A Computer-free Construction of the Lyons Group

U. Meierfrankenfeld Department of Mathematics Michigan State University

East Lansing, Michigan 48824

USA meier@math.msu.edu

and

C.W. Parker School of Mathematics and Statistics University of Birmingham Birmingham B15 2TT

UK

cwp@for.mat.bham.ac.uk

July 11, 2003

Abstract

A finite group L is of Ly-type provided that L is simple and has an involution z with C_{L}(z) isomorphic to the double cover of Alt(11), 2.Alt(11). In this paper we give a computer-free construction of a group of Ly-type.

1 Introduction

The first evidence for the existence of a sporadic simple group with centralizer of an involution isomorphic to the double cover of the alternating group of degree 11 was given by Lyons [Ly] in 1972. Subsequently, Sims [Si] constructed such a group and proved its uniqueness. His proof provides generators and relations for the group and uses a computer to determine the index of a subgroup which is isomorphic to G_{2}(5). The linear representations of the Lyons group such as the Meyer–Neutsch–Parker 111-dimensional GF(5) representation [MNP] or the Jansen–Wilson 651-dimensional GF(3) representation [JW] ultimately depend on the relations that Sims produced to show that the group their matrices generate is the Lyons group. In [CFYT] the 111-dimensional GF(5) matrix group was used to obtain permutations which generate the Lyons group and using these permutations a further computer based existence proof has been given by Gollan [Gol]. A more elementary, though still computer dependent, uniqueness proof can be found in Wilson [Wi]. In [AschS] Aschbacher and Segev gave the first computer-free uniqueness proof of a group of Ly-type. The purpose of this paper is to give a hand construction of a group of Ly-type. Thus our main theorem is

Theorem 1.1 There exists a group of Ly-type.

Formally, for a natural number n, we follow Ivanov [Iv] and define an amalgam A of rank n to be a set M such that for each 1 ≤ i ≤ n there are subsets M_{i} and a binary operation ∗_{i} defined upon M_{i} such that the following hold:

(i) M = ∪_{i=1}^{n}M_{i};

(ii) ∩_{i=1}^{n}M_{i} ≠ ∅;

(iii) (M_{i}, ∗_{i}) is a group; and

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

Generally, A is denoted by the n-tuple of groups (M_{1}, …, M_{n}). And, for J ⊆ {1, …, n}, we use M_{J} to denote the subset ∩_{j∈J}M_{j}. A completion of the amalgam (M_{1}, …, M_{n}) is a group G and a mapping θ from ∪_{i=1}^{n}M_{i} to G such that

(i) G = ⟨θ(M)⟩; (ii) for 1 ≤ i ≤ n, θ|_{M_{i}} is a group homomorphism.

Note that every amalgam has the trivial group as a completion. Every amalgam also has a universal completion, this completion maps surjectively on to every other completion (see [Iv, 1.3.2]). Note that a non-trivial amalgam may have the trivial group as its universal completion. A completion is called faithful provided that θ is injective.

Our approach to the proof of Theorem 1.1 is similar to that followed by Ivanov and Meierfrankenfeld in [IM] when they constructed J_{4}: we show that the universal completion of a certain amalgam of groups is a group of Ly-type or is trivial. Then we prove that the given amalgam exists in GL_{111}(5).

The amalgams that we shall be concerned with are defined as follows:

Definition 1.2 An amalgam (M_{1}, M_{2}, M_{3}) is called a Ly-amalgam provided that

1. M_{1} ∼ 3.McL.2, M_{2} ∼ 3^{6}.2^{3}.Sym(5) and M_{3} ∼ 3^{5}.2.Mat_{11}.

2. |M_{2} : M_{12}| = 2, |M_{2} : M_{23}| = 10 and |M_{3} : M_{13}| = 11.

3. |M_{23} : M_{123}| = 2.

4. No non-trivial subgroup of M_{123} is normal in M_{1}, M_{2} and M_{3}.

Notice that, a priori, there may be a multitude of Ly-amalgams up to isomorphism.

Our main theorem follows from the more specific result

Theorem 1.3 Suppose that T is a Ly-amalgam. Then

(a) T is unique up to isomorphism.

(b) There exists an amalgam isomorphic to T in GL_{111}(5).

(c) The universal completion of T is a group of Ly-type.

In particular, there exists a group of Ly-type.

The definition of a Ly-amalgam is motivated by [Ly, Propositions 2.3, 2.5 and 2.7] from which it can be deduced that, if there exists a group of Ly-type, then it is a completion of a Ly-amalgam. Using this fact and Theorem 1.3 we obtain Corollary 1.4 There exists a unique group of Ly-type.

The first five sections of this article contain the preparatory results needed for the proof of 1.3. Specifically, Section 2 contains general results about amalgams, information about the 5-dimensional GF(3)Mat_{11}-modules and facts about representations of extraspecial groups. Section 3 contains a characterization of the alternating groups, 3.2, which will be used in Section 7 to show that, if there is a faithful completion of a Ly-amalgam, then it contains a subgroup isomorphic to 2.Alt(11). In Section 4 we gather a myriad of facts about McL and Aut(McL). Section 5 is given over to the construction of a group H ≅ 3.McL and a 90-dimensional representation for 3.Aut(McL) over GF(5). This section is necessary because the current proof of the existence of the triple cover of McL relies on the existence of the Lyons group.

In Section 6 we get to grips with Ly-amalgams. In 6.3 we prove that a Lyamalgam exists and in 6.4 we prove that, up to isomorphism, there is only one. Note that this is not the same thing as showing that the universal completion of a Ly-amalgam is not trivial. We need the results of Section 6 for the construction of a Ly-amalgam in GL_{111}(5) in Section 8. In Section 7 we suppose that M is a faithful completion of a Ly-amalgam and analyse the coset graph Γ of M_{1}, M_{2} and M_{3} in M; our aim is to determine the orbits of M_{1} on its cosets in M. We quickly find five orbits, 7.3, and the remainder of the section is devoted to proving that these are the only orbits. Of particular consequence is 7.4, where a subgroup H of M isomorphic to 2.Alt(11) is uncovered (though, of course, at this stage we do not know that it is the centralizer of an involution in the whole of M). Not only is this positive progress towards showing that M is of Ly-type, but this subgroup is very easy to calculate in! We add the cosets of this subgroup to our coset graph with incidence defined in the usual way. Setting Z = Z(H), in 7.7, we show a certain connected component of the subgraph of Γ which is fixed by Z, in fact has H as its stabilizer in M. This then allows us to prove that if M_{1} and M_{1}h are in the same connected component of the fixed graph of a conjugate of Z, then M_{1}h lies in one of the orbits that we have already found. Next we focus on the normalizer of a certain subgroup F of order five. We find that M contains a subgroup N of shape 5^{1+4}.4.Sym(6), a connected component of the fixed graph of F has stabilizer N and that, once again, if M_{1} and M_{1}h are in the same connected component of the fixed graph of F, then M_{1}h is already accounted for in one of our five orbits. Finally in 7.15 we show that there are just the five orbits found in the very beginning of the endeavour. The detailed theorem of Section 7 is

Theorem 1.5 Suppose that M is a faithful completion of a Ly-amalgam of groups (M_{1}, M_{2}, M_{3}). Then

(a) M_{1} has five orbits on the right cosets of M_{1} in M. Moreover, the orbit stabilisers have shape 3.Aut(McL), 3^{2+4}.4.Sym(5), 2.Sym(7), 4.Sym(6) and 5^{1+2}.Sym(3).

(b) M has order 2^{8} · 3^{7} · 5^{6} · 7 · 11 · 31 · 37 · 67.

(c) M has a unique class of involutions and C_{M}(t) ≅ 2.Alt(11) for any involution t in M.

(d) For i = 1, 2, 3 , M_{i} is a maximal 3-local subgroup of M.

(e) M has a subgroup F of order five with N_{M}(F) of shape 5^{1+4}.4.Sym(6).

(f) M is a group of Ly-type.

In our final section we construct a Ly-amalgam in GL_{111}(5). Once this is done we have completed the proof of 1.3.

We close the introduction with some comments on our notation. Suppose that Γ is a graph with varying types of vertices. If e ∈ Γ, then Γ(e) is the set of neighbours of e. If i is a type of a vertex of Γ, then Γ_{i}(e) = {x ∈ Γ(e) | x has type i}. If A operates as a group of automorphisms on Γ, then Γ^{A} represents the induced subgraph of Γ consisting of the vertices fixed by A. Also for G ≤ Aut(Γ) and e ∈ Γ, G_{e} will be the stabilizer in G of e.

The alternating, symmetric and Mathieu groups of degree n are represented by Alt(n), Sym(n) and Mat_{n} respectively. We shall use Mat_{9} to represent the point-stabiliser in Mat_{10}. We use D_{n}, Q_{n} and SD_{n} to represent the dihedral, quaternion and semidihedral groups of order n respectively. We have been terribly inconsistent with our notation for cyclic groups; we either use C_{n} or when appropriate simply n (rarely seen alone) for a cyclic group of order n. For a prime p, we use p^{a+b+c} to represent a p-group of order p^{a+b+c}. Often the summands in the superscript indicate that a group operating on the p-group has a chief factor of that order on the group. Beware the group p^{a+b+c} may be abelian so, for example, p^{1+4} does not necessarily mean an extraspecial group of order p^{5}. We write G ∼ A.B. . . . .Z or say that G has shape A.B. . . . .Z when G has a normal series with factors isomorphic to A, B, …, Z. Thus for example M_{2} ∼ 3^{6}.2^{3}.Sym(5) indicates that M_{2} contains normal subgroups of order 3^{6} and 3^{6}.2^{3} and a factor group isomorphic to Sym(5). It does not specify the structure of the normal 3-subgroup or that of the Sylow 2-subgroup of the normal subgroup of order 3^{6}.2^{3}. However, M_{2} ∼ 3^{6}.C_{8}.Sym(5) does specify the structure of Sylow 2-subgroup of the subgroup of order 3^{6}.2^{3}. We follow Atlas notation and write ‘.’ to mean non-split extension; thus 2.Alt(5) ≅ SL_{2}(5). Also, on some occasions, we shall use the symbol ‘:’ to denote a split extension such as 2^{2} : Sym(3) ≅ Sym(4). Hopefully the remainder of our notation is standard and can be found in [Go].

Acknowledgement We thank Gerhard Hiss for his contribution to this project. Particularly, he provided the original proof on which the presented proof of Theorem 5.2 is based. Moreover, he has read the entire manuscript and has suggested a number of clarrifications.

2 Preliminaries

Lemma 2.1 Suppose that Γ is a graph, G is a group acting on Γ and A⊴H ≤ G. Assume that F is a non-empty connected subset of Γ^{A} such that

1. N_{G_{e}}(A) = H_{e} for all e ∈ F.

2. If e in F and d ∈ Γ^{A}(e), then d is H-conjugate to some element in F.

3. H = ⟨H_{e} | e ∈ F⟩.

4. No two distinct elements of F are conjugate under N_{G}(A).

Let Ψ be the connected component of Γ^{A} containing F. Then Ψ = {e^{h} | e ∈ F, h ∈ H} and H = N_{G}(A) ∩ Stab_{G}(Ψ).

Proof: Let Ψ^{∗} = {e^{h} | e ∈ F, h ∈ H}. Since Ψ is a connected component of Γ^{A}, each H_{e}, e ∈ F, stabilizes Ψ and so (3) implies that H stabilizes Ψ. Thus Ψ^{∗} ⊆ Ψ. Suppose now that a ∈ Ψ and d(F, a) = n. If n = 1, then there is an e ∈ F such that a ∈ Ψ(e) and (2) implies that Ψ(e) ⊆ Ψ^{∗}. Hence a ∈ Ψ^{∗}. We now proceed by induction on n. Let b ∈ Ψ(a) with d(F, b) = n − 1. Then by induction there exists h ∈ H such that b^{h} ∈ F. But then either a^{h} ∈ F (so a ∈ Ψ^{∗}) or Ψ(a^{h}) ∩ F is non-empty. In the latter case (2) shows that there is an h_{1} in H such that a^{hh_{1}} ∈ F. Therefore in any case a ∈ Ψ^{∗} and so Ψ = Ψ^{∗}. Let e ∈ F and g ∈ N_{G}(A) ∩ Stab_{G}(Ψ). Then, as Ψ = Ψ^{∗}, there exist d ∈ F and h ∈ H such that e^{g} = d^{h}. But then (4) implies e = d and (1) gives gh^{−1} ∈ H. Therefore, g ∈ H and so N_{G}(A) ∩ Stab_{G}(Ψ) ≤ H. Since trivially H ≤ N_{G}(A) ∩ Stab_{G}(Ψ), we have N_{G}(A) ∩ Stab_{G}(Ψ) = H as claimed.

Remark 2.2 A set F as in 2.1 can often be found using (variations of) the following procedure:

0. Pick some e ∈ Γ^{A} and put F = {e} = N.

1. For all e ∈ N verify that N_{G_{e}}(A) = H_{e}. This for example can be done by finding a subset X of F such that N_{G_{x}}(A) = H_{x} for all x ∈ X and N_{G_{e}}(A) = ⟨N_{G_{ex}}(A) | x ∈ X⟩.

2. For all e ∈ N determine a set of representatives for the orbits of N_{G_{e}}(A) on Γ^{A}(e).

3. Replace N by the orbit representatives found in (2) which are not Hconjugate to a member of F.

4. Replace F by F ∪ N. Finally, if elements of F are conjugate under H, remove all but one of them from F.

5. If N ≠ ∅ go back to (1).

6. Verify that no two distinct elements of F are conjugate under N_{G}(A).

The above procedure will be followed in 7.7 and 7.12 in Section 7.

Lemma 2.3 Let G and T be groups, L ⊴ G and H ≤ G with G = LH. Assume ϕ : H → T is a homomorphism and suppose that there exists a unique homomorphism ψ : L → T with ψ |_{H∩L}= ϕ |_{H∩L}. Then there exists a unique homomorphism Φ : G → T with Φ |_{H}= ϕ.

Proof: Clearly if Φ exists then it is unique and Φ(lh) = ψ(l)ϕ(h) for all l ∈ L and h ∈ H. It remains to verify that Φ is well-defined and is a homomorphism.

So suppose that lh = l̃h̃. Define k = hh̃^{−1} = l^{−1}l̃and observe that k ∈ H∩L. Then

Φ(l̃h̃) = ψ(l̃)ϕ(h̃) = ψ(lk)ϕ(k^{−1}h)
= ψ(l)ψ(k)ϕ(k^{−1})ϕ(h) = ψ(l)ψ(k)ψ(k^{−1})ϕ(h)
= ψ(l)ϕ(h) = Φ(hl).

Thus Φ is well-defined.

An elementary calculation shows that Φ is an homomorphism if and only if (2.3.1) for all l ∈ L, h ∈ H we have

ψ(l) = ϕ(h)^{−1}ψ(l^{h^{−1}})ϕ(h).

Fix h ∈ H and define ψ^{∗} : L → T by ψ^{∗}(l) = ϕ(h)^{−1}ψ(l^{h^{−1}})ϕ(h). Then ψ^{∗} is the composition of three homomorphisms, namely conjugation by h^{−1} followed by ψ and finally conjugation by ϕ(h). Therefore, ψ^{∗} is a homomorphism from L to T. Clearly ψ^{∗} equals ϕ when restricted to H ∩L and thus, by the uniqueness of ψ, ψ = ψ^{∗} and (2.3.1) is demonstrated. Hence Φ is a homomorphism and the lemma is proved.

The next lemma is extracted from [IM, 2.2]. It will be used in Section 8 when we construct a representation of a Ly-amalgam.

Lemma 2.4 Let (M_{1}, M_{2}, M_{3}) be an amalgam, H a group and A a subgroup of Aut(H). Suppose that, for each i ∈ {1, 2, 3}, there exist homomorphisms

α_{i} : M_{i} → H

and elements a_{i} ∈ A such that the homomorphisms

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

Put M_{23}^{∗} = M_{23}^{α_{3}a_{2}^{−1}}, M_{13}^{∗} = M_{13}^{α_{1}}, M_{12}^{∗} = M_{12}^{α_{1}} and B^{∗} = M_{123}^{α_{1}}. Then

(a) The following two statements are equivalent:

(a1) There exist b_{1}, b_{2}, b_{3} ∈ A such that for all {i, j} ⊂ {1, 2, 3}

α_{i}b_{i}|_{M_{ij}} = α_{j}b_{j}|_{M_{ij}}.

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

(b) B^{∗} ≤ ⋂_{{i,j}⊂{1,2,3}}M_{ij}^{∗} and a_{2}a_{1}a_{3} ∈ C_{A}(B^{∗}). In particular, (a1) and (a2)

hold if

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

(c) Assume that (a1) holds and that for each i ∈ {1, 2, 3}, α_{i} is one-to-one. For each i ∈ {1, 2, 3} set M_{i}^{∗} = M^{α_{i}b_{i}}. If M_{i}^{∗}∩M_{j}^{∗} = M_{ij}^{α_{i}b_{i}} for all 1 ≤ i < j ≤ 3, then (M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗}) is an amalgam isomorphic to (M_{1}, M_{2}, M_{3}).

Next we present some well-known facts about the GF(3)-representations of Mat_{10} and Mat_{11}.

Lemma 2.5 Suppose that H ≅ Mat_{11}, K ≤ H with K ≅ Mat_{10} and E is the field with 3 elements.

(a) There is a unique irreducible EH-module, V, of dimension 5 in which K leaves invariant a 1-space.

(b) There are two irreducible EK′-modules of dimension 4. One is obtained from the other by tensoring with the −1-representation.

(c) When restricted to K, V is an indecomposable EK-module with composition factors of dimension 1 and 4.

(d) The orbits of H on the 1-dimensional subspaces of V have lengths 11 and 110 with stabilisers K and 3^{2}:Q_{8} respectively. Moreover, K operates non-trivially on the 1-space fixed by K.

(e) K has orbits of length 10 and 30 on the 1-spaces and hyperplanes of either of the 4-dimensional irreducible modules. The 1-space stabilisers are Mat_{9} ≅ 3^{2}:Q_{8} and Sym(4) respectively.

(f) Suppose that W is an irreducible 4-dimensional EK-module. Then, up to isomorphism, there is a unique indecomposable EK-module of dimension 5 with a quotient isomorphic to W and a submodule of dimension 1 which is inverted by K. Furthermore, dim_{E} H^{1}(K, W) = 1.

(g) Let f be an element of order 4 in H. Then dim_{E}(C_{V} (f)) = 1 and dim_{E}(C_{V} (f^{2})) = 3.

Proof: ¿From [Jam, Theorem 7.1] we see that the 11-dimensional EH-module W = (−1)_{K}^{H} has composition factors 1, 5 and ¯5 (using the notation from [Jam]). Moreover, from [Jam, Section 7] we have that 5 and ¯5 are the only 5-dimensional irreducible modules; they are dual to one another. Clearly 1 is not a quotient of W as K centralizes it and, similarly, W has no submodule centralized by K. Hence W is a uniserial module 5/1/¯5. This proves (a).

¿From [Be, pg. 208] or [MATLAS] we have that K′ ≅ Alt(6) has a unique irreducible E-module of dimension 4. This module extends to K and so (b) holds.

Let V be as in part (a). Then V restricted to K has at least one 1dimensional composition factor. Since Mat_{10} has no E-modules of dimension less than 4, we have (c).

¿From (a), we see that H has an orbit of length 11 on the 1-spaces of V. Let {v_{1}, …, v_{11}} be this orbit. Then the normalizer of the subspace U = ⟨v_{1}, v_{2}⟩ is L ≅ Mat_{9} ∼ 3^{2}:Q_{8} and L inverts U. Hence L fixes the 1-space ⟨v_{1} + v_{2}⟩. It follows that ⟨v_{1} + v_{2}⟩ is a representative for an orbit of length 110. Since there are only 121 subspaces of dimension 1 in V, we conclude that (d) holds.

For part (e) we set U = V/⟨v_{1}⟩. Since (⟨v_{1} + v_{2}⟩ + ⟨v_{1}⟩)/⟨v_{1}⟩ is a 1-space fixed by L, we conclude that K has an orbit of length 10 on W. Since Sym(4) fixes a 1-space, we also have an orbit of length 30.

Let R be the 4-dimensional EAlt(6)-module. Then it is easy to show that H^{1}(Alt(6), R) has dimension at most 2. Since the 5-dimensional submodule of the permutation module for Sym(6) over E has centralisers of elements of order 3 and cycle shape (1^{3}, 3) of dimension 3 and of elements of cycle shape (3^{2}) of dimension 2, we see that this 5-dimensional module does not admit K. Since Aut(Alt(6)) acts on H^{1}(Alt(6), R), we conclude that H^{1}(Alt(6), R) has dimension 2 and that Out(Alt(6)) acts non-trivially on this space. We deduce that H_{1}(K, W) has dimension 1. Tensoring with the −1-representation of K now delivers (f).

We obtain part (g) from the ordinary character table of Mat_{11} [ATLAS] or [Frob] and the decomposition matrix [Jam, Theorem 7.1].

Lemma 2.6 Suppose that p is an odd prime, P is an extraspecial p-group of exponent p and order p^{1+2n}. Let K = C_{Out(P)}(Z(P)), H = P:K ∼ p^{1+2n}:Sp_{2n}(p), E ≤ Q be elementary abelian of maximal order and E be the set of maximal subgroups of E which complement Z(P). Assume that r is a prime with r ≠ p and let V be an irreducible GF(r)P -module of dimension dp^{n} where d is the smallest integer such that GF(r^{d}) contains a pth root of unity. Then

(i) V = ⊕_{E^{∗}∈E}C_{V} (E^{∗}) where, for each E^{∗} ∈ E, C_{V} (E^{∗}) has dimension d and admits Z(P) irreducibly.

(ii) V extends uniquely to a GF(r)H-module.

(iii) Suppose that E^{∗} ∈ E and let L = N_{K}(E^{∗})/C_{K}(E^{∗}). Then L ≅ GL_{n}(p) and acts on C_{V} (E^{∗}). Furthermore, for x ∈ L, either det_{E^{∗}}(x) is a square in GF(p) and centralizes C_{V} (E^{∗}), or x inverts C_{V} (E^{∗}).

Proof: Part (i) is well known. Since the irreducible GF(r)P -representations are uniquely determined by the action of Z(P) and H centralizes Z(P), the irreducible modules for P all extend uniquely to H. Thus (ii) holds.

This brings us to (iii). We have N_{K}(E) is a subgroup of K isomorphic to p^{\binom{n}{2}+n}:GL_{n}(p). Set K^{∗} = N_{K}(E^{∗}) and L = K^{∗}/C_{K^{∗}}(E^{∗}). Then L ≅ GL_{n}(p). Now L acts on C_{V} (E^{∗}) and, as K^{∗} commutes with Z(P), the only non-trivial action comes from the cyclic group L/L′ ≅ GF(p)^{∗} of order p − 1. Select an element λ of L which projects to a generator of L/L′ and such that P_{1} = [P, λ] is an extraspecial group of order p^{3}. Then P_{2} = C_{P} (λ) is an extraspecial pgroup of order p^{1+2(n−1)}. Note that P_{1}⟨λ⟩ commutes with P_{2}. Thus Clifford’s Theorem implies that V restricted to P_{1}⟨λ⟩ is a single homogeneous component. So V restricted to P_{1}⟨λ⟩ is a direct sum of p^{n−1} isomorphic irreducible modules of dimension dp. Let V_{1} = ⟨C_{V} (E^{∗})^{P_{1}}⟩. Then V_{1} is a submodule of V of dimension dp. Applying (i) to V_{1}, we see that V_{1} is a sum of p d-dimensional spaces T and on these subspaces ⟨λ⟩ acts with orbits of length 1 and p − 1. So λ induces a p − 1-cycle π_{λ} on T \ {C_{V} (E^{∗})}. As π_{λ} is an odd permutation, det_{V_{1}}(λ) = −det_{C_{V} (E^{∗})}(λ). Since V is a direct sum of p^{n−1} P_{1}⟨λ⟩-modules isomorphic to V_{1} and det_{V} (λ) = 1, we deduce that det_{V_{1}}(λ) = −1. This proves (iii).

Lemma 2.7 Suppose that Q is an extraspecial group of order 3^{5}, K ≤ Aut(Q) with K ≅ 2.Alt(5) ≅ SL_{2}(5), H = Q : K. Let B be the normalizer of a Sylow 3-subgroup of H and i ∈ {18, 36}. Then

(a) There exists a unique irreducible faithful i-dimensional GF(5)H-module V_{i}.

(b) V_{36} ≅ V_{18} ⊗ V_{2}, where V_{2} is the unique 2-dimensional GF(5)H-module.

(c) V_{i} is uniquely determined by its restriction to B. More precisely, if ϕ and ψ are two irreducible embeddings of H into GL_{i}(5) with ϕ_{B} = ψ_{B}, then ϕ = ψ.

(d) Let E be the unique normal subgroup of order 3^{3} in B and let E^{∗} be a complement to Z(Q) in E. Assume that T ≅ C_{4} is a Sylow 2-subgroup of N_{B}(E) and let t ∈ T be an involution. Then T centralizes C_{V_{18}}(E^{∗}) and

t inverts C_{V_{36}}(E^{∗}).

Proof: We first prove parts (a) and (b). Note that, by 2.6 (ii), H has a faithful irreducible 18-dimensional GF(5)H-module. Tensoring this module by the natural GF(5)K-module V_{2} gives an irreducible GF(5)H-module of dimension 36. Our task is to show that these are the unique irreducible modules of these dimensions. To do this we show that there is a unique (up to conjugacy) embedding of H in GL(V_{i}). Suppose now that i ∈ {18, 36} and V_{i} is an irreducible GF(5)H-module of dimension i. Then V_{i}|_{Q} is the direct sum of j = \frac{i}{18} (isomorphic) irreducible GF(5)Q-modules of dimension 18. In particular, V_{i} is uniquely determined, up to isomorphism, as a GF(5)Q-module. Note that N_{GL(V_{i})}(Q) ∼ (3^{1+4} : Sp_{4}(3)∘GL_{j}(25)).2 (where ∘ denotes central product) and so N_{GL(V_{i})}(Q)^{∞} = L_{1} × L_{2}, with L_{1} ∼ 3^{1+4}.Sp_{4}(3) and L_{2} ∼ SL_{j}(25). Since H is perfect and C_{H}(Q) ≤ Z(H), the Three Subgroups Lemma implies that H has no non-trivial automorphism centralizing Q. Thus the embedding of Q into L_{1} can be extended uniquely to an embedding of H into L_{1}. Moreover, if j = 2 there exists a non-trivial homomorphism of H/Q ≅ SL_{2}(5) into L_{2} and this homomorphism is unique up to conjugation in GL_{2}(25). Thus (a) and (b) hold.

We now move on to part (c). By (a) ψ = ϕρ for some ρ ∈ Inn(GL(V_{i})). Since, by hypothesis, ψ_{B} = ϕ_{B}, ρ centralizes B^{ϕ}. As B acts irreducibly on V_{i}, End_{B}(V_{i}) = End_{H}(V_{i}) ≅ GF(25). Thus ρ ∈ End_{H}(V_{i}) and so ρ centralizes H^{ϕ}. Therefore, ϕ = ψ.

Finally we prove part (d). By 2.6 (iii), T centralizes C_{V_{18}}(E). Moreover, t inverts V_{2} and so by (b), t inverts C_{V_{36}}(E).

3 A Characterization of the non-Abelian Simple Alternating Groups

In this section we show that certain amalgams of groups characterize the alternating groups. We will employ this characterization in Section 7 to identify a subgroup isomorphic to 2.Alt(11) in a faithful completion of a Ly-amalgam of groups. We begin with a definition.

Definition 3.1 Let k and l be integers greater than 2, n = k + l, Ω be a set of size n and for i ∈ {k − 1, k, k + 1}, Ω_{i} be a subset of Ω of size i such that Ω_{k−1} ⊂ Ω_{k} ⊂ Ω_{k+1}. Then an amalgam of type (k, l)-Alt(n) is an amalgam of groups isomorphic to the amalgam (Stab_{Alt(Ω)}(Ω_{i}) | k − 1 ≤ i ≤ k + 1) in Alt(Ω).

Observe that Stab_{Alt(Ω)}(Ω_{i}) = Stab_{Alt(Ω)}(Ω \ Ω_{i}) and thus the definition of an amalgam of type (k, l)-Alt(n) is symmetric in k and l.

Theorem 3.2 Suppose that M is a faithful completion of an amalgam (M_{1}, M_{2}, M_{3}) of type (k, l)-Alt(n). If n ≥ 5, then M ≅ Alt(n).

Proof: Without loss of generality we may assume that M is the universal completion of the amalgam. Then there exists a homomorphism ^{−} : M → Alt(Ω) such that

(\overline{M_{j}} | 1 ≤ j ≤ 3) = (Stab_{Alt(Ω)}(Ω_{i}) | k − 1 ≤ i ≤ k + 1),

where Ω and Ω_{i} are as in 3.1.

Suppose first that k = 2. Then M_{1} ≅ Alt(n − 1), M_{2} ≅ Sym(n − 2) ∧^{|}Sym(2) and M_{3} ≅ Sym(n − 3) ∧^{|}Sym(3). Observe that M_{12} ≅ Alt(n − 2) has index 2 in M_{2} and is normal therein. Also M_{123} ≅ Alt(n − 3), M_{13}/M_{123} and M_{23}/M_{123} are two different groups of order two in M_{3}/M_{123}. Thus, as M_{3}/M_{123} ≅ Sym(3), M_{3} = ⟨M_{13}, M_{23}⟩. Therefore, M = ⟨M_{1}, M_{2}⟩.

Choose an involution t ∈ M_{23}\M_{1}. Then, as M_{12}⊴M_{2}, M_{12} ≤ M_{1}∩M_{1}^{t} and, as \overline{M_{1}}∩\overline{M_{1}^{t}} = \overline{M_{12}}, M_{12} = M_{1}∩M_{1}^{t}. Let Γ be the graph with vertices the cosets of M_{1} in M and edges the translates of (M_{1}, M_{1}t) by right multiplication with elements from M. Then M operates faithfully on Γ and, as M = ⟨M_{1}, M_{2}⟩ = ⟨M_{1}, t⟩, Γ is connected.

Since |M_{3} : M_{13}| = 3, M_{1}M_{3} contains three cosets of M_{1}: M_{1}, M_{1}t and M_{1}s where s and t are involutions (t being the one above in M_{23}). Moreover, M_{3} acts 2-transitively on these three cosets and so M_{1}s is adjacent to both M_{1} and M_{1}t; in particular, Γ(M_{1}t) ∩ Γ(M_{1}) ≠ ∅. Now M_{1}^{t} is the stabilizer in M of the vertex M_{1}t and (as n ≥ 5) M_{1}^{t} operates 2-transitively on Γ(M_{1}t). Hence, as Γ(M_{1}t) ∩ Γ(M_{1}) ≠ ∅, every neighbour of M_{1}t is a neighbour of M_{1}. Since M_{1} has exactly n − 1 neighbours, we find that |Γ| ≤ 1 + (n − 1) = n. Thus |M| ≤ n · |M_{1}| = \frac{n!}{2} and, as Alt(n) is an image of M, we finally conclude that M ≅ Alt(n) in this case.

We now assume that k ≥ 3 and by symmetry also that l ≥ 3. Let Ω_{k−2} be a set of size k − 2 contained in Ω_{k−1}. For i = 1, 2, 3 let P_{i} be the subgroup of M_{i} which maps to the subgroup of \overline{M_{i}} which fixes Ω_{k−2} pointwise. Then (P_{i}|1 ≤ i ≤ 3) is of type (2, l)-Alt(l + 2). Let P = ⟨P_{i} | 1 ≤ i ≤ 3⟩. As l ≥ 3, l + 2 ≥ 5 and so, by the k = 2 case, P ≅ Alt(l + 2). Again using l ≥ 3, we have Stab_{M_{123}}(Ω_{k−2}) induces Sym(k − 2) on Ω_{k−2} and normalizes every generating subgroup of P and hence P. Put M_{0} = Stab_{M_{123}}(Ω_{k−2})P. Then M_{0} ∼ Alt(l + 2).Sym(k − 2), M_{0} ≅ \overline{M_{0}} = Stab_{Alt(Ω)}(Ω_{k−2}), and for i = 1, 2, M_{0} ∩ M_{i} = Stab_{M_{123}}(Ω_{k−2})P_{i} ≅ \overline{M_{0}} ∩ \overline{M_{i}}. Thus (M_{i} | 0 ≤ i ≤ 2) is of type (k − 1, l + 1)-Alt(n) and so, by induction on k, M^{∗}=⟨M_{i} | 0 ≤ i ≤ 2⟩ ≅ Alt(n).

Since M_{3} = ⟨M_{13}, M_{23}⟩ ≤ M^{∗}, we finally have M = M^{∗} and the lemma is proved.

Lemma 3.3 Let M be a group, A a finite subgroup of M, D ∈ Syl_{3}(A) and L_{D} a subgroup of N_{M}(D) containing D. Suppose that

1. A ≅ Alt(5).

2. L_{D} ≅ Sym(3) ∧^{|}Sym(n), n ≥ 2.

3. L_{D} ∩ A = N_{A}(D) ≅ Sym(3).

4. For g ∈ A and E = D^{g} define L_{E} = L_{D}^{g} (well-defined by (3)). Then

L_{D} ∩ L_{E} ≅ \{ \begin{matrix}Sym(n − 1) & if ⟨D, E⟩ ≅ Alt(4) \\ C_{2} × Sym(n − 2) & if ⟨D, E⟩ = A \end{matrix} .

5. M = ⟨A, L_{D}⟩.

Then M ≅ Alt(n + 3).

Proof: We shall show that M is a completion of a (4, n−1)-Alt(n+3) amalgam of groups. The result will then follow from 3.2.

Select E, F ∈ Syl_{3}(A) with B = ⟨D, E⟩ ≅ Alt(4) and A = ⟨D, F ⟩ and define

M_{1} = L_{D},
M_{2} = B(L_{D} ∩ L_{E}) and
M_{3} = A(L_{D} ∩ L_{F}).

Notice that, by (3), for X ∈ Syl_{3}(A),

L_{D} ∩ L_{X} ∩ A = N_{A}(X) ∩ N_{A}(D).

In particular, by (1), L_{D} ∩ L_{E} ∩ A = 1 and |L_{D} ∩ L_{F} ∩ A| = 2 and L_{D} ∩ L_{E} and L_{D} ∩ L_{F} are uniquely determined up to automorphisms of L_{D} in L_{D}. Also we see that L_{D} ∩ L_{E} inverts D and L_{D} ∩ L_{F} contains an element which centralizes F and inverts D. It now follows from (4) that M_{2} ≅ Sym(4) ∧^{|}Sym(n − 1) and M_{3} ≅ Sym(5) ∧^{|}Sym(n − 2). We now calculate that

M_{12} = M_{1} ∩ M_{2} = L_{D} ∩ B(L_{D} ∩ L_{E}) = (L_{D} ∩ B)(L_{D} ∩ L_{E})
= D(L_{D} ∩ L_{E}) ≅ Sym(3) ∧^{|}Sym(n − 1).

and, as |L_{D} ∩ L_{F} ∩ A| = 2,

M_{13} = (L_{D} ∩ A)(L_{D} ∩ L_{F}) = D(L_{D} ∩ L_{F}) ≅ Sym(3) × Sym(n − 2).

Furthermore, we now determine M_{123} in M_{1} and get that M_{123} = M_{12} ∩ M_{13} ≅ Sym(3) ∧^{|}Sym(n − 2). Finally, since M_{2} > M_{23} ≥ BM_{123} we have M_{23} = BM_{123} ≅ Sym(4) ∧^{|}Sym(n − 2). It follows that (M_{1}, M_{2}, M_{3}) of type (4, n − 1)- Alt(n + 3). Hence M ≅ Alt(n + 3).

4 Properties of McL

In this section M = Aut(McL), U ≤ M with U ∼ U_{4}(3).2, K ≤ M with K ∼ U_{3}(5).2 and F ≤ K, with |F | = 5 and C_{M}(F) divisible by 3. Set L = N_{M}(F) ∼ 5^{1+2}.3.8.2. We define Γ_{2} to be the set of 3-central groups of order three in M, Γ_{3} the set of elementary abelian groups of order 3^{4} in M and Γ_{4} the set of 2-central groups of order 2 in M. Let Γ = Γ_{2} ∪ Γ_{3} ∪ Γ_{4}. We make Γ into a graph by stipulating that for i ∈ {2, 3, 4} and a_{i} ∈ Γ_{i}, a_{2} is adjacent to a_{3} if a_{2} ≤ a_{3}, a_{2} is adjacent to a_{4} if [a_{2}, a_{4}] = 1 and a_{3} is adjacent to a_{4} if a_{4} normalizes a_{3}. Let Λ the left cosets of U in M. We make Λ into a graph by stipulating that x, y ∈ Λ are adjacent if and only if M_{x} ∩ M_{y} ∼ 3^{4}.Mat_{10}. Thus Λ is the graph, discovered by McLaughlin, which admits M as a primitive rank 3 permutation group and has parameters k = 112, l = 162, λ = 30 and μ = 56 [McL]. Finally we let Δ be the set of cyclic subgroups of order 5 which are not equal to F in N_{M}(F). We have gleaned much of the the information about the subgroup structure of M from the Atlas [ATLAS] and [Fink].

Lemma 4.1 (a) A Sylow 3-subgroup of M contains a unique elementary abelian subgroup, Q, of order 3^{4}.

(b) M operates transitively on Γ_{i} for each i ∈ {2, 3, 4}.

(c) N_{M}(Q) operates naturally as Mat_{10} on Γ_{2} ∩ Q which has order 10.

(d) Let x ∈ Γ_{2}. Then Γ_{4} ∩N_{M}(x) ⊂ O_{2,3}(N_{M′}(x)) ∼ 3^{1+4} : 2; in particular, if z ∈ Γ_{4} ∩ N_{M}(x), then z normalizes all the Sylow 3-subgroups with centre x.

Proof: Let S be a Sylow 3-subgroup of M. Then S contains an elementary abelian subgroup Q of order 3^{4} with normalizer 3^{4} : (Mat_{10} × C_{2}). Let Q_{1} be a further elementary abelian subgroup of S of order 3^{4}. Then, as Mat_{10} does not possess transvections when operating on a 4-dimensional GF(3)-space, we have S = QQ_{1} and Z(S) ≥ Q_{1} ∩ Q which has order 9. However, the centre of S has order 3 and so we have a contradiction; no such Q_{1} exists. Hence (a) holds and (c) follows from 2.5 (e) as N_{M}(Q) ≥ S. Part (b) follows immediately from (a) and the fact that the centres of a Sylow 2-subgroup of M and a Sylow 3-subgroup of M have orders 2 and 3 respectively. Suppose next that x ∈ Γ_{2} and let X = N_{M′}(x). Then X ∼ 3^{1+4}.2.Sym(5). Let K be a complement to O_{3}(X). Then K centralizes O_{2}(K) := ⟨z⟩ ∈ Γ_{4} and so K ≤ C_{M′}(z) ≅ 2.Alt(8) and normalizes a 3-cycle in the quotient Alt(8). Thus we deduce that all the involutions of K/⟨z⟩ are products of exactly two transpositions. Each of these involutions therefore lifts to an element of order 4 in K and so z is the unique involution of K and the final part of the lemma holds.

Lemma 4.2 (a) The cliques of maximal size in Γ have three elements, any clique in Γ lies in a clique of maximal size and M′ acts transitively on the set of cliques of maximal size.

(b) Let {a, b, c} be a maximal clique with a, b, c of type 2, 3, 4 respectively. Then (ba) M_{a} ∼ 3^{1+4}.4.Sym(5), M_{b} ∼ 3^{4}.(C_{2} × Mat_{10}) and M_{c} ∼ 2.Sym(8). (bb) M_{ab} ∼ 3^{1+2+1}.(C_{2} × 3^{2}.Q_{8}), M_{ac} ∼ 2.(Sym(3) × Sym(5)) and M_{bc} ∼

3^{2}.(C_{2} × SD_{16}). (bc) M_{abc} ∼ 2.(Sym(3) × C_{2} × Sym(3)).

(c) Let c ∈ Γ_{4}. Then M_{c} acts transitively on (Γ_{4})^{c}\{c} and, for d ∈ (Γ_{4})^{c}\{c}, M_{cd} ∼ 2.2^{1+2+2}.Sym(3).

(d) Let c ∈ Γ_{4} and i ∈ {2, 3}. Then Γ_{i}^{c} = Γ_{i}(c). In particular, M_{c} acts transitively on Γ_{i}^{c}.

Proof: Let S be a Sylow 3-subgroup of M and x = Z(S) ∈ Γ_{2}. By 4.1 (a), x is incident to a unique y ∈ Γ_{3} which is contained in S and by 4.1 (d) the normalizer of S contains z ∈ Γ_{4} which normalizes y and centralizes x. Thus {x, y, z} is a clique and, as the maximal size of a clique is three, {x, y, z} is a maximal clique. Moreover, we have shown that any clique with an element of type 2 and an element of type 3 is contained in a maximal clique with three elements. If we have a clique {y, z} of type 3, 4, then z ≤ N_{M′}(y). By 4.1 (c), N_{M′}(y) acts as Mat_{10} on Γ_{2} ∩ y and z induces a cycle of type 2^{4}. It follows that z centralizes exactly two elements of Γ_{2} ∩ y. Thus {y, z} extends to a clique with three elements. Now suppose that {x, z} is a clique of type 2, 4 with x = Z(S). Then 4.1 (d) indicates that z ≤ N_{M}(S) and so 4.1 (a) implies that there is a y ∈ Γ_{3} which is normalized by z and centralized by x. Thus {x, z} also extends to a clique with three elements. This shows that all maximal cliques have size three. Now suppose that {a, b, c} and {a_{1}, b_{1}, c_{1}} are cliques with a, a_{1} ∈ Γ_{2}, b, b_{1} ∈ Γ_{3} and c, c_{1} ∈ Γ_{4}. Without loss of generality we may assume that b = b_{1}. Then, as M_{b} operates transitively on the 10 elements of type 2 in b we can conjugate a to a_{1} by an element which fixes b. Hence we may assume a = a_{1}. Now M_{ab} ∼ 3^{4}.3^{2}(C_{2} × Q_{8}) ∼ 3^{4}(2 × Mat_{9}) by 2.5 (e). Hence, by 4.1 (d), and Sylow’s theorem, M_{ab} operates transitively on Γ_{4} ∩ M_{ab} and so M acts transitively on maximal cliques. We obtain part (ba) directly from [ATLAS]. Also we have M_{ab} ∼ 3^{4}.(C_{2} × Mat_{9}) ∼ 3^{1+2+1}.(C_{2} × 3^{2}.Q_{8}) and M_{bc} = C_{M_{b}}(c) ∼ 3^{2}.(C_{2} × SD_{16}). Additionally, as a projects to a 3-cycle in M_{c}/c, M_{ac} = N_{M_{c}}(a) ∼ 2.(Sym(3)×Sym(5)). Finally we calculate in 2.Sym(8) that M_{abc} ∼ 2.(Sym(3) × Sym(3) × C_{2}). Therefore, parts (bb) and (bc) hold. Let c ∈ Γ_{4}. Then Γ_{4}^{c} are the conjugates of c which lie in M′_{c} ∼ 2.Alt(8). As there is a unique class of involutions in 2.Alt(8) (those corresponding to a product of 4 commuting transpositions), we conclude that part (c) holds. Finally, part (d) follows directly from the definition of incidence in Γ.

Lemma 4.3 Suppose that t ∈ M \ M′ is an involution. Then C_{M}(t) ≅ C_{2} × Mat_{11} and, for i = 2, 4, C_{M}(t) acts transitively on Γ_{i}^{t} and has two orbits on Γ_{3}^{t}. Moreover, setting C_{t} = C_{M}(t) we have

(a) If x ∈ Γ_{3}^{t}, then either M_{x} ∩ C_{t} ∼ C_{2} × Mat_{10} and t inverts O_{3}(x) or M_{x} ∩ C_{t} ∼ C_{2} × 3^{2} : SD_{16}.

(b) If x ∈ Γ_{4}^{t}, then M_{x} ∩ C_{t} ∼ 2.((Sym(2) ∧^{|}2^{3}) : Sym(3)).

Proof: That C_{t} = C_{M}(t) ≅ C_{2}×Mat_{11} can be read from the Atlas [ATLAS]. Since Mat_{11} contains a unique class of involutions, we see that C_{t} is transitive on Γ_{4}^{t}. Let x ∈ Γ_{2}^{t}. Then x is inverted by t and there is a z ∈ Γ_{4}^{t} ∩ N_{M}(x). Thus ⟨t, x⟩ ≤ C_{M}(z) ∼ 2.Sym(8). Since C_{t} contains no subgroup isomorphic to 2.Alt(6), we see that t projects to an involution of cycle shape (2, 2, 2) in C_{M}(z)/⟨z⟩. It follows that C_{C_{M} (z)}(t) ∼ 2.((Sym(2) ∧^{|}2^{3}) : Sym(3)) which is transitive on all the subgroups of order 3 which are normalized by t and generated by elements which project to 3-cycles in C_{M}(z)/⟨z⟩. As C_{t} acts transitively on Γ_{4}^{t}, we conclude that C_{t} acts transitively on Γ_{2}^{t}. Now suppose that x ∈ Γ_{3}^{t}. Then t is an involution in N_{M}(x) ∼ 3^{4} : (2 × Mat_{10}) which is not in M′. Thus either t inverts O_{3}(x) or t is a diagonal involution. In the former case we have that K = C_{t} ∩ N_{M}(x) ∼ C_{2} × Mat_{10} with O_{2}(K) = ⟨t⟩. Since N_{M}(x) has a unique class of complements to O_{3}(N_{M}(x)) and C_{t} has a unique class of subgroups isomorphic to C_{2} × Mat_{10}, we conclude that C_{t} acts transitively on x ∈ Γ_{3}^{t} with O_{3}(x) inverted. In the case that t is a diagonal involution in M_{x}, we have C_{t} ∩ M_{x} ∼ C_{2} × 3^{2} : SD_{16}. This subgroup is the normaliser in C_{t} of a Sylow 3-subgroup and so is also unique up to conjugacy in C_{t} and once again ⟨t⟩ = O_{2}(C_{t} ∩ M_{x}). It follows that C_{t} acts transitively on Γ_{3}^{t} in this case as well.

Lemma 4.4 Let a ∈ Γ_{2}. Then M_{a} has five orbits on Γ_{2}, X_{i}(a), 0 ≤ i ≤ 4. Moreover, the orbits and some of their properties may be described as follows:

(a) X_{0}(a) = {a}.

(b) If b ∈ X_{1}(a), then [a, b] = 1, M_{ab} ∼ 3^{4}.(C_{2} ×Q_{8}) and there exists a unique c ∈ Γ_{3}(a) ∩ Γ_{3}(b).

(c) If b ∈ X_{2}(a), then ⟨a, b⟩ ∼ 2.Alt(4), M_{ab} ∼ 4.Sym(4) and |X_{1}(a) ∩ X_{1}(b)| = 4.

(d) If b ∈ X_{3}(a), then ⟨a, b⟩ ∼ 2.Alt(5), M_{ab} ∼ (Q_{8} × C_{3}).2 and |X_{1}(a) ∩ X_{1}(b)| = 1.

(e) If b ∈ X_{4}(a), then ⟨a, b⟩ ∼ 5^{1+2}.3, M_{ab} ∼ D_{10} and X_{1}(a) ∩ X_{1}(b) = ∅.

Proof: Let t be an involution in M′ (so an element of Γ_{4}). Then C_{M}(t) ∼ 2.Sym(8) and a group of order three in C_{M}(t) is in Γ_{2} if and only if it projects to a 3-cycle in Sym(8). Furthermore, as alluded to in the description of F above, we can find elements from Γ_{2} in C_{M}(F). Thus we see that two elements from Γ_{2} may generate the groups indicated in parts (b)-(e) of the lemma. Hence M_{a} has at least 5 orbits on Γ_{2}. We next determine the stabilisers of these configurations. Assume that a, b ∈ Γ_{2} with a ≠ b and suppose that c ∈ Γ_{3} and ⟨a, b⟩ ≤ c. Then, by 4.1 (c), M_{ab} ∼ 3^{4} : (C_{2} × Q_{8}) and c = O_{3}(M_{ab}). In particular, Γ_{3}(a) ∩ Γ_{3}(b) = {c}.

In the cases when ⟨a, b⟩ = 2.Alt(4) or 2.Alt(5), M_{ab} centralizes the central involution of ⟨a, b⟩ and we may therefore calculate in 2.Sym(8) to find the structure of M_{ab}. Finally, in case (e), we have M_{ab} ≤ L ∼ 5^{1+2}.3.8.2, from which we calculate M_{ab} ≅ D_{10}. Thus we have the orders of each X_{i}(a). Since |Γ_{2}| = 15400 = ∑_{i=0}^{4} |X_{i}(a)|, we deduce that {X_{i}(a) | 0 ≤ i ≤ 4} is the set of orbits of M_{a} on Γ_{2}. Observing that X_{1}(a) ∩ X_{1}(b) = (M_{ab} ∩ Γ_{2}) \ {a, b} now completes all the parts of the lemma.

Our next lemma helps us to recognize the orbits of M_{a} on Γ_{2} in terms of the action of elements of Γ_{2} on Λ.

Lemma 4.5 (a) Every maximal clique in Λ has size 5 and is of the form Λ^{a} for some a ∈ Γ_{2}. Every clique of size three lies in a unique maximal clique and every clique of size 2 lies in 10 maximal cliques.

(b) Let Φ be a maximal clique in Λ and x ∈ Λ \ Φ. Then x is adjacent to exactly two elements of Φ.

(c) Suppose that a, b ∈ Γ_{2}. Then

(ca) Λ^{a} = Λ^{b} if and only if a = b;

(cb) |Λ^{a} ∩ Λ^{b}| = 2 if and only if b ∈ X_{1}(a); (cc) |Λ^{a} ∩ Λ^{b}| = 1 if and only if b ∈ X_{2}(a); (cd) if Λ^{a} ∩ Λ^{b} = ∅ and there exists a clique of size four intersecting Λ^{a}

and Λ^{b} in sets of size two, then b ∈ X_{3}(a); and (ce) if Λ^{a} ∩ Λ^{b} = ∅ and no clique of size four intersects Λ^{a} and Λ^{b} in sets

of size two, then b ∈ X_{4}(a).

Proof: Let x ∈ Λ. Then from [McL] we know that Λ(x) consists of 112 vertices and for y ∈ Λ(x), |Λ(x) ∩ Λ(y)| = 30. Hence the induced subgraph on Λ(x) is isomorphic to the graph on the singular 1-spaces in a 6-dimensional non-degenerate orthogonal space of “−”-type over GF(3) in which two 1-spaces are incident if and only if they are perpendicular. Since maximal cliques in the graph Λ(x) have just 4 members, a maximal clique in Λ has exactly 5-vertices. Furthermore, since every clique of size two in Λ(x) is contained in a unique maximal clique in Λ(x) we have that every clique of size three is contained in a unique maximal clique. Similarly, we have that every clique of size two is contained in exactly ten maximal cliques.

Since the elements of Γ_{2} ∩ M_{x} fix a clique of size 4 in Λ(x), we have that for a ∈ Γ_{2}, Λ^{a} contains a clique Φ = {a} ∪ Λ(x)^{a} of size 5. Assume that there is a y ∈ Λ^{a} \ Φ. Then y ∉ Λ(x). From the structure of Λ [McL], y is incident to exactly 56 elements of Λ(x)^{a}. Since 56 is not divisible by 3, y is incident to some z ∈ Λ(x)^{a} ⊂ Φ. But then |Λ(z)^{a}| = 5, a contradiction. Using the fact that for a, b ∈ Γ_{2} ∩ U, ⟨a, b⟩ ∼ 3^{2} or 2.Alt(4) we deduce that the first three statements in (c) are also true.

Let a, b ∈ Γ_{2} with Λ^{a} ∩ Λ^{b} = ∅. Note that X_{1}(a) ∩ X_{1}(b) ≠ ∅ if and only if there exists a clique of size five, Λ^{c} intersecting Λ^{a} and Λ^{b} in sets of size two. Thus, using 4.4 (e), the last two statements in (c) hold.

To prove (b) we note first that by (a), x is adjacent to at most two elements of Φ. Let P be one of the ten sets of size two in Φ. Then P lies in 9 cliques of size 5 different to Φ. Each of these cliques contains three elements not contained in Φ. Hence we have found 10 · 9 · 3 = 270 distinct elements in Λ which are adjacent to exactly two elements of Φ. Since |Λ \ Φ| = 275 − 5 = 270, (b) is proved.

Recall that L = N_{M}(F) ∼ 5^{1+2}.3.8.2 and Δ is the set of subgroups of O_{5}(L) of order 5 which are not equal to F.

Lemma 4.6 L has two orbits Ξ and Θ on Λ. Furthermore,

(a) |Ξ| = 125, Ξ = ⋃_{d∈Γ_{2}^{F}} Λ^{d} and O_{5}(L) acts regularly on Ξ; and

(b) |Θ| = 150, Θ = ⋃_{e∈Δ} Λ^{e}, O_{5}(L) has 6 orbits of length 25 on Θ and O_{5,2}(L) has three orbits Λ_{i}(L), 1 ≤ i ≤ 3 each of length 50 on Θ.

Proof: Let E be a group of order five in U. Then, as L contains a Sylow 5subgroup of M, without loss of generality we may assume that E ≤ L. Notice also that F does not fuse into U, and hence F operates semiregularly on Λ. Since there are 6 choices for ⟨E, F ⟩ in O_{5}(L) and L acts transitively on such subgroups, it follows that O_{5}(L) has a multiple of 6 orbits of length 25 on Λ and the remaining orbits are regular. We conclude that O_{5}(L) has one regular orbit and 6 orbits of length 25; the latter orbits are all fused into a single orbit by L. Hence L has two orbits on Λ. Let Ξ be the orbit of length 125 and Θ be the orbit of length 150. Notice that ⋃_{a∈Γ_{2}^{F}} Λ^{a} and ⋃_{e∈Δ} Λ^{e} are both L-invariant.

By 4.5 (c), for a, b ∈ Γ_{2}^{F}, Λ^{a} ∩ Λ^{b} = ∅ and so |⋃_{a∈Γ_{2}^{F}} Λ^{a}| = 5|Γ_{2}^{F} | = 5 · 25 = 125. Therefore, (a) holds. Since also ⋃_{e∈Δ} Λ^{e} is L-invariant and O_{5}(L) is regular on Ξ, we must have Θ = ⋃_{e∈Δ} Λ^{e}. Thus (b) also holds.

Remember that K ∼ U_{3}(5).2 and that the Hoffman-Singleton graph H may be constructed on the cosets K′/A where A is any subgroup of K′ isomorphic to Alt(7) and two cosets x, y ∈ K′/A are incident if and only if K′_{xy} ≅ Alt(6). Note that, as F ≤ K, O_{5}(L) ≤ K.

Lemma 4.7 Let A_{i}, 1 ≤ i ≤ 3, be representatives for the three conjugacy classes of subgroups isomorphic to Alt(7) in K′ and let H be the Hoffman-Singleton graph with vertex stabilisers the conjugates of A_{1}. Then

(a) H has diameter 2, has no cycles of length less than five and K′ acts transitively on paths of length three and on cycles of length five.

(b) The orbits of A_{1} on H have lengths 1, 7 and 42 and the orbits of A_{2} and A_{3} on H have lengths 15 and 35.

(c) (ca) O_{5}(L) has two orbits L^{1} and L^{2} on H.

(cb) For i = 1, 2, F has five orbits L_{k}^{i}, 1 ≤ k ≤ 5 on L^{i}. Each L_{k}^{i} is, as a

subgraph of H, a cycle of length five. (cc) For a fixed i ∈ {1, 2} and for j, k ∈ {1, …, 5} with j ≠ k, no two

elements of L_{j}^{i} and L_{k}^{i} are adjacent.

(cd) Let 1 ≤ j ≤ k ≤ 5. Then L_{j}^{1} ∪ L_{k}^{2} is a Petersen graph.

Proof: Part (a) is well-known and also easily verified. The first part of (b) follows from (a) so to prove (b) it remains to determine the orbits of A_{2} and A_{3} on H. Note that A_{1} and A_{2} both contain a Sylow 3-subgroup and a Sylow 7subgroup of K′. Hence, as no proper subgroup of Alt(7) contains both a Sylow 3-Subgroup and a Sylow 7-subgroup, there exist orbits of A_{2} on H (respectively A_{3} on H) one which has length divisible by 3 and not by 7 and one which has length divisible by 7 and not by 3. On the other hand, the Sylow 5-subgroups of A_{1} and A_{2} are not conjugate in K′ and so all orbits of A_{2} on H (respectively A_{3} on H) have length divisible by 5. This gives (b).

Since |H| = 50, we see that O_{5}(L) has 2 orbits L^{1} and L^{2} on H. So (ca) holds. Let t be an involution in C_{K′}(F). Then O_{5}(L)⟨t⟩ operates transitively on H. Moreover, C_{K′}(t) ∼ 2.Sym(5), C_{A_{1}}(t) ∼ 2.(Sym(3) × Sym(2)) and C_{Alt(6)}(t) ∼ 2.(Sym(2) × Sym(2)). Thus Fix_{H}(t) is a Petersen graph, and, as F ≤ C_{K′}(t), t fixes pointwise one F -orbit in L^{1} and one F -orbit in L^{2}. As there are exactly 25 involutions in O_{5}(L)⟨t⟩ and no two involutions in O_{5}(L)⟨t⟩ have the same fixed points we deduce that (cb) and (cd) are true. Finally, since the valency of every vertex is 7, and 7 neighbours to each vertex have been accounted for in (cb) and (cd), we deduce that (cc) also holds.

Lemma 4.8 Suppose that H is a Hoffman-Singleton graph for K′ and K is the graph with vertices the edges of H and two vertices incident if and only if they are the first and last edges of a path of length 3 in H.

(a) Every clique of size 2 in K is contained in 10 maximal cliques.

(b) Let Ψ be a maximal clique in K. Then either |Ψ| = 5 or |Ψ| = 3 and there exists a cycle (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}) of length 6 in H such that Ψ corresponds to {(a_{1}, a_{2}), (a_{3}, a_{4}), (a_{5}, a_{6})}.

Proof: Suppose that c = {c_{1}, c_{2}} is a clique of size 2 in K. Then c corresponds to a path of length 3 in H and, as such, by 4.7 (a), is unique up to conjugation. Thus, by 4.7 (cb), we may suppose that c is a path of length 3 in L_{1}^{1}. Now, by 4.7 (cc), no element of c is incident to any other edge in L_{1} \ L_{1}^{1}. Thus any maximal clique containing c must be contained in P = L_{1}^{1} ∪ L_{k}^{2} for some k ∈ {1, …, 5}, which, by 4.7 (cd), is a Petersen graph. Calculation in P shows that c is contained in exactly two maximal cliques: one of size 5 and one which consists of every other edge in a cycle of length 6. Hence parts (a) and (b) hold.

Lemma 4.9 (a) On Λ, K has two orbits Λ_{0}(K) and Λ_{1}(K) of lengths 100 and 175 respectively. K′ acts transitively on Λ_{1}(K) and has two orbits of length 50, Λ_{2}(K) and Λ_{3}(K), on Λ_{0}(K).

(b) As a K-set Λ_{1}(K) is isomorphic to the edges in a Hoffman-Singleton graph H. Two such edges are adjacent in Λ_{1}(K) if they are the first and the last edge of some path of length three in H.

(c) Λ_{2}(K) and Λ_{3}(K) are Hoffman-Singleton graphs and as K′-sets neither isomorphic to each other nor to H.

(d) Λ_{1}(K) = Ξ ∪ Λ_{i}(L) for some i ∈ {1, 2, 3}.

(e) Suppose that Ψ is a maximal clique in Λ. Then |Ψ ∩ Λ_{1}(K)| = 1, 3 or 5.

Proof: It follows from 4.6 (a) and (b) that L ∩ K has orbits of length 125, 50 and 100 on Λ. As M = ⟨L, K⟩, K does not normalize the orbit of length 125. Furthermore, since 275 does not divide |K|, K does not act transitively on Λ. Also K does not have a subgroup of index 225 and so we deduce that the orbits of K must have length 175 and 100. In particular, 4.6 implies that part (d) holds. Since K has a unique conjugacy class of subgroups of index 175, Λ_{1}(K) is as a K-set isomorphic to the set of edges of a Hoffman-Singleton graph H. Therefore, K′ is transitive on Λ_{1}(K). Moreover, K′ has no subgroup of index 100 and hence K′ has two orbits each of length 50 on Λ_{0}(K). So part (a) holds. Since elements of order five have at most five fixed points on Λ, Λ_{2}(K), Λ_{3}(K) and H are as K′-sets pairwise non-isomorphic. Let i ∈ {2, 3}. Then Λ_{i}(K) is, as a graph, either isomorphic to the Hoffman-Singleton graph or its complement. But the latter has cliques of size larger than five which is against 4.5 (a). So (c) holds.

To prove the second statement of (b) let x, y be two distinct elements of Λ_{1}(K). Define δ(x, y) to be the minimal distance of a vertex of H on x (remember x corresponds to an edge of H) to a vertex of H on y. Assume now in addition that x, y are adjacent in Λ. If δ(x, y) = 0, the seven edges on a vertex would form a clique of size 7 in Λ, which is impossible as a maximal clique has size 5. So assume that δ(x, y) = 2. We use the notation of 4.7 (c). Let x_{k} be an edge in L_{k}^{1} and x_{0} some edge in L_{1}^{1} different from x_{1}. Then {x_{1}, x_{2}, x_{3}, x_{4}, x_{5}} and {x_{0}, x_{2}, x_{3}, x_{4}, x_{5}} are cliques of size 5 in Λ which intersect in a clique of size 4, a contradiction to 4.5 (a). Thus x and y are adjacent if and only if δ(x, y) = 1 and so (b) holds.

Finally, suppose that Ψ is a maximal clique in Λ. As, by 4.5 (a), each clique of size 2 in Λ is contained in exactly 10 maximal cliques and as a clique of size 2 in Λ_{1}(K) is contained in 10 maximal cliques of Λ_{1}(K) by 4.8, we see that Ψ ∩ Λ_{1}(K) has size 0, 1, 3 or 5. Because, by (c), Λ_{0}(K) is the union of two Hoffman-Singleton graphs, 4.7 (a) implies that Ψ ⊈ Λ_{0}(K). Hence |Ψ ∩ Λ_{1}(K)| = 0 is impossible. Therefore, part (e) holds.

The next proposition is the ultimate aim of our preliminary calculations within Aut(McL).

Proposition 4.10 Suppose that a ∈ Γ_{2}. Then Γ_{2}^{F} ⊈ X_{4}(a).

Proof: We use the notation of 4.6 and 4.9. If there exists u ∈ Ξ ∩ Λ^{a}, then, by 4.6 (a), there exists b ∈ Γ_{2}^{F} with u ∈ Λ^{b}. Thus Λ^{a} ∩ Λ^{b} ≠ ∅ and 4.5 (c) implies that b ∉ X_{4}(a) and the proposition holds.

Therefore we may assume that Ξ∩Λ^{a} = ∅ and so 4.6 implies Λ^{a} ⊆ Θ. Choose notation in accordance with 4.9 so that Λ_{1}(K) = Ξ ∪ Λ_{1}(L) and Λ_{0}(K) = Λ_{2}(L) ∪ Λ_{3}(L). Because Λ^{a} ∩ Θ = ∅, 4.5 (a) and 4.9 (e) imply that |Λ^{a} ∩ Λ_{1}(L)| = |Λ^{a} ∩ Λ_{1}(K)| = 1, 3 or 5. Conjugation by elements of L permutes {Λ_{i}(L) | 1 ≤ i ≤ 3} and yields |Λ^{a} ∩ Λ_{i}(L)| = 1, 3 or 5 for all 1 ≤ i ≤ 3. Hence Λ^{a} intersects one of the Λ_{i}(L) in a set of size 3 and the other 2 in sets of size one. Without loss of generality we may assume that Λ^{a} ∩ Λ_{1}(L) = Λ^{a} ∩ Λ_{1}(K) has size 3. Set Φ = Λ^{a} ∩ Λ_{1}(L) and recall that, by 4.9 (b), Λ_{1}(K) is the set of edges of a Hoffman-Singleton graph H with two edges adjacent if they are the ends of a path of length three in H. Moreover, the edges of H which lie in L^{1} ∪ L^{2} correspond to the vertices of Λ_{1}(L) and the vertices of Ξ correspond to edges which join L^{1} to L^{2}. As, by 4.5 (a), Φ is a maximal clique in Λ_{1}(K), 4.8 (b) implies that Φ is every second edge on a cycle Π of length six in H. Let Ψ consist of the three edges on Π which are not contained in Φ. Then Ψ is a clique of size 3 in Λ. Let b ∈ Ψ. Then b is not adjacent to any of the elements in Φ and so by 4.5 (b) is adjacent to both of the elements in Λ^{a} \ Φ. Hence Ψ ∪ (Λ^{a} \ Φ) is a clique of size 5 and thus, by 4.5 (a), Ψ ∪ (Λ^{a} \ Φ) = Λ^{c} for some c ∈ Γ_{2}. Since Π is a closed path, an even number of edges of Π join L^{1} to L^{2}. Thus, as Φ ∩ Ξ = ∅, |Ξ ∩ Ψ| = |Ξ ∩ Λ^{c}| = 2. Therefore, without loss of generality we may assume that Ψ ∩ L^{1} = {e} and e ∈ X where X ⊆ L^{1} is an F -orbit. Since Φ ∩ Ξ = ∅, we deduce from 4.7 (c) that the two edges of Φ which are incident as edges in H to e also lie in X. Let Y ⊆ L^{2} be the F -orbit on H which contains the remaining element of Φ and C be the clique corresponding to the edges joining X to Y in the Petersen graph X ∪ Y. Then C is an F -orbit on Ξ and so C = Λ^{b} for some b ∈ Γ_{2}^{F}. But then |Λ^{b} ∩ Λ^{c}| ≥ 2 ≤ |Λ^{c} ∩ Λ^{a}| which, by 4.5 (c), implies that c ∈ X_{1}(a) ∩ X_{1}(b). Finally, 4.4, implies that b ∉ X_{4}(a) and we conclude that Γ_{2}^{F} ⊈ X_{4}(a).

Lemma 4.11 (a) Let Φ be a maximal clique in Λ and c ∈ Λ \ Φ. Then M′_{cΦ} induces Sym(3) × Sym(2) on Φ.

(b) There exists a unique class of tuples (Φ, a, b, c), where Φ is a maximal clique in Λ, a ≠ b ∈ Φ and c is an element of Λ neither adjacent to a nor to b.

(c) Let (Φ, a, b, c) be any tuple as in (b) and let {d, e} = Φ(c). Then M′_{abc} ≅ Alt(6) and O_{3}(M_{de}) ∩ M_{abc} is a Sylow 3-subgroup of M_{abc}.

Proof: By 4.5 (b), c is incident to exactly two elements of Φ. Let {e, d} = Φ(c). Put Q = O_{3}(M_{de}) and let Ψ be the unique maximal clique in Λ containing d,e and c. Then O_{3}(M_{Φ}) acts transitively on the nine maximal cliques containing {a, b} but different from Φ. Also Q = Q ∩ O_{3}(M_{Φ})(Q ∩ O_{3}(M_{Ψ})) and so, as Q preserves every clique containing {d, e}, O_{3}(M_{Φ}) acts transitively on Ψ \ {d, e}. Hence we see that the pair (Φ, c) is unique up to conjugation under M′ and M_{Φc} = M_{Φ{de}}O_{3}(M_{Φ}). Thus (a) and (b) hold. Moreover, |M′_{abc} ∩ Q| = 3^{2} and M′_{abcΦ} ∼ 3^{2}.4. By (b), M′_{abc} acts transitively on the ten maximal cliques containing a and b and so |M′_{abc}/M′_{abcΦ}| = 10. Thus M′_{abc} is a subgroup of M′_{ab} of order 360 and so M′_{abc} ≅ Alt(6).

5 3.Aut(McL) and its 90-dimensional representation over GF(5)

In this section we determine the Schur multiplier of McL, construct a unique non-split group 3.Aut(G) and prove that it has a unique 90-dimensional representation over GF(5).

Lemma 5.1 The following hold:

(i) The Schur multiplier of McL has order 3.

(ii) There exists a group G = 3.Aut(McL) and O_{3}(G) is inverted by G.

Proof: Let M = Aut(McL), G = M′ and m(G) = |H^{2}(G, ℂ^{∗})| be the order of the Schur multiplier of G. Then, since |G| = 2^{7} · 3^{6} · 5^{3} · 7 · 11, m(G) is prime to 7 and 11 [Asch, 33.14]. Moreover, as the Schur multiplier of 2.Alt(8) is trivial and the Schur multiplier of U_{3}(5) is a 3-group, m(G) is a power of 3. Suppose that G̃ is the universal covering group of G and let A = Z(G̃). We shall show first that |A| ≤ 3. Recall that G and hence G̃ acts on the graph Λ of Section 4. Let Φ, a, b, c, d, e be as in 4.11 and set Q_{ab} = O_{3}(G_{ab}). Since both G_{a} and G_{ab} contain Sylow 3-subgroups of G, Gaschütz’ Theorem implies that G̃_{a} and G̃_{ab} are also perfect central extensions. Since G_{ab} does not fix a symplectic form on Q_{ab}, we conclude that Q̃_{ab} is abelian. In particular, 4.11 (c) implies that G̃_{abc} has abelian Sylow 3-subgroups. Therefore, G̃_{abc} splits over A and is thus a complement to Q̃_{ab} in G̃_{ab}. Hence, as G̃_{ab} is perfect, we must have A ≤ [Q̃_{ab}, G̃_{ab}]. Since, by 2.5 (f), |H^{1}(G_{abc}, Q_{ab})| = 3, we conclude that |A| ≤ 3 as claimed. To complete the lemma it only remains to prove that m(G) ≥ 3. To achieve this we employ what Griess calls the stable cocycle method [Gr]. Let B be a cyclic group of order 3 which is centralized by G and inverted by M. We recall from [CE, p. 257] that a cocycle in H^{2}(M_{a}, B) is M-stable if the conjugation map c_{g} : H^{2}(M_{a} ∩ M_{a}^{g}, B) → H^{2}(M_{a}^{g^{−1}} ∩ M_{a}, B) makes the triangular diagram made from the restriction maps H^{2}(M_{a}, B) → H^{2}(M_{a} ∩ M_{a}^{g}, B) and H^{2}(M_{a}, B) → H^{2}(M_{a}^{g^{−1}} ∩ M_{a}, B) commute. Clearly, it suffices to check this for representatives of the double cosets of M_{a} in M. In our special case we can choose x_{0}, x_{1}, x_{2} double M_{a}-coset representatives so that x_{0} = 1, x_{1} ∈ Z(M_{{a,b}c})^{#} and a^{x_{2}} = c. Because M_{a} inverts B, we have |H^{2}(M_{a}, B)|_{3} = 3 from [ATLAS]. Let r be a non-trivial cocycle in |H^{2}(M_{a}, B)|. To show that r is M-stable it suffices to show that, for i ∈ {0, 1, 2}, the restricted cocycle z_{i} in H^{2}(M_{a} ∩M_{a}^{x_{i}}, B) is (M_{a} ∩M_{a}^{x_{i}})⟨x_{i}⟩-stable. We continue to use the notation of the first part of the lemma. Since M_{abc} ≅ Mat_{10} and M_{abc} inverts B, H^{2}(M_{abc}, B) = 0 [ATLAS]. Thus the restriction of r to M_{abc} is trivial. As M_{abc} contains a Sylow 3-subgroup of M_{ac}, z_{2} is trivial and so clearly M_{{a,c}}- stable. Next we recall that M_{ab} = M_{abc}Q_{ab}, x_{1} centralizes M_{abc} and x_{1} inverts Q_{ab}. But this clearly implies z_{1} is M_{ab}⟨x_{1}⟩-stable. We conclude that r is indeed M-stable and so the lemma follows from [CE, Proposition 9.4].

Our next objective is

Theorem 5.2 Suppose that G ≅ 3.McL. Then G has a faithful irreducible representation of degree 45 in characteristic 5.

For the proof of 5.2 we set H = 3_{2}.U_{4}(3) ( [ATLAS] notation), G = 3.McL and let k denote the algebraic closure of the field with 5 elements. We follow the Atlas [ATLAS] for conjugacy class names.

In Table 1 we give the faithful part of the ordinary character table of H = 3_{2}.U_{4}(3). The elements from classes 3B, 3C, 3D, 9A, 9B, 9C and 9D are not listed as their character values are all zero. We also recall that b7 = \frac{−1+\sqrt{−7}}{2} and b7^{∗∗} = \frac{1+\sqrt{−7}}{2}.

1A 2A 3A 4A 4B 5A 6A 6B 6C 7A B** 8A 12A
36 4 9 4 0 1 1 -2 -2 1 1 0 1
45_{1} -3 -9 1 1 0 3 0 0 b7 ** -1  1
45_{2} -3 -9 1 1 0 3 0 0 ** b7 -1  1
126 14 -9 2 2 1 -1 2 2 0  0 0 -1
189 -3 27 5 1 -1 3 0 0 0  0 1 -1
315_{1} 11 18 -1 -1 0 2 2 2 0  0 1  2
315_{2} -5 18 3 -1 0 -2 -2 4 0  0 -1  0
315_{3} -5 18 3 -1 0 -2 4 -2 0  0 -1  0
630 6 -45 2 -2 0 3 0 0 0  0 0 -1
720 16 18 0 0 0 -2 -2 -2 -1 -1 0  0
729 9 0 -3 1 -1 0 0 0 1 1 -1 0
756 -12 27 -4 0 1 3 0 0 0  0 0 -1
945 -15 -27 1 1 0 -3 0 0 0  0 1  1

Table 1: Ordinary faithful irreducible characters of 3_{2}.U_{4}(3)

Lemma 5.3 The faithful non-projective kH-modules form a unique block with Brauer tree as follows:

36 ∘ 36 _{189} ∘ 153_{756} ∘ 603_{729} ∘ 126_{126} ∘.

Proof: All the character values in Table 1 which are divisible by 5 remain irreducible when reduced modulo 5 and are projective. So they form blocks of size 1. This leaves the five characters of degrees 36, 126, 189, 729 and 756. By [HL, 2.1.5], these characters form a single block B. Furthermore, [HL, 2.1.22] tells us that the node of highest degree is not an end node. Thus the node labelled 756 is internal.

Since, by [HL, 2.1.5], two consequetive nodes correspond to a 5-modular projective indecomposable module, the sum of the degrees at those nodes has to be a multiple of 5. It follows that we have four possible Brauer trees for B. On the other hand, from the ordinary character table of U_{4}(3) [ATLAS], we see that H has two non-faithful, irreducible, projective 5-modular representation of degree 35, 35_{1} and 35_{2}. Thus we may form the faithful, projective kH-module 35_{1} ⊗ 36. Decomposing the corresponding complex character we have that

35_{1} ⊗ 36 = 36 + 189 + 315_{3} + 720

and this establishes the existence of a projective 5-modular character of dimension 225. This now leaves two possible Brauer trees for B. The false tree indicates that there should be a faithful irreducible kH-module of dimension

1A 2A 3A 4A 4B 6A 6B 6C 7A B** 8A 12A
36 4 9 4 0 1 -2 -2 1 1 0 1
45_{1} -3 -9 1 1 3 0 0 b7 ** -1  1
45_{2} -3 -9 1 1 3 0 0 ** b7 -1  1
126 14 -9 2 2 -1 2 2 0  0 0 -1
153 -7 18 1 1 2 2 2 -1 -1 1 -2
315_{1} 11 18 -1 -1 2 2 2 0  0 1  2
315_{2} -5 18 3 -1 -2 -2 4 0  0 -1  0
315_{3} -5 18 3 -1 -2 4 -2 0  0 -1  0
603 -5 9 -5 -1 1 -2 -2 1  1 -1  1
630 6 -45 2 -2 3 0 0 0  0 0 -1
720 16 18 0 0 -2 -2 -2 -1 -1 0  0
945 -15 -27 1 1 -3 0 0 0  0 1  1

Table 2: 5-modular faithful irreducible characters of 3_{2}.U_{4}(3)

27. This is immediately shown to be impossible by considering the subgroup of shape H_{3} ∼ 3^{1+4}.Alt(6) of H. Indeed, this subgroup has orbits of length 36 and 45 on the maximal subgroups of O_{3}(H_{3}) which do not contain the centre of H. This shows that the minimal conceivable dimension for a faithful kHmodule is 36. Hence the correct Brauer tree is the one described in the lemma. Alternatively, consult [MATLAS].

As a consequence of 5.3 we obtain the faithful part of the 5-modular character table of 3_{2}.U_{4}(3). This is recorded in Table 2.

Lemma 5.4 (a) U_{4}(3) has a unique faithful irreducible ordinary character of degree less than 22. Partial information for this character is as follows:

1A 2A 3A 4A 4B 5A 6A 6B 6C 7A B** 8A 12A
21 5 -6 1 1 1 2 -1 -1 0  0 -1 -2

(b) McL has a unique irreducible ordinary character of degree 22. Partial character information is as follows:

1A 2A 3A 3B 4A 5A 5B 6A 6B 7A B** 8A 12A
22 6 -5 4 2 -3 2 3 0 1  1 0  -1

(c) The character of degree 21 for U_{4}(3) remains irreducible after reduction modulo 5.

(d) The character of degree 22 for McL has composition factors 1 + 21 when reduced modulo 5.

Proof: (a) and (b) follow from the ordinary character table of U_{4}(3) and McL respectively (see [ATLAS]).

Let T be the subgroup of U_{4}(3) of shape 2^{4}.Alt(6). Then, by considering the eigenspaces of O_{2}(T) and noting that T acts transitively on the maximal subgroups of O_{2}(T), we see that any faithful module for U_{4}(3) in odd characteristic has dimension at least 15. Hence either the 21-dimensional complex representation remains irreducible when reduced modulo 5 or it has a trivial composition factor. However, by [ATLAS], the multiplicity of the trivial module for the 5′-subgroup C_{3} × Q_{8} (contained in 2^{1+4}2.Sym(4)) on the 21-space is

\frac{1}{24}(21 + 5 + 2 · −6 + 6 · 1 + 2 · 2 + 12 · −2) = 0

and hence (c) holds. Alternatively consult [MATLAS].

For (d) we note that McL is the elementwise stabilizer of a triangle of type 223 in the Leech lattice, [CS, Table 10.4]. Hence McL stabilizes vectors a, b such that (a, a) = (b, b) = 4 and (a + b, a + b) = 6. Thus (a, b) = −1, and (a, a − b) = 5 and (b, a − b) = −5. Hence in Λ̄ = Λ/5Λ we see that McL acts on the 21-space M = ⟨ā, b̄⟩^{⊥}/⟨\overline{a − b}⟩. By (c), U_{4}(3) acts irreducibly on M and so (d) holds.

Lemma 5.5 Partial information on the fusion of conjugacy classes for H in G is as follows:

H 1A 2A 3A 4A 4B 5A 6A 6B 6C 7A B** 8A 12A
G 1A 2A 3A 4A 4A 5B 6A 6B 6B 7A B** 8A 12A

Proof: By 5.4 (a), McL has a character of degree 22. By 5.4 (b) this character decomposes as 1 + 21 when restricted to U_{4}(3). Comparing character values we see that the fusion must be as claimed.

¿From the character table of 3.McL [ATLAS] we have:

Lemma 5.6 3.McL has (among others) the following faithful irreducible ordinary characters:

1A 2A 3A 4A 5B 6A 6B 7A B** 8A 12A
1980 -36 9 4 0 9 0 -1 -1 0  1
2376_{1} -24 -56 0 1 6 0 b7 ** 0  0
2376_{2} -24 -56 0 1 6 0 ** b7 0  0
8019_{2} -45 0 3 -1 0 0 -b7 ** -1  0
8019_{1} -45 0 3 -1 0 0 ** -b7 -1  0

Lemma 5.7 Restricted to 3_{2}.U_{4}(3) we have (a) 1980 = 45_{1} + 45_{2} + 189 + 756 + 945.

(b) For i = 1, 2, 2376_{i} = 45_{i} + 630 + 756 + 945.

(c) For i = 1, 2, 45_{i} is a summand of 8019_{i}.

Proof: For (a) and (b) one verifies that character values on the right hand side add up to the character values on the left. For (c) we claim that (45_{i}, 8019_{i}) is positive. Indeed for all x not in 7A or 7B ∗ ∗, 45_{i}(x) · 8019_{i}(x^{−1}) is a nonnegative integer. Furthermore b7 = \frac{1}{2}(−1 + \sqrt{−7}) and so b7 · (−b7) has 3/2 as it real part. Thus |3_{2}.U_{4}(3)| · (45_{i}, 8019_{i}) is the sum of positive integers.

We are now ready to begin our proof of 5.2. For i = 1, 2 we let λ_{i} be the ℂG-module induced from the ℂH-module corresponding to the character 45_{i}. Similarly, let λ̂_{i} be the kG-module induced from the kH-module corresponding to the 5-modular character ˆ45_{i}. More generally we will use “hats” to indicate modules over k.

Lemma 5.8 For i = 1, 2 we have λ_{i} = 1980 + 2376_{i} + 8019_{i}.

Proof: By 5.7, we know that 1980, 2376_{i} and 8019_{i} are summands of λ_{i}. Since λ_{i} has degree 45 · 275 = 12375 = 1980 + 2376 + 8019, the lemma holds.

Lemma 5.9 For i = 1, 2 we have λ̂_{i} is a projective indecomposable kG-module.

Proof: Since ˆ45_{i} is a projective kH-module, λ̂_{i} is a projective kG-module. Suppose that λ̂_{i} is the sum of two proper projective kG-submodules ρ̂_{1} and ρ̂_{2}. As, for j = 1, 2, ρ̂_{j} is projective, ρ̂_{j} is the reduction of some ℂG-module ρ_{j}. By 5.8, ρ_{1} +ρ_{2} is the sum of three irreducibles, and so either ρ_{1} or ρ_{2} is irreducible. But none of 1980, 2376 and 8019 is divisible by 125, which is a contradiction. It follows that λ̂_{i} is indecomposable.

Lemma 5.10 Restricted to 3_{2}.U_{4}(3),

(a) ˆ1980 has the following composition factors:

{ ˆ45_{1}, ˆ45_{2}, ˆ36, ˆ153, ˆ153, ˆ603, ˆ945}; and

(b) for i = 1, 2, ˆ2376_{i} has the following composition factors:

{ ˆ45_{i}, ˆ153, ˆ603, ˆ630, ˆ945}.

Proof: Use 5.7 and 5.3.

Let μ̂_{i} be the unique irreducible quotient of λ̂_{i} [Alp, Theorem 5.3]. Then, as ˆ45_{i} is a summand of ˆ1980 |_{H}, ˆ1980 is a quotient of λ̂_{i} and consequently μ̂_{i} is a quotient of ˆ1980. Hence every composition factor of μ̂_{i} |_{H} is also a composition factor of ˆ1980 |_{H} and, using exactly the same argument, is a composition factor of { \atop 2376_{i}} |_{H}. From 5.10 we conclude that μ̂_{i} |_{H}= ˆ45_{i} + a_{i} · ˆ153 + b_{i} · ˆ603 + c_{i} · ˆ945, for some {a_{i}, b_{i}, c_{i}} ∈ {0, 1}. Since, by 5.5, 4A and 4B fuse in 3.McL the values of μ̂_{i} |_{H} on 4A and 4B elements must agree. As the character values on 4A and 4B elements agree on ˆ45_{i}, ˆ153 and ˆ945 and do not agree on ˆ603, we have b_{i} = 0. Moreover, μ̂_{1} and μ̂_{2} are also both composition factors of ˆ1980 and so c_{1} + c_{2} ≤ 1. So we may assume without loss of generality that c_{1} = 0. Put

μ̂ = μ̂_{1}.

Then we have demonstrated that either

μ̂ |_{H}= ˆ45_{1}

or

μ̂ |_{H}= ˆ45_{1} ⊕ ˆ153.

In the first case μ̂ has degree 45 and 5.2 is proven. Therefore, we assume from now on that

μ̂ |_{H}= ˆ45_{1} ⊕ ˆ153

and seek a contradiction.

Because we will now only be considering 5-modular modules, we drop the “hats” which were used to distinguish between ordinary and 5-modular representations. By 5.4 (c), kG has a (non-faithful) 21-dimensional representation. Set

σ = 21 ⊗ μ.

Then σ is a faithful kG-module. We have

Lemma 5.11 σ |_{H}= (153 \ (36 ⊕ 603) \ 153) ⊕ 45_{1} ⊕ 45_{2} ⊕ 945 ⊕ 945 ⊕ 603. In particular, as a kH-module, 21 ⊗ 45_{1} is the projective cover of the 153.

Proof: Since 45_{1} is projective, μ |_{H}= 45_{1} ⊕ 153. Hence we need to compute the kH-modules 21 ⊗ 45_{1} and 21 ⊗ 153. Their Brauer characters are:

1A 2A 3A 4A 4B 6A 6B 6C 7A B** 8A 12A
21 ⊗ 45_{1} 945 -15 54 1 1 6 0 0 0  0 1  -2
21⊗ 153 3213 -35 -108 1 1 4 -2 -2 0  0 -1  4

Now it is easy to verify that 21 ⊗ 45_{1} has composition factors 36, 603 and 153 twice. Since 45_{1} is projective, 21⊗45_{1} is projective. Since 36 and 603 occur only once in 21 ⊗ 45_{1}, their projective cover cannot be a summand of 21 ⊗ 45_{1}. By 5.3 the projective cover of 153 is 153\(36⊕603)\153. Therefore 21⊗45_{1} is the projective cover of 153. Finally we note that 21 ⊗ 153 has composition factors 45_{1}, 45_{2}, 945 twice and 603. All of these modules except 603 are projective kH-modules and so 21 ⊗ 153 is completely reducible and 5.11 holds.

Let α be an outer automorphism of G normalizing H. Then for each kGmodule δ, we define δ^{α} to be the kG-module with G-action given by v · g = v^{α(g)}. We will be concerned with the kG-modules δ^{∗α} where ∗ as usual denotes duality. Since there is a unique kG-module of dimension 21 we have 21^{∗α} ≅ 21. Moreover, the ℂH-module λ_{1} = 1980 + 2376_{1} + 8019_{1} is also invariant under ∗α. Thus ∗α fixes both μ and σ = 21 ⊗ μ. In particular, there exists a k-linear isomorphism

β : σ → σ^{∗}

such that

β(m^{g}) = β(m)^{α(g)}

for all m ∈ σ and all g ∈ G. For γ ≤ σ, set

γ̃ = {m ∈ σ|ϕ(m) = 0 for all ϕ ∈ β(γ)}.

Then, if γ is a kG-submodule of σ, we have γ̃ is a kG-submodule of σ and σ/γ̃ ≅ γ^{∗α}.

Let ρ be the unique kH-submodule of σ isomorphic to 45_{1} and let δ be the kG-submodule of σ spanned by ρ. Notice that, since ρ is unique in σ, any kGsubmodule of σ that contains a composition factor isomorphic to 45_{1} contains δ. Let ϵ be a maximal kG-submodule of δ. We claim that ϵ is unique and δ/ϵ ≅ μ. First note that by the definition of δ, ρ does not lie in any proper kG-submodule of δ. In particular, ρ ≰ ϵ and so δ/ϵ is an irreducible kG-module which restricted to H has (ρ+ϵ)/ϵ ≅ ρ ≅ 45_{1} as a submodule. Therefore, δ/ϵ ≅ μ. Furthermore, if τ is any proper kG-submodule of δ, then τ ≤ ϵ. For otherwise, since δ/ϵ is irreducible, δ = τ + ϵ and so τ/(τ ∩ ϵ) ≅ (ϵ + τ)/ϵ = δ/ϵ ≅ μ. This implies that τ has a kH-composition factor isomorphic to 45_{1}. But then ρ ≤ τ, which is a contradiction. Hence τ ≤ ϵ and ϵ is unique.

Since μ is ∗α-invariant and δ/ϵ ≅ μ, we have ϵ̃/δ̃ ≅ μ^{∗α} ≅ μ. Therefore, (ϵ̃/δ̃)|_{H} has 45_{1} as a composition factor. Hence ρ ≤ ϵ̃, ρ ≰ δ̃ and δ ≤ ϵ̃. As ϵ̃/δ̃ is irreducible, we conclude that ϵ̃ = δ + δ̃ and δ/(δ ∩ δ̃) ≅ ϵ̃/δ̃ ≅ μ. Thus δ ∩ δ̃ is a maximal submodule of δ and so δ ∩ δ̃ = ϵ. So

ϵ̃/ϵ ≅ δ/ϵ ⊕ δ̃/ϵ = (δ + δ̃)/ϵ ≅ μ ⊕ δ̃/ϵ.

Recall that, by hypothesis, μ|_{H} = 45_{1} ⊕153. So, as 153 is ∗α-invariant, we have 153 is a composition factor of ϵ|_{H} if and only if it is a composition factor of (σ/ϵ̃)|_{H}. Hence as, by 5.11, σ|_{H} only has two composition factors isomorphic to 153, we see that the composition factors isomorphic to 153 only appear in ϵ̃/ϵ. But then 153 must occur as a composition factor of (δ̃/ϵ)|_{H}. It follows that σ|_{H} has a section isomorphic to the direct sum of two copies of 153. This is the final straw, as by 5.11, σ|_{H} contains no composition factor isomorphic to 153 ⊕ 153. This contradiction draws to a close the proof of 5.2.

Corollary 5.12 3.Aut(McL) has a faithful irreducible representation of degree 90 over GF(5).

Suppose now that G ≅ 3.Aut(McL) and let

G_{2} ∼ 3^{1+1+4}.4.Sym(5)

and

G_{3} ∼ 3^{1+4}.(C_{2} × Mat_{10})

be subgroups of G with

G_{23} := G_{2} ∩ G_{3} ∼ 3^{1+1+2+1}(C_{2} × 3^{2}.Q_{8}) = 3^{1+1+2+1}(C_{2} × Mat_{9})

Set Z = O_{3}(G), Q_{2} = O_{3}(G_{2}) and Q_{3} = O_{3}(G_{3}). Now let (G_{2}, G_{3}) be the abstract amalgam isomorphic to the amalgam just described in G.

Lemma 5.13 Any two faithful irreducible 90-dimensional GF(5)-modules for the amalgam (G_{2}, G_{3}) are isomorphic. In particular, if ϕ is a faithful irreducible representation of the amalgam (G_{2}, G_{3}) into GL_{90}(5), then ⟨G_{2}^{ϕ}, G_{3}^{ϕ}⟩ ≅ 3.Aut(McL).

Proof: Let V be any faithful irreducible 90-dimensional GF(5)-module for the amalgam (G_{2}, G_{3}), and view V as a 45-dimensional semilinear representation over GF(25). Note that both the normalizer of a Sylow 5-subgroup and a Sylow 2-subgroup are maximal, 3′-subgroups of Mat_{10}. It follows that G_{3} has two orbits L and H of lengths 36 and 45 on the maximal subgroups of Q_{3} not containing Z. Since [V, Q_{3}] is a direct sum of the centralizers in V of the maximal subgroups of Q_{3} and Z fixes only 0 in V, we deduce that the Wedderburn components for Q_{3} on V = [V, Q_{3}] = [V, Z] are 1-dimensional and are transitively permuted by G_{3}. Moreover, if C_{V} (H) ≠ 0 for H maximal in Q_{3}, then H ∈ H and N_{G_{3}}(H) ∼ 3^{5}.(C_{2} × SD_{16}).

Let X_{2} be the group of order three normal in G_{2} and different from Z and A, B be the other two maximal subgroups of ⟨X_{2}, Z⟩. Note that O^{2}(G_{2})/Y ∼ 3^{1+4}.2.Alt(5) and Q_{2}/Y is extraspecial for each Y ∈ {Z, X_{2}, A, B}. Furthermore, we know that V has an O^{2}(G_{2})-invariant decomposition

V = C_{V} (X_{2}) ⊕ C_{V} (A) ⊕ C_{V} (B).

Since G_{23} has orbits of length 9 and 36 on H, V decomposes as a G_{23}-module into a direct sum of irreducible modules Z_{4} and Z_{5} of dimensions 9 and 36 respectively. Since C_{V} (A) has dimension at least 9, we have Z_{4} = C_{V} (X_{2}) and Z_{5} = [V, X_{2}]. In particular, Z_{4} and Z_{5} are also invariant under G_{2}. Thus, by 2.6 and 2.7 (a), Z_{4} and Z_{5} are uniquely determined when restricted to O^{2}(G_{2}).

Suppose that H ∈ H and define V^{H} = C_{V} (H). So V^{H} is 1-dimensional. To complete the determination of V as a G_{3}-module, we need to determine the action of N_{G_{3}}(H) on V^{H}. Let S^{∗} be a Sylow 2-subgroup of N_{G_{3}}(H) and S = C_{S^{∗}}(Z). Then S ≅ SD_{16} and it suffices to determine C_{S}(V^{H}). For i = 2, 3, let K_{i} = S ∩ G′_{i} and set K_{0} = Ω_{1}(K_{2}). Then K_{3} ≅ Q_{8}, K_{2} ≅ C_{4} and K_{0} ≅ C_{2}. We clearly have S′ ≤ C_{S}(V^{H}). First, select H ∈ H such that X_{2} ≤ H. Then V^{H} ≤ C_{Z_{4}}(H ∩ Q_{2}) and so, by 2.7 (d), K_{2} centralizes V^{H}. Thus S′K_{2} centralizes V^{H} and we know that S′K_{2} ≅ Q_{8}. Thus C_{S}(V^{H}) ≅ Q_{8} or SD_{16}. Next choose H ∈ H so that X_{2} ≰ H. Then V^{H} ≤ C_{Z_{5}}(H ∩ Q_{2}) and, using 2.7 (d) again, K_{0} inverts V^{H}. Thus C_{S}(V^{H}) ≅ Q_{8}. Therefore, V is uniquely determined as a G_{3}-module. It particular, Z_{4} and Z_{5} are uniquely determined as G_{23}-modules and then, by 2.3 and 2.7, Z_{4} and Z_{5} are also uniquely determined as G_{2}-modules. Since End_{GF(5)G_{2}}(V) = End_{GF(5)G_{23}}(V), V is also uniquely determined as module for the amalgam (G_{2}, G_{3}). Finally, as by 5.12, 3.Aut(McL) has an irreducible representation of degree 90 over GF(5), the lemma holds.

6 Existence and Uniqueness of Ly-amalgams

In this section we establish the existence and uniqueness of Ly-amalgams. The proof of our next lemma is taken from [IM, Lemma 6.3].

Lemma 6.1 Let (M_{1}, M_{2}, M_{3}) and (M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗}) be amalgams. Suppose that

1. (i) There exists L ≤ M_{123} such that L ⊴ M_{2}, C_{M_{2}}(L) = 1 and for i ∈

{1, 3}, M_{i2} = N_{M_{i}}(L). (ii) There exists L^{∗} ≤ M_{123}^{∗} such that L^{∗} ⊴ M_{2}^{∗}, C_{M_{2}^{∗}}(L^{∗}) = 1 and for

i ∈ {1, 3}, M_{i2}^{∗} = N_{M_{i}^{∗}}(L^{∗}).

2. For i ∈ {1, 3} there exist isomorphisms ϕ_{i} : M_{i} → M_{i}^{∗} such that ϕ_{1} |_{M_{13}}= ϕ_{3} |_{M_{13}}, M_{13}^{ϕ_{i}} = M_{13}^{∗} and L^{ϕ_{i}} = L^{∗}.

3. M_{2} = ⟨M_{12}, M_{23}⟩ and M_{2}^{∗} = ⟨M_{12}^{∗}, M_{23}^{∗}⟩.

Then there exists an isomorphism ϕ_{2} : M_{2} → M_{2}^{∗} such that for i ∈ {1, 3}, ϕ_{2} |_{M_{2i}}= ϕ_{i} |_{M_{2i}} and M_{2i}^{ϕ_{2}} = M_{2i}^{∗}. In particular, (M_{1}, M_{2}, M_{3}) and (M_{1}^{∗}, M_{2}^{∗}, M_{3}^{∗}) are isomorphic.

Proof: Let ϕ be the restriction of ϕ_{1} or ϕ_{3} to L. Then by assumption L^{ϕ} = L^{∗}. Define λ : M_{2} → Aut(L^{∗}) by m^{λ} : l^{ϕ} → (l^{m})^{ϕ} for all m ∈ M_{2} and l ∈ L. Furthermore, define λ^{∗} : M_{2}^{∗} → Aut(L^{∗}) by m^{λ^{∗}} : l → l^{m} for all m ∈ M_{2}^{∗}, l ∈ L^{∗}. Since C_{M_{2}}(L) = 1 = C_{M_{2}^{∗}}(L^{∗}), λ and λ^{∗} are monomorphisms. Let i ∈ {1, 3}. Then M_{i2}^{ϕ_{i}} = N_{M_{i}}(L)^{ϕ_{i}} = N_{M_{i}^{∗}}(L^{∗}) = M_{i2}^{∗}. We claim that λ and ϕ_{i}λ^{∗} agree on M_{i2}. Indeed let l ∈ L and m ∈ M_{i2}. Then

m^{ϕ_{i}λ^{∗}} : l^{ϕ} → (l^{ϕ})^{m^{ϕ_{i}}} = (l^{ϕ_{i}})^{m^{ϕ_{i}}} = (l^{m})^{ϕ_{i}} = (l^{m})^{ϕ},

proving the claim. In particular,

M_{2}^{λ} = ⟨M_{12}^{λ}, M_{23}^{λ}⟩ = ⟨M_{12}^{ϕ_{1}λ^{∗}}, M_{23}^{ϕ_{3}λ^{∗}}⟩ = ⟨M_{12}^{∗λ^{∗}}, M_{23}^{∗λ^{∗}}⟩ = M_{2}^{∗λ^{∗}}.

Set ϕ_{2}=λλ^{∗−1}. Then ϕ_{2} is an isomorphism between M_{2} and M_{2}^{∗} which agrees on M_{12} with ϕ_{1} and on M_{23} with ϕ_{3}, completing the proof of the lemma.

We now recall the following definition from the introduction:

Definition 6.2 An amalgam of groups (M_{1}, M_{2}, M_{3}) is called a Ly-amalgam provided that

1. M_{1} ∼ 3.McL.2, M_{2} ∼ 3^{6}.2^{3}.Sym(5) and M_{3} ∼ 3^{5}.2.Mat_{11}.

2. |M_{2} : M_{12}| = 2, |M_{2} : M_{23}| = 10 and |M_{3} : M_{13}| = 11.

3. |M_{23} : M_{123}| = 2.

4. No non-trivial subgroup of M_{123} is normal in M_{1}, M_{2} and M_{3}.

The next proposition establishes the existence of a Ly-amalgam of groups.

Proposition 6.3 There exists a Ly-amalgam of groups (M_{1}, M_{2}, M_{3}) such that

(a) M_{1} ∼ 3.Aut(McL), M_{2} ∼ 3^{2+4}.8.Sym(5) and M_{3} ∼ 3^{5} : (C_{2} × Mat_{11}).

(b) M_{12} ∼ 3^{1+1+4}.4.Sym(5), M_{13} ∼ 3^{1+4}.(C_{2} × Mat_{10}) and M_{23} ∼ 3^{2+2+1}.(C_{2} × 3^{2}.SD_{16}).

(c) M_{123} ∼ 3^{1+1+2+1}.(C_{2} × 3^{2}.Q_{8}).

(d) M_{2} = M_{23}M_{21}.

Proof: By 5.1 there exists a unique group of shape 3.Aut(McL). Denote this group by M_{1}. By 2.5 (a), there is a unique group X of shape 3^{5} : (C_{2} × Mat_{11}) such that C_{X}(O_{3}(X)) = O_{3}(X), O_{3}(X) is an irreducible GF(3)Mat_{11}-module and C_{2} × Mat_{10} inverts a cyclic subgroup in O_{3}(X). Denote this group by M_{3}. Then both M_{1} and M_{3} contain subgroups of shape X ∼ 3^{(1+4)}:(C_{2} × Mat_{10}) with O_{3}(X) an indecomposable GF(3)Mat_{10}-module (see 2.5 (d) and the proof of Theorem 5.1). By 2.5 (f), a subgroup which has the shape of X is unique up to isomorphism. Hence we may arrange for M_{1} and M_{3} to intersect in a common subgroup M_{13} of shape 3^{(1+4)} : (C_{2} × Mat_{10}). Let S be a Sylow 3subgroup of M_{13}, Z_{2} be the centre of S and set Z_{1} = O_{3}(M_{1}). So Z_{2} has order 9 and Z_{1} order 3. Let T ∈ Syl_{2}(N_{M_{13}}(S)). Then T ≅ Q_{8} × C_{2}. Write T = A × B where |A| = 2 and AO_{3}(M_{3})/O_{3}(M_{3}) centralizes M_{3}/O_{3}(M_{3}) and, without loss of generality, B ≅ Q_{8} is a subgroup of O^{2}(M_{3}). Then, as M_{3}/O_{3}(M_{3}) operates irreducibly on O_{3}(M_{3}), A inverts every non-trivial element of Q_{3} and hence of Z_{2}. As Z(B) = B′ ≤ M′_{13} ≤ M′_{1} = C_{M_{1}}(Z_{1}) and Z_{2} admits B, C_{Z_{2}}(Z(B)) = Z_{2} and B operates on Z_{2} as a Klein four group. Thus, by 2.5 (g), there are distinct cyclic subgroups, B_{1} and B_{2}, of order 4 in B such that Z_{2} = C_{Z_{2}}(B_{1}) × C_{Z_{2}}(B_{2}). In particular, the third cyclic subgroup of order 4, B_{3}, in B inverts every element of Z_{2}. Let b be a generator of B_{3} and a be an element of order 2 in A. Then C = ⟨ab⟩ is of order 4 and centralizes Z_{2}. Now ⟨A, B_{3}⟩ ≅ C_{4} × C_{2} inverts every element of Z_{2} while the other two over-groups of C are quaternion and each centralizes a cyclic subgroup of Z_{2} . As there are only two cyclic subgroups of order 4 in C_{4} × C_{2} we conclude that there is a unique cyclic subgroup of order 4 in T which inverts Z_{2}, namely B_{3}.

We define

L = ⟨B_{3}^{N_{M_{13}}(S)}⟩ = B_{3}[B_{3}, S] ≤ M_{13}.

Then, L ∼ 3^{2+4}.4 and C_{M_{1}}(L) = C_{M_{3}}(L) = 1. For i = 1, 3, and 13, define M_{i2} = N_{M_{i}}(L). Then T ≤ M_{12} and T ≤ M_{23}. Let X=N_{M_{1}}(Z_{2}). As C_{O_{3}(L)}(b^{2}) = Z_{2}, M_{12} ≤ X ∼ 3^{2+4}.4.Sym(5). Now since O_{2,3}(X) inverts Z_{1}, O_{2,3}(X) ≰ M′_{1}. Hence, from the structure of McL, O_{2,3}(X) also inverts Z_{2}/Z_{1}. Since Sp_{4}(3) does not contain a subgroup 4.Alt(5), we see that O_{2,3}(X) inverts Z_{2}. Thus O_{2,3}(X) and L are equal. Hence M_{12} = X and M_{12} ∼ 3^{1+1+4}.4.Sym(5). Because M_{23} ≥ N_{M_{13}}(S) and [N_{M_{3}}(S) : N_{M_{13}}(S)] = 2 the maximality of N_{M_{3}}(S) in M_{3} implies that

M_{23} = N_{M_{3}}(S) ∼ 3^{2+2+1}.(C_{2} × 3^{2}.SD_{16}) ∼ 3^{2+4}.8.(Sym(3) × Sym(2)).

Finally we get

M_{123} = N_{M_{13}}(S) ∼ 3^{2+2+1}.(C_{2} × 3^{2}.Q_{8}) ∼ 3^{2+4}.4.(Sym(3) × Sym(2)).

Since, for i ∈ {1, 3}, C_{M_{i}}(L) = 1, we identify M_{i2} with their images in Aut(L). Then M_{12}∩M_{23} = M_{123}. Let M_{2} be the subgroup of Aut(L) generated by M_{12} and M_{23}. We claim that M_{2} ∼ 3^{2+4}.8.Sym(5).

Let T ∈ Syl_{2}(L) and D = C_{M_{2}}(T). Then, as |T | = 4, we have |M_{2}/DL| = 2, by the Frattini Argument, and, as D centralizes the element of order 2 in T, the Three Subgroup Lemma implies that D acts faithfully on O_{3}(L)/Z_{2}. Thus, since D centralizes T, D is isomorphic to a subgroup of GL_{2}(9). Let D_{i} = M_{i2}∩D for i = 1, 3, 13. Then D_{1}/T ≅ Alt(5), D_{3}/T ≅ C_{2} × Sym(3) and D_{13}/T ≅ Sym(3). Since all the 2-elements centralizing an element of order three in GL_{2}(9) are contained in Z(GL_{2}(9)) we conclude that D_{3} = D_{13}X where X centralizes D_{1}. Thus D = ⟨D_{1}, D_{2}⟩ = ⟨D_{1}, X⟩ ∼ 8.Alt(5) and so M_{2} ∼ 3^{2+4}.8.Sym(5) as claimed. Hence the amalgam (M_{1}, M_{2}, M_{3}) is a Ly-amalgam of groups and statements (a), (b) and (c) hold.

Theorem 6.4 Up to isomorphism, there is a unique Ly-amalgam of groups (M_{1}, M_{2}, M_{3}).

Proof: Let (M_{1}, M_{2}, M_{3}) be any Ly-amalgam of groups and, for i = 1, 2, 3, set Q_{i} = O_{3}(M_{i}). Because |M_{3}/M_{13}| = 11, we have M_{13} ∼ 3^{5}.(2 × Mat_{10}). Also, since elements of order five in McL are not centralized by a group of order 3^{4}, M′_{13}/Q_{3}, and hence also M′_{3}/Q_{3}, acts faithfully on Q_{3}. Thus, as M_{3} operates non-trivially on Q_{3} and M_{3} possesses no 4-dimensional irreducible GF(3)-modules, M_{3} operates irreducibly on Q_{3}. Hence, by 2.5 (a), Q_{3} is the unique irreducible GF(3)M′_{3}-module in which M_{13} ∩ M′_{3} stabilizes a 1-space. In particular, 2.5 (c) implies that Q_{3} does not split over Q_{1} as an M_{13}-module and, by 2.5 (d), M_{13} ∩ M′_{3} inverts Q_{1}. Thus M_{1} ∼ 3.Aut(McL). It follows that C_{M_{13}}(Q_{3}) = Q_{3} and therefore, M_{3}/Q_{3} ≅ C_{2} × Mat_{11} acts faithfully on Q_{3}. Thus M_{1} and M_{3} are unique up to isomorphism. Since the outer automorphism group of M_{13} is trivial (the group 3^{4} does not admit PGL_{2}(9)), the pair of groups (M_{1}, M_{3}) is unique up to isomorphism. As M_{12} ∼ 3^{6}.2^{x}.Alt(5).2^{y} with x + y = 3, the structure of M_{1} implies M_{12} ∼ 3^{1+1+4}.4.Sym(5). Let L = O_{3,2}(M_{12}). As M_{12} is normal in M_{2}, L is normal in M_{2}. Clearly M_{12} = N_{M_{1}}(L). Since |M_{13}/M_{123}| = 10, M_{123} has a unique Sylow 3-subgroup S and M_{123} = N_{M_{1}}(S). In particular, L ≤ M_{123} ≤ M_{3}. Moreover, |M_{23}/M_{123}| = 2 implies that S ⊴ M_{23} and so M_{23} ∼ 3^{2+2+1}.(C_{2} × 3^{2}.SD_{16}), M_{23} = N_{M_{3}}(L) and M_{123} ∼ 3^{1+1+2+1}.(C_{2} × 3^{2}.Q_{8}). The argument concluding the proof of 6.3 shows that M_{2} ∼ 3^{2+4}.8.Sym(5) and M_{2} = M_{12}M_{23}. Finally we apply 6.1 to get that (M_{1}, M_{2}, M_{3}) is unique up to isomorphism. This in conjunction with 6.3 gives the theorem.

We finish this section with a corollary to the proof of 6.3.

Corollary 6.5 Suppose that (M_{1}, M_{2}, M_{3}) is a Ly-amalgam of groups and let K^{∗} ∈ Syl_{2}(M_{23}). Then

(a) C_{M_{2}}(Z(Q_{2})) = O^{2}(M_{2}) = O^{2}(M_{12}).

(b) M_{2}/O^{2}(M_{2}) ≅ D_{8}.

(c) C_{K^{∗}Q_{2}/Q_{2}}(O^{2}(M_{2})/Q_{2}) ≅ C_{8}.

(d) O_{2,3}(M_{2}) ∼ 3^{1+1+2+2}.C_{8}.

7 Completions of Ly-amalgams

Throughout this section M is assumed to be a faithful completion of a Lyamalgam of groups (M_{1}, M_{2}, M_{3}). We identify M_{1}, M_{2} and M_{3} with their images in M. For the moment let Γ = Γ(M; M_{1}, M_{2}, M_{3}) be the coset graph of M_{1}, M_{2} and M_{3} in M with two cosets incident if and only if they have nontrivial intersection. Then M acts on Γ by right multiplication. For i = 1, 2 and 3 set Γ_{i} = M/M_{i}. Our goal is to determine all the orbits of M_{1} on Γ_{1}. For a, b ∈ Γ_{1} we write a ⋊ ⋉ b if a and b are adjacent to a common vertex of type 2. We denote the distance between a, b ∈ Γ_{1} in the graph (Γ_{1}, ⋊ ⋉) by d(a, b). For a ∈ Γ set

Q_{a} = O_{3}(M_{a})

and

Z_{a} = Z(Q_{a}).

We now assume that a ∈ Γ_{1} is a fixed vertex and change notation so that M_{a} = M_{1}. Then we set

X_{0}(a) = {a}

and

X_{1}(a) = {b ∈ Γ_{1} | a ⋊ ⋉ b}.

A path a, b, c, … in Γ is said to be of type i − j − k … if a ∈ Γ_{i}, b ∈ Γ_{j}, c ∈ Γ_{k}, . . .. We freely use the information about the structure of M_{1}, M_{2} and M_{3} guaranteed by 6.4 and given in 6.3.

Notice that the for a ∈ Γ_{1}, the results of Section 4 apply to describe the action of M_{a} on Γ_{i}(a) for 2 ≤ i ≤ 4. Also note that if a ⋊ ⋉ b and c ∈ Γ_{2}(a)∩Γ_{2}(b), then {c} = Γ_{2}(a) ∩ Γ_{2}(b) and {a, b} = Γ_{1}(c). So, if a ⋊ ⋉ b, then M_{ab} = M_{abc} ∼ 3^{2+4}.4.Sym(5) has index 2 in M_{c}.

Lemma 7.1 There exist unique M-classes of paths of type 1−2−1 and 1−3−1 in Γ.

Proof: Because |M_{2} : M_{12}| = 2, paths of type 1 − 2 − 1 are in one-to-one correspondence with paths of type 1 − 2. Thus there is a unique class of these paths. That there is a unique class of paths of type 1 − 3 − 1 follows from the 2-transitivity of M_{3} on the cosets of M_{13}.

Lemma 7.2 Suppose that b ∈ Γ_{2} and c ∈ Γ_{3}(b). Then

(a) Γ_{1}(b) ⊆ Γ_{1}(c).

(b) if f ∈ Γ_{1}(c) and Z_{f} ≤ Q_{c} ∩ Q_{b}, then f ∈ Γ_{1}(b) and Z_{f} ≤ Z_{b}.

Proof: By 6.3 (a) and (b), |Γ_{1}(b)| = 2. Let {a, e} = Γ_{1}(b). We may assume that a ∈ Γ_{1}(c). Since M_{b} = M_{ba}M_{bc}, by 6.3 (d), M_{bc} acts transitively on Γ_{1}(b). Therefore, e ∈ Γ_{1}(c) and so (a) holds.

Now note that M_{c} acts three-transitively on Γ_{1}(c) and M_{cb} is the stabiliser of a set of two points under this action. Hence M_{cb} has an orbit of length 9 and an orbit of length 2 on Γ_{1}(c), the latter is {a, e}. Suppose that f ∈ Γ_{1}(c) and Z_{f} ∈ Q_{c} ∩ Q_{b}. Since ⟨Z_{a}, Z_{e}⟩ = Z_{b} ≤ Q_{c} ∩ Q_{f}, M_{cf} normalizes Q_{c} ∩ Q_{f} and ⟨Z_{f}^{M_{c}}⟩ = Q_{c}, the action of M_{cb} now shows us that Z_{f} ≤ Z_{b} and f ∈ {a, e} as claimed.

Lemma 7.3 There exist four M-classes of paths a, b, c, d, e of type 1 − 2 − 1 − 2 − 1 in Γ. These classes are distinguished by the isomorphism type of the subgroup of M_{c} generated by Z_{a} and Z_{e} as follows:

⟨Z_{a}, Z_{e}⟩ ∈ {3^{2}, 2.Alt(4), 2.Alt(5), 5^{1+2}.3}.

Moreover,

(a) If [Z_{a}, Z_{e}] = 1, then e ∈ X_{1}(a) and there exists a unique f ∈ Γ_{3}(a) ∩ Γ_{3}(c) ∩ Γ_{3}(e).

(b) If ⟨Z_{a}, Z_{e}⟩ ∼ 2.Alt(4), then M_{ae} ∼ 2.Sym(7).

(c) If ⟨Z_{a}, Z_{e}⟩ ∼ 2.Alt(5), then M_{ae} ∼ 2.(C_{2} × Sym(6)).

(d) If ⟨Z_{a}, Z_{e}⟩ ∼ 5^{1+2}.3, then M_{ace} ≅ D_{30}.

Proof: By 7.1 we may assume that the path a − b − c is fixed. We work inside M_{c}. By 4.4, M_{bc} has four orbits on Γ_{2}(c) \ {b} and since a vertex of type 2 has only two neighbours of type 1, there exist exactly four paths of type 1 − 2 − 1 − 2 − 1. Moreover, 4.4 (b), (c), (d) and (e) together show that the four classes are distinguished by the isomorphism type of ⟨Z_{a}, Z_{e}⟩Z_{c}/Z_{c} as described.

Suppose that [Z_{a}, Z_{e}] ≤ Z_{c}. Then, by 4.4 (b), there exists a unique f ∈ Γ_{3}(b) ∩ Γ_{3}(c) ∩ Γ_{3}(d). Then a, e ∈ Γ_{1}(f) and the two transitivity of M_{f} on Γ_{1}(f) indicates that there is a h ∈ Γ_{2}(f) with {a, e} = Γ_{1}(h). So, by definition, e ∈ X_{1}(a) and (a) holds.

Notice that if ⟨Z_{a}, Z_{e}⟩ is non-abelian, then, Z_{a} and Z_{e} are conjugate in ⟨Z_{a}, Z_{e}⟩.

Suppose that ⟨Z_{a}, Z_{e}⟩Z_{c}/Z_{c} ∼ 2.Alt(i), i = 4, 5. Then 4.4 (c) and (d) (allowing for the central 3 in M_{c} which is inverted) give M_{ace} ∼ 2.(Sym(3) × Sym(4)) when i = 4 and M_{ace} ≅ 2.(C_{2} ×Sym(3)×Sym(3)) when i = 5. Let t be the involution in ⟨Z_{a}, Z_{e}⟩. Then clearly M_{ae} ≤ C_{M_{a}}(t) ∼ 6.Sym(8). Therefore as Z_{a} ∩ M_{ae} = 1, M_{ae}/⟨t⟩ is isomorphic to a subgroup of Sym(8).

We shall show that all the 3-cycles of M_{ae}/⟨t⟩ are conjugate. So let r⟨t⟩/⟨t⟩ be such 3-cycle with |⟨r⟩| = 3. If ⟨r, Z_{c}⟩Z_{a}/Z_{a} ≅ 2.Alt(4) or 2.Alt(5), ⟨r⟩ and Z_{c} are conjugate in ⟨r, Z_{c}⟩ ≤ M_{ae}. So suppose that [Z_{c}, r] = 1 and pick y ∈ X_{1}(c) with r ∈ Z_{y}. Then y has distance at most two from a in (Γ_{1}, ⋊ ⋉) and so, by (a), a ⋊ ⋉ y and as [Z_{y}, Z_{e}] = 1, the same argument yields y ⋊ ⋉ e. Moreover, the path (a, y, e) in (Γ_{1}, ⋊ ⋉) is unique up to conjugacy in M and so conjugate to (a, c, e). Hence y and c are conjugate in M_{ae}. Thus M_{ae}/⟨t⟩ contains a unique class of 3-cycles and the normalizer in M_{ae}/⟨t⟩ of such 3-cycles is conjugate to M_{ace}/⟨t⟩ and so is isomorphic to Sym(3) × Sym(4) if i = 4 and C_{2} × Sym(3) × Sym(3) if i = 5. Furthermore, we note that M_{ae} > M_{ace} and ⟨Z_{a}, Z_{e}⟩ ∩ Z_{c} = 1, which gives the first parts of (b) and (c).

If i = 4, this implies M_{ae} ∼ 2.Sym(7) or M_{ae} ∼ 2.(Sym(4) × Sym(4)).2. Let x ∈ Γ_{2}(c) be chosen so that ⟨Z_{a}, Z_{e}⟩ normalizes Z_{x}. Then N_{M_{x}}(⟨Z_{a}, Z_{e}⟩) ∼ 3^{2}.8.Sym(4) and so M_{xae} ∼ 3^{2}.8.2. Moreover, the elements of order eight in M_{xae} centralizes Z_{a} while M_{xae} inverts Z_{a}. Therefore, we conclude that M_{axe}Z_{a}/Z_{a}⟨t⟩ is a subgroup of Sym(8) = N_{M_{a}}(t)/Z_{a}⟨t⟩ of shape 3^{2}.D_{8} which intersects Alt(8) in 3^{2}.C_{4}. Now a subgroup of shape (Sym(4) × Sym(4)).2 has a unique conjugacy class of subgroups of shape 3^{2}.D_{8} and this class of subgroups intersects Alt(8) in a subgroup of shape 3^{2}.(C_{2} × C_{2}). Thus when i = 4 we cannot have M_{ae} ∼ 2.(Sym(4) × Sym(4)).2. Hence when i = 4 we have M_{ae} ∼ 2.Sym(7) and so (b) holds.

If i = 5, we calculate that M_{ae} ∼ 2.(C_{2} × Sym(6)) or M_{ae} ∼ 2.(C_{2} × 3^{2}.D_{8}). Let f ∈ X_{1}(c) with Z_{f} ≤ ⟨Z_{a}, Z_{e}⟩ and ⟨Z_{a}, Z_{f}⟩ ∼ 2.Alt(4) ∼ ⟨Z_{e}, Z_{f}⟩. Then, by (b), M_{af} ∼ 2.Sym(7) ∼ M_{fe}. Both of these group are in C_{M_{f}} (t) ∼ 6.Sym(8) and thus M_{afe} contains a subgroup 2.Alt(6). Therefore, M_{ae} ∼ 2.(C_{2}×Sym(6)) and so (c) holds.

Finally, if ⟨Z_{a}, Z_{e}⟩Z_{c}/Z_{c} ∼ 5^{1+2}.3, then, 4.4 gives M_{ace} ≅ D_{30} and, as Z_{e} and Z_{c} are conjugate in ⟨Z_{a}, Z_{e}⟩, (d) also holds.

Notice that 7.3 completely determines those vertices in (Γ_{1}, ⋊ ⋉) at distance 2 from a. Moreover, in all but the case of a − b − c − d − e as in 7.3 (d), we know M_{ae} and thus the action of M_{a} on such paths. Define

X_{2}(a) = {e ∈ Γ_{1} | d(a, e) = 2 and ⟨Z_{a}, Z_{e}⟩ ∼ 2.Alt(4)},
X_{3}(a) = {e ∈ Γ_{1} | d(a, e) = 2 and ⟨Z_{a}, Z_{e}⟩ ∼ 2.Alt(5)}

and

X_{4}(a) = {e ∈ Γ_{1} | d(a, e) = 2 and ⟨Z_{a}, Z_{e}⟩ ∼ 5^{1+2}.3}.

Suppose that e ∈ X_{3}(a), a − b − c − d − e is a path as in 7.3 joining a to e and set Z = Z(⟨Z_{a}, Z_{e}⟩). Note that Z ≤ M_{a}. We define

H = ⟨C_{M_{a}}(Z), Z_{e}⟩.

Notice that, by 7.3, C_{M_{a}}(Z) operates transitively on the set

{f ∈ X_{3}(a) | Z = Z(⟨Z_{a}, Z_{f}⟩)}.

Since we claim that any faithful completion of a Ly-amalgam of groups is a group of Ly-type, the relevance of the next lemma is clear.

Lemma 7.4 We have H ≅ 2.Alt(11).

Proof: Let A = ⟨Z_{a}, Z_{e}⟩. Then A/Z ≅ Alt(5). Furthermore, we have C_{M_{a}}(Z)/Z ≅ Sym(3) ∧^{|}Sym(8). Since A ≤ M_{c}, we have

N_{A}(Z_{a}) ≤ N_{M_{c}}(Z_{b}) = M_{cb} = M_{ab} ≤ M_{a}.

Therefore, as A ≰ M_{a},

C_{M_{a}}(Z) ∩ A = N_{A}(Z_{a}) ∼ 3.4.

Suppose that f ∈ X_{1}(c) with Z_{f} ∈ Syl_{3}(A), then from 7.3 we have

C_{M_{a}}(Z) ∩ C_{M_{f}} (Z) = \{ \begin{matrix}2.Sym(7) & ⟨Z_{a}, Z_{f}⟩ ∼ 2.Alt(4) \\ 2.(C_{2} × Sym(6)) & ⟨Z_{a}, Z_{f}⟩ = A \end{matrix} .

We now apply 3.3 with H/Z, A/Z, C_{M_{a}}(Z)/Z and 8, in place of M, A, L_{D} and n, to get H/Z ≅ Alt(11). Finally, because H ≥ C_{M_{a}}(Z) ≅ (2.Sym(8)) ∧^{|}Sym(3), we conclude that H ≅ 2.Alt(11).

We now adjust our notation, if necessary, so that Z_{2} ≤ H and |Z_{3} ∩H| = 3^{3}. So with this choice of H we have

C_{M_{1}}(Z(H)) ≅ Sym(3) ∧^{|}2.Sym(8),

C_{M_{2}}(Z(H)) ∼ 3^{2}.8.Sym(5)

and

C_{M_{3}}(Z(H)) ∼ 3^{3}.(C_{2} × GL_{2}(3)) ∼ 3^{3}(C_{2} × 2.Sym(4)).

We now extend Γ by adding vertices of type 4 corresponding to the cosets of H in M. Thus for the rest of this section

Γ = Γ(M; M_{1}, M_{2}, M_{3}, H),

Γ_{4} = M/H and once again two vertices in Γ are incident if and only if the cosets intersect non-trivially. Often we will denote H by M_{4}. Observe that at this stage we do not know the structure of M_{j} ∩ H for j ∈ {2, 3}, though we do know that H ∩ M_{1} = C_{M_{1}}(Z). For t ∈ Γ_{4}, set Z_{t} = Z(M_{t}) and let Ω(t) be a set of size 11 admitting M_{t}/Z_{t} naturally. Lastly, given d, e ∈ Γ_{4} we write d ⋆ e provided that Γ_{1}(d) ∩ Γ_{1}(e) ≠ ∅ and [Z_{d}, Z_{e}] = 1.

Lemma 7.5 Let b, a, c, d be a path of type 4−1−2−1 in Γ such that Z_{b} ≤ M_{abcd}. Then c and d are adjacent to b in Γ.

Proof: Let t be an involution in M′_{a}. Then, as M′_{a} has a unique class of involutions, there exists x ∈ Γ_{4}(a) with t ∈ Z_{x}. Since M_{ax} = C_{M_{a}}(t), x is uniquely determined by a and t. As, by 4.1(d), M′_{a} ∩ M_{c} also has a unique class of involutions, we conclude that the path (b, a, c) and so also (b, a, c, d) is unique up to conjugation. Since (M_{4}, M_{1}, M_{2}) is a path like (b, a, c), we have that c is adjacent to b.

Let y ∈ X_{3}(a) with Z_{b} = Z(⟨Z_{a}, Z_{y}⟩). Then, by the construction of H, ⟨Z_{a}, Z_{y}⟩ ≤ M_{b} and so y is adjacent to b.

We note, by calculating in the centralizer of t in M_{a}, that there exists u, v, w ∈ X_{1}(a) fixed by t such that [Z_{u}, Z_{w}] = 1 and ⟨Z_{u}, Z_{v}⟩ ∼ ⟨Z_{v}, Z_{w}⟩ ∼ 2.Alt(5). Select r ∈ Γ_{4}(u) with t ∈ Z_{r}. Then by the previous paragraph applied to (u, v, r) in place of (a, y, b), r is adjacent to v. Applying the same trick once more this time to (v, w, r) yields that r is adjacent to w. As (b, a, d) is conjugate to (r, u, w), we conclude that d is adjacent to b.

We now clarify the structure of the edge stabilisers in Γ.

Lemma 7.6 Suppose that d ∈ Γ_{4} and {a, b, c, d} is a clique in Γ. Then, for γ ∈ {a, b, c}, M_{γd} = C_{M_{γ}}(Z_{d}). In particular, for i ∈ {1, 2, 3}, we have M_{i}∩H = C_{M_{i}}(Z(H)).

Proof: Since {a, b, c} is a clique in Γ \ Γ_{4}, we may assume that a ∈ Γ_{1}, b ∈ Γ_{2} and c ∈ Γ_{3}. By the construction of H and 7.4 we have, C_{M_{a}}(Z_{d}) = N_{M_{d}}(Z_{a}) and so M_{da} = C_{M_{a}}(Z_{d}).

Let {a, x} = Γ_{1}(b). By 7.5, x is incident to d and so there exists h ∈ M_{d} with a^{h} = x and x^{h} = a. Thus h ∈ M_{db}. Since [C_{M_{b}}(Z_{d}) : C_{M_{ab}}(Z_{d})] = 2 and h ∉ C_{M_{ab}}(Z_{d}), we have C_{M_{b}}(Z_{d}) = C_{M_{ab}}(Z_{d})⟨h⟩ ≤ M_{d} and hence C_{M_{b}}(Z_{d}) = M_{bd}. This in turn implies that C_{M_{c}}(Z_{d}) = ⟨C_{M_{ac}}(Z_{d}), C_{M_{bc}}(Z_{d})⟩ = ⟨M_{acd}, M_{bcd}⟩ = M_{cd}. Finally, as M_{1}, M_{2}, M_{3}, H is a clique, we have the last statement of the lemma as well.

Let d ∈ Γ_{4}. Then Γ_{1}(d) corresponds to subsets of Ω(d) of size 3, Γ_{2}(d) corresponds to partitions of subsets of Ω(d) of size 6 into two parts of size 3 and the elements of Γ_{3}(e) to partitions of subsets of Ω(d) of size 9 into three parts each of size 3. Incidence between elements of Γ(d) is then recognised by symmetrized containment. The set {e ∈ Γ_{4} | d ∗ e} corresponds to partitions of subsets of Ω of size 8 into four parts of size 2.

Proposition 7.7 Let d ∈ Γ_{4} and let Γ^{d} be the connected component of Γ^{Z_{d}}, the fixed points of Z_{d} on Γ, containing d. Then M_{d} = Stab_{M}(Γ^{d}). Moreover, for every x ∈ Γ^{d}, M_{xd} = C_{M_{x}}(Z_{d}).

Proof: Let Λ = Γ^{Z_{d}}. We follow the algorithm laid out in 2.2 so that we can apply 2.1. Let {a, b, c, d} be a clique in Γ containing d with a ∈ Γ_{1}, b ∈ Γ_{2} and c ∈ Γ_{3}. Then a, b, c are representatives for the orbits of M_{d} on Γ(d). Moreover, by 7.6, C_{M_{γd}}(Z_{d}) = M_{γd} for γ ∈ {a, b, c}. Let e ∈ Λ_{4}(a) \ {d}. Since, by assumption, Z_{d} fixes e, [Z_{e}, Z_{d}] = 1 and so

Z_{e} ≤ C_{M_{a}}(Z_{d}) = M_{ad} ≤ M_{d}.

In particular, by 4.2 (c) and (d), we have b, c, d, e is a complete set of representatives for the orbits of M_{ad} on Λ(a). Now, Z_{d} is an involution in M_{e} ∼ 2.Alt(11) and so we know that the non-trivial element of Z_{e} has cycle shape (2, 2, 2, 2) when it acts on Ω(e). Thus Z_{a} corresponds to the unique 3-cycle in M_{e}/Z_{e} centralized by Z_{d}. Hence C_{M_{e}}(Z_{d}) normalizes Z_{a}, and therefore

C_{M_{e}}(Z_{d}) = C_{M_{ae}}(Z_{d}) = M_{ade} = M_{de} ∼ 2.(Sym(3) ∧^{|}2^{3}).Sym(4).

Observe that M_{bd} ∼ 3^{2}.8.Sym(5) acts transitively on Λ_{1}(b) with representative a and on Λ_{3}(b)(= Γ_{3}(b)) with representative c and that Λ_{4}(b) = ∅. Next we consider the action of M_{cd} on Λ(c). Now M_{cd} = C_{M_{c}}(Z_{d}) ∼ (3^{3}.(2×2.Sym(4))) and O_{3}(M_{cd}) fixes every element of Γ_{i}(c) for i = 1, 2. Thus, for i = 1, 2, M_{cd} operates as 2 × 2.Sym(4) on Λ_{i}(c). Since Z_{d} acts with exactly 3 fixed points and 4 pairs of interchanged elements on the 11 elements in Γ_{1}(c), we infer that M_{cd} is transitive on Λ_{1}(c) (which has 3 elements and representative a) and has two orbits on Λ_{2}(c) one of length 3 and one of length 4. Furthermore, M_{cd} has two orbits on Λ_{4}(c) with representatives e and d. Let b and f be the respective representatives for these last two orbits. Notice that Z_{d} interchanges the two elements of Γ_{1}(f) and so Λ_{1}(f) = ∅. Also, as Z_{d} centralizes Z_{b} and, by 7.2 (b), Z_{b} ≰ Q_{c} ∩ Q_{f}, Z_{b}Q_{f}/Q_{f} = Q_{c}Q_{f}/Q_{f} is centralized by Z_{d}. It follows that C_{M_{f}} (Z_{d}) normalizes Q_{c}Q_{f} and so C_{M_{f}} (Z_{d}) ≤ M_{fc}. Therefore,

C_{M_{f}} (Z_{d}) ≤ C_{M_{c}}(Z_{d}) ≤ M_{d}

and so

M_{df} = M_{dcf} ∼ 3^{3}.(C_{2} × C_{2} × Sym(3)).

We will now investigate Λ(e). Choose notation such that on Ω(e), a corresponds to [1, 2, 3] and Z_{d} acts as ⟨(4, 5)(6, 7)(8, 9)(10, 11)⟩. Then we can identify the image of C_{M_{e}}(Z_{d}) in M_{e}/Z_{e} acting on Ω(e) with

⟨(1, 2, 3), (1, 2)(5, 7)(4, 6)(8, 9), (6, 8, 10)(7, 9, 11), (4, 6)(5, 7)(8, 10)(9, 11)⟩.

It is now straightforward to calculate that M_{de}(= C_{M_{e}}(Z_{d})) has four orbits on Λ_{3}(e) with representatives

\begin{matrix}c̃ & ↔ & [1, 2, 3 | 4, 6, 8 | 5, 7, 9] \\ g & ↔ & [1, 10, 11 | 4, 6, 8 | 5, 7, 9] \\ h & ↔ & [1, 10, 11 | 2, 4, 5 | 3, 6, 7] \\ i & ↔ & [1, 10, 11 | 2, 4, 5 | 3, 8, 9] \end{matrix}

where we remark that the last two orbits are fused under N_{M_{e}}(⟨Z_{d}, Z_{e}⟩). We also calculate that M_{de} has one orbit on Λ_{1}(e) \ {a} with representative

j ↔ [1, 10, 11]

(note that in this case Z_{d} normalizes, but does not centralize, Z_{j}) and two orbits on Λ_{2}(e) with representatives

\begin{matrix}f̃ & ↔ & [4, 6, 8 | 5, 7, 9] \\ k & ↔ & [1, 10, 11 | 2, 4, 5] \end{matrix} .

Recall that incidence of x, y ∈ Γ(e) is recognized in Ω(e) by one the objects being a refinement of the other. Since c̃ is adjacent to a we may assume that c = c̃ and noting that f̃ is adjacent to c̃ = c we may also assume that f = f̃. Because Z_{d} inverts Z_{j}, 4.3 gives us that

C_{M_{j}}(Z_{d}) ≅ C_{2} × Mat_{11}.

By determining the stabilizer of j (which represents an orbit of length 12 under M_{de}) in M_{de} we calculate

M_{dej} ≅ C_{2} × Q_{8} : Sym(3) ≅ C_{2} × GL_{2}(3).

As Z_{d} inverts Z_{j} but not Q_{g}, Z_{d}Q_{g}/Q_{g} is a diagonal involution in M_{g}/Q_{g} ∼ Mat_{11} × C_{2} and so we deduce that

C_{M_{g}}(Z_{d}) ∼ 3^{2}.(C_{2} × GL_{2}(3)).

Since we already know C_{M_{f}} (Z_{d}) = M_{df} ≤ M_{d} and g ∈ Λ_{3}(f) is in an orbit of length 3, we conclude that

M_{dfg} = C_{M_{g}}(Z_{d}) ∩ M_{f} ∼ C_{2} × 3^{2}.(C_{2} × Sym(3)).

Furthermore, as (using M_{3} is 3-transitive on its neighbours of type 1) M_{fgj} ∼ 3^{5}(SD_{16} × C_{2}),

M_{dfgj} ∼ C_{2} × 3^{2}.2^{2}.

In particular, we note M_{dfgj} ≰ M_{dej} and so, as M_{dej} is a maximal subgroup of C_{M_{j}}(Z_{d}),

C_{M_{j}}(Z_{d}) = ⟨M_{dej}, M_{dfgj}⟩ = M_{dj} ∼ C_{2} × Mat_{11}.

Since Z_{d} inverts Z_{k}, 6.5 (a) and (b) indicate that the non-trivial element z of Z_{d} is a square in M_{k}/Q_{k}. It follows that z acts as an Alt(5)-type involution on Γ_{3}(k). Hence, |Λ_{3}(k)| = 2 and so Λ_{3}(k) = {h, i} (each adjacent to e). Now 2.5 (d) indicates that Z_{d} inverts exactly one of Q_{h} and Q_{i}. We choose notation so that Z_{d} inverts Q_{i}. Thus C_{M_{i}}(Z_{d}) ≅ C_{2} × Mat_{11}, M_{dei} = C_{M_{i}}(Z_{d}) ∩ C_{M_{i}}(Z_{e}) ≅ C_{2} × GL_{2}(3), and M_{dji} = C_{M_{j}}(Z_{d}) ∩ M_{i} ≅ C_{2} × Mat_{10} by 4.3 (a). Hence,

C_{M_{i}}(Z_{d}) = ⟨M_{dei}, M_{dji}⟩ = M_{di} ∼ C_{2} × Mat_{11}.

Furthermore, by 4.3 (a), M_{djg} = C_{M_{j}}(Z_{d}) ∩ M_{g} = C_{M_{jg}}(Z_{d}) ∼ C_{2} × 3^{2}.SD_{16} and so

C_{M_{g}}(Z_{d}) = ⟨M_{dfg}, M_{djg}⟩ = M_{dg} ∼ C_{2} × 3^{2}.GL_{2}(3).

Also, because C_{M_{k}}(Z_{d}) ≤ M_{i},

C_{M_{k}}(Z_{d}) = M_{dik} = M_{dk} ∼ C_{2} × 3^{2}.SD_{16}.

Finally, we claim that h^{r} = g for some r ∈ M_{d}. Indeed, by 4.3, such an r can be found in M_{dj}.

Put F = {a, b, c, d, e, f, g, i, j, k}. We have demonstrated

(7.7.1) If x ∈ F, then C_{M_{x}}(Z_{d}) = M_{dx}.

Also note that for x ≠ y ∈ F having the same type, x, y are distinguished by the action of Z_{d} on Z_{x} and Z_{y}. Indeed, Z_{d} centralizes Z_{a} but inverts Z_{j}; Z_{d} centralizes Z_{b}, inverts Z_{k} and neither centralizes nor inverts Z_{f}; Z_{d} centralizes a 3^{3} in Z_{c}, a 3^{2} in Z_{g} and inverts Z_{i}; and Z_{d} equals Z_{d} but not Z_{e}. Thus F is a set of representatives of distinct orbits of M_{d} on Γ^{d}.

We next show that if x ∈ F and y ∈ Λ(x), then y is M_{d}-conjugate to an element of F. We do this by using (7.7.1) to show that, for x ∈ F, there is a set Ξ of representatives for the orbits of M_{xd} on Λ(x) with Ξ ⊂ F ∪ {h, k^{r}} where r ∈ M_{dj} is the element that achieves h^{r} = g. We find

Ξ = \{ \begin{matrix}\{b, c, d, e\} & when x = a \\ \{a, c, d\} & when x = b \\ \{a, b, d, e, f\} & when x = c \\ \{a, b, c\} & when x = d \\ \{a, c, f, g, h, i, j, k\} & when x = e \\ \{c, e, g\} & when x = f \\ \{e, f, j, k^{r}\} & when x = g \\ \{e, j, k\} & when x = i \\ \{e, g, k, i\} & when x = j \\ \{e, h, i, j\} & when x = k \end{matrix} .

Indeed, for x ∈ {a, b, c, d, e}, the claim is transparent from our previous discussion. Suppose that x = f. Then M_{df} ∼ 3^{2}.(C_{2} × C_{2} × Sym(3)) which acts transitively on Γ_{1}(f), so that Λ_{1}(f) = ∅. Now on Λ_{3}(f), M_{df} has orbits of length 1, 3 and 6, the orbits of lengths 1 and 3 corresponding to the fixed points of Z_{d} with representatives c and g, respectively. Finally, we also have that M_{df} acts transitively on Λ_{4}(f) this time with representative e. Next assume that x = g, then M_{dg} ∼ C_{2} × 3^{2}.GL_{2}(3). Since GL_{2}(3) has one class of non-central involutions, M_{dg} operates transitively on Λ_{4}(g) with representative e. Also Λ_{1}(g) has three members and M_{dg} acts transitively on these, the representative is j. Finally Λ_{2}(g) has two orbits under M_{dg}; one has representative f, the other k^{r}. Now assume that x = i. Then M_{di} ∼ C_{2} × Mat_{11}, a complement to O_{3}(M_{i}) with Z_{d} inverting O_{3}(M_{i}). It follows that Λ_{1}(i) = Γ_{1}(i) and Λ_{2}(i) = Γ_{2}(i) both of which admit M_{di} transitively with representatives k and j. Furthermore, as Mat_{11} has a unique class of involutions, M_{di} is also transitive on Λ_{4}(i). For x = j we refer the reader to 4.3 to obtain representatives {e, j, k, i}. Finally we assume that x = k. Then M_{dk} ∼ C_{2} × 3^{2}.SD_{16}. Since the elements of Λ_{4}(k) correspond to involutions which invert O_{3}(M_{dk}), we have a single class here with representative e. The elements of Λ_{3}(k) are h and k as discussed earlier and, as Z_{d} inverts Z_{k}, Λ_{1}(k) has order 2 and representative j.

Hence, we have shown that the hypothesis of 2.1 holds. Therefore, Γ^{d} = F^{M_{d}} and M_{d} = Stab_{M}(Γ_{D}) ∩ N_{M}(Z_{d}), as claimed.

Notice that one consequence of 7.7 is that for d ∈ Γ_{4}, M_{d} has exactly two orbits on Γ_{1} ∩ Γ^{d}. One orbit consists of those elements a ∈ Γ^{d} of type 1 with [Z_{d}, Z_{a}] = 1, the other those b ∈ Γ^{d} of type 1 with [Z_{d}, Z_{b}] = Z_{b} and M_{db} ∼ C_{2} × Mat_{11}. We now incorporate this fact into our notation. Let d ∈ Γ_{4} and x ∈ Γ_{1}^{d} we write x † d if Z_{d} inverts Z_{x}. Thus we have d † x if and only if x ∈ Γ^{d} \ Γ_{1}(d). Another consequence of the proof of 7.7 is

Corollary 7.8 Suppose that d ∈ Γ_{4} and f ∈ Γ_{2} ∩ Γ^{d}. If Z_{d} neither inverts nor centralizes Z_{f}. Then there exists a unique c ∈ Γ_{3}(d) ∩ Γ_{3}(f).

Proof: This follows as M_{df} = M_{dcf} for any c ∈ Γ_{3}(d) ∩ Γ_{3}(f).

Lemma 7.9 Let d ∈ Γ_{4} and x, y ∈ Γ_{1}^{d}. Then d(x, y) ≤ 2.

Proof: If x, y ∈ Γ_{1}(d), then, since Z_{x} and Z_{y} are generated by 3-cycles in M_{d}/Z_{d}, there is a u ∈ Γ_{1}(d) with x ⋊ ⋉ u ⋊ ⋉ y. Thus d(x, y) ≤ 2. Suppose that x ∉ Γ_{1}(d) and y ∈ Γ_{1}(d). Then M_{xd} ≅ C_{2} × Mat_{11}, acts 3-transitively on Ω(d) and, in particular, M_{xd} acts transitively on Γ_{1}(d). Hence we only need to show that d(x, y) ≤ 2 for some x, y fulfilling the assumptions of the current case. Pick e ∈ Γ_{4}(y) with e ⋆ d. We may suppose that the action of y and d on Ω(e) are described as follows: y ↔ [1, 2, 3] and d ↔ [4, 5][6, 7][8, 9][10, 11]. Select x̃ ∈ Γ_{1}(e) with x̃ ↔ [1, 4, 5]. Then x̃ ∈ Γ_{1}^{d} \ Γ_{1}(d) and d(x̃, y) = 2 and we are done in this case.

Suppose finally that neither x nor y are adjacent to d. Then as there are only two orbits of M_{d} on Γ_{1}^{d}, y ∈ {x^{M_{d}}}. Recall that permutation rank for M_{d} on {x^{M_{d}}} is that of Alt(11) on Mat_{11} is five (see [ATLAS]) and so there are exactly 5 distinct classes of triples (x, y, d) with the property that neither x nor y is adjacent to d. Let a, b ∈ Γ_{1} with a ⋊ ⋉ b. We now work in M_{b}. For 0 ≤ i ≤ 4, let Y_{i}(a) be the five distinct orbits of M_{ab} on the neighbours of b in (Γ_{1}, ⋊ ⋉)(see 4.4). Then for 0 ≤ i ≤ 4, select c_{i} ∈ Y_{i}(a). By 4.4, there exists d_{i} ∈ Γ_{4} with d_{i} † b and Z_{d_{i}} ≤ M_{abc_{i}}. Then a † d_{i} † c_{i} and we have found representatives for all five classes of triples (x, y, d) with x † d † y and in each case d(x, y) = 2.

Lemma 7.10 Let b ∈ X_{4}(a). Then M_{ab} ∼ 5^{1+2}Sym(3) and |X_{1}(a) ∩ X_{1}(b)| = 25. In particular, O_{5}(M_{ab}) acts transitively on X_{1}(a) ∩ X_{1}(b) and Z(O_{5}(M_{ab})) fixes X_{1}(a) ∩ X_{1}(c) pointwise.

Proof: Let c ∈ X_{1}(a) ∩ X_{1}(b) and put A = ⟨Z_{a}, Z_{b}⟩. Then, by 7.3 (d), A ∼ 5^{1+2}.3, C_{5} ≅ Z(A) ≤ M_{abc} ≅ D_{30} and M_{ab} ≤ N_{M_{a}}(Z(A)) ∼ 3.5^{1+2}.3.SD_{16}. Pick e ∈ Γ_{4} with e † a and Z_{e} ≤ M_{abc}. Then e † b, M_{abce} = Z_{e} and M_{ae} ∼ C_{2} × Mat_{11} ∼ M_{be}. In particular, by orders, M_{abe} = M_{ae} ∩ M_{be} ≠ Z_{e} = M_{abce} and thus M_{ab} ≠ M_{abc}. Since M_{abc} = N_{M_{ab}}(Z_{c}), the structure of M_{abc} and N_{M_{a}}(Z(A)) imply that M_{ab} ∼ 5^{1+2}.Sym(3), as required.

Suppose that b ∈ X_{4}(a), put A = ⟨Z_{a}, Z_{b}⟩ and define F = Z(A). Let c ∈ X_{1}(a) ∩ X_{1}(b) and let f = Γ_{2}(c) ∩ Γ_{2}(a). Notice that M_{abc} inverts both Z_{a} and Z_{b}. Pick d ∈ Γ_{4}(a) so that Z_{d} centralizes both Z_{f} and F (so we can choose M_{a} = M_{1}, M_{d} = H and M_{f} = M_{2}). Then we have

N_{M_{d}}(F) ∼ 2.(5.4 ∧^{|}Sym(6))

N_{M_{a}}(F) ∼ 3.5^{1+2}.3.SD_{16}

and

N_{M_{ad}}(F) = 6.(5 × 3).4.

Define

R = ⟨N_{M_{d}}(F), N_{M_{a}}(F)⟩.

Also, just as in 7.10, fix e ∈ Γ_{4} with a † e, b † e, d ⋆ e and so that Z_{e} inverts Z_{f}. (So Z_{e} is an involution in M_{abc} ∼ D_{30}). Notice that Z_{e} inverts F. Since Z_{d} ≤ C_{M_{a}}(F) and acts fixed-point-freely on O_{5}(C_{M_{a}}(F))/F. Hence Z_{d} fixes a unique complement to O_{5}(C_{M_{a}}(F))/F in N_{M_{a}}(F)/F and so we infer that Z_{d} centralizes a unique group of order 3^{2} in C_{M_{a}}(F). In particular, c is the unique member of X_{1}(a) ∩ X_{1}(b) fixed by Z_{d}.

We exploit this notation in the following lemma where we uncover the structure of the subgroup R of M.

Lemma 7.11 We have R ∼ 5^{1+4}.4.Sym(6) and for x ∈ {a, f, d, e}, N_{M_{x}}(F) ≤ R.

Proof: Set A = X_{1}(c)^{F}, A_{0} = O_{5}(⟨Z_{x} | x ∈ A⟩), B = X_{1}(a)^{F}, B_{0} = O_{5}(⟨Z_{x} | x ∈ B⟩) and Y = {a, f, d, e}.

From the structure of N_{M_{a}}(F) ∼ 3.5^{1+2}.3.SD_{16}, we have |B| = 25. On the other hand, we have both X_{1}(a) ∩ X_{1}(b) ⊆ B and |X_{a}(a) ∩ X_{1}(b)| = 25 from 7.10. Thus B = X_{1}(a) ∩ X_{1}(b) and B corresponds to the set of Sylow 3-subgroups of M_{ab}.

In particular, we have B_{0} = O_{5}(M_{ab}) is extraspecial of order 5^{3} with centre F and for x ≠ y ∈ B we have x ∈ X_{4}(y).

Similarly, as Z_{a} and Z_{b} are both subgroups of M_{c} and centralize F, A ≤ M_{c} and A corresponds to the set of Sylow 3-subgroups of A. Therefore, A_{0} = O_{5}(A) is also extraspecial of order 5^{3} with centre F. Because A_{0} centralizes B_{0} and A_{0} ∩ B_{0} = Z(A_{0}) = Z(B_{0}) = F, A_{0}B_{0} is extraspecial of order 5^{5}. Notice also that, by 7.10, B_{0} operates transitively on B. Since A corresponds to the set of Sylow 3-subgroups of A and B corresponds to the set of Sylow 3-subgroups of M_{ab}, we have A ⊆ X_{1}(x) for all x ∈ B.

For each x ⊆ Y define R_{x} = M_{x} ∩ N_{M}(F). Then, by definition, R = ⟨R_{d}, R_{a}⟩. Since F ∈ Syl_{5}(M_{ac}) and M_{ac} has index 2 in M_{f} the Frattini argument implies M_{f} = R_{f}M_{ac} and so R_{f} acts non-trivially on Γ_{1}(f) = {a, c}. Therefore, the definitions of A and B imply that R_{f} interchanges A_{0} and B_{0} and so normalizes A_{0}B_{0}. Put R_{0} = A_{0}B_{0}R_{f}. Then, from 6.5 (i) and (ii) we get

(7.11.1) R_{f} ∼ 3^{2}.5.4.D_{8} and R_{0} ∼ 5^{1+4}.4.3^{2}.D_{8}.

From the structure of 2.Alt(11), we get

(7.11.2) R_{d} ∼ 2.(5.4 ∧^{|}Sym(6)).

Our goal is to show that R_{d} normalizes A_{0}B_{0} and R = A_{0}B_{0}R_{d}. Using 7.7 and 7.11.1, we get

R_{fd} = M_{fd} ∩ N_{M}(F) = C_{R_{f}} (Z_{d}) = R_{f} ∼ 3^{2}.4.5.D_{8}.

In particular, R_{f} is a maximal subgroup of R_{d}.

We now investigate R_{e}, so set R_{0e} = C_{R_{0}}(Z_{e}). Then, as Z_{e} normalizes both A_{0} and B_{0} we have

R_{0e} = C_{A_{0}}(Z_{e})C_{B_{0}}(Z_{e})C_{R_{f}} (Z_{e}).

Now, recalling that Z_{e} inverts F, we have C_{A_{0}}(Z_{e}) ∼ C_{5} ∼ C_{B_{0}}(Z_{e}) and, as A_{0} and B_{0} commute, C_{A_{0}}(Z_{e})C_{B_{0}}(Z_{e}) is an elementary abelian subgroup of order 25 which intersects F trivially. On the other hand, as R_{f} ≤ R_{d} and Z_{e} ≤ R_{f}, C_{R_{f}} (Z_{e}) can be calculated in R_{d}. We find that C_{R_{f}} (Z_{e}) ∼ 2^{4}. Therefore,

R_{0e} ∼ 5^{2}.2^{4}.

In particular, R_{0e} is a subgroup of the normalizer of a Sylow 5-subgroup of M_{e}; that is a subgroup of index 2 in a group of shape 2.(5 : 4 ∧^{|}5 : 4).2. Furthermore, in R_{d} we compute

R_{de}=C_{R_{d}}(Z_{e}) ∼ 4.2^{3} and C_{R_{de}}(F) = C_{R_{def}} ≅ Q_{8}.

We conclude that C_{R_{de}}(F) is a Sylow 2-subgroup of C_{R_{e0}}(F). Notice also that as N_{M}(F)/C_{M}(F) has order at most 4, we have R_{e} = R_{de}C_{R_{e}}(F) and R_{e}/C_{R_{e}}(F) = 4. Furthermore, as Z_{e} inverts F and so Z_{e} ≰ C_{M}(F)∩R_{e}. Hence C_{R_{e}}(F) splits over Z_{e}. Recalling that C_{R_{0e}}(F) ∼ 5^{2} : Q_{8}, it follows from the structure of 2.Alt(11) that the overgroups in M_{e} of R_{de}R_{0e} are M_{e}, 2.Alt(10), or 2.(Alt(5) × Alt(5)) : 4 (using R_{e}/C_{R_{e}}(F) has order 4). Since C_{R_{e}}(F) splits over F, we conclude that R_{e} is the normalizer of Sylow 5-subgroup in M_{e} and hence

(7.11.3) R_{e} ∼ 5^{2}.2^{5}, R_{0e} is a normal subgroup of R_{e} of index 2 and R_{e} = C_{R_{de}}(F)R0e.

Let t ∈ C_{R_{de}}(F) \ R_{0}. Then t^{2} ∈ R_{0e}. In R_{d} we calculate that R_{fd} ∩ R_{fd}^{t} ∼ 5.4.D_{8} and, by 7.11.3, R_{0e} = R_{0e}^{t}. In R_{0} we see that ⟨R_{fd} ∩ R_{fd}^{t}, R_{0e}⟩ ∼ 5^{1+4}.4.D_{8} and thus A_{0}B_{0} ≤ R_{0} ∩ R_{0}^{t}. It follows that A_{0}B_{0} = O_{5}(R_{0} ∩ R_{0}^{t}) and so t normalizes A_{0}B_{0}. As R_{d} = ⟨R_{df}, t⟩, R_{d} normalizes A_{0}B_{0}. Now for x ∈ Y we have R_{x} = (R_{x} ∩ R_{d})(R_{x} ∩ A_{0}B_{0}). Hence R_{x} ≤ A_{0}B_{0}R_{d} and so, by 7.11.2, R = A_{0}B_{0}R_{d} ∼ 5^{1+4}.4.Sym(6) as claimed.

We continue with the notation of 7.11 in the next lemma.

Lemma 7.12 Let Γ^{R} be the connected component of Γ^{F} containing a. Then R = Stab_{M}(Γ^{R}).

Proof: Let X = {a, d, f}. Then by 7.11 N_{M_{x}}(F) ≤ R for all x ∈ X. We claim that X is a complete set of representatives for the orbits of R on Γ^{R}. But this is clear, as for all x ∈ X, X \{x} is a complete set of representatives for the orbits of R∩M_{x} on Γ(x)^{F}. Indeed, if x = a, then, as we saw in the first few lines of 7.11, X_{1}(a) ∩ X_{1}(b) admits R_{a} transitively with representative c. Since each element of X_{1}(a)∩X_{1}(b) determines a unique element of Γ_{2}(a)^{F}, R_{a} is transitive on Γ_{2}(a)^{F} with representative f. Now Γ_{3}(a)^{F} is empty and Γ_{4}(a)^{F} corresponds to the conjugates of Z_{d} in R_{f} and so we have a single orbit with representative d. As seen just before (7.11.1), R_{f} is transitive on Γ_{1}(f)^{F} with representative a. Also Γ_{3}(f)^{F} = ∅ and Γ_{4}(f)^{F} = {d}. Finally, since the neighbours of type 1, 2 and 3 of d correspond to a subgroups or M_{d}/Z_{d} generated by a 3-cycle, two commuting 3-cycles and 3 commuting 3-cycles respectively, we see that Γ_{3}(d)^{F} = ∅ and Γ_{2}(d)^{F} and Γ_{1}(d)^{F} admit R_{d} transitively with representatives a and f respectively.

Hence we may apply 2.1 to obtain R = Stab(Γ^{R}) ∩ N_{M}(F). Note that F is the elementwise stabilizer of Γ^{R} and so Stab_{M}(Γ^{R}) ≤ N_{M}(F). Therefore, R and Stab_{M}(Γ^{R}) have the same orbits and stabilizers on Γ^{R}. Thus, by a Frattini argument, R = Stab_{M}(Γ^{R}).

Lemma 7.13 Let x ≠ y ∈ Γ_{1}^{R} and put B = ⟨Z_{x}, Z_{y}⟩. Then either Z_{x}O_{5}(R) = Z_{y}O_{5}(R) and y ∈ X_{4}(x) or Z_{x}O_{5}(R) ≠ Z_{y}O_{5}(R) and B ≅ BO_{5}(R)/O_{5}(R). In particular, d(x, y) ≤ 2.

Proof: This follows from the structure of R as follows. Set R̄ = R/F and let x ∈ Γ_{1}^{R}. Then Z_{x} maps to a 3-cycle in R/O_{5,2}(R) and NR̄(Z̄_{x}) ∼ 3 × 5^{2}.3.SD_{16}. Thus in R̄ there are 500 conjugates of Z̄_{x}. Let C be a complement to O_{5}(R̄) which contains Z̄_{x}. Since C_{O_{5}(R̄)}(Z̄_{x}) ∼ 5^{2}, Z̄_{x} is contained in exactly 25 complements. Let C_{1} be a further complement containing Z̄_{x} and let t and t_{1} be the respective central involutions of C and C_{1}. Suppose that y ∈ Γ_{1}^{R}, x ≠ y and Z̄_{y} ≤ C_{1} ∩ C. Put B = ⟨Z̄_{x}, Z̄_{y}⟩. Since B ≤ C, Z_{x}O_{5}(R) ≠ Z_{y}O_{5}(R). Thus BO_{5}(R)/O_{5}(R) ≅ C_{3} ×C_{3}, 2.Alt(4) or 2.Alt(5). In any case B acts fixedpoint freely on O_{5}(R̄). On the other hand B centralizes ⟨t, t_{1}⟩ ∩ O_{5}(R̄) ∼ C_{5}, a contradiction. Therefore, any two complements that contain Z̄_{x} intersect in a subgroup which normalizes Z̄_{x}. So the complements which contain Z̄_{x} account for 19.25 + 1 conjugates of Z̄_{x}. Moreover, in \overline{Z_{x}O_{5}(R)} there are 25 conjugates of Z̄_{x}. Thus we have accounted for all the conjugates of Z̄_{x} and any pair lie in \overline{Z_{x}O_{5}(R)} and generate a group of shape 5^{2}.3 or live in a complement and thus centralize an involution. In the latter case the distance between the corresponding vertices is at most 2 by 7.9 while in the former case there is a u ∈ Γ_{1}^{R} for which Z̄_{u} centralizes ⟨Z̄_{x}, Z̄_{y}⟩ and so y ∈ X_{4}(x) and this completes the proof.

Lemma 7.14 Suppose that e, f ∈ Γ_{4} with e ⋆ f, d ∈ Γ_{1}(f) and for 1 ≠ t ∈ Z_{e} assume that d^{t} ∈ X_{2}(d). Let T = ⟨h ∈ M_{def} | h^{2} = 1⟩. Then

(a) T ≅ D_{24}.

(b) |M_{def} : T | = 2;

(c) T has three orbits O_{1}, O_{2} and O_{3} on Ω(b) of lengths 3, 4 and 4 respectively. T induces Sym(3) on O_{1}, and D_{8} on O_{2} and O_{3}. The orbits O_{2} and O_{3} are not isomorphic as T -sets and O_{2} ∪ O_{3} is an orbit for M_{def}.

Proof: We first compute the structure of T and M_{def} inside M_{f}. Because d^{t} ∈ X_{2}(d) we may choose notation such that in Ω(e),

d ↔ [1, 2, 3] and e ↔ [1, 2][3, 4][5, 6][7, 8].

Then

M_{def}/Z_{f} ↔ ⟨(9, 10, 11), (1, 2)(5, 7)(6, 8)(9, 10), (5, 6)(7, 8), (1, 2)(5, 6)⟩

and so

T/Z_{f} ↔ ⟨(9, 10, 11), (1, 2)(5, 7)(6, 8)(9, 10), (56)(78)⟩.

In particular T ≅ D_{24}, |M_{def}/T | = 2 and M_{def} fuses the two classes of fours groups in T. Let g ∈ Γ_{1}(e) ∩ Γ_{1}(f). Then Z_{g} = O_{3}(T). Note also that Z_{e} ≰ M_{def} and so M_{def} operates faithfully on Ω(e). The statements about the orbits of T on Ω(e) are now readily verified.

Proposition 7.15 We have d(x, y) ≤ 2 for all x, y ∈ Γ_{1}.

Proof: Suppose that the statement is false. Then there is a pair a, d ∈ Γ_{1} with d(a, d) = 3. Let a, b, c, d be a path in (Γ_{1}, ⋊ ⋉) connecting a and d. Then c ∈ X_{i}(a) and b ∈ X_{j}(d) for some i, j ∈ {2, 3, 4}. We begin with

(7.15.1) Suppose that c ∈ X_{2}(a). Let e ∈ Γ_{4}(a) ∩ Γ_{4}(c) and let 1 ≠ t ∈ Z_{e}, then d^{t} ∉ X_{1}(d).

If d^{t} ∈ X_{1}(d), then, by 7.3 (a), there exists a unique f in Γ_{3}(c) ∩ Γ_{3}(d) ∩ Γ_{3}(d^{t}). By the uniqueness of f, e ∈ Γ_{4}(f) and so |Γ_{1}(f) ∩ Γ_{1}(e)| = 3. As c ∈ X_{2}(a) we see by calculating in Ω(e), that a ⋊ ⋉ g for at least one g ∈ Γ_{1}(f)∩Γ_{1}(e). But g ⋊ ⋉ d and so d(a, d) ≤ 2, which is a contradiction.

(7.15.2) Suppose that f ∈ Γ_{4} and a ∈ Γ^{f}. Then Γ_{2}(d) ∩ Γ^{f} = ∅.

Suppose that there exists g ∈ Γ_{2}(d) and f ∈ Γ_{4} with a, g ∈ Γ^{f}. If Z_{f} centralizes or inverts Z_{g}, then d ∈ Γ^{f} and d(a, d) ≤ 2 by 7.9. So assume that Z_{f} neither inverts nor centralizes Z_{g}. Then, by 7.8, there exists a unique h ∈ Γ_{3}(f) ∩ Γ_{3}(g). So |Γ_{1}(f) ∩ Γ_{1}(h)| = 3. Let i ∈ Γ_{1}(f) ∩ Γ_{1}(h). Then i ⋊ ⋉ d. Since a, i ∈ Γ^{f}, d(a, i) ≤ 2, by 7.9. Hence d(a, i) = 2. As Z_{f} centralizes Z_{i}, the the three possible structures for ⟨Z_{a}, Z_{i}⟩ indicate that a ∈ Γ_{1}(f) and a ∈ X_{2}(i) ∪ X_{3}(i). Let 1 ≠ t′ ∈ Z_{f}. Then d^{t′} ∈ X_{1}(d). Hence 7.15.1 implies that a ∈ X_{3}(i) for each i ∈ Γ_{1}(f) ∩ Γ_{1}(h). But then using the shape of these elements on Ω(f) we see that there is an involution t_{1} in M_{fh} inverting Q_{h} and centralizing Z_{a}. Let q ∈ Γ_{4} with Z_{q} = ⟨t_{1}⟩. Then a, d ∈ Γ^{q} and so 7.9 implies that d(a, d) ≤ 2, a contradiction. Thus 7.15.2 holds.

(7.15.3) Up to changing the roles of a and d we may assume that c ∉ X_{4}(a).

Suppose that c ∈ X_{4}(a) and set F = Z(⟨Z_{a}, Z_{c}⟩). Then, by 7.10, X_{1}(a) ∩ X_{1}(c) has order 25 and is fixed pointwise by F. Now considering F as a subgroup of M_{c} and applying 4.10 we see that X_{1}(a) ∩ X_{1}(c) ⊄ X_{4}(d). Hence there exists b_{1} ∈ X_{1}(a) ∩ X_{1}(c) such that b_{1} ∉ X_{4}(d) which is precisely what we required.

By 7.15.3 we may assume that c ∈ X_{2}(a) ∪ X_{3}(a). Let e ∈ Γ_{4}(a) ∩ Γ_{4}(c) and let 1 ≠ t ∈ Z_{e}. We focus our attention on the path d, c, d^{t} in (Γ, ⋊ ⋉).

(7.15.4) d ≠ d^{t}.

If d = d^{t}, then a, d ∈ Γ^{e} and so d(a, d) ≤ 2 by 7.9. Thus d ≠ d^{t}.

(7.15.5) d^{t} ∉ X_{1}(d).

Suppose that d^{t} ∈ X_{1}(d). Then there exists g ∈ Γ_{2}(d)∩Γ_{2}(d^{t}) with g, a ∈ Γ^{e}, this is against 7.15.2. Thus d^{t} ∉ X_{1}(d).

(7.15.6) d^{t} ∉ X_{3}(d).

Let f ∈ Γ_{4}(c) ∩ Γ_{4}(d) ∩ Γ_{4}(d^{t}). Then f ∗ e and calculating in Ω(f) shows that it is impossible for d^{t} ∈ X_{3}(d).

(7.15.7) d^{t} ∉ X_{2}(d).

Aiming for another contradiction we suppose that d^{t} ∈ X_{2}(d). Let f be in Γ_{4}(c) ∩ Γ_{4}(d) ∩ Γ_{4}(d^{t}). Then f ⋆ e. Using 7.14 we compute in M_{e} that M_{edf} contains an involution s with [Z_{a}, Z_{a}^{s}] = 1. But this contradicts 7.15.2 applied with the roles of a and d interchanged.

By 7.15.4 through 7.15.7, we have d^{t} ∈ X_{4}(d). Set T = M_{cdd^{t}}. Then, by 7.3 (d), T ≅ D_{30}. Put F = O_{5}(T). Because t ∈ M′_{c} and normalizes F, t centralizes both F and Z_{c} and hence also T. Therefore T ≤ M_{e}. Since a ∈ X_{2}(c) ∪ X_{3}(c) we see in M_{e} acting on Ω(e), that either F fixes a or there exist involutions s in T with [Z_{a}, Z_{a}^{s}] = 1. We then obtain contradictions via 7.13 and 7.15.2 (applied with the roles of a and d interchanged) respectively.

We now prove the main theorem of this section.

Theorem 7.16 Suppose that M is a faithful completion of a Ly-amalgam of groups (M_{1}, M_{2}, M_{3}). Then

(a) M_{1} has five orbits on M/M_{1}. Moreover, the orbit stabilisers have shape 3.Aut(McL), 3^{2+4}.4.Sym(5), 2.Sym(7), 4.Sym(6) and 5^{1+2}.Sym(3).

(b) M has order 2^{8} · 3^{7} · 5^{6} · 7 · 11 · 31 · 37 · 67.

(c) M has a unique class of involutions and C_{M}(t) ≅ 2.Alt(11) for any involution t in M.

(d) For i = 1, 2, 3 , M_{i} is a maximal 3-local subgroup of M.

(e) M has a subgroup F of order five with N_{M}(F) ∼ 5^{1+4}.4.Sym(6).

(f) M is a group of Ly-type.

Proof: Part (a) follows from 7.3 (a), (b) and (c), 7.10 and 7.15 and part (b) is a direct consequence of (a). Let H be the subgroup of M found in 7.4. Then H ≅ 2.Alt(11) and so, by (b), 1 ≠ t ∈ Z(H) is a 2-central involution. Since H has a unique class of involution different from t and these involutions are conjugate to t, M contains a unique class of involutions. If g ∈ Γ_{1}^{Z(H)}, then 7.15 implies that d(M_{1}, g) ≤ 2 and so Γ_{1}^{Z(H)} is connected. Thus N_{M}(Γ_{1}^{Z(H)}) = C_{M}(t) = H by 7.7. Hence (c) holds. That M_{1} is a maximal subgroup follows immediately from parts (a) and (b). As for M_{i}, i = 2, 3, M_{i} acts transitively on the M-conjugates of Z_{1} in Q_{i} and so by a Frattini argument N_{M}(Q_{i}) = M_{i}(N_{M}(Q_{i}) ∩ N_{M}(Z_{1})) = M_{i}N_{M_{1}}(Q_{i}) = M_{i}. Hence M_{2} and M_{3} are also maximal 3-locals. Let F be the cyclic group of order 5 introduced just before 7.11. Then 7.15 shows that Γ^{F} is connected and so (e) follows from 7.12. Finally we come to part (f). If N is a normal subgroup of M and N ∩ M_{i} = 1 for i = 1, 2, 3, then M/N is also a faithful completion of (M_{1}, M_{2}, M_{3}). Since, by (a), |M| = |M/N| we get N = 1. So suppose that N_{i} = M ∩M_{i} ≠ 1 for some i ∈ {1, 2, 3} and fix i. It follows that Z_{1} ≤ N_{i} ≤ N. Thus Z_{2} = ⟨Z_{1}^{M_{2}}⟩ ≤ N and M′_{1} = ⟨Z_{2}^{M_{1}}⟩ ≤ N. Then M_{3} = ⟨(M′_{1} ∩ M_{3})^{M_{3}}⟩ ≤ N, M_{1} = M′_{1}M_{13} ≤ N and M_{2} = M_{12}M_{23} ≤ N. Thus N = M and M is simple. Finally (c) implies that M is of Ly-type.

8 A Ly-amalgam in GL_{111}(5)

In this section we show that there exists a Ly-amalgam of groups, (M_{1}, M_{2}, M_{3}), in GL_{111}(5). This then establishes the existence of a faithful completion of a Ly-amalgam of groups. Our calcuations depend on the details of the structure of such an amalgam which is given in 6.3. Let F be the field with 5 elements. The following notation is used to describe the action of the groups M_{1}, M_{2} and M_{3} on the various modules as we build the 111-dimensional F -space. Let K^{∗} ∈ Syl_{2}(M_{23}) and define

B = M_{123},
S = O_{3}(B) and
K = K^{∗} ∩ B.

Also, for i ∈ {1, 2, 3, 12, 13, 23}, we define

K_{i} = K ∩ O^{2}(M_{i}),
Q_{i} = O_{3}(M_{i}) and
Z_{i} = Z(Q_{i}).

Recall that S ∈ Syl_{3}(B). Finally we let K_{0} be the subgroup of K_{2} of order two, for i = 2, 3, D_{i} be the maximal subgroup of K satisfying [D_{i}, O^{2}(M_{i})] ≤ Q_{i}, and D_{2}^{∗} be the maximal subgroup of K^{∗} with [D_{2}^{∗}, O^{2}(M_{2})] ≤ Q_{2}.

In the following technical lemma we detail the structure of, and relationship between, many of the subgroups that we have just introduced.

Lemma 8.1 (a) B = SK, M_{23} = SK^{∗} and M_{2} = O^{2}(M_{2})K^{∗}.

(b) K^{∗} ≅ C_{2} × SD_{16}.

(c) K ≅ C_{2} × Q_{8}.

(d) K_{1} ≅ Q_{8} and K_{1} = C_{K}(Q_{1}).

(e) D_{3} ≅ C_{2}.

(f) D_{2} ≅ C_{4} and D_{2}^{∗} ≅ C_{8}.

(g) K_{2} ≅ C_{4}, K_{0} ≅ C_{2} and K_{2} = K_{12} = C_{K^{∗}}(Z(Q_{2})) ≤ K_{1}.

(h) D_{2}K_{2} ≅ C_{4} × C_{2} and D_{3} ≤ D_{2}K_{2}.

(i) K_{3} ≅ Q_{8} and K_{2} ∩ K_{3} = K_{0}.

(j) D_{2} ≤ K_{3} and D_{2} = K^{∗′}.

(k) D_{2}^{∗} ≰ M′_{3}.

(l) C_{K^{∗}∩M′_{3}}(S/Q_{2}) ≅ D_{8}.

Proof: Parts (a), (b), (c), (e) and K_{3} ≅ Q_{8} follow from 6.3 and (f) follows from 6.5. Since O^{2}(M_{12}) = O^{2}(M_{2}) = C_{M_{2}}(Z_{2}), by 6.5 (a), and O^{2}(M_{12})/Q_{2} ≅ 2.Alt(5), we have K_{2} = K_{12} = C_{K}(Z_{2}) and so (g) holds. As, by definition, D_{2} centralizes K_{2}, D_{2}K_{2} is the unique normal subgroup of K^{∗} isomorphic to C_{2} × C_{4}. In particular, K^{∗′} = D_{2}K_{2} ∩ K_{3} ≅ C_{4} and so, as D_{2}K_{2} contains only two cyclic groups of order 4, D_{2}K_{2} ∩ K_{3} ∈ {K_{2}, D_{2}}. Since, by 6.5 (b), M_{2} and K^{∗} induce D_{8} on Z(Q_{2}), K^{∗}/K_{2} ≅ D_{8} and thus we infer that K_{2} ≠ K^{∗′}. Hence D_{2} = D_{2}K_{2} ∩ K_{3} = K^{∗′} and K_{2} ∩ K_{3} = K_{0}. So (h), (i) and (j) hold. It only remains to prove (k) and (l). Let X be the unique conjugate of Z_{1} in Q_{3} fixed by K^{∗}. Then, by 2.5 (d), C_{M′_{3}}(X) ∼ 3^{5}.Alt(6), Since Q_{3} = X(Q_{3} ∩ Q_{2}), we have S = XQ_{2}. Therefore, C_{K^{∗}∩M′_{3}}(S/Q_{2}) is a subgroup of 3^{5}.Alt(6) of order 8. This proves (l). Moreover, D_{2}^{∗} also centralizes S/Q_{2} and so X. But by (f), D_{2}^{∗} ≅ C_{8} and and so by (l) D_{2}^{∗} ≰ M′_{3}. Thus (k) is also proven and the lemma is complete.

Let 1 ≤ i, j ≤ 3 with i ≠ j. If X is an F M_{i}-module, then R_{ij}(X) denotes the restriction of X to M_{ij}. If, on the other hand, W is an F M_{ij}-module, then I^{i}(W) = W ⊗_{F M_{ij}} F M_{i} denotes the F M_{i}-module induced from W and R_{B}(W) denotes the restriction of W to B. Finally, if U is an F B-module, then I^{ij}(U) = U ⊗_{F B} F M_{ij}.

In the ensuing construction X_{t}(i) will always denote an F M_{i}-module, W_{t}(ij) an F M_{ij}-module and U_{t} an F B-module. If G is a group, H ≤ G, U is an F H-module and W is an F G-module we write U → W provided that U is isomorphic to an F H-submodule of W restricted to F H. (We remark that in all cases below the F H-submodule of W isomorphic to U will be unique).

Recall that Aut(McL) has two irreducible 21-dimensional representation over F arising from the Leech-lattice modulo 5; they differ by the −1 representation (see 5.4 (d)). For the moment let X_{1}(1) be either of the irreducible 21-dimensional F M_{1}/Q_{1}-modules. So

(1) X_{1}(1) is irreducible of dimension 21.

Put

W_{1}(13) = C_{X_{1}(1)}(Q_{3}) and W_{2}(13) = [X_{1}(1), Q_{3}].

Then X_{1}(1) = W_{1}(13) ⊕ W_{2}(13). By 2.5 (e), M_{13} has two orbits one of length 10 and one of length 30 on the maximal subgroups of Q_{3}/Z_{1}. Set Q_{3}^{∗} = Q_{3}D_{3}. Then, since Q_{3}^{∗} inverts Q_{3}, it follows that

(2) (a) W_{1}(13) is 1-dimensional and W_{1}(13) → X_{1}(1).

(b) W_{2}(13) is irreducible of dimension 20 and W_{2}(13) → X_{1}(1).

(c) W_{2}(13) is, as an F Q_{3}^{∗}-module, the direct sum of ten 2-dimensional Wedderburn components.

The two 21-dimensional F M_{1}-modules can be distinguished by the action of D_{3} on W_{1}(13). We choose X_{1}(1) so that

(3) D_{3} inverts W_{1}(13).

Set

U_{1} = R_{B}(W_{1}(13)).

Since B permutes the ten Wedderburn components for Q_{3}^{∗} in W_{2}(23) as Mat_{9}, it follows from (2) (c) that

(4) Restricted to B, W_{2}(13) is the direct sum of irreducible modules U_{2} and U_{3} of dimension 2 and 18 respectively.

Put

W_{1}(12) = C_{X_{1}(1)}(Q_{2}) and W_{2}(12) = [X_{1}(1), Q_{2}].

Then W_{1}(12) is 3-dimensional and, by 2.7, W_{2}(12) is 18-dimensional. Since X_{1}(1) is irreducible, M′_{12} does not normalize W_{1}(13). Hence, as R_{B}(W_{1}(12)) = U_{1} + U_{2}, we conclude that W_{1}(12) is irreducible as an M_{13}-module. Because Q_{2}Q_{3} is contained in M′_{13}, C_{W_{1}(12)}(Q_{2}Q_{3}) = W_{1}(13) and so (using facts about the 3-dimensional F (2.Alt(5))-module, which is in fact the irreducible constituent of the degree 5 permutation module), K_{2} inverts W_{1}(13). Now (3), 8.1 (h) and (j) imply that D_{2} centralizes W_{1}(13) and hence, as W_{1}(12) is irreducible, D_{2} centralizes W_{1}(12). Therefore, C_{M_{12}}(W_{1}(12)) = Q_{2}D_{2}. Pulling these facts together we have

(5) (a) Restricted to M_{12}, X_{1}(1) is the direct sum of irreducible modules W_{1}(12) and W_{2}(12) of dimension 3 and 18 respectively.

(b) C_{M_{12}}(W_{1}(12)) = Q_{2}D_{2}.

(c) C_{M_{12}}(W_{2}(12)) = Q_{1}.

(d) Restricted to B, W_{1}(12) is isomorphic to the direct sum U_{1} and U_{2}.

(e) Restricted to B, W_{2}(12) is isomorphic to U_{3}.

Since M_{2}/Q_{2}D_{2} ≅ Sym(5) × C_{2}, there exists a unique F M_{2}-module X_{1}(2) such that

(6) (a) X_{1}(2) is irreducible of dimension 3.

(b) Restricted to M_{12}, X_{1}(2) is isomorphic to W_{1}(12).

(c) D_{2}^{∗} inverts X_{1}(2) and C_{M_{2}}(X_{1}(2)) = Q_{2}D_{2}.

We have

X_{1}(2) = C_{X_{1}(2)}(S) ⊕ [X_{1}(2), S].

Put

W_{1}(23) = C_{X_{1}(2)}(S) and W_{2}(23) = [X_{1}(2), S].

Then it is straightforward to verify

(7) (a) W_{2}(23) is irreducible of dimension 2.

(b) Restricted to B, W_{2}(23) is isomorphic to U_{2}.

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

By 8.1 (j), D_{2} ≤ K_{3} ≤ M′_{3}. Thus D_{2} ≤ M_{13} ∩ M′_{3}. Since D_{2} ≰ M′_{1} whereas M′_{13} ≤ M′_{1}, we get D_{2}M′_{13} = M_{13} ∩ M′_{3}. Hence, by (5) (b), M_{13} ∩ M′_{3} centralizes W_{1}(13). Let X_{1}(3) be the 1-dimensional F M_{3}-module X_{1}(3) which is centralized by M′_{3} and inverted by M_{3}. Then the restriction of X_{1}(3) to M_{13} is isomorphic to W_{1}(13). Then as F B-modules R_{23}(X_{1}(3)) and W_{1}(23) are both isomorphic to U_{1}. On the other hand as, by 8.1 (k), D_{2}^{∗} ≰ M′_{3}, D_{2}^{∗} inverts both W_{1}(23) and C_{X_{1}(2)}(S). Thus R_{23}(X_{1}(3)) and W_{1}(23) are isomorphic as F M_{23}-modules. We have proved:

(8) (a) X_{1}(3) is 1-dimensional.

(b) Restricted to M_{13}, X_{1}(3) is isomorphic to W_{1}(13).

(c) Restricted to M_{23}, X_{1}(3) is isomorphic to W_{1}(23).

(d) Restricted to B, X_{1}(3) is isomorphic to U_{1}.

Put

X_{2}(2) = I^{2}(W_{2}(12)).

Then X_{2}(2) is 36-dimensional and when restricted to M_{12}, X_{2}(2) is the direct sum of W_{2}(12) and W_{3}(12), where W_{3}(12) is irreducible of dimension 18. As Q_{1} centralizes W_{2}(12) but not W_{3}(12), W_{2}(12) and W_{3}(12) are not isomorphic. Therefore, X_{2}(2) is irreducible of dimension 36. Put

U_{4} = R_{B}(W_{3}(12)) and W_{3}(23) = R_{23}(X_{2}(2)).

We have

(9) (a) W_{3}(12) is irreducible of dimension 18 and is not isomorphic to W_{2}(12) as an F M_{12}-module.

(b) X_{2}(2) is irreducible of dimension 36.

(c) Restricted to M_{12}, X_{2}(2) is isomorphic to the direct sum of W_{2}(12) and W_{3}(12).

(d) U_{4} is irreducible of dimension 18 and not isomorphic to U_{3}.

(e) Restricted to B, W_{3}(12) is isomorphic to U_{4}.

(f) W_{3}(23) ≅ I^{23}(U_{3}) is irreducible of dimension 36.

(g) Restricted to B, W_{3}(23) is isomorphic to the direct sum of U_{3} and U_{4}.

Let

X_{2}(3) = I^{3}(W_{2}(23)).

Then X_{2}(3) is 110-dimensional. Since M_{3} operates on M_{3}/M_{23} as it does on duads in the S(4,5,11)-Steiner system, following back through (7) and (2) c) delivers

(10) (a) W_{2}(23) is a Wedderburn component for Q_{3} on X_{2}(3).

(b) X_{2}(3) is irreducible of dimension 110.

By (2) b) and (4), we have W_{2}(13) ≅ I^{13}(U_{2}) and U_{3} → W_{2}(13). Thus W_{2}(13) → X_{2}(3) and so U_{3} → X_{2}(3). Since, M_{3} acts on M_{3}/M_{23} as it does on duads in the Steiner system and since M_{3} permutes the Wedderburn components in X_{2}(3) as is permutes the cosets M_{3}/M_{23}, we can use (10) to calculate (11) (a) Restricted to M_{13}, X_{2}(3) is isomorphic to the direct sum of W_{2}(13)

and W_{3}(13) where W_{3}(13) is irreducible of dimension 90 = 45 · 2.

(b) Restricted to M_{23}, X_{2}(3) is isomorphic to the direct sum of W_{2}(23), W_{3}(23) and W_{4}(23), where W_{4}(23) is irreducible of dimension 72.

(c) Set U_{5} = R_{B}(W_{4}(23)). Then U_{5} is irreducible of dimension 72.

(d) R_{B}(W_{3}(13)) is the direct sum of U_{4} and U_{5}.

(e) Restricted to B, X_{2}(3) is isomorphic to the direct sum of U_{2}, U_{3}, U_{4} and U_{5}.

We now claim that there is a 72-dimensional F M_{2}-module which when restricted to M_{23} is isomorphic to W_{4}(23). Let E_{1} and E_{2} be the two subgroups of order three in Z_{2} not conjugate to Q_{1} in M_{2} and set R = O^{2}(M_{2}). Then, by 6.5, R centralizes Z_{2} and so, in particular, normalizes both E_{1} and E_{2}. Now for i = 1, 2, we let W_{i} be the unique F R/E_{i}-module of dimension 36 and W = W_{1} ⊕ W_{2} (see 2.7). Then W is a faithful F R-module. Next let R_{3} = R ∩ M_{3}. We will show that W is isomorphic to W_{4}(23) as an F R_{3}- module. Thus in GL_{72}(F) we may arrange that the images of M_{23} and R intersect in a subgroup isomorphic to R_{3}. Denote the images by M_{23}^{∗}, R^{∗} and R_{3}^{∗} respectively. Following the same methods as in the end of the proof of 6.4 we see that ⟨R^{∗}, M_{23}^{∗}⟩ ≅ M_{2}. Therefore, to prove our claim it suffices to show that W and W_{4}(23) are isomorphic as F R_{3}-modules. Set Y = W_{4}(23) and, for i = 1, 2, set Y_{i} = C_{Y} (E_{i}). Since W_{3}(13) admits M_{13} faithfully, 0 = C_{W_{3}(13)}(Q_{1}) = C_{U_{5}}(Q_{1}) = C_{W_{4}(23)}(Q_{1}). Thus Y = Y_{1} ⊕ Y_{2}. Since both Y_{1} and Y_{2} are F R_{3}-modules, our claim will be proved when we show that, for i = 1, 2, the F R_{3}-modules Y_{i} and W_{i} are isomorphic. It clearly suffices to show Y_{1} and W_{1} are isomorphic.

Let E be the preimage of the complement to Z_{2}/E_{1} in (Q_{2} ∩ Q_{3})/E_{1} which is normalized by K_{2} (so E = [Q_{2} ∩ Q_{3}, K_{2}]E_{1}/E_{1}) and set W_{E} = C_{W_{1}}(E) and Y_{E} = C_{Y_{1}}(E) and consider them as F N_{R_{3}}(E)-modules. From 2.7 and 2.6 we have dim_{F} W_{E} = 4 = dim_{F} Y_{E}. Note that the F R_{3}-modules W_{1} and Y_{1} are induced from the F N_{R_{3}}(E)-modules W_{E} and Y_{E} respectively. Hence to show that W_{1} is isomorphic to Y_{1} as an F R_{3}-module it suffices to show that W_{E} and Y_{E} are isomorphic as N_{R_{3}}(E)-modules.

Note that N_{R_{3}}(E) = Q_{3}K_{2} acts as Q_{3}K_{2}/E ∼ 3^{2}.C_{4} (by 8.1 (e)) on both W_{E} and Y_{E}. Now K_{2} normalizes Z_{2}E and so, as K_{0} does not centralize Q_{3}/E, we see that K_{2} has an orbit of length 2 and two orbits of length 1 on the maximal subgroups of Q_{3} which contain E. Since Z_{2} acts with out fixing any non-zero vectors on both W_{E} and Y_{E}, we conclude that

Y_{E} = C_{Y_{E}}(A_{1}) ⊕ C_{Y_{E}}(A_{2})
W_{E} = C_{W_{E}}(A_{1}) ⊕ C_{W_{E}}(A_{2})

where A_{1} and A_{2} are the maximal subgroups of Q_{3} which contain E and are interchanged by K_{2}. Thus to conclude the proof that Y_{E} and W_{E} are isomorphic as F N_{R_{3}}(E)-modules it only remains to show that they are isomorphic as F K_{0}- modules. We will show that K_{0} in fact inverts both spaces. This follows directly from 2.7 (d) for W_{E}. Thus we concentrate on Y_{E}. Let t ∈ C_{K^{∗}∩M′_{3}}(S/Q_{2}) be an involution (see 8.1 (l)). Then, as t centralizes S/Q_{2}, (4) and (7) (b) imply that t either centralizes or inverts W_{2}(23). On the other hand, by (8) and (6) (c), t centralizes X_{1}(3), but not X_{1}(2). Thus we see that t inverts W_{2}(23). Since t is conjugate into K_{0} in M′_{3}, we see that C_{X_{2}(3)}(E) is not centralized by K_{0}. Now, as mentioned earlier, E is contained in exactly four maximal subgroups of Q_{3}; extend our notation above by denoting the other two maximal subgroups of Q_{3} which contain E by A_{3} and A_{4} and assume that A_{3} = EZ_{2}. Then

C_{X_{2}(3)}(E) = C_{Y_{E}}(A_{1}) ⊕ C_{Y_{E}}(A_{2}) ⊕ C_{X_{2}(3)}(A_{3}) ⊕ C_{X_{2}(3)}(A_{4}),

where each summand is 2-dimensional. Now we plainly have C_{X_{2}(3)}(A_{3}) ≅ W_{2}(23) and C_{X_{2}(3)}(A_{4}) ≅ C_{X_{2}(2)}(E) and these are both centralized by K_{0}. Since A_{1} and A_{2} are interchanged by K_{2} we know that K_{0} operates in the same way on both C_{Y_{E}}(A_{1}) and C_{Y_{E}}(A_{2}). Therefore, K_{0} inverts Y_{E} as claimed. Thus we have shown that there exists an F M_{2}-module, X_{3}(2), which satisfies

(12) X_{3}(2) is irreducible of dimension 72 and is restricted to M_{23} isomorphic to W_{4}(23).

Let W_{4}(12) = R_{12}(X_{3}(2)). Note that both W_{3}(13) and W_{3}(12) ⊕ W_{4}(12) are, when restricted to B, isomorphic to U_{4} ⊕ U_{5}. Thus taking G_{2} = M_{12}, G_{3} = M_{13} and G_{23} = M_{123}, we can apply 5.13 to get an F M_{1}-module X_{2}(1) so that

(13) (a) X_{2}(1) is irreducible of dimension 90.

(b) Restricted to M_{13}, X_{2}(1) is isomorphic to W_{3}(13).

(c) Restricted to M_{12}, X_{2}(1) isomorphic to the direct sum of W_{3}(12) and W_{4}(12).

Let X be a 111-dimensional vector space over F. We are now able to construct a completion of the Ly-amalgam in GL(X). Set H = GL(X) and A = Inn(GL(X)). Then, by (1)-(13), for i = 1, 2, 3 there exist monomorphism α_{i} : M_{i} → H such that as an F M_{1}^{α_{1}}-module X = X(1) = X_{1}(1) ⊕ X_{2}(1), as an F M_{2}^{α_{2}}-module X = X(2) = X_{1}(2) ⊕ X_{2}(2) ⊕ X_{3}(2) and as an F M_{3}^{α_{3}}-module X = X(3) = X_{1}(3) ⊕ X_{2}(3). Let {i, j} ⊂ {1, 2, 3}. Then using (1)-(13), X(i) and X(j) are isomorphic as F M_{ij}-modules. Thus there exist a_{1}, a_{2}, a_{3} ∈ A such that

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

Therefore, the assumptions of 2.4 are fulfilled. Set M_{12}^{∗} = M_{12}^{α_{1}}, M_{23}^{∗} = M_{23}^{α_{3}a_{2}^{−1}}, M_{13}^{∗} = M_{13}^{α_{1}}, and B^{∗} = B^{α_{1}}. If N is one of B^{∗}, M_{23}^{∗}, M_{13}^{∗} and M_{12}^{∗}, then X is the direct sum of pairwise non-isomorphic, absolutely irreducible F Nmodules and so C_{H}(N) consists of exactly those linear transformations which act as non-zero scalars on each of the irreducible F N-submodules. By 2.4 (b), B^{∗} ≤ N and so for each choice of N, C_{A}(N) ≤ C_{A}(B^{∗}). Since, on restriction to M_{ij} (respectively B), X is a direct sum of pairwise non-isomorphic F M_{ij}- modules, it is now easy to verify that C_{A}(B^{∗}) has order 4^{4} and that C_{A}(B^{∗}) = C_{A}(M_{23}^{∗})C_{A}(M_{13}^{∗})C_{A}(M_{12}^{∗}). Thus by 2.4 (a), 2.4 (a1) holds. Put M_{i}^{∗∗} = M_{i}^{α_{i}b_{i}}. Then by 2.4 (c) we get

Theorem 8.2 Suppose that X is a 111-dimensional vector space over F. Then there exists a Ly-amalgam in GL(X).

Theorem 1.3 now follows by combining 6.4, 8.2 and 7.16.

References

[Alp] J.L. Alperin, Local Representation Theory, Cambridge University Press, Cambridge, 1986.

[Asch] M. Aschbacher, Finite Group Theory, Cambridge University Press, Cambridge, 1986.

[Asch1] M. Aschbacher, Sporadic Groups, Cambridge University Press, Cambridge, 1994.

[AschS] M. Aschbacher and Y. Segev, The uniqueness of groups of Lyons type, J. AMS, Vol 5, No. 1, 75-98 1992.

[Be] D. Benson, Modular Representation Theory: New Trends and Methods, Springer-Verlag, Berlin, New York, Tokyo, (1984).

[ATLAS] J.H. Conway, R.T. Curtis, S.N. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups, Clarendon Press, Oxford, 1985.

[CE] H. Cartan, S. Eilenberg, Homological Algebra, OUP, 1956.

[CS] J.H.Conway, N.J.A.Sloane, Sphere Packings, Lattices and Groups, Springer Verlag, New York, 1988.

[CFYT] G. Cooperman, L. Finkelstein, B. York and M. Tselman, Constructing Permutation Representations for Large Matrix Groups, Proc. Int. Symposium on Symbolic and Algebraic Manipulation (ISSAC-94), ACM Press, 1194.

[Fink] L. Finkelstein, The Maximal Subgroups of Conway’s Group C_{3} and McLaughlin’s groups, J. Alg. 25 (1973) 58-89.

[Frob] G. Frobenius, Über die Charaktere der mehrfach transitiven Gruppen, Sitzung berichte Ak. Berlin, (1904), 558-571.

[Gol] H. W. Gollan, A New Existence Proof for Ly, The Sporadic Simple Group of R. Lyons, Habilitationsschrift, Universität GH Essen, FB Mathematik und Informatik, 1995.

[Go] D. Gorenstein, Finite Groups, Harper & Row, 1968.

[Gr] R.L. Griess, Schur Multipliers of Finite Simple Groups of Lie Type, Trans. A.M.S. Vol 183, 355- 421, 1973.

[HL] G. Hiss, K. Lux, Brauer Trees of Sporadic Groups, Clarendon Press, Oxford, 1989.

[HLP] G. Hiss, K. Lux, R. Parker, The 5-Modular Characters of the McLaughlin Group and its Covering Group, Manuscripta Math. 73, 91-114 (1991).

[IM] A.A. Ivanov, U. Meierfrankenfeld, A Computer Free Construction of J_{4}, J. Alg. 219 (1999), no. 1, 113–172.

[Iv] A.A. Ivanov, Geometry of Sporadic Groups I, Petersen and Tilde Geometries, Encyclopedia of Mathematics and its Applications, CUP, 1999.

[Jam] G.D. James, The Modular Characters of the Mathieu Groups, J. Alg. 27, 57-111 (1973.)

[MATLAS] Ch. Jansen, K. Lux, R.A. Parker and R.A. Wilson, An Atlas of Brauer Characters, Clarendon Press, Oxford, 1995.

[JW] Ch. Jansen and R.A. Wilson, The Minimal 3-Modular Representation for the Lyons Group, Comm. Alg. 24(3), 873-879 (1996).

[Ly] R. Lyons, Evidence for a New Finite Simple Group, J. Alg. 20 540-569, (1972); Errata, J. Alg. 34, 188-189, (1975).

[McL] McLaughlin, J, A Simple Group of Order 898, 128,000, Theory of Finite Groups (Symposium, Harvard, Univ., Cambridge, Mass., 1968), Ed, R. Brauer and C. Sah, pp. 109-111, Benjamin, New York, 1969.

[MNP] W. Meyer, W. Neutsch, R. Parker, The Minimal 5-Representations of Lyons’ Sporadic Group, Math. Ann. 272 (1985), no. 1, 29–39.

[Si] C. C. Sims, The Existence and Uniqueness of Lyons’ Group, in Finite groups ’72 (eds. T. Gagen, M. P. Hale and E. E. Shult), North Holland, 1973, pp.138-141.

[Wi] R.A. Wilson, A Representation for the Lyons Group in GL_{2480}(4), and a New Uniqueness Proof, Arch. Math. (Basel) 70 (1998), no. 1, 11–15.