Semiclassical parabolic systems related to M_{24}

A. Fukshansky^{∗}and G. Stroth^{∗} Martin-Luther-Universität Halle-Wittenberg Fachbereich Mathematik und Informatik Institut für Algebra und Geometrie 06099 Halle, Germany

April 9, 1997

There is a long history of parabolic systems generalizing the buildings associatied to finite groups of Lie type. One of the main results until now might be the classification of so called parabolic systems of classical type in [Ni, Tim1, Tim2, Tim3, Str1, Str2, Str3, Str4, Str5, Hei1, Mei1]. See [Mei2] for a survey. These parabolic systems mainly are related to groups of Lie type again. There is a second class of parabolic systems which has a large variety of sporadic examples. These are called parabolic systems of semiclassical type (see [Str6]). In this paper we give one important step in the classification of semiclassical parabolic systems. Let us start with the definition.

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

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

(2) For B = ⋂_{i=1}^{n} P_{i} we have P_{i}/B_{P_{i}} is a rank one group of Lie type in characteristic two and B = P_{i} ∩ P_{j} for each pair P_{i}, P_{j}.

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

(α) P_{ij} is a rank two Lie group in characteristic two, (β) P_{ij} = P_{i}P_{j}, (γ) P_{ij} is the triple cover of Σ_{6}.

(4) B_{G} = 1

If in (3) the case (γ) never occurs the system is called of classical type. In this paper we assume that the case (3)(γ) occurs for at least one choice i, j. Then the system is called semiclassical.

To each parabolic system we can associate a diagram Δ in the following way. The nodes of the diagram correspond to the groups P_{1}, ..., P_{n}, denoted by {1, ..., n}. For i, j we connect the nodes i and j by the corresponding Dynkin

∗This work was supported by DFG, Str 203/10-1

diagram if we have (3)(α) and by ∘  ∘^{∼}  if we have (3)(γ). If we will allow Σ_{6}

and 3Σ_{6} at the same place we also write ∘ ∘^{( ∼)} . In the case (3)(β) the nodes

are not connected at all and by ∘ ∘ ......... we denote any possible connection of the given kinds.

Since non-connected diagrams correspond to the product of the groups belonging to the connected parts we may assume

(5) Δ is connected.

According to [Tim3] the semiclassical diagrams of rank three (i.e. n = 3)

are of the shape ∘  ∘  ∘^{∼} , ∘  ∘  ∘^{∼}  ^{∼} , ∘  ∘  ∘^{∼} . The dia-
gram ∘  ∘  ∘^{∼}  is very interesting because it belongs to sporadic groups.

According to [Hei2] and [Row] a semiclassical parabolic system with diagram

∘  ∘  ∘^{∼}  has as group G one of the following : He, M_{24} or 3^{7}Sp_{6}(2). In

the first two cases |B| = 2^{10} where |B| = 2^{9} in the third. Furthermore in all rank three cases above we have P_{i}/B_{P_{i}} ≅ Σ_{3}.

So for this paper we make the following additional assumption

(6) P_{i}/B_{P_{i}} ≅ Σ_{3}

Then an easy exercise shows B_{P_{i}} = O_{2}(P_{i}) (see [StrWo]). The aim of this paper is to finish the classificaion of this type of semiclassical systems of rank four. The general case will appear in a forthcoming paper.

So from now on we have n = 4, i.e. G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩. For i ∈ {1, 2, 3, 4} set G_{i} = ⟨P_{j}|j ≠ i⟩, and K_{i} = B_{G_{i}}. Then G_{i}/K_{i} is a group which belongs to the corresponding rank three diagram. There is a natural subdivision concerning

the rank three diagram of type ∘  ∘  ∘^{∼} . In [Wie2] the case was treated

that for any such subdiagram we have that G_{i}/K_{i} ≅ 3^{7}Sp_{6}(2). Hence we will assume that G_{i}/K_{i} ≅ He or M_{24} occurs at least once. So we will choose the numbering such that

(7) G_{4}/K_{4} ≅ He or M_{24}.

These parabolic systems we call parabolic systems of type M_{24} of rank four. There are two known diagrams with corresponding groups and parabolic

systems of type M_{24} : First  ∘  ∘  ∘  ∘^{∼}  which represents the
group Co_{1} and second ∘  ∘  ∘  ∘^{∼}  ^{(∼)}  the corresponding group being

one of M(24), M(24)′, 3M(24) or 3M(24)′ . In this paper we will prove Theorem 1 Let {P_{1}, P_{2}, P_{3}, P_{4}} be a parabolic system of rank four and of type M_{24}, and Δ be the corresponding diagram. 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)′, and there is a normal subgroup K of G such that G/K ≅ M(24)′or M(24).

In fact in (iv) we cannot say much more as the universal group G is always infinite.

Combining our result with results due to G. Gillespie [Gil] and C. Wiedorn [Wie1], [Wie2] we have

Theorem 2 Let G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩, {P_{1}, P_{2}, P_{3}, P_{4}} be a semiclassical para-

bolic system with diagram Δ and a subdiagram of type ∘  ∘  ∘^{∼}  . Then

one of the following holds

(i) {P_{1}, P_{2}, P_{3}, P_{4}} is of M_{24}-type.

(ii) 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).

(iii) 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 lemma 1.7 (iv) (⋆).

(iv) Δ : ∘  ∘  ∘  ∘^{∼}  , and there is a normal subgroup K of
G with G/K ≅ F_{4}(2).
(v) Δ : ∘  ∘  ∘  ∘^{∼} , G ≅ 3^{35}Sp_{8}(2).

In fact in case (iv) there is a nonsplit extension of an elementary abelian group of order 3^{833} by F_{4}(2) which belongs to this diagram [Wie3]. It is an open question whether G is isomorphic to this group. It is true if K is abelian [Wie4].

A few words on the proof might be helpful. The first step is to show that the diagrams are exactly those of the theorem. Furthermore, which will be done at the same time, we show that the K_{i} and G_{i}/K_{i} are as in the theorem. For this there is a well established method, the amalgam method. This relates the diagram with certain small modules (so called F_{1}–modules) for the groups G_{i}/K_{i}. The interplay of these modules then provides us with the results mentioned before. In our special case a great part of the work was already done by Chr. Parker [Par1], [Par2], who described the structure of K_{4}. After this we basically

are left with the diagrams ∘  ∘  ∘  ∘^{∼}  and ∘  ∘  ∘  ∘^{∼}  ^{(∼)} .

Here we are going to give the exact structure of G_{i}, i = 1, 2, 3, 4, i.e. the multiplication table. For this we start with the parabolic G_{4} given by Parker’s paper and set up generators and relations for this group. Those who are used to work with generators and relations will realize that our systems are made up by very special principles. Mostly one is interested in systems of generators and relations which use not too many of them. This is not our aim. In fact we use much to many generators. The point is that our set of generators and relations at the end should represent the geometry and one should be able to work in this geometry. So for the group G_{4} for example we not just like to define this group. At the same time certain subsets of generators and relations should define the intersections of the other parabolics G_{i} with G_{4}. So for the group B, a group of order 2^{21}, we use 21 generators, where maybe 6 would suffice. But for our purpose it is much better to work with 21 generators. On the other hand to handle such a great set of generators with a computer becomes difficult. So we have to use short relations to overcome this problem. After we have G_{4} in this way, we see G_{i} ∩ G_{4} for i ≠ 4, and then we add one further generator a_{i} for each i to get G_{i} = ⟨G_{i} ∩ G_{4}, a_{i}⟩ and the necessary relations. We show that this all can be done in a unique way, which then shows that the amalgam is uniquely determined. Using the Todd-Coxeter algorithm in case of

∘  ∘  ∘  ∘^{∼} , we see that our generators and relations define Co_{1}. In
case of ∘  ∘  ∘  ∘^{∼}  ^{(∼)}  this is not true, as we can show using some

arguments due to A. Pasini.

Part of this work was done when the authors visited A. Pasini in 1996. We like to thank the university of Siena and CNR for their support and hospitality.

We also like to thank A. Stein, who did many of the computer calculations for this paper.

1 Preliminaries

In this chapter we investigate the rank three case and prove some results on modules which will be of help later on. We use notations and assumptions as introduced before. For the proof of the main theorem we will distinguish two cases whether the following assumption is valid or not:

(ASS) C_{G_{4}}(O_{2}(G_{4})) ≤ O_{2}(G_{4}).

If (ASS) is true, then we can use the results of Ch. Parker [Par1], [Par2] and know that G_{4}/K_{4} ≅ He is not possible, so G_{4}/K_{4} ≅ M_{24} and we also know the structure of K_{4}, which is elementary abelian of order 2^{11} or 2^{12} and is the Golay code– , the extended Golay code– or the Todd–module.

Definition: Let G be a group and V a faithful GF (2)G-module. If there exists a nontrivial elementary abelian 2-group A with |V : C_{V} (A)| ≤ |A| then V is called an F -module. If |V : C_{V} (A)| ≤ 2|A| is true, V is called an F_{1}-module. In both cases A is called an offending subgroup.

Lemma 1.1 M_{24} does not possess an F-module.

Proof: See [As]. □

Lemma 1.2 Assume (ASS). If X_{i} ≤ K_{i}, X_{i} ≠ K_{4}, i = 1, 2 or 3, X_{i} elementary abelian. Then |X_{i}| < |K_{4}|.

Proof: Otherwise |X_{i} : X_{i} ∩ K_{4}| ≥ |K_{4} : X_{i} ∩ K_{4}| and so |X_{i}K_{4}/K_{4}| ≥ |K_{4} : C_{K_{4}}(X_{i})|. But this contradicts lemma 1.1. □

Lemma 1.3 Assume (ASS). Then {P_{i}, P_{j}, P_{k}} does not belong to the A_{7} - ge-

ometry with diagram i  j  k ∘  ∘  ∘ or i  j  k ∘  ∘  ∘ .

Proof: We have ⟨P_{i}, P_{k}⟩/O_{2}(⟨P_{i}, P_{k}⟩) ≅ (Z_{3}×Z_{3})·2 and in the first system we have ⟨P_{j}, P_{k}⟩/O_{2}(⟨P_{j}, P_{k}⟩) ≅ A_{6}. This shows that the possible diagrams are

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

4  2  1  3 ∘  ∘  ∘  ∘^{∼}  .

We have chosen notation such that G_{4}/K_{4} ≅ M_{24} and G_{3}/K_{3} ≅ A_{7} . So we have K_{4} ≤ K_{3}. As K_{4} ≮G_{3} , we get some x ∈ G_{3} with K_{4} ≠ K_{4}^{x} ≤ K_{3} . But this contradicts lemma 1.2. □

Lemma 1.4 Let V be one of the following modules for M_{24} : the Golay code of order 2^{11}, the extended Golay code of order 2^{12} or the dual of the Golay code (the Todd code) of order 2^{11}. Let S ∈ Syl_{2}(M_{24}). Then |C_{V} (S)| = 2. If V contains the Golay code then C_{M_{24}}(C_{V} (S)) = 2^{6}3Σ_{6} while C_{M_{24}}(C_{V} (S)) = 2^{4} : A_{8} for the Todd–module V. In any case if t ∈ M_{24} is an involution then |V : C_{V} (t)| ≥ 2^{4}.

Proof : See [Par1], [Par2]. □

For the next lemma choose the numbering in the M_{24}-subdiagram as follows:

∘  ∘  ∘^{∼} 1  2  3

Lemma 1.5 Assume (ASS). Then G_{3}/K_{3} ≅ (3^{7})Sp_{6}(2) and also G_{1}/K_{1} ≅ (3^{7})Sp_{6}(2).

Proof: Assume first G_{1}/K_{1} ≅ (3^{7})Sp_{6}(2). By [Par1], [Par2] we know |K_{4}| ≤ 2^{12}. As |S : K_{4}| = 2^{10} and |S : K_{1}| = 2^{9}, we see |K_{1}| ≤ 2^{13}. Let ρ be an element of order three such that N_{⟨P_{2},P_{3}⟩}(⟨ρ⟩) involves Σ_{6}. Then [ρ, K_{1}] ≠ 1 and then [O_{3}(G_{1}/K_{1}), K_{1}] ≠ 1 but that would force |K_{1}| ≥ 2^{14}, a contradiction. So we have G_{3}/K_{3} ≅ (3^{7})Sp_{6}(2). By lemma 1.2 we know that the Frattini subgroup Φ(K_{3}) is not trivial. Suppose first Φ(K_{3}) ⊆ K_{4}. Then [K_{4}, Φ(K_{3})] = 1 and so [Φ(K_{3}), ⟨K_{4}^{G_{3}}⟩] = 1. As K_{4} ≰ K_{3} and K_{3}K_{4}/K_{4} is abelian, we see |K_{3}K_{4}/K_{4}| = 16. Further ⟨P_{1}, P_{2}⟩ = K_{3}C_{⟨P_{1},P_{2}⟩}(Φ(K_{3})). As we know that C_{⟨P_{1},P_{2}⟩}(Φ(K_{3}))/K_{4} ≤ ⟨P_{1}, P_{2}⟩/K_{4}, we see [⟨P_{1}, P_{2}⟩, Φ(K_{3})] = 1. Hence we get [K_{3}, Φ(K_{3})] = 1 and so Φ(K_{3}) ⊆ Z(G_{3}). As |Ω_{1}(Z(S))| = 2, we see |Φ(K_{3})| = 2. Further C_{G_{4}}(Ω_{1}(Z(S))) involves L_{3}(2) and so by lemma 1.4 |K_{4}| = 2^{11} and |K_{3}| = 2^{12}. As |K_{3}K_{4} : K_{4}| = 2^{4} we get |K_{3} ∩ K_{4}| = 2^{8} and Z(K_{3}) ⊆ K_{3} ∩ K_{4}. So [G_{3}, Z(K_{3})] = 1. Hence we have Z(K_{3}) = K′_{3} = Φ(K_{3}) and then K_{3} is extraspecial, a contradiction. So we are left with Φ(K_{3}) ⊈ K_{4}. As K_{3}K_{4}/K_{4} is nonabelian we have |Φ(K_{3})K_{4}/K_{4}| = 2. Then also |Φ(K_{3}) : Φ(K_{3}) ∩ K_{4}| = 2. As [Φ(K_{3}), K_{4}] ≠ 1, we see that K_{4} induces transvections on Φ(K_{3}) to the hyperplane Φ(K_{3}) ∩ K_{4}. But then |K_{4} : K_{4} ∩ K_{3}| = 2, a contradiction. □

Lemma 1.6 Assume (ASS). For all i ∈ {1, ..., 4} are equivalent:

(i) Ω_{1}(Z(S)) ≮G_{i}

(ii) Z_{i} is an F_{1}-module

Proof: This is [Wie1, (2.1),(2.2)].□

The next lemma gives some information about rank three diagrams. The proofs are e.g. in [Hei2], [Tim3].

Lemma 1.7 Let G = ⟨P_{1}, P_{2}, P_{3}⟩ be a parabolic system of classical or semiclassical type with diagram Δ, such that P_{i}/O_{2}(P_{i}) ≅ Σ_{3} for all i. Then we have:

(i) If Δ = ∘  ∘  ∘ , then G ≅ A_{7} or A_{8}.
(ii) If Δ = ∘  ∘  ∘^{∼} , then G ≅ M_{24}, He or 3^{7}Sp_{6}(2)
(iii) If Δ = ∘  ∘  ∘ , then G ≅ A_{7} or Sp_{6}(2).
(iv) If Δ = 1  2  3 ∘  ∘  ∘^{(∼)}  ^{(∼)} , then there are two possibilities:
(⋆) G_{i}/K_{i} ≅ A_{6}, Σ_{6}, 3A_{6} or 3Σ_{6}, (i ∈ {1, 3}), 2^{4} ≤ |K_{i}| ≤ 2^{5}, K_{i}
is an undecomposable 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} ≅ (Z_{3} × Z_{3})Z_{2}.

If we have the situation of lemma 1.7(iv)(⋆), we will call such a parabolic system a system of type 3U_{4}(3) or U_{4}(3) according to G_{i}/K_{i} involves 3A_{6} for at least one i or not.

The following lemma is well known. For convenience of the reader we will give a proof.

Lemma 1.8 Let G ≅ Ω^{+}(8, 2) and V be a 9-dimensional GF (2)G-module such that [V, G] is the natural module. Then [V, G] is a direct summand.

Proof: Let P ≤ G be the parabolic fixing a singular vector v_{1} ∈ [V, G]. Then P is an extension of E, an elementary abelian group of order 2^{6}, by Ω^{+}(6, 2). Now we have V_{1} ⊆ V such that V_{1}/⟨v_{1}⟩ is the orthogonal module for P/E and [E, V_{1}] = ⟨v_{1}⟩. As |V/V_{1}| = 4, we see that [P, V] = V_{1}. Let S ∈ Syl_{2}(P). Then |C_{V_{1}/⟨v_{1}⟩}(S)| = 2 and so C_{V/⟨v_{1}⟩}(E) > V_{1}/⟨v_{1}⟩. Let V_{2} be the preimage of C_{V/⟨v_{1}⟩}(E). Then for e ∈ E^{♯} we have |V_{2} : C_{V_{2}}(e)| = 2 and so |C_{V_{2}}(E)| ≥ 4. As C_{V_{1}}(E) = ⟨v_{1}⟩, we get |C_{V_{2}}(E)| = 4. As Ω^{+}(6, 2) is perfect, we see [C_{V_{2}}(E), P] = 1. So we get that S centralizes some vector v ∈ V \ [V, G]. Now by Gaschütz [Hu, (I.17.4)] [V, G] is a direct summand. □

2 Diagrams of rank four

In this chapter we are going to prove the following

Proposition 2.1 Let Γ be a parabolic system of M_{24}-type of rank four. Then one of the following holds :

(i) Δ is a string,

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

We establish proposition 2.1 in a series of lemmas. For the rest of this chaper we are working under the assumptions of proposition 2.1.

Lemma 2.1 Suppose the diagram Δ of Γ is a star, then (ASS) does not hold.

Proof: Suppose (ASS) holds. Then due to [Tim3] we have two possible diagrams

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

Now in the first case we see with lemma 1.7, [Par1] and [Par2] that G_{3}/K_{3} ≅ M_{24}, (3^{7})Sp_{6}(2) or A_{7}. In the first case K_{3} is elementary abelian with |K_{3}| = |K_{4}|. But this contradicts lemma 1.2. The latter two cases contradict lemma 1.5 and lemma 1.3. In the second diagram we see with lemma 1.7, [Par1] and [Par2] that G_{1}/K_{1} ≅ M_{24} or 3^{7}Sp_{6}(2) so we get a contradicion with lemma 1.2 and lemma 1.5 again.

Lemma 2.2 If Δ is a star then Δ is of shape [figure: ∘; ∘; ∘; ∘; ∼; ∼; 1; 2; 4; 3] , where G_{4} ≅ M_{24} or He and G_{3} ≅ M_{24} or He.

Proof: By lemma 2.1 we have G_{4} = K_{4}C_{G_{4}}(K_{4}). There are again two possible diagrams:

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

As O_{2}(⟨P_{2}, P_{3}⟩) involves exactly one nontrivial 3Σ_{6}-module which is 6-di-mensional, we see that G_{1}/K_{1} ≅ M_{24}, He in case (1). As Z(⟨P_{2}, P_{3}⟩/K_{1}) = 1, we get K_{4} ≤ K_{1}. Now Ω_{1}(Z(S)) ≤ ⟨P_{1}, P_{2}⟩ and Ω_{1}(Z(S)) ≤ ⟨P_{2}, P_{4}⟩, so Ω_{1}(Z(S)) ≤ G_{3}. By lemma 1.7 we have G_{3}/K_{3} ≅ A_{7} or A_{8}. Now in the latter case we get G_{3} = C_{G_{3}}(K_{3})K_{3}, which contradicts the action of ⟨P_{1}, P_{2}⟩ on K_{3}. So we have G_{3}/K_{3} ≅ A_{7}. Now K_{3}/Z(K_{3}) is the natural A_{7}–module. But ⟨P_{1}, P_{2}⟩ induces orbits of length 7, 7, 21, 28 on K_{3}/Z(K_{3}) and the orbits of length 7 generate a subgroup of order 8 in K_{3}/Z(K_{3}), while in the action of A_{7} on the natural module there is just one orbit of length 7 generating the whole module, a contradiction. Suppose we have the second diagram. Let G_{3}/K_{3} ≅ A_{7}. Then we obtain ⟨P_{2}, P_{4}⟩/O_{2}(⟨P_{2}, P_{4}⟩) ≅ A_{6} . By [Tim3] we get that G_{1}/K_{1} is parabolic of type U_{4}(3). But then O_{2}(⟨P_{2}, P_{3}⟩) involves a 4-dimensional Σ_{6}-module, a contradiction. Hence G_{3}/K_{3} ≅ A̸_{7}. As ⟨P_{1}, P_{2}⟩ induces a direct sum of two L_{3}(2)-modules on O_{2}(⟨P_{1}, P_{2}⟩/C_{O_{2}(⟨P_{1},P_{2}⟩)}(O^{2}(⟨P_{1}, P_{2}⟩)), we see that G_{3}/K_{3} ≅ (3^{7})Sp_{6}(2). So G_{3}/K_{3} ≅ M_{24} or He. Now K_{4}K_{3}/K_{3} ≤ Z(⟨P_{1}, P_{2}⟩/K_{3}) and then |K_{3}K_{4}/K_{3}| ≤ 2. As K_{4} ∩ K_{3} < ⟨G_{4}, G_{3}⟩ = G, we see K_{4} ∩ K_{3} = 1. Hence |K_{4}| = |K_{3}| ≤ 2. Suppose |K_{4}| = |K_{3}| = 2. From G_{4}/K_{4} ≅ M_{24} we see that |O_{2}(⟨P_{1}, P_{3}⟩)| = 2^{9} and O_{2}(⟨P_{1}, P_{3}⟩)/K_{4} possesses no subgroup of index two normal in ⟨P_{1}, P_{3}⟩. As K_{4} ≤ O_{2}(⟨P_{1}, P_{4}⟩) too, we see that K_{4} ≤ K_{2}. Hence |K_{2}| = 2^{9} and |Z(K_{2})| = 2^{3}. Furthermore G_{2}/K_{2} = (Z_{3} × Z_{3} × Z_{3})V_{4}. As 3^{2} |̸ |GL(3, 2)| we see that 3^{2}| |C_{G_{2}}(Z(K_{2}))|. Hence ⟨P_{1}, P_{3}⟩ acts in the same way on Z(K_{2}) as ⟨P_{1}, P_{4}⟩. But C_{Z(K_{2})}(⟨P_{1}, P_{3}⟩) = K_{4} while C_{Z(K_{2})}(⟨P_{1}, P_{4}⟩) = K_{3} and so K_{4} = K_{3}, a contradiction. So we have K_{3} = K_{4} = 1 and the lemma is proved. □

Lemma 2.3 If Δ is a square, then (ASS) does not hold.

Proof: Suppose false. Then we get with [Tim3] that Δ has one of the following shapes:

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

In case (1) we have G_{3}/K_{3} ≅ M_{24} or 3^{7}Sp_{6}(2) by lemma 1.7, [Par1] and [Par2]. But this contradicts lemma 1.2 and lemma 1.5. In case (2) we have G_{1}/K_{1} ≅ M_{24}, (3^{7})Sp_{6}(2) or A_{7} by lemma 1.7 and [Par1, Par2]. Again the lemmas 1.2, 1.5 and 1.3 yield a contradiction. So we have that Δ is of type (3). Application of lemmas 1.2 and 1.3 yields G_{2}/K_{2} ≅ (3^{7})Sp_{6}(2). Furthermore by lemma 1.3 G_{1}/K_{1} ≅ A_{8}. By lemma 1.2 K_{4} ⊈ K_{1} and so |K_{4}K_{1}/K_{1}| = 2^{3} and then |K_{4}K_{1}/K_{4}| = 2^{7} . If t ∈ Z(K_{1}), then |K_{4} : C_{K_{4}}(t)| ≤ 2^{3}, as |K_{4} : K_{4} ∩K_{1}| = 8. By lemma 1.4 we get [t, K_{4}] = 1, i.e. [Z(K_{1}), K_{4}] = 1 and so [Z(K_{1}), K_{1}⟨K_{4}^{G_{1}}⟩] = 1, i.e. Z(K_{1}) ≤ Z(G_{1}) and then Ω_{1}(Z(K_{1})) = Ω_{1}(Z(S)). This shows Ω_{1}(Z(K_{2})) ≠ Ω_{1}(Z(S)) and then |K_{4} : K_{4} ∩ K_{2}| ≥ 2^{4} by lemma 1.4 again. But O_{2}(⟨P_{1}, P_{3}⟩)/K_{2} ≅ 2^{1+4} × V_{4}, where O_{2}(⟨P_{1}, P_{3}⟩/K_{2})/Z(O_{2}(⟨P_{1}, P_{3}⟩)/K_{2}) is an irreducible ⟨P_{1}, P_{3}⟩ - module. As K_{4} is abelian, this shows |K_{4} : K_{4} ∩ K_{2}| ≤ 2^{3}, a contradiction. □

Lemma 2.4 Δ is never a square.

Proof: By lemma 2.3 we may assume G_{4} = K_{4}C_{G_{4}}(K_{4}). Let Δ be ∘  ∘ ∘  ∘ 4  3 1  2 ......... ^{∼} ..........
Let first  ......... ∘  ∘ 4  3 = ∘  ∘ 4  3 . Then by lemma 1.7 G_{1}/K_{1} ≅ M_{24},

He or 3^{7}Sp_{6}(2). As O_{2}(⟨P_{2}, P_{3}⟩) involves exactly one nontrivial 3Σ_{6}-module, the 6-dimensional one, we get G_{1}/K_{1} ≅ M_{24} or He. Hence we get G_{1} = K_{1}C_{G_{1}}(K_{1}) by lemma 2.3. As Z(⟨P_{2}, P_{3}⟩)/K_{1}) = 1, we see K_{1} = K_{4} = 1. But now O_{2}(⟨P_{1}, P_{2}⟩) is the only extraspecial subgroup of order 2^{7} in S, the same applies for O_{2}(⟨P_{3}, P_{4}⟩). Hence O_{2}(⟨P_{1}, P_{2}⟩) < ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩ = G, a contradiction.

So we have  ......... ∘  ∘ 4  3 = ∘  ∘ 4  3^{(∼)}  by [Tim3] applied to G_{1}. Let

G_{3}/K_{3} ≅ He or M_{24}. Then by lemma 2.3 we also get G_{3} = K_{3}C_{G_{3}}(K_{3}). As both O_{2}(⟨P_{2}, P_{3}⟩) and O_{2}(⟨P_{1}, P_{4}⟩) involve just one nontrivial 3Σ_{6}-module, the 6-dimensional one, we see with lemma 1.7 that both G_{1}/K_{1} and G_{2}/K_{2} are not the 3U_{4}(3) or U_{4}(3) amalgam. Hence we have K_{1} = O_{2}(⟨P_{2}, P_{3}⟩) = O_{2}(⟨P_{3}, P_{4}⟩) = O_{2}(⟨P_{1}, P_{4}⟩) = K_{2}. But then K_{1} = 1, a contradiction as |S| ≥ 2^{10}.

Let G_{3}/K_{3} ≅ (3^{7})Sp_{6}(2). Then O_{2}(⟨P_{1}, P_{2}⟩)/K_{3} is an indecomposable L_{2}(7)-module, while O_{2}(⟨P_{1}, P_{2}⟩)/K_{4} involves a direct sum of two L_{2}(7)–modules as K_{4} just involves trivial L_{2}(7)–modules, a contradiction.

Let G_{3}/K_{3} ≅ A_{7}. Then O_{2}(⟨P_{1}, P_{2}⟩) = K_{4}K_{3}. Now |K_{4}Z(K_{3}) : K_{4}| ≤ 2 and so [⟨P_{1}, P_{2}⟩^{(∞)}, Z(K_{3})] = 1. Hence [G_{3}, Z(K_{3})] = 1 as G_{3}/K_{3} is simple. Now we see that Z(K_{3})∩K_{4} is normalized by ⟨G_{3}, G_{4}⟩ = G, and so Z(K_{3})∩K_{4} = 1. As C_{⟨P_{1},P_{2}⟩}(O_{2}(⟨P_{1}, P_{2}⟩)) ≤ O_{2}(⟨P_{1}, P_{2}⟩), we have C_{G_{3}}(K_{3}) ≤ K_{3} and so Ω_{1}(Z(S)) ≤ Z(K_{3}). As Z(K_{3}) ∩ K_{4} = 1, we get K_{4} = 1. So K_{3} ≅ 2^{1+3+3}. But the action of A_{7} on the extraspecial group does not fit with the action of L_{3}(2) in M_{24} or He.

So we have by [Tim3] G_{3}/K_{3} ≅ A_{8}. We have K_{4} ⊆ K_{3} as Z(⟨P_{1}, P_{2}⟩/K_{3}) = 1. Hence K_{3}/K_{4} is elementary abelian. As K_{3} involves exactly one nontrivial irreducible ⟨P_{1}, P_{2}⟩/O_{2}(⟨P_{1}, P_{2}⟩)-module, the natural one, we see that K_{3} involves exactly one natural L_{4}(2)-module V. As G_{1}/K_{1} is not of type U_{4}(3), we get with [Tim3] that ⟨P_{3}, P_{4}⟩/O_{2}(⟨P_{3}, P_{4}⟩) ≅ Σ_{6} or 3Σ_{6}. Hence G_{2}/K_{2} ≅ A̸_{7}.

Assume G_{2}/K_{2} ≅ M_{24} or He. Then by lemma 2.3 G_{2} = K_{2}C_{G_{2}}(K_{2}) and |K_{2}| = |K_{4}|. As Z(⟨P_{1}, P_{3}⟩/K_{4}) = 1, we see that K_{2} ≤ K_{4} and so K_{2} = K_{4} = 1. As S/K_{2} contains exactly one extraspecial subgroup of order 2^{7}, we see that O_{2}(⟨P_{1}, P_{2}⟩) = O_{2}(⟨P_{1}, P_{4}⟩) < ⟨P_{1}, P_{2}, P_{4}⟩ = G_{3}, contradicting |S : K_{3}| = 2^{6}.

So we have G_{2}/K_{2} ≅ (3^{7})Sp_{6}(2). As |K_{2}| ≤ 4, we have K_{2} ⊆ Z(G_{2}). As |Z(S)| = |K_{1}| · 2 = |K_{2}|, we see that K_{2} = Z(S) and so K_{1} = 1, |K_{2}| = 2. Hence G_{2}/O_{3}(G_{2}) is a covering group of Sp_{6}(2). Now K_{3} is the natural module for L_{4}(2). In Sp_{6}(2) we see that S does not split over K_{3}. So by Gaschütz G_{3} does not split over K_{3}. From M_{24} or He, we see, that S splits over K_{3}, and so by Gaschütz G_{3} splits over K_{3}, a contradiction. □

Now lemmas 2.1 to 2.4 prove proposition 2.1. □

3 String diagrams

In this chapter we will prove the following

Proposition 3.1 Let {P_{1}, P_{2}, P_{3}, P_{4}} be a parabolic system of M_{24}–type. If the diagram Δ is a string then one of the following holds:

(1) Δ : 4  3  2  1 ∘  ∘  ∘  ∘^{∼}

(2) Δ : 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  ^{(∼)}

(3) Δ : 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  ^{∼}  , and G_{1} ≅ G_{4} ≅ He.

In the cases (1), (2) (ASS) is true but not in case (3).

We will prove proposition 3.1 in two lemmas 3.1 and 3.2.

Lemma 3.1 If Δ is a string diagram, then (ASS) holds or G_{1} ≅ G_{4} ≅ He

and Δ is of type 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  ^{∼}

Proof: Assume that (ASS) does not hold. By [Tim3] Δ has one of the following shapes:

(1) 4  3  2  1 ∘  ∘  ∘  ∘^{∼}

(2) 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  ^{(∼)}

(3) 4  3  2  1 ∘  ∘  ∘  ∘^{(∼)}  ^{∼}

(4) 4  3  2  1 ∘  ∘  ∘  ∘^{∼}

(5) 4  3  2  1 ∘  ∘  ∘  ∘^{∼}  .

Suppose, we have (1), (3), or (4). Then C_{K_{3}}(⟨P_{1}, P_{2}⟩′) ⊆ K_{4}. As ⟨P_{1}, P_{2}⟩′ ≤ G_{3}, we have C_{K_{3}}(⟨P_{1}, P_{2}⟩′)≤G_{3}. As G_{4} = K_{4}C_{G_{4}}(K_{4}), we see that C_{K_{3}}(⟨P_{1}, P_{2}⟩′) = 1. As G_{3}/K_{3} ≅ Σ_{3} × 3Σ_{6} or (Z_{3} × 3A_{6}) · 2, we get |K_{4}| ≤ 2. Now in G_{1}/K_{1} we see C_{P_{4}}(O_{2}(P_{4})) ⊆ O_{2}(P_{4}). This shows that G_{3}/K_{3} ≅ (Z_{3} × 3A_{6}) · 2 and so K_{4} = 1. In particular |S| = 2^{10}. Then choose ρ ∈ ⟨P_{1}, P_{2}⟩ with ρO_{2}(⟨P_{1}, P_{2}⟩) ∈ O_{3}(⟨P_{1}, P_{2}⟩/O_{2}(⟨P_{1}, P_{2}⟩)) and o(ρ) = 3. Set P_{5} = ⟨ρ, S⟩. Inside of M_{24} or He one calculates that ⟨P_{2}, P_{3}, P_{5}⟩ ≅ 2^{4}A_{8} or He, respectively, where ⟨P_{3}, P_{5}⟩ ≅ 2^{6}3Σ_{6} in the latter. Hence if we are in (1) then we see that

W := ⟨P_{5}, P_{2}, P_{3}, P_{4}⟩ is a parabolic system with the diagram [figure: ∘; ∘; ∘; ∘; 4; 3; 2; 5]

or [figure: ∘; ∘; ∘; ∘; ∼; 4; 3; 2; 5] in case of G_{4} ≅ He. By proposition 2.1 we have a

contradiction in the latter. Hence W has a factor Ω^{+}(8, 2) by [Tim2]. But then |S| ≥ 2^{12}, a contradiction. If we are in (3) and G_{4} ≅ M_{24}, we get a dia-

gram of type [figure: ∘; ∘; ∘; ∘; (∼); 4; 3; 2; 5] , where W_{4}/O_{2}(W_{4}) ≅ A_{8}.

But then by proposition 2.1, [Str3] or [Wie1] we have P_{5} ⊆ ⟨P_{4}, P_{3}, P_{2}⟩. So we see that G_{1} contains a subgroup 2^{4} : A_{8}. Hence we have G_{1} ≅ M_{24} as He does not contain 2^{4} : A_{8}. Now W_{5} ≅ W_{2} ≅ M_{24}. But in W_{2} ≅ M_{24} we have W_{4,2} ≅ 2^{1+6}L_{3}(2) ≅ W_{5,4} in W_{5} ≅ M_{24}. So W_{5,4} = W_{4,2}. Then W_{4} = ⟨W_{5,4}, W_{4,2}⟩ ≅ 2^{1+6}L_{3}(2), a contradiction.

If we are in (3) and G_{4} ≅ He we get a diagram of type [figure: ∘; ∘; ∘; ∘; ∼; (∼); 5; 3; 4; 2] .

If P_{5} ≤ ⟨P_{2}, P_{3}, P_{4}⟩, we have G_{1} ≅ He. If P_{5} ≰ ⟨P_{2}, P_{3}, P_{4}⟩, then by proposition 2.1 we see that G_{1} ≅ M_{24} or He. If G_{1} ≅ M_{24}, we may interchange 1 and 4 and argue as before to achieve a contradiction. So we have G_{1} ≅ He ≅ G_{4}.

Let Δ be of type (4). As |S| = 2^{10} and ⟨P_{3}, P_{4}⟩ involves U_{3}(3), we see that 2^{4} ≤ |K_{1}| ≤ 2^{5}. Now have that G_{1} = K_{1}C_{G_{1}}(K_{1}). In G_{1}/K_{1} we see that ⟨P_{2}, P_{3}⟩/K_{1} is an extension of an elementary abelian group of order eight by L_{3}(2). Hence O_{2}(⟨P_{2}, P_{3}⟩) involves exactly one nontrivial L_{3}(2)–module. But from G_{4}, we see that there are exactly two of them, a contradiction.

Assume next that Δ is of type (5). Then G_{1}/K_{1} ≅ He or M_{24} too, as 3^{7}Sp_{6}(2) does not involve 2^{6}3Σ_{6}. As Z(⟨P_{2}, P_{3}⟩/K_{4}) = 1, we see that K_{1} = K_{2} = 1. Now S contains exactly one extraspecial subgroup W of order 2^{7} which yields W = O_{2}(⟨P_{1}, P_{2}⟩) = O_{2}(⟨P_{3}, P_{4}⟩), a contradiction.

Let finally Δ be of type (2). We have G_{3}/K_{3} ≅ L_{3}(2) × Σ_{3}. Again by lemma 1.7(iv) we get K_{4} ⊆ K_{1} and so ⟨K_{4}^{P_{4}}⟩ ⊆ K_{1}. As G_{3} = ⟨P_{1}, P_{2}, P_{4}⟩, we see K_{4} ⊆ K_{3}. But then G_{3} ∩ G_{4}/K_{3} ≅ L_{3}(2) × Z_{2} but ⟨P_{1}, P_{2}⟩/K_{4} possesses no subgroup of index two, a contradiction. □

Lemma 3.2 Let Δ be a string diagram and assume (ASS). Then Δ is of type

∘  ∘  ∘  ∘^{∼}  or ∘  ∘  ∘  ∘^{∼}  ^{(∼)} .

Proof: The possible other diagrams are

(1) 4  3  2  1 ∘  ∘  ∘  ∘^{(∼)}  ^{∼}

(2) 4  3  2  1 ∘  ∘  ∘  ∘^{∼}

(3) 4  3  2  1 ∘  ∘  ∘  ∘^{∼}  .

By lemmas 1.3 and 1.5 and [Par1], [Par2] we have in case (1) and (3) that G_{1}/K_{1} ≅ M_{24} too. As |K_{4}| = |K_{1}| this contradicts lemma 1.2. So we are left with (2).

Let ρ ∈ ⟨P_{1}, P_{2}⟩ with o(ρ) = 3 and ρO_{2}(⟨P_{1}, P_{2}⟩) ∈ Z(⟨P_{1}, P_{2}⟩′/O_{2}(⟨P_{1}, P_{2}⟩)). Set P_{5} = ⟨ρ, S⟩. Then W := ⟨P_{5}, P_{2}, P_{3}, P_{4}⟩ is a parabolic system with the dia-

gram [figure: ∘; ∘; ∘; ∘; 4; 3; 2; 5] where W_{4}/O_{2}(W_{4}) ≅ A_{8}. As a Sylow 2-subgroup of

⟨P_{3}, P_{2}, P_{4}⟩ and ⟨P_{3}, P_{4}, P_{5}⟩ is as in G_{2}(2) which is not isomorphic to a Sylow 2-subgroup of A_{8}, we get P_{4} ⊈ ⟨P_{3}, P_{2}, P_{5}⟩, P_{2} ⊈ ⟨P_{4}, P_{3}, P_{5}⟩, and P_{5} ⊈ ⟨P_{3}, P_{2}, P_{4}⟩. Now by [Str5] such a parabolic system does not exist, a contradiction. □

So we have proved proposition 3.1. In the remainder of this chapter we investigate the two diagrams (1) and (2) in proposition 3.1.

Lemma 3.3 Let {P_{1}, P_{2}, P_{3}, P_{4}} be a parabolic system of M_{24}–type with dia-

gram Δ : ∘  ∘  ∘  ∘^{∼} 1  2  3  4^{(∼)} . Then |K_{4}| = 2^{11} or 2^{12}, where K_{4} is the Go-

lay code– or extended Golay code–module, and G′_{4} does not split over K_{4} ∩ G′_{4}. Furthermore G_{1} is an extension of K_{1}, an extraspecial group of order 2^{13}, by G_{1}/K_{1}, which is parabolic isomorphic to a subgroup of 3(·3) · Aut(U_{4}(3)) containing 3(·3) · U_{4}(3) and the action on K_{1}/Z(K_{1}) is uniquely determined.

Proof: By proposition 3.1 (ASS) holds. Now we have |K_{4}| = 2^{11} or 2^{12} by [Par1] and [Par2]. If G_{1}/K_{1} is not parabolic isomorphic to a subgroup of 3(·3) · Aut(U_{4}(3)), then K_{4} ⊆ K_{1} by lemma 1.7. But this contradicts lemma 1.2. So we have that G_{1}/K_{1} ≅_{p} K < ∼ 3(·3) · Aut(U_{4}(3)). Let K_{4} be the Todd– module. Then |C_{K_{4}}(K_{1})| = 2^{4}. As |S| = 2^{21} and 2^{9} ≥ |S : K_{1}| ≥ 2^{8} we see that 2^{12} ≤ |K_{1}| ≤ 2^{13} and so |K_{1} ∩ K_{4}| ≤ 2^{7}. Now for the group 3Σ_{6} we have a 4–, a 1–, and a 6–dimensional module involved in K_{4}. So we see that K_{1} ∩ K_{4} involves the 6–dimensional module. As C_{K_{4}}(K_{1}) is 4–dimensional we now get that |K_{1} ∩ K_{4}| ≥ 2^{10}, a contradiction. So we have that K_{4} involves the Golay–code–module. Now we see |K_{1}K_{4}/K_{4}| = 2^{6}. The action of M_{24} on the extended Golay–code–module shows that |[K_{1}, K_{4}]| = 2^{7} and that G_{1} ∩ G_{4} acts indecomposably on K_{4}/[K_{1}, K_{4}]. As |K_{1}K_{4}/K_{4}| ≥ 2^{4}, this shows that K_{1} ∩ K_{4} = [K_{1}, K_{4}] and [K_{1}, K_{4}, K_{1}] = Ω_{1}(Z(S)). Hence |Z(K_{1})| = 2 and so K_{1} is extraspecial of order 2^{13}. Application of [Wie5, Theorem 2] gives that G_{1}/C_{G_{1}}(K_{1}/Z(K_{1})) ≅ 3U_{4}(3) · 2 or 3U_{4}(3) · 4.

Let us choose notation as in [Wie5], see also proposition 5.1. Then ⟨s, q, t, x, y⟩ ≅ 3A_{6} and is contained in G_{1} ∩ G_{4}. Using the matrices given there we can see that K_{1} ∩ K_{4}/Z(K_{1}) does not possess a complement in K_{1}/Z(K_{1}) invariant under ⟨s, q, t, x, y⟩. Hence K_{1}⟨s, q, t, x, y⟩/Z(K_{1}) does not split over K_{1} ∩ K_{4}/Z(K_{1}). In particular a Sylow 2–subgroup of K_{1}⟨s, q, t, x, y⟩ does not split over K_{1} ∩ K_{4}. Then also S does not split over K_{4}. By Gaschütz we get that G_{4} does not split.□

Lemma 3.4 Let {P_{1}, P_{2}, P_{3}, P_{4}} be a parabolic system of M_{24}–type with dia-

gram Δ : ∘  ∘  ∘  ∘ 4  3  2  1^{∼} , then |K_{4}| = 2^{11}, K_{4} is the Todd–module, and

G_{4} splits over K_{4}.

Proof: By proposition 3.1 (ASS) holds. We have G_{3}/K_{3} ≅ Σ_{3} × 3Σ_{6} or (Z_{3} × 3A_{6}) · 2. By lemma 1.2 K_{4} ≰ K_{3} and so G_{3}/K_{3} ≅ Σ_{3} × 3Σ_{6}. Let x ∈ P_{4} \ S. Then |(K_{4} ∩ K_{3})^{x}(K_{4} ∩ K_{3}) : K_{4} ∩ K_{3}| = 2^{6}, which shows that 2^{5} ≥ |K_{4} ∩ K_{4}^{x}| ≥ 2^{4} As K_{4} ∩ K_{4}^{x} is invariant under ⟨P_{1}, P_{2}⟩^{(∞)}, we see that K_{4} has to be the Todd–module. Then by lemma 1.4 we have |K_{4}| = 2^{11} and so |K_{4} ∩ K_{4}^{x}| = 2^{4}.

Let ρ ∈ P_{4}, o(ρ) = 3. Then ρ acts fixed point freely on K_{3}/K_{4} ∩ K_{4}^{x} and trivially on K_{4} ∩ K_{4}^{x}. As N_{K_{4}}(⟨ρ⟩) is an indecomposable submodule of K_{4} of dimension five on which N_{⟨P_{1},P_{2}⟩}(⟨ρ⟩) induces Σ_{6}, and any A_{6} over the fourdimensional module splits, we get that N_{G_{3}}(⟨ρ⟩) ≅ 2^{4}3Σ_{6} · Σ_{3} contains a subgroup 3Σ_{6}. Furthermore K_{4} contains a 3Σ_{6} submodule V of order 2^{6}. Then V^{ρ} is 3Σ_{6}-invariant and V ∩ K_{4} = 1. Hence there is a subgroup 2^{6}3Σ_{6} intersecting K_{4} trivially. This shows by Gaschütz that G_{4} splits over K_{4}. □

Lemma 3.5 Let {P_{1}, P_{2}, P_{3}, P_{4}} be as in lemma 3.4 and ρ ∈ ⟨P_{1}, P_{2}⟩ of order three with ρO_{2}(⟨P_{1}, P_{2}⟩ ∈ O_{3}(⟨P_{1}, P_{2}⟩/O_{2}(⟨P_{1}, P_{2}⟩)). Set P_{5} = ⟨ρ, S⟩. Then ⟨P_{2}, P_{3}, P_{4}, P_{5}⟩ is an extension of an extraspecial group of order 2^{9} by Ω^{+}(8, 2).

Proof: The structure of M_{24} shows that ⟨P_{3}, P_{2}, P_{5}⟩ ≅ 2^{11}2^{4}A_{8}. As P_{4} ⊈ ⟨P_{3}, P_{2}, P_{1}⟩ we see that ⟨P_{4}, P_{3}, P_{2}, P_{5}⟩ =: W is a parabolic system with diagram

[figure: ∘; ∘; ∘; ∘; 4; 3; 2; 5]. By [Tim2] we have that W/S_{W} ≅ Ω^{+}(8, 2). Furthermore by

lemma 3.4 we have |S_{W} | = 2^{9}. Now either S_{W} is extraspecial or elementary abelian. Suppose the latter. Then S_{W} involves the natural module V. By lemma 1.8 H_{1}(V, Ω^{+}(8, 2)) = 0. So S_{W} is a direct sum of the natural module and the trivial module. Hence |Ω_{1}(Z(S))| = 4. But this contradicts the structure of G_{4}.

4 The Co_{1} - case

In this chapter we finish our investigations of the first diagram in proposition 3.1. More precisely we are going to prove the following

Proposition 4.1 If {P_{1}, P_{2}, P_{3}, P_{4}} is a parabolic system of M_{24}–type with di-

agram Δ : ∘  ∘  ∘  ∘ 4  3  2  1^{∼} , then G ≅ Co_{1}.

Proof: By lemma 3.4 we can give generators and relations for G_{4}. For this we just have to construct a split extension of the Todd–module with M_{24}. Let {u_{1}, ..., u_{11}} be a basis of K_{4} and ⟨v_{1}, ..., v_{6}⟩ an elementary abelian subgroup of M_{24} where ⟨s, t, q, p, x, y, z⟩ ≅ 3Σ_{6}, acts on. Then M_{24} is generated by this group and an element u, where ⟨v_{1}, ..., v_{6}, p, s, t, q, u, y⟩ ≅ 2^{1+6}L_{3}(2) and ⟨v_{1}, ..., v_{6}, p, s, t, q, u, y, z⟩ ≅ 2^{4}A_{8}. Then this group is given by generators and relations as follows.

First we have the defining relations for the elementary abelian groups ⟨u_{1}, ..., u_{11}⟩ and ⟨v_{1}, ..., v_{6}⟩. The action of the v_{i} on ⟨u_{1}, ..., u_{11}⟩ is given by [v_{1}, u_{1}], u_{2}^{v_{1}}u_{2}u_{1},[u_{3},v_{1}]u_{1},[u_{4},v_{1}],[u_{5},v_{1}],u_{6}^{v_{1}}u_{2}u_{3}u_{6}u_{1},u_{7}^{v_{1}}u_{2}u_{3}u_{4}u_{7}u_{1}, [u_{8},v_{1}],u_{9}^{v_{1}}u_{4}u_{5}u_{9},u_{10}^{v_{1}}u_{5}u_{10},u_{11}^{v_{1}}u_{4}u_{5}u_{11}, [u_{1},v_{2}],[u_{2},v_{2}]u_{1},[u_{3},v_{2}],[u_{4},v_{2}]u_{1},[u_{5},v_{2}]u_{1},u_{6}^{v_{2}}u_{3}u_{4}u_{5}u_{6},u_{7}^{v_{2}}u_{4}u_{5}u_{7}u_{1}, u_{8}^{v_{2}}u_{2}u_{3}u_{4}u_{8},u_{9}^{v_{2}}u_{2}u_{3}u_{5}u_{9},u_{10}^{v_{2}}u_{2}u_{5}u_{10},u_{11}^{v_{2}}u_{2}u_{3}u_{4}u_{11}, [u_{1},v_{3}],u_{2}^{v_{3}}u_{3}u_{5},u_{3}^{v_{3}}u_{2}u_{5},[u_{4},v_{3}],[u_{5},v_{3}],u_{6}^{v_{3}}u_{5}u_{7}u_{8}u_{10}u_{11}u_{1},u_{7}^{v_{3}}u_{5}u_{6}u_{9}u_{10}, [u_{8},v_{3}]u_{1},u_{9}^{v_{3}}u_{5}u_{6}u_{7}u_{10},u_{10}^{v_{3}}u_{5}u_{6}u_{7}u_{8}u_{11}u_{1}, u_{11}^{v_{3}}u_{5}u_{6}u_{7}u_{9}u_{10}u_{11}u_{1}, [u_{1},v_{4}],u_{2}^{v_{4}}u_{3}u_{5},[u_{3},v_{4}],u_{4}^{v_{4}}u_{2}u_{3}u_{4}u_{5},u_{5}^{v_{4}}u_{2}u_{3},u_{6}^{v_{4}}u_{3}u_{5}u_{6}u_{7}u_{8}u_{9},u_{7}^{v_{4}}u_{2}u_{3}u_{6}u_{9}u_{10}, u_{8}^{v_{4}}u_{2}u_{3}u_{6}u_{7}u_{8}u_{9}u_{10},u_{9}^{v_{4}}u_{2}u_{3}u_{5}u_{9},u_{10}^{v_{4}}u_{2}u_{5}u_{6}u_{8},u_{11}^{v_{4}}u_{2}u_{3}u_{6}u_{7}u_{9}u_{10}u_{11}u_{1}, [u_{1},v_{5}],u_{2}^{v_{5}}u_{3}u_{4}u_{1},u_{3}^{v_{5}}u_{2}u_{4},[u_{4},v_{5}]u_{1},[u_{5},v_{5}]u_{1},u_{6}^{v_{5}}u_{2}u_{4}u_{5}u_{7}u_{8}u_{9}u_{10}, u_{7}^{v_{5}}u_{4}u_{5}u_{7}u_{1},u_{8}^{v_{5}}u_{2}u_{3}u_{4}u_{8},u_{9}^{v_{5}}u_{2}u_{3}u_{4}u_{5}u_{6}u_{7}u_{10}, u_{10}^{v_{5}}u_{3}u_{4}u_{5}u_{8}u_{10},u_{11}^{v_{5}}u_{2}u_{3}u_{6}u_{7}u_{9}u_{10}u_{11}u_{1}, [u_{1},v_{6}],u_{2}^{v_{6}}u_{3}u_{4},u_{3}^{v_{6}}u_{3}u_{1},u_{4}^{v_{6}}u_{2}u_{3}u_{1},u_{5}^{v_{6}}u_{2}u_{3}u_{4}u_{5}u_{1},u_{6}^{v_{6}}u_{4}u_{5}u_{6}u_{7}u_{9}u_{11}, u_{7}^{v_{6}}u_{5}u_{6}u_{9}u_{10},[u_{8},v_{6}]u_{1},u_{9}^{v_{6}}u_{4}u_{6}u_{7}u_{10},u_{10}^{v_{6}}u_{2}u_{3}u_{4}u_{7}u_{9}u_{10}u_{11}u_{1}, u_{11}^{v_{6}}u_{4}u_{5}u_{11}.

The relations for 3Σ_{6} are

s^{2},q^{2},t^{2}, (st)^{2}q,p^{2},[p,s],[p,t], x^{3}, q^{x}s,s^{x}qs, (xt)^{2},[p,x] y^{3},q^{y}t,t^{y}qt,(ys)^{2},p^{y}pqt, (xy)^{5} z^{−1}(xptypqs)^{4}, [z,y], [z, x], z^{3}, [z, s], [z, t], [z, q], (zp)^{2}.

To see these relations we use the following identification with elements in Σ_{6} : q = (12)(34), s = (13)(24), t = (12)(56), x = (123), (y = (153)(264), p = (56).

The action of this group on ⟨u_{1}, ..., u_{11}⟩ is given by

[u_{1},s],[u_{2},s],u_{3}^{s}u_{2}u_{5},[u_{4},s],u_{5}^{s}u_{2}u_{3},u_{6}^{s}u_{2}u_{6}u_{7}u_{8}u_{9}u_{1},u_{7}^{s}u_{2}u_{11},u_{8}^{s}u_{9}u_{11},u_{9}^{s}u_{2}u_{7}u_{8}, u_{10}^{s}u_{2}u_{7}u_{10}u_{11},u_{11}^{s}u_{2}u_{7}, [u_{1},t],[u_{2},t]u_{1},[u_{3},t]u_{1},[u_{4},t],[u_{5},t]u_{1},[u_{6},t]u_{1},u_{7}^{t}u_{2}u_{3}u_{7}u_{1}, u_{8}^{t}u_{2}u_{3}u_{4}u_{8},u_{9}^{t}u_{3}u_{5}u_{9}u_{1},u_{10}^{t}u_{3}u_{4}u_{5}u_{10}u_{1},u_{11}^{t}u_{2}u_{5}u_{11}u_{1}, [u_{1},q],[u_{2},q],[u_{3},q]u_{1},[u_{4},q],[u_{5},q]u_{1},u_{6}^{q}u_{3}u_{4}u_{5}u_{6}u_{1},u_{7}^{q}u_{2}u_{7}, u_{8}^{q}u_{2}u_{3}u_{4}u_{5}u_{8}u_{1},u_{9}^{q}u_{3}u_{4}u_{5}u_{9}u_{1},u_{10}^{q}u_{3}u_{5}u_{10}u_{1},u_{11}^{q}u_{2}u_{11}, [u_{1},p], u_{2}^{p}u_{2}u_{4}u_{1},u_{3}^{p}u_{3}u_{4},u_{4}^{p}u_{4},[u_{5},p]u_{1},u_{6}^{p}u_{2}u_{4}u_{5}u_{7}u_{8}u_{9}u_{10}, u_{7}^{p}u_{3}u_{6}u_{8}u_{9}u_{10}u_{1},u_{8}^{p}u_{2}u_{3}u_{4}u_{5}u_{8},u_{9}^{p}u_{3}u_{6}u_{7}u_{9}u_{10}u_{11}u_{1}, u_{10}^{p}u_{3}u_{4}u_{6}u_{7}u_{11},u_{11}^{p}u_{2}u_{4}u_{5}u_{6}u_{7}u_{10}u_{1}, [u_{1},y],u_{2}^{y}u_{2}u_{4}u_{5},u_{3}^{y}u_{2}u_{4},[u_{4},y],u_{5}^{y}u_{2}u_{3}u_{4}u_{5},u_{6}^{y}u_{2}u_{4}u_{5}u_{6}u_{7}u_{8}u_{9},u_{7}^{y}u_{2}u_{4}u_{6}u_{7}u_{10}, u_{8}^{y}u_{4}u_{5}u_{9}u_{11}u_{1},u_{9}^{y}u_{2}u_{4}u_{6}u_{10},u_{10}^{y}u_{2}u_{7}u_{8}u_{10},u_{11}^{y}u_{2}u_{4}u_{5}u_{7}u_{8}u_{11}, u_{1}^{x}u_{1}u_{2}u_{3}u_{5}, u_{2}^{x}u_{1}u_{3}u_{5}u_{7}u_{11}, u_{3}^{x}u_{1}u_{2}u_{6}u_{8}u_{10}, u_{4}^{x}u_{8}u_{9}u_{11}, u_{5}^{x}u_{1}u_{2}u_{3}u_{6}u_{7}u_{8}u_{10}u_{11}, u_{6}^{x}u_{1}u_{2}u_{3}u_{5}u_{6}, u_{7}^{x}u_{1}u_{5}u_{6}u_{7}u_{8}u_{10}, u_{8}^{x}u_{2}u_{3}u_{7}u_{9}, u_{9}^{x}u_{1}u_{2}u_{4}u_{6}u_{10}u_{11}, u_{10}^{x}u_{1}u_{2}u_{6}u_{7}u_{9}, u_{11}^{x}u_{1}u_{3}u_{6}u_{7}u_{8}u_{10}, [u_{1},z],u_{2}^{z}u_{2}u_{4}u_{1},u_{3}^{z}u_{5}u_{1},u_{4}^{z}u_{3}u_{5},u_{5}^{z}u_{3}u_{4},u_{6}^{z}u_{2}u_{8}u_{10}u_{11},u_{7}^{z}u_{2}u_{4}u_{5}u_{6}u_{7}u_{10}u_{1}, u_{8}^{z}u_{3}u_{4}u_{7}u_{8}u_{9}u_{11}u_{1},u_{9}^{z}u_{3}u_{4}u_{5}u_{6}u_{10},u_{10}^{z}u_{4}u_{5}u_{6}u_{8}u_{9}u_{11}u_{1},u_{11}^{z}u_{3}u_{6}u_{8}u_{9}u_{10}u_{1}.

The action on ⟨v_{1}, ..., v_{6}⟩ is given by

[v_{1},q],[v_{2},q],[v_{3},q],[v_{4},q],v_{5}^{q}v_{5}v_{1},v_{6}^{q}v_{6}v_{2}, [v_{1},s],[v_{2},s],[v_{3},s],[v_{4},s],v_{5}^{s}v_{5}v_{3},v_{6}^{s}v_{6}v_{4}, [v_{1},t],[v_{2},t],v_{3}^{t}v_{1}v_{3},v_{4}^{t}v_{4}v_{2},[v_{5},t],[v_{6},t], [v_{1},p],v_{2}^{p}v_{1}v_{2},[v_{3},p],v_{4}^{p}v_{4}v_{3},[v_{5},p],v_{6}^{p}v_{5}v_{6}, [v_{1},y],[v_{2},y],v_{3}^{y}v_{2}v_{5},v_{4}^{y}v_{1}v_{2}v_{6},v_{5}^{y}v_{3}v_{5},v_{6}^{y}v_{4}v_{6}, v_{1}^{x}v_{3}, v_{2}^{x}v_{4}, v_{3}^{x}v_{1}v_{3}, v_{4}^{x}v_{2}v_{4}, [v_{5}, x], [v_{6}, x], v_{1}^{z}v_{1}v_{2},v_{2}^{z}v_{1},v_{3}^{z}v_{3}v_{4},v_{4}^{z}v_{3},v_{5}^{z}v_{5}v_{6},v_{6}^{z}v_{5}.

These are generators and relations for 2^{11}2^{6}3Σ_{6}. Now we add u with u^{3} = 1 and the following relations:

[u_{1}, u], u_{2}^{u}u_{3}u_{1}, u_{3}^{u}u_{3}u_{4}, u_{4}^{u}u_{2}u_{3}u_{4}u_{1}, u_{5}^{u}u_{2}u_{3}u_{4}u_{5}u_{1}, u_{6}^{u}u_{4}u_{5}u_{6}u_{7}u_{9}u_{11}, u_{7}^{u}u_{2}u_{3}u_{5}u_{7}u_{10}u_{11}u_{1}, u_{8}^{u}u_{9}u_{11}, u_{9}^{u}u_{5}u_{8}u_{10}, u_{10}^{u}u_{5}u_{7}u_{9}u_{10}u_{11}, u_{11}^{u}u_{6}u_{7}u_{8}u_{11}u_{1}, [v_{1}, u], v_{2}^{u}v_{2}q, [v_{3}, u], v_{4}^{u}v_{4}s, v_{5}^{u}v_{5}p, (v_{6}u)^{2}, q^{u}v_{2}, [t, u], p^{u}v_{5}, s^{u}v_{4}, (ysv_{2}u)^{3}, [x, u], (zu)^{4}.

These are the relations for 2^{11}M_{24}. One easily checks that ⟨v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, p, q, s, t, x, y, z, u⟩ ≅ M_{24}. Maybe the easiest way is to see that ⟨v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, p, q, s, t, x, y, z⟩ is an extension of an elementary abelian group of order 2^{6} by Σ_{6}, ⟨v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, p, q, s, t, y, u⟩ is an extension of a group of order 2^{7} by L_{3}(2) and ⟨v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, p, q, s, t, x, u⟩ is an extension of a group of order 2^{8} by Σ_{3} × Σ_{3}. Now we have a parabolic system of type M_{24} and rank three and so we may quote [Hei2].

As u_{1} is centralized by ⟨v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}, p, q, s, t, x, y, z⟩ we see that ⟨u_{1} , …, u_{11}⟩ is the module from lemma 3.4. So we have the structure of G_{4}, which is uniquely determined.

By lemma 3.5 we have ⟨P_{4}, P_{3}, P_{2}, P_{5}⟩ ≅ 2^{1+8}Ω_{8}^{+}(2), where ⟨P_{3}, P_{2}, P_{5}⟩ is generated by {u_{1}, ..., u_{11}, v_{1}, ..., v_{6}, s, t, q, p, u, y, z}. Now choose some v ∈ P_{4} with o(v) = 3 such that ⟨P_{2}, P_{3}, P_{5}, v⟩ = ⟨P_{2}, P_{3}, P_{5}, P_{4}⟩. As G_{3}/K_{3} ≅ Σ_{3}×3Σ_{6}, we may choose v such that [v, z] = 1 = [v, y]. As (K_{4} ∩ K_{3})^{v}(K_{4} ∩ K_{3}) = K_{3}, we get |C_{K_{3}}(v)| = 2^{4} and so C_{G_{3}}(v) ≅ ⟨v⟩ × 2^{4}3Σ_{6}. This implies that we may choose v with [v, x] = 1.

The extraspecial normal 2-subgroup of 2^{1+8}Ω_{8}^{+}(2) is given by X := ⟨u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, v_{1}, v_{2}, q, t⟩, which can be seen inside of ⟨P_{2}, P_{3}, P_{5}⟩. Using the relations obtained so far we see that v acts on X/⟨u_{1}⟩ as the following matrix

( \begin{matrix}1 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} )

where the basis is {u_{2}, u_{3}, u_{4}, u_{3}u_{5}, v_{1}, v_{2}, q, t}. In ⟨P_{2}, P_{3}, P_{4}, P_{5}⟩/X we get the following relations:

[s, v], [p, v], [v, y], [v, z], [u_{6}u_{7}u_{9}u_{10}, v], v_{3}^{v}v_{3}u_{8}u_{9}u_{11}, (u_{8}u_{9}u_{11})^{v}v_{3}, v_{4}^{v}u_{7}u_{8}u_{9}v_{4}, v_{5}^{v}u_{6}u_{7}u_{8}u_{9}u_{10}v_{5}, (u_{7}u_{8}u_{9})^{v}v_{4}, (u_{6}u_{7}u_{8}u_{9}u_{10})^{v}v_{5}, v_{6}^{v}v_{6}u_{7}u_{9}u_{11}, (u_{7}u_{9}u_{11})^{v}v_{6}, v^{u_{6}u_{7}u_{9}u_{11}}v, (vu)^{7}, (u^{2}v)^{4}.

These can easily calculated as we know the action of all elements of ⟨P_{2}, P_{3}, P_{4}, P_{5}⟩/X on X, so we just have to do matrix multiplications.

As the extension of X/⟨u_{1}⟩ by Ω_{8}^{+}(2) is determined by the extension of the Sylow 2-subgroup and this is known from G_{4}, we can calculate modulo ⟨u_{1}⟩ :

[v, y], [v, z], (vu)^{7}, (u^{2}v)^{4}, [s, v], [p, v], v_{3}^{v}u_{8}u_{9}u_{11}v_{3}, (u_{8}u_{9}u_{11})^{v}v_{3}u_{2}u_{3}u_{5}, v_{4}^{v}u_{2}u_{7}u_{8}u_{9}v_{4}, v_{5}^{v}u_{2}u_{4}u_{5}u_{6}u_{7}u_{8}u_{9}u_{10}v_{5}, (u_{2}u_{7}u_{8}u_{9})^{v}v_{4}u_{2}u_{3}u_{5}, (u_{4}u_{6}u_{7}u_{8}u_{9}u_{10})^{v}v_{5}v_{2}u_{2}u_{3}u_{4}, v_{6}^{v}v_{6}u_{4}u_{5}u_{7}u_{9}u_{11}, (u_{2}u_{7}u_{9}u_{11})^{v}v_{6}v_{2}v_{1}u_{2}u_{3}u_{4}, v^{u_{6}u_{7}u_{9}u_{11}}v, (u_{6}u_{7}u_{9}u_{10})^{v}u_{4}u_{6}u_{7}u_{9}u_{10}v_{1}.

As can be check with MAGMA this defines modulo ⟨u_{1}⟩ a nonsplit extension of an elementary abelian group of order 2^{8} by Ω^{+}(8, 2).

As [x, v] = 1 and u_{1}^{x} = u_{1}u_{2}u_{3}u_{5}, we get

(u_{1}u_{2}u_{3}u_{5})^{v} = (u_{1}u_{2}u_{3}u_{5})^{x^{−1}vx} = u_{1}^{vx} = u_{1}^{x} = u_{1}u_{2}u_{3}u_{5}.

In this way we can calculate all relations involving v, which then yields :

v^{3}, [v, y], [v, z], (vu)^{7}, (u^{−1}v)^{4}, u_{2}^{v}u_{2}u_{3}u_{5}v_{1}v_{2}u_{1}, u_{3}^{v}u_{4}u_{5}v_{2}, u_{4}^{v}v_{1}u_{1}, (u_{3}u_{5})^{v}v_{1}v_{2}u_{1}, v_{1}^{v}u_{4}v_{1}, v_{2}^{v}u_{3}u_{4}u_{5}v_{2}u_{1}, [v, q], [t, v], [s, v], [p, v], (u_{6}u_{7}u_{9}u_{10})^{v}v_{1}u_{4}u_{1}u_{6}u_{7}u_{9}u_{10}, (u_{6}u_{7}u_{9}u_{11}v)^{2}, v_{3}^{v}v_{3}u_{8}u_{9}u_{11}, (u_{8}u_{9}u_{11})^{v}u_{1}u_{2}u_{3}u_{5}v_{3}, v_{4}^{v}v_{4}u_{2}u_{7}u_{8}u_{9}u_{1}, v_{5}^{v}v_{5}u_{2}u_{4}u_{5}u_{6}u_{7}u_{8}u_{9}u_{10}, (u_{2}u_{7}u_{8}u_{9})^{v}u_{2}u_{3}u_{5}v_{4}, (u_{4}u_{6}u_{7}u_{8}u_{9}u_{10})^{v}u_{2}u_{3}u_{4}v_{2}v_{5}, v_{6}^{v}u_{4}u_{5}u_{7}u_{9}u_{11}v_{6}, (u_{2}u_{7}u_{9}u_{11})^{v}u_{1}u_{2}u_{3}u_{4}v_{1}v_{2}v_{6}.

This then provides us with generators and relations for G = ⟨P_{1}, P_{2}, P_{3}, P_{4}⟩, where all these come uniquely from the diagram.

Now we calculate that ⟨u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6}u_{7}, u_{6}u_{8}, u_{6}u_{9}, u_{6}u_{10}, u_{11}⟩⟨v_{1}, v_{2}, v_{3}, v_{4}v_{5}, v_{6}, t, q, ps, yz^{−1}, u^{v_{5}}, y^{x^{−1}}s(y^{−1}s)^{x^{−1}}⟩ is an extension of an elementary abelian group of order 2^{10} byAut(M_{22}) in G_{4}. We generate this group by two parabolics containing the Sylow 2-subgroup ⟨u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6}u_{7}, u_{6}u_{8}, u_{6}u_{9}, u_{6}u_{10}, u_{11}⟩⟨v_{1}, v_{2}, v_{3}, v_{4}v_{5}, v_{6}, t, q, ps⟩ =: T, where P̃_{1} = ⟨T, yz^{−1}, y^{x^{−1}}s(y^{−1}s)^{x^{−1}}⟩ , P̃_{1}/O_{2}(P̃_{1}) ≅ Σ_{5} and P̃_{2} = ⟨T, u^{v_{5}}⟩, P̃_{2}/O_{2}(P̃_{2}) ≅ Σ_{3}. Now choose P̃_{3} = ⟨T, v^{v_{5}}⟩ ⊆ ⟨P_{2}, P_{3}, P_{4}, P_{5}⟩. Then T ∈ Syl_{2}(P̃_{3}) and P̃_{3}/O_{2}(P̃_{3}) ≅ Σ_{3}. Furthermore ⟨P̃_{2}, P̃_{3}⟩/O_{2}(⟨P̃_{2}, P̃_{3}⟩) ≅ L_{3}(2) as ⟨u, v⟩ generate L_{3}(2) and ⟨P̃_{3}, P̃_{1}⟩ = P̃_{3}P̃_{1} as ⟨P̃_{3}, P̃_{1}⟩ ⊆ G_{3}, where P̃_{3} ⊆ P_{4} and P̃_{1} ⊆ ⟨P_{2}, P_{1}⟩. So we have

that ⟨P̃_{1}, P̃_{2}, P̃_{3}⟩ possesses a diagram of type ∘  ∘  ∘^{M_{22}} . Application of

[Str6, Theorem 4] shows that ⟨P̃_{1}, P̃_{2}, P̃_{3}⟩ ≅ Co_{2}. Now using the Todd-Coxeter algorithm we get |G : ⟨P̃_{1}, P̃_{2}, P̃_{3}⟩| = 98280. Hence |G| = |Co_{1}|. By [RoStr] Co_{1} possesses a geometry of this type, so G ≅ Co_{1}.

5 The M(24)-case

In this chapter we investigate parabolic systems whose diagram is as in proposition 3.1 (2). First we prove that the corresponding amalgam is uniquely determined.

Proposition 5.1 Let {P_{1}, P_{2}, P_{3}, P_{4}} be a parabolic system of M_{24}–type with

diagram Δ : 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  ^{(∼)}  , then the amalgam is uniquely determined

and isomorphic to one of M(24)′ or M(24), 3M(24) or 3M(24)′.

Proof: By lemma 3.3 we know that K_{1} is an extraspecial group of order 2^{13} and (G_{1}/C_{G_{1}}(K_{1}/Z(K_{1})))^{(∞)} ≅ 3U_{4}(3). In particular if ⟨P_{3}, P_{4}⟩/O_{2}(⟨P_{3}, P_{4}⟩) ≅ 3Σ_{6}, then we see that C(K_{1}) ≰ O_{2}(⟨P_{3}, P_{4}⟩). Hence we have an element of order 3, which centralizes K_{1} and so it does not affect the structure of the amalgam, as it also centralizes K_{4} ∩ G′_{4}. So we may assume that we have ⟨P_{3}, P_{4}⟩/O_{2}(⟨P_{3}, P_{4}⟩) ≅ Σ_{6}. If we can show that this amalgam is as in M(24) or M(24)′, then the other is as in the triple cover of these groups. So we now get G_{1}/C_{G_{1}}(K_{1}/Z(K_{1})) ≅ 3U_{4}(3) · 2 or 3U_{4}(3) · 4 depending on |K_{4}| = 2^{11} or 2^{12}.

Let {u_{1}, ..., u_{11}} be a basis for K_{4} in case of |K_{4}| = 2^{11} and {v_{1}, ..., v_{6}} be an elementary abelian group of order 2^{6} such that {u_{1}, ..., u_{7}, v_{1}, ..., v_{6}} = K_{1} where [u_{i+1}, v_{i}] = u_{1}, i = 1, ..., 6 and [u_{j}, v_{i}] = 1 for j ≠ i + 1. Choose ⟨s, t, q, x, y, p⟩ ≅ 3Σ_{6} as in proposition 4.1. Hence we have the following relations

s^{2} ,t^{2} ,q^{2}, (st)^{2}q, q^{x}s, s^{x}qs, x^{t}x, q^{y}t, t^{y}qt, y^{s}y, (xy)^{5}, x^{3}, y^{3}, p^{2}, [p, s], [p, t], [x, p], p^{y}pqt.

One should have in mind that these are relations modulo ⟨u_{1}⟩. As shown in [Wie5] there is a subgroup 3Σ_{6} in G_{1}/Z(K_{1}), so we may in fact use the relations above for 3Σ_{6} in G_{1}/Z(K_{1}).

Then ⟨u_{1}, ..., u_{11}⟩⟨v_{1}, ..., v_{6}⟩⟨s, t, q, x, y, p⟩/⟨u_{1}, ..., u_{11}⟩ is a subgroup isomorphic to 2^{6}3Σ_{6} in M_{24}. The action of v_{1}, ..., v_{6}, s, t, q, x, y, p on ⟨u_{1}, ...u_{11}⟩ is given by

u_{8}^{v_{1}}u_{5}u_{8}, u_{9}^{v_{1}}u_{1}u_{3}u_{6}u_{7}u_{9}, u_{10}^{v_{1}}u_{5}u_{7}u_{10}, u_{11}^{v_{1}}u_{7}u_{11}, u_{8}^{v_{2}}u_{4}u_{8}, u_{9}^{v_{2}}u_{1}u_{2}u_{7}u_{9}, u_{10}^{v_{2}}u_{4}u_{6}u_{10}, u_{11}^{v_{2}}u_{6}u_{11}, u_{8}^{v_{3}}u_{1}u_{3}u_{5}u_{6}u_{7}u_{8}, u_{9}^{v_{3}}u_{1}u_{5}u_{6}u_{7}u_{9}, u_{10}^{v_{3}}u_{1}u_{3}u_{5}u_{6}u_{10}, u_{11}^{v_{3}}u_{11}, u_{8}^{v_{4}}u_{1}u_{2}u_{4}u_{7}u_{8}, u_{9}^{v_{4}}u_{1}u_{4}u_{7}u_{9}, u_{10}^{v_{4}}u_{1}u_{2}u_{4}u_{6}u_{7}u_{10}, u_{11}^{v_{4}}u_{11}, u_{8}^{v_{5}}u_{4}u_{8}, u_{9}^{v_{5}}u_{2}u_{4}u_{9}, u_{10}^{v_{5}}u_{3}u_{4}u_{5}u_{10}, u_{11}^{v_{5}}u_{3}u_{11}, u_{8}^{v_{6}}u_{4}u_{5}u_{8}, u_{9}^{v_{6}}u_{2}u_{3}u_{4}u_{5}u_{9}, u_{10}^{v_{6}}u_{2}u_{5}u_{10}, u_{11}^{v_{6}}u_{2}u_{11}, [q, u_{1}], u_{2}^{q}u_{2}u_{6}, u_{3}^{q}u_{3}u_{7}, [u_{4}, q], [u_{5}, q], [u_{6}, q], [u_{7}, q], u_{8}^{q}u_{10}u_{11}, u_{9}^{q}u_{1}u_{9}u_{11}, u_{10}^{q}u_{8}u_{11}, [u_{11}, q], [u_{1}, s], [u_{2}, s], [u_{3}, s], u_{4}^{s}u_{4}u_{6}, u_{5}^{s}u_{5}u_{7}, [u_{6}, s], [u_{7}, s], u_{8}^{s}u_{1}u_{10}, u_{9}^{s}u_{1}u_{8}u_{9}u_{10}u_{11}, u_{10}^{s}u_{1}u_{8}, [u_{11}, s], [u_{1}, t], u_{2}^{t}u_{2}u_{4}, u_{3}^{t}u_{3}u_{5}, [u_{4}, t], [u_{5}, t], [u_{6}, t], [u_{7}, t], [u_{8}, t], u_{9}^{t}u_{8}u_{9}, u_{10}^{t}u_{8}u_{11}, u_{11}^{t}u_{8}u_{10}, [p, u_{1}], u_{2}^{p}u_{2}u_{3}, [u_{3}, p], u_{4}^{p}u_{4}u_{5}, [u_{5}, p], u_{6}^{p}u_{6}u_{7}, [u_{7}, p], u_{8}^{p}u_{11}u_{10}, u_{9}^{p}u_{8}u_{10}u_{9}, u_{10}^{p}u_{11}u_{8}, [u_{11}, p], [u_{1}, x], u_{2}^{x}u_{2}u_{4}, u_{3}^{x}u_{3}u_{5}, u_{4}^{x}u_{2}, u_{5}^{x}u_{3}, [u_{6}, x], [u_{7}, x], u_{8}^{x}u_{8}u_{9}, u_{9}^{x}u_{8}, u_{10}^{x}u_{9}u_{10}u_{11}, u_{11}^{x}u_{8}u_{10}, [y, u_{1}], [y, u_{2}]u_{5}u_{7}, [y, u_{3}]u_{4}u_{5}u_{6}u_{7}, [y, u_{4}]u_{6}, [y, u_{5}]u_{7}, u_{6}^{y}u_{4}, u_{7}^{y}u_{5}, u_{8}^{y}u_{1}u_{10}, [u_{9}, y], u_{10}^{y}u_{11}, u_{11}^{y}u_{1}u_{8}.

As the action of ⟨s, t, q, x, y, p⟩ on K_{1}/Z(K_{1}) is uniquely determined by lemma 3.3 we get, using the description in [Wie5], that modulo ⟨u_{1}⟩ we have [v_{1}, q], [v_{2}, q], [v_{3}, q]u_{7}, [v_{4}, q]u_{6}, v_{5}^{q}v_{5}v_{1}u_{5}u_{7}, v_{6}^{q}v_{2}v_{6}u_{6}u_{4}, [v_{1}, s], [v_{2}, s], [v_{3}, s], [v_{4}, s], [v_{5}, s]v_{3}, [v_{6}, s]v_{4}, [v_{1}, t], [v_{2}, t], v_{3}^{t}v_{1}v_{3}u_{5}u_{7}, v_{4}^{t}v_{2}v_{4}u_{4}u_{6}, [v_{5}, t]u_{5}, [v_{6}, t]u_{4}, [v_{1}, p], [v_{2}, p]v_{1}, [v_{3}, p], [v_{4}, p]v_{3}u_{7}, [v_{5}, p]u_{7}, v_{6}^{p}u_{5}u_{6}u_{7}v_{5}v_{6}, v_{1}^{x}v_{3}u_{3}, v_{2}^{x}v_{4}u_{2}u_{6}, v_{3}^{x}v_{1}v_{3}u_{5}u_{7}, v_{4}^{x}v_{2}v_{4}u_{4}u_{6}, [v_{5}, x]u_{5}, [v_{6}, x]u_{2}, [y, v_{1}], [y, v_{2}], v_{3}^{y}v_{2}v_{5}, v_{4}^{y}v_{1}v_{2}v_{6}, v_{5}^{y}v_{3}v_{5}, v_{6}^{y}v_{4}v_{6}.

All these relations also hold in ⟨u_{1}, ..., u_{11}, v_{1}, ..., v_{6}, s, t, q, x, y, p⟩ besides possibly the relations involving s, t, q, x, y, p. If s^{2} = u_{1}, then we see that s⟨u_{1}, …, u_{11}⟩ does not contain involutions. But in G_{4} we have that s⟨u_{1}, …, u_{11}⟩ is conjugate to some element in ⟨u_{1}, …, u_{11}, v_{1}, …, v_{6}⟩/⟨u_{1}, …, u_{11}⟩. So we must have a subgroup 3A_{6} in G_{1} and then all relations not involving p are left unchanged. From the remaining relations there are four which have to be changed to [v_{2}, p] = v_{1}u_{1}, [v_{4}, p] = v_{3}v_{7}u_{1}, [v_{5}, p] = u_{7}u_{1}, v_{6}^{p} = v_{6}v_{5}u_{1}u_{5}u_{6}u_{7}. So we have defining relations for ⟨u_{1}, ..., u_{11}, v_{1}, ..., v_{6}, s, t, q, x, y, p⟩.

To define G_{4} we need some further element u of order three. We choose u such that ⟨S, u⟩ = P_{1}. We get the following action of u on K_{4}:

u_{1}^{u}u_{1}u_{7}, u_{2}^{u}u_{2}u_{11}, [u_{3}, u], u_{4}^{u}u_{4}u_{8}u_{10}u_{11}, [u_{5}, u], [u_{6}, u], u_{7}^{u}u_{1}, u_{8}^{u}u_{8}u_{4}, u_{9}^{u}u_{2}u_{4}u_{9}, u_{10}^{u}u_{2}u_{8}, u_{11}^{u}u_{2}.

Modulo ⟨u_{1}, ..., u_{11}⟩ we have, using the matrices for the action on the 11-dimen-sional module obtained before, that

[v_{1}, u] = 1 = [t, u] = [v_{3}, u], u^{v_{6}} = u^{−1}, v_{2}^{u} = qv_{2}, q^{u} = v_{2}, v_{4}^{u} = v_{4}s, s^{u} = v_{4}, [x, u] = 1, (v_{1}v_{2}v_{5})^{u} = v_{2}v_{5}qp, (v_{1}qp)^{u} = v_{1}v_{2}v_{5}, (ysv_{2}u)^{3} = 1.

We may choose u such that 1 = [x, u] holds in G_{4}. Now we have v_{1}^{u} = v_{1}r_{1}, t^{u} = tr_{2}, where r_{1}, r_{2} ∈ [K_{4}, u]. Furthermore r_{2} ∈ C_{[K_{4},u]}(x), which shows

r_{2} ∈ ⟨u_{1}, u_{7}⟩.

Now t^{ysv_{2}u} = tr_{2}. Let (ysv_{2}u)^{3} = r ∈ K_{4}. Then t^{r} = tr_{2}^{(ysv_{2}u)^{2}}r_{2}^{ysv_{2}u}r_{2}. This yields

t^{r} = u_{5}t and r_{2} ∈ ⟨u_{7}⟩u_{1}, or t^{r} = t and r_{2} ∈ ⟨u_{7}⟩.

Next we determine r_{4} with u^{v_{6}u_{2}u_{4}} = u^{−1}r_{4}. As [v_{6}u_{2}u_{4}, x] = 1 , we see r_{4} ∈ ⟨u_{1}, u_{7}⟩. Now u = u^{(v_{6}u_{2}u_{4})^{2}} = r_{4}ur_{4}^{v_{6}u_{2}u_{4}} implies r_{4} ∈ ⟨u_{1}u_{7}⟩ . If r_{4} ≠ 1 we may replace u by uu_{1}u_{7}. So we may assume

r_{4} = 1.

Then t^{v_{6}u_{2}u_{4}uv_{6}u_{2}u_{4}} = t^{u^{2}} implies r_{2}r_{2}^{u} = r_{2}^{v_{6}u_{2}u_{4}} and so r_{2} ∈ ⟨u_{7}⟩. In particular r ∈ C_{K_{4}}(t) = ⟨u_{1}, u_{4}, u_{5}, u_{6}, u_{7}, u_{8}, u_{10}u_{11}⟩. We have v_{1}^{u} = v_{1}r_{1} with r_{1} ∈ C_{K_{4}}(v_{1}) ∩ [K_{4}, u] = ⟨u_{1}, u_{7}, u_{4}, u_{8}u_{10}u_{11}⟩. Then v_{1}^{v_{6}u_{2}u_{4}uv_{6}u_{2}u_{4}} = v_{1}^{u^{2}} implies

r_{1} ∈ ⟨u_{7}⟩u_{1}.

Now [x, u] = 1 shows v_{3}^{u} = v_{3}r_{1}. Then v_{3}^{ut} = v_{1}v_{3}u_{5}u_{7}r_{1}^{t} = v_{3}^{tur_{2}} = v_{1}v_{3}u_{1}u_{5} and so r_{1} = u_{1}u_{7}. Set v_{2}^{u} = qv_{2}r_{5},q^{u} = v_{2}r_{6}, where r_{5} ∈ C_{K_{4}}(qv_{2}). Then v_{2}^{ut} = v_{2}^{tur_{2}} and v_{2}^{uv_{3}} = v_{2}^{v_{3}uu_{1}u_{7}} show r_{5}^{v_{3}} = u_{7}r_{5} and r_{5} ∈ ⟨u_{1}, u_{4}, u_{5}, u_{6}, u_{7}⟩u_{8}u_{10}u_{11}. From v_{2}^{v_{6}u_{2}u_{4}uv_{6}u_{2}u_{4}} = v_{2}^{u^{2}} we get r_{6} = r_{5}^{u} and r_{5} ∈ ⟨u_{1}, u_{4}, u_{5}, u_{6}⟩u_{8}u_{10}u_{11} or r_{6} = r_{5}^{u}u_{1} and r_{5} ∈ ⟨u_{1}, u_{4}, u_{5}, u_{6}⟩u_{7}u_{8}u_{10}u_{11}. From v_{2} = v_{2}^{u^{3}} we get r_{6} = r_{5}^{u}. As r ∈ C_{K_{4}}(t) we see v_{2}^{(ysv_{2}u)^{3}} = v_{2} or v_{2}u_{4}. As v_{2}^{(ysv_{2}u)^{3}} = v_{2}r_{5}^{u}r_{2}^{(ysv_{2}u)}r_{5}^{u(ysv_{2}u)}r_{5}^{(ysv_{2}u)^{2}} we get by checking the possibilities for r_{5} and r_{2} that r_{2} = 1 and

r_{5} ∈ ⟨u_{1}, u_{4}, u_{5}⟩u_{8}u_{10}u_{11} if v_{2}^{ysv_{2}u} = v_{2} or
r_{5} ∈ ⟨u_{1}, u_{4}, u_{5}⟩u_{6}u_{8}u_{10}u_{11} if v_{2}^{ysv_{2}u} = v_{2}u_{4}.

Now q^{v_{6}u_{2}u_{4}uv_{6}u_{2}u_{4}} = q^{u^{2}} implies

r_{5} ∈ ⟨u_{1}, u_{5}⟩u_{8}u_{10}u_{11} or r_{5} ∈ ⟨u_{1}, u_{5}⟩u_{6}u_{8}u_{10}u_{11}.

As [x, u] = 1 we get v_{4}^{u} = v_{4}su_{11}r_{5}^{x} and s^{u} = v_{4}r_{6}^{x}u_{2}u_{6}. As s^{ut} = s^{tu} we then get r_{5} ∈ ⟨u_{5}⟩u_{6}u_{8}u_{10}u_{11} . The relation v_{4}^{ut} = v_{4}^{tu} then shows

r_{5} = u_{6}u_{8}u_{10}u_{11}.

Set (v_{1}v_{2}v_{5})^{u} = v_{2}v_{5}qpr_{9} and (v_{1}qp)^{u} = r_{10}v_{1}v_{2}v_{5}. We have r_{10} ∈ ⟨u_{1}, u_{2}u_{3}, u_{2}u_{6}, u_{4}, u_{5}, u_{7}, u_{10}u_{11}⟩ = C_{K_{4}}(v_{1}v_{2}v_{5}). As [u, t] = 1 we get r_{10}^{t} = r_{10}u_{5} and so r_{10} ∈ ⟨u_{1}, u_{4}, u_{5}, u_{7}, u_{10}u_{11}⟩u_{3}u_{6} . As [u, v_{1}] = u_{1}u_{7}, we get

r_{10} ∈ ⟨u_{1}, u_{4}, u_{5}, u_{7}⟩u_{3}u_{6} .

As [u, v_{3}] = u_{1}u_{7}, we get r_{10}^{v_{3}} = u_{1}r_{10} and so

r_{10} ∈ ⟨u_{1}, u_{5}, u_{7}⟩u_{3}u_{4}u_{6}.

As [u, x] = 1, we get

r_{10} ∈ ⟨u_{1}, u_{7}⟩u_{3}u_{4}u_{6}.

Applying the same relations to (v_{1}v_{2}v_{5})^{u} , we see

r_{9} ∈ ⟨u_{1}, u_{7}⟩u_{8}u_{10}u_{11}.

From v_{1}qp = (v_{1}qp)^{u^{3}} we get r_{10}^{u^{2}}r_{9}r_{10}r_{9}^{u} = 1 and so we have the following pairs for (r_{9}, r_{10}):

(u_{8}u_{10}u_{11}, u_{3}u_{4}u_{6}); (u_{7}u_{8}u_{10}u_{11}, u_{1}u_{3}u_{4}u_{6}); (u_{1}u_{7}u_{8}u_{10}u_{11}, u_{7}u_{3}u_{4}u_{6}) and (u_{1}u_{8}u_{10}u_{11}, u_{1}u_{7}u_{3}u_{4}u_{6}).

As (v_{1}qp)^{v_{6}u_{2}u_{4}u} = (v_{1}qp)^{u^{2}v_{6}u_{2}u_{4}}, we get by checking these pairs that just (u_{7}u_{8}u_{10}u_{11}, u_{1}u_{3}u_{4}u_{6}) and (u_{1}u_{7}u_{8}u_{10}u_{11}, u_{7}u_{3}u_{4}u_{6}) are possible. Now finally (v_{1}qp)^{(ysv_{2}u)^{3}} = (v_{1}qp)^{r} gives

r_{9} = u_{7}u_{8}u_{10}u_{11} and r_{10} = u_{1}u_{3}u_{4}u_{6}.

Now also r is determined and we have

r = u_{4}u_{8}.

So we have the following relations for u:

v_{1}^{u}u_{7}u_{1}v_{1}, [t, u], v_{3}^{u}u_{7}u_{1}v_{3}, u^{v_{6}}u_{10}u_{8}u_{4}u_{2}u, v_{2}^{u}u_{6}u_{11}u_{10}u_{8}qv_{2}, q^{u}v_{2}u_{4}u_{6}, v_{4}^{u}v_{4}su_{6}, s^{u}v_{4}, [x, u], (v_{1}v_{2}v_{5})^{u}u_{7}u_{8}u_{10}u_{11}pqv_{5}v_{2}, (v_{1}qp)^{u}v_{1}v_{2}v_{5}u_{3}u_{4}u_{6}u_{1}, (ysv_{2}u)^{3}u_{8}u_{4}.

So we have generators and relations for G_{4} in case |K_{4}| = 2^{11}. If |K_{4}| = 2^{12}, we have a further basis element u_{12} where

u_{12}^{v_{1}} = u_{3}u_{6}u_{12}, u_{12}^{v_{2}} = u_{1}u_{2}u_{6}u_{7}u_{12}, u_{12}^{v_{3}} = u_{12}, u_{12}^{v_{4}} = u_{12}, u_{12}^{v_{5}} = u_{2}u_{3}u_{7}u_{12}, u_{12}^{v_{6}} = u_{3}u_{6}u_{12}, u_{12}^{s} = u_{12}, u_{12}^{t} = u_{4}u_{10}u_{11}u_{12}, u_{12}^{q} = u_{1}u_{6}u_{11}u_{12}, u_{12}^{p} = u_{1}u_{3}u_{7}u_{11}u_{12}, u_{12}^{x} = u_{4}u_{10}u_{11}u_{12}, u_{12}^{y} = u_{1}u_{4}u_{6}u_{8}u_{11}u_{12}, u_{12}^{u} = u_{12}.

By lemma 3.3 we know that the action of G_{1} on K_{1}/Z(K_{1}) is uniquely determined. We choose v ∈ P_{4}, o(v) = 3. Adopting the notation of [Wie5] to our basis {u_{2}, ..., u_{7}, v_{1}, ..., v_{6}} of K_{1}/Z(K_{1}) we may choose v such that

[u_{6}, v] = 1 = [u_{7}, v] = [v, u_{3}v_{5}v_{2}] = [u_{5}, v] = [v, v_{1}u_{2}v_{2}v_{6}] = [v_{3}, v] = [u_{4}, v] = [v_{4}, v], u_{2}^{v} = v_{2}u_{6}u_{7}, v_{2}^{v} = v_{2}u_{2}u_{7}, v_{5}^{v} = u_{6}u_{7}v_{5}u_{2}v_{1}u_{3}, v_{6}^{v} = u_{7}v_{6}v_{2}u_{3}.

We have that ⟨S, v, x⟩ ≅ ⟨P_{3}, P_{4}⟩ and S/Z(K_{1}) splits over K_{1}/Z(K_{1}). So we have a complement in ⟨P_{3}, P_{4}⟩ to K_{1}/Z(K_{1}). Hence we may assume that we have the following further relations:

[u_{8}u_{10}u_{11}, v] = 1, u_{11}^{v} = qu_{11}u_{10}u_{8}, q^{v} = qu_{11}, s^{v} = s, u_{8}^{v} = t, t^{v} = u_{8}t, (u_{9}v)^{2} = 1, p^{v} = u_{11}p, (xv)^{4} = 1.

All these relations are modulo Z(K_{1}) = ⟨u_{1}⟩. We have ⟨S, y, v⟩ = ⟨P_{2}, P_{4}⟩. Calculating the action of [v, y] on K_{1}/Z(K_{1}) we see that this is the same as the action of u_{10}qt. As ⟨P_{2}, P_{4}⟩/K_{1} acts faithfully on K_{1}/Z(K_{1}) we have [v, y]qtu_{10} ∈ K_{1}. Applying [v, y]qtu_{10} to X := ⟨u_{8}u_{10}u_{11}, u_{11}, q, s, u_{8}, t, u_{9}, p, v, y⟩ we see that [v, y]qtu_{10} centralizes X. So [v, y]qtu_{10} is contained in C_{K_{1}/Z(K_{1})}(X) = 1. Now we have defining relations for G_{1}/Z(K_{1}).

Next we determine the relations involving v in G_{1}. If for some r ∈ G_{1} we have [v, r] ∈ Z(K_{1}) then [v, r] = 1. This shows that we just have to treat u_{2}^{v}, v_{2}^{v}, v_{5}^{v}, v_{6}^{v}, u_{11}^{v}, q^{v}, u_{8}^{v}, t^{v}, p^{v}, [v, y]. Working in ⟨P_{2}, P_{4}⟩ we check all possiblities by computer. This shows that we have the following relations:

v^{3}, [u_{1}, v], [u_{6}, v], [u_{7}, v], [v, u_{3}v_{5}v_{2}], [u_{5}, v], [v, v_{1}u_{2}v_{2}v_{6}], [v_{3}, v], u_{2}^{v}u_{6}u_{7}v_{2}u_{1}, [u_{4}, v], v_{2}^{v}u_{7}u_{2}v_{2}u_{1}, [v_{4}, v], v_{5}^{v}u_{3}v_{1}u_{2}v_{5}u_{7}u_{6}u_{1}, v_{6}^{v}u_{3}v_{2}v_{6}u_{7}, [u_{8}u_{10}u_{11}, v], u_{11}^{v}u_{8}u_{10}u_{11}qu_{1}, q^{v}u_{11}qu_{1}, [s, v], u_{8}^{v}t, t^{v}u_{8}t, (u_{9}v)^{2}, p^{v}pu_{11}u_{1}, [v, u_{1}u_{2}u_{6}u_{7}u_{8}u_{9}u_{10}u_{12}], [v, y]qtu_{10}u_{1}, (xv)^{4}.

To get some relation between u and v we have to move to G_{2}. We have (G_{2}/K_{2})^{(∞)} ≅ Z_{3}×A_{6} and so [v, u] ∈ K_{2}. We calculate [v, u] ∈ C_{K_{2}}(K_{4}∩K_{2}) = ⟨u_{1}, u_{2}, u_{3}, u_{5}, u_{6}, u_{7}, u_{11}⟩.

Now, as [v, u] ∈ [K_{2}, v] ∩ [K_{2}, u] we have [v, u] ∈ ⟨u_{1}u_{7}u_{2}, u_{1}u_{11}⟩. Checking the order of G_{2} by Todd-Coxeter we get [v, u] = u_{1}u_{2}u_{7}.

So we have shown that the amalgam is uniquely determined if |K_{4}| = 2^{11}.

So let |K_{4}| = 2^{12}. To prove uniqueness we just have to calculate u_{12}^{v}. From the known action on K_{1}/Z(K_{1}) we see that (vu_{12})^{2} = qu_{8}u_{10}u_{11} modulo K_{1}. Now again by applying the word (vu_{12})^{2}u_{11}u_{10}u_{8}q to the elements of X, we see that (vu_{12})^{2}u_{11}u_{10}u_{8}q ∈ C_{K_{1}/Z(K_{1})}(⟨u_{8}u_{10}u_{11}, q, s, t, y⟩) = ⟨v_{1}, v_{2}⟩Z(K_{1})/Z(K_{1}). As ((vu_{12})^{u})^{2} = (vu_{1}u_{7}u_{2}u_{12})^{2} we get [(vu_{12})^{2}, u] = [v_{2}qu_{8}u_{10}u_{11}, u]. But we can distinguish the elements in ⟨u_{1}, v_{1}, v_{2}⟩ by their commutator with u, so

(vu_{12})^{2} = v_{2}qu_{11}u_{10}u_{8}.

Hence proposition 5.1 is proved. □

The following proposition has been shown by A. Ivanov [Iv]. We present a proof using our generators and relations.

Proposition 5.2 The geometry for M(24) with diagram ∘  ∘  ∘  ∘^{∼}

is simply connected, i.e. if the maximal parabolic subgroups of G are as in M(24) then G ≅ M(24).

Proof: It is sufficient to prove it for the M(24)′ parabolic system. Now the additional information is (G_{1}/K_{1})^{(∞)} ≅ 3U_{4}(3). We add a further relation (s^{(xyx^{−1}sxy^{−1}sx^{−1})^{−1}}vu_{9})^{u} = 1. This is the relation, which comes from the Weyl group of U_{4}(2) inside of U_{4}(3). Then we can show that G_{1}^{∞} ≅ 2^{1+12}3U_{4}(3) just by using the Todd-Coxeter algorithm, and the group generated by {u_{1}, ..., u_{11}}{v_{1}, v_{2}, q, t, v_{3}v_{4}v_{5}, v_{3}v_{6}, ps, u^{v_{5}}, y, xyx^{−1}sxy^{−1}sx^{−1}} is a subgroup isomorphic to 2^{11}M_{23} in G_{4}. Furthermore the group generated by {u_{1}, ..., u_{11}} {v_{1}, v_{2}, q, t, v_{3}v_{4}v_{5}, v_{3}v_{6}, ps, y, xyx^{−1}sxy^{−1}sx^{−1}, v} is a subgroup isomorphic to 2^{2}2^{1+8}(U_{4}(2) × Z_{3}) · 2 in G_{1}. Hence these two groups generate a group H which belongs to the following diagram

∘  ∘  ∘  ∘^{M_{22}} ^{U_{4}(2)·2} .

Application of [Str6, Theorem 9] shows H ≅ M(23). Now using Todd-Coxeter we get |G : H| = |M(24)′ : M(23)| and so G ≅ M(24)′. □

The following lemma is due to A. Pasini.

Lemma 5.1 The geometry ∘  ∘  ∘  ∘^{∼}  for M(24) is not 2-simply

connected.

Proof: We have to show that the universal 2-cover of the given diagram is infinite. The diagram supplies us with the information about 2-residues, and they are like those in M(24). Let Γ be such a geometry. Let us choose the type

numbers like this: Δ = 1  2  3  4 ∘  ∘  ∘  ∘^{∼}  . Note, that all the 3-residues
which contain type 1, these are 1  3  4 ∘  ∘  ∘ , 1  2  3 ∘  ∘  ∘^{∼}  , and
1  2  4 ∘  ∘  ∘ are simply connected, so by [Pas, theorem 1] the universal

cover of the 1-truncation of Γ (that is, the geometry, that is left, if we drop all elements of Γ of type 1) is equal to the 1-truncation of the universal cover of Γ: T̃ r_{1}(Γ) = T r_{1}(Γ̃). So we are done if we can show that T̃ r_{1}(Γ) is infinite.

Now we want to get information about the diagram Δ_{1} of T r_{1}(Γ). Since the

diagram of Res(3) in Γ is 1  2  4 ∘  ∘  ∘ , we have Δ_{1} = ∘  ∘  ∘ 2  3  4 ......... ......... .
Res(2) in Γ has the diagram 1  3  4 ∘  ∘  ∘ , and thus Δ_{1} = ∘  ∘  ∘ 2  3  4 .........  .

We now want to determine the gonality g of the residue of type 4 in T r_{1}(Γ).

(1) g ≥ 3.

Suppose g = 2. Then there are e_{1}, e_{2} of type 3 and l_{1}, l_{2} of type 2, such that e_{1} and e_{2} are incident with both, l_{1} and l_{2}.

[figure: ∘; ∘; ∘; ∘; l_{1}; l_{2}; e_{1}; e_{2}]
Consider Res(4) in Γ with diagram 1  2  3 ∘  ∘  ∘^{∼} . We find p of type 1

in the residue of e_{2}, such that p is incident with l_{1} and l_{2}, since the residue is a projective plane.

[figure: ∘; ∘; ∘; ∘; ∘; l_{1}; l_{2}; e_{1}; e_{2}; p]

But now there is a generalized 2-gon in the residue of p. This is not possible in

Res(4), since the gonality of 2  3 ∘  ∘^{∼}  is 5. So we have g ≥ 3.

Let us define a property of diagram geometries (PL):

(PL) is true in a geometry of rank 3, with the types 1,2,3, if for any p of type 1, l of type 2, and u, v of type 3, where l and p are incident with u and v, then necessarily p is incident with l:

[figure: ∘; ∘; ∘; ∘; u; v; l; p; .........]
(2) (PL) holds in the geometry of M_{24} with diagram 1  2  3 ∘  ∘  ∘^{∼}  .

In this geometry for M_{24} in terms of the 11- or 12-dimensional module over GF (2) the points, lines, and planes are certain 1-, 2-, and 3-spaces, where two different spaces are incident if one is included in the other (See [RoStr] or the action of G_{4} on K_{4} in proposition 5.1). So if we have two different 3-spaces, which intersect in a 2-space and a 1-space, the 1-space must lie inside the 2space. Hence (2) is proved.

(3) g ≥ 4.

Suppose g = 3. Then we can find elements m, n, l of type 2 and elements u, v, w of type 3, such that their incidence graph is a generalized 3-gon:

[figure: ∘; ∘; ∘; ∘; ∘; ∘; n; l; v; w; m; u; ∘; ∘; ∘; ∘; ∘; ∘; ∘; n; l; v; w; m; u; p]

Consider Res(4) again. Here we can find an element p of type 1 in the residue of w such that p is incident with m and l. Now p is incident with u and v, since it is in the residue of m and l.

But because (PL) holds in the M_{24} - geometry, p is incident with n and we have found a generalized 3-gon in the residue of p in the geometry with diagram

1  2  3 ∘  ∘  ∘^{∼}  , but this is a contradiction. So g ≥ 4. We have seen, that

the gonality graph of T r_{1}(Γ) looks like this: ∘  ∘  ∘ ≥ 4  4  . Note that T r_{1}(Γ)

is thick, since it is a truncation of a thick geometry. Now we apply [Pas, theorem 2] and T̃ r_{1}(Γ) is infinite.□

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