} is adjacent to a_{k} if and only if Z_{a_{k}} ≤ Q_{a_{2}} ∩ Q_{a} and if and only if Q_{a_{k}} ∩ Q_{a} ≤ M_{a_{2}} ∩ Q_{a}.

(ii) a_{3} is adjacent to a_{4} if and only if Z_{a_{4}} ≤ Z_{a_{3}} and if and only if Q_{a_{4}} ∩ Q_{a} ≤ Q_{a_{3}} ∩ Q_{a}.

(iii) Z_{a_{4}} ≤ Q_{a_{3}} if and only if ^{}a̸_{3}aa_{4} ≠ 1344.

Proof. (i) Without loss a = M_{1} and a_{2} = M_{2}. Suppose that Z_{a_{k}} ≤ Q_{2} ∩Q_{1}. Then Q_{2} centralizes Z_{a_{k}}. Since M_{1k} is maximal in M_{1}, N_{M_{1}}(Z_{k}) = M_{1k} and thus Q_{2} fixes a_{k}. Now Q_{2}Q_{1} = O_{2}(M_{12}Q_{1}) and so O_{2}(M_{12}Q_{1}) fixes a_{k}. But this easily implies that a_{k} is adjacent to M_{2}. Suppose next that Q_{a_{k}} ∩ Q_{1} ≤ M_{2} ∩ Q_{1}. Since [Q_{1}, Q_{a_{k}}] ≤ Q_{1} ∩ Q_{a_{k}} we conclude that Q_{a_{k}} normalizes M_{2} ∩ Q_{1} and so Q_{a_{k}} ≤ M_{12}Q_{1}. As above a_{k} is adjacent to M_{2}. Hence one direction of (i) is proved. The other direction is obvious.

(ii) is proved similar to (i).

For (iii) assume that ^{}a̸_{3}aa_{4} ≠ 1344. Then by D_{a_{3}}(a) there exists a path 3^{a_{3}} − 4^{b} − 3^{c} − 4^{a_{4}} in Γ(a). Hence Z_{a_{4}} ≤ Z_{c} ≤ Q_{b} ∩ Q_{1} ≤ Q_{a_{3}} ∩ Q_{1}. □

Lemma 8.4 (i) Given a path 3^{b} − 1^{c} − 3^{d} with b ≠ d. Then (Q_{b}∩Q_{c})(Q_{d}∩Q_{c}) = Q_{c} if Γ_{2}(bcd) =Ø, and (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) = Q_{c} ∩ M_{x}, if x ∈ Γ_{2}(bcd).

(ii) Given a path 4^{b} − 1^{c} − 3^{d} with Γ_{2}(bcd) =Ø. Then (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) = Q_{c}.

(iii) Given a path 1^{a} − 4^{b} − 3^{c} with ^{}a̸bc = 96. Then Q_{a} ∩ M_{c} = Q_{a} ∩ Q_{b}, M_{ac} = M_{abc}, (Q_{a} ∩ Q_{b})(Q_{c} ∩ Q_{b}) = Q_{b}, Q_{a} ∩ Z_{c} = Z_{b} and |(Q_{a} ∩ Q_{c})Z_{c}/Z_{c}| = 2^{4}.

Proof. (i) Since M_{bc} is maximal in M_{c}, N_{M_{c}}(Q_{b} ∩Q_{c}) = M_{bc} and so Q_{b} ∩Q_{c} ≠ Q_{d} ∩Q_{c}. Suppose that (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) ≠ Q_{c}. Since the three hyperplanes of Q_{c} containing Q_{b} ∩ Q_{c} are conjugated under M_{bc}, we conclude that (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) ≤ Q_{c} ∩ M_{x} for some x ∈ Γ_{2}(bc). Thus by 8.3 x ∈ Γ(bcd). Since |Q_{c}/Q_{b} ∩ Q_{c}| = 4, (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) = Q_{c} ∩ M_{x}.

(ii) Similar to (i).

(iii) Since ^{}a̸bc = 96, it is easy to see that Q_{a} ∩ M_{c} ≤ Q_{b}. Since Q_{a} does not fix c, Q_{a} does not normalize Q_{b} ∩ Q_{c}. Thus Q_{a} ∩ Q_{b} ≰ Q_{b} ∩ Q_{c}. Since Q_{b}^{∗} acts irreducibly on Q_{b}/Q_{b} ∩ Q_{c}, Q_{b} = (Q_{a} ∩ Q_{b})(Q_{c} ∩ Q_{b}). In particular, Q_{a} ∩ Q_{c} has order |Q_{a} ∩ Q_{b}|/|Q_{b}/Q_{c} ∩ Q_{b}| = 2^{7}/2^{2} = 2^{5}.

Suppose that Q_{a} ∩ Z_{c} ≠ Z_{b}. Since Q_{b}^{∗} acts irreducibly on Z_{c}/Z_{b} we conclude that Z_{c} ≤ Q_{a}. But then Q_{a} centralizes Z_{c} and Q_{a} ≤ M_{c}, a contradiction. Hence Q_{a} ∩ Z_{c} = Z_{b}, which implies that M_{ac} = M_{abc} and also that |(Q_{a} ∩ Q_{c})Z_{c}/Z_{c}| = |(Q_{a} ∩ Q_{c})/Z_{b}| = 2^{5}/2 = 2^{4}. □

Lemma 8.5 Given a path 1^{b} −3− 4^{c} or 2^{b} −3− 4^{c}. Then b is adjacent to c.

Proof. Let i ∈ {1, 2}. Then M_{i3}M_{43} = M_{3}. So if g ∈ M_{3}, then M_{i3}g = M_{i3}h for some h ∈ M_{43}. Thus M_{i}g is adjacent to M_{4}. □

Lemma 8.6 (i) There exists a unique class of nd-paths 1^{a} − 3^{b} − 1^{c}. Moreover, for any such path M_{abc}/Q_{b} ≅ L_{3}(2) × Sym(3), Q_{a}M_{abc} = M_{ab}, M_{ac} = M_{abc} and Q_{a} ∩ Q_{c} = Z_{b}.

(ii) M_{a} acts transitively on X_{1}(a) and (Γ_{1}, ∼) is connected.

(iii) |X_{1}(a)| = 2^{2} · 3 · 5 · 11 · 23 = 15, 180.

Proof. (i) Q_{a} acts transitively on the four elements of Γ_{1}(b) \ {a} and so all but the last two statements of (i) follow. Since M_{ab} is maximal in M_{a}, N_{M_{a}}(Q_{a}∩Q_{b}) = M_{ab}. Since Q_{a}∩Q_{b} = Q_{a}∩M_{c}, M_{ac} ≤ N_{M_{a}}(Q_{a} ∩Q_{b}) = M_{ab}. Since ⟨Q_{a}, Q_{c}⟩Q_{b}/Q_{b} ≅ Alt(5) and since ⟨Q_{a}, Q_{c}⟩ centralizes Q_{a} ∩Q_{c}, Q_{a} ∩ Q_{c} = Z_{b}.

(ii) By (i) M_{a} is transitive on X_{1}(a). Let c ∈ X_{1}(a) and b ∈ Γ_{3}(ac). Then ⟨M_{ab}, M_{bc}⟩ = M_{b} and so ⟨M_{a}, M_{c}⟩ = ⟨M_{a}, M_{b}⟩ = M and hence (Γ_{1}, ∼) is connected.

(iii) |X_{1}(a)| = 4 · |Γ_{3}(a)| = 2^{2} · 3 · 5 · 11 · 23. □

Lemma 8.7 There exists a unique class of nd-paths 1^{a} − 3^{b} − 2^{c}. Moreover for any such path, M_{abc}/Q_{b} ≅ L_{3}(2)×Sym(3), M_{abc}Q_{a} = M_{ab}, M_{abc}Q_{c} = M_{bc}, Q_{a}∩M_{c} = Q_{a}∩Q_{b}, Q_{c}∩M_{a} = Q_{c}∩Q_{b}, Q_{a} ∩ Q_{c} = Z_{b}, (Q_{a} ∩ M_{c})Q_{c} = O_{2}(M_{bc}), (Q_{c} ∩ M_{a})Q_{a} = O_{2}(M_{ab}) and M_{ac} = M_{abc}.

Proof. Q_{a} acts transitively on the four elements of Γ_{2}(b) \ Γ_{2}(a) and Q_{c} acts transitively on the two elements of Γ_{1}(b) \ Γ_{1}(c). Furthermore ⟨Q_{a}, Q_{c}⟩Q_{b}/Q_{b} ≅ Sym(5) and since Q_{a} ∩ Q_{c} is centralized by ⟨Q_{a}, Q_{c}⟩ we conclude that Q_{a} ∩ Q_{c} = Z_{b}. In particular, M_{ac} ≤ N_{M_{a}}(Z_{b}) = M_{ab}. Moreover, by an order argument Q_{b} = (Q_{b} ∩ Q_{a})(Q_{b} ∩ Q_{c}) and the remaining statements are readily verified. □

Lemma 8.8 Given a path i^{b} − 2^{c} − 3^{d} with i = 3 or 4. Then Z_{b} ≰ Q_{d} if and only if V_{b}(c) ∩ V_{d}(c) = 0.

Proof. Note that Q_{c} ≰ Q_{d}, Q_{c} = ⟨Z_{x} | x ∈ Γ_{i}(c)⟩ and M_{cd} acts transitively on {x ∈ Γ_{i}(c) | V_{x}(c)∩V_{d}(c) = 0}. Hence it suffices to show that Z_{x} ≤ Q_{d} whenever x ∈ Γ_{i}(c) with V_{x}(c)∩V_{d}(c) ≠ 0. Pick y ∈ Γ_{1}(c) with V_{y}(c) ≤ V_{x}(c) ∩ V_{d}(c), i.e. y ∈ Γ_{1}(xcd). Then by 8.3 Z_{x} ≤ Q_{y} ∩ Q_{c} ≤ Q_{d} ∩ Q_{c}.□

Lemma 8.9 There exists a unique class of nd-paths 1^{a} − 3^{b} − 2^{c} − 3^{d} − 1^{e} with V_{b}(c) ∩ V_{d}(c) = 0. Moreover, for any such path |M_{ae}| = |M_{abcde}| = 2^{14} · 3^{2}, |Z_{b} ∩ M_{e}| = 4, Q_{a} ∩ Q_{e} = 1, M_{ae}Q_{a} = N_{M_{a}}(Z_{b} ∩ M_{e}), M_{ae}Q_{c} = M_{bcd} and M_{ae}Q_{c}/Q_{c} ≅ Sym(4) × Sym(4).

Proof. By 8.7 M_{abc}Q_{c} = M_{bc} and so there exists a unique class of nd-paths 1^{a} − 3^{b} − 2^{c} − 3^{d} with V_{b}(c) ∩ V_{d}(c) = 0. By 8.8 Z_{b} ≰ Q_{d} and so Z_{b} acts transitively on the two elements of Γ_{1}(d) \ Γ_{1}(c). Thus the uniqueness assertion in the lemma is proved and Z_{b} ∩ M_{e} = 8/2 = 4. In particular, |M_{abcde}| = |M_{abc}|/(^{}b̸cd ·^{}c̸de) = 2^{19} · 3^{2} · 7/(2^{4} · 7 · 2) = 2^{14} · 3^{2}. Moreover, M_{abcde}Z_{b} = M_{abcd}. By 8.7, (Q_{a} ∩ M_{c})Q_{c} = O_{2}(M_{bc}) and thus Q_{a} ∩ M_{c} acts transitively on the 2-spaces in V_{b}(c) + V_{d}(c) which intersect V_{b}(c) trivially (there are 16 such 2-spaces). Hence M_{bcd} = M_{abcd}Q_{c} = M_{abcde}Q_{c}, |Q_{a} ∩ M_{c}/Q_{a} ∩ M_{cd}| = 2^{4}, |Q_{a} ∩ M_{cd}| = 2^{9}/2^{4} = 2^{5} and |Q_{a} ∩ M_{cde}| = 2^{4}. It follows that |M_{abcde}Q_{a}/Q_{a}| = 2^{10} · 3^{2}. Since |M_{ab}/Q_{a}| = 2^{10} · 3^{2} · 7 and M_{ab} acts transitively on the 7 subgroups of order 4 in Z_{b}, |N_{M_{ab}}(Z_{b} ∩ M_{e})/Q_{a}| = 2^{10} · 3^{2}. So M_{abcde}Q_{a} = N_{M_{ab}}(Z_{b} ∩ M_{e}).

Since V_{b}(c) ∩ V_{d}(c) = 0, Γ_{4}(bcd) = Ø and so Z_{b} ∩ Z_{d} = 1. By 8.7, Q_{c} ∩ Q_{e} = Z_{d} and so Z_{b} ∩ Q_{e} = Z_{b} ∩ Z_{d} = 1. In particular, Q_{a} ∩ M_{e} ≰ Q_{e} and since M_{ae} normalizes Q_{a} ∩ M_{e} we conclude M_{ae}Q_{e} ≠ M_{e}. By symmetry M_{ae}Q_{a} ≠ M_{a}. By 3.4 the only group between N_{M_{ab}}(Z_{b}∩M_{e}) and M_{a} is M_{ab}. Hence M_{ae} ≤ M_{b}. By symmetry M_{ae} ≤ M_{d}. Since Z_{b}Q_{d} = Q_{c}Q_{d}, M_{bd} ≤ N_{M_{d}}(Q_{c}Q_{d}) = M_{cd}, M_{bd} = M_{bcd} and M_{ae} = M_{abcde}. Hence M_{ae}Q_{c}/Q_{c} = M_{abcde}Q_{c}/Q_{c} = M_{bcd}/Q_{c} ≅ Sym(4)×Sym(4). It remains to prove that Q_{a} ∩ Q_{e} = 1. As V_{b}(c) ∩ V_{d}(c) = 0, Q_{b} ∩ Q_{d} ≤ Q_{c}. By 8.7, Q_{a} ∩ M_{c} ≤ Q_{b} and so

Q_{a} ∩ Q_{e} ≤ Q_{a} ∩ Q_{b} ∩ Q_{d} ∩ Q_{e} ≤ (Q_{a} ∩ Q_{c}) ∩ (Q_{e} ∩ Q_{c}) = Z_{b} ∩ Z_{d} = 1. □

Lemma 8.10 Given an nd-path 4^{b} − 3^{c} − 4^{d}. Then Q_{b}^{∗} acts transitively on Γ_{4}(c)\{b}, M_{bcd}Q_{d}^{∗} = M_{cd}, |Q_{b} ∩ Q_{d}/Z_{c}| = 2^{4}, Q_{b} ∩ Q_{d} ≤ Q_{c}, Q_{c} = (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}), (Q_{b} ∩ M_{d})Q_{d}^{∗}/Q_{d}^{∗} = O_{2}(M_{dc}/Q_{d}^{∗}) and M_{bcd} = C_{M_{b}}(Z_{d}) = M_{bd}.

Proof. Since Z_{b} ≤ Z_{c} ≤ Q_{b} and Q_{b}^{∗} acts fixed-point freely on Q_{b}/Z_{b}, Q_{b}^{∗} acts transitively on Z_{c}/Z_{b}. Now [Z_{c}, Q_{b}] = Z_{b} and so Q_{b}^{∗} acts transitively on Z_{c} \ Z_{b} and so also on Γ_{4}(c) \ {b}. In particular, M_{bcd}Q_{b}^{∗} = M_{bc} and by symmetry, M_{bcd}Q_{d}^{∗} = M_{cd}. Since |Q_{b} ∩ Q_{c}/Z_{c}| = 2^{13}/2^{2+3} = 2^{8} and |Q_{c}/Z_{c}| = 2^{12}, 2^{4} ≤ |Q_{b} ∩ Q_{c} ∩ Q_{d}/Z_{c}| ≤ 2^{8}, where 2^{4} occurs exactly then Q_{c} = (Q_{c} ∩ Q_{b})(Q_{c} ∩ Q_{d}). Since M_{bc} is a maximal subgroup of M_{c}, Q_{b} ∩ Q_{c} ≠ Q_{d} ∩ Q_{c}. Moreover, the elements of order 5 in M_{bcd} act fixed-point freely on Q_{c}/Z_{c} and so also on Q_{b} ∩ Q_{c} ∩ Q_{d}/Z_{c}. Thus |Q_{b} ∩ Q_{c} ∩ Q_{d}/Z_{c}| = 2^{4} and (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) = Q_{c}. Since Q_{c}Q_{d} = O_{2}(M_{cd}) we conclude (Q_{b} ∩ Q_{c})Q_{d} = O_{2}(M_{dc}). Clearly, (Q_{b} ∩ M_{d})Q_{d}^{∗}/Q_{d}^{∗} ≤ O_{2}(M_{dc}/Q_{d}^{∗}). Now since Q_{b} is extraspecial, [Z_{c}, Q_{b} ∩ M_{d}] = [Z_{c}, C_{Q_{b}}(Z_{d})] = Z_{b} ≰ Z_{d} and so Q_{b} ∩ M_{d} inverts Q_{d}^{∗}/Q_{d}. Since ⟨Z_{d}^{Q_{b}^{∗}}⟩ = Z_{c}, C_{M_{b}}(Z_{d}) ≤ N_{M_{b}}(Z_{c}) = M_{bc} and so M_{bd} ≤ C_{M_{b}}(Z_{d}) ≤ C_{M_{bc}}(Z_{d}) = M_{bcd}. □

Lemma 8.11 There exists a unique class of nd-paths 1^{a} − 4^{b} − 3^{c} − 4^{d} with ^{}a̸bc = 96. Moreover, for any such path M_{abcd}Q_{d}^{∗} = M_{cd}, M_{abcd} = C_{M_{a}}(Z_{d}) = M_{ad}, M_{abcd}Q_{a}/Q_{a} = C_{M_{a}/Q_{a}}(Z_{d}Q_{a}/Q_{a}) and Z_{d}Q_{a}/Q_{a} is in the class of non 2-central involutions of M_{a}/Q_{a}.

Proof. Let 1^{a} − 4^{b} − 3^{c} be an nd-path with ^{}a̸bc = 96. By 8.10 Q_{b}^{∗} acts transitively on Γ_{4}(c) \ {b}. In particular, the existence and uniqueness statements hold with |M_{abcd}| = |M_{ab}|/(^{}a̸bc ·^{}b̸cd) = 2^{21} · 3^{3} · 5/(96 · 6) = 2^{15} · 3 · 5.

By 8.10 (Q_{b} ∩ M_{d})Q_{d}^{∗}/Q_{d}^{∗} = O_{2}(M_{dc}/Q_{d}^{∗}) and by 4.7 Q_{b}Q_{c}/Q_{b} acts regularly on {a | a ∈ Γ_{1}(b),^{}a̸bc = 96}. Hence M_{abcd}Q_{d}^{∗} = M_{cd}. Note that Z_{d} ≤ Z_{c} and Z_{d} ≠ Z_{b}. By 8.4 (iii), Q_{a} ∩ Z_{c} = Z_{b}. Thus Z_{d} ≰ Q_{a} and d is not adjacent to a. Since M_{abcd} centralizes Z_{d} and has order divisible by 5, Z_{d}Q_{a}/Q_{a} is in the class of non 2-central involutions in M_{a}/Q_{a} ≅ Mat_{24} (see 3.1). Thus |C_{M_{a}/Q_{a}}(Z_{d}Q_{a}/Q_{a})| = 2^{9} · 3 · 5. Since Q_{a} ∩ M_{c} ≤ Q_{b} by 8.4 (iii) and C_{M_{b}}(Z_{d}) = M_{bcd} by 8.10 we have C_{Q_{a}}(Z_{d}) = Q_{a} ∩ M_{bcd} = C_{Q_{a}∩Q_{b}}(Z_{d}) and as Q_{b} is extraspecial |Q_{a} ∩ M_{bcd}| = |Q_{a} ∩ Q_{b}|/2 = 2^{6}. Thus |M_{abcd}Q_{a}/Q_{a}| = |M_{abcd}|/2^{6} = 2^{9} · 3 · 5, M_{abcd}Q_{a}/Q_{a} = C_{M_{a}/Q_{a}}(Z_{d}Q_{a}/Q_{a}) and M_{abcd} = C_{M_{a}}(Z_{d}) = M_{ad}. □

Lemma 8.12 There exists a unique class of nd-paths 1^{a} − 4^{b} − 3^{c} − 4^{d} − 1^{e} with ^{}a̸bc = 96 = ^{}e̸dc. Moreover, for any such path M_{ae} = M_{abcde}, Q_{a}∩Q_{e} = 1, M_{ae}/(Q_{b}∩Q_{d}) ≅ Sym(5), |(Q_{b}∩Q_{d})/Z_{c}| = 2^{4} and |M_{ae}| = 2^{10} · 3 · 5.

Proof. The uniqueness statement follows from 8.11. By 8.10, |(Q_{b} ∩ Q_{d})/Z_{c}| = 2^{4}. By 8.4 (iii), |(Q_{a} ∩ Q_{c})Z_{c}/Z_{c}| = 2^{4}. By 8.10 with the rôles of b and d interchanged, (Q_{d} ∩ M_{b})Q_{b}^{∗}/Q_{b}^{∗} = O_{2}(M_{bc}/Q_{b}^{∗}) and so by 4.7 Q_{d} ∩ M_{b} acts transitively on the 32 elements x in Γ_{1}(b) with ^{}x̸bc = 96. Thus M_{bc} = M_{abc}(Q_{d} ∩ M_{b}). Suppose (Q_{a} ∩ Q_{c})Z_{c} = Q_{b} ∩ Q_{d}. Then M_{bc} = M_{abc}(Q_{d} ∩ M_{b}) normalizes (Q_{a} ∩ Q_{c})Z_{c}/Z_{c}. A contradiction, since by 8.1 M_{bc} does not normalize a subgroup of order 2^{4} in Q_{c}/Z_{c}. So (Q_{a} ∩ Q_{c})Z_{c} ≠ Q_{b} ∩ Q_{d}. Since 5 divides |M_{abcd}|, Q_{a} ∩ Q_{c} ∩ Q_{d} ≤ Z_{c}.

Since O_{2}(M_{cd}/Q_{d}^{∗}) ∩ M_{de}/Q_{d}^{∗} = 1, Q_{a} ∩ M_{cde} ≤ Q_{d}. Similarly Q_{a} ∩ M_{c} ≤ Q_{b}. By 8.10 Q_{b} ∩ Q_{d} ≤ Q_{c}. Hence Q_{a} ∩ M_{cde} ≤ Q_{a} ∩ Q_{c} ∩ Q_{d} ≤ Z_{c}. By 8.4 (iii), Q_{a} ∩ Z_{c} ≤ Z_{b} and thus Q_{a} ∩ M_{cde} = Z_{b}. By symmetry Q_{e} ∩ M_{abc} = Z_{d}. By 8.11 M_{abcd} = M_{ad} and so Q_{e} ∩ M_{a} = Q_{e} ∩ M_{abc} = Z_{d}. Thus M_{ae} ≤ C_{M_{e}}(Z_{d}) = M_{ed} and M_{ae} = M_{abcde}. By symmetry, Q_{a} ∩ M_{e} = Z_{b} and so Q_{a} ∩ Q_{e} = 1. Note that M_{ae}/O_{2}(M_{ae}) ≅ Sym(5) and O_{2}(M_{ae}) ≤ Q_{b} ∩ Q_{d} and so M_{ae}/(Q_{b} ∩ Q_{d}) ≅ Sym(5). □

Lemma 8.13 Γ has five classes of nd-paths 1^{a} − 3^{b} − 1^{c} − 3^{d} − 1^{e}. The classes can be described as follows:

Class 1: ^{}b̸cd = 42, e ∈ X_{1}(a) and |Γ_{1}(d) ∩ X_{1}(a)| = 3.

Class 2: ^{}b̸cd = 42, e ∉ X_{1}(a) and there exists a unique f ∈ Γ_{4}(a) ∩ Γ_{4}(e). For f we have a̸fe = 16, M_{ae}/Q_{f}^{∗} ≅ Sym(6) and M_{ae}Q_{a} = M_{af}. n

Class 3: ^{}b̸cd = 56 and Γ_{1}(d) ⊂ X_{1}(a).

Class 4: ^{}b̸cd = 1008 and there exists an nd-path 1^{a} − 3^{l} − 2^{j} − 3^{m} − 1^{e} with V_{l}(j) ∩ V_{m}(j) = 0.

Class 5: ^{}b̸cd = 2688 and there exists an nd-path 1^{a} − 4^{f} − 3^{g} − 4^{h} − 1^{e} with f, g, h ∈ Γ(c) and a̸fg = 96 = ^{}e̸hg.

Proof. By 8.6 (i), M_{ac}Q_{c} = M_{bc} and so by D_{b}(c), ^{}b̸cd determines the nd-path (a, b, c, d) up to conjugacy.

Since Q_{c}Q_{d}/Q_{d} acts regularly on the four elements of Γ_{1}(d) \ {c} we conclude that Q_{b} ∩ Q_{c} acts transitively on Γ_{1}(d) \ {c} provided that (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) = Q_{c} and has two orbits if (Q_{b} ∩ Q_{c})(Q_{d} ∩ Q_{c}) is a hyperplane in Q_{c}. Thus 8.4 (i) implies:

(*) For r = 56, 1008 and 2688 there exists exactly one class of nd-paths 1^{a} − 3^{b} − 1^{c} − 3^{d} − 1^{e} with b̸cd = r, and for r = 42 there exist at most two classes of such paths.

Assume now that ^{}b̸cd = 42. Then by D_{b}(c) there exist f ∈ Γ_{4}(bcd) and g ∈ Γ_{2}(bcd). Note that f and g are adjacent. By 8.5, f is adjacent to a and e. Replacing g by R_{c}(g) if necessary, we may assume that g is adjacent to a (see 2.6 applied to Γ(b)). Then by D_{b}(c), f and g are uniquely determined by (a, b, c, d) and so M_{abcd} ≤ M_{fg}. Consider D_{a}(f). Exactly three of the five elements in Γ_{1}(d) are adjacent to g. If e is adjacent to g then ^{}a̸fe = 60 and e ∈ X_{1}(a). If e is not adjacent to g, then ^{}a̸fe = 16, M_{afe}/Q_{f}^{∗} ≅ Sym(6), Q_{a} ∩ M_{e} = Q_{a} ∩ Q_{f}, Q_{a}M_{afe} = M_{af} and, since M_{af} is maximal in M_{a}, M_{ae} ≤ N_{M_{a}}(Q_{a} ∩ M_{e}) = N_{M_{a}}(Q_{a} ∩ Q_{f}) = M_{af}. Thus (a, b, c, d, e) is in Class 1 if e is adjacent to g and in Class 2 if e is not adjacent to g.

Assume next that ^{}b̸cd = 56. Then by D_{b}(c) there exists f ∈ Γ_{4}(bcd) and a and e are adjacent to f. Since Γ_{2}(bcd) = Ø, D_{a}(f) shows that ^{}a̸fd = 96 and Γ_{1}(d) ⊂ X_{1}(a). So (a, b, c, d, e) is in Class 3.

Assume now that ^{}b̸cd = 1008. Then by D_{b}(c) there exists an nd-path 3^{b} − 2^{f} − 3^{g} − 2^{h} − 3^{d} in Γ(c) with R_{c}(f) ≠ h. Using 2.6 and replacing f by R_{c}(f) and h by R_{c}(h), if necessary, we assume that a is adjacent to f and e is adjacent to h. Note c ∈ Γ_{1}(fgh) and f ≠ R_{c}(h), which

means that Ω_{f}(g) ∩ Ω_{h}(g) ≠Ø. In Γ(g) we find a unique nd-path 2^{f} − 1^{i} − 2^{j} − 1^{k} − 2^{h} with R_{i}(f) = j = R_{k}(h) ( indeed if Ω_{f}(g) = {1, 2} and Ω_{h}(g) = {1, 3}, then Ω_{j}(g) = {4, 5}, Ω_{i}(g) = 3 and Ω_{k}(g) = 2). In Γ(f) there exists a unique vertex l of type 3 adjacent to a and i (indeed, l is defined by V_{l}(f) = V_{a}(f) + V_{i}(f)) and similarly there exists a unique m ∈ Γ_{3}(ehk). Since R_{i}(f) = j = R_{k}(h), l and m are both adjacent to j. Furthermore, a is adjacent to f and j = R_{i}(f). Thus by 2.6 applied to Γ(l), a and j are not adjacent and by symmetry e and j are not adjacent. Suppose that there exists x ∈ Γ_{4}(ljm). By 8.5 x is adjacent to i and k and we see in Γ(j) that x is adjacent to g and so also to f and c. Moreover, a and x are adjacent to l and thus a and x are adjacent. So x is adjacent to a and c, V_{b}(f) = V_{a}(f) + V_{c}(f) ≤ V_{x}(f) and b is adjacent to x. By symmetry x is adjacent to d and so x ∈ Γ_{4}(bcd), a contradiction to ^{}b̸cd = 1008 and D_{b}(c). Thus

no such x exists and we found an nd-path 1^{a} − 3^{l} − 2^{j} − 3^{m} − 1^{e} with V_{l}(j) ∩ V_{m}(j) = 0. Hence (a, b, c, d, e) is in Class 4.

Assume finally that ^{}b̸cd = 2688. Then by D_{b}(c) there exists an nd-path 3^{b} − 4^{f} − 3^{g} − 4^{h} − 3^{d} in

Γ(c) with ^{}b̸ c^{f} g = 8 = ^{}d̸ c^{h} g. By 8.5, a is adjacent to f and e is adjacent to h. Since ^{}b̸ c^{f} g = 8 we conclude from D_{a}(f) that ^{}a̸fg = 96 and by symmetry, ^{}e̸hg = 96. Hence (a, b, c, d, e) is in Class 5. □

Lemma 8.14 M_{a} has exactly three orbits X_{2}(a), X_{3}(a) and X_{4}(a) on {e ∈ Γ_{1} | d(a, e) = 2}. Moreover we can choose notation so that

(i) |X_{2}(a)| = 2^{4} · 7 · 11 · 23 = 28, 336 and M_{ae} ∼ 2^{1+6+6}3 · Sym(6) if e ∈ X_{2}(a),

(ii) |X_{3}(a)| = 2^{7} · 3 · 5 · 7 · 11 · 23 = 3, 400, 320 and M_{ae} ∼ 2^{12}(Sym(3) × Sym(3)) if e ∈ X_{3}(a),

(iii) |X_{4}(a)| = 2^{11} · 3^{2} · 7 · 11 · 23 = 32, 643, 072 and M_{ae} ∼ 2^{3+4}Sym(5) if e ∈ X_{4}(a).

Proof. 8.13, 8.9 and 8.12. □

Lemma 8.15 Given a path 1^{a} − 4^{b} − 1^{c} − 3^{d} − 1^{e} with d(a, e) = 3. Then ^{}a̸bc = 16 and ^{}b̸cd = 2880.

Proof. Clearly d(a, c) = 2 and so ^{}a̸bc = 16. If d is adjacent to b, b is adjacent to e and by D_{a}(b), d(a, e) ≤ 2, a contradiction. Thus ^{}b̸cd ≠ 15.

Let x ∈ Γ_{3}(bc) and suppose that ^{}x̸cd = 56. Then by 8.13, Γ_{1}(x) ≤ X_{1}(e). On the other hand by D_{a}(b), X_{1}(a) ∩ Γ_{1}(x) ≠Ø. Thus X_{1}(a) ∩ X_{1}(e) ≠ Ø and d(a, e) ≤ 2, a contradiction.

Thus ^{}x̸cd ≠ 56 for all x ∈ Γ_{3}(bc). Suppose that ^{}b̸cd = 720. Then by D_{b}(c) there exists an

nd-path 4^{b} − 3^{x} − 4^{y} − 3^{d} in Γ(c) with ^{}x̸ y^{c} d = 8. Thus by D_{x}(c), ^{}x̸cd = 56, a contradiction.

Suppose that ^{}b̸cd = 180. Then there exists an nd-path 4^{b} − 2^{x} − 3^{d} in Γ(c). Since ^{}a̸bc = 16, replacing x by R_{c}(x), if necessary, we may assume that ^{}a̸bx = 60 (compare D_{a}(b)), in which case

there exists y ∈ Γ_{3}(abx). So we found an nd-path 1^{a} − 3^{y} − 2^{x} − 3^{d} − 1^{e}. Since d(a, e) = 3, Γ_{1}(yxd) =Ø, and so V_{y}(x) ∩ V_{d}(x) = 0. By 8.9 and 8.13 Class 4, d(a, e)=2, a contradiction.

Thus ^{}b̸cd = 2880 and the lemma is proved. □

Lemma 8.16 There exists a unique class of nd-paths 1^{a} − 4^{b} − 1^{c} − 3^{d} − 1^{e} with ^{}a̸bc = 16 and

b̸cd = 2880. Moreover, for any such nd-path there exists an nd-path 1^{a} − 3^{h} − 2^{f} − 4^{g} − 1^{e} with V_{h}(f) ∩ V_{g}(f) = 0 and ^{}f̸ge = 56.

Proof. By 8.13 Class 2, M_{ac}Q_{c} = M_{bc} and so there exists a unique class of nd-paths 1^{a} − 4^{b} − 1^{c} − 3^{d} with^{}a̸bc = 16 and^{}b̸cd = 2880. Moreover, Γ_{2}(bcd) =Ø and so by 8.4 (ii), (Q_{b}∩Q_{c})(Q_{d}∩Q_{c}) = Q_{c}. Since Q_{c} acts transitively on Γ_{1}(d) \ {c}, the uniqueness part of the lemma is proved.

By D_{b}(c) there exists an nd-path 4^{b} − 2^{f} − 4^{g} − 3^{d} in Γ(c). Replacing f by R_{c}(f), if necessary, we may assume that ^{}a̸bf = 60 (compare D_{a}(b)), in which case there exists h ∈ Γ_{3}(abf). Note that d is adjacent to e and g and so e and g are adjacent by 8.5. Since ^{}b̸cd = 2880 we see from D_{b}(c) that Γ_{3}(bcg) =Ø and so V_{b}(f) ∩ V_{g}(f) = V_{c}(f). Moreover, since ^{}a̸bc = 16, c and h are not adjacent. So V_{c}(f) ≰ V_{h}(f) ≤ V_{b}(f) and V_{h}(f) ∩ V_{g}(f) = 0.

Since ^{}b̸cd = 2880, d is not adjacent to f. On the other hand both d and f are adjacent to c and we see from the D_{f}(g) that ^{}f̸gd = 84 and hence ^{}f̸ge = 56. □

Lemma 8.17 Given a path 2^{a} − 1^{b} − 1^{c} with c = R_{b}(a). Then

(i) Q_{a} ∩ Q_{c} = Q_{b} ∩ Q_{c} and Q_{a} ∩ Q_{c} is maped to ⋀^{2} V_{b}^{∗}(c) under the isomorphism Q_{c} → ⋀^{2} V^{∗}(c).

(ii) [Q_{c}, V (a)] = V_{b}(a).

Proof. Since Q_{a}∩Q_{b} is centralized by Q_{b} and c = R_{b}(a) we have Q_{a}∩Q_{b} = Q_{c}∩Q_{b} = Q_{a}∩Q_{b}∩Q_{c}. Note that ⋀^{2} V_{b}^{∗}(c) is the unique proper M_{bc}-submodule in ⋀^{2} V^{∗}(c). Thus Q_{b} ∩ Q_{c} is maped to ⋀^{2} V_{b}^{∗}(c). Moreover, since M_{bc} = M_{abc} we also conclude that Q_{a} ∩ Q_{c} = Q_{b} ∩ Q_{c}. Thus (i) holds.

(ii) holds since V_{b}(a) is the unique proper M_{abc}-submodule in V (a). □

Lemma 8.18 (i) There exists a unique class of nd-paths 1^{a} − 3^{b} − 2^{c} − 4^{d} − 1^{e} with V_{b}(c)∩V_{d}(c) = 0 and ^{}c̸de = 56. Moreover, for any such path d(a, e) = 3, |M_{ae}| = 2^{10} · 3^{2}, M_{ae} = M_{abcde},

Q_{a} ∩ M_{e} = 1 and there exists an nd-path 1^{e} − 3^{h} − 2^{g} − 4^{i} − 1^{a} with V_{h}(g) ∩ V_{i}(g) = 0 and g̸ia = 56.

(ii) Put X_{5}(a) = e^{M_{a}}. Then |X_{5}(a)| = 2^{11} · 3 · 5 · 7 · 11 · 23 = 54, 405120.

(iii) Given a path 1^{a} − 4^{j} − 1^{k} − 3^{l} − 1^{e}. Then e ∈ ⋃_{i=0}^{5} X_{i}(a).

Proof. (i) and (ii) By 8.7, (Q_{a} ∩ M_{c})Q_{c} = O_{2}(M_{bc}) and so Q_{a} ∩ M_{c} acts transitively on the 64 elements of {x ∈ Γ_{4}(c) | V_{b}(c) ∩ V_{x}(c) = 0}. Hence there exists a unique class of nd-paths

1^{a} − 3^{b} − 2^{c} − 4^{d} with V_{b}(c) ∩ V_{d}(c) = 0. Moreover, we see in M_{c}/Q_{c} that M_{bcd}/Q_{c} is a complement to O_{2}(M_{cd}/Q_{c}) = Q_{d}Q_{c}/Q_{c} in M_{cd}/Q_{c}. Thus M_{cd} = M_{bcd}Q_{d}. By 8.7 M_{bc} = M_{abc}Q_{c} and so M_{cd} = M_{bcd}Q_{d} = ((M_{abc}Q_{c}) ∩ M_{d})Q_{d} = M_{abcd}Q_{c}Q_{d}. We claim that Z_{b}(Q_{c} ∩ Q_{d}) = Q_{c}. Indeed, identifying Q_{c} with ⋀^{2} V (c)^{∗} we have Z_{b} = ⋀^{2} V_{b}(c)^{∗} and Q_{c} ∩ Q_{d} = V (c)^{∗} ⋀ V_{d}(c)^{∗} and the claim follows from V (c)^{∗} = V_{b}(c)^{∗} ⊕ V_{d}(c)^{∗}. Since Z_{b} ≤ M_{abcd}, M_{cd} = M_{abcd}Q_{c}Q_{d} = M_{abcd}Q_{d}. In particular, the uniqueness statement is proved.

Put K = M_{abcde}. Let x be the number of paths as in (i) starting with a. Then x = |Γ_{3}(a)| · a̸bc ·^{}b̸cd ·^{}c̸de = (3 · 5 · 11 · 23) · 4 · 64 · 56 = 2^{11} · 3 · 5 · 7 · 11 · 23 and so |K| = |M_{a}|/x = 2^{10} · 3^{2}.

Since ^{}c̸de = 56, ^{}e̸dc = 240 and so by D_{e}(d) there exists a unique f ∈ Γ_{1}(cd) with ^{}e̸df = 16. Moreover, if we define g = R_{f}(c) then there exists h ∈ Γ_{3}(gde). Since V_{f}(c) ≤ V_{d}(c), V_{f}(c) ≰ V_{b}(c) and so there exists a unique element i ∈ Γ_{4}(bcf), namely i is determined by V_{i}(c) = V_{f}(c) + V_{b}(c). Then i is adjacent to a by 8.5 and, since g = R_{f}(c), to g as well. Since V_{b}(c) ∩ V_{d}(c) = 0 and V_{b}(c) ≤ V_{i}(c) we have V_{i}(c) ∩ V_{d}(c) = V_{f}(c). Conjugation under Q_{f} yields V_{i}(g) ∩ V_{d}(g) = V_{f}(g). Since ^{}f̸de = 16, f is not adjacent to h and so since V_{h}(g) ≤ V_{d}(g), V_{i}(g) ∩ V_{h}(g) = 0. Consider

the nd-path 2^{g} − 1^{f} − 2^{c} − 3^{b} − 1^{a} in Γ(i). By D_{g}(i) since R_{f}(g) = c, we have ^{}g̸ic = 7 and ^{}g̸ib = 28. Now one can see from D_{c}(i) that Γ_{1}(ic) and Γ_{3}(ic) are points and lines of the projective plane of order 2 with the natural incidence relation. In view of this observation and indexes in D_{g}(i), every element from {x | x ∈ Γ_{1}(ib),^{}g̸ix = 14} is adjacent to c. Hence ^{}g̸ia = 56. In particular, the path (e, h, g, i, a) has all the properties stated in the lemma.

Since [Q_{i} ∩ M_{h}, V (g)] = [Q_{i} ∩ M_{h}, V_{i}(g) + V_{h}(g)] ≤ V_{i}(g) ∩ V_{h}(g) = 0 we have

Q_{i} ∩ M_{h} ≤ Q_{g}

and so Z_{b} ∩ M_{h} ≤ Q_{c} ∩ Q_{g}. Identify Q_{c} with ⋀^{2} V (c)^{∗}. By 8.17 (i), Q_{c} ∩ Q_{g} = ⋀^{2} V_{f}^{∗}(c) and so

Z_{b} ∩ Q_{c} ∩ Q_{g} = ⋀^{2} V_{b}(c)^{∗} ∩ ⋀^{2} V_{f}(c)^{∗} = ⋀^{2}(V_{b}(c)^{∗} ∩ V_{f}(c)^{∗}) = ⋀^{2} V_{i}(c)^{∗} = Z_{i}.

Hence Z_{b} ∩ M_{h} = Z_{i}. Since V_{i}(g) ∩ V_{h}(g) = 0, Z_{i} = ⋀^{2} V_{i}(g)^{∗} ≰ V (g)^{∗} ⋀ V_{h}(g)^{∗} = Q_{g} ∩ Q_{h}. Moreover, Q_{g} ∩ M_{e} ≤ Q_{h} and thus Z_{i} ≰ M_{e}. Since f, g, h and i are uniquely determined in terms of (a, b, c, d, e), K ≤ M_{fghi} and, in particular, Z_{b} ∩ M_{e} = Z_{b} ∩ K ≤ Z_{i} ∩ M_{e} = 1. From Q_{a} ∩ M_{c} ≤ Q_{b}, Q_{b} ∩ M_{d} ≤ Q_{c} and (see 8.7) Q_{a} ∩ Q_{c} = Z_{b} we conclude that Q_{a} ∩ K = 1. Recall that |K| = 2^{10} · 3^{2}. Since K ≤ M_{abi} and |M_{abi}/Q_{a}| = |M_{ab}/Q_{a}|/7 = 2^{10} · 3^{2} we have KQ_{a} = M_{abi}.

Suppose, that Q_{a} ∩ M_{e} ≠ 1 and pick a Sylow 2-subgroup R of K. Then C_{Q_{a}∩M_{e}}(R) ≠ 1 and since C_{Q_{a}}(R) = C_{Q_{a}}(RQ_{a}) = Z_{i} we get Z_{i} ≤ M_{e}, a contradiction.

Hence Q_{a} ∩ M_{e} = 1. Since Q_{a} ∩ M_{y} ≠ 1 for all y ∈ Γ_{1} with d(a, y) ≤ 2, d(a, e) = 3.

Suppose K ≠ M_{ae}. Then by 3.4, M_{ae}Q_{a} is equal to one of M_{a}, M_{ab} or M_{ai}. Suppose M_{abe} ≠ K. Then M_{abe}Q_{a} = M_{ab}. In particular, |M_{abe}/K| = 7 and so M_{abe} = O^{2,3}(M_{abe})K. From ^{}a̸bc = 4 we conclude O^{2,3}(M_{b}) ≤ M_{c}. Now also K ≤ M_{c} and so M_{abe} ≤ M_{c}. Note that Q_{c} ∩ M_{e} = Z_{b}(Q_{c} ∩ Q_{d}) ∩ M_{e} = Q_{c} ∩ Q_{d} and so M_{ce} ≤ N_{M_{c}}(Q_{c} ∩ Q_{d}) = M_{cd}. Thus M_{ce} = M_{cde} and M_{abe} = M_{abce} = M_{abcde} = K, a contradiction.

Thus M_{abe} = K and M_{ae}Q_{a} = M_{ai}. On the other hand, by 8.7 M_{ge} ≤ M_{ghe} and as seen above Q_{i} ∩ M_{h} ≤ Q_{g}. In particular, Q_{i} ∩ M_{e} ≤ Q_{i} ∩ M_{h} = Q_{i} ∩ Q_{g}. Since Z_{i} acts transitively on the two elements in Γ_{1}(h) \ Γ_{1}(gh), Q_{i} ∩ Q_{g} = Z_{i}(Q_{i} ∩ M_{e}). Thus M_{ae} = M_{aie} ≤ N_{M_{i}}(Q_{i} ∩ Q_{g}) = M_{ig}, a contradiction, since 3^{3} divides |M_{ae}| = |M_{ai}/Q_{a}| but not |M_{ig}|.

Thus M_{ae} = M_{abcde} = M_{abcdefghi} and (i) and (ii) hold.

(iii) follows from 8.16, 8.15, 8.14 and (i). □

Lemma 8.19 Given an nd-path 1^{a} −4 − 1 − 4− 1^{e}. Then e ∈ X_{l}(a) for some 0 ≤ l ≤ 5.

Proof. Consider first a path 1^{a} − 2^{b} − 4^{c} − 2^{d} − 1^{e}. Then by D_{b}(c) there exists a path 2^{b} − 3^{f} − 1^{g} − 2^{d} in Γ(c). Pick i ∈ Γ_{4}(abf) and h ∈ Γ_{3}(gde). Then by 8.5 i is adjacent to g and we found a path

1^{a} − 4^{i} − 1^{g} − 3^{h} − 1^{e}. So by 8.18 (iii), e ∈ X_{l}(a), for some 0 ≤ l ≤ 5.

Consider now an nd-path 1^{a} − 4^{b} − 1^{c} − 4^{d} − 1^{e}. By D_{b}(c) there exists a path 4^{b} − 3^{f} − 4^{g} − 3^{h} − 4^{d} in Γ(c). If d(a, c) ≤ 1 or d(c, e) ≤ 1 we are done by 8.18 (iii). So suppose that ^{}a̸bc = 16 = ^{}c̸de. Then by D_{a}(b) and D_{e}(d) we have ^{}a̸bf ≠ 96 ≠ ^{}e̸dh and there exist i ∈ Γ_{2}(abf) and j ∈ Γ_{2}(hde). Then by 8.5 both i and j are adjacent to g and we are done by the first paragraph of the proof. □

Lemma 8.20 Let 4^{b} − 3^{c} − 4^{d} − 3^{e} − 4^{f} be an nd-path with [Z_{b}, Z_{f}] ≠ 1. Then ^{}c̸de = 160, Q_{b} ∩ Q_{f} = Z_{d}, Q_{b} ∩ M_{e} acts transitively on the four elements in {α ∈ Γ_{4}(e) | [Z_{b}, Z_{α}] ≠ 1}, M_{bdf}Q_{d}/Q_{d} ≅ Sym(3) × C_{2} and (Q_{b} ∩ Q_{d} ∩ M_{f})Q_{f}^{∗}/Q_{f}^{∗} = (Q_{d} ∩ M_{f})Q_{f}^{∗}/Q_{f}^{∗} = O_{2}(M_{ef}/Q_{f}^{∗}).

Proof. By D_{c}(d), if ^{}c̸de ≠ 160, then there exists α ∈ Γ_{1}(cde). Hence Z_{c}Z_{e} ≤ Q_{α} and [Z_{c}, Z_{e}] = 1, a contradiction.

Thus ^{}c̸de = 160 and by D_{c}(d) there exists a unique nd-path 3^{c} − 2^{g} − 1^{h} − 2^{i} − 3^{e} in Γ(d), such that R_{h}(g) = i, in which case h is not adjacent to c. Moreover, by 8.5 b is adjacent to g and f is adjacent to i. Notice that V_{b}(g) ∩ V_{d}(g) = V_{c}(g), V_{h}(g) ≰ V_{c}(g) and V_{h}(g) ≤ V_{d}(g). Thus V_{h}(g) ≰ V_{b}(g) and so h is not adjacent to b. By symmetry, h is not adjacent to f. Hence Z_{f} ≰ Q_{h} ∩ Q_{i} = Q_{h} ∩ Q_{g}. Put R = Q_{g} ∩ Q_{h}. We compute in Q_{g}:

R ∩ Q_{b} = (Q_{b} ∩ Q_{g}) ∩ (Q_{h} ∩ Q_{g}) = (V_{b}(g)^{∗} ∧ V (g)^{∗}) ∩ (V_{h}(g)^{∗} ∧ V_{h}(g)^{∗}) = (V_{b}(g)^{∗} ∩ V_{h}(g)^{∗}) ∧ V_{h}(g)^{∗}.

Since V_{b}(g)^{∗} ∩ V_{h}(g)^{∗} has order two we conclude that the subgroups of order two in R ∩ Q_{b} are all of the form Z_{δ} for some δ ∈ Γ_{4}(gh) = Γ_{4}(ghi) with V_{b}(g)^{∗} ∩ V_{h}(g)^{∗} ≤ V_{δ}(g)^{∗} ≤ V_{h}(g)^{∗}. Notice that for any such δ there exists γ ∈ Γ_{3}(bgδ) and so Z_{b} ≤ Z_{γ} ≤ Q_{δ}.

Suppose that R ∩ Q_{b} ∩ Q_{f} ≠ Z_{d} and pick δ ∈ Γ_{4}(ghi) \ {d} with Z_{δ} ≤ R ∩ Q_{b} ∩ Q_{f}. Then Z_{b} ≤ Q_{δ} and similarly, Z_{f} ≤ Q_{δ}. Thus Z_{b} and Z_{f} are both contained in the elementary abelian group Q_{d} ∩ Q_{δ}, a contradiction.

Thus R∩Q_{b}∩Q_{f} = Z_{d}. Since |R| = 2^{6} and |R∩Q_{b}| = 2^{3} = |R∩Q_{f}| we get |R/(R∩Q_{b})(R∩Q_{f})| = 2. Since [R ∩ Q_{b}, Q_{b} ∩ Q_{f}] ≤ Z_{b} ∩ R = 1 we have [(R ∩ Q_{b})(R ∩ Q_{f}), Q_{b} ∩ Q_{f}] = 1. Now R is a natural Ω_{6}^{+}(2)-module for M_{ig}/Q_{i}Q_{g} ≅ L_{4}(2) and so no element of M_{ig} acts as a transvection on R. Thus Q_{b} ∩ Q_{f} ≤ C_{M_{ig}}(R) = Q_{i}Q_{g}. Now by 8.17 (ii) [Q_{i}Q_{g}, V (i)] ≤ V_{h}(i), [Q_{f}, V (i)] ≤ V_{f}(i), [Q_{i}Q_{g} ∩ Q_{f}, V (i)] ≤ V_{f}(i) ∩ V_{h}(i) = 1 and so Q_{g}Q_{i} ∩ Q_{f} ≤ Q_{i}. By symmetry, Q_{g}Q_{i} ∩ Q_{b} ≤ Q_{g} and thus

Z_{d} ≤ Q_{b} ∩ Q_{f} ≤ (Q_{i}Q_{g} ∩ Q_{b}) ∩ (Q_{i}Q_{g} ∩ Q_{f}) = Q_{b} ∩ Q_{g} ∩ Q_{i} ∩ Q_{f} = Q_{b} ∩ R ∩ Q_{f} = Z_{d}.

Since |Q_{b} ∩ Q_{d}| = 2^{7} = |Q_{f} ∩ Q_{d}| and |Q_{d}| = 2^{13} we conclude

Q_{d} = (Q_{b} ∩ Q_{d})(Q_{f} ∩ Q_{d}) ≤ (Q_{b} ∩ Q_{d})Q_{e}.  (∗)

By 8.10 (Q_{b} ∩ M_{d})Q_{d}^{∗}/Q_{d}^{∗} = O_{2}(M_{cd}/Q_{d}^{∗}) and M_{bcd}Q_{d}^{∗} = M_{cd}. Thus M_{bcde}Q_{d}^{∗} = M_{cde}. Also ((Q_{b} ∩M_{d})Q_{d}^{∗} ∩M_{bcde}Q_{d}^{∗}) = ((Q_{b} ∩M_{d})∩(M_{bcde}Q_{d}^{∗}))Q_{d}^{∗} = (Q_{b} ∩M_{e})Q_{d}^{∗} and and (Q_{b} ∩M_{e})Q_{d}^{∗}/Q_{d}^{∗} = O_{2}(M_{cd}/Q_{d}^{∗}) ∩ M_{cde}/Q_{d}^{∗}. Thus by 4.8, M_{bde}Q_{d}/(Q_{b} ∩ M_{e})Q_{d} ≅ Sym(3) × C_{2} and (Q_{b} ∩ M_{e})Q_{d}/Q_{d} has order two and inverts Q_{d}^{∗}/Q_{d}. Since Q_{d}^{∗}Q_{e}/Q_{e} ≅ Alt(4) and since by (*), Q_{d}Q_{e}/Q_{e} = (Q_{b} ∩ Q_{d})Q_{e}/Q_{e} we conclude that Q_{b} ∩ M_{e} acts as a dihedral group of order eight on Z_{e} with Z_{b} mapping onto the centre of the D_{8}. Hence Q_{b} ∩ M_{e} acts transitively on {α ∈ Γ_{4}(e) | [Z_{b}, Z_{α}] ≠ 1}, [Q_{b} ∩ M_{f}, Z_{e}] = [Q_{b} ∩ M_{f}, C_{Z_{e}}(Z_{b})Z_{f}] ≤ Z_{d}, Q_{b} ∩ M_{f} ≤ Q_{d} and M_{bdf}Q_{d}/Q_{d} ≅ Sym(3) × C_{2}.

Since by (*) (Q_{b} ∩ M_{f})Q_{f}^{∗} = (Q_{b} ∩ Q_{d} ∩ M_{f})Q_{f}^{∗} = (Q_{d} ∩ M_{f})Q_{f}^{∗}, the last statement follows from 8.10. □

Lemma 8.21 There exists a unique class of nd-paths 1^{a} − 4^{b} − 3^{c} − 4^{d} − 3^{e} − 4^{f} − 1^{g} such that a̸bc = 96 = ^{}g̸fe and [Z_{b}, Z_{f}] ≠ 1. Moreover, for any such path, C_{M_{ag}}(Z_{d}) = M_{adg} = M_{abcdefg}, |M_{adg}| = 24, M_{ag} ∩Q_{a} = 1, Q_{d} ∩M_{adg} = Z_{d}, M_{adg}Q_{d}/Q_{d} ≅ Sym(3)×C_{2}, g ∉ ⋃_{i=0}^{5} X_{i}(a) and there

exists an nd-path p =1^{a} −3 − 2 − 3− 1^{α} − 3^{β} − 1^{g} with |M_{adg}/M_{dp}| = 3, α ∈ Γ_{1}(d) and Z_{d} ≤ Q_{α} ∩ M_{dp}.

Proof. By 8.11 there exists a unique class of nd-paths 1^{a} − 4^{b} − 3^{c} − 4^{d} with ^{}a̸bc = 96. Moreover, by the same lemma M_{abcd}Q_{d}^{∗} = M_{cd}, M_{abcd}Q_{a}/Q_{a} = C_{M_{a}/Q_{a}}(Z_{d}) and M_{abcd} = C_{M_{a}}(Z_{d}) = M_{ad}. Thus C_{M_{ag}}(Z_{d}) = M_{adg} = M_{abcdefg}.

By 8.20^{}c̸de = 160, Q_{b}∩Q_{f} = Z_{d} and Q_{b}∩M_{e} acts transitively on Z_{e}\M_{b}. Thus our path from b to f is unique up to conjugation and M_{bdf}Q_{d}/Q_{d} ≅ Sym(3)×C_{2}. By 4.7 O_{2}(M_{ef}/Q_{f}^{∗}) acts regularly on {α ∈ Γ_{1}(f) | ^{}e̸fα = 32}. By 8.20 O_{2}(M_{ef}/Q_{f}^{∗}) = (Q_{b} ∩ Q_{d} ∩ M_{f})Q_{f}^{∗}/Q_{f}^{∗} = (Q_{d} ∩ M_{f})Q_{f}^{∗}/Q_{f}^{∗}, so Q_{b} ∩ Q_{d} ∩ M_{f} is transitive on the same set and Q_{d} ∩ M_{fg} ≤ Q_{f}. Similarly Q_{d} ∩ Q_{f} ∩ M_{b} is transitive on the set {α ∈ Γ_{1}(b) | ^{}c̸bα = 32} and Q_{d} ∩M_{ba} ≤ Q_{b}. Since (Q_{b} ∩Q_{d})∩(Q_{f} ∩Q_{d}) = Z_{d} the uniqueness follows and M_{adg}Q_{d}/Q_{d} ≅ Sym(3) × C_{2}. Notice that C_{Q_{b}∩M_{g}}(Z_{d}) ≤ Q_{b} ∩ M_{gf} ≤ Q_{b} ∩ Q_{f} = Z_{d} and so Q_{b} ∩ M_{g} = Z_{d}. Thus Q_{d} ∩ M_{adg} = Q_{b} ∩ Q_{d} ∩ Q_{f} = Z_{d}. Furthermore, |M_{adg}| = |M_{ab}|/(96·6·160·4·32) = 24, Q_{a}∩M_{c} ≤ Q_{b} and so Q_{a}∩M_{dg} = Q_{a}∩Q_{b}∩M_{g} = Q_{a}∩Z_{d} = 1. Hence C_{Q_{a}∩M_{g}}(Z_{d}) = 1 and Q_{a} ∩ M_{g} = 1.

Suppose that g ∈ X_{i}(a) for some 0 ≤ i ≤ 5. Since |Q_{a} ∩ M_{g}| = 1, i = 5. It follows from 8.18 that M_{ag} has an elementary abelian normal subgroup A of order 2^{6}. If Z_{d} is in A, then 2^{6} divides |C_{M_{ag}}(Z_{d})| and if Z_{d} ≰ A, 2^{3} divides |C_{A}(Z_{d})| and so 2^{4} divides |C_{A}(Z_{d})Z_{d}|, and in any case we get contradiction to |M_{adg}| = 24. Thus g ∉ X_{i}(a) for all 0 ≤ i ≤ 5.

By D_{c}(d) there exist three nd-paths of type 3 − 2 − 1 − 3 from c to e in Γ(d). Moreover they are transitively permuted by M_{cde}. Since the elements of order three in Q_{d}^{∗} act fixed-point freely on Q_{d}/Z_{d} and since Z_{b} ≤ Q_{d}, Q_{d}^{∗} ∩ M_{b} ≤ Q_{d}. Thus O^{3}(M_{adg}) ≰ Q_{d}^{∗} and |M_{cde}/Q_{d}^{∗}|_{3} = 3 implies that M_{adg} acts transitively on those three nd-paths. Let (c, h, i, e) be one of them. Then h is adjacent

to b and since ^{}a̸bc = 96, D_{a}(b) yields a unique nd-path 1^{a} − 3^{l} − 2^{k} − 1^{j} − 2^{h} in Γ(b) with R_{j}(h) = k. Let m be the unique vertex of type 3, adjacent to j, h and i. Since k = R_{j}(h), m is adjacent to k. Since ^{}g̸fe = 96 and i is adjacent to e and f, D_{g}(f) shows that there exists a unique n ∈ Γ_{3}(gfi). Put p = (a, l, k, m, i, n, g). Then M_{pd} = M_{adghi} and so |M_{adg}/M_{pd}| = 3. Since h and i are adjacent to d, Z_{d} ≤ M_{adghi} = M_{pd} and Z_{d} ≤ Q_{i}. Thus the lemma holds (with α = i and β = n). □

Lemma 8.22 There exists a unique class of nd-paths q = 1^{a} − 3^{b} − 2^{c} − 3^{d} − 1^{e} − 3^{f} − 1^{g} with g ∉ ⋃_{i=0}^{5} X_{i}(a). Moreover, for any such path |M_{q}| = 24 and M_{q}/O_{2}(M_{q}) ≅ Sym(3).

Proof. The existence of such a path has been established in 8.21.

Suppose there exists x ∈ Γ_{4}(bcd). Then by 8.5 x is adjacent to a and e, a contradiction to 8.19.

So V_{b}(c) ∩ V_{d}(c) = 0 and the path 1^{a} − 3^{b} − 2^{c} − 3^{d} − 1^{e} is as in 8.9.

Suppose that Z_{d} ∩ M_{a} ∩ Q_{f} ≠ 1 and pick x ∈ Γ_{4}(de) with Z_{x} ≤ Z_{d} ∩ M_{a} ∩ Q_{f}. Then x is adjacent to c and Z_{x} ≤ Q_{c}. Since Q_{c} ∩ M_{a} ≤ Q_{b} we get Z_{x} ≤ Q_{b}. Thus by 8.8, V_{x}(c) ∩ V_{b}(c) ≠ 0. In particular, there exists y ∈ Γ_{1}(bcx). Since Z_{x} ≤ Q_{f}, 8.3 (iii) implies also that ^{}f̸ex ≠ 1344 and so

by D_{f}(e) there exists a path 4^{x} − 3^{z} − 4^{u} − 3^{f} in Γ(e). Then g is adjacent to u. Put v = R_{y}(c). Since c is not adjacent to a, 2.6 applied to Γ(b) implies that v is adjacent to a. Clearly v is also adjacent

to x. By D_{z}(x) there exists a path 2^{v} − 3^{w} − 1^{t} − 3^{z} in Γ(x). Then u is adjacent to t and g. Pick

s ∈ Γ_{4}(avw). Then s is adjacent to t and we found a path 1^{a} − 4^{s} − 1^{t} − 4^{u} − 1^{g}. Since g ∉ ⋃_{i=0}^{5} X_{i}(a) this must be an nd-path, a contradiction to 8.19.

Hence Z_{d} ∩ M_{a} ∩ Q_{f} = 1. If ^{}d̸ef ≠ 2688, then D_{f}(e) and 8.3 (iii) imply |Z_{d} ∩ Q_{f}| ≥ 4. By 8.9, |Z_{d} ∩ M_{a}| = 4. Since |Z_{d}| = 8, the latter means that Z_{d} ∩ Q_{f} ∩ M_{a} ≠ 1. So ^{}d̸ef = 2688, |Z_{d} ∩ Q_{f}| = 2 and by 8.9, M_{ae}Q_{e} = N_{M_{de}}(Z_{d} ∩ M_{a}). Put X = C_{M_{de}}(Z_{d}). Then by 3.5 M_{def}X = N_{M_{de}}(Z_{d} ∩ Q_{f}) and M_{def}/Q_{e} ≅ Sym(4). It follows that M_{def} acts transitively on {A ≤ Z_{d} | |A| = 4, A ∩ Q_{f} = 1}. Moreover, N_{M_{def}} (A)/Q_{e} ≅ Sym(3) for any such A. Thus both N_{M_{de}}(Z_{d} ∩ M_{a}) and M_{ae} act transitively on {f ∈ Γ_{3}(e) | ^{}d̸ef = 2688, Z_{d} ∩ M_{a} ∩ Q_{f} = 1} and M_{aef}Q_{e}/Q_{e} ≅ Sym(3) ≅ M_{aef}/O_{2}(M_{aef}). Moreover, Q_{e} = (Z_{d} ∩ M_{a})(Q_{e} ∩ Q_{f}) and so Z_{d} ∩ M_{a} acts transitively on Γ_{1}(f) \ {e}. Thus our nd-path from a to g is unique up to conjugacy and M_{q}Q_{e}/Q_{e} ≅ Sym(3) ≅ M_{q}/O_{2}(M_{q}). Finally, |M_{abcdefg}| = |M_{ae}|/(\frac{4}{7} · 2688 · 4) = 24 and the lemma is proved. □

Let X_{6}(a) = g^{M_{a}}, where g is as in 8.21 or equally well as in 8.22.

Lemma 8.23 (i) Let g ∈ X_{6}(a). Then |M_{ag}| = 2^{3} · 3 · 11 · 23, M_{ag} has two orbits on Γ_{2}(g) and acts transitively on {{ρ, ρ^{Q_{g}}} | ρ ∈ Γ_{2}(g)}.

(ii) |X_{6}(a)| = 2^{18} · 3^{2} · 5 · 7 = 82, 575, 360.

Proof. Let (a, b, c, d, e, f, g) and p be as in 8.21. Then by 8.21 and 8.22, |M_{adg}| = 24 = |M_{p}|, M_{adg}/Z_{d} ≅ Sym(3) × C_{2}, |M_{pd}| = 8, C_{M_{ag}}(Z_{d}) = M_{adg}, M_{p}/O_{2}(M_{p}) ≅ Sym(3) and M_{ag} ∩ Q_{g} = 1. Put A = M_{p} ∩ Q_{α}, then by 8.21, Z_{d} ≤ A. Hence A is a nontrivial normal 2-subgroup of M_{p} and C_{M_{p}}(A) ≤ M_{pd} is a 2-group. Since |O_{2}(M_{p})| = 4 we get that A is elementary abelian of order 4 and that M_{p} ≅ Sym(4). Thus M_{dp} is a dihedral group of order 8 and so N_{M_{ag}}(M_{dp}) ≤ C_{M_{ag}}(Z_{d}). In particular, M_{pd} is a Sylow 2-subgroup of M_{ag}. Moreover, there exists t in M with (a, b, c, d, e, f, g)^{t} = (g, f, e, d, c, b, a). Notice that t ∈ M_{d} and so t normalizes M_{adg}. Thus we may assume that M_{dp}^{t} = M_{dp}.

We claim that A ∩ A^{t} = Z_{d}. Clearly Z_{d} ≤ A ∩ A^{t}. By 8.21 α is adjacent to d and since t ∈ M_{d}, also α^{t} is adjacent to d. Since d(a, g) > 2, α ≠ α^{t}. We claim that Q_{α} ∩ Q_{α}^{t} ≤ Q_{d}. Indeed if Γ_{3}(αdα^{t}) = Ø, i.e. if^{}α̸dα^{t} = 16, then Q_{α}∩Q_{α}^{t} ≤ O_{2}(M_{αdα^{t}}) ≤ Q_{d}, and if δ ∈ Γ_{3}(αdα^{t}) 8.6 implies Q_{α} ∩ Q_{α}^{t} = Z_{δ} ≤ Q_{d}. By 8.21, Q_{d} ∩ M_{ag} = Z_{d} and so A ∩ A^{t} ≤ M_{ag} ∩ Q_{α} ∩ Q_{α}^{t} ≤ M_{ag} ∩ Q_{d} ≤ Z_{d}.

In particular, A ≠ A^{t}. Put E = O_{2}(M_{adg}). Since M_{adg}/Z_{d} ≅ Sym(3) × C_{2}, |E| = 4. Moreover, t normalizes E and so A ≠ E ≠ A^{t}, E ≅ C_{4} and M_{adg} is a dihedral group of order 24. Let D = O_{3}(M_{adg}) and note that ED = C_{M_{ag}}(Z_{d}D). Since D centralizes Z_{d}, 3.1 implies C_{M_{g}/Q_{g}}(D) ≅ C_{3} × L_{3}(2). Now a subgroup of L_{3}(2) with a centralizer of an involution a cyclic group of order four clearly is a cyclic group of order four and so C_{M_{ag}}(D) = DE. In particular, D is a Sylow 3-subgroup of M_{ag}. Note that all involutions in M_{pd} are contained in A ∪ A^{t} and so conjugate into Z_{d} under M_{p} and M_{p}^{t}, respectively. Thus M_{ag} has a unique class of involutions. Let z be any involution in M_{ag} and put C(z) = M_{adg} ∩C_{M_{ag}}(z). If 3 divides |C(z)|, z ∈ C_{M_{ag}}(D) = DE and z ∈ Z_{d}. Hence exactly one of the following holds: z ∈ M_{adg}, |C(z)| = 2 or C(z) = 1. Moreover, if C(z) = ⟨y⟩ for one of the twelve involutions y ∈ M_{adg} \ Z_{d}, then z is one of the ten involutions in C_{M_{ag}}(y) \ M_{adg}. Thus, if r is the number of involutions in M_{ag}, i.e. r = |M_{ag}/M_{adg}|, then r = 13 + 12 · 10 + 24s = 133 + 24s for some non-negative integer s. On the other hand, since |M_{g}/Q_{g}| = 2^{10} · 3^{3} · 5 · 7 · 11 · 23, r divides 5 · 7 · 11 · 23 and we conclude that r = 11 · 23 or r = 5 · 7 · 23. The latter case is impossible by Burnside’s p-complement theorem for p = 23 and so r = 11 · 23.

Hence |M_{ag}| = 2^{3}·3·11·23. In particular M_{adg} and M_{p} are maximal {2, 3, 5, 7}-subgroups of M_{ag}. Since both M_{fg} and M_{βg} are {2, 3, 5, 7}-groups we conclude that M_{afg} = M_{adg} and M_{aβg} = M_{p}.

Since |Γ_{2}(αβg)| = 3 we can choose x ∈ Γ_{2}(αβg) with M_{dp} ≤ M_{x}. Since the non-trivial elements of odd order in M_{ag} act fixed-point freely on Γ_{2}(g) we conclude that M_{agx} = M_{dp} = M_{agy} = M_{ag{x,y}}, where y = R_{g}(x). In particular, |M_{ag}/M_{agx}| = 759 and the lemma is proved. □

We remark that using the list of maximal subgroups of Mat_{24} or the classification of groups with dihedral Sylow 2-subgroups it is not difficult to see that M_{ag} for g ∈ X_{6}(a) is isomorphic to L_{2}(23). But we will not need this fact.

Lemma 8.24 Let g ∈ Γ_{1} with d(g, a) = 3. Then g ∈ X_{5}(a) ∪ X_{6}(a).

Proof. Pick e ∈ Γ_{1} with d(e, a) = 2 and e ∼ g. If e ∈ X_{2}(a), then g ∈ X_{5}(a) by 8.18 (iii). If e ∈ X_{3}(a) then g ∈ X_{5}(a) ∪ X_{6}(a) by 8.22.

So we may assume that e ∈ X_{4}(a). In particular, by 8.13 there exists an nd-path 1^{a} − 4^{b} − 3^{c} − 4^{d}

− 1^{e} − 3^{f} − 1^{g} with ^{}a̸bc = 96 = ^{}e̸dc. By D_{d}(e) there exists a path 4^{d} − 2^{h} − 3^{i} − 2^{j} − 3^{f} in Γ(e). Note that ^{}c̸de = 32 and so by 4.7 there exists k ∈ Γ_{3}(deh) with ^{}c̸dk = 40. Thus by D_{c}(d) there exists l ∈ Γ_{1}(cdk). Then l is adjacent to b by 8.5. By D_{a}(b) and ^{}a̸bc = 96 there exists m ∈ Γ_{3}(abl) and

so a ∼ l. Considering the path 3^{k} − 2^{h} − 3^{i} − 2^{j} − 3^{f} in Γ(e) we see in D_{k}(e) that ^{}k̸ef ≠ 2688. Thus by 8.13 applied to (g, f, e, k, l), l ∈ X_{r}(g) for some 0 ≤ r ≤ 3. Thus, by the first paragraph of the proof ( applied to (g, l, a)) in place of (a, e, g), a ∈ X_{5}(g) ∪ X_{6}(g) and the lemma is established. □

Lemma 8.25 Let z ∈ Γ_{1}. Then d(a, z) ≤ 3. In particular, Γ_{1} = ⋃_{i=0}^{6} X_{i}(a).

Proof. Suppose not. Since (Γ_{1}, ∼) is connected, there exists z ∈ Γ_{1} with d(a, z) = 4.

We claim that there does not exist an nd-path 1^{a} − 4^{b} − 1^{c} − 3^{d} − 1^{e} − 3^{f} − 1^{z}. If such a path exists then d(a, e) = 3 and d(c, z) = 2. By 8.15 and 8.13 this means that ^{}a̸bc = 16,^{}b̸cd = 2880 and c ∈ X_{2}(z) ∪ X_{3}(z) ∪ X_{4}(z).

Suppose first that c ∈ X_{2}(z). Then there exists ρ ∈ Γ_{4}(cz) and we found an nd-path 1^{a} − 4^{b} − 1^{c}

− 4^{ρ} − 1^{z}. Hence by 8.19, d(a, z) ≤ 3, a contradiction.

Suppose next that c ∈ X_{3}(z) and choose an nd-path 1^{c} − 3^{g} − 2^{h} − 3^{i} − 1^{z} with V_{g}(h) ∩ V_{i}(h) = 0. By D_{b}(c) there exists j ∈ Γ_{2}(cg) with ^{}b̸cj ≠ 384. Replacing j by R_{c}(j) if necessary, we may assume that j = R_{k}(h) for some k ∈ Γ_{1}(gh). (Indeed, we may assume Ω_{c}(g) = 1, Ω_{h}(g) = {1, 2} and Ω_{j}(g) = {2, 3} or {4, 5}. Replacing j by R_{c}(j) in the first case we may assume that the second case holds and so k ∈ Γ_{1}(g) with Ω_{k}(g) = 3 does the trick.) Pick l ∈ Γ_{4}(khi). Then l is adjacent to j and z. Since c ∈ X_{3}(z), Γ_{4}(cz) =Ø thus l is not adjacent to c. Let Λ = {V_{x}(j)/V_{c}(j) | x ∈ Γ_{3}(cj), V_{x}(j) ≤ V_{c}(j) + V_{l}(j)} and Θ = {V_{x}(j)/V_{c}(j) | x ∈ Γ_{3}(cj),^{}b̸cx ≠ 2880}. Then Λ is the set 1-spaces in a 3-subspace of V (j)/V_{c}(j). Moreover, since ^{}b̸cj ≠ 384 we get from D_{b}(c), that Θ is the set of 1-spaces in a 3- or 4-subspace of V (j)/V_{c}(j). Thus |Θ∩Λ| ≥ 3 and there exists m ∈ Γ_{3}(cj) with m ≠ g, ^{}b̸cm ≠ 2880 and V_{m}(j) ≤ V_{c}(j) + V_{l}(j). In particular, V_{m}(j) ∩ V_{l}(j) = V_{n}(j) for some n ∈ Γ_{1}(mjl). If n = k, V_{g}(j) = V_{k}(j) + V_{c}(j) = V_{m}(j), a contradiction to m ≠ g. Thus n ≠ k and there exists a unique o ∈ Γ_{3}(kjn). Since V_{o}(j) = V_{k}(j) + V_{n}(j) ≤ V_{l}(j), o is adjacent to l. Since h = R_{k}(j), o and i are both adjacent to l and h. Hence there exists p ∈ Γ_{1}(ihol). Put q = R_{p}(h). Since n is adjacent to j and o, n is not adjacent to h = R_{k}(j) by 2.6 applied to Γ(o). Since also p ∈ Γ(o), n is adjacent to q = R_{p}(h). Similarly, since z is not adjacent to h, we see in Γ(i) that z is adjacent to q. Hence there exists r ∈ Γ_{3}(nqz) and we found an nd-path 1^{a} − 4^{b} − 1^{c} − 3^{m} − 1^{n} − 3^{r} − 1^{z} with ^{}b̸cm ≠ 2880, a contradiction to the second paragraph of the proof.

Suppose finally that c ∈ X_{4}(z) and choose an nd-path 1^{c} − 4^{g} − 3^{h} − 4^{i} − 1^{z} with^{}c̸gh = 96 = ^{}z̸ih. Regard Γ_{3}(cg) as 1-spaces and Γ_{2}(cg) as the isotropic 2-spaces of a four dimensional symplectic space S over GF (2). With the help of D_{b}(c) we will show that there exists y ∈ Γ_{2}(cg) such that b̸cm ≠ 2880 for all m ∈ Γ_{3}(cgy). Indeed, if ^{}b̸cg ≠ 1440 choose y such that y is adjacent to b. If b̸cg = 1440, there exists u ∈ Γ_{2}(cg) such that v ∈ Γ_{3}(cg) is perpendicular to u in S if and only if b̸cv ≠ 2880. Choose y = u in this case. By 4.7 there exists m ∈ Γ_{3}(cgy) with ^{}g̸hm = 40. Hence by D_{h}(g) there exists n ∈ Γ_{1}(mgh). Then n is adjacent to h and i and since ^{}z̸ih = 96, there exists

r ∈ Γ_{3}(hnz) and again we found an nd-path 1^{a} − 4^{b} − 1^{c} − 3^{m} − 1^{n} − 3^{r} − 1^{z} with ^{}b̸cm ≠ 2880, a contradiction.

This completes the proof of the claim. Pick g ∈ Γ_{1} with g ∼ z and d(g, a) = 3. By 8.24,

g ∈ X_{5}(a) ∪ X_{6}(a). By the claim g ∉ X_{5}(a) and so g ∈ X_{6}(a). Pick an nd-path 1^{e} − 3^{f} − 1^{g} − 3^{h} − 1^{z} with d(e, a) = 2. Let j ∈ Γ_{2}(ghz). Then by 8.23 there exists t ∈ M_{ag} such that j^{t} is adjacent to f and so f^{t^{−1}} is adjacent to j. Replacing (e, f) by (e^{t^{−1}}, f^{t^{−1}}) we may assume that f is adjacent to j. Since V_{g}(j) ≤ V_{f}(j) ∩ V_{h}(j), there exists k ∈ Γ_{4}(fjh). Then k is adjacent to e and z and we get a contradiction to the claim applied with the rôles of a and z interchanged. □

Theorem 8.26 Let M be a faithful completion of the J_{4}-triangle of groups (M_{1}, M_{2}, M_{3}) and let M_{4} be as above. Then

(i) M_{1} has seven orbits on M/M_{1}. The lengths of these orbits are 1; 2^{2}·3·5·11·23 = 15, 180; 2^{4}·7· 11·23 = 28, 336; 2^{7}·3·5·7·11·23 = 3, 400, 320; 2^{11}·3^{2}·7·11·23 = 32, 643, 072; 2^{11}·3·5·7·11·23 = 54, 405, 120 and 2^{18} · 3^{2} · 5 · 7 = 82, 575, 360.

(ii) |M| = 2^{21} · 3^{3} · 5 · 11^{3} · 23 · 29 · 31 · 37 · 43 = 86, 775, 571, 046, 077, 562, 880.

(iii) M is simple.

(iv) Let 1 ≠ z ∈ Z(M_{4}). Then C_{M}(z) = M_{4}.

Proof. (i) follows from 8.25, 8.6, 8.14, 8.18 and 8.23 and (ii) follows from (i).

(iii) Let N be a normal subgroup of M. If N ∩ M_{i} ≠ 1 for some 1 ≤ i ≤ 3 we conclude that Z_{4} ≤ Z_{i} ≤ N (notice that Z_{1} = Q_{1} and Z_{2} = Q_{2}). Hence Q_{2} = ⟨Z_{4}^{M_{2}}⟩ ≤ N, M_{1} = ⟨Q_{2}^{M_{1}}⟩ ≤ N and for j = 2, 3, M_{j} = ⟨M_{1j}^{M_{j}}⟩ ≤ N. Thus M = N. So suppose that M_{i} ∩ N = 1 for all i. Let 1 ≤ i < j ≤ 3. Then M_{ij} is a maximal subgroup of M_{j} and M_{i} is not isomorphic to an overgroup of M_{ij} in M_{j}. Thus M_{i}N ∩ M_{j}N = M_{ij}N and so M/N is a faithful completion of a J_{4}-triangle. By (ii) |M| = |M/N| and so N = 1.

(iv) Let t ∈ C_{M}(z) and put a = M_{1}, b = M_{1}t = a^{t} and c = M_{4}. Then z ∈ Q_{a} ∩ Q_{b} and so by 8.21, 8.18, 8.12 and 8.9, b ∈ X_{i}(a) for some 0 ≤ i ≤ 2. It is now easy to check that {d ∈ Γ_{4}(b) | Z_{d} ≤ Q_{a} ∩ Q_{b}} = Γ_{4}(a, b). Thus c, c^{t} ∈ Γ_{4}(a) and Z_{c} = Z_{c}^{t} implies c = c^{t} and t ∈ M_{4}.□

Acknowledgement

The authors are sincerely thankful to Sergey V. Shpectorov and to Corinna Wiedorn who have read very carefully the previous version of the paper and made a number of very helpful suggestions on how to improve the exposition.

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Department of Mathematics Imperial College London SW7 2BZ, UK a.ivanov@ic.ac.uk

Department of Mathematics Michigan State University East Lansing, Michigan 48824 meier@math.msu.edu