The semiclassical parabolic systems

A. Fukshansky, G. Stroth, C. Wiedorn Martin-Luther-Universität Halle-Wittenberg Fachbereich Mathematik und Informatik Institut für Algebra und Geometrie 06099 Halle, Germany

The classification of parabolic systems was launched for generalizing the notion of a building related to groups of Lie type and intended for the application to structure and classification of the finite simple groups. One of the first major results in this area was the classification of the so called parabolic systems of classical type due to S. Heiss [Hei1], Th. Meixner [Mei1], R. Niles [Ni], G. Stroth [Str1, Str2, Str3, Str4, Str5] and F. Timmesfeld [Tim1, Tim2, Tim3]. See [Mei2] for a survey.

For the convenience of the reader we give the definition of a classical parabolic system. (Recall that for a group G and a subgroup X the subgroup ⋂_{g∈G} X^{g} is denoted ty X_{G}.)

Let G be a group generated by finite subgroups P_{1}, …, P_{n} such that

(1) P_{i} ∉ ⟨P_{j} | j ∈ {1, …, n} \ {i}⟩, i = 1, …, n.

(2) If B = ⋂_{i=1}^{n} P_{i}, then P_{i}/B_{P_{i}} is a rank one group of Lie-type in character-

istic p for some prime p, B/B_{P_{i}} is the Borel subgroup in P_{i}/B_{P_{i}}, and B = P_{i} ∩ P_{j} for any pair P_{i}, P_{j}. Further B_{G} = 1.

(3) For each pair i, j ∈ {1, …, n} let P_{i,j} = ⟨P_{i}, P_{j}⟩ be one of the following:

(α) P_{i,j} = P_{i}P_{j}

(β) P_{i,j}/B_{P_{i,j}} is a rank two Lie type group in characteristic p.

In fact all groups of Lie type have a classical parabolic system. It turns out that among the sporadic simple groups we find McL, LyS and Suz with such a parabolic system. A systematical search [RoStr] lead to a wider class of parabolic systems including more sporadic groups. This is the class of semiclassical systems which is the subject of this paper. For a semiclassical system we just have to fix p = 2 in the definition above and add in (3) the further possibility

(γ) P_{i,j}/B_{P_{i,j}} ≅ 3Σ_{6} or 3A_{6}.

Furthermore we assume that there is some pair i, j realizing (γ).

The sporadic groups showing up in this context are M_{24}, He, Co_{1}, M(24)′ and F_{1}.

Let now G = ⟨P_{1}, …, P_{n}⟩ be a semiclassical parabolic system. We call n the rank of the system. To each system of rank n we associate a diagram Δ in the following way. The nodes of the diagram are {1, …, n} and i, j is connected by the Dynkin diagram corresponding to P_{i,j}/B_{P_{i,j}}, if this is a

rank two Lie type group, by _{i} _{j} ∘ ∘^{∼} if P_{i,j} is as in (3)(γ) and not connected otherwise.

If we will allow ∘ ∘ and ∘ ∘^{∼} at the same place we also write ∘ ∘^{(∼)} . We will denote any possible connection between two nodes by ∘ ∘ ........ . Since non connected diagrams correspond to a product of groups we may assume

(4) Δ is connected.

According to [Tim3] the semiclassical diagrams of rank three correspond to diagrams ∘ ∘ ∘^{(∼)} ^{∼} , or ∘ ∘ ∘^{∼} . For the latter according to [Hei2] and [Row] we have G ≅ He, M_{24} or 3^{7}Sp_{6}(2). The first step in a classification of semiclassical systems in general is the classification of systems of rank four. Of course then one of the above diagrams is involved in Δ. The systems of rank four in which any subdiagram of shape ∘ ∘ ∘^{∼} belongs to 3^{7}Sp_{6}(2) have been classified by G. Gillespie [Gil] and C. Wiedorn [Wie1]. If some subdiagram of type ∘ ∘ ∘^{∼} belongs to M_{24} or He the rank four systems are classified by [FuStr], provided P_{i}/B_{P_{i}} ≅ Σ_{3} for all 1 ≤ i ≤ 4. In (1.2) we will see that this always holds. The rank four diagrams involving ∘ ∘ ∘^{(∼)}^{∼} are handled in [Wie1].

The purpose of this paper is to finish the classification of semiclassical parabolic systems, i.e. to give a classification of rank at least five.

To state the main result we need some notation and definitions. Let G = ⟨P_{i} | i ∈ I⟩. For J ⊆ I set G_{J} = ⟨P_{i} | i ∈ I\J⟩, hence G = G_{ϕ}. Set K_{J} = B_{G_{J}}.

For Δ and J ⊆ I let Δ_{J} be the subdiagram obtained from Δ by dropping all nodes in J and all edges having one vertex in J. Then G_{J}/K_{J} is a parabolic system with diagram Δ_{J}.

Now we will give a list of some parabolic systems.

(i) Δ is a complete bipartite graph all of whose edges are of type ∘ ∘^{(∼)} . There exists J ⊆ I such that Δ_{J} = ∘ ∘ ∘^{(∼)} ^{(∼)} , K_{J} = 1 and G_{J} is one of the groups in theorem (1.1)(a)(iv)(⋆)(⋆⋆).

(ii) The systems He(n, m), n, m ≥ 1 :

The diagram Δ is of type [figure: ∘; ∘; ∘; ∘; ∘; ∘; ∼; ∼; ∼; ∼; 1; 2; b_{m}; b_{1}; a_{n}; a_{1}; .; .; .; .; .; .; .]

Let J ⊆ I such that Δ_{J} is of type ∘ ∘ ∘^{∼} , then K_{J} = 1 and G_{J} ≅ He.

(iii) The systems X(2, m), m ≥ 0 :

The diagram Δ is of type  [figure: ∘; ∘; ∘; ∘; ∘; ∼; (∼); 3; 4; m+4; 2; 1; .; .; .] .

Let J ⊆ I such that Δ_{J} is of type ∘ ∘ ∘^{(∼)} , then K_{J} = 1 and G_{J} ≅ 3^{7}Sp_{6}(2) or Sp_{6}(2). Set J′ = {4, …, m + 4}, then G_{J′} ≅ 2^{6}A_{7}.

(iv) The systems Y (n, m), n, m ≥ 1 :

The diagram Δ is of type  [figure: ∘; ∘; ...; ∘; ∘; ∘; ∘; ∼; (∼); n+1; n; n+2; n+m+1; 2; 1; .; .; .].

Let J ⊆ I such that Δ^{J} corresponds to ∘ ∘... ∘ ∘ ∘^{(∼)}_{n+1}_{n}_{2}_{1}  ,

then K_{J} = 1 or G_{J} ≅ 2^{6} · A_{7}. In the latter n = 1 and there is at most one J ⊆ I such that G_{J} ≅ 2^{6}A_{7} and (α) below holds for all other J. Further one of the following holds

(α) For all J of that type such that K_{J} = 1 we have

G_{J} ≅ 3^{\frac{1}{3} (2^{n+2}−1)(2^{n+1}−1)}Sp_{2(n+2)}(2) or Sp_{2(n+2)}(2). (β) n = 3 and for all J we have K_{J} = 1 and G_{J} ≅ F_{1} (γ) n = 2 and for all J we have K_{J} = 1 and G_{J} ≅ Co_{1}. (δ) n = 1 and for all J we have K_{J} = 1 and exactly one of the

following occurs: G_{J} ≅ M_{24} or G_{J} ≅ He.

This also includes m = 1. Then J = ϕ and in (α)−(δ) we have G_{J} = G is one of these groups.

(v) The systems M(n, m), n ≥ 2, m ≥ 1 : We have a partition I = {1} ∪ I_{n} ∪ I_{m}, where |I_{n}| = n and |I_{m}| = m. The diagram induced on I_{n} ∪ I_{m} is as in (i), i.e. a complete bipartite graph all of whose edges are of type ∘ ∘^{(∼)} . There are no edges from 1 to any node in I_{m} and there is exactly one edge from 1 to t ∈ I_{n}. Let u ∈ I_{m}, v ∈ I_{n}\{t}, and set J=I \{1, u, v, t}. Then

Δ_{J} = _{1}  _{t}  _{u}  _{v} ∘ ∘ ∘ ∘^{∼}  ^{(∼)} .

We have K_{J} = 1 and G_{J} is locally M(24), M(24)′, 3M(24) or 3M(24)′. (In the sense of [Gold].)

(vi) System of type F̃_{4} : Δ is of shape ∘ ∘ ∘ ∘^{∼} . This system has been described in [Wie3].

(vii) Square system:

Δ is of shape ∘ ∘ ∘ ∘_{1}  _{2}^{3}  ^{4}^{∼}  , G_{3} ≅ G_{4} ≅ A_{7} and G_{1,2} ≅ 3A_{6}.

In this paper we prove:

Theorem. Let G = ⟨P_{1}, …, P_{n}⟩ be a semiclassical parabolic system of rank n with connected diagram Δ. Then it is one of the systems from the list above.

The remaining notations and definitions are standard. For convenience of the reader we state the definitions around the coset graph used in this paper.

Let Γ = Γ(G_{1}, G_{2}) be the coset graph for G = ⟨G_{1}, G_{2}⟩, i.e. the vertices are the cosets of G_{1} and G_{2} and (G_{1}g, G_{2}h) is an edge iff G_{1}g ∩ G_{2}h ≠ ϕ. Then G acts on Γ by right multiplication. Let us identify the vertices G_{1}, G_{2} with 1, 2 respectively. Then G_{i} is the stabilizer in G of the vertex i. For δ ∈ Γ denote by G_{δ} the stabilizer of δ in G. Then G_{δ} is conjugate to G_{1} or G_{2}.

Assume that G_{1} ∩ G_{2} contains a Sylow 2-subgroup S of G_{1} and G_{2}. Then set for δ ∈ Γ :

Z_{δ} = ⟨Ω_{1}(Z(T)) | T ∈ Syl_{2}(G_{δ})⟩ and
b_{δ} = min{d(δ, λ) | λ ∈ Γ, Z_{δ} ⊈ G_{Δ(λ)}},

where G_{Δ(λ)} is the pointwise stabilizer of the neighborhood of λ, and set

b = min{b_{δ} | δ ∈ Γ}.

As G acts edge transitively on Γ, we get only two parameters b_{1} and b_{2} and so b = min{b_{1}, b_{2}}.

By a critical pair we mean a pair (δ, δ′) of vertices of Γ such that

(i) Z_{δ} ⊈ G_{Δ(δ′)},

(ii) d(δ, δ′) = b.

This coset graph will play an important role in the proof of the main theorem.

By [ShStr] a diagram of shape ∘ ∘... ∘ ∘ ∘^{∼}_{n-1}  _{n}_{2}_{1}  corresponds to a

non-split extension of Sp_{2n}(2) by an abelian group of order 3^{\frac{1}{3} (2^{n}−1)(2^{n−1}−1)} if G_{{1,...,n−3}} ≅ 3^{7}Sp_{6}(2). If we do not care about the tilde, i.e. the diagram

is of shape ∘ ∘... ∘ ∘ ∘^{(∼)}_{n-1}  _{n}_{2}_{1}  , then we will denote the corresponding

group by (3^{\frac{1}{3} (2^{n}−1)(2^{n−1}−1)}) Sp_{2n}(2).

§1 Preliminaries

The proof of the main theorem will be done by induction on the rank of the parabolic system. For convenience of the reader we first state Timmesfeld’s result [Tim3] on rank three systems.

(1.1) Theorem. Let G = ⟨P_{1}, P_{2}, P_{3}⟩ be a parabolic system of classical or semiclassical type with connected diagram Δ. Then one of the following holds:

(a) For all i we have P_{i}/O_{2}(P_{i}) ≅ Σ_{3} and

(i) Δ = ∘ ∘ ∘ and G ≅ A_{7} or A_{8}.

(ii) Δ = ∘ ∘ ∘^{∼} and G ≅ M_{24}, He or 3^{7}Sp_{6}(2)

(iii) Δ = ∘ ∘ ∘ and G ≅ A_{7} or Sp_{6}(2).

(iv) Δ = _{1} _{2} _{3} ∘ ∘ ∘^{(∼)} ^{(∼)} , G_{i}/K_{i} ≅ A_{6}, Σ_{6}, 3A_{6}, or 3Σ_{6}, i ∈ {1, 3}, and there are two possibilities:

(⋆) 2^{4} ≤ |K_{i}| ≤ 2^{5}, K_{i} is an indecomposable G_{i}-module and C_{K_{i}}(G_{i}) = 1, for i = 1, 3. Moreover Ω_{1}(Z(S)) < G_{2} and

G_{2}/K_{2} ≅ Σ_{3} × Σ_{3}.
(⋆⋆) K_{1} = K_{3} = 1, G_{2}/K_{2} ≅ (ZZ_{3} × ZZ_{3})ZZ_{2}.
(v) Δ = ∘ ∘ ∘_{1}  _{2}  _{3} , ∘ ∘ ∘_{1}  _{2}  _{3} , and G_{1} ≅ G_{2}(2) or G_{2}(2)′.

(b) There exists at least one i with P_{i}/O_{2}(P_{i}) ≅ Σ̸_{3}, we have a classical parabolic system and one of the following holds:

(i) the diagram is of type C_{3} and G contains a normal subgroup isomorphic to the corresponding group of Lie type.

(ii) the diagram is a triangle and B = 1.

(iii) Δ = ∘ ∘ ∘ n  n_{2}  and P_{2}/O_{2}(P_{2}) ≅ Σ̸_{3}.
(iv) Δ = ∘ ∘ ∘_{1}  or ∘ ∘ ∘_{1}  and P_{1}/O_{2}(P_{1}) ≅ L_{2}(4).

One of the facts which follows from (1.1) is that a diagram for a semiclassical system with P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for all i never contains a triangle. We will use this fact freely in what follows.

Next we will establish the classification of rank four semiclassical systems with connected diagram. For this first we prove that P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for any such system.

(1.2) Lemma. Let {P_{1}, P_{2}, P_{3}, P_{4}} be a semiclassical parabolic system of rank four with connected diagram. Then P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for i = 1, 2, 3, 4.

Proof. By (1.1) the parabolic subgroups P_{i} in a semiclassical rank-three-system satisfy P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for all i. So we get that Δ is a string and we may choose notation such that Δ has the shape

∘ ∘ ∘ ........ ........_{1}  _{2}  _{3}  _{4}∘^{∼}  ,

where P_{1}/O_{2}(P_{1}) ≅ Σ̸_{3}. As P_{2}/O_{2}(P_{2}) ≅ Σ_{3} we get P_{1}/O_{2}(P_{1}) ≅ L_{2}(4) by (1.1) and Δ is of the form

∘ ∘ ∘_{L_{2}(4)}  ∘^{∼}  or ∘ ∘ ∘_{L_{2}(4)}  ∘^{∼}  .

The first diagram was shown not to exist by [Wie2; (4.13)]. So assume that

we have ∘ ∘ ∘_{1}  _{2}  _{3}  _{4}∘^{∼} .

Now we have G_{1}/K_{1} ≅ M_{24}, He or 3^{7}Sp_{6}(2). In the latter the assertion follows with [Wie2; (4.12)] again. So assume that G_{1}/K_{1} ≅ M_{24} or He. As G_{2}/K_{2} ≅ L_{2}(4) × 3Σ_{6}, we see that | K_{1} : O_{2}(K_{1}) |= 3 and K_{2} = O_{2}(K_{2}). Now we apply [ParRow; Theorem A] to (P_{1}, G_{1}). This shows I C_{G_{1}}(O_{2}(K_{1})) ⊈ O_{2}(K_{1}). As P_{1}/O_{2}(P_{1}) ≅ L_{2}(4), B has no normal Sylow-3-subgroup, so I C_{K_{1}}(O_{2}(K_{1})) ⊆ O_{2}(K_{1}). Thus G_{1} = K_{1} I C_{G_{1}}(O_{2}(K_{1})). As Z(G_{1} ∩ G_{4}/K_{4}) = 1, we see O_{2}(K_{1}) ⊆ K_{4}. But for a Sylow 2-subgroup S of G_{1} ∩ G_{4} we have | S : K_{4} |= 2^{12} while | S : O_{2}(K_{1}) |= 2^{10} (notice that G_{4}/K_{4} ≅ Ω_{8}^{−}(2)). This contradiction proves the lemma. □

Now (1.1) and (1.2) together with the results of [FuStr], [Gil], [Wie2], and [Wie3] yield

(1.3) Theorem. Let G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩, {P_{1}, P_{2}, P_{3}, P_{4}} be a semiclassical parabolic system with connected diagram Δ. Then one of the following holds

(i) Δ : [figure: ∘; ∘; ∘; ∘; ∼; ∼; 1; 2; 3; 4] G_{4}, G_{3} ≅ He or M_{24}

(ii) Δ : _{1}  _{2}  _{3}  _{4} ∘ ∘ ∘ ∘^{∼}  ^{∼}  G_{1} ≅ G_{4} ≅ He
(iii) Δ : _{4}  _{3}  _{2}  _{1} ∘ ∘ ∘ ∘^{∼}  G ≅ Co_{1}
(iv) Δ : _{1}  _{2}  _{3}  _{4} ∘ ∘ ∘ ∘^{∼}  ^{(∼)}  the groups G_{i} are isomorphic to the

corresponding groups in M(24), 3M(24), M(24)′ or 3M(24)′.

(v) Δ : [figure: ∘; ∘; ∘; ∘; ∼; 3; 4; 2; 1] G_{3}/K_{3} ≅ A_{7}, K_{1} = K_{2} = 1, K_{3} is elementary abelian of order 2^{6} and G_{1} ≅ G_{2} ≅ 3^{7}Sp_{6}(2).

(vi) Δ : [figure: ∘; ∘; ∘; ∘; (∼); ∼; 3; 4; 2; 1] if G_{2}/K_{2} ≅ 3^{7}Sp_{6}(2) or Sp_{6}(2) then K_{1} = K_{2} = 1, if G_{2}/K_{2} ≅ A_{7}

then K_{1} = 1, G_{2} = 2^{6}A_{7}, and
G_{3}/K_{3} as in (1.1)(a)(iv) (⋆).

(vii) Δ : ∘ ∘ ∘ ∘^{∼} there is a normal subgroup K of G with G/K ≅ F_{4}(2).

(viii) Δ : ∘ ∘ ∘ ∘^{∼} G ≅ 3^{35}Sp_{8}(2).

(ix) Δ :  ∘ ∘ ∘ ∘_{3}  _{4}^{1}  ^{2}^{∼}  K_{i} = 1 for all i, G_{1} ≅ G_{2} ≅ A_{7} and

G_{3} and G_{4} are as in (1.1)(a)(iv)(⋆⋆).

(x) Δ :  ∘ ∘ ∘ ∘_{3}  _{4}^{1}  ^{2}^{∼}_{(∼)}_{(∼)}  _{(∼)}  one of the following holds:

(1) K_{i} = 1, G_{i} is as in (1.1)(a)(iv)(⋆⋆) for all i,

(2) we can choose numeration such that K_{1} =K_{3} =1, G_{1} and G_{3} are as in (1.1)(a)(iv)(⋆) and G_{2}/K_{2} and G_{4}/K_{4} are of type (1.1)(a)(iv)(⋆⋆).

(xi) Δ : [figure: ∘; ∘; ∘; ∘; (∼); (∼); (∼); 3; 2; 4; 1] one of the following holds:

(1) K_{i} = 1, G_{i} is as in (1.1)(a)(iv)(⋆⋆) for i = 1, 3, 4,

(2) we can choose numeration such that K_{1} = K_{3} = 1, G_{1} and G_{3} are as in (1.1)(a)(iv)(⋆) and G_{4}/K_{4} is of type (1.1)(a)(iv) (⋆⋆).

In the remainder of this section we prove some results on representations of the groups which will be involved in the proof of the main theorem.

First of all we need some facts about representations of Chevalley groups. We do not state results in full generality as we will just need them for untwisted groups over GF (2). The reader who is interested in the general theory should consult the fundamental paper of R. Steinberg [Stein].

(1.4) Lemma. [Sm] Let G = G(q), q = p^{f}, be a Chevalley group (twisted or not) and V be a nontrivial irreducible GF (q)–module for G. Let P be a parabolic subgroup of G and V_{p} = I C_{V} (O_{p}(P)). Then V_{p} is an irreducible module for P.

Let now G = G(2) be untwisted of rank at least two, V be an irreducible GF (2)G–module and {P_{1}, …, P_{n}} be the set of minimal parabolic subgroups containing a given Sylow 2-subgroup S. Set V_{i} = I C_{V} (O_{2}(P_{i})). As P_{i}/O_{2}(P_{i}) ≅ Σ_{3} we have with (1.4) that V_{i} is either trivial or a two– dimensional Σ_{3}–module. Let Λ be the n–dimensional vectorspace over GF (2) with basis {λ_{1}, …, λ_{n}}. Now set a_{i} = 0 if V_{i} is trivial and a_{i} = 1 else. By

V ⟶ λ = ∑_{i=1}^{n} a_{i}λ_{i}

we get a mapping from the set of isomorphism types of irreducible modules to the set of elements of Λ. We denote V by V (λ). Then we have the following fundamental theorem due to R. Steinberg.

(1.5) Theorem. Let G = G(2) be an untwisted Chevalley group over GF (2) of rank at least two and V be an irreducible GF (2)–module for G. Then the mapping V ⟶ V (λ) above is a bijective mapping between the isomorphism classes of modules and the vectors of Λ.

The module V (λ), λ = ∑_{i=1}^{n} λ_{i}, is called the Steinberg module for G. Its dimension is the order of a Sylow 2-subgroup of G.

The theorem (1.5) will be used to identify modules. We just have to check the action of the minimal parabolics on the centralizers of their unipotent radical. Then the module is uniquely determined.

(1.6) Definition. Let G be some group acting faithfully on some GF (p)- module V.

(a) The module V is called an F -module if |V : I C_{V} (A)| ≤ |A| for some nontrivial elementary abelian p-subgroup A of G.

(b) The module V is called an F_{1}-module if |V : I C_{V} (A)| ≤ p · |A| for some nontrivial elementary abelian p-subgroup A of G.

(c) The group A in (a) and (b) is called an offending subgroup.

(1.7) Lemma. Let G be one of Sp_{2n}(q) or Ω_{2n}^{+}(q), n ≥ 3, q = 2^{f} and V be an F_{1}-module with offending subgroup A. Then one of the following holds:

(i) G ≅ Sp_{2n}(q) and V is the natural module, V is the 8-dimensional spin module and n = 3, or V is the 16-dimensional spin module, n = 4, and q = 2. In the latter V is not an F -module. If V is the natural module, A ⊆ O_{2}(G_{v}), v ∈ V^{#}, then |V : I C_{V} (A)| = |A| and A contains a transvection.

(ii) G ≅ Ω_{2n}^{+}(q), V is the natural module or n = 4 or 5 and V is one of two 8-dimensional, resp. 16-dimensional halfspin modules.

Proof. [Coop; (2.1)], [McCl; (6.40)]. □

(1.8) Lemma. Let G = Ω_{2n}^{+}(q), q = 2^{f}, V an irreducible module over GF (q). Suppose G = ⟨P_{1}, …, P_{n}⟩ where the P_{i}’s are the minimal parabolic subgroups corresponding to the diagram

[figure: ∘; ∘; ∘; n; n-1; …; ∘; ∘; 2; 1].

Let S ∈ Syl_{2}(⋂_{i=1}^{n} P_{i}).

(i) If V in the natural module, then N_{G}( I C_{V} (S)) = ⟨P_{1}, …, P_{n−1}⟩

(ii) If V is a halfspin module, then N_{G}( I C_{V} (S)) = ⟨P_{1}, P_{3}, …, P_{n}⟩ or ⟨P_{2}, P_{3}, …, P_{n}⟩.

If q = 2 the converse of (i) and (ii) is also true.

Proof. (i) and (ii) are well known. The last assertion follows from (1.5). □

(1.9) Lemma. Let G = Sp_{2n}(q), q = 2^{f}, V an irreducible GF (q)- module, G = ⟨P_{1}, …, P_{n}⟩ where the P_{i}’s are the minimal parabolic subgroups corresponding to the following diagram

∘ ∘ ∘_{1}  _{2}  … ∘ ∘_{n-1}  _{n}

Let S ∈ Syl_{2}(⋂_{i=1}^{n} P_{i}).

(i) If V is the natural module, then N_{G}( I C_{V} (S)) = ⟨P_{1}, …, P_{n−1}⟩

(ii) If V in the spin module, then N_{G}( I C_{V} (S)) = ⟨P_{2}, …, P_{n}⟩.

If q = 2 the converse of (i) and (ii) is also true.

Proof. Well known, the last assertion follows from (1.5). □

(1.10) Lemma. Let G ≅ Sp_{2n}(2), t ∈ G an involution and V a nontrivial irreducible GF (2)G–module.

(i) If n = 3 and |V : I C_{V} (t)| ≤ 8 then V ≅ V (λ_{1}) or V (λ_{3}).

(ii) Let n = 3 and t a transvection in the natural representation of G. If |V : I C_{V} (t)| ≤ 16 then V ≅ V (λ_{i}), i = 1, 2, 3.

(iii) Let n = 4 and t an involution of type a_{2} in the natural representation. If |V : I C_{V} (t)| ≤ 2^{8} then V ≅ V (λ_{1}) or V (λ_{4}).

(iv) Let n = 4 and |V : I C_{V} (t)| ≤ 16, then V ≅ V (λ_{1}) or V (λ_{4}).

Proof. First of all we prove

(∗) Let G ≅ Sp_{4}(2), t ∈ G be an involution and V an irreducible GF (2)G– module with |[V, t]| = 2. Then |V | = 16 and V is one of the two natural modules.

We have Sp_{4}(2) ≅ Σ_{6}. As t cannot invert an element of order 5 in A_{6}, we get t ∉ A_{6}. So by suitable numbering we have [t, P_{1}] = 1, P_{1} being one of the two parabolic subgroups containing a given Sylow 2-subgroup. Now the assertion follows with (1.9).

For the remainder of the proof choose numbering of the minimal parabolic subgroups according to the diagram

∘ ∘_{2}_{1}  ∘ ∘ ∘ …  _{n-1}  _{n} .

(i) Set V_{1} = I C_{V} (O_{2}(G_{1})). By (1.9) we may assume that V_{1} is a nontrivial module. Let ⟨i⟩ = Z(G_{1}). We may assume t ∈ G_{1} \ O_{2}(G_{1}). Hence t is nontrivial on [V, i] and V/ I C_{V} (i). This shows that t induces a transvection on V_{1}. By (∗) V_{1} is one of the natural modules for Sp_{4}(2). As [V, i] = V_{1} by (1.4) we now know that t is not of type c_{1} or c_{2} in the natural representation.

Let t be of type c_{3}. Then t inverts some element ν of order 7 in G. This shows |[V, t]| = 8 and |[V, ν]| = 64. Set V_{3} = I C_{V} (O_{2}(G_{3})). By (1.9) we may assume that V_{3} is nontrivial. If V_{3} is the eight–dimensional Steinberg module, then [V, ν] ⊆ V_{3} and so [V, G_{3}] = V_{3}. But as O_{2}(G_{3}) ≅ V̸_{3} as G_{3}/O_{2}(G_{3})– module, this is impossible. So by (1.5) V_{3} is the natural module or its dual. As O_{2}(G_{3}) is indecomposable, we get | I C_{V/V_{3}}(G_{3})| ≤ 2. Hence |[V, G_{3}]| ≤ 2^{7} and so |V | ≤ 2^{8}. In particular V/V_{1} is an irreducible G_{1}/O_{2}(G_{1})–module, which implies [V, O_{2}(G_{1}), O_{2}(G_{1})] = 1. As G = ⟨ I C_{G}(x) | x ∈ O_{2}(G_{1})^{#}⟩, we see that there is some x ∈ O_{2}(G_{1})^{#} with I C_{V} (x) > V_{1}. So we have found x ∈ O_{2}(G_{1}) of type a_{2} with |V : I C_{V} (x)| ≤ 8. Hence we may assume that t is of type a_{2} right from the beginning. So we may assume that t is in the center of G_{2}. Now G_{2} induces on [V, t] a subgroup of L_{3}(2). The largest factor group of G_{2} which can be embedded into L_{3}(2) is Σ_{3}, so we get [V, t, O_{2}(G_{2})] = 1. Application of [Str6; Theorem] gives the assertion.

(ii) We may choose t such that t ∈ Z(G_{1}). Hence [V, t, O_{2}( I C_{G}(t))] = 1. The assertion follows with [Str6; Theorem].

(iii) We choose notation such that G_{2} = I C_{G}(t). Set V_{i} = I C_{V} (O_{2}(G_{i})) for i = 1, 2. As V_{2} ≤ [V, t] by (1.4) we get |V_{2}| ≤ 2^{8}. Let |V_{2}| = 2, then by (1.5) we have V = V (λ_{2}). Now I C_{V} (Z(G_{1})) and V/ I C_{V} (Z(G_{1})) are nontrivial G_{1}–modules. As t is conjugate to some involution in G_{1} \O_{2}(G_{1}), we see that |[V, t]| ≥ 4. In particular V_{2} ≠ [V, t]. As W = O_{2}(G_{2})/Z(O_{2}(G_{2})) is an irreducible G_{2}/O_{2}(G_{2})–module we get that [V, t]/V_{2} involves a module isomorphic to W. Hence |[V, t]| ≥ 2^{9}, a contradiction. So we have |V_{2}| > 2, Let |V_{2}| = 4. By (1.5) we may assume V = V (λ_{1} +λ_{2}). In particular by (1.9) V_{1} = I C_{V} (O_{2}(G_{1})) is the natural Sp_{6}(2)–module. Further we see that [V, t]/V_{2} involves a nontrivial Σ_{6}–module. In particular I C _{I C_{G}(t)}([V, t]) ⊆ O_{2}( I C_{G}(t)).

Let H_{1} be a complement of O_{2}(G_{1}) in G_{1} with N_{G}(H_{1}) ≅ H_{1} × Σ_{3} and g ∈ G with s = t^{g} ∈ H_{1} of type a_{2} in the natural representation. Suppose that V just involves natural modules as non-trivial H_{1}–factors. Then there is some ρ ∈ I C_{H_{1}}(s), o(ρ) = 3, such that [ρ, [V, s]] = 1. But this contradicts the fact that I C _{I C_{G}(t)}([V, t]) ⊆ O_{2}( I C_{G}(t)). So V involves at least one non-trivial H_{1}–module, which is not the natural module.

Let i be an involution in I C_{G}(H_{1}), then we have that [V, i] and V/ I C_{V} (i) both involve the natural module. Let U = ⟨V_{2}^{G_{1}}⟩. As |V_{2} : V_{2} ∩ V_{1}| = 2, we see with (1.9) and Gaschütz ([Hu; I.17.4]) that either |U : V_{1}| = 2 or U/V_{1} involves the natural module. In the former we would have U = V_{2}V_{1} and so [O_{2}(G_{1}), U] = [V_{2}, O_{2}(G_{1})] ⊆ V_{2}, a contradiction. So we have at least three natural modules involved. As there has to be a module involved that is not the natural one, by (i) this has to be the spin module and so |[V, t]| = 2^{8}.

Let ν ∈ I C_{G}(H_{1}), o(ν) = 3. We have |V/ I C_{V} (ν)| ≤ |V/ I C_{V} (i)|^{2} ≤ 2^{12}. As H_{1} acts on I C_{V} (ν) and |[V, ν]| ≤ 2^{12} we see that [V, ν] involves two natural modules and I C_{V} (ν) involves one natural module and one spin module. We know [V, t] ≅ [V, s] = W > W_{1} > W_{2} > 1, |W_{2}| = 4 = |W/W_{1}| and |W_{1}/W_{2}| = 16, W_{1}/W_{2} is a nontrivial Σ_{6}–module. Since we have Z(O_{2}( I C_{H_{1}}(s))) = Z(O_{2}( I C_{G}(s))) we get [ν, Z(O_{2}( I C_{G}(s)))] = 1 and so [W_{2}, ν] = 1 and [W, ν] ⊆ W_{1}. Hence W_{1} = W_{2}⊕[W_{1}, ν], since |[[V, s], ν]| = 16. But now

[[W_{1}, ν], O_{2}( I C_{H_{1}}(s))] ⊆ W_{2} ∩ [W_{1}, ν] = 1.

This shows [W_{1}, O_{2}( I C_{G}(s))] = 1, a contradiction.

Assume next |V_{2}| = 16. Let [ I C_{G_{2}}(Z(O_{2}(G_{2}))), [V, t]] = [V, t]. Then there is some ρ ∈ G_{2}, o(ρ) = 3, with [[V, t], ρ] = 1 and [O_{2}(G_{2}), ρ] = O_{2}(G_{2}). Then [[V, t], O_{2}(G_{2})] = 1 and the assertion follows with [Str6; Theorem].

So we may assume V_{2} = [[V, t], I C_{G_{2}}(Z(O_{2}(G_{2})))] ≠ [V, t]. In particular [Z(O_{2}(G_{2})), [V, t]] = 1. Application of [Str6; Corollary 2] implies V = V (λ_{i}), i = 1, 2, 3, 4. By (1.5) obviously V ≅ V̸ (λ_{2}). So we just have to consider V ≅ V (λ_{3}). Then V_{1} is not the natural or spin module. Let H_{1} and s be as before. We see |[V_{1}, s]| ≥ 16 by (i) and so |[V, t]| = 2^{8}. This then shows [ I C_{V} (Z(G_{1})), G_{1}] = V_{1}. Since O_{2}(G_{1}) ≅ V̸_{1} as G_{1}/O_{2}(G_{1})–module, we get V_{1} = I C_{V} (Z(G_{1})) and so [V, t] ⊆ V_{1} as t ∈ O_{2}(G_{1}). In particular [[V, t], O_{2}(G_{1})] = 1. Now [[V, t], ⟨O_{2}(G_{1})^{G_{2}}⟩] = 1. But then we have ⟨O_{2}(G_{1})^{G_{2}}⟩ = I C_{G_{2}}(Z(O_{2}(G_{2}))), a contradiction.

We are left with |V_{2}| = 2^{8}. Now V_{2} = [V, t] and the assertion follows with [Str6; Theorem].

(iv) We may assume t ∈ G_{1} \ O_{2}(G_{1}). Suppose there is some s ∈ O_{2}(G_{1}) of type a_{2} in the natural representation with t ∼ ts. Then |[V, s]| ≤ 2^{8} and the assertion follows with (iii). So we may assume that there is no such s. Then we see that t has to induce a transvection on O_{2}(G_{1}). So t is not of type c_{2}. Moreover t is not of type a_{2}, otherwise we are done by (iii). As we may assume that V ≅ V̸ (λ_{1}), we see | I C_{V} (O_{2}(G_{1}))| ≥ 2^{6} and so |[V, Z(G_{1})]| ≥ 2^{6}, i.e. t is not of type c_{1}. As t has to induce a transvection on O_{2}(G_{1}), we see that t is of type c_{3}. But there are always two commuting elements of type c_{3} which just differ by an element of type a_{2}. This can easily be seen in O_{6}^{+}(2) ≅ Σ_{8} where c_{3}–elements correspond to involutions of type (12)(34)(56) and a_{2}–elements correspond to involutions of type (13)(24). □

(1.11) Lemma. Let G ≅ 3^{7}Sp_{6}(2) and V be a faithful 56- dimensional GF (2)G-module. Let t be an involution of Sp_{6}(2) of type c_{1}, c_{2} or a_{2} (on the natural Sp_{6}(2)-module). Then |V : I C_{V} (t) |≥ 2^{16}.

Proof. By [Sh] we have V = V_{1} ⨁ ⋯ ⨁ V_{28}, where V_{i} = I C_{V} (E_{i}), E_{i} a hyperplane in O_{3}(G) with N_{G}(E_{i})/O_{3}(G) ≅ O_{6}^{−}(2). Let t ∈ N_{G}(E_{i}) be of type c_{1} then t ∈ N_{G}(E_{i})\N_{G}(E_{i})′. In that case | [V_{i}, t] |= 2. As | I C_{G/O_{3}(G)}(t) : I C_{N_{G}(E_{i})/O_{3}(G)}(t) |= 16, we have 16 E_{i}’s with [E_{i}, t] ⊆ E_{i} and so | [V, t] |≥ 2^{16}. Let t ∈ N_{G}(E_{i}) be of type c_{2} or a_{2} and so t ∈ N_{G}(E_{i})′. Hence [V_{i}, t] = 1. As | I C_{G/O_{3}(G)}(t) : I C_{N_{G}(E_{i})/O_{3}(G)}(t) |= 8 or 4 respectively we get that there are at least 20 of the E_{i}’s which are not normalized by t. Since |V_{i}| = 4 this yields | [V, t] |≥ 2^{20}. □

(1.12) Lemma. (a) Let t ∈ Sp_{2n}(2), n ≥ 3, be an involution. Then Sp_{2n}(2) is generated by 2n + 1 conjugates of t.

(b) Let G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ be a semiclassical parabolic system with diagram ∘ ∘ ∘_{1} _{2} _{3} _{4}∘^{∼} . Let t be an involution in O_{2}(G_{1}). Then there are nine conjugates of t which generate G.

Proof. (a) Let G_{1} be the stabilizer of a point in the natural representation. It is enough to prove that G_{1} is generated by 2n conjugates of t. This will be done by induction on n.

First we investigate Sp_{4}(2) ≅ Σ_{6}. If t is an involution in Σ_{6} \ A_{6} then t is conjugate to (12) or (12)(34)(56) and we see that Σ_{6} is generated by 5 conjugates. If t = (12)(34) then ⟨(12)(34), (13)(45), (13)(24), (13)(56)⟩ = A_{6}.

We have that O_{2}(G_{1}) contains three classes of involutions which are of type c_{1}, c_{2}, or a_{2} in the natural representation. As n ≥ 3, all these elements are conjugate into G_{1} \ O_{2}(G_{1}), so we may assume t ∈ G_{1} \ O_{2}(G_{1}). Now by induction and the remark about Σ_{6}, we get that G_{1}/O_{2}(G_{1}) is generated by 2n−1 conjugates of tO_{2}(G_{1}) or n = 3 and these conjugates just generate A_{6}. Now there is s ∈ O_{2}(G_{1})\Z(G_{1}) with t ∼ ts in G_{1}. Hence G_{1} is generated by 2n conjugates or n = 3 and these conjugates just generate G′_{1}. So assume the latter. Choose some conjugate t^{g} ∈ G \ G_{1} and set H = ⟨G′_{1}, t^{g}⟩. Let V be the natural module for G. On V^{#} we have G′_{1}–orbits of length 1, 30, and 32. Hence as neither 11 nor 31 divides |Sp_{6}(2)| we see that H acts transitively on V^{#}. So

2^{8} · 3^{4} · 5 · 7 = |G′_{1}| · 63 | |H|.

As Sp_{6}(2) is simple we get H = Sp_{6}(2). So in any case Sp_{6}(2) is generated by at most 7 conjugates of t. The assertion now follows by induction.

Let now G, t be as in (b). By (1.3)(vii) G has a homomorphic image F_{4}(2). Hence there is some conjugate t^{g} ∈ G_{4}\O_{2}(G_{4}). By (a) G_{4}/O_{2}(G_{4}) is generated by seven conjugates of t^{g}. As t^{g} does not act trivially on O_{2}(G_{4})/Z(O_{2}(G_{4})) and O_{2}(G_{4})/Z(G_{4}) is an indecomposable G_{4}–module, we see that G_{4} is generated by eight conjugates of t. As any involution in O_{2}(G_{1}) is conjugate to some involution in O_{2}(G_{1} ∩ G_{4}) \ O_{2}(G_{1}), we see that there is some t^{h} ∈ G_{1}\O_{2}(G_{1}). Hence there is even some t^{x} ∈ G_{1}\G_{1} ∩ G_{4}. But then G = ⟨G_{4}, t^{x}⟩ and so G is generated by nine conjugates of t. □

(1.13) Lemma. Let G = F_{4}(2) with diagram ∘ ∘ ∘_{1}  _{2}  _{3}  _{4}∘, and

let G_{1}, G_{4} be the two maximal parabolics sharing a Sylow 2-subgroup with G_{1}/O_{2}(G_{1}) ≅ Sp_{6}(2) ≅ G_{4}/O_{2}(G_{4}). Set ⟨i⟩ = Z(G_{1}) and ⟨j⟩ = Z(G_{4}).

(a) If V is an irreducible GF (2)G-module with quadratically acting group A ≤ O_{2}(G_{1}), |A| ≥ 2^{5}, then V is one of the two 26-dimensional modules.

(b) Let V = V (λ_{1}) be one of the two 26-dimensional irreducible GF (2)- modules. Then

(i) | [V, i] |≥ 2^{9} and [V, i], V/ I C_{V} (i) both involve the 8-dimensional spin module for G_{1}/O_{2}(G_{1}) and I C_{V} (i)/[V, i] involves the natural module. Further [V, j] and V/ I C_{V} (j) are natural modules for G_{4}/O_{2}(G_{4}), the module I C_{V} (j)/[V, j] is isomorphic to V (λ_{2}), the 14-dimensional irreducible module for Sp_{6}(2) ≅ G_{4}/O_{2}(G_{4}), and [[V, j], O_{2}(G_{4})] = 1. If x is a transvection on [V, j], then | I C_{V} (j) : I C _{I C_{V} (j)}(x)| ≥ 2^{5}.

(ii) Let W ⊆ O_{2}(G_{1}) be a quadratic fours group on V with i ∈ W,

then W ⊆ Z(O_{2}(G_{1})). (iii) If W ⊆ O_{2}(G_{1}) is a quadratic fours group on V, then we have

|W : W ∩ Z(O_{2}(G_{1}))| ≤ 2. (iv) If W ⊆ O_{2}(G_{1}) is a quadratic fours group with j ∈ W, then

W ⊆ O_{2}(G_{4}).

(v) Let A ⊆ Z(O_{2}(G_{1})) be a quadratic group on V such that for all a, b ∈ A^{♯} we have |V : I C_{V} (a)| ≤ 2^{6} and |V : I C_{V} (a)∩ I C_{V} (b)| ≤ 2^{7}, then |A| ≤ 8.

Proof. We first prove that (a) is true if O_{2}(G_{1}) contains a quadratically acting fours group ⟨x, y⟩ with ⟨x, y⟩ ∩ Z(O_{2}(G_{1})) = 1. There is some element g ∈ O_{2}(G_{1}) with x^{g} = x, y^{g} = yi. Hence also ⟨x, i⟩ acts quadratically. And so [V, i, ⟨x^{I C_{G}(i)}⟩] = 1. But ⟨x^{I C_{G}(i)}⟩ = O_{2}(G_{1}) and hence the assertion follows with [Str6]. The same applies for G_{4}.

Set B = A ∩ Z(O_{2}(G_{1})). So we may assume |A : B| ≤ 2. Hence |B ∩ O_{2}(G_{4})| ≥ 8. Again we may assume |B ∩ O_{2}(G_{4}) : B ∩ Z(O_{2}(G_{4}))| ≤ 2. Hence 4 ≤ |B ∩ Z(O_{2}(G_{4}))| ≤ |Z(O_{2}(G_{1})) ∩ Z(O_{2}(G_{4}))| = 4. So we get ⟨i, j⟩ = B ∩ Z(O_{2}(G_{4})). Now either A ⊈ Z(O_{2}(G_{1})) or A ⊈ Z(O_{2}(G_{4})), and so either O_{2}(G_{1}) ⊆ ⟨A^{G_{1}}⟩ or O_{2}(G_{4}) ⊆ ⟨A^{G_{4}}⟩. So either [V, i, O_{2}(G_{1})] = 1 or [V, j, O_{2}(G_{4})] = 1. Now the assertion follows with [Str6].

(b) (i) We have [G_{1}, I C_{V} (O_{2}(G_{1}))] = 1 while [G_{4}, I C_{V} (O_{2}(G_{4}))] ≠ 1. Further I C_{V} (O_{2}(G_{4})) is a six-dimensional Sp_{6}(2)-module by (1.4) and the definition of V (λ_{1}). As G_{4} acts indecomposably on O_{2}(G_{4})/⟨j⟩, we see that either [V, j] = I C_{V} (O_{2}(G_{4})) or [V, j] also involves an eight-dimensional module. But then dim [V, j] ≥ 14, which is more than half of the dimension of V.

We have I C_{V} (O_{2}(G_{1})) is one-dimensional and so by the same argument as above we see that [V, i] involves an eight-dimensional module, which also shows |[V, i]| ≥ 2^{9} and that [V, i] and V/ I C_{V} (i) both involve the spin module.

Now dim( I C_{V} (i)/[V, i]) ≤ 26 − 2 · 9 = 8. So [ I C_{V} (i), O_{2}(G_{1})] ⊆ [V, i]. Assume [V, j][V, i] is G_{1}-invariant. Then also [[V, j][V, i], O_{2}(G_{1})] is G_{1}-invariant. But the order of [[V, j], O_{2}(G_{1})] = [V, j] ∩ [V, i] is 2^{5}, a contradiction. So ⟨[V, j]^{G_{1}}⟩/[V, i] has to be a nontrivial module which must involve the natural module by (1.5) because [[V, j], G_{1} ∩ G_{4}] ≤ [V, i].

We have j is conjugate to a transvection x in G_{4}/O_{2}(G_{4}). So we see |[ I C_{V} (j)/[V, j], x]| ≤ 2^{4}. Now by (1.10)(ii) I C_{V} (j)/[V, j] involves just V (λ_{i}), i = 1, 2, 3. As for the spin module W = V (λ_{3}) we would have |[W, x]| = 2^{4} and there are exactly three G_{1} ∩ G_{4}/O_{2}(G_{1} ∩ G_{4}) - composition factors in I C_{V} (j)/[V, j], we see that V (λ_{3}) is not involved. As | I C_{V} (j)/[V, j]| = 2^{14} there cannot be involved three natural modules neither. So we must have I C_{V} (j)/[V, j] ≅ V (λ_{2}). Now the assertion (i) concerning the action of G_{1} and G_{4} is proved. The assertion about I C _{I C_{V} (j)}(x) follows from (1.10)(i).

(ii) As just seen we have [[V, i], O_{2}(G_{1})] ≠ 1. But as [⟨W^{G_{1}}⟩, [V, i]] = 1, we see ⟨W^{G_{1}}⟩ = Z(O_{2}(G_{1})).

(iii) Let W = ⟨a, b⟩ with W ∩ Z(O_{2}(G_{1})) = 1. Then there is t ∈ O_{2}(G_{1}) with a^{t} = a, b^{t} = bi. Now also ⟨a, i⟩ is a quadratic fours group, contradicting (ii).

(iv) If W ⊈ O_{2}(G_{4}), then ⟨W^{G_{4}}⟩ = G_{4} and so [[V, j], G_{4}] = 1, a contradiction.

(v) Suppose | A |≥ 2^{4}. As A ⊆ Z(O_{2}(G_{1})), we have for B = A ∩ O_{2}(G_{4}), |B| ≥ 2^{3}. As | Z(O_{2}(G_{1})) ∩ Z(O_{2}(G_{4}))| = 4, and i ∉ A by (i), we see

|B : B ∩ Z(O_{2}(G_{4}))| ≥ 4.

By conjugation in G_{1} we may assume ⟨x⟩ = B ∩ Z(O_{2}(G_{4})) ≠ 1. Let j ∉ B. Then we choose b ∈ B\Z(O_{2}(G_{4})) and a ∈ O_{2}(G_{4}) with b^{a} = bj. Hence |V : I C_{V} (⟨b, x⟩)| ≤ 2^{7} ≥ |V : I C_{V} (⟨bj, x⟩)| and so |V : I C_{V} (⟨j, x⟩)| ≤ 2^{8}. As i ∈ ⟨j, x⟩, this contradicts (i). So we have j ∈ B. This means that for u ∈ Z(O_{2}(G_{1})) in any case ⟨i⟩u ∩ A just contains one involution conjugate to j or the intersection is empty. Now by (iv) A ⊆ ⋂_{a∈A^{♯}} O_{2}( I C_{G}(a)). But this

implies that A corresponds to a totally isotropic subspace in the symplectic six-space Z(O_{2}(G_{1}))/⟨i⟩ and so |A| ≤ 8. □

Next we show that G = F_{4}(2) does not possess an F_{1}-module. In fact P. McClurg [McCl] has shown that there is just one candidate namely the module V (λ_{1}). For proving that V (λ_{1}) is not an F_{1}-module for F_{4}(2) he embeds F_{4}(2) in E_{6}(2) and quotes that V (λ_{1}) for E_{6}(2) is not an F_{1}-module, unfortunately the latter is. So we give a proof using the results of [McCl] concerning the codimensions of centralizers of involutions which reduces to consider V (λ_{1}). An independent proof can be found in a forthcoming paper due to U. Meierfrankenfeld and G. Stroth where F_{1}-modules for all Lie type groups are determined.

(1.14) Lemma. G = F_{4}(2) does not possess an F_{1}-module.

Proof. As has been said before it is enough to treat the case V = V (λ_{1}). Let A be an offending subgroup in G, i.e. 1 ≠ A is elementary abelian and |V : I C_{V} (A)| ≤ 2|A|. By [GoLySo; Theorem 3.3.3] we have |A| ≤ 2^{11} and so |V : I C_{V} (A)| ≤ 2^{12}. We adopt the notation from (1.13).

Any involution in O_{2}(G_{1}) is conjugate to some involution in G_{4} \ O_{2}(G_{4}), hence we see with (1.13)(b)(i) that |V : I C_{V} (t)| ≥ 2^{6} for any involution t ∈ G.

This shows |A| ≥ 2^{5}.

(1) A ∩ Z(O_{2}(G_{4})) ≠ 1.

Suppose that A ∩ Z(O_{2}(G_{4})) = 1. Then |A ∩ O_{2}(G_{4})| ≤ 2^{4} and so |A : A ∩ O_{2}(G_{4})| ≥ 2. As [A, I C_{V} (O_{2}(G_{4}))] ≠ 1 and so |V : I C_{V} (A)| ≥ 2^{7}, we see |A : O_{2}(G_{4}) ∩ A| ≥ 4. Now by (1.13)(b)(i) and (1.7) we get

2|A| ≥ |V : I C_{V} (A)| ≥ 4 · 4 · 4 · |A : O_{2}(G_{4}) ∩ A|.

Hence |A ∩ O_{2}(G_{4})| ≥ 2^{5}, a contradiction.

(2) If i ∈ A, then A ⊆ O_{2}(G_{1}).

Let x ∈ A\O_{2}(G_{1}). By (1.13)(b)(i) we have that I C_{V} (i) involves the spin module W_{1} and the natural module W_{2}. If |W_{2} : I C_{W_{2}}(x)| = 2, then |W_{1} : I C_{W_{1}}(x)| = 16. If |W_{2} : I C_{W_{2}}(x)| > 2, so |W_{1} : I C_{W_{1}}(x)| ≥ 4. Thus, if i ∈ A then by (1.13)(b)(i)

|V : I C_{V} (A)| ≥ |V : I C_{V} (i)| · | I C_{V} (i) : I C_{V} (⟨i, x⟩)| ≥ 2^{9} · 2^{4} = 2^{13}

which is impossible.

(3) If j ∈ A, then A ≤ O_{2}(G_{4}).

Suppose A ⊈ O_{2}(G_{4}). If |A ∩ Z(O_{2}(G_{4}))| ≥ 2^{5} then there are elements i_{1}, i_{2} ∈ A ∩ Z(O_{2}(G_{4})), such that i_{1} ∼ i_{2} ∼ i and i_{2} ∉ O_{2}( I C_{G}(i_{1})). This contradicts (2). So we have |A ∩ O_{2}(G_{4})| ≤ 2^{8}. Now

|V : I C_{V} (A)| = |V : I C_{V} (j)|| I C_{V} (j) : I C _{I C_{V} (j)}(A)|

and I C_{V} (j) involves a natural module and V (λ_{2}). If A induces a transvection on the natural module then the codimension of the centralizer in V (λ_{2}) is four. So in any case we get | I C_{V} (j) : I C _{I C_{V} (j)}(A)| ≥ 16|A/A ∩ O_{2}(G_{4})| with (1.7). So |A ∩ O_{2}(G_{4})| ≥ 2^{9}, a contradiction.

Now suppose that we have j ∈ A. Then A ≤ O_{2}(G_{4}) by (3) and |A| ≥ 2^{5} . If A ∩ Z(O_{2}(G_{4})) = ⟨j⟩, then |A| = 2^{5} and I C_{V} (j) = I C_{V} (A). But we have ⟨A^{G_{4}}⟩=O_{2}(G_{4}) and so

I C_{V} (O_{2}(G_{4})) = I C_{V} (⟨A^{G_{4}}⟩) = ⟨ I C_{V} (A)^{G_{4}}⟩ = ⟨ I C_{V} (j)^{G_{4}}⟩ = I C_{V} (j)

in contradiction to I C_{V} (O_{2}(G_{4})) = [V, j].

So A ∩ Z(O_{2}(G_{4})) ≠ ⟨j⟩ and by conjugation in G_{4} we may assume that i ∈ A. Now (1.13)(b)(i) implies that |V : I C_{V} (⟨i, j⟩)| ≥ 2^{10}, |A| ≥ 2^{9}, and |A ∩ Z(O_{2}(G_{2}))| ≥ 2^{5}. Again there are i_{1}, i_{2} ∈ A with i_{1} ∼ i ∼ i_{2} but i_{1} ∉ O_{2}( I C_{G_{2}}(i_{2})), a contradiction to (2).

Hence A does not contain any conjugate of j. Suppose i ∈ A. Then by (2) A ⊆ O_{2}(G_{1}) and by (1.13)(b) |A| ≥ 2^{8}. So A ∩ Z(O_{2}(G_{1})) ≠ ⟨i⟩ and A contains a conjugate of j, a contradiction.

Therefore, A does not contain any conjugate of i or j and A is not an elementary abelian subgroup of G of maximal rank, i.e. |A| ≤ 2^{10}. Suppose ij ∈ A. There is H ≤ G, H ≅ Sp_{8}(2) such that Z(O_{2}(G_{1})) ⊆ H and (H ∩G_{1})O_{2}(G_{1}) = G_{1}. Choose x ∈ H an involution of type c_{2} on the natural module of H. Then x is conjugate to some element of type c_{2} in G_{1}/O_{1}(G_{1}) and so |W : I C_{W} (x)| = 2^{4} for the spin module W of G_{1}/O_{2}(G_{1}) ≅ Sp_{6}(2). So by (1.13)(b)(i) we get |V : I C_{V} (x)| ≥ 2^{10}. We are going to prove that ij is of type c_{2}. As |[V, j]| = 2^{6} and as any involution in Z(O_{2}(G_{1}))\⟨i⟩ is either of type c_{2} or a_{2}, we get that j is of type a_{2} and then ij is of type c_{2}. So |V : I C_{V} (ij)| ≥ 2^{10}. Hence, |A| ≥ 2^{9} and | I C_{V} (ij) : I C_{V} (A)| ≤ 2. Since I C_{V} (O_{2}(G_{4})) ⊆ I C_{V} (ij) and I C_{V} (O_{2}(G_{4})) is the natural module for G_{4}/O_{2}(G_{4}), we get |A : A ∩ O_{2}(G_{4})| ≤ 2. But now |A ∩ O_{2}(G_{4})| ≥ 2^{8} and |A ∩ Z(O_{2}(G_{4}))| ≥ 2^{4}. Then we may assume that A ∩ Z(O_{2}(G_{4})) contains some x ∈ O_{2}(G_{1})\Z(O_{2}(G_{1})). So xij ∼ xj in G_{1}. Hence x ≁ xij in G_{4} and so x or xij is conjugate to i in G_{4}. This proves the lemma. □

(1.15) Lemma. Let G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ be a semiclassical parabolic system

with diagram ∘ ∘ ∘_{1}  _{2}  _{3}  _{4}∘^{∼}  . Let V be a non-trivial GF (2)G–module.

(a) If for some involution t ∈ O_{2}(G_{1}) we have |V : I C_{V} (t)| ≤ 2^{12}, then [O_{3}(G_{1}), V] = 1.

(b) The module V is not an F_{1}–module.

Proof. We may assume V to be irreducible. Let A be an offending subgroup for an F_{1}–module V. By [Wie1] and [GoLySo; Theorem 3.3.3] we get that |A| ≤ 2^{11}, hence |V : I C_{V} (a)| ≤ 2^{12} for any a ∈ A^{♯}. So |V : I C_{V} (a)| ≤ 2^{12} for some involution a in both (a) and (b). By (1.12)(b) we may generate G with 9 conjugates of a and so |V | ≤ 2^{108}.

(∗) If [O_{3}(G_{1}), V] ≠ 1 ≠ [O_{3}(G_{4}), V] then |V : I C_{V} (a)| ≥ 2^{16} for any a ∈ O_{2}(G_{1}).

Set V_{4} = [V, O_{3}(G_{4})]. Then V_{4} = ⨁_{E∈H} I C_{V_{4}}(E), H the set of hyperplanes of O_{3}(G_{4}). Hence if I C_{V_{4}}(E) ≠ 1, we have at most 54 conjugates of E and so U = N(G_{4}(E)/O_{2,3}(G_{4})) must be isomorphic to a subgroup of index at most 54 in Sp_{6}(2). This gives that U ≅ O_{6}^{±}(2), so |E^{Sp_{6}(2)}| = 28 or 36. As 2^{3}L_{3}(2) is not contained in GL_{6}(3) we get that A_{8} ≅ Ω_{6}^{+}(2) cannot act on a hyperplane. Hence |E^{Sp_{6}(2)}| = 28 and so |V_{4}| = 2^{56}. As a is conjugate to some b ∈ O_{2}(G_{4} ∩ G_{1})\O_{2}(G_{1}) in G we see with (1.11) that |[V_{4}, b]| ≥ 2^{16} and so |[V, a]| ≥ 2^{16}.

As O_{3}(G_{1}) ∩ O_{3}(G_{4}) ≠ 1 and O_{3}(G_{i}) is an irreducible G_{i}–module, either both O_{3}(G_{i}), i = 1, 4, act trivially or non-trivially on V. Hence (∗) proves (a).

For (b) suppose first [O_{3}(G_{1}), V] = 1. Then G/ I C_{G}(V) is an F_{4}–amalgam. By [Str3; (2.1)] G/ I C_{G}(V) ≅ F_{4}(2) But by (1.14) F_{4}(2) does not possess an F_{1}–module.

Now [O_{3}(G_{1}), V] ≠ 1. Then by (∗) A ∩ O_{2}(G_{1}) = 1. This shows |A| ≤ 2^{6}. Now choose b ∈ A^{#}. Then [b, Z(O_{2}(G_{1}))] ≠ 1, so there is an x ∈ Z(O_{2}(G_{1})), 1 ≠ x, such that xb ∼ b in G. We get

|V : I C_{V} (b)| = |V : I C_{V} (bx)| ≤ |V : I C_{V} (A)| ≤ 2^{7}.

This shows |V : I C_{V} (x)| ≤ 2^{14}, a contradiction to (∗). □

(1.16) Lemma. Let G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ be a semiclassical parabolic system with diagram ∘ ∘ ∘ ∘^{(∼)}^{∼}_{1} _{2} _{3} _{4} and V be a nontrivial GF (2)G- module. Then we have

(i) W = O_{2}(G_{4}^{(∞)}) does not act quadratically on V.

(ii) For ⟨i⟩ = Z(S), S ∈ Syl_{2}(P_{1}) we have

(α) There is no fours group ⟨i, x⟩ ⊆ S acting quadratically on V.

(β) [[V, i], G_{1}^{(∞)}] ≠ 1.
(γ) |[V, i]| > 2^{7}.

Proof. Let U = S_{G_{1}}. By [FuStr] U is an extraspecial group of order 2^{13} and |W ∩ U| = 2^{7}.

Suppose that W acts quadratically on V. Then also W ∩ U acts quadratically. We have some element g ∈ G_{1} with (W ∩ U)^{g}(W ∩ U) = U. Hence |(W ∩ U)^{g}W/W | = 2^{6}, so M_{24} possesses a quadratic subgroup of order 2^{6} on some module. But this contradicts [MeiStr]. This is (i).

Suppose (ii)(α) is false. Let first x ∉ U. Then [x, U] ⊈ ⟨i⟩. Hence by [Wie4] x does not induce a transvection on U/⟨i⟩. So there is a g ∈ U with x^{g} = xy ≠ x, y^{2} = 1, and y ∉ ⟨i⟩. In particular, ⟨i, xy⟩ acts quadratically and so ⟨i, y⟩ acts quadratically. So without loss we may assume ⟨i, x⟩ ≤ U. Hence ⟨x^{G_{1}}⟩ = U and so [V, i, U] = 1. This shows that any fours group in U containing i acts quadratically. Set R = ⟨P_{2}, P_{3}⟩. Then R induces on (W ∩ U)/⟨i⟩ the six-dimensional 3Σ_{6}-module and on W/W ∩ U the fourdimensional Σ_{6}-module. If y ∈ W \W ∩ U, we have |U/⟨i⟩ : I C_{U/⟨i⟩}(y)| = 16. Hence all R-orbits on W \W ∩ U have length divisible by 240. The G_{4}-orbits on W^{♯} have length 1771 and 276. In particular there are representatives for both orbits in W ∩ U\⟨i⟩. So choose j ∈ W ∩ U such that I C_{G_{4}}(j) involves Aut(M_{22}). We know that ⟨i, j⟩ acts quadratically. So [V, j, ⟨i^{I C_{G_{4}}(j)}⟩] = 1. Since W/⟨j⟩ is an irreducible I C_{G_{4}}(j)-module we get ⟨j, i^{I C_{G_{4}}(j)}⟩ = W. So [V, j, W] = 1. But now [V, ⟨j^{G_{4}}⟩, W] = 1. As ⟨j^{G_{4}}⟩ = W we get [V, W, W] = 1 contradicting (i). So (α) is proved.

Now (β) and (γ) follow from the fact that [V, i, U] ≠ 1 by (α) and that G_{1} acts irreducibly on U/⟨i⟩. This is (ii). □

(1.17) Lemma. Assume (1.16). Then V involves more than one nontrivial irreducible G_{4}-factor.

Proof. Set V_{1} = I C_{V} (W) and H_{4} = G′_{4}. If we have [V, H_{4}] ≤ V_{1}, then [V, W, W] ≤ [V, H_{4}, W] ≤ [V_{1}, W] = 1, contradicting (1.16)(i). Hence it must be [V, H_{4}] ≰ V_{1}. Assume now that V involves exactly one nontrivial irreducible H_{4}-module. Then [V_{1}, H_{4}] = 1, and as H_{4}′ = H_{4}, we see I C_{V/V_{1}}(H_{4}) = 1. So let V_{2}/V_{1} be the nontrivial irreducible H_{4}-module in V. As [W, V] ⊈ V_{1} by (1.16)(i), we see V_{2}/V_{1} = [W, V]V_{1}/V_{1}. In particular V_{2}/V_{1} ≅ W as H_{4}/W -module. As H^{1}(M_{24}, W) is one-dimensional (see [Asch2; 22.8]) we see |V : V_{2}| ≤ 4. Let i ∈ W be as in (1.16)(ii). Then I C_{H_{4}}(i)/W ≅ 2^{6} · 3Σ_{6}. Let φ ∈ I C_{H_{4}}(i), o(φ) = 3, with |[φ, W]| = 2^{6}. Then also |[φ, V]| = 2^{6} and [φ, V] ∩ V_{1} = 1. As [V_{2}, W] ≤ V_{1}, we have [[φ, V], i] ⊆ V_{1} ∩ [φ, V] = 1. This shows | I C_{V} (i) : V_{1}| ≥ 2^{6}. As |V : V_{1}| ≤ 2^{13}, we see |V : I C_{V} (i)| ≤ 2^{7}, contradicting (1.16)(ii)(γ). □

(1.18) Lemma. Assume (1.16). Then there exists no subgroup A ≠ 1 in O_{2}(G_{4}) with |V : I C_{V} (A)| ≤ 2|A|.

Proof. Suppose there is some 1 ≠ A ≤ O_{2}(G_{4}) with |V : I C_{V} (A)| ≤ 2|A|. Suppose further i ∈ A. By (1.16)(ii) (γ) we have |[V, i]| ≥ 2^{8}. Set H_{1} = G_{1}^{(∞)}. We have W ≤ H_{1}. By (1.16)(i) we know [[V, i], W] ≠ 1 and so [[V, i], H_{1}] ≠ 1. As |A| ≤ 2^{12}, we have |[V, i]| ≤ 2^{13}. If |[V, i]| = 2^{13}, then A = O_{2}(G_{4}) and I C_{V} (A) = I C_{V} (i) is normal in ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ = G, a contradiction.

We have |[V, i]| ≤ 2^{12} and so as H_{1} acts irreducibly on U/⟨i⟩, U = S_{H_{1}}, we see [[V, i], U] = 1, contradicting (1.16)(ii)(α).

The G_{4}–orbits on W^{#} have lengths 1771 and 276 and now, as A does not contain any G_{4}–conjugate of i, all elements of (A ∩ W)^{♯} are in the orbit of length 276, in particular |A ∩ W | ≤ 2^{8}. If |A ∩ W | = 2^{8}, choose g ∈ G_{4} with (A ∩ W)^{g} ≠ A ∩ W. Then |A ∩ W : A ∩ A^{g} ∩ W | ≥ 2 and so

|(A ∩ W)^{g} ∪ (A ∩ W)| ≥ 1 + 2 · 255 − 127 = 384 > 276 + 1,

a contradiction. Hence we have |A ∩ W | ≤ 2^{7} and so |A| ≤ 2^{8}. In particular |[V, j]| ≤ 2^{9} for any j ∈ A ∩ W. So we have: If A ∩ W ≠ 1, then there is some j ∈ A^{♯} such that I C_{G_{4}}(j) involves M_{22} and |[V, j]| ≤ 2^{9}. Let now H̃_{4} = I C_{G_{4}}(j).We have H̃_{4}/W ≅ Aut(M_{22}). As W/⟨j⟩ is an irreducible H̃_{4}- module and |[V, j]| < |W/⟨j⟩|, we see [W, [V, j]] = 1. As W = ⟨j^{g} | g ∈ G_{4}⟩, we get [V, W, W] = 1 contradicting (1.16)(i).

So we are left with A ∩ W = 1, i.e. |A| = 2 and |V : I C_{V} (A)| ≤ 4. Let ⟨a⟩ = A. Then there is some conjugate ab ≠ a with b ∈ W. Hence |V : I C_{V} (b)| ≤ 16. By (1.16)(ii)(γ) b ≁ i in G_{4}. So b ∼ j and we get again [V, W, W] = 1, a contradiction. □

(1.19) Lemma. Assume (1.16). Set U = S_{G_{1}}. Then for 1 ≠ A ⊆ U, A elementary abelian, we get |V : I C_{V} (A)| > |A|.

Proof. Suppose there is some A with |V : I C_{V} (A)| ≤ |A|. We have |A| ≤ 2^{7}, as A is elementary abelian. In particular |[V, t]| ≤ 2^{7} for any t ∈ A^{♯}. There is some element g ∈ G with t^{g} ∈ G_{4} \ O_{2}(G_{4}). By [Par; (3.7)] we have |[Ṽ , t^{g}]| ≥ 2^{4} for any nontrivial irreducible G_{4}/O_{2}(G_{4})-module Ṽ involved in V. This shows that V involves exactly one nontrivial irreducible G_{4}/O_{2}(G_{4})- module, contradicting (1.17). □

(1.20) Lemma. Let G = L_{5}(2) with diagram ∘ ∘ ∘ ∘_{1}  _{2}  _{3}  _{4} and mod-

ule V = V (λ_{2}). Let t ∈ G be an involution. If t is a transvection in the natural representation, then |[V, t]| = 8. Else |[V, t]| = 16. If t_{1} ≠ t_{2} are transvections in the natural representations then I C_{V} (t_{1}) ≠ I C_{V} (t_{2}).

Proof. We have that V (λ_{2}) is the exterior square of the natural module (see [GoLySo; Example 1.14.9 (e)]). So |V | = 2^{10}. Further by (1.4) we have that V_{4} = I C_{V} (O_{2}(G_{4})) is the orthogonal G_{4}/O_{2}(G_{4})–module. Hence |V : V_{4}| = 16 and so |[V, t]| ≤ 16 for t ∈ O_{2}(G_{4}). As t is centralized by some L_{3}(2) in G_{4} which itself has two nontrivial irreducible modules in V_{4}, we see |[V, t]| = 8. Let now t ∈ G not be a transvection. Then t inverts some element ω of order 5. As |[V, ω]| = 2^{8}, we get |[V, t]| ≥ 16. We may assume ω, t ∈ G_{4} \ O_{2}(G_{4}). Then |[V_{4}, t]| = 4 and |[V/V_{4}, t]| = 4. Hence | I C_{V} (t)| ≥ 64. This shows |[V, t]| = 16.

Suppose I C_{V} (t_{1}) = I C_{V} (t_{2}). We may assume t_{1} ∈ Z(G_{1} ∩ G_{4}). Then X = ⟨ I C_{V} (t_{1}), I C_{V} (t_{2})⟩ ≅ G_{1}, G_{4}, or G as it has to be a parabolic subgroup. So X acts on I C_{V} (t_{1}) and V/ I C_{V} (t_{1}). Since the latter is of order 8 this is a contradiction. □

For the convenience of the reader we collect some well known basic results on parabolic systems.

(1.21) Lemma. Let G = ⟨G_{1}, G_{2}⟩, K_{i}≤ G_{i}, i = 1, 2, and K_{1} ⊆ G_{2}, K_{2} ⊆ G_{1}. Suppose (G_{1} ∩ G_{2})_{G} = 1. Let p be a prime number such that p |̸ |K_{1}K_{2}/K_{1}| and p |̸ |K_{1}K_{2}/K_{2}|, then p |̸ |K_{1}K_{2}|.

Proof. [Str3; (1.14)] □

(1.22) Lemma. Let {P_{1}, …, P_{n}} be a parabolic system with B/B_{P_{i}} is a 2-group for all i. Then B is a 2-group.

Proof. [Str3; (1.15)] □

The next lemma is basically [StrWo; (2.1),(2.2)]. But as the assumptions differ a little bit see [Wie1; 1.4] for a proof.

(1.23) Lemma Let G = ⟨G_{1}, G_{2}⟩. Set B = G_{1} ∩ G_{2} and K_{i} = B_{G_{i}} for i=1,2. Assume B_{G_{1}} is a 2-group and the following holds.

(i) There is S ⊆ G_{1}∩G_{2}, S ∈ Syl_{2}(G_{2}) and set Z_{i} = ⟨Ω_{1}(Z(S))^{G_{i}}⟩, i = 1, 2, then Z_{1} ⊆ K_{1} and moreover any 2-element in G_{1} ∩ G_{2} centralizing Z_{1} is in K_{1}.

(ii) |G_{2} : G_{1} ∩ G_{2}| = 3, G_{2} is finite and Z_{2} ≤ O_{2}(G_{2}) = S_{G_{2}}.

(iii) B_{G} = 1.

Then Z_{1} is an F_{1}-module for G_{1} with offending subgroup in S_{G_{2}}K_{1}/K_{1}, or for φ ∈ ⟨K_{1}^{G_{2}}⟩, φ^{3} ∈ G_{1} ∩G_{2}, we have |[φ, O_{2}(G_{2})]| = 4 and |[O_{2}(⟨S, φ⟩), φ]| = 4.

(1.24) Lemma ([Wie1; 1.5]). Let {P_{1}, …, P_{n}} be a semiclassical parabolic system. Assume (1.23) with G_{2} = P_{1} and G_{1} = ⟨P_{2}, …, P_{n}⟩. Let further ⟨P_{1}, P_{2}⟩/B_{⟨P_{1},P_{2}⟩} be a rank two Lie group or 3Σ_{6}, 3A_{6} and B_{⟨P_{1},P_{2}⟩} contain a Sylow 2-subgroup of I C_{⟨P_{1},P_{2}⟩}(B_{⟨P_{1},P_{2}⟩}). Then Z_{1} is an F_{1}–module with offending subgroup in S_{G_{2}}.

§2 Extension of ∘ ∘ ∘ ∘^{∼} .

In this chapter we will assume that {P_{1}, …, P_{n}} is a semiclassical parabolic system involving an F̃_{4}-subdiagram. The aim is to show that n = 4. Hence it is enough to show that there is no system {P_{1}, P_{2}, P_{3}, P_{4}, P_{5}} that involves an F̃_{4}-subdiagram. We choose notation such that P_{2}, P_{3}, P_{4}, and P_{5} generate the F̃_{4}-group. We recall that G_{1,2}/K_{1} ≅ G_{1,5}/K_{1} ≅ 2^{1+6+8}3^{7}Sp_{6}(2) and G_{1,3}/K_{1,3} ≅ G_{1,4}/K_{1,4} ≅ L_{3}(2) × Σ_{3}.

(2.1) Lemma. There is no system with diagram Δ of shape

∘ ∘ ∘_{1}  _{2}  _{3}  _{4}  _{5} ∘ ∘^{∼}  .

Proof. By [ShStr] we have G_{5}/K_{5} ≅ 3^{35}Sp_{8}(2). By (1.22) and (1.2) we have that B is a 2-group and so K_{i} is a 2-group for all i. Assume first I C_{S}(K_{i}) ⊆ K_{i} for i = 1 and 5. By (1.15) G_{1}/K_{1} does not possess an F_{1}-module. Application of (1.24) shows Ω_{1}(Z(S)) = Z_{1}.

We have G_{2}/K_{2} ≅ Σ_{3} × 3^{7}Sp_{6}(2) and |G_{2} : G_{2} ∩ G_{1}| = 3. Now we investigate the coset graph Γ(G_{1}, G_{2}). Let (2, α) be a critical pair. Assume α ∼ 2. According to [StrWo; (1.8)] we may choose notation such that Z_{2} is an F -module with offending subgroup Z_{α}/ I C_{Z_{α}}(Z_{2}). Since Z(G_{2}) = 1 we get |Z_{2}| = 4. Set V_{1} = ⟨Z_{2}^{G_{1}}⟩. As [Z_{2}, K_{1}] ⊆ Z_{1} we have [V_{1}, K_{1}] ⊆ Z_{1}.

Assume first V′_{1} = 1. Let β ∈ Δ(2) with d(β, α) = d(2, α) − 1. Then V_{β} ⊆ G_{α} but V_{β} ⊈ K_{α}. Hence |V_{β} : V_{β} ∩ G_{α+1}| = 2. As Z_{β} = [Z_{2}, Z_{α}] we have [V_{α+1}, Z_{β}] = 1 and so V_{α+1} ⊆ G_{β}. Now [V_{β} ∩ K_{α}, V_{α+1}] ⊆ V_{β}. As [Z_{2}, Z_{α+1}] ≠ 1 and [V_{β} ∩ K_{α+1}, V_{α+1}] ⊆ Z_{α+1}, we get [V_{β} ∩ K_{α+1}, V_{α+1}] = 1. Hence for t ∈ V_{α+1}, we get |[V_{β}, t]| ≤ 2^{12}. Suppose V_{α+1} ⊆ K_{β}. Then, as Z_{β} ⊆ V_{α+1} we get [V_{β}, V_{α+1}] ⊆ V_{α+1} and so V_{α+1} < ⟨V_{β}, G_{α} ∩ G_{α+1}⟩ = G_{α}, a contradiction.

So we have V_{α+1} ⊈ K_{β}. Now the application of (1.15)(a) shows that G_{β}/K_{β} I C_{G_{β}}(V_{β}) is an F_{4}(2)-amalgam and so by [Str3; (2.1)] is isomorphic to F_{4}(2). By [Coop] we have |[V_{β}, t]| ≥ 2^{6} and so |V_{β} ∩ G_{α+1} : V_{β} ∩ K_{α+1}| ≥ 2^{5}. If there are two nontrivial irreducible modules involved, then |[V_{β}, t]| = 2^{12} and I C_{V_{β}}(t) = I C_{V_{β}}(V_{α+1}) for all t ∈ V_{α+1}\K_{β}. But this would imply that one of them is an F_{1}-module, which contradicts (1.14). So we have that there is just one nontrivial irreducible module involved. Now as V_{β} ∩ G_{α+1} acts quadratically on V_{α+1} we get with (1.13)(a) that V_{α+1} and so also V_{β} just involves a 26-dimensional and trivial F_{4}(2)-modules. Let V be the 26dimensional module. As V_{1} = ⟨Z_{2}^{G_{1}}⟩ and |Z_{2}| = 4, we see that

V_{β}/ I C_{V_{β}}(G_{β}/K_{β}) ≅ V ≅ V_{α+1}/ I C_{V_{α+1}}(G_{α+1}/K_{α+1}).

Now I C_{V} (G_{α+1} ∩ G_{α}/K_{α+1}) ≠ 1 and so this module is one-dimensional. Hence for any i ∈ Z(G_{α+1} ∩ G_{α}/K_{α+1}) with i ≠ 1 we have by (1.13)(b) [[V, i], O_{2}(G_{α+1} ∩ G_{α}/K_{α+1})] ≠ 1. As V_{β} ∩ G_{α+1} acts quadratically we see with (1.13)(b)(ii),(iii) that |V_{β} ∩ G_{α+1} : V_{β} ∩ K_{α+1}| ≤ 2^{7}. Suppose equality. Then by (1.13)(b)(iv) (V_{β} ∩G_{α+1})K_{α+1}/K_{α+1} ∩Z(O_{2}(G_{α+1} ∩G_{α}/K_{α+1})) does not cover Z(O_{2}(G_{α+1} ∩ G_{α}/K_{α+1})) modulo i which is impossible by (1.13)(b)(ii),(iii). Hence |V_{β} ∩ G_{α+1} : V_{β} ∩ K_{α+1}| ≤ 2^{6}. Now as V is not an F -module, we have |V_{α+1} : V_{α+1} ∩ K_{β}| ≤ 2^{6}, too. Hence by (1.13)(b)(i) i ∉ (V_{β} ∩ G_{α+1})K_{α+1}/K_{α+1}. Set

U = (V_{β} ∩ G_{α+1})K_{α+1}/K_{α+1} ∩ Z(O_{2}(G_{α+1} ∩ G_{α}/K_{α+1})).

Suppose that |V_{β} ∩ G_{α+1} : V_{β} ∩ K_{α+1}| = 2^{6}. Then for a suitable s we get (V_{β} ∩G_{α+1})K_{α+1}/K_{α+1} = U⟨s⟩. Now there is some s_{1} ∈ O_{2}(G_{α+1}∩G_{α}/K_{α+1}) with s^{s_{1}} = si. So also ⟨i⟩U acts quadratically on V. As any coset of ⟨i⟩ in ⟨i⟩U contains some conjugate of the element j in the notation of (1.13)(b), we see that there is some conjugate j^{g} such that ⟨i⟩U ⊈ O_{2}( I C_{G_{α+1}/K_{α+1}}(j^{g})), contradicting (1.13)(b)(iv).

So we have |V_{β} ∩ G_{α+1} : V_{β} ∩ K_{α+1}| = 2^{5}. Now |V_{α+1} : V_{α+1} ∩ K_{β}| ≤ 2^{5}. By (1.13)(b)(iii) we have |U| ≥ 16 but by (1.13)(b)(v) we would get |U| ≤ 8, a contradiction.

So let V′_{1} ≠ 1. Then b = 2 and V_{1} = Z(V_{1})X where X is extraspecial of width n for some n. Obviously n ≥ 13. Let g ∈ G_{2} with Z_{1} ≠ Z_{1}^{g} ≤ Z_{2} ≤ V_{1}. Then |X^{g} : X^{g} ∩ K_{2}| = 2. As [K′_{1}, V_{1}] = 1 we get Z_{1}^{g} ⊈ K′_{1}. In particular X^{g} ∩ K_{1} is elementary abelian and so |X^{g} ∩ G_{1} : X^{g} ∩ K_{1}| ≥ 2^{12}. On the other hand Z_{1}^{g} ⊆ Z_{2} ⊆ K_{1} implies (X^{g} ∩ G_{1}/X^{g} ∩ K_{1})′ = 1, a contradiction as O_{2}(G_{1} ∩ G_{2}/K_{1}) does not contain elementary abelian subgroups of order 2^{12}.

So we are left with α ∼ 1. Then there is a fours group Z̃_{2} in Z_{2} with Z̃_{2} < G_{2} and Z̃_{2} ⊈ K_{α}. Let first b > 1. Set again V_{1} = ⟨Z̃_{2}^{G_{1}}⟩. Let β ∈ Δ(2) with d(β, α) = b − 1. Then [V_{β}, V_{α}] ⊆ V_{β} ∩ V_{α} and again |V_{α} : I C_{V_{α}}(t)| ≤ 2^{12} for t ∈ V_{β}. As above G_{1}/K_{1} I C_{G_{1}}(V_{1}) ≅ F_{4}(2). Further V_{1} involves just one nontrivial irreducible F_{4}(2) - module, the 26-dimensional one. As V_{β} acts quadratically on V_{α}, we again see with (1.13)(b) that |V_{β} : V_{β} ∩ K_{α}| ≤ 2^{6} and a central involution i in G_{α−1} ∩ G_{α}/K_{α} is not in V_{β}. By symmetry we also have |V_{α} : V_{α} ∩ K_{β}| ≤ 2^{6}, and therefore, again by (1.13)(b)(v), we can conclude that |V_{β} : V_{β} ∩ K_{α}| = 2^{6} = |V_{α} : V_{α} ∩ K_{β}| and V_{β}K_{α}/K_{α} ⊈ Z(O_{2}(G_{α−1} ∩ G_{α}/K_{α})). But then we may argue as above to get a quadratically acting subgroup in Z(O_{2}(G_{α−1} ∩ G_{α}/K_{α})) of order at least 2^{6}, which again contradicts (1.13)(b)(iv).

Finally let b = 1. Then Z_{2} ⊈ K_{1} and Z̃_{2} induces a transvection on K_{1}. By (1.12)(b) G_{1}/K_{1} centralizes a subgroup of index 2^{9} in K_{1}, contradicting I C_{S}(K_{1}) ⊆ K_{1}.

So assume now I C_{S}(K_{1}) ⊈ K_{1}, hence G_{1} = K_{1} I C_{G_{1}}(K_{1}). As K_{1}K_{5}/K_{5} is in the center of G_{1} ∩ G_{5}/K_{5}, we see |K_{1}K_{5}/K_{5}| ≤ 2. Now from G_{5}/K_{5} we see G_{1} ∩ G_{5}/K_{1}K_{5} ≅ 3^{7}2^{7}Sp_{6}(2) or 3^{7}2^{6}Sp_{6}(2), but there are no such factors of G_{1} ∩ G_{5}/K_{1}. So we have I C_{S}(K_{1}) ⊆ K_{1}. As G_{1} ∩ G_{5}/K_{1} is 2constrained we get from I C_{S}(K_{5}) ⊈ K_{5}, that K_{5} ⊆ K_{1}. But |S : K_{5}| = 2^{16}, while |S : K_{1}| = 2^{24}. □

(2.2) Lemma. There is no semiclassical parabolic system with diagram Δ of type

[figure: ∘; ∘; ∘; ∘; ∘; 5; 1; 4; 3; 2; ∼] .

Proof. By (1.2) we see that P_{i}/B_{P_{i}} ≅ Σ_{3} for all i. So by (1.22) B is a 2-group.

Assume first I C_{S}(K_{i}) ⊆ K_{i} for all i. Let Z_{1} ≤ Z(G_{1}). Then by (1.24) Z_{3} and Z_{4} both are F_{1}-modules. As G_{1} ∩G_{i} is the centralizer of Ω_{1}(Z(S)) in G_{i}, for i = 3, 4, we see with (1.5) that 10-dimensional L_{5}(2)-modules V (λ_{2}) or V (λ_{3}) are involved. Since by (1.20) on this module V we have |[V, t]| = 2^{3} for transvections in L_{5}(2) and |[V, t]| = 2^{4} else, we see that just one such module is involved. As Z_{3} is generated by conjugates of Ω_{1}(Z(S)), we see with a result of Gaschütz [Hu; I 17.4] that |Z_{3}| = 2^{10} = |Z_{4}|. Hence |Z_{3} ∩ Z_{4}| = 2^{6}.

Let Γ = Γ(G_{3}, G_{4}) be the coset graph and b = b_{Γ}. Let (α, β) be a critical pair in Γ. Choose δ ∈ Δ(α) and γ ∈ Δ(β) with d(γ, α) = b − 1. By (1.20) we have |Z_{α}K_{β}/K_{β}| = 2^{4} = |Z_{β}K_{α}/K_{α}| and so Z_{α}K_{β}/K_{β} = O_{2}(G_{γ} ∩ G_{β}/K_{β}). In particular Z_{α} ∩ Z_{β} ⊇ [Z_{α}, Z_{β}] = I C_{Z_{β}}(O_{2}(G_{γ} ∩ G_{β})) = Z_{β} ∩ Z_{γ}.

Assume b > 1. Then [Z_{α}, Z_{δ}] = 1. Thus [Z_{δ}, Z_{β} ∩ Z_{γ}] ⊆ [Z_{δ}, Z_{α}] = 1. This means Z_{δ} ⊆ I C_{G_{γ}}(Z_{β} ∩ Z_{γ}) ⊆ G_{γ} ∩ G_{β} ⊆ G_{β}. Now Z_{δ}Z_{α}K_{β} = Z_{α}K_{β} and so [Z_{δ}, Z_{β}] ⊆ [Z_{α}, Z_{β}] ⊆ Z_{α}. Hence Z_{α}Z_{δ} < ⟨G_{α} ∩G_{δ}, Z_{β}⟩. Choose δ with Z_{β} ⊈ G_{δ}, then we have Z_{α}Z_{δ} < G_{α}. But then Z_{α} ∩ Z_{δ} = [Q_{α}, Z_{α}Z_{δ}] < G_{α}, a contradiction.

So assume now b = 1. Then Z_{3} ≰ K_{4} and as O_{2}(G_{3}∩G_{4}/K_{4}) is elementary abelian, we have Z_{3} ⊈ Φ(K_{3}). As Z_{3} is an irreducible G_{3}-module, we have Z_{3} ∩ Φ(K_{3}) = 1. Then K_{3} is elementary abelian. The same argument also shows that K_{4} is elementary abelian. Now [K_{4}, K_{3}] = [Z_{4}, Z_{3}] = Z_{4} ∩ Z_{3}. In particular [G_{3}, K_{3}] = Z_{3}. Let L be the L_{3}(2) involved in G_{3} ∩ G_{1}. Then from G_{3}, we see that L induces at most five 3-dimensional irreducible modules in O_{2}(G_{1} ∩ G_{3}), while from G_{1}/K_{1} in which G_{1} ∩ G_{3}/K_{1} corresponds to a parabolic subgroup in F_{4}(2), we see that there are at least six such Lmodules involved, a contradiction.

So assume next Z_{1} ≰ Z(G_{1}). Application of (1.24) shows that Z_{1} is an F_{1}-module contradicting (1.15)(b).

So we are left with I C_{S}(K_{i}) ⊈ K_{i} for some i. If i = 3, then because of Z(G_{3} ∩ G_{4}/K_{4}) = 1, we see K_{3} = K_{4} = 1, contradicting |S| > 2^{10}. Hence I C_{S}(K_{3}) ⊆ K_{3} and I C_{S}(K_{4}) ⊆ K_{4}. Let I C_{S}(K_{2}) ⊈ K_{2}. Then |K_{2}K_{1}/K_{1}| ≤ 2 and we get a contradiction to |S : K_{2}| = 2^{16} and |S : K_{1}| = 2^{24}. So also I C_{S}(K_{2}) ⊆ K_{2} and I C_{S}(K_{5}) ⊆ K_{5}. It remains I C_{S}(K_{1}) ⊈ K_{1}. Then again |K_{1}K_{2}/K_{2}| ≤ 2. But G_{1} ∩ G_{2}/K_{1} does not possess a factor group isomorphic to 2^{6}Sp_{6}(2). □

(2.3) Lemma. There is no semiclassical parabolic system with diagram Δ of shape

[figure: ∘; ∘; ∘; ∘; ∘; ∼; 2; 1; 3; 4; 5; .]

Proof. First of all by (1.22) and (1.2) B is a 2-group. Let I C_{S}(K_{i}) ⊆ K_{i} for i = 1, 2. Then by (1.24) we may assume that Z_{1} is an F_{1}-module, a contradiction to (1.15)(b).

Hence we may assume I C_{S}(K_{1}) ⊈ K_{1}. Now |K_{1}K_{2}/K_{2}|=|K_{1}K_{2}/K_{1}| ≤2 and we get 2′-elements in G_{1}∩G_{2} centralizing K_{1} and so also K_{2}. This implies that I C_{S}(K_{2}) ⊈ K_{2}, too. In particular K_{1} ∩K_{2} = 1 and |K_{1}| = |K_{2}| ≤ 2. Let X = G_{4,5}. Then by (1.3)(v) we have K_{5} = O_{2}(G_{1} ∩ G_{5}) and X/K_{5} ≅ 2^{6}A_{7}. Further the action of G_{1,4,5}/K_{1} on K_{5}/K_{1} shows 2^{5} ≥ |Z(O_{2}(X))| ≥ 2^{4} and Z(O_{2}(X)) ⊆ Z(K_{5}). Now we have ⟨Z(O_{2}(X))^{G_{1,5}}⟩ ⊆ Z(K_{5}). Hence |Z(K_{5})| = 2^{3}|Z(O_{2}(X))|. But X has to act nontrivially on Z(K_{5})/Z(O_{2}(X)), a contradiction. □

(2.4) Proposition. If {P_{1}, …, P_{n}} is a semiclassical parabolic system with connected diagram Δ of rank at least five, then there is no subdiagram of type

∘ ∘ ∘_{2}  _{3}  _{4}  _{5}∘^{∼}  .

Proof. Suppose false. It is enough to prove the theorem for rank (Δ) = 5. Let the node 1 be connected with 2. Then Δ_{5} is a connected diagram having a residue belonging to 3^{7}Sp_{6}(2). By (1.3) we get that Δ_{5} is of shape

∘ ∘ ∘_{1}  _{2}  _{3}  _{4}∘^{∼} .

If 1 is not connected with 5, then Δ is as in (2.1), a contradiction. So let 1 be connected with 5, then also Δ_{2} is of shape

∘ ∘ ∘_{3}  _{4}  _{5}  _{1}∘^{∼}  .

and so Δ is as in (2.2), a contradiction.

So assume that 1 is connected neither to 2 nor to 5. Then we may assume that 1 is just connected to 3. Again by (1.3) Δ_{2} is of type

∘ ∘ ∘_{1}  _{3}  _{4}  _{5}∘^{∼}  .

and so Δ is as in (2.3), a contradiction. □

§3 Diagrams containing ∘ ∘ ∘ ^{(∼)} ∘^{∼} .

In this paragraph we will determine the possible semiclassical parabolic systems {P_{1}, P_{2}, P_{3}, P_{4}, P_{5}} of rank 5 whose diagram Δ involves a subdiagram

of the shape ∘ ∘ ∘ ^{(∼)} ∘^{∼}_{1}  _{2}  _{3}  _{4} . We have seen in (1.3) that the subdia-

gram ∘ ∘ ∘^{∼} belongs to M_{24}. Further G_{4,5}/K_{5} ≅ 2^{11}M_{24} or 2^{12}M_{24}, K_{4,5}/K_{5} is elementary abelian and is the Golay code module for G_{4,5}/K_{4,5}, K_{1,5}/K_{5} is an extraspecial 2-group of order 2^{13} and G_{1,5}/K_{1,5} is a 3U_{4}(3)- amalgam.

(3.1) Lemma.The diagram Δ is one of the following

(1) ∘ ∘ ∘  ^{(∼)} ∘ ∘_{5}  _{1}  _{2}  _{3}  _{4}^{∼}
(2) ∘ ∘ ∘^{∼}  ∘ ∘_{1}  _{2}  _{3}  _{4}  _{5}^{∼}

(3) [figure: ∘; ∘; ∘; ∘; ∘; ∼; (∼); (∼); 1; 2; 3; 5; 4] (4) [figure: ∘; ∘; ∘; ∘; ∘; ∼; ∼; (∼); (∼); 1; 2; 5; 3; 4]

(5) ∘  ∘ ∘ ∘ ∘_{1}  _{5}^{2}  ^{3}  ^{4}^{∼}  ^{∼}

Proof. We choose the numbering as follows:

∘  ∘ ∘  ^{(∼)} ∘_{1}  _{2}  _{3}  _{4}^{∼}  ,

where 5 is the fifth vertex of the diagram.

First let us assume that the diagram is a string. If 1 is connected with 5 then the residue of 4 has a diagram of rank 4 which contains an M_{24}- subdiagram. So by (1.3) the residue of 4 has the diagram

∘ ∘ ∘ ∘_{5}  _{1}  _{2}  _{3}^{∼}  ,

that is case (1).

Let 5 be connected with 4. Then the diagram Δ_{1} of the residue of 1 is a string of rank 4, containing ∘ ∘ ∘^{(∼)}^{∼} . By (1.2) we have P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for all i. And so by (1.3) we get that Δ_{1} is of the shape

∘ ∘ ∘^{∼}  ∘^{∼}  ,

and this is case (2).

So we may assume that our diagram is not a string. Let 2 be connected with 5. The residue of 4 contains an M_{24}-subdiagram and by (1.3) it has to have the diagram

[figure: ∘; ∘; ∘; ∘; ∼; ∼; 1; 2; 5; 3; .]

By (1.1) and (1.3) this forces the diagram of the residue of 1 to be a square and so of type

∘ ∘ ∘ ∘_{5}  _{4}^{2}  ^{3}^{∼}^{∼}  ^{(∼)} or  ∘ ∘ ∘ ∘_{(∼)}^{∼}  ^{∼}_{5}  _{4}^{2}  ^{3}^{(∼)} .

Because of (1.3) there are no diagrams of type ∘ ∘ ∘ ∘^{∼}^{∼}  ^{(∼)} . So we get
that Δ looks like [figure: ∘; ∘; ∘; ∘; ∘; ∼; ∼; (∼); (∼); 1; 2; 5; 3; 4] and this is case (4).

Now let 3 be connected with 5. Then again by (1.3) the residue of 4 has to have the diagram ∘ ∘ ∘ ^{(∼)} ∘^{∼} , so we have case (3).

At last let 1 and 4 be connected with 5, that is the diagram is a circle. But now by (1.3) the residues of 1 and of 2 have string diagrams and we get case (5). □

Not all of these possible diagrams do exist, as the next two lemmas will show.

(3.2) Lemma. The cases (1) and (5) in (3.1) do not occur.

Proof. In both cases we have G_{4}/K_{4} ≅ Co_{1} by (1.3). First let us assume I C_{G_{4}}(K_{4}) ⊆ K_{4}. Using [ParRow; Theorem B] we know that K_{4} is extraspecial of order 2^{25}. If we consider case (5), application of (1.3) to G_{1} and G_{2} shows G_{2}/K_{2} ≅ Co_{1} as well. Now I C_{G_{2}}(K_{2}) ⊆ K_{2}, too, and so again K_{2} is extraspecial of order 2^{25}. But then

Z(K_{2}) = Ω_{1}(Z(S)) = Z(K_{4}) < ⟨G_{2}, G_{4}⟩ = G,

a contradiction.

So let us consider case (1) and the diagram

∘ ∘ ∘  ^{(∼)} ∘ ∘_{5}  _{1}  _{2}  _{3}  _{4}^{∼}  .

We have G_{3}/K_{3} ≅ L_{4}(2) × Σ_{3}. Choose g ∈ G_{3}\G_{4} with g^{2} ∈ G_{4}. Then we have |K_{4}^{g} : K_{4}^{g} ∩ K_{3}| = 2 and K_{4}^{g} ∩ K_{4} is elementary abelian. Therefore |(K_{4}^{g} ∩ K_{3})K_{4} : K_{4}| ≥ 2^{11}. By [Asch1] M_{24} does not possess an F-module, so O_{2}(G_{4} ∩ G_{5}/K_{4}) = J(S/K_{4}) and then this is the only elementary abelian subgroup of order 2^{11} in S/K_{4}. Hence X = (K_{4}^{g} ∩ K_{3})K_{4} = O_{2}(G_{4} ∩ G_{5}). As N_{G_{3,4}}(X)K_{3,4} = G_{3,4} we get X < ⟨G_{4,5}, G_{3,4}⟩ = G_{4}, a contradiction.

Only the case G_{4} = K_{4} I C_{G_{4}}(K_{4}) is left to be considered. We know that Z(G_{4} ∩ G_{5}/K_{5}) = 1 and so K_{4} ⊆ K_{5}. Since we have |S : K_{4}| = 2^{21} and 2^{21} ≤ |S : K_{5}| ≤ 2^{22} we see that |K_{4}| ≥ |K_{5}|. So we get K_{4} = K_{5} = 1. But now O_{2}(G_{4} ∩ G_{5}) is the Todd module in G_{4} and the Golay code module in G_{5} by [FuStr; 3.1], a contradiction. □

(3.3) Lemma. Case (2) in (3.1) does not occur.

Proof. Suppose there is case (2), so we have the diagram

∘ ∘ ∘^{∼}  ∘ ∘_{1}  _{2}  _{3}  _{4}  _{5}^{∼}  .

Let first I C_{S}(K_{5}) ⊆ K_{5} and I C_{S}(K_{1}) ⊆ K_{1}. W.l.o.g. Z_{1} ≠ Ω_{1}(Z(S)). Let us consider the graph Γ = Γ(G_{1}, G_{2}). We see with (1.24) that Z_{1} is an F_{1}- module with offending subgroup contained in O_{2}(G_{1} ∩ G_{2}/K_{1}), but this fact contradicts (1.18). So now w.l.o.g. we have I C_{S}(K_{1}) ⊈ K_{1}. As S_{G_{1}∩G_{5}}/K_{5} is extraspecial we get |K_{1}K_{5}/K_{5}| ≤ 2, and (G_{1} ∩ G_{5} ∩ G_{4})^{(∞)} lies in the centralizer of the subgroup K_{1} ∩ K_{5} which has an index smaller or equal to 4 in K_{5}. But since (G_{1} ∩ G_{5} ∩ G_{4})^{(∞)} has no factorgroup isomorphic to a subgroup of Σ_{3}, it even lies in the centralizer of K_{5}. So I C_{S}(K_{5}) ⊈ K_{5}, |K_{5}K_{1}/K_{1}| ≤ 2 and hence K_{5} ∩ K_{1} = 1. In particular |K_{1}| ≤ 2 ≥ |K_{5}|. As |S_{G_{1}∩G_{5}}/K_{1}| = |S_{G_{1}∩G_{5}}/K_{5}| we have |K_{1}| = |K_{5}| ≤ 2. So O_{2}(G_{4} ∩ G_{5}) is abelian and has the order 2^{11}, 2^{12}, or 2^{13} and O_{2}(G_{1} ∩G_{2}) has the same order and is abelian as well.

There is only one abelian subgroup of order |O_{2}(G_{4}∩G_{5}/K_{5})| in S/K_{1} and S/K_{5} resp., and we get O_{2}(G_{1} ∩G_{2}) = O_{2}(G_{4} ∩G_{5})< ⟨G_{1} ∩G_{2}, G_{4} ∩G_{5}⟩ = G, but this is a contradiction. □

Now we will describe the groups that belong to the remaining possible diagrams. We change numbering a little bit.

(3.4) Lemma Let the diagram Δ be of type

[figure: (1); ∘; ∘; ∘; ∘; ∘; ∼; (∼); (∼); 1; 2; 5; 3; 4; or; (2); ∘; ∘; ∘; ∘; ∘; ∼; ∼; (∼); (∼); 1; 2; 4; 3; 5]

Then K_{3} = 1 = K_{4}, and G_{3} and G_{4} are locally M(24), M(24)′, 3M(24), or 3M(24)′.

Proof. Assume first I C_{S}(K_{3}) ⊆ K_{3} and I C_{S}(K_{4}) ⊆ K_{4}. W.l.o.g. assume Z_{3} ≠ Ω_{1}(Z(S)). Consider (2). By (1.3)(i) K_{5} = O_{2}(G_{4} ∩ G_{5}) = O_{2}(G_{5} ∩ G_{3}). In particular K_{3} ⊆ K_{5}. But by (1.3)(x) if K_{3} ⊈ K_{1} then K_{5} ⊆ K_{1}. So in any case K_{3} ⊆ K_{1}. By (1.19) Z_{3} is not an F -module with offending subgroup in K_{1}/K_{3}, so J(K_{1}) ⊆ K_{3} and we have J(K_{3}) = J(K_{1}) < ⟨G_{1}, G_{3}⟩, a contradiction.

Consider now (1). By (1.3)(xi) either K_{3} ⊆ K_{1} or K_{4} ⊆ K_{1}. So if Ω_{1}(Z(S)) ≠ Z_{4} then by symmetry we may assume K_{3} ⊆ K_{1} and get a contradiction as above.

So let Ω_{1}(Z(S)) = Z_{4}, then Ω_{1}(Z(S)) ≠ Z_{1}. Again by (1.19) we have J(K_{1}K_{3}) ⊆ K_{3} and so J(K_{3}) = J(K_{1}K_{3}). But G_{14}/K_{1} is a (3)U_{4}(3)- amalgam and by [Wie4] this amalgam does not have an F -module. So J(S) ⊆ K_{1}. In particular J(K_{1}K_{3}) ⊆ K_{1} and so J(K_{1}) = J(K_{3}). So in any case J(K_{1}) = J(K_{3}) < G, a contradiction.

Now let I C_{S}(K_{3}) ⊈ K_{3}. Then K_{3} ⊆ K_{4} because Z(G_{3} ∩ G_{4}/K_{4}) = 1. We know |K_{4} : K_{3}| ≤ 2, so I C_{S}(K_{4}) ⊈ K_{4} and K_{4} = K_{3} = 1. □

We have proved the following

(3.5) Proposition. Let {P_{1}, …, P_{5}} be a semiclassical parabolic system with connected diagram Δ which involves a subdiagram ∘ ∘ ∘ ^{(∼)} ∘^{∼} . Then

[figure: Δ =; ∘; ∘; ∘; ∘; ∘; ∼; (∼); (∼); 1; 2; 5; 3; 4; or; Δ =; ∘; ∘; ∘; ∘; ∘; ∼; ∼; (∼); (∼); 1; 2; 4; 3; 5]

and K_{3} = K_{4} = 1.

§4 Diagrams containing ∘ ∘ ∘ ∘^{∼} .

The next goal is to determine all the possible semiclassical parabolic systems {P_{1}, …, P_{k}} whose corresponding diagram Δ has a subdiagram of type ∘ ∘ ∘ ∘^{∼} . We are basically interested in k = 5. But some of the results of this paragraph also concern diagrams of rank greater than five.

(4.1) Lemma. There is no semiclassical parabolic system {P_{1}, …, P_{k}} with k ≥ 5 and diagram

∘ ∘ ∘^{∼}_{k}  [figure: ∘; ∘; …; ∘; ∘; 2; 1; 3; 4]

and G_{1}/K_{1} ≅ G_{2}/K_{2} ≅ 3^{\frac{1}{3} (2^{k−1}−1)(2^{k−2}−1)}Sp_{2k−2}(2).

Proof. Suppose false. Let at first I C_{G_{1}}(K_{1}) ⊆ K_{1} and I C_{G_{2}}(K_{2}) ⊆ K_{2}.

(1) We may assume Z_{1} = Ω_{1}(Z(S)).

Suppose Z_{1} ≠ Ω_{1}(Z(S)) ≠ Z_{2}. We may assume that Z_{1} is an F - module with an offending subgroup contained in O_{2}(G_{1} ∩ G_{2}). Then by [Wie2; 2.8] O_{3}(G_{1}/K_{1}) acts trivially on Z_{1}, so Z_{1} is the natural module by (1.7). In particular ⟨P_{3}, …, P_{k}⟩ ⊆ I C_{G}(Ω_{1}(Z(S))) by (1.9). Further we conclude P_{1} ⊈ I C_{G}(Ω_{1}(Z(S))) ⊉ P_{2}. So Z_{k} ≠ Ω_{1}(Z(S)) and Z_{k} is an F_{1}–module for G_{k}/K_{k} ≅ Ω_{2k−2}^{+}(2) by (1.24). Now by (1.7) and (1.8) we get I C_{G_{k}}(Ω_{1}(Z(S))) = ⟨P_{k−2}, …, P_{2}, P_{1}⟩, ⟨P_{k−1}, …, P_{3}, P_{2}⟩, or ⟨P_{k−1}, …, P_{3}, P_{1}⟩, a contradiction.

By symmetry we may choose notation such that Ω_{1}(Z(S)) < G_{1}.

(2) k = 5.

Assume k ≥ 6. By (1.24), (1.7), and (1.8) Z_{k} contains a halfspin module Z̃_{k} of Ω_{2k−2}^{+}(2) and k = 6. Consider the graph Γ(G_{6}, G_{5}). Let (6, α) be a critical pair. If α ∼ 6, Z_{6} has to be an F -module with offending subgroup Z_{α}K_{6}/K_{6} ≤ O_{2}(G_{5} ∩ G_{6}/K_{6}) and hence Z_{6} is the halfspin module. But this contradicts [Str3; (1.10)]. So α ∼ 5.

Let b_{Γ} > 1. Set V_{5} = ⟨Z̃_{6}^{G_{5}}⟩. We have V′_{5} = 1 because b_{Γ} ≥ 3. Now we choose notation such that d(5, α) = d(6, α) − 1. So V_{5} ⊆ G_{α} and V_{α} ⊆ G_{5}. Let α + 1 ∈ Δ(α) be such that Z_{6} ⊈ G_{α+1}. Then |Z_{α+1} : I C_{Z_{α+1}}(Z_{6})| ≤ 2^{9}, since |Z_{α+1} : Z_{α+1} ∩ G_{6}| ≤ 2 and

|Z_{α+1} ∩ G_{6} : Z_{α+1} ∩ K_{6}| ≤ |K_{5}K_{6}/K_{6}| = |O_{2}(G_{6} ∩ G_{5}/K_{6})| = 2^{8}.

Now I CZ̃_{α+1}(Z_{6})< ⟨Z_{6}, K_{α+1}⟩ and the fact that ⟨Z_{6}, K_{α+1}⟩ acts transitively

on Δ(α) implies I CZ̃_{α+1}(Z_{6}) ⊆ ⋂_{g∈G_{α}} Z̃_{α+1}^{g} = ⋂_{g∈⟨Z_{6},K_{α+1}⟩} Z̃_{α+1}^{g}.

We know that the dimension of Z̃_{α+1} is 2^{4}, hence |Z̃_{α+1} : I CZ̃_{α+1}(Z_{6})| ≥ 2^{8}. This yields Z̃_{6} = Z_{6}. Now we have |Z_{α+1} : Z_{α+1} ∩ K_{6}| ≥ 2^{8} and in the same way |Z_{6} : Z_{6} ∩ K_{α+1}| ≥ 2^{8}. Let x ∈ Z_{6}\K_{α}. Then Z_{α−1} ∩ Z_{α+1} < ⟨K_{α+1}, x⟩ and so Z_{α−1} ∩ Z_{α+1} < G_{α}.

We know that Z_{α+1}/Z_{α} is an irreducible (G_{α} ∩ G_{α+1})- module, and this yields Z_{α} = Z_{α−1} ∩ Z_{α+1} has order 2^{8}. Hence we get |[Z_{α+1}, x]| = 2^{8} and Z_{α} = I C_{Z_{α+1}}(Z_{6}). Finally [Z_{α+1}, x] ∩ Z_{α} = 1.

We have |[Z_{α+1} ∩ G_{6}, x]| ≥ 2^{7} and [Z_{α+1} ∩ G_{6}, x] ⊆ Z_{6}. Furthermore |Z_{6} ∩ G_{α+1} : Z_{6} ∩ K_{α+1}| ≥ 2^{7}. So there is some y ∈ Z_{6} ∩ G_{α+1} such that [Z_{α+1}, y] = I C_{Z_{α+1}}(y) = Z_{α}. Then |[Z_{α+1} ∩ G_{6}, y]| ≥ 2^{7}, thus |Z_{α} ∩ Z_{6}| ≥ 2^{7} and [Z_{α+1} ∩G_{6}, y]∩[Z_{α+1} ∩G_{6}, x] = 1. Hence |V_{α} ∩Z_{6}| ≥ 2^{14}. So Z_{α+1} lies in the centralizer of a subgroup of index 4 in Z_{6}, but this contradicts the action on the halfspin module.

So we are left to deal with b_{Γ} = 1. Let Z̃_{6} ⊆ K_{5}. Then for β ∈ Δ(5) we have [Z̃_{6}, Z̃_{β}] = 1 according to [Str3; (1.10)]. Hence Z̃_{6} ⊆ K_{β} and so [Z̃_{6}, Z_{β}] = 1. Thus even [Z̃_{6}, ⟨Z_{β}^{G_{5}}⟩] = 1. Since G_{5} = (G_{6} ∩ G_{5})⟨Z_{β}^{G_{5}}⟩, we have Z̃_{6} < ⟨G_{5}, G_{6}⟩ = G, but this is a contradiction. Let now Z̃_{6} ⊈ K_{5}. Then G_{6} ∩ G_{5} normalizes a subgroup of index 2 in Z̃_{6}, and this is a contradiction to the action on the halfspin module. We have proved (2).

The following statement is true:

(3) Let Γ be the graph Γ(G_{1}, G_{2}) and (2, α) a critical pair. Then α ∼ 1.

Suppose α ∼ 2. We may choose notation such that Z_{α} ⊈ G_{1}. Since Z_{2} is an F -module, by (1.7) Z_{2} has a submodule Z̃_{2}, which is a natural module. Furthermore, Z̃_{2} has a subgroup ˜ Z̃_{2} of index two, such that [Z_{α}, ˜ Z̃_{2}] ⊆ Ω_{1}(Z(S)) ∩ Z̃_{2}. Since |Z̃_{2} : ˜ Z̃_{2}| = 2 and there is no fours group of transvections in Sp_{8}(2), we have |Z̃_{2} : Z̃_{2} ∩ K_{α}| ≤ 4. So |Z_{α} : Z_{α} ∩ K_{2}| ≤ 4 and then also |Z_{2} : Z_{2} ∩ K_{α}| ≤ 4. If Z_{2} involves two natural modules, Z_{α} has to induce transvections on both. In particular |Z̃_{2} : Z̃_{2} ∩ K_{α}| = 2. But now we argue as before to get |Z_{α} : Z_{α} ∩ K_{2}| = 2 = |Z_{2} : Z_{2} ∩ K_{α}|, a contradiction. So Z_{2}/Z̃_{2} is trivial. As Z(G_{2}) = 1, we get with Gaschütz ([Hu; I 17.4])

Z_{2} = Z̃_{2}. Now |Z_{1}| = 2 and |Z_{3}| = 4. Set E_{1} = ⟨Z_{3}^{G_{1}}⟩. As [S, Z_{3}] = Z_{1} we have [E_{1}, K_{1}] ⊆ Z_{1}.

We want to show, that [E_{1}, Z_{2}] = 1. Let us assume, that [E_{1}, Z_{2}] ≠ 1. As E_{1}K_{2} is a normal subgroup in G_{1} ∩ G_{2} this implies |E_{1} : E_{1} ∩ K_{2}| = 2 or 2^{7}. Let |E_{1} : E_{1}∩K_{2}| = 2 at first. Then we have E_{1}K_{2}/K_{2} = Z(G_{1}∩G_{2}/K_{2}). As Z_{3} < S, Z_{3} ⊆ Z_{2} and E_{1} induces transvections on Z_{2}, we see that [E_{1}, Z_{3}] = 1 and so E′_{1} = 1. In particular Z_{2} ⊈ E_{1}.

Let β ∈ Δ(2) with [Z_{β}, E_{1}] ≠ 1. So [E_{1} ∩ K_{β}, E_{β} ∩ G_{1}] ⊆ E_{1} ∩ Z_{β} = 1, since E′_{1} = 1. We have [E_{β} ∩ K_{1}, E_{1} ∩ G_{β}] ⊆ E_{β} ∩ Z_{1}.

Suppose Z_{1} ⊆ E_{β}. Then E_{β} ⊆ G_{1} ∩ G_{2}. But [E_{β}, Z_{2}] ⊆ [E_{β}, K_{β}] = Z_{β}. On the other hand [Z_{2}, G_{1} ∩ G_{2}] ⊉ Z_{β}, so [E_{β}, Z_{2}] = 1, a contradiction to [E_{1}, Z_{2}] ≠ 1. Therefore we have [E_{β} ∩K_{1}, E_{1} ∩G_{β}] = 1. Without loss we may assume that |E_{1} ∩ G_{β} : E_{1} ∩ K_{β}| ≤ |E_{β} ∩ G_{1} : E_{β} ∩ K_{1}|. Since we know that |E_{1} : E_{1} ∩ G_{β}| ≤ 2 ≥ |E_{β} : E_{β} ∩ G_{1}|, we get that E_{1} is an F_{1}-module for G_{1} with offending subgroup (E_{β} ∩ G_{1})K_{1}/K_{1} and by (1.7) it must be the spin or the natural module. In any case by (1.9) P_{3} centralizes Z_{3} but this is not true. So we have |E_{1} : E_{1} ∩ K_{2}| = 2^{7}. Now [K_{1}, E_{1}] = Z_{1} implies Z_{2} ⊈ K_{1} and this contradicts α ∼ 2.

We have proved that [E_{1}, Z_{2}] = 1, in particular [E_{1}, Z_{3}] = 1 and E′_{1} = 1. Furthermore we know that E_{1} is no F_{1} - module. Let E_{1} ⊆ K_{α−1} ⊆ G_{α}. If [E_{1}, Z_{α}] = [Z_{2}, Z_{α}], then E_{1}Z_{2} < ⟨G_{1} ∩ G_{2}, Z_{α}⟩ = G_{2}. Hence Z_{2} ∩ [K_{2}, E_{1}Z_{2}] is a normal subgroup of G_{2}, so we have Z_{2} ⊆ E_{1} or E_{1} ⊆ Z(K_{2}). Since I C_{S}(E_{1}) ⊆ K_{1}, E_{1} ⊆ Z(K_{2}) would lead to the contradiction K_{1} = K_{2} = 1. So we have Z_{2} ⊆ E_{1}, this implies E_{1} < ⟨G_{1}, G_{2}⟩ = G, a contradiction. Now [E_{1}, Z_{α}] ≠ [Z_{2}, Z_{α}]. If |Z_{2} : Z_{2} ∩ K_{α}| ≠ 2, then there exists an x ∈ Z_{α} such that [Z_{2}, x] ≠ 1 and [x, O_{2}(G_{α} ∩ G_{α−1})] = Z_{α−1}. Now we conclude [E_{1}Z_{2}, x] = Z_{α−1} = [x, Z_{2}] ⊆ [Z_{α}, Z_{2}] ⊆ Z_{2} and again E_{1}Z_{2} < ⟨G_{1} ∩ G_{2}, x⟩. But as [x, Z_{2}] ≠ Z_{1} and x induces a transvection on Z_{2} we have x ∉ G_{1}. So G_{2} = ⟨G_{1} ∩ G_{2}, x⟩ and we get a contradiction as before.

Now we have proved |Z_{2} : Z_{2} ∩K_{α}| = 2. Then I C_{Z_{α}}(Z_{2}) induces transvections on E_{1}. But Z_{1} ⊈ Z_{α}, hence I C_{Z_{α}}(Z_{2}) ⊈ K_{1} and so E_{1} is an F -module and we get a contradiction.

Thus E_{1} ⊈ K_{α−1}. Choose β ∈ Δ(2) such that d(β, α) = d(2, α)−1. There is some x ∈ Z_{2} which induces a transvection on Z_{α} as Z_{2} offends on Z_{α}. So Z_{α−1} = ⋂_{y∈Z_{2}\K_{α}}[y, Z_{α}]. By symmetry we get Z_{β} = ⋂_{y∈Z_{α}\K_{2}}[y, Z_{2}], hence Z_{α−1} = Z_{β}. We have [E_{α−1} ∩ K_{1}, E_{1}] ⊆ E_{1} ∩ E_{α−1} ∩ [K_{1}, E_{1}] ⊆ E_{α−1} ∩ Z_{1} = 1 since [Z_{1}, Z_{α}] ≠ 1 but [E_{α−1}, Z_{α}] = 1.

If Z_{2} ⊆ E_{1}, then |K_{1}K_{2} : K_{2}| = 2 as [K_{1}, E_{1}] ⊆ Z_{1}. Also |K_{1}K_{2} : K_{1}| = 2. Since Z_{α−2} ⊆ E_{α−1} ∩ K_{1} we get E_{1} ⊂ K_{α−2} and so |E_{1} : E_{1} ∩ K_{α−1}| = 2. Since we know that |K_{β}K_{2}/K_{2}| = 2, we have |E_{α−1} : E_{α−1} ∩ K_{2}| ≤ 2 and |E_{α−1} : E_{α−1}∩K_{1}| ≤ 4. Hence for e ∈ E_{1}\K_{α−1} we get |E_{α−1} : I C_{E_{α−1}}(e)| ≤ 4, and E_{1} is an F_{1}-module, a contradiction.

Therefore Z_{2} ⊈ E_{1} and then Z_{β} ⊈ E_{1}. Now we have

[E_{α−1} ∩ G_{1}, E_{1} ∩ K_{α−1}] ⊆ Z_{α−1} ∩ E_{1} = Z_{β} ∩ E_{1} = 1.

Thus E_{1} is an F_{1}-module again, because of |E_{α−1} : E_{α−1} ∩ G_{1}| ≤ 2. This is a contradiction and we have proved (3).

By (1.24) Z_{2} is an F_{1}–module. By (1) Z(G_{2}) = 1 and so by (1.7) and (1.9) a natural module Z̃_{2} is a submodule of Z_{2}. Define Z̃_{3} = Z_{3} ∩ Z̃_{2} and E_{1} = ⟨Z̃_{3}^{G_{1}}⟩. As [Z̃_{3}, K_{1}] ⊆ Z_{1} and I C_{G_{1}}(K_{1}) ⊆ K_{1} we get E_{1} ⊆ K_{1}.

(4) b_{Γ} > 1

Suppose b_{Γ} = 1. First let Z̃_{2} ⊆ K_{1}. Then |E_{1} : E_{1} ∩ K_{2}| ≤ 2. Since [Z_{2}, E_{1}] ≠ 1 it follows that |E_{1} : E_{1} ∩ K_{2}| = 2. But then Z_{2} induces transvections on E_{1} and E_{1} would be an F –module and so [P_{3}, Z̃_{3}] = 1, a contradiction.

So Z̃_{2} ⊈ K_{1}. Since Φ(K_{2}) ⊆ K_{1} and because we could assume Z̃_{2} ⊆ Φ(K_{2}) if Φ(K_{2}) ≠ 1, it follows that Φ(K_{2}) = 1. Clearly [E_{1} ∩ K_{2}, K_{2}] = 1. Since |K_{1}K_{2}/K_{1}| = |K_{1}K_{2}/K_{2}| ≥ |E_{1} : E_{1} ∩ K_{2}|, E_{1} is an F -module, a contradiction.

Using (1) - (4) we want to derive a contradiction to our assumption I C_{G_{1}}(K_{1}) ⊆ K_{1} and I C_{G_{2}}(K_{2}) ⊆ K_{2} at the beginning of this proof.

Since by (4) and (3) b_{Γ} ≥ 3 we get E′_{1} = 1 = [E_{1}, Z_{2}]. Let β ∈ Δ(2) again, with d(β, α) = d(2, α)−1. Then [E_{β}, E_{α}] ⊆ E_{α} ∩E_{β}. Assume that E_{α} ⊆ K_{β}. Thus E_{α} ⊆ G_{2}, and then [E_{α}, Z_{2}] ⊆ E_{α} ∩ Z_{2}. As E_{α} acts quadratically it follows that |E_{α} : E_{α} ∩ K_{2}| ≤ 16. Hence for any t ∈ Z_{2}\K_{α} we have |E_{α} : I C_{E_{α}}(t)| ≤ 16. By [Wie2; 2.8] O_{3}(G_{α}/K_{α}) acts trivially on E_{α}. Thus by (1.10)(iv) E_{α} involves only natural and spin modules. In particular by (1.9) P_{3} ⊆ I C_{G}(Z̃_{3}), a contradiction.

So we have to consider the case E_{α} ⊈ K_{β}. Let t ∈ E_{β}. Then we have |E_{α} ∩ K_{β} : I C_{E_{α}∩K_{β}}(t)| ≤ 2 and hence |E_{α} : I C_{E_{α}}(t)| ≤ 2|E_{α} : E_{α} ∩ K_{β}| ≤ 2^{8}.

Again by [Wie2; 2.8] O_{3}(G_{α}/K_{α}) acts trivially. Thus E_{α}/Z_{α} is an Sp_{8}(2)- module.

If only the natural and the spin module are involved, we get with (1.9) [P_{3}, Z̃_{3}] = 1 again, and this would be a contradiction.

Let V_{α} be an nontrivial module involved in E_{α} which is not a natural or a spin module. By (1.10)(iii) we have that E_{α} cannot contain an element t which is of type a_{2} on V_{β}. The same is true for E_{β}. So |E_{α}/E_{α} ∩ K_{β}| ≤ 8. But now we get |[E_{α}, t]| ≤ 16, a contradiction by (1.10)(iv).

So our first assumption was wrong. Let now w.l.o.g. I C_{G_{1}}(K_{1}) ⊈ K_{1}. It follows that |K_{1}K_{2}/K_{2}| ≤ 2. We have [(G_{1} ∩ G_{2})^{∞}, K_{1} ∩ K_{2}] = 1 and |K_{2} : K_{1} ∩ K_{2}| ≤ 2, therefore I C_{G_{2}}(K_{2}) ⊈ K_{2} and K_{1} ∩ K_{2} = 1. So we get |K_{1}| = |K_{2}| ≤ 2. As G_{k}/K_{k} ≅ Ω_{2k−2}^{+}(2) we get |S| = 2^{(k−1)^{2}} or 2^{(k−1)^{2}+1} and this yields |K_{k}| = 2^{k−1} or 2^{k}. So I C_{G_{k}}(K_{k}) ⊈ K_{k}. But as Z(G_{k} ∩ G_{1}/K_{1}) = 1 = Z(G_{k} ∩ G_{2}/K_{2}) we have K_{k} ⊆ K_{1} ∩ K_{2} = 1, again a contradiction and the proof of (4.1) is completed. □

(4.2) Lemma. Let {P_{1}, …, P_{k}} be a semiclassical parabolic system with diagram

∘ ∘_{k}  ∘ ∘ ∘ …  ^{∼} _{1}_{2}_{3}

where k ≥ 5. Then G ≅ 3^{\frac{1}{3} (2^{k}−1)(2^{k−1}−1)}Sp_{2k}(2) or k = 5 and G ≅ F_{1}.

Proof. Let G_{{4,5,...,k}}/K_{{4,5,...,k}} ≅ 3^{7}Sp_{6}(2). Then the claim follows with [ShStr]. So by (1.3) we may assume that G_{{5,6,...,k}}/K_{{5,6,...,k}} ≅ Co_{1}. First let k = 5 and assume I C_{G_{5}}(K_{5}) ⊆ K_{5}. According to [ParRow; Theorem B], K_{5} is extraspecial of order 2^{25}. Using [Iv] we get G ≅ F_{1}. Hence we may assume I C_{G_{5}}(K_{5}) ⊈ K_{5}. If k = 6 [ParRow] implies that I C_{G_{6}}(K_{6}) ⊈ K_{6}. Let now k = 5 or 6 and I C_{G_{k}}(K_{k}) ⊈ K_{k}. Since G_{k−1,k}/K_{k} is perfect and G_{k−1}/K_{k−1} ≅ Σ_{3} × M_{24} or Σ_{3} × Co_{1}, we see K_{k} ⊈ K_{k−1}. But by general arguments K_{k} ∩ K_{1} ⊆ K_{k−1}. As Z(G_{1,k}/K_{1}) = 1, we have K_{k} ⊆ K_{1}. And so K_{k} ⊆ K_{k−1}, a contradiction. □

(4.3) Lemma. There is no semiclassical parabolic system {P_{1}, …, P_{k}} with diagram of type

∘ ∘ ∘^{∼}_{k}  [figure: ∘; ∘; …; ∘; ∘; 2; 1; 3; 4; ,]

where G_{1}/K_{1} ≅ Co_{1} or F_{1}.

Proof. Assume false. First let k = 5. Then by (1.3) G_{1}/K_{1} ≅ G_{2}/K_{2} ≅ Co_{1}. If I C_{G_{1}}(K_{1}) ⊆ K_{1}, then by [ParRow; Theorem B] we get K_{1} is extraspecial of order 2^{25}. In the same way we get K_{2} is extraspecial of order 2^{25}. But then Ω_{1}(Z(S)) < ⟨G_{1}, G_{2}⟩ = G and this cannot hold.

Thus I C_{G_{1}}(K_{1}) ⊈ K_{1}. As G_{1} ∩ G_{2}/K_{2} is isomorphic to 2^{11}M_{24} we get K_{1} = K_{2} = 1. Hence |S| = 2^{21}.

We have G_{3}/K_{3} < ∼ 3Σ_{6} × Σ_{3} × Σ_{3}. From G_{1} we know that G_{1,2,3} induces in K_{1,2,3} exactly one four-dimensional, two six-dimensional and one trivial module for 3Σ_{6}. So all the nontrival modules must be involved in K_{3}. This shows |K_{3}| ≥ 2^{16}. As in G_{1} = Co_{1} we see G_{1,3}/K_{1,3} ≅ 3Σ_{6} × Σ_{3} we get that |K_{3}| = 2^{16} and so G_{3}/K_{3} ≅ (3A_{6} × ZZ_{3} × ZZ_{3}) · V_{4}. Further |Z(K_{3})| = 16 and K_{3}/Z(K_{3}) is a direct sum of two six-dimensional modules for 3A_{6}. Since also G_{3,2}/K_{3,2} ≅ 3Σ_{6} × Σ_{3} and G_{5,4,3}/K_{5,4,3} ≅ Σ_{3} × Σ_{3} we conclude that there are exactly 3 subgroups of order 3 in O_{3}(G_{3}/K_{3}), which are normal in G_{3}/K_{3}. None of them acts trivially on K_{3}. But since K_{3}/Z(K_{3}) ≅ 2^{6} ⨁ 2^{6} there exists an element ϱ ∈ G_{3} with o(ϱ) = 3 and [ϱ, K_{3}] ⊆ Z(K_{3}) = Φ(K_{3}). So [ϱ, K_{3}] = 1 and this is a contradiction.

We are left with the case k = 6. Then

Δ = [figure: ∘; ∘; ∘; ∘; ∼; ∘; ∘; ,]

but this amalgam contains a subamalgam of type

[figure: ∘; ∘; ∘; ∘; ∘; ∘; ,]

([FuStr], Lemma 3.3) which does not exist by [Tim1]. □

The lemmas (4.1), (4.2), and (4.3) prove the following

(4.4) Lemma. There is no semiclassical parabolic system {P_{1}, …, P_{k}}, with k > 4 and diagram Δ of type

∘ ∘ ∘^{∼}_{k}  [figure: ∘; ∘; …; ∘; ∘; 2; 1; 3; 4; .]
□

(4.5) Lemma. There is no semiclassical parabolic system {P_{1}, …, P_{k}}, with k > 4 and diagram Δ of type

∘ ∘ ∘^{(∼)}_{1}  _{2}  ∘ ∘ ∘ …  ^{∼} _{k}_{k-1}  .

Proof. Suppose false. If we have G_{{1,...,k−3}}/K_{{1,...,k−3}} ≅ 3^{7}Sp_{6}(2) and G_{{4,...,k}}/K_{{4,...,k}} is a group of Lie type or 3^{7}Sp_{6}(2) then the claim follows from [Wie2; (4.12)]. So G_{{1,...,k−3}}/K_{{1,...,k−3}} ≅ M_{24} or He. By (4.2) and (1.3) k = 5 or 6 and G_{1}/K_{1} ≅ Co_{1} or F_{1}. So by [FuStr; Lemma 3.3] we find a subamalgam of type

[figure: ∘; ∘; ∘; (∼); ∘; ∘; or; ∘; ∘; ∘; ∘; (∼); ∘; ∘; .]

But these diagrams do not exist according to (4.4) or [Str3]. □

(4.6) Lemma. Let {P_{1}, …, P_{k}}, k > 4, be a semiclassical parabolic system with diagram Δ of type

∘ ∘_{k}  …  [figure: ∘; ∘; ∘; ∘; ∼; (∼); 4; 3; 1; 2] .

Then K_{1} = K_{2} = 1, G_{1} ≅ G_{2} ≅ Co_{1} or F_{1} and k = 5 or 6, resp., or G_{1} ≅ 3^{\frac{1}{3} (2^{k−1}−1)(2^{k−2}−1)}Sp_{2k−2}(2) and G_{2} ≅ (3^{\frac{1}{3} (2^{k−1}−1)(2^{k−2}−1)}) Sp_{2k−2}(2).

Proof. We will prove K_{1} = K_{2} = 1 by induction on k. By (1.3)(i)(vi) we have for a parabolic system with diagram

[figure: ∘; ∘; ∘; ∘; ∼; (∼); 1; 2]

that the kernels K_{1} and K_{2} are trivial or G_{2} ≅ 2^{6}·A_{7}. The second case cannot occur as a residue of {5, …, k} since by [Str4] there is no parabolic system

with diagram ∘ ∘ ∘_{5}  _{4}  _{3}  _{1}∘ where the 5-residue is the A_{7}–system. So

by induction we have K_{k} = K_{1,k} = K_{2,k} and thus K_{1} ⊆ K_{k} ⊇ K_{2}. We have K_{2}K_{1}/K_{1} < G_{1} ∩ G_{2}/K_{1}.

Let (G_{1}/K_{1})/O_{3}(G_{1}/K_{1}) ≅ Sp_{2k−2}(2). By (1.3)(vi) we conclude that G_{{5,...,k}}/K_{{5,...,k}} ≅ (3^{7})Sp_{6}(2). Then, applying [ShStr], we deduce that (G_{2}/K_{2})/O_{3}(G_{2}/K_{2}) ≅ Sp_{2k−2}(2). We know that K_{k} = K_{1,k}, therefore K_{k}/K_{1} ⊈ O_{2}((G_{1} ∩ G_{2})/K_{1}). We have that O_{2}(G_{1} ∩ G_{2})/K_{1} involves exactly two irreducible G_{1} ∩ G_{2}/O_{2}(G_{1} ∩ G_{2})-modules, both of them being not trivial. If K_{2} ⊈ K_{1} then K_{2} involves at least one of these modules in particular [K_{2}, K_{k}] ⊈ K_{1}. But since K_{k}/K_{1} is abelian we have [K_{k}, K_{2}] ⊆ K_{1}, a contradiction. Thus K_{2} ⊆ K_{1}. We have |S/K_{1}| = |S/K_{2}| and so K_{1} = K_{2} = 1.

By (4.2) and (1.3) we are left with G_{1}/K_{1} ≅ Co_{1} or F_{1}. Applying (1.3) we see G_{2}/K_{2} ≅ Co_{1} or F_{1} as well. If I C_{G_{1}}(K_{1}) ⊆ K_{1}, then by [ParRow] G_{1}/K_{1} ≅ Co_{1} and K_{1} ≅ K_{2} is extraspecial of order 2^{25}. Thus we have Z(S) < ⟨G_{1}, G_{2}⟩ = G, a contradiction.

So I C_{G_{1}}(K_{1}) ⊈ K_{1} and I C_{G_{2}}(K_{2}) ⊈ K_{2}. In particular K_{1} ∩ K_{2} = 1. Since Z(G_{1} ∩ G_{2}/K_{1}) = 1 if G_{1}/K_{1} ≅ F_{1}, in this case we have K_{1} = K_{2} = 1.

Now let G_{1}/K_{1} ≅ Co_{1} and k = 5. Then |Z(G_{1} ∩ G_{2}/K_{1})| = 2 and |K_{1}| = |K_{2}| ≤ 2. We have that G_{4,5}/K_{4,5} belongs to ∘ ∘ ∘^{∼} ^{∼} . By (1.3)(i) K_{4,5,2} = K_{4,5,1}. So (1.1)(a)(iv) implies G_{5,4,3}/K_{5,4,3} ≅ (ZZ_{3} ×ZZ_{3})·ZZ_{2}. Then we have G_{3}/K_{3} ≅ L_{3}(2) × [(ZZ_{3} × ZZ_{3}) · ZZ_{2}] and |Z(K_{3}/K_{1})| = 2^{2}. If K_{1} ≠ 1 then K_{1} ≮G_{3}. Hence |Z(K_{3})| = 2^{3}. Since G_{1} ∩ G_{3} and G_{2} ∩ G_{3} both induce Σ_{3} on Z(K_{3}), G_{3} acts on Z(K_{3}) in the same way as G_{1} ∩ G_{3}, so [K_{1}, G_{3}] = 1. This is a contradiction and we get again K_{1} = 1. □

(4.7) Lemma. There is no semiclassical parabolic system {P_{1}, …, P_{5}} with diagram Δ of type

[figure: ∘; ∘; ∘; ∘; ∘; ∼; 5; 1; 4; 3; 2; or]  ∘ ∘ ∘ ∘ ∘_{2}  _{3}^{1}  ^{4}  ^{5}^{∼} .

Proof. Assume false. Consider G_{2}. According to (1.3)(v) G_{2,5}/K_{2,5} ≅ A_{7} and G_{1,2}/K_{1,2} ≅ G_{2,3}/K_{2,3} ≅ 3^{7}Sp_{6}(2). In the first diagram G_{5}/K_{5} ≅ L_{5}(2), so G_{2,5}/K_{2,5} ≅ A_{8}, a contradiction. In the second diagram using [Tim1] applied to G_{5} we conclude that G_{1,5}/K_{1,5} ≅ A_{7} as well. By (1.3)(viii) we have G_{1}/K_{1} ≅ 3^{35}Sp_{8}(2) and so G_{1,5}/K_{1,5} ≅ A_{8}, a contradiction again. □

(4.8) Proposition. Let {P_{1}, …, P_{5}} be a semiclassical parabolic system with connected diagram Δ containing a subdiagram of type ∘ ∘ ∘ ∘^{∼} . Then one of the following holds:

(i) Δ = ∘ ∘ ∘ ∘ ∘^{∼} , G ≅ F_{1} or 3^{155}Sp_{10}(2)

(ii) Δ = [figure: ∘; ∘; ∘; ∘; ∘; ∼; (∼); 1; 2] , K_{1} = K_{2} = 1 and G_{1} ≅ G_{2} ≅ Co_{1} or

G_{1} ≅ 3^{35}Sp_{8}(2) and G_{2} ≅ 3^{35}Sp_{8}(2) or Sp_{8}(2).

Proof. We choose notation such that we have a subdiagram

∘ ∘ ∘ ∘_{4}_{5}  _{3}  _{2}^{∼}  .
Let us consider the residue of 5. It contains the subdiagram ∘ ∘ ∘_{4}  _{3}  _{2}^{∼} .

Suppose that the diagram of the residue of 5 is connected.

If this is a string then by (1.3)(iii) it has one of the following shapes:

∘ ∘ ∘ ∘_{3}_{2}  _{4}  _{1}^{∼}  ,  ∘ ∘ ∘ ∘_{3}_{2}  _{4}  _{1}^{∼}  ^{∼}  ,
∘ ∘ ∘ ∘_{3}_{4}  _{2}  _{1}^{∼}  , or ∘ ∘ ∘ ∘_{3}_{4}  _{2}  _{1}^{∼}  ^{(∼)}  .

The second diagram is not possible since according to (1.3)(ii) G_{1}/K_{1} ≅ He in this case but G_{1}/K_{1} has to be isomorphic to M_{24} or 3^{7}Sp_{6}(2). The last two diagrams are not possible because of (2.4) and (3.5). If the diagram is a star it is of shape

[figure: ∘; ∘; ∘; ∘; ∼; (∼); 1; 2; 4; 3; or] [figure: ∘; ∘; ∘; ∘; ∼; 2; 4; 3; 1] .

By (1.3) it is never a square. Thus the possible diagrams are:

[figure: ∘; ∘; ∘; ∘; ∘; ∼; (∼); 1; 2; 5; 4; 3; ,; ∘ ∘ ∘ ∘ ∘_{5}  _{1}^{4}  ^{3}  ^{2} ........ _{(∼)}^{∼} ,; ∘; ∘; ∘; ∘; ∘; ∼; 1; 2; 5; 4; 3; ,]

∘ ∘ ∘ ∘ ∘_{5}  _{1}^{4}  ^{3}  ^{2} ........ ^{∼} , and [figure: ∘; ∘; ∘; ∼; ∘; ∘; 2; 3; 4; 5; 1; .]

In the first case we are done with (4.6). The third and the last case is not possible by (4.7) and (4.4) respectively. Considering in the fourth diagram the residue of 4 we conclude with (1.3)(ii) that it is of type ∘ ∘ ∘ ∘_{1}_{5} _{2} _{3}^{∼} and so Δ is of the shape

5  _{1}^{4}  ^{2}  ^{3} ∘ ∘ ∘ ∘ ∘^{∼} ,

but this contradicts (4.7).

There is the second diagram left. Considering the residue of 2 we get by (1.3)(ix)(x)

∘ ∘ ∘ ∘ ∘_{5}  _{1}^{4}  ^{3}  ^{2} ........ ^{∼} .

Considering G_{4}/K_{4} we get a contradiction to (1.3) again.

Let us now assume that Δ_{5} is not connected. Then by [Str5] Δ is of type

∘ ∘ ∘_{1}  _{5}  _{4}  _{3}  _{2} ∘ ∘^{∼}  or ∘ ∘ ∘^{∼}_{1}  _{5}  _{4}  _{3}  _{2} ∘ ∘^{∼}  .

The second diagram is out by (4.5), the first yields case (i) by (4.2). Now the proposition is proved. □

(4.9) Proposition. Let {P_{1}, …, P_{n}}, n ≥ 5, be a semiclassical parabolic system with connected diagram Δ containing a subdiagram

∘ ∘_{1}  _{2}  ∘ ∘ ∘ …  _{n-1}_{n-2}^{∼}  .

Then Δ is one of the following

(i) ∘ ∘ ∘_{1}  _{2}_{n}  ∘ ∘ ∘ …  ^{∼} _{n-1}_{n-2}  or

(ii) ∘ ∘ ∘_{2}  _{3}_{1}  … [figure: ∘; ∘; ∘; ∘; ∼; (∼); n; n-1; n-2].

Proof. We prove the assertion by induction on n. The case n = 5 is Proposition (4.8). So we may assume that n > 5. We have to determine how the node n is connected to the given subdiagram. By induction applied to Δ_{1} the node n can just be connected with the nodes 1, 2, or n − 2. Suppose n is not connected with the nodes 1 and n − 2. Then n is just connected with 2. By induction we get that Δ_{1} has the shape

∘ ∘ ∘_{2}  _{3}_{n}  ∘ ∘ ∘ …  ^{∼}_{n-2}  _{n-1} ,

but this is a contradiction to (4.4).

So n is connected with 1 or n − 2. Suppose n is just connected with 1. By [Str5] applied to Δ_{{4,...,n−1}} and (4.5) we get that Δ must be of the shape

∘ ∘ ∘_{1}  _{2}_{n}  ∘ ∘ ∘ …  ^{∼}_{n-2}  _{n-1} ,

this is case (i).

Let finally n be connected with n − 2. By induction we see that Δ_{1} is of the shape

∘ ∘ ∘_{2}  _{3}  … [figure: ∘; ∘; ∘; ∘; ∼; (∼); n; n-1; n-2].

If n is not connected to 1 this is case (ii). So suppose that n and 1 are

connected as well. Then Δ_{{2,...,n−3}} has the form ∘ ∘ ∘ ∘ ........ ^{(∼)}  ^{∼}_{1}  _{n}  _{n-2}  _{n-1} . By
(1.3) we have that Δ_{{2,...,n−3}} must be ∘ ∘ ∘ ∘^{∼}  ^{∼}_{1}  _{n}  _{n-2}  _{n-1} . Now Δ_{n−1} is a

circle which has a subdiagram of the shape

∘ ∘ ∘_{3}_{2}  ∘ ∘ ∘ …  ^{∼} _{n}_{n-2}  .

So we get a contradiction by induction. □

§5 Proof of the theorem

In this chapter we are going to prove the main theorem. This will be done by induction on the rank of the diagram. In (2.4) we have shown that Δ has no residue of F̃_{4}-type. Now we will show that Δ cannot contain a residue of square type.

(5.1) Lemma. There is no parabolic system {P_{1}, P_{2}, P_{3}, P_{4}, P_{5}} such that the diagram for {P_{1}, P_{2}, P_{3}, P_{4}} is of type

∘ ∘ ∘ ∘_{1}  _{2}^{3}  ^{4}^{∼}

Proof. Suppose false. We have that {P_{1}, P_{2}, P_{3}, P_{4}} is as in (1.3)(ix). Further P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for 1 ≤ i ≤ 5 by (1.2). In particular we know that G_{1,5}/K_{1,5} is of type (1.1)(a)(iv)(⋆⋆) and G_{3,5}/K_{3,5} ≅ A_{7}. Let Δ be the diagram for {P_{1}, P_{2}, P_{3}, P_{4}, P_{5}}. As Δ does not contain triangles by (1.1) we may assume that 5 is not connected with 1. In particular Δ_{1} is connected.

By (1.3) we get that Δ_{1} is one of the following three diagrams:

(1)  [figure: ∘; ∘; ∘; ∘; ∼; (∼); 3; 4; 5; 2; ,] (2) ∘ ∘ ∘ ∘_{5}  _{2}^{3}  ^{4}^{∼} , ^{or (3)} ∘ ∘ ∘ ∘_{5}  _{2}^{3}  ^{4}^{∼}^{(∼)}  _{(∼)}

In case (1) Δ_{3} has the shape

∘ ∘ ∘ ∘^{(∼)}_{5}  _{4}  _{2}  _{1},

this cannot happen by (1.3) and [Str2]. In case (2) Δ_{3} has the shape

[figure: ∘; ∘; ∘; ∘; 4; 2; 5; 1] .

Since G_{3,5}/K_{3,5} ≅ A_{7} it follows with [Str2] that this cannot occur. There is case (3) left. Now Δ_{4} has the shape

∘ ∘ ∘ ∘_{5}  _{2}^{3}  ^{1}^{(∼)}  _{(∼)} .

Application of (1.3) resp. [Str2] shows that G_{3,4}/K_{3,4} ≅ A_{7} . But now Δ_{3} does not exist by [Str2]. □

(5.2) Lemma. If Δ is one of the diagrams of the theorem then the corresponding statements about the groups hold as well.

Proof. We prove the claim by induction on the rank of Δ. If rank(Δ) = 4 then this is (1.3). So let rank(Δ) ≥ 5.

Let first Δ be a complete bipartite graph all of whose edges are of type ∘ ∘^{(∼)} . Choose x ∈ I such that Δ_{x} is connected, contains ∘ ∘^{(∼)} , and |K_{x}| is minimal. By induction there exists a subset J, x ∈ J ⊆ I, such that Δ_{J} is of shape ∘ ∘ ∘^{(∼)} ^{(∼)} and K_{J} = K_{x}. As |I| ≥ 5 there is y ≠ x, y ∈ J. Now |K_{y}| ≥ |K_{x}| and K_{y} ⊆ K_{J}, hence K_{y} = K_{x} = 1.

Let Δ be of type He(n, m) and J ⊆ I be any subset such that Δ_{J} is of shape ∘ ∘ ∘^{∼} . Then again because of |I| ≥ 5 there are x, y ∈ J, x ≠ y. By induction applied to Δ_{x} resp. Δ_{y} we get K_{x} = K_{J} = K_{y} = 1.

Let Δ be of type X(2, m) and J ⊆ I such that Δ_{J} is of shape ∘ ∘ ∘^{(∼)} . By [Str1; 4.3.1] applied to Δ_{J\{1,2}} we have G_{J}/K_{J} ≅ A_{7}. Hence by (1.3) G_{J}/K_{J} ≅ (3^{7})Sp_{6}(2). Let J′ = {1, 4, …, 4 + m − 1}. Then K_{J′} = K_{4} by induction applied to Δ_{4}. On the other hand Δ_{1} is of type Y (1, m + 1) with G_{J′}/K_{J′} ≅ 3^{7}Sp_{6}(2). So K_{J′} = K_{1}, hence K_{J′} = 1. Now |S| = 2^{9} and K_{J} = 1 for any J with G_{J}/K_{J} ≅ (3^{7})Sp_{6}(2).

Finally, if J′ = {4, …, m + 4} we see G_{J′} ≅ 2^{6}A_{7} by (1.3)(v) applied to Δ_{I\{1,2,3,m+4}}.

Let Δ be of type Y (n, m). If m = 1, we get the assertion with (1.3) and (4.2), up to the fact that it was not stated in (1.3) that for the diagram

[figure: ∘; ∘; ∘; ∘; ∼; ∼; 1; 2; 3; 4]

with G_{3}, G_{4} ≅ M_{24}, He always G_{3} ≅ G_{4} holds. So suppose G_{3} ≅ M_{24} and G_{4} ≅ He. Then by [FuStr; (3.3)] we would get a system with diagram

∘ ∘ ∘ ∘_{x}  _{1}  _{2}  _{3}^{∼}  or ∘ ∘ ∘ ∘_{x}  _{3}^{1}  ^{2} ........ _{∼}

and G_{x}/K_{x} ≅ He, a contradiction to (1.3).

If m = 2 and n > 1, we may apply (4.6). So let m > 2. Let J ⊂ I belong to the diagram ∘ ∘... ∘ ∘ ∘^{(∼)}_{n+1}_{n}_{2}_{1} . Then by induction we get

that G_{J} is as in the definition of a Y (n, m)-structure. It just remains to show that there are no J_{1}, J_{2}, J_{1} ≠ J_{2} such that G_{J_{1}} ≅ 2^{6}A_{7} = G_{J_{2}}. But this is equivalent to the following: There is no rank four classical system with diagram

[figure: ∘; ∘; ∘; ∘; 1; 2; 3; 4]

such that G_{4}/K_{4} ≅ G_{3}/K_{3} ≅ A_{7}. But this follows with [Str4].

Let finally Δ be of type M(n, m). Let J = I \ {1, t, u, v} such that Δ_{J} is of shape ∘ ∘ ∘ ∘^{(∼)}^{∼}_{1} _{t} _{u} _{v} . Application of (1.3)(iv) shows that G_{J}/K_{J} is locally isomorphic to M(24), M(24)′, 3M(24) or 3M(24)′. If rank(Δ) = 5 we get K_{J} = 1 with (3.5). Otherwise there are x, y ∈ J, x ≠ y, and we get K_{x} = K_{J} = K_{y} = 1 by induction.

This proves the lemma. □

Now we are going to show that a connected diagram Δ such that any subdiagram of Δ which contains ∘ ∘^{∼} is one of the diagrams of the theorem, is itself a diagram of the theorem. For this we choose some x ∈ I such that Δ_{x} is connected. As there are at least two possible choices we may assume that Δ_{x} contains ∘ ∘^{∼} .

(5.3) Lemma. If Δ_{x} is a complete bipartite graph all of whose edges are of type ∘ ∘^{(∼)} , then Δ is one of the diagrams of the theorem.

Proof. Let I_{1} ∪ I_{2} be the partion of the nodes of Δ_{x}, n = |I_{1}|, and m = |I_{2}|. Assume first I_{1} = {1}, i.e. Δ_{x} is a star. Suppose x is connected to 1 in Δ. As Δ does not contain triangles by (1.2) and (1.1) we get that x is not connected with any y ∈ I_{2}. Hence the diagram Δ has the shape of a star. If we have the subdiagram ∘ ∘_{1} _{x}^{(∼)} , then Δ is as in the theorem (i). If we have the subdiagram ∘ ∘_{1} _{x} , then Δ is of the shape Y (1, m). So assume now that x is not connected with 1 in Δ but with some 2 ∈ I_{2}. Assume there is some 3 ∈ I_{2}, such that 3 is not connected to x in Δ. Then Δ_{I\{1,2,3,x}} is of shape

∘ ∘ ∘ ∘^{(∼)}  ^{(∼)} ........_{3}  _{1}  _{2}  _{x}.

By (1.3) and [Str2] we see that Δ_{I\{1,2,3,x}} is of shape

∘ ∘ ∘ ∘^{(∼)}  ^{∼}_{3}  _{1}  _{2}  _{x}.

Let 4 ∈ I_{2}, 4 ≠ 2, such that x is connected with 4 in Δ. Then we get that Δ_{I\{1,2,4,x}} is of type

∘ ∘ ∘ ∘^{(∼)} ........_{4}  _{x}^{1}  ^{2}^{∼}

contradicting (1.3). Hence x is just connected with 2 and Δ is of shape M(m, 1).

Assume now that x is connected with all nodes in I_{2}. Let a ∈ I_{2}. As |I_{2}| ≥ 3 we may choose b ∈ I_{2} \ {a} such that Δ_{I\{1,a,b,x}} contains ∘ ∘_{1} _{c}^{∼} for some c ∈ {a, b}. As Δ_{I\{1,a,b,x}} has the shape of a square we get with

(1.3) and (5.1) that we have the subdiagram ∘ ∘_{1} _{x}^{(∼)} and so Δ is a complete

bipartite graph all of whose edges are of type ∘ ∘^{(∼)} .

Assume now |I_{1}| ≥ 2 ≤ |I_{2}|. Then we may assume that x is connected with 1 in I_{1}. As we have no triangles in Δ, x is not connected to any element in I_{2}. Let 2 ∈ I_{1}, such that 2 is not connected to x. Then for a ∈ I_{2}, we have by [Str2] and (1.3) that Δ_{I\{1,2,a,x}} is of type

∘ ∘ ∘ ∘^{(∼)}^{∼}_{x}  _{1}  _{a}  _{2}.

As before we see that x is not connected to any element in I_{1} \ {1} and so Δ is of type M(n, m).

If x is connected to all elements in I_{1}, then as above we see that for b ∈ I_{1}, we have a subdiagram ∘ ∘_{x} _{b}^{(∼)} , and so Δ is a complete bipartite graph all of

whose edges are of type ∘ ∘^{(∼)} . □

(5.4) Lemma. If Δ_{x} is of type He(n, m) then Δ is as in the theorem.

Proof. Let a be some end node of Δ_{x} connected with x. Then Δ_{I\{1,2,a,x}} is of shape

∘ ∘ ∘ ∘^{∼} ........_{x}  _{a}  _{1}  _{2} or ∘ ∘ ∘ ∘_{x}^{a}  _{2}^{1} ..................^{∼}

but this contradicts (1.3) as G_{I\{a,1,2}} ≅ He. So we have that x is not connected to any end node. We may assume that x is connected to 1 but not to 2. Let b be an end node not connected with 1. By induction and the shape of Δ_{bx} we get Δ_{b} is of type He(n + 1, m − 1) or Y (1, n + 1). So in any case we have a subdiagram ∘ ∘_{x} _{1}^{∼} and then Δ is of type He(n + 1, m). □

(5.5) Lemma. If Δ_{x} is of type X(2, m), then Δ is one of the diagrams of the theorem.

Proof. Let x be connected with m + 4. Now Δ_{I\{2,3,m+4,x}} is a rank four diagram with a 3^{7}Sp_{6}(2)-residue. By (1.3) we get that the diagram is of shape

∘ ∘ ∘ ∘^{∼}_{2}  _{3}  _{m+4}  _{x} ,

contradicting (2.4).

Let x be connected with y, 4 ≤ y < m + 4. By the same arguments as before we may assume that the subdiagram ∘ ∘_{3} _{y} has no tilde. Now Δ_{I\{3,y,m+4,x}} is of shape

∘ ∘ ∘ ∘^{∼}  ........_{m+4} _{3}  _{y}  _{x},

contradicting (1.3). Hence x is not connected to any y with 4 ≤ y ≤ m + 4. Let first m ≠ 0. Then Δ_{4} is connected and has a subdiagram of shape

[figure: ∘; ∘; ∘; ∘; ∼; m+4; 3; 2; 1] .

Hence by induction we have that Δ_{4} is of type X(2, m) and so Δ is of type X(2, m + 1).

Let now m = 0. We may assume that Δ_{1} is connected. As Δ_{1} contains a 3^{7}Sp_{6}(2)-residue, we get with (1.3) that Δ_{1} has one of the following shapes.

(1) ∘ ∘ ∘ ∘^{∼}_{x}  _{2}  _{3}  _{4} ,

(2) [figure: ∘; ∘; ∘; ∘; ∼; 4; 3; x; 2]

(3) [figure: ∘; ∘; ∘; ∘; ∼; (∼); 2; 3; x; 4]

The first case contradicts (4.8). In (2) and (3) x cannot be connected to 1 since there are no triangles in Δ by (1.1). Hence for Δ we get the following two possibilities.

(1) [figure: ∘; ∘; ∘; ∘; ∘; ∼; 1; x; 2; 4; 3]

(2) [figure: ∘; ∘; ∘; ∘; ∘; ∼; (∼); x; 4; 1; 2; 3]

The second diagram is of type X(2, 2). So assume, that we have the first. Then Δ_{4} is a D_{4}-diagram and so by [Tim1] G_{4}/K_{4} ≅ Ω_{8}^{+}(2). Then G_{x,4}/K_{x,4} ≅ A_{8}, but in the residue of x we conclude with (1.3)(v) that G_{x,4}/K_{x,4} ≅ A_{7}, a contradiction. □

(5.6) Lemma. If Δ_{x} is of type Y (n, m), then Δ is one of the diagrams in the theorem.

Proof. By (4.9) we may assume m > 1. Let x be connected with m + n + 1. Application of (4.8) to Δ_{{1,...,n−2,n+2,...,n+m}} shows n = 1 and (2.4) shows that Δ_{{3,...,m+1}} is of the shape

∘ ∘ ∘ ∘^{(∼)}^{∼}_{1}  _{2}  _{m+2} _{x}.

Let 3 ≤ y < m + 2 be not connected with x. Then Δ_{I\{2,x,y,m+2}} is of the shape

∘ ∘ ∘ ∘^{(∼)}  ^{∼}_{y}  _{2} _{m+2}  _{x},

contradicting (1.3). So we have that x is connected with all y, 3 ≤ y ≤ m+2. Further all these edges are of type ∘ ∘^{(∼)} . As we know from Δ_{{3,...,m+1}} that Δ_{{3,...,m+1,x}} is a M_{24}-residue, we have that Δ_{{3,...,m+2,x}\{y}} is a M_{24}- residue for all y with 3 ≤ y ≤ m + 2 by (1.3)(i). By (1.3)(iv) Δ_{{3,...,m+2}\{y}} is locally M(24), M(24)′, 3M(24)′ or 3M(24) and so Δ is of type M(2, m).

So assume now that the node x is not connected with n + m + 1. Let then n + 2 ≤ y < n + m + 1 be connected with x. Then we may assume that the subdiagram ∘ ∘_{n+1} _{y} has no tilde. Hence the shape of Δ_{I\{n+m+1,n+1,y,x}}

is ∘ ∘ ∘ ∘^{∼}  ........_{n+m+1} _{n+1}  _{y}  _{x}. Application of (1.3) gives a contradiction.

Hence x is not connected to any y with n+2 ≤ y ≤ n+m+1. Now Δ_{n+2} is connected and contains the subdiagram ∘ ∘ ∘^{∼} . By induction we get that Δ_{n+2} is of type Y (n, m), X(2, m−1), He(1, m−1) or Y (n+1, m−1). Hence we get that Δ is of type Y (n, m+1), X(2, m), He(1, m) or Y (n+1, m), respectively. □

(5.7) Lemma. If Δ_{x} is of type M(n, m), then Δ is one of the diagrams of the theorem.

Proof. Let y ∈ I_{m} connected with x. As Δ does not contain triangles we have that x is not connected with any element in I_{n}. Now Δ_{I\{1,t,y,x}} is of the

shape ∘ ∘ ∘ ∘^{(∼)}^{∼}_{1}  _{t}  _{y}  _{x} by (1.3) where t is the unique element connected

to 1. Let u ∈ I_{m} not connected with x, then Δ_{I\{t,y,u,x}} is of the shape

∘ ∘ ∘ ∘^{(∼)}^{∼}^{∼}_{u}  _{t}  _{y}  _{x} , contradicting (1.3). So we have that x is connected

with all elements in I_{m}. Now Δ is of type M(n + 1, m).

Now let x be not connected with any element in I_{m}. Let x be connected with t. Choose v ∈ I_{n}\{t}. Then we have for u ∈ I_{m} that Δ_{I\{t,v,u,x}} is of shape

∘ ∘ ∘ ∘^{(∼)}^{∼} ........_{x}  _{t}  _{u}  _{v} or  ∘ ∘ ∘ ∘_{x}^{t}  _{v}^{u} ..........  ^{(∼)} ........^{∼}  .

Let x not be connected with v. Then by (1.3) we have that Δ_{I\{t,v,u,x}} is

of type ∘ ∘ ∘ ∘^{(∼)}^{∼}_{x}  _{t}  _{u}  _{v} . Now Δ_{I\{1,t,u,x}} if of type

[figure: ∘; ∘; ∘; ∘; ∼; u; t; x; 1; ,]

with M_{24}-residues, which contradicts (1.3).

So we have that x is connected with all v ∈ I_{n}. By (1.3) we see that Δ_{I\{t,v,u,x}} has the shape

∘ ∘ ∘ ∘_{x}^{t}  _{v}^{u}^{(∼)}^{(∼)}  ^{∼}_{(∼)}  .

Furthermore by (1.3) we have that Δ_{I\{1,t,x,v}} is of shape ∘ ∘ ∘ ∘^{(∼)}_{1}  _{t}  _{x}  _{v}^{∼}

and that Δ is of type M(n, m + 1).

Now suppose that x is not connected with t but with some y ∈ I_{n} \ {t}. Choose u ∈ I_{m}. Then Δ_{I\{t,y,u,x}} is of shape

∘ ∘ ∘ ∘^{∼}^{(∼)} ........_{x}  _{y}  _{u}  _{t}

This is a contradiction to (1.3) and (3.5).

We are left with the case that just 1 is connected with x. Then Δ_{I\{1,t,x,u}},

u ∈ I_{m} has the shape ∘ ∘ ∘ ∘ ........  ^{∼}_{x}  _{1}  _{t}  _{u} with an M_{24}-residue ∘ ∘ ∘^{∼}_{1}  _{t}  _{u} .

Application of (1.3) shows that Δ_{I\{1,t,x,u}} is of type

∘ ∘ ∘ ∘^{∼}_{x}  _{1}  _{t}  _{u} .

Now choose v ∈ I_{n}. Then Δ_{I\{1,x,t,u,v}} is of type

∘ ∘ ∘ ∘ ∘^{∼}  ^{(∼)}_{x}  _{1}  _{t}  _{u}  _{v}

contradicting (4.8). □

Proof of the theorem. Let {P_{1}, …, P_{n}} be a minimal counterexample to the theorem with connected diagram Δ. Then by induction any subdiagram containing ∘ ∘^{∼} fulfills the theorem. So we may apply (5.3) – (5.7) to show that Δ has one of the shapes of the theorem. But now by (5.2) the corresponding statements about the groups also hold, a contradiction.

This proves the theorem. □

References

[Asch1] M. Aschbacher, GF (2)-representations of finite groups, Amer. J. of Mathematics 104, 1982, 683-771.

[Asch2] M. Aschbacher, 3-Transposition Groups, Cambridge University Press 1997,

[Coop] B. Cooperstein, An enemies list for factorization theorems, Comm. Algebra 6, 1978, 1239-1288.

[FuStr] A. Fukshansky, G. Stroth, Semiclassical parabolic systems related to M_{24}, Geom. Dedicata, to appear.

[Gil] G. Gillespie, Die Nichtexistenz einer bestimmten Klasse parabolischer Systeme mit quadratischem Diagramm, Diplomarbeit FU Berlin 1993.

[Gold] D. M. Goldschmidt, Automorphisms of trivalent graphs, Ann. of Math., 111(1980), 377-404.

[GoLySo] D. Gorenstein, R. Lyons, R. Solomon, The classification of the finite simple groups, Number III, Mathematical Service and Monographs 40.3. AMS 1998.

[Hei1] S. Heiss, Extensions for chamber systems of type A_{3} for A_{7}, J. Algebra 123, 1989, 120-150.

[Hei2] S. Heiss, On a minimal parabolic system of type M_{24}, J. Algebra 142, 1991, 188-200.

[Hu] B. Huppert, Endliche Gruppen I, Springer 1967.

[Iv] A. Ivanov, A geometric characterization of the monster, In: Groups, Combinatorics and Geometry, London Math. Soc. LN 165 (eds. M. Liebeck, J. Saxl), 1992, 46-62.

[McCl] P. McClurg, The classification of F_{1}-pairs, PhD.- thesis, Univ. of California at Santa Cruz, 1980.

[Mei1] Th. Meixner, Parabolic systems: The GF (3)-case, Trans. Amer. Math. Soc. 338, 1993, 615-637

[Mei2] Th. Meixner, Groups acting transitively on locally finite classical Tits Chamber Systems, In: Finite geometries, buildings and related topics (Pingree Park 1988), 45-65, Oxford Univ. Press 1990.

[MeiStr] U. Meierfrankfenfeld, G. Stroth, Quadratic GF (2)- modules for sporadic groups and alternating groups, Comm. Algebra 18, 1990, 2099-2140.

[Ni] R. Niles, BN-pairs and finite groups with parabolic type subgroups, J. Algebra 75, 1982, 484-492

[Par] Chr. Parker, Groups containing a subdiagram ∘ ∘ ∘^{(∼)} , Proc. London Math. Soc. 65, 1992, 85-120.

[ParRow] Chr. Parker, P. Rowley, Amalgams which involve sporadic simple groups I, II, J. Alg. 193, 1997, 441-528.

[RoStr] M. Ronan, G. Stroth, Minimal parabolic systems for the sporadic groups, Europ. J. Comb. 5, 1984, 59-91.

[Row] P. Rowley, On the minimal parabolic system related to M_{24}, J. London Math. Soc. 40, 1989, 40-56.

[Sh] S. Shpectorov, Natural representations of some tilde and Petersentype geometries, Geom. Dedicata 54, 1995, 87-102.

[ShStr] S. Shpectorov, G. Stroth, Classification of certain types of tilde geometries, Geom. Dedicata 49, 1994, 155-172.

[Sm] S. Smith, Irreducible modules and parabolic subgroups, J. Algebra 75, 1982, 286-289.

[Stein] R. Steinberg, Lectures on Chevalley groups, Lecture Notes, Yale University, 1967.

[Str1] G. Stroth, One node extensions of buildings, Geom. Dedicata 25, 1988, 71-120

[Str2] G. Stroth, Chamber systems, geometries and parabolic systems whose diagram contains only bonds of strength 1 and 2, Invent.Math. 102, 1990, 209-234

[Str3] G. Stroth, The nonexistence of certain Tits geometries with affine diagram, Geom. Dedicata 28, 1988, 277-319

[Str4] G. Stroth, Geometries of type M related to A_{7}, Geom. Dedicata 20, 1986, 265-293.

[Str5] G. Stroth, A local classification of finite classical Tits geometries of characteristic ≠ 3, Geom. Dedicata 28, 1988, 93-105.

[Str6] G. Stroth, Strong quadratic modules, Israel J. Math. 79, 1992, 257-279.

[StrWo] G. Stroth, S.K. Wong, Minimal parabolic systems and small representations, Comm. Algebra 15, 1987, 751-790.

[Tim1] F. Timmesfeld, Tits geometries and parabolic systems in finitely generated groups, I, II, Math. Z. 184, 1983, 377-396, 449-487.

[Tim2] F. Timmesfeld, Locally finite classical Tits chamber systems of large order, Invent. Math. 87, 1987, 603-641.

[Tim3] F. Timmesfeld, Classical locally finite Tits chamber systems of rank 3, J. Algebra 124, 1989, 9-59.

[Wie1] C. Wiedorn, A Tilde Geometry related to F_{4}, J. Alg. to appear.

[Wie2] C. Wiedorn, Some semiclassical parabolic systems of rank 4, J. Alg. to appear.

[Wie3] C. Wiedorn, A Tilde Geometry for F_{4}(2), Beiträge zur Alg. und Geom., 38, 1997, 337-342.

[Wie4] C. Wiedorn, An F_{1}–Module for the 3U_{4}(3)–Amalgam, J. Alg. 195, 1997, 337-356. preprint.