Ru < E_{7}(5) and HS < E_{7}(5)

P. B. Kleidman 30 West 10th Street New York, NY 10011

U. Meierfrankenfeld Michigan State University East Lansing, MI 48824

A. J. E. Ryba Marquette University Milwaukee, WI 53233

In this paper we describe our discovery that the sporadic simple groups Ru, HS and M_{22} are contained in the simple Chevalley group E_{7}(5).

The work of [9] produces a short list of the possibilities for a sporadic simple subgroup of an exceptional group of Lie type. Apart from possible embeddings of M_{22}, HS and Ru in groups of type E_{7} in characteristic 5, all of the embeddings of [9] are already known to occur. Thus our paper completes the classification of sporadic simple subgroups of exceptional groups of Lie type.

We offer two proofs of the embedding Ru < E_{7}(5). The first is a computer proof, and the second is totally by hand. In particular, the second proof provides the only known computerfree construction of Ru. Moreover, our computer proof includes the first published presentation of Ru and thus gives the first easily verifyable computer construction of Ru. Similarly we give a hand proof and a computer proof of the embedding HS < E_{7}(5). As a step in our hand proof of HS < E_{7}(5) we establish the embedding M_{22} < E_{7}(5): of course, since M_{22} is a subgroup of HS, this result also follows as a consequence of our computer proof of HS < E_{7}(5).

We were led to conjecture the inclusions Ru < E_{7}(5) and HS < E_{7}(5) for the following reasons. The double cover 2.Ru has a faithful 28-dimensional character χ, and the character values of χ + χ^{∗} are all compatible with the character values of groups of type E_{7} acting on their natural 56-dimensional module. Similarly the double cover 2.HS has a faithful 56dimensional character, whose values are compatible with the character values of groups of type E_{7} acting on their natural 56-dimensional module. Now Ru, 2.Ru, HS and 2.HS all contain a subgroup 5^{2}:20, an elementary abelian group of order 25 extended by a cyclic group of order 20 acting faithfully on the 5^{2}. Since 20 is not the order of an element in the Weyl group W (E_{7}) = 2 × S_{6}(2), it can be shown that 5^{2}:20 does not embed in groups of type E_{7}(K), where K is a field of characteristic prime to 5. Thus Ru and HS embed in E_{7}(q) only if 5 | q. On the other hand, all local subgroups of 2.Ru and 2.HS embed in 2.E_{7}(5), whence our conjectures Ru < E_{7}(5) and HS < E_{7}(5).

Throughout, G denotes the double cover 2.E_{7}(5), \overline{G} denotes the simple group E_{7}(5) and V is the natural 56-dimensional module for G over GF (5). Most of our notation follows that of the ATLAS[At]. The four sections of our paper are independent and are arranged in chronological order. A later, independent, proof of Ru < E_{7}(5) appears in [7].

Figure 1: Generating 2Ru

[figure: 2Ru; 29:7; 7; D_{14}]

1 A Computer Construction of Ru as a Subgroup of E_{7}(5).

We used R. A. Parker’s meataxe system (see [10]) to work with 56 × 56 matrices over GF (5), representing G. Since Ru contains 2^{6}.G_{2}(2), where the G_{2}(2) acts transitively on the 63 hyperplanes in the normal 2^{6}, it follows that any faithful representation of Ru over a field in odd characteristic has dimension at least 63. In particular Ru can not act on V and so Ru ≰ G. Thus we seek to classify subgroups 2.Ru in G.

Any subgroup 2.Ru can be generated as in Figure 1. There is a unique class of groups of order 29 in G, with normalizer

(6 × (29 × 449).7).2 < (6 × U_{7}(5)).2 < (2 × U_{8}(5)).2.

Moreover, an element of order 7 in N_{G}(29) satisfies

C_{G}^{∗}(7) = (SU_{3}(5).3 × 7).2 × SL_{2}(125).

(Here C^{∗} denotes the invertilizer: that is the set of elements which either invert or centralize a given element.) An involution inverting the 7 extends SU_{3}(5).3 to SU_{3}(5).Sym(3). There are precisely 9450 involutions in SU_{3}(5).Sym(3)\SU_{3}(5), and so the 7 is contained in precisely 2 × 9450 groups D_{14} in G. The factor of 2 comes from the involution in SL_{2}(125), which lies in the center of G. Now G has just 3 classes of involutions, with representatives Z, X, ZX. Here ⟨Z⟩ = Z(G), and X and ZX have respective traces +8 and −8 on V. Thus we call the involutions conjugate to X and ZX plus involutions and minus involutions, respectively. Obviously the 2 × 9450 groups D_{14} come in pairs – one containing plus involutions, the other containing minus involutions.

Thus to classify subgroups isomorphic to 2.Ru in G, it suffices to check each of these 9450 plus D_{14}’s, and determine which ones together with the 29 generate 2.Ru. On the computer, we found all 9450 such D_{14} groups, called X_{1}, …, X_{9450} say, and investigated the groups Y_{i} = ⟨29, X_{i}⟩. We discarded any Y_{i} that contains an element with an order which is not the order of an element of 2.Ru. Precisely 6 remained, Y_{1}, …, Y_{6} say. In order to identify these six groups, we use the presentation of Ru given by the following Theorem.

Figure 2: The Coxeter group X

[figure: a; b; g; c; d; e; f; w; 12]

Let X denote the Coxeter group with the Coxeter diagram given in Figure 2. (Thus X is generated by involutions a, b, c, d, e, f, g and w whose pairwise products have orders 2, 3, 4, or 12 according as the corresponding nodes of the diagram are unjoined, joined by an unmarked single edge, joined by a double edge, or joined by an edge marked 12.) Let R be the quotient of X obtained by adjoining the following additional relations: c = g^{fed}, w = c^{gbag}c^{gabg}c, d = (bfg)^{8}, e = (bc)^{6} = (abc)^{4}, (be)^{d}(be)^{dfbef} = db^{a}g^{bfabgbfg} = (ec)^{de}(ec)^{dw} = (cw)^{bcba}(cw)^{bwc} = c^{bcdewb}(db)^{wedcbcbcdewd} = 1.

Theorem 1.1 The group R is isomorphic to the Rudvalis group.

Proof: Let T (respectively L) denote the group presented by those generators and relations of R that do not mention f or g, (respectively f, g, or w). Computer coset enumeration shows that the image of T in R has index 8120. Moreover, standard permutation group computations show that in the resulting permutation representation of R the images of T and L have sizes 11232 = |L_{3}(3):2| and 1600 × |L_{3}(3):2| = |^{2}F_{4}(2)′|. Moreover, the image of T has orbits of lengths 1, 1755 and 2304 in the permutation representation of R.

A second enumeration (of the cosets of the trivial subgroup in L) shows that |L| = |L_{3}(3):2|. It is routine to show that L_{3}(3):2 is a quotient of L, and thus the group L and its images in T and R are copies of L_{3}(3):2. A final coset enumeration shows that there are 1600 cosets of the image of L in T, and hence |T | = |^{2}F_{4}(2)′|. Our earlier computation of the size of the image of T in a permutation representation of R proves that the image of T in R has size |^{2}F_{4}(2)′| and thus |R| = 8120|T | = |Ru|.

We now use a standard argument (see for example [17]) to show that the groups T and R must be simple. We illustrate this argument for the group T. The image of L in the permutation representation of T on the 1600 cosets of the image of L in T has orbits of sizes 1, 312, 351 and 936. In particular this faithful permutation representation of T must be primitive. Thus any minimal non-identity normal subgroup, N say, in T must be transitive on the cosets of the image of L. Moreover N ∩ L is isomorphic to a normal subgroup of L_{3}(3):2 and therefore |N ∩ L| ∈ {1, |L_{3}(3)|, |L_{3}(3):2|}. We deduce that the characteristically simple group N has order 1600, |^{2}F_{4}(2)′/2|, or |^{2}F_{4}(2)′|. The classification of finite simple group shows that only the last of these orders is possible, and in particular T must be simple. Another appeal to the classification of finite simple groups gives T ≅ ^{2}F_{4}(2)′. A similar (slightly easier) argument now shows that R is simple and thus since the simple group Ru is characterized by its order, R ≅ Ru. QED

We now apply Theorem 1.1 to identify the six subgroups Y_{1}, …, Y_{6} of G. For each of these groups we know generating matrices x, y, z of orders 29, 7, 2 (from the subgroups 29:7, 7 and D_{14} of Figure 1). We may assume that the element y is replaced by one of its powers so that x^{y} = x^{16}. Moreover, in each of the groups Y_{1}, …, Y_{6}, experiment shows that we may replace x by one of its powers so that, modulo scalar matrices, xz has order 15 and xzx has order 20. (In fact there are two mutually inverse choices for x with these properties. We can make either choice of x.) We compute the following matrices in the group generated by x, y and z. Let j = (x^{3}zxz)^{10}z, a = [z, j]^{6}, b = [z, j^{2}]^{4}, c = (b[z, j^{3}]^{3})^{6}[z, j^{3}]^{3}, d = ([z, j^{3}]^{3}(ab^{j^{2}})^{2})^{2}([z, j^{3}]^{3}(ab^{j^{2}})^{2})^{z^{j}}, e = z, k = [a, x^{4}]^{(b[a,x^{4}]^{5})^{6}ab}, l = c^{k^{4}}, f = l^{cd}l^{dc}l, g = c^{def}, and w = c^{gbag}c^{gabg}c. Modulo scalar matrices, each of the groups Y_{1}, …, Y_{6} is generated by its elements a, b, c, d, e, f, g, w — since in each case the following words recover the generators x, y, z (modulo scalars): z = e, m = aba(abgc)^{5}abg, y = [(mf)^{13}e(md)^{12}e(md)^{12}, e], n = e[b, aw]^{(abgc)^{5}gc}, p = (ewg(abgc)^{5}a(abgc)^{5}bg(abgc)^{5}bawg(abgc)^{5}cbg)^{fw(ne^{fw})^{3}}, q = (pa)^{7}, r = (pb)^{6}, s = (pc)^{10}, x = (y^{−1}ps^{r}q^{rsqsqr})^{−2}. Moreover, modulo scalar matrices, for each of the groups Y_{1}, …, Y_{6} the generators a, b, c, d, e, f, g, w satisfy the relations of R of Theorem 1.1. Therefore, Z(G)Y_{i}/Z(G) ≅ Ru, for i = 1, …, 6. Since we have already observed that Ru can not be a subgroup of G, we must have Y_{i} ≅ 2.Ru for i = 1, …, 6.

Now N_{G}(29) ∩ N_{G}(7) ≅ 12 acts on Y_{1}, …, Y_{6}. Furthermore, N_{2.Ru}(29) ∩ N_{2.Ru}(7) ≅ 4, and so each of Y_{1}, …, Y_{6} contains the subgroup of order 4 in N_{G}(29) ∩ N_{G}(7). Moreover, the group of order 3 in N_{G}(29) ∩ N_{G}(7) cannot normalize one of Y_{1}, …, Y_{6}, for this group of order 3 centralizes the 29 (in G) and yet there is no group of order 3 in the 29−centralizer of 2.Ru. Thus N_{G}(29) ∩ N_{G}(7) has two orbits of size 3 on Y_{1}, …, Y_{6}. Now 2.Ru contains a unique class of 29:7, and we have

C_{2.Ru}^{∗}(7) = Q_{8} × D_{14} ≤ Q_{8} × Sz(8).

Thus the 7 is contained in precisely two groups D_{14}. Consequently each 2.Ru in G can be generated in just two ways as in (^{∗}), one way with a plus D_{14} and the other with the minus D_{14}. We have therefore proved that there are just two classes of 2.Ru in G.

Therefore, there are just two classes of subgroup isomorphic to Ru in \overline{G}. The non-abelian composition factors of centralizers of involutions in Aut(\overline{G}) are L_{2}(5), O_{12}^{+}(5), L_{8}(5), U_{8}(5), E_{6}(5), and ^{2}E_{6}(5), none of which contain Ru. Consequently the outer automorphism of \overline{G} must interchange the two classes of Ru in \overline{G}.

We have now proved assertions (A) and (B), as well as part of (D), in the Theorem below. Our construction gives an explicit matrix action of 2.Ru on the natural 56-dimensional GF (5)-module for 2.E_{7}(5) and a standard application of the meataxe (see [10]) provides the decomposition of this module given in (D). The 133−dimensional adjoint module for the Lie algebra associated with E_{7}(5) is a constituent of the symmetric square of the 56-dimensional module. We used the meataxe to determine that 133 is the smallest degree of a non-trivial constituent of the action of Ru on the symmetric square of the 56−dimensional module. (The 56−dimensional module has two irreducible 28−dimensional constituents, thus we analyzed the two 406−dimensional modules obtained as symmetric squares of 28−dimensional representations, and the 784−dimensional tensor product of the two 28−dimensional representations. The last of these computations is close to the size limit for our implementation of the Meataxe.) The statements in (C) follow.

Theorem 1.2 (A) The simple Chevalley group E_{7}(5) contains precisely two classes of subgroups isomorphic to Ru.

(B) The outer automorphism of E_{7}(5) fuses the two classes.

(C) Each subgroup Ru acts irreducibly on the 133-dimensional Lie algebra associated with E_{7}(5).

(D) In the double cover 2.E_{7}(5), each Ru lifts to 2.Ru, and acts indecomposably with two irreducible constituents of dimension 28 on the natural 56-dimensional GF (5)-module for 2.E_{7}(5).

2 A Computer-free Construction of the Rudvalis Group as a Subgroup of E_{7}(5).

In this chapter we will give a computer-free proof of the following Theorem.

Theorem 2.1 Suppose that E_{6}(5) has a subgroup ^{2}F_{4}(2) which acts irreducibly on the 27dimensional GF (5)E_{6}(5)-modules, then E_{7}(5) has a subgroup isomorphic to the Rudvalis group.

We remark that an unpublished paper of M. Aschbacher on the maximal subgroups of E_{6} contains a computer free proof that E_{6}(5) indeed has a subgroup isomorphic to ^{2}F_{4}(2) which acts irreducibly on the 27−spaces for E_{6}(5).

Lemma 2.2 Let X be a 2-dimensional vector space over GF (5), S a Sylow 2-subgroup of GL(X), A the unique subgroup of S isomorphic to C_{4} × C_{4}, ϕ an automorphism of S such that X^{ϕ} ≅ X^{∗}, the dual module of X. Then ϕs inverts A for some s in S.

Proof: Note first that there exist 1-dimensional subspaces X_{1}, X_{2} of X with X = X_{1} ⊕ X_{2}, S = N_{GL(X)}({X_{1}, X_{2}}) and A = N_{GL(X)}(X_{1}) ∩ N_{GL(X)}(X_{2}). Let ψ be the automorphism of A given by inversion. Then X^{∗} and X^{ψ} are isomorphic as A−modules. Hence also X^{ϕ} and X^{ψ} are isomorphic as A−modules and there exists s ∈ N_{GL(X)}(A)) such that ϕs and ψ agree on A. Since N_{GL(X)}(A) = S, s ∈ S and the lemma is proved. QED

Let E be the parabolic subgroup of G such that E = QL with Q = O_{5}(E), |Q| = 5^{27} and L ≅ C_{4} × E_{6}(5). By assumption, L has ^{2}F_{4}(2) as a subgroup. But it can be proved that such a ^{2}F_{4}(2) cannot be extended to a 2.Ru in G. Instead we will look for a different class of complements with respect to Q. For this we first have to study the action of ^{2}F_{4}(2) on its 27-dimensional irreducible module over GF (5).

Let F be a group with^{2}F_{4}(2)′ ≤ F ≤ C_{4}×^{2}F_{4}(2). Let S be a Sylow 2-subgroup of F and let P_{1} and P_{2} be the two maximal subgroups of F containing S, ordered so that Z(P_{1}/Z(F)) ≠ 1. Let Γ_{0} be the coset-graph of F with respect to P_{1} and P_{2}. Then Γ_{0} is the generalized octagon associated with ^{2}F_{4}(2). For γ ∈ Γ_{0}, let △^{k}(γ) be the set of vertices in Γ_{0} at distance exactly k from γ. Further put α = P_{1} and β = P_{2} and note that α and β are vertices of Γ_{0}. Put T = F′, L_{i} = P_{i} ∩ T, Q_{i} = O_{2}(L_{i}), Z_{i} = Z(Q_{i}) and V_{1} = ⟨Z_{2}^{P_{1}}⟩. Let 1 ≠ z_{1} ∈ Z_{1}. If P_{i} normalizes a subgroup R_{i} in F or in some F −module, and if δ = P_{i}g ∈ Γ_{0}, put R_{δ} = R_{i}^{g}.

We assume that the reader is familiar with the structures of P_{1}, P_{2} and S (see for example [11] or [5]). We remark here that |S ∩ T | = 2^{11}, L_{1}/Q_{1} is a Frobenius group of order 20, Z_{1} has order 2, V_{1}/Z_{1} is the unique irreducible L_{1}/Q_{1} module of order 16, V_{1} is elementary abelian, L_{2}/Q_{2} ≅ Sym(3) and Z_{2} is the unique irreducible L_{2}/Q_{2}−module of order 4.

We pay special attention to groups F such that

F/Z(F) ≅ ^{2}F_{4}(2), |Z(F)| = 2, F′ ≅ (^{2}F_{4}(2))′ and F/F′ ≅ C_{4}.  (+)

Lemma 2.3 Let W be a faithful irreducible 27−dimensional GF (5)F -module. (a) Let U_{1} = C_{W} (V_{1}).

(aa) W = C_{W} (V_{1}) ⊕ [C_{W} (Z_{1}), V_{1}] ⊕ [W, Z_{1}] ,

(ab) [C_{W} (Z_{1}), V_{1}] = ⨁_{γ∈△^{2}(α)} U_{γ},

(ac) [W, Z_{1}] = ∑_{γ∈△^{4}(α)} U_{γ} = ⨁_{γ∈△^{3}(β)\△^{2}(α)} U_{γ},

(ad) U_{1} = C_{W} (V_{1}) is 1-dimensional, C_{W} (V_{1}) = C_{W} (Q_{1}), P_{1}/C_{P_{1}}(U_{1}) ≅ C_{4} and U_{1} is not isomorphic to its dual GF (5)P_{1} module,

(ae) [C_{W} (Z_{1}), V_{1}] is irreducible of dimension 10 and if ϕ is any automorphism of P_{1}, then [C_{W} (Z_{1}), V_{1}]^{ϕ} and the dual of [C_{W} (Z_{1}), V_{1}] are not isomorphic as GF (5)P_{1}−modules,

(af) [W, Z_{1}] is irreducible of dimension 16. If (+) holds [W, Z_{1}] is isomorphic to its dual GF (5)P_{1}−module.

(b)

(ba) W = C_{W} (Z_{2}) ⨁ [W, Z_{2}].

(bb) [W, Z_{2}] = ⨁_{γ∈△^{3}(β)} U_{γ},

(bc) C_{W} (Z_{2}) = ⨁_{γ∈△^{1}(β)} U_{γ},

(bd) [W, Z_{2}] is irreducible of dimension 24. If (+) holds [W, Z_{2}] is isomorphic to its dual GF (5)P_{2}−module,

(be) C_{W} (Z_{2}) is irreducible of dimension 3 and not isomorphic to its dual GF (5)P_{2} module.

(c) Let X ≤ Y be GF (5)P_{1}−modules such that Y and Y/X are isomorphic to [W, Z_{1}] as P_{1}−modules. Then Y splits over X.

(d) F′ has two classes of involutions. If z_{1} and i are representatives of these classes then C_{W} (z_{1}) is 11-dimensional while C_{W} (i) is 15-dimensional. Moreover, |C_{F} (i)| = 2^{11} · 3.

(e) F has a unique class of elements d of order three such that C_{F′}(d) has even order. Moreover, for any such d and any t ∈ C_{F′}(d) with |t| = 2, we have N_{F′}(D) ∩ C_{F′}(t) ≅ D_{24}. If (+) holds then |C_{F} (d)| = 2^{5} · 3^{3}

(f) Let d be an element of order five in F. Then C_{F} (d) is a {2, 5}−group.

Proof: (aa) and (ba) follow from the fact that V_{1}, Z_{2} and Z_{1} are 2−groups and so coprime action applies.

Let R_{1} = [W, Z_{1}] and Σ = {H ≤ V_{1} | V_{1} = Z_{1} ⊕ H} . Note that Q_{1} and P_{1} act transitively on Σ. By co-prime action

R_{1} = ⊕_{H∈Σ} C_{R_{1}}(H).

It follows that dim R_{1} is a multiple of 16. Since dim W = 27 we conclude that dim R_{1} = 16, hence dim C_{R_{1}}(H) = 1 and thus Q_{1} acts irreducibly on R_{1}. In particular the first half of (af) holds. Let γ ∈ △(β) with γ ≠ α. Since there exist exactly 8 elements of Σ containing Z_{γ}, we have dim C_{R_{1}}(Z_{γ}) = 8. Further, dim C_{W} (Z_{γ}) = dim C_{W} (Z_{1}) = 27 − 16 = 11 and so

dim C_{W} (Z_{2}) = dim(C_{W} (Z_{1}) ∩ C_{W} (Z_{γ})) = 11 − 8 = 3.

Let Σ_{1} = {H ≤ V_{1} | Z_{1} ≤ H, |V_{1}/H| = 2} and Y_{1} = [C_{W} (Z_{1}), V_{1}]. Then

Y_{1} = ⨁_{H∈Σ_{1}} C_{Y_{1}}(H).

Note that P_{1} has two orbits Σ_{2} and Σ_{3} on Σ_{1} of lengths 5 and 10, respectively. So the dimension of Y_{1} is a multiple of 5. Since dim [Y_{1}, Z_{γ}] = dim [C_{W} (Z_{1}), Z_{γ}] = dim [C_{W} (Z_{γ}), Z_{1}] = dim C_{R_{1}}(Z_{γ}) = 8 and dim C_{W} (Z_{1}) = 11, we get dim Y_{1} = 10. Suppose that C_{Y_{1}}(H) ≠ 0 for H ∈ Σ_{3}. Since Z_{γ} lies in exactly one member of Σ_{2}, the group Z_{γ} lies in exactly 6 members of Σ_{3} and thus C_{Y_{1}}(Z_{γ}) is 6-dimensional, a contradiction to dim C_{Y_{1}}(Z_{γ}) = 10 − 8 = 2. So

Y_{1} = ⨁_{H∈Σ_{2}} C_{Y_{1}}(H) and dim C_{Y_{1}}(H) = 2 for H ∈ Z_{2}.

Recall that U_{1} = C_{W} (V_{1}). Then C_{W} (Z_{1}) = U_{1} ⊕ Y_{1} and hence dim U_{1} = 11 − 10 = 1. As W is irreducible, U_{1} ≠ U_{γ}, and so we have [U_{γ}, V_{1}] ≠ 1. Observe that C_{L_{γ}}(U_{γ}) ≥ O^{2}(L_{γ})

(since L_{γ} acts as a subgroup of GL_{1}(5) ≅ C_{4} on U_{γ}). However, V_{1} ≤ C_{T} (Z_{γ}) = L_{γ}, and V_{1} does not centralize U_{γ}. As V_{1}O^{2}(L_{γ})/O^{2}(L_{γ}) ≅ C_{2} is the unique proper subgroup of L_{γ}/O^{2}(L_{γ}) ≅ C_{4} we deduce that C_{T} (U_{γ}) = C_{L_{γ}}(U_{γ}) = O^{2}(L_{γ}), thus L_{γ}/C_{L_{γ}}(U_{γ}) ≅ C_{4} ≅ P_{γ}/C_{P_{γ}}(U_{γ}) and hence U_{1} is not self-dual as a GF (5)P_{1}−module.

Since V_{1} inverts U_{γ} and Z_{1} centralizes U_{γ}, U_{γ} lies in Y_{1}. Moreover, since V_{1} ≤ L_{γ}, the group V_{1} acts on U_{γ}, thus C_{V_{1}}(U_{γ}) is the hyperplane in Σ_{2} that contains Z_{γ}. Now C_{V_{1}}(U_{γ}) = V_{1} ∩ O^{2}(L_{γ}) = V_{1} ∩ Q_{γ}. Moreover the single hyperplane of Σ_{2} that contains Z_{γ} also contains Z_{γ}Z_{1} = Z_{β} and so also contains Z_{γ′}, where △^{1}(β) = {α, γ, γ′}. Hence V_{1} ∩ Q_{γ} = V_{1} ∩ Q_{γ′}, and Σ_{2} = {V_{1} ∩ Q_{δ} | δ ∈ △^{2}(α)}. So C_{Y_{1}}(V_{1} ∩ Q_{γ}) = U_{γ} + U_{γ′}, and

Y_{1} = ⨁_{δ∈△^{2}(α)} U_{δ}.

In particular (ab) holds. Now C_{T} (U_{γ} + U_{γ′}) = O^{2}(L_{γ}) ∩ O^{2}(L_{γ′}) = Q_{γ} ∩ Q_{γ′} and |S ∩ T /Q_{γ} ∩ Q_{γ′} | = 32. Since a Sylow 2−subgroup of GL_{2}(5) has order 32 and is isomorphic to C_{4} wreath C_{2} we conclude that S∩T /Q_{γ} ∩Q_{γ′} ≅ C_{4} wreath C_{2}, that S = (S∩T) C_{S} (U_{γ} + U_{γ′}) and that the action of S on U_{γ} + U_{γ′} is irreducible, but not self-dual. Thus Y_{1} is irreducible of dimension 10. Suppose there exists an automorphism ϕ of P_{1} so that [C_{W} (Z_{1}), V_{1}]^{ϕ} is isomorphic to the dual module of [C_{W} (Z_{1}), V_{1}]. As F rob_{20} has no outer automorphisms, we may assume without loss that ϕ centralizes P_{1}/O_{2}(P_{1}). Then (U_{γ} + U_{γ′})^{ϕ} is isomorphic, as an S−module, to the dual module of U_{γ} + U_{γ′}. In particular, ϕ normalizes C_{S}(U_{γ} + U_{γ′}). Now O_{2}(P_{2})/C_{S}(U_{γ} + U_{γ′}) is the unique subgroup of S/C_{S}(U_{γ} + U_{γ′}) isomorphic to C_{4} × C_{4} and so by Lemma 2.2, up to an inner automorphism of S, the automorphism ϕ inverts O_{2}(P_{2})/C_{S}(U_{γ} + U_{γ′})). As O_{2}(P_{2})O_{2}(P_{1}) = S and C_{S}(U_{γ} + U_{γ′}) ≤ O_{2}(P_{1}) we conclude that ϕ inverts S/O_{2}(P_{1}) ≅ C_{4}, a contradiction to [P_{1}, ϕ] ≤ O_{2}(P_{1}). Thus no such ϕ exists and (ad) is proved.

Let U_{2} = C_{W} (Z_{2}). Then

U_{2} = U_{α} + U_{γ} + U_{γ′}.

and so (bc) holds. For X ≤ P_{2}, let X̃ = XC_{P_{2}}(U_{2})/C_{P_{2}}(U_{2}). As C_{L_{2}}(U_{α}) = O^{2}(L_{1}) ∩ L_{2} = Q_{1} = ⋂_{ρ∈△^{1}(α)} L_{ρ}, we have C_{L_{2}}(U_{2}) = ⋂_{ρ∈△^{2}(β)} L_{ρ} and so by [5, 9.4.3]

L̃_{2} ≅ (C_{4} × C_{4}).Sym(3)

In particular, (be) holds.

We will now determine P̃_{2}, which does depend on the precise structure of F. Since, by assumption, F is irreducible on W and since C_{W} (Q_{1}) is 1−dimensional, F acts absolutely irreducibly on W. Hence all elements of Z(F) act as scalars on W and so also on U_{2}. Now no element of L̃_{2} acts as a scalar and so Z̃(F)L̃_{2} ≅ Z(F) × L̃_{2}. As the full monomial subgroup of GL(U_{2}) is isomorphic to C_{4} × L̃_{2} we see that at least one of the following holds

• Z(F) = 1 and P̃_{2} ≅ (C_{4} × C_{4}).Sym(3)

• |Z(F)| ≤ 2, F ≅ Z(F) ×^{2}F_{4}(2) and P̃_{2} ≅ C_{2} × (C_{4} × C_{4}).Sym(3)

• F/F′ ≅ C_{4} or C_{2} × C_{4} and P̃_{2} ≅ C_{4} ≀ Sym(3)

We remark that P̃_{2} is uniquely determined by the structure of F except when F ≅ ^{2}F_{4}(2). In this case its easy to see that F has two different irreducible 27−dimensional representations giving rise to the two different possibilities for P̃_{2}.

Note that Q_{1} is a normal subgroup of S generated by involutions. Furthermore we have C_{S∩T} (U_{γ} + U_{γ′}) ≤ Q_{1} and so Q_{1}/C_{Q_{1}}(U_{γ} + U_{γ′}) has order eight. Hence by the structure of C_{4} ≀ C_{2}, Q_{1} acts as a D_{8} on U_{γ} + U_{γ′}. In particular O_{2}(P_{1}) acts irreducibly on U_{γ} + U_{γ′}.

Suppose in this paragraph that (+) holds. We wish to show that U_{γ} + U_{γ′} is self–dual as an O_{2}(P_{1})−module. Let B be the set of elements in P_{2} that act as scalars on U_{2}, let C = C_{P_{2}}(U_{γ} + U_{γ′}), let D = O_{2}(P_{1}) ∩ O_{2}(P_{2}) and let E = [O_{2}(P_{2}), S]. As B is normal in P_{2} with O_{2}(P_{2})/B ≅ C_{4} × C_{4}, [5, 9.4.3] implies B ≰ O_{2}(P_{1}) and so B ≰ D. On the other hand, by the structure of S̃, there are exactly two subgroups X̃ of Õ_{2}(P_{2}) such that Z̃(F)Ẽ ≤ X̃ and Õ_{2}(P_{2})/X̃ ≅ C_{4}, namely C̃Ẽ and B̃Ẽ. Since S/O_{2}(P_{1}) ≅ C_{4}, the group D̃ meets the conditions on X̃ and so D̃ = C̃Ẽ and D = CE. But E ≤ Q_{1} ∩ Q_{2} and thus D = C(Q_{1} ∩ Q_{2}). Since O_{2}(P_{1}) = Q_{1}D we conclude that O_{2}(P_{1}) = CQ_{1}. In particular, O_{2}(P_{1}) acts as a D_{8} and so self–dually on U_{γ} + U_{γ′}.

Back to the general case. Recall (ab) and put Y_{2} = ⨁_{δ∈△^{2}(α)\△^{1}(β)} U_{δ}. Let g ∈ L_{1} \ L_{2}. Then P_{2}^{g} ∩S = O_{2}(P_{1}). Further (U_{γ} +U_{γ′})^{g} is a Wedderburn component for O_{2}(P_{1}) on Y_{2} (and is self–dual if (+) holds). It follows that the S−module Y_{2} is isomorphic to the S−module induced from the O_{2}(P_{1})−module (U_{γ} + U_{γ′})^{g} and is irreducible (and is self dual if (+) holds).

Pick δ ∈ △^{4}(α). By [5, 7.4,7.5], Z_{1} ≤ L_{δ} and Z_{1} ≰ Q_{δ}. So Z_{1} inverts U_{δ} and U_{δ} ≤ [W, Z_{1}]. As seen above Q_{1} acts irreducibly on R_{1} = [W, Z_{1}]. Further, Q_{1} acts transitively on the subset △^{3}(β) \ △^{2}(α) of △^{4}(α), dim[W, Z_{1}] = 16 and | △^{3} (β) \ △^{2}(α)| = 16. So

R_{1} = [W, Z_{1}] = ⨁_{δ∈△^{3}(β)\△^{2}(α)} U_{δ} = ∑_{δ∈△^{4}(α)} U_{δ}.

Let R_{2} = [W, Z_{2}]. From its definition, Y_{2} is 8−dimensional and from our earlier calculation of dim[W, Z_{2}], the space C_{[W,Z_{2}]}(Z_{1}) is 8−dimensional. Hence Y_{2} = [C_{W} (Z_{1}), Z_{2}] = C_{[W,Z_{2}]}(Z_{1}). So by coprime action, R_{2} = ⨁_{δ∈△^{1}(β)} C_{R_{2}}(Z_{d}) = Y_{2} ⊕ Y_{2}^{d} ⊕ Y_{2}^{d^{2}}, where d is an element of order 3 in L_{2}. It follows that

R_{2} = ⨁_{δ∈△^{3}(β)} U_{δ}.

Moreover, as Y_{2} is irreducible (and self-dual if (+) holds) as an S−module, R_{2} is irreducible (and self-dual if (+) holds) as a P_{2}−module. In particular, (bb) and (bd) hold.

So to complete the proof of (a) and (b), it remains to show that R_{1} is self-dual as a P_{1}−module if (+) holds. Pick H ∈ Σ. Then O^{2} (N_{P_{1}}(H)) centralizes C_{R_{1}}(H). As R_{2} is self-dual as a P_{2}−module, S acts self-dually on R_{2} and on R_{1}. This implies that N_{P_{1}}(H) = SO^{2}(N_{P_{1}}(H)) acts self-dually on C_{R_{1}}(H). Since R_{1} = ⨁_{H∈Σ} C_{R_{1}}(H), the group P_{1} is self-dual on R_{1} and (a) and (b) are proved.

To prove (c) let H ∈ Σ. Then Y (respectively X) is induced from the 2 (1)-dimensional N_{P_{1}}(H) module C_{Y} (H), (C_{X}(H)). As Q_{1} acts transitively on Σ, we have N_{P_{1}}(H)Q_{1} = P_{1}. Thus N_{P_{1}}(H)/O_{2}(N_{P_{1}}(H)) ≅ F rob_{20} and N_{P_{1}}(H) is generated by its 2-elements. Since C_{X}(H) and C_{Y} (H)/C_{X}(H) are isomorphic as N_{P_{1}}(H)−modules, all the 2−elements and so all the elements of N_{P_{1}}(H) act as scalars on C_{Y} (H). Hence C_{Y} (H) splits over C_{X}(H) and so also Y splits over X.

That F′ has two classes of involutions is well known (see for example [3]). Clearly C_{W} (z_{1}) is 11-dimensional. We can choose i ∈ V_{1}. Namely, choose i ∈ V_{1} but i ∉ Z_{δ} for δ ∈ △^{1}(α). Then i centralizes U_{1}, moreover i lies in exactly three elements of Σ_{2} and in eight elements of Σ. Hence dim C_{W} (i) = 1 + 2 · 3 + 8 = 15 and so (d) holds.

(e) and (f) are well known and are easily deduced from [3]. QED Lemma 2.4 Let K = D_{8} ≀ Sym(5) and let t be an automorphism of order 2 which centralizes K/O_{2}(K) but not O_{2}(K)/Z(O_{2}(K)). Let A be a non-abelian subgroup of O_{2}(O^{2}(K)) such that |N_{K}(A)| is divisible by five and Z(K) ≰ A. Then |A| = 2^{8}, N_{K⟨t⟩}(A)/A ∼ D_{8}.F rob_{20} and N_{K⟨t⟩}(A) induces an outer automorphism on O_{2}(N_{K⟨t⟩}(A))/A.

Proof: Let x be an element of order five in K normalizing A. Put X = ⟨x⟩, I = O_{2}(O^{2}(K)) and J = [Z(I), K]. Then I/Z(K) has order 2^{12}, the group X acts fixed point freely on I/Z(K), also |J| = 2^{4} and Z(I) = I′ = Z(K)J. As A is not abelian and Z(K) ≰ A, A′ = J = Z(A). Also AZ(I) ≠ I since otherwise Z(K) ≤ I′ ≤ A′ ≤ A. Hence |A/J| = 2^{4} and |A| = 2^{8}. Let A^{∗} = AZ(I) and note that Φ(A^{∗}) = Φ(A) = J. Let D_{1}, …, D_{5} be the five D′_{8}s that are naturally permuted by K, ordered so that D_{i+1} = D_{i}^{x}. Let D_{1} = ⟨a_{1}^{∗}, b_{1}^{∗}⟩, with a_{1}^{∗} and b_{1}^{∗} of order two. Inductively define a_{i+1}^{∗} = a_{i}^{x} and b_{i+1}^{∗} = b_{i}^{x}. Let z_{i}^{∗} = [a_{i}^{∗}, b_{i}^{∗}] ∈ Z(D_{i}). For c ∈ {a, b, z} let c_{i} be the product of the c_{j}^{∗} with 1 ≤ j ≤ 5 and j ≠ i. If a_{1} ∈ A^{∗} then A^{∗} ≤ ⟨a_{1}^{X}⟩Z(I) = ⟨a_{1}, … a_{5}⟩Z(I) and A^{∗} is abelian, a contradiction. Similarly b_{1} ∉ A^{∗} and a_{1}b_{1} ∉ A^{∗}. Thus A^{∗} has an element s of the form a_{1}b_{i} with 2 ≤ i ≤ 5 or of the form a_{1}b_{i}b_{j} with 1 ≤ i < j ≤ 5. Note that (a_{1}b_{2})^{2} = [a_{1}, b_{2}] = z_{3}^{∗}z_{4}^{∗}z_{5}^{∗} ∉ J and so the case s = a_{1}b_{2}, and more generally the case s = a_{1}b_{i}, is impossible. Note also that (a_{1}b_{1})^{2} = z_{1} ∈ J. So we get i ≠ 1 in the second case. Suppose that i = 2 and j = 3. Then s^{x} = a_{2}b_{3}b_{4} and [s, s^{x}] = [a_{1}, b_{2}b_{3}][b_{2}, a_{2}][b_{3}, a_{2}]. Since the first two of these factors are in J but the last one is not, this case is impossible. Similarly, the cases (i, j) = (2, 4), (3, 5) and (4, 5) are ruled out. Thus (i, j) = (2, 5) or (3, 4). Since [A^{∗}, X] = A, there exist exactly two choices for A for a given X. Note that an element in the normalizer of X which acts as (2354) on D_{1}, …, D_{5} interchanges these two choices. Let L = K⟨t⟩. Then |N_{L}(X)/N_{L}(X) ∩ N_{L}(A)| = 2. Let D_{0} = C_{O_{2}(K)}(X). Then [D_{0}, I] ≤ J and so D_{0} ≤ N_{K}(A) and N_{O_{2}(K)}(A) = D_{0}A. Since O^{2}(K) does not normalize A^{∗} we conclude that N_{L}(A) ≤ N_{L}(X)O_{2}(K) and we have to decide whether N_{L}(A)/D_{0}A is isomorphic to C_{2} × D_{10} or to F rob_{20}. In the first case we may assume that t normalizes A. As X acts irreducibly on A/J we conclude that [A, t] ≤ J and so A^{∗}/Z(I) = C_{I/Z(I)}(t). But then K normalizes A^{∗}/Z(I), a contradiction. The very last statement follows as N_{K⟨t⟩}(A) ≰ K. QED

Lemma 2.5 (a) E_{6}(5) has two conjugacy classes of involutions. There exist representatives r and s of these classes such that

C_{E_{6}(5)}(r) ∼ 4.D_{5}(5).4 and C_{E_{6}(5)}(s) ∼ 2.(L_{2}(5) × L_{6}(5)).2.

Moreover, if U is a 27-dimensional GF (5)−module for E_{6}(5), then C_{U}(r) is 11-dimensional and C_{U}(s) is 15-dimensional. As a C_{E_{6}(5)}(r)−module U is the direct sum of three irreducible submodules of dimensions 1, 10 and 16. The kernel of the action of C_{E_{6}(5)}(r) on the invariant 1-space is C_{E_{6}(5)}(r)′.

(b) 2.E_{7}(5) has three conjugacy classes of involutions. There exist representatives z, z_{0} and i of these classes such that z_{0} ∈ Z(2.E_{7}(5)), i = z · z_{0} and

C_{2.E_{7}(5)}(z) ∼ 2^{2}.(L_{2}(5) × D_{6}(5)).2

Moreover, C_{V} (z) is a tensor product of natural modules for SL_{2}(5) and Ω_{12}^{+}(5), and [V, z] is a half-spin module for 2.D_{6}(5).

Proof: The conjugacy classes of involutions and their centralizers are well known and easily deduced, for example by the methods found in [2]. The information about U and V is easily computed using the subgroup 5^{27}E_{6}(5) of E_{7}(5), the subgroup 5^{1+56}2E_{7}(5) of E_{8}(5), the Steinberg relations and the weight theory of modules for groups of Lie type. QED

Recall that E is a parabolic subgroup QL ∼ 5^{27}.(4 × E_{6}(5)) in G. Let X_{1} = C_{V} (Q), X_{3} = [V, Q] and X_{2} = [X_{3}, Q]. Then

0 < X_{1} < X_{2} < X_{3} < V

is the unique chief series for E on V, moreover E = N_{G}(X_{1}), the modules X_{1} and V/X_{3} are 1-dimensional and X_{2} and X_{3}/X_{2} are 27−dimensional mutually dual E−modules.

By assumption L′ contains a subgroup ^{2}F_{4}(2) acting irreducibly on Q. Hence L contains a subgroup F̃ fulfilling (+), and acting irreducibly on Q.

Define S̃, P̃_{1} and P̃_{2} (as subgroups of F̃) in a way analogous to the above. The reader should notice that Z(F̃) = Z(G) centralizes Q, thus Q is irreducible, but not faithful as an F̃−module. But we still can apply Lemma 2.3 to F̃/Z(G) and conclude that P̃_{1} normalizes a non-trivial subgroup ⟨f_{−}⟩ of Q. Let f_{+} ∈ P̃_{1} be of order 5 and r ∈ P̃_{1} ∩ F̃′ of order 4 with f_{+}^{r} = f_{+}^{2}. As seen in the proof of Lemma 2.3, ⟨r⟩ acts faithfully on ⟨f_{−}⟩. So either f_{−}^{r} = f_{−}^{2} or f_{−}^{r} = f_{−}^{3}. Note that there exists σ in N_{G}(L) inducing a graph automorphism on L (indeed such a σ can be chosen to invert the Cartan subgroup of G). Then the action of L on Q^{σ} is dual to the action of L on Q and replacing F̃ and f by F̃^{σ} and f^{σ} if necessary, we may assume that f_{−}^{r} = f_{−}^{2}. Put f = f_{+} f_{−}. Then f^{r} = f^{2}. Let u be an element of order 4 in P̃_{1} centralized by f_{+}. Let Z_{1} = Z(P̃_{1} ∩ F̃′).

Lemma 2.6 (a) The element u centralizes f_{−}.

(b) f_{−} is a root element of G and C_{G}(f_{−}) ∼ 5^{1+32}.2^{2}.D_{6}(5). The C_{G}(f_{−})−module [V, f_{−}] is a natural orthogonal 12−space for Ω_{12}^{+}(5) and the subspace X_{1} is a singular 1-space. The quotient, ([V, f_{−}] + X_{2})/X_{2} is 1−dimensional and [V, f_{−}] ∩ X_{2} = X_{1} ⊕ [C_{X_{2}}(Z_{1}), Q_{1}].

(c) C_{G}(f_{−}) ∩ C_{G}(Z_{1}) ∼ ⟨f_{−}⟩ × 2^{2}.D_{6}(5)

(d) C_{V} (f_{−}) ≤ [X_{3}, Z_{1}] + [V, f_{−}] + X_{2}

Proof: (a): Pick s ∈ Z(L) and e ∈ L′ with u = se. Note that e ∈ F̃ and so e^{2} ∈ F̃′ = F′ and e^{2} ∈ Q_{1}. In particular, u^{2}F′ = s^{2}F′. Since F/F′ ≅ C_{4} and F = ⟨u⟩Z(F)F′ we get s^{2} ≠ 1 and |s| = 4. Since s inverts Q, it is enough to show that e inverts f_{−}. Let D = C_{L′}(Z_{1}). By Lemma 2.3, C_{Q}(Z_{1}) has order 5^{11} and ⟨f_{−}⟩ is the unique cyclic subgroup of Q normalized by P̃_{1} and so by Lemma 2.5a, D has shape 4.D_{5}(5).4, moreover D normalizes ⟨f_{−}⟩ and C_{D}(f_{−}) = D′. As a D−module, Q is the direct sum of irreducible modules of dimensions 1, 10 and 16. Using Lemma 2.3ab we conclude, by the action of P̃_{1} on the 10-dimensional space, that P̃_{1} is contained in a subgroup Y of D, such that Y/Z_{1} is isomorphic to the subgroup of index 4 in D_{8} ≀ Sym(5). In addition, C_{Y} (f_{+}) ∩ D′ = Z_{1}, and so su ∉ D′. Thus e does not centralize f_{−}. As e^{2} ∈ Q_{1}, the element e^{2} does centralize f_{−} and so e inverts f_{−}. Thus (a) holds.

(b): As D lies in a parabolic P (of shape 5^{16}D in E_{6}(5)) that fixes a 1−space in Q, the group ⟨f_{−}⟩ is normalized by the parabolic QZ(L)P of G and thus is a root group. Hence also the second, third and fourth statements in (b) hold. As Q centralizes V/X_{3} and X_{3}/X_{2} but not V/X_{2} and as V/X_{3} is one dimensional, ([V, f_{−}] + X_{2})/X_{2} is 1-dimensional. Now [X_{2}, f_{−}] = X_{1} and ([X_{2}, Q_{1}] + X_{1})/X_{1} is the unique 10-dimensional subspace of X_{2}/X_{1} invariant under P_{1} and thus also the last statement in (b) holds.

(c): By (b), Z_{1} centralizes [V, f_{−}] and so (c) holds.

(d): As Z_{1} centralizes [V, f_{−}], the element f_{−} centralizes [V, Z_{−}]. Now both C_{V} (f_{−})+X_{2}/X_{2} and [X_{3}, Z_{1}] + [V, f_{−}] + X_{2}/X_{2} are 17-dimensional and (d) holds. QED

Since S̃ = O_{2}(O^{2}(P̃_{1}))⟨u⟩⟨r⟩, the group S̃ normalizes O^{2}(P̃_{1})⟨f⟩. Put S = S̃, P_{1} = ⟨S, f⟩, P_{2} = P̃_{2} and F = ⟨P_{1}, P_{2}⟩.

Lemma 2.7 (a) F normalizes a complement M_{0} in X_{3} to X_{2}. Put M = M_{0} + X_{1}. Then N_{E}(M) ∩ Q = 1. In particular, F ∩ Q = 1 and F fulfills (+).

(b) P_{1} normalizes exactly two 1-spaces in V namely X_{1} and U_{1}, where U_{1} ≤ M_{0}. Moreover, C_{V} (O^{2}(P_{1})) = X_{1} + U_{1}.

(c) [V, f_{−}] = X_{1} + U_{1} + [C_{X_{2}}(Z_{1}), Q_{1}]

Proof: Put U_{1} = [C_{V} (Q_{1}), f, r]. Note that Q_{1} centralizes a 1-space in each of the modules X_{i}/X_{i−1}, and C_{V} (Q_{1}) is 4-dimensional. As [V, Q] ≰ X_{2}, [V, f_{−}] ≰ X_{2}. Since [X_{3}, f_{−}] ≤ X_{2} and V = C_{V} (Q_{1})X_{3}, we conclude [C_{V} (Q_{1}), f_{−}] ≰ X_{2}. On the other hand as L acts completely reducibly on V, we have [C_{V} (Q_{1}), f_{+}] = 1. Thus [C_{V} (Q_{1}), f] ≰ X_{2}. By Lemma 2.3ad applied to F′ and the modules X_{3}/X_{2} and X_{2}/X_{1}, the element r^{2} inverts C_{X_{3}/X_{1}}(Q_{1}). In particular, [r, f] centralizes X_{3} and, as r inverts f and f has odd order, we conclude that f centralizes C_{X_{3}/X_{1}}(Q_{1}). It follows that ([C_{V} (Q_{1}), f] + X_{1})/X_{1} is 1−dimensional and not contained in X_{2}/X_{1}. Hence U_{1} is 1−dimensional, U_{1} ≤ X_{3}, U_{1} ≰ X_{2} and P_{1} normalizes U_{1}.

Let U_{2} = ⟨U_{1}^{P_{2}}⟩. As |P_{2}/S| = 3, the space U_{2} is at most 3-dimensional. By Lemma 2.3 applied to W = X_{3}/X_{2}, we know that (U_{2} + X_{2})/X_{2} is 3-dimensional. Thus U_{2} is a 3-space and U_{2} ∩ X_{2} = 1. Let U_{3} = ⟨[U_{2}, Q_{1}]^{P_{1}}⟩. Since [U_{2}, Q_{1}] is a 2-space and |P_{1}/S| = 5, we similarly have that U_{3} is 10-dimensional and U_{3} ∩X_{2} = 1. Let U_{4} = ⟨[U_{3}, Z_{2}]^{P_{2}}⟩. Since [U_{3}, Z_{2}] is eight dimensional, U_{4} is 24-dimensional and U_{4} ∩ X_{2} = 1. Let U_{5} = [U_{4}, Z_{1}]. Since X_{3}/X_{2} and X_{2}/X_{1} are dual as E−modules, Lemma 2.3af implies that [V, Z_{1}]X_{2}/X_{2} and [X_{2}, Z_{1}] are isomorphic and absolutely irreducible as P_{1}−modules and as S−modules. (Note here that Z_{1} is trivial on X_{1} and V/X_{3}.) By Lemma 2.3c, [V, Z_{1}] splits over [X_{2}, Z_{1}] as a P_{1}−module. It is now easy to see that every S−submodule of [V, Z_{1}] is invariant under P_{1}. In particular P_{1} normalizes U_{5}. Put M_{0} = U_{2} + U_{4}. Note that by Lemma 2.3, (M_{0} + X_{2})/X_{2} = X_{3}/X_{2} and that M_{0} has dimension at most 3 + 24 = 27. Thus M_{0} ∩ X_{2} = 0 and M_{0} is a complement to X_{2} in X_{3}. Then

M_{0} = U_{1} + (U_{2} ∩ U_{3}) + (U_{3} ∩ U_{4}) + U_{5} = U_{1} + U_{3} + U_{5}.

and so F = ⟨P_{1}, P_{2}⟩ normalizes M_{0}.

It follows that [X_{3}, N_{E}(M) ∩ Q] ≤ M ∩ X_{2} = X_{1}. Since Q does not centralize X_{3}/X_{1} and E is irreducible on Q, we have N_{E}(M) ∩ Q = 1. Since F Q = F̃Q, the last statement of (a) also holds.

As f_{+} centralizes C_{V} (Q_{1}) the last statement in (b) holds once we prove that C_{C_{V} (Q_{1})}(f_{−}) = X_{1} + U_{1}. As seen above [C_{V} (Q_{1}), f_{−}] ≰ X_{2} and by a dual argument [C_{V} (Q_{1}), f_{−}] ≠ 0. Hence C_{C_{V} (Q_{1})}(f_{−}) is at most 2-dimensional and so C_{C_{V} (Q_{1})}(f_{−}) = X_{1} + U_{1}. As r centralizes X_{1} but not U_{1}, the P_{1}-modules X_{1} and U_{1} are not isomorphic. Thus (b) holds.

As seen above U_{1} ≤ [V, f_{−}]. Thus (c) follows from Lemma 2.6b. QED

Let χ_{0} = X_{1}^{G} and ζ_{0} = Z_{1}^{G}.

Lemma 2.8 U_{1} ∈ χ_{0} and C_{M}(Z_{1}) = ⨁ {X | X ∈ χ_{0}, X ≤ C_{M}(Z_{1})}.

Proof: We claim that E acts transitively on {Z ∈ ζ_{0} | [X_{1}, Z] = 1}. Indeed, for any such Z, we have dim[V, Z] = 32 = dim[X_{3}/X_{1}, Z] = 2 · dim[X_{2}/X_{1}, Z]. Thus the claim follows from Lemma 2.5a. By this claim, G acts transitively on {(Z, X) ∈ ζ_{0} × χ_{0} | [X, Z] = 1} and thus C_{G}(Z_{1}) acts transitively on {X ∈ χ_{0} | [X, Z_{1}] = 1}. By Lemma 2.5b, C_{G}(Z_{1}) ∼ 2^{2}.(L_{2}(5) × D_{6}(5)).2 and C_{V} (Z_{1}) is the tensor product of natural modules for SL_{2}(5) and Ω_{12}^{+}(5). By Lemma 2.6c, f_{−} lies in the factor SL_{2}(5) and so [V, f_{−}] is one of six 12-spaces invariant under 2^{2}.D_{6}(5). By Lemma 2.6b, X_{1} is a singular point in [V, f_{−}] and it follows that {X ∈ χ_{0} | [X, Z_{1}] = 1} is precisely the union of the sets of singular points in the six 12-spaces. By Lemma 2.7c, X_{1} +U_{1} = C_{[V,f_{−}]}(Q_{1}) and so X_{1} +U_{1} is a non-degenerate subspace of [V, f_{−}]. As X_{1} is singular in [V, f_{−}], the space X_{1} + U_{1} is of “+”-type and so contains exactly one member of χ_{0} distinct from X_{1}. By Lemma 2.5c, U_{1} is the only 1−space in X_{1} + U_{1}, which is invariant under P_{1} and distinct from X_{1}. So U_{1} ∈ χ_{0}. By the proof of Lemma 2.3ab, [C_{M_{0}}(Z_{1}), V_{1}] is the direct sum of 5 pairwise non-isomorphic irreducible Q_{1}−modules, each of which is the direct sum of two conjugates of U_{1} under F. Moreover, Q_{1} normalizes each of the six 12-spaces and it is now easy to see that C_{M}(Z_{1}) intersects each of the six 12-spaces in a 2-space of “+”- type. This clearly implies the lemma. QED

Lemma 2.9 N_{E}(M) ∩ C_{G}(Z_{1}) = P_{1} and F contains a Sylow 2-subgroup of N_{E}(M).

Proof: Put N = N_{E}(M) and Y = C_{N}(Z_{1}). By Lemma 2.7, N ∩ Q = 1. By Lemma 2.5a,

C_{E}(Z_{1}) ∼ 5^{11}.(C_{4} × 4.D_{5}(5).4)

and C_{E}(Z_{1}) normalizes a 10-space in X_{3}/X_{2}. By the proof of Lemma 2.8, Y normalizes X_{1}+U_{1} and a decomposition of this 10-space into an orthogonal sum of five 2-spaces of “+”-type. Let K be the full normalizer in C_{E}(Z_{1}) of this decomposition. Then

K ∼ 5^{11}(C_{4} × 2.\frac{1}{2} (D_{8} ≀ Sym(5)).2).

Let M_{1} be the complement in X_{3} to X_{2} normalized by L. We claim that U_{3} ≰ M_{1} and U_{5} ≰ M_{1}. As U_{1} is S−invariant and S ≤ L, we have U_{1} ≤ M_{1}. Since P_{2} ≤ L we get U_{2} ≤ M_{1}. Suppose that U_{3} ≤ M_{1}. Then P̃_{1} normalizes U_{3} and since f_{−} ∈ P_{1}P̃_{1} we conclude that f_{−} normalizes U_{3}. Note that f_{−} centralizes X_{3}/X_{2} and U_{3} ≤ X_{3}. Hence [U_{3}, f_{−}] ≤ X_{2} ∩ U_{2} = 0 and so f_{−} centralizes U_{3}, a contradiction to Lemma 2.6d. Thus U_{3} is not contained in M_{1}. It follows that neither U_{3} ∩ U_{4} nor U_{4} are contained in M_{1}. As P_{2} ≤ L acts irreducibly on U_{4} we conclude that U_{4} ∩ M_{1} = 0 and so finally U_{5} ≰ M_{1}.

As Y ∩Q = 1, Y is isomorphic to a subgroup of K/O_{3}(K) and so Y/O_{2}(Y) is isomorphic to a section of Sym(5). As 5 divides the order of P_{1} we conclude that O_{2}(P_{1}O_{2}(Y)/O_{2}(Y)) = 1 and so O_{2}(P_{1}) ≤ O_{2}(Y). As C_{Q}(O_{2}(P_{1})) = ⟨f_{−}⟩, any conjugate of L in E containing O_{2}(P_{1}) is of the form L^{x} for some x ∈ ⟨f_{−}⟩. By Sylow’s theorem O_{2}(Y) lies in some conjugate of L in E and so O_{2}(Y) ≤ L^{x} for some x ∈ ⟨f_{−}⟩. Thus O_{2}(Y) normalizes [M_{1}, Z_{1}]^{x}. By Lemma 2.6b, x centralizes [M_{1}, Z_{1}] and so O_{2}(Y) normalizes [M_{1}, Z_{1}], [X_{2}, Z_{1}] and U_{5} = [M, Z_{1}]. Since all of these Y −modules are irreducible and U_{5} ≤ [V, Z_{1}] = [X_{2}, Z_{1}] ⊕ [M_{1}, Z_{1}] we conclude that these O_{2}(Y)−modules are pairwise isomorphic and so O_{2}(Y) acts self-dually on [X_{2}, Z_{1}].

For each X ≤ E, we write X^{∗} for C_{X}(X_{1}). Since Z(L) acts transitively on X_{1}^{#}, the groups F̃, F and P_{1} also act transitively on X_{1}^{#}. Thus Y = Y^{∗}P_{1}. Let C (respectively Ĉ) be the largest subgroup of K^{∗} centralizing (centralizing or inverting) (U_{3}+X_{2})/X_{2}. Then ĈQ/Q ≅ C_{4}. Since O_{2}(Y) acts self-dually on [X_{2}, Z_{1}] and the elements of Ĉ act as scalars we conclude that Ĉ ∩ Y centralizes or inverts [X_{2}, Z_{1}]. Hence Ĉ ∩ Y^{∗} ≤ C and Y^{∗}C ∩ Ĉ = (Y^{∗} ∩ Ĉ)C = C. Thus no element of Y^{∗}C/C inverts (U_{3} + X_{2})/X_{2}. We are now in a position to apply Lemma 2.4 (with K⟨t⟩ = K^{∗}/C and A = O_{2}(Y^{∗})C/C ∩ O_{2}(O^{2}(K^{∗}/C)). Note that A is not abelian as Q_{1}C/C ≤ A and Q_{1} ∩ C = Z_{1}. As r ∈ P_{1}^{∗}, the group P_{1}^{∗} has F rob_{20} as a quotient. Since Y^{∗}C normalizes A we conclude that Y^{∗}C ≤ O_{2}(Y^{∗})P_{1}^{∗} and A = Q_{1}C/C. From the structure of D_{8}, from the last two statements in Lemma 2.4 and from O_{2}(Y^{∗}) ∩ Ĉ ≤ C we conclude that O_{2}(Y^{∗})C = Q_{1}C. Thus Y^{∗}C = P_{1}^{∗}C. Since C ∩ Y^{∗} ∩ Q = 1, we have C ∩ Y^{∗} = Z_{1} ≤ P_{1}^{∗}. So Y^{∗} = P_{1}^{∗} and Y = P_{1}, proving the first statement of the lemma.

Since Z_{1} = Z(S) ∩ S′, N_{G}(S) ≤ C_{G}(Z_{1}) and so S a Sylow 2-subgroup of N. QED.

Lemma 2.10 There exists t ∈ N_{G}(P_{1}) ∩ N_{G}(M) with X_{1} ≠ X_{1}^{t}.

Proof: Let t ∈ N_{G}(P_{1}). By Lemma 2.7b, C_{V} (O^{2}(P_{1})) = U_{1} + X_{1} and so t normalizes U_{1} + X_{1}. Also by Lemma 2.3ae, the isomorphism types of the two 10−dimensional modules in V invariant under P_{1} are not conjugate under an automorphism of P_{1}. So t normalizes U_{3}. Hence t ∈ N_{G}(M) if and only if t ∈ N_{G}(U_{5}).

Note that P_{1} is contained in ⟨f_{−}⟩ C_{L}(Z_{1}). Let ⟨t_{1}⟩ = Z(C_{L′}(Z_{1})). Then by Lemma 2.5a, t_{1} is of order 4 and [t_{1}, P_{1}] ≤ Z_{1}. So t_{1} ∈ N_{G}(P_{1}). Further C_{G}(Z_{1}) is of shape 2^{2}.(L_{2}(q)×D_{6}(5)).2. Note that [V, f_{−}] is 12-dimensional and there exists a quadratic form on [V, f_{−}] invariant (up to scalar multiplication) under the action of N_{G}(⟨f_{−}⟩). Moreover, [V, f_{−}] is equal to the sum of X_{1}, U_{1} and the 10-dimensional subspace of X_{2}, which is normalized by P_{1}. In particular P_{1} normalizes a decomposition of [V, f_{−}] into an orthogonal sum of six 2-dimensional subspaces of “+”-type. Let T_{0} be the largest subgroup of 2^{2}.D_{6}(5), which normalizes this decomposition and normalizes X_{1} + U_{1}. Then P_{1} normalizes T_{0} and T_{0}/Z_{1} is isomorphic to a subgroup of index 4 in (D_{8} × D_{8} ≀ Σ_{5}).

Let T = C_{T_{0}}(f). Then T/Z_{1} ≅ D_{8} × C_{2} and T normalizes O^{2}(P_{1}) and so also Q_{1} and V_{1} = Q′_{1}. Moreover, t_{1} ∈ T, [t_{1}, T] ≤ Z_{1}, T′ = Z(G)Z_{1} and T ∩ P_{1}/Z_{1} ≅ C_{4}. Pick t_{2} ∈ T with t_{2}^{2} ∈ Z_{1} and [t_{2}, T] ≰ Z_{1}. We will show that either t_{2} or t_{1}t_{2} fulfills the conclusion of the lemma. It is easy to check in T_{0} that t_{2} centralizes a 1-space in each of the six 2-spaces and that these six 1-spaces form a 6-space of “+”-type. As the central involution of Ω_{6}^{+}(5) lifts to an element of order four in 2.Ω_{6}^{+}(5) ≅ SL_{4}(5) we conclude that t_{2} and by symmetry t_{1}t_{2} are elements of order 4. It follows that

⟨t_{1}, t_{2}⟩ ≅ Q_{8} and ⟨t_{1}, t_{2}⟩ ∩ P_{1} = Z_{1}.

We will now examine the action of ⟨t_{1}, t_{2}⟩ on the set Π of 16-spaces in V invariant under O^{2}(P_{1}). Note that |Π| = 5 + 1 = 6. Put H = [V_{1}, f]. Then ⟨t_{1}, t_{2}⟩ normalizes H and acts faithfully on the 2-dimensional space C_{[V,Z_{1}]}(H). So the orbits of ⟨t_{1}, t_{2}⟩ on the 1-dimensional subspaces of C_{[V,Z_{1}]}(H), and hence also on Π, are all of length 2. Now by Lemma 2.9, t_{1} does not normalize M and so [M, Z_{1}] ≠ [M, Z_{1}]^{t_{1}}. Hence [M, Z_{1}] = [M, Z_{1}]^{t} for t = t_{2} or for t = t_{1}t_{2}. Recall that U_{5} = [M, Z_{1}] and thus the selected t is in N_{G}(M).

It remains to show that t normalizes P_{1}. Since t normalizes O^{2}(P_{1}), it is enough to prove that [t_{1}, N_{S}(⟨f⟩)] ≤ P_{1}. Since T = ⟨t_{1}, t_{2}⟩ (T ∩ P_{1}), T ∩ P_{1} is the largest subgroup of T acting trivially on Π. As N_{S}(⟨f⟩) acts trivially on Π, the same is true for [t_{1}, N_{S}(⟨f⟩)] and so [t_{1}, N_{S}(⟨f⟩)] ≤ T ∩ P_{1} ≤ P_{1}, completing the proof of the Lemma. QED

Lemma 2.11 Let R = ⟨F, F^{t}⟩. Then F has three orbits on the right cosets of F in R. The orbit stabilizers are F, P_{1} and a group of order 2^{5}·3·5^{2}·13. In particular |R| = 2^{15}·3^{3}·5^{3}·7·13·29.

Proof: Note first that R normalizes M. We will divide the proof into several steps.

1 Let χ = X_{1}^{R} and ζ = z_{1}^{R}. Then R acts transitively on (a) {(X, z) | X ∈ χ, z ∈ ζ and z centralizes X},

(b) {(X, z) | X ∈ χ, z ∈ ζ and z inverts X}.

Clearly a Sylow 2-subgroup of N_{R}(X_{1}) contains representatives of each class of involutions in N_{R}(X_{1}). By Lemma 2.9, S is Sylow 2−subgroup of N_{R}(X_{1}) and so F contains representatives of each class of involutions in N_{R}(X_{1}). As F/F′ ≅ C_{4}, all the involutions in F are contained in Z(G) × F′. By Lemma 2.3, F′ has two classes of involutions with representatives z_{1} and i. Moreover C_{W} (i) is 15-dimensional and so dim C_{V} (i) = 2 + 2 · 15 = 32. Hence F has two orbits on F ∩ ζ with representatives z_{1} and z_{0}i, where z_{0} is the central involution in G. Now z_{1} centralizes X_{1} and z_{0}i inverts X_{1}. Thus (1) holds.

2 (a) C_{R}(z_{1}) acts transitively on C_{M}(z_{1}) ∩ χ and |C_{M}(z_{1}) ∩ χ| = 12, (b) C_{R}(z_{1}) and P_{1} act transitively on [M, z_{1}] ∩ χ and |[M, z_{1}] ∩ χ| = 80,

(c) |C_{R}(z_{1})| = 2^{15} · 3 · 5,

(d) C_{N_{R}(X_{1})}(i) ≤ F.

The two transitivity statements for C_{R}(z_{1}) follow from (1). By Lemma 2.8, |C_{M}(z_{1}) ∩ χ_{0}| = 12. Moreover, by Lemma 2.3, U_{3} contains 10 elements of U_{1}^{F} and so C_{M}(z_{1}) ∩ χ contains at least the 11 elements of C_{M}(z_{1}) ∩ χ_{0} \ {X_{1}}. Conjugation by t shows that C_{M}(z_{1}) ∩ χ also contains the 11 elements of C_{M}(z_{1})∩χ_{0} \{U_{1}}. Thus C_{M}(z_{1})∩χ = C_{M}(z_{1})∩χ_{0} and (a) holds. By Lemma 2.9, C_{R}(z_{1}) ∩ N_{R}(X_{1}) = P_{1} and so

|C_{R}(z_{1})| = 12 · |C_{R}(z_{1}) ∩ N_{R}(X_{1})| = 12 · |P_{1}| = 2^{15} · 3 · 5.

Now |[M, z_{1}] ∩ χ| = |C_{R}(z_{1})| / |C_{R}(z_{1}) ∩ N_{R}(X)|, where X ∈ χ is inverted by z_{1}. By Lemma 2.3d, |C_{F} (i)| = 2^{11} · 3. Further, |C_{R}(z_{1}) ∩ N_{R}(X)| ≥ |C_{F} (i)| and so |[M, z_{1}] ∩ χ| ≤ 80. Finally, P_{1} has an orbit of length 80 on [M, z_{1}]∩χ. Indeed, there are 80 points at distance 4 from a in Γ_{0} (the generalized octagon associated to F), these 80 points correspond to 80 elements in U_{0}^{F} and U_{0}^{F} is a subset of χ. So |[M, z_{1}] ∩ χ| = 80, moreover |C_{R}(z_{1}) ∩ N_{R}(X)| = |C_{F} (i)| and P_{1} acts transitively on [M, z_{1}] ∩ χ. This completes the proof of (2).

3 N_{R}(X_{1}) = F = N_{R}(F). In particular, the actions of R on X_{1}^{R} and on R/F are isomorphic.

Let N = C_{R}(X_{1}). Then, as in (1), N has two classes of involutions, with representatives z_{1} and i. By (2)(d) and Lemma 2.8, F′ contains the centralizers of z_{1} and i in N. Hence by a standard argument, see for example [6, 9.2.1], F′ = N. Thus N_{R}(X_{1}) = F. Now X_{1} is the unique 1-space in M normalized by F and so N_{R}(F) ≤ N_{R}(X_{1}).

Let Γ be the graph with vertices χ and edges {X_{1}, U_{1}}^{R}. Since R = ⟨F, F^{t}⟩ we have

4 Γ is connected.

5 Let a, b ∈ χ. Suppose there exists z ∈ ζ, such that z normalizes a and b. Then a and b have distance at most 2 in Γ. If z centralizes a or b, then a and b have distance at most 1.

Suppose first that z centralizes a. Then we may assume without loss that a = X_{1} and z = z_{1}. Then (2) implies that b = X_{1} or b ∈ U_{1}^{F} and so a and b are at distance at most 1.

In the general case pick c ∈ χ so that z centralizes c. Then a and b are at distance at most 1 from c, and (5) is proved.

For a vertex a, put R_{a} = N_{R}(a), and for an edge {a, b}, let 1 ≠ z(a, b) ∈ Z(R′_{a} ∩ R′_{b}). Note that, if (a, b) = (X_{1}, U_{1})^{g}, then z(a, b) = z_{1}^{g} = z(b, a). For g ∈ F, we identify α^{g} ∈ Γ_{0} with U_{1}^{g} ∈ Γ.

6 (a) R acts transitively on geodesics of length 2 in Γ. Moreover, the stabilizer of a geodesic of length 2 is isomorphic to C_{2} × F rob_{20}

(b) Let d and e be at distance 2 in Γ. Then R_{d} ∩ R_{e} acts transitively on the set of pairs (a, b) such that {a, b} is an edge with z(a, b) ∈ R_{d} ∩ R_{e}. Moreover, R_{a} ∩ R_{b} ∩ R_{d} ∩ R_{e} is isomorphic to C_{2} × C_{4}.

Let a and b be in α^{F}. Suppose that a and b are at distance less than or equal to 6 in Γ_{0}. Then there exists c ∈ α^{F}, such that c is (in Γ_{0}) at distance 2 from a and at distance at most 4 from b. Put z = z(X_{1}, c). Then by Lemma 2.3ab&ac, z centralizes a and normalizes b. Thus (5) implies that either a = b or a is adjacent to b in Γ. Suppose that every pair of elements in α^{F} are adjacent in Γ. Then every pair of elements in α^{F} ∪ {X_{1}} are adjacent. Since Γ is connected, we conclude that α^{F} ∪{X_{1}} is the set of vertices of Γ. Hence |R| = |F |·(|F |+1) and so |R|_{2} = 2^{14}, a contradiction to (2)(c). So there are two elements of α^{F}, that have distance 8 in Γ_{0}, and have distance 2 in Γ. Since P_{1} is transitive on △^{8}(α) and since every geodesic of length two in Γ is conjugate to one with X_{1} as its midpoint, we conclude that R is transitive on geodesics of length 2 in Γ. Moreover, the stabilizer in F of two elements of distance 8 in α^{F} is a C_{2} × F_{20}. Hence (a) is proved.

To prove (b) we assume without loss that a = X_{1} and b = U_{1}. Since z(a, b) normalizes d and e, we get by (5) that d, e ∈ α^{F} and that d and e are at distance less than or equal to 4 from b in Γ_{0}. Since d and e are at distance 8 from each other, b lies on a geodesic from d to e in Γ_{0} and is at distance 4 from both d and e. Now F acts transitively on paths of length 8 in Γ_{0} starting with a vertex in α^{F} and the stabilizer of such a path is a C_{2} × C_{4}. This proves (b).

7 Let d and e be at distance 2 in Γ. Then R_{d} ∩ R_{e} acts transitively on R_{d} ∩ R_{e} ∩ ζ. Moreover, if z ∈ R_{d} ∩ R_{e} ∩ ζ, then R_{d} ∩ R_{e} ∩ C_{R}(z) has order 2^{5}.3 and has a normal Sylow 3−subgroup.

Let z ∈ R_{d} ∩ R_{e} ∩ ζ. By (6), R_{d} ∩ R_{e} acts transitively on R_{d} ∩ R_{e} ∩ ζ and R_{d} ∩ R_{e} ∩ C_{R}(z) acts transitively on all pairs (a, b) such that {a, b} is an edge with z = z(a, b). Put A = R_{a} ∩ R_{b} ∩ R_{d} ∩ R_{e} and B = R_{d} ∩ R_{e} ∩ C_{R}(z). Then |A| = 8. Next we show that there are exactly 12 choices for (a, b). Indeed a is in C_{V} (z) ∩ χ and so by (2)(a), there are exactly twelve choices for a. Moreover, as R_{a} ∩ R_{b} = C_{R_{a}}(z), b is uniquely determine by z and a. It follows that |B| = 8 · 12 = 2^{5} · 3. We claim that A is normal in B. For this let {ā, b̄} be an edge different from {a, b} with z = z(ā, b̄). Again choose notation so that a = X_{1}. Since z = z(ā, b̄), z centralizes ā and b̄ and so ā and b̄ are at distance 2 from b in Γ_{0}. It follows from [5] that A normalizes ā and b̄, and so A = Rā ∩ Rb̄ ∩ R_{d} ∩ R_{e}. Thus A is independent of the choice of (a, b) and therefore normal in B. Let B̃ = C_{B}(d) and Ã = C_{A}(d). Let g ∈ A with |g| = 4. As ⟨g⟩ acts faithfully on the group of order five in R_{a} ∩ R_{d} ∩ R_{e}, g^{2} ≠ z_{0}. As g ∉ Z(G)F′, g^{2} ∉ F′ and so g^{2} ≠ z. Thus g^{2} = z_{0}z. Since z and z_{0} invert d, g^{2} centralizes d. Thus A/Ã is elementary abelian, |A/Ã| = 2 and Ã ≅ C_{4}. Let i be the involution in Ã. Suppose that the Sylow 3−subgroups of B are not normal in B. Since 3 divides B̃ and |B̃/Ã|_{2} ≤ 4, B̃/Ã ≅ A_{4}. Let Y = O_{2}(B̃) and Y^{∗} = [Y, O^{2}(B̃)]. Then Y = Y^{∗}Ã and [Y^{∗}, Ã] = 1. Hence Ã ≤ Z(Y) and Y/Z(Y) is elementary abelian. It follows that |Y′| ≤ 2. Hence either Y is abelian, Y^{∗} ∩ Ã = 1 and Y ≅ C_{4} × C_{2} × C_{2} or Y is not abelian, Y^{∗} ≅ Q_{8} and Y ≅ C_{4} ∘ Q_{8} ≅ C_{4} ∘ D_{8}. In particular, ⟨i⟩ = Φ(Y). Note that Ã = C_{R}(d) ∩ R_{e} ∩ C_{R}(i). We claim that all involutions in C_{R}(d) ∩ R_{e} are conjugate under R_{d} ∩ R_{e}. Indeed let j be any such involution. Since d and e have distance 2, (5) implies that j ∉ ζ. Thus z_{0}j ∈ ζ and by the first part of (7), the conjugacy class of z_{0}j in R_{d} ∩ R_{e} is uniquely determined. Thus the claim holds. Moreover, by the structure of Y there exists an involution j in B̃ different from i. Then by the claim j = i^{h} for some h ∈ R_{d} ∩ R_{e}. Now [Ã, j] = 1 and Ã ≤ C_{R}(d) ∩ R_{e} ∩ C_{R}(j) = B̃^{h}. As B̃^{h}/Y^{h} has order three, Ã ≤ Y^{h}. Hence

i ∈ Φ(Ã) ≤ Φ(Y^{h}) = ⟨i^{h}⟩ = ⟨j⟩

a contradiction, which proves (7).

8 Let a be at distance 2 from X_{1} in Γ. Then |F ∩ R_{a}| = 2^{5} · 3 · 5^{2} · 13 and F ∩ R_{a} has exactly two orbits on the neighbors of X_{1} in Γ.

Let z ∈ F ∩ R_{a} ∩ ζ and D be the Sylow 3−subgroup of C_{F ∩R_{a}}(z). By Lemma 2.3e |N_{F} (D)| = 3^{3} · 2^{5} and so by (7), C_{F ∩R_{a}}(z) contains a Sylow 2-subgroup of N_{F} (D). Put K = F′ ∩ R_{a}. Then by Lemma 2.3, C_{K}(z) ≅ D_{24}. Since F ∩ R_{a} acts transitively on the involutions in K, we conclude that the Sylow 2-subgroups of K are dihedral groups of order 8 and that K has exactly one class of involutions. By (6), |K| is divisible by 5, and since |F′| = 2^{11} · 3^{3} · 5^{2} · 13, we have |K| = 2^{3} · 3^{1+u} · 5^{1+v} · 13^{w}, where u is 0, 1 or 2, and v and w are 0 or 1. Since C_{K}(z) is a maximal subgroup of N_{F′}(D), u is 0 or 2. We claim that K has an orbit on α^{F} with orbit stabilizer C_{2} × F rob_{20} and an orbit with an orbit stabilizer of order 2^{5}. Indeed let b be at distance 1 from X_{1} and a in Γ. Then by (6), F ∩ R_{b} ∩ R_{a} ≅ C_{2} × F rob_{20}. Moreover, any 2-subgroup of F fixes a point in α^{F} and so there exists c in α^{F} so that 2^{5} divides |F ∩ R_{a} ∩ R_{c}|. Suppose 5 divides |F ∩ R_{a} ∩ R_{c}|. Then 2 ≤ |O_{2}(F′ ∩ R_{a} ∩ R_{c})| ≤ 8 and so O^{2}(F′ ∩ R_{a} ∩ R_{c}) centralizes O_{2}(F′ ∩ R_{a} ∩ R_{c}), a contradiction since the involutions in K are not centralized by elements of order 5 in K. So |F ∩ R_{a} ∩ R_{c}| = 2^{5}. In particular

1755 = |α^{F} | ≥ |b^{K}| + |c^{K}| = 2^{2} · 3^{1+u} · 5^{v} · 13^{w} + 3^{1+u} · 5^{1+v} · 13^{w},

and so

(∗)  65 ≥ 3^{u} · 5^{v} · 13^{w}.

Now |K/N_{K}(D)| = 5^{1+v} ·13^{w} and |K/N_{K}(D)| is congruent to 1 modulo 3, so we must have v = 1. By Lemma 2.3, the centralizers of elements of order 5 in F are {2, 5}−groups and no involution in K is centralized by an element of order five . Thus the centralizers of elements of order five in K are 5-groups. In particular the Sylow 5-subgroups of K are T I−sets, and so the number of Sylow 5-subgroups in K is congruent to 1 modulo 25. Since no divisor of 2^{3} · 3^{3} is 1 mod 25, w ≠ 0. Thus w = 1 and by (∗), u = 0 and the equal sign holds in (∗). This means that F ∩ R_{b} has no further orbit on α^{F} and (8) is proved.

We remark that using the list of maximal subgroups of the Tits group or the classification of groups with dihedral Sylow 2-subgroups it is not difficult to see that K ≅ L_{2}(25), but we will not need this fact.

9 F has three orbits on R/F with lengths 1, 1755 and 2304.

In view of (8) it is enough to prove that there exist no points at distance 3 from X_{1} in Γ. One easily checks in Γ_{0} that there exist points b, c, d in α^{F} such that b has distance 8 from both c and d in Γ_{0} and c and d are at distance 2 in Γ_{0}. Then c and d are adjacent in Γ and b is at distance 2 from c and d in Γ. Let a be at distance 2 from X_{1}. By (8), F ∩ R_{a} has two orbits on the neighbors of a in Γ. One orbit is the set of common neighbors of X_{1} and a. By (6) there exists g ∈ R with b^{g} = X_{1} and c^{g} = a. Then d^{g} lies in the second orbit and has distance 2 from X_{1} In particular every point adjacent to a is at distance at most 2 to X_{1} in Γ. This completes the proof of (9) and of Lemma 2.11. QED

It now follows from Lemma 2.11 that R has the following properties:

(a) R has a subgroup F with F/Z(G) ≅ ^{2}F_{4}(2).

(b) F has 3 orbits on R/F with lengths 1, 1755 and 2304.

By [12] we conclude that \overline{R} is isomorphic to the Rudvalis group. This completes the proof of Theorem 2.1.

3 A Computer-free Construction of the Higman-Sims Group as a Subgroup of E_{7}(5).

In this chapter we will prove the following Theorem:

Theorem 3.1 E_{7}(5) contains subgroups isomorphic to M_{22} and the Higman-Sims group.

We start with some of the properties of M_{22} we will need in the proof of the theorem.

Lemma 3.2 Let M = M_{22} act faithfully on Ω = {1, 2, …, 22}, let ω ∈ Ω and let D be the stabilizer of a hexad H in Ω.

(a) M_{ω} acts transitively on Ω \ {ω} and M_{ω} ≅ L_{3}(4).

(b) D ∼ 2^{4}.Alt(6), D acts transitively on Ω \ H and any subgroup of shape 2^{4}Alt(6) in M is conjugate to D.

(c) Let E ≤ M_{ω} with E ≅ 2^{4}Alt(5). Then E has orbits of lengths 1, 5 and 16 or of lengths 1, 1 and 20 on Ω. In the first case E stabilizes a hexad, and in the second case N_{M}(E) ∼ 2^{4}Sym(5).

(d) Let A ≤ B ≤ M with A ≅ Alt(6) and B ≅ Alt(7). Then A is not contained in a conjugate of D.

(e) Let A ≤ D such that A ≅ Alt(5) and such that O_{2}(D) is a natural SL_{2}(4)−module for A. Then A has orbits of lengths 1 and 5 on H and orbits of lengths 1 and 15 or of lengths 6 and 10 on Ω \ H.

(f) Let A ≤ D such that A ≅ Alt(5) and such that O_{2}(D) is a natural Ω_{4}^{−}(2)−module for A. Then A acts transitively on H and has orbits of lengths 1, 5 and 10 on Ω \ H.

(g) If A ≤ D with A ≅ Alt(6), then A has orbits of lengths 1 and 15 or of lengths 6 and 10 on Ω \ H.

(h) If A ≤ M_{ω} with A ≅ Alt(6), then A has orbits of lengths 1, 6 and 15 on Ω.

(i) M has no subgroup of index 56.

Proof: The maximal subgroups of M_{22} and their orbits on Ω are listed in Table 10.3 on page 285 of [4]. We use this table without further reference. In particular, (a) and (b) hold. ¿From the definition of a Steiner system, the set Ω \ {ω} together with the set of hexads containing ω form a projective plane of order 4. In particular both the two point stabilizer and the stabilizer of an incident point hexad pair have shape 2^{4}SL_{2}(4) = 2^{4}Alt(5) where the 2^{4} is a natural SL_{2}(4)−module. Let E^{∗} be the normalizer of a pair of points. Then E^{∗} ≅ 2^{4}Sym(5) and (c) holds.

To prove (h), let f be an element of order five in A. Then f has exactly one fixed point η on Ω \ {ω}. Clearly A does not fix η and |η^{A}| divides |A| and is congruent to 1 modulo 5. Thus |η^{A}| = 6. As f acts fixed-point freely on the remaining 15 points and A has no orbit of length 5, (h) holds.

(g): By (h) we may assume that A has no fixed points on Ω. Then by the same argument as in the proof of (h), A has orbits of length 6 and 10 on Ω \ H.

For (e) and (f) we note that if A fixes a point η outside H, then the action of A on Ω \ (H ∪ {ω}) is isomorphic to the action of A on O_{2}(D)^{#}. Since in case (f) there exists a unique class of subgroups Alt(5) in O_{2}(D)A, (f) holds. In case (e), we need to rule out the possibility that the orbits of A on Ω \ H have lengths 5, 5 and 6. In this case, the elements of order three in A would have 4 fixed-points on Ω \ H, but only one fixed-point on O_{2}(D), a contradiction.

(d) Note that B is unique up to conjugation and has an orbit Ξ of length 7. Hence A has orbits of length 1 and 6 on Ξ and so is transitive on Ω \ Ξ by (h). As no two hexads can intersect in a set of size 5, the orbit of length 6 is not a hexad. Thus (d) holds.

Since M has no maximal subgroup of index dividing 56, (i) holds. QED

For the convenience of the reader we recall the definition of the HS group as found [8]. Let (S, B) be the Steiner System of type (3, 6, 22). Let G be the (undirected) graph with vertex set {∗} ∪ S ∪ B, where ∗ is a new symbol. In G

(a) The vertex ∗ is joined to each point in S.

(b) Each point α ∈ S is joined to the 21 hexads containing α.

(c) Two hexads are joined if and only if they are disjoint.

Then by [8], Aut(G) is transitive on G and Aut(G) has a simple subgroup of index 2 and order 44,352,000, now called the Higman Sims group HS. We refer to G as the Higman-Sims graph.

The next two lemmas characterize M_{22} and HS in terms of certain subgroups.

Lemma 3.3 Let M be a group and L, M_{1} and M_{2} subgroups of M such that L ≅ L_{3}(4), M_{1} ∼ 2^{4}Sym(5), M_{2} ∼ 2^{4}Alt(6), L∩M_{1} ≅ L∩M_{2} ∼ 2^{4}Alt(5), M_{1}∩M_{2} ∼ 2^{4}Sym(4), L∩M_{1}∩M_{2} ∼ 2^{4}Alt(4) and M = ⟨L, M_{1}, M_{2}⟩. Then M ≅ M_{22}.

Proof: Put Γ = M/L and α = L ∈ Γ. Let t ∈ M_{1} ∩ M_{2} \ L. Since M_{1} ∩ L is normal in M_{1}, we have M_{1} ∩ L ≤ L^{t}. As M_{1} ∩ M_{2} does not normalize L ∩ M_{2}, the element t does not normalize L and so L ∩ L^{t} = L ∩ M_{1}. Put Γ_{0} = {α} ∪ α^{tL}. Then |Γ_{0}| = 1 + 21 = 22.

Now |α^{M_{2}}| = |M_{2}/M_{2} ∩ L| = 6 and α^{M_{2}} = {α} ∪ α^{t(M_{2}∩L)} ⊂ Γ_{0}. Further, as M_{1} = ⟨t⟩(M_{1} ∩ L), we have

α^{M_{2}M_{1}} = α^{M_{2}⟨t⟩ (M_{1}∩L)} = α^{M_{2}(M_{1}∩L)} = α ∪ α^{t(M_{2}∩L)(M_{1}∩L)}.

Since M_{1} ∩ L acts transitively on α^{tL} \ {α} we get that α^{M_{2}M_{1}} = Γ_{0}. Hence M_{1} and L normalize Γ_{0}. Note that M_{2} = ⟨M_{2} ∩ L, M_{2} ∩ M_{1}⟩ and so M = ⟨M_{1}, M_{2}, L⟩ = ⟨M_{1}, L⟩. Thus M normalizes Γ_{0}, also Γ = Γ_{0} and |M/L| = 22. Put B = α^{M_{2}} and B = {B^{m}|m ∈ M}. We claim that (Γ, B) is a Steiner System of type (3, 6, 22). Since L is doubly transitive on Γ \ {α}, M is triply transitive on Γ. Hence each set of three elements in Γ lies in e elements of B where e is a positive integer independent of the set of three. Counting tuples (H, a, b, c) such that H ∈ B and a, b, c are pairwise different elements of H we get

|B| · 6 · 5 · 4 = 22 · 21 · 20 · e.

As |B| = |M/M_{2}| = 22 · |L|/|M_{2}| = 77 we get e = 1 and the claim is established.

Since M ≤ Aut(Γ, B) ≅ Aut(M_{22}) and |M| = 22 · |L| = |M_{22}|, we deduce that M ≅ M_{22}. QED

Lemma 3.4 Let H be a group, M and D subgroups of H, and L a subgroup of M. Suppose that each of the following holds:

(i) M ≅ M_{22}, L ≅ L_{3}(4) and D ∼ 2^{4}Sym(6),

(ii) There exists t ∈ N_{H}(L) with t^{2} ∈ L such that D∩M ∼ 2^{4}Sym(5), D∩M^{t} ∼ 2^{4}Alt(6),

and D ∩ D^{t} ∼ 2^{4}Sym(4).

(iii) H = ⟨M, L, D, t⟩.

Then H is isomorphic to HS, C_{2} × HS or Aut(HS).

Proof: Let Γ be the graph whose vertices are the right cosets of M in H, and whose edges are the sets {Mh, Mth} for h ∈ H. Put α = M and β = Mt.

Note that ⟨H_{α}, H_{{α,β}}⟩ = ⟨M, L⟨t⟩ ⟩ and D = ⟨D ∩ M, D ∩ M^{t}⟩. So ⟨H_{α}, H_{{α,β}}⟩ = H and Γ is connected.

Let △^{i}(α) be the set of vertices at distance exactly i from α and put △(α) = △^{1}(α). Then | △ (α)| = 22, L = H_{αβ} and L acts transitively on △(β) \ {α}. Let r ∈ D ∩ M \ L and put γ = α^{trt}. Then {β, γ} = {α, β}^{rt} and {β, γ} is an edge. Since r ∈ D and D ∩ M^{t} is normal in D, r normalizes D ∩ M^{t}. Moreover, t^{2} ∈ L ≤ M and so r^{t} normalizes D^{t} ∩ M. Thus D^{t} ∩ M ≤ H_{α} ∩ H_{α}^{r^{t}} = H_{αγ}. In particular γ is not adjacent to α and thus γ ∈ △^{2}(α).

Suppose that H_{αγ} ≠ D^{t} ∩ M. As D^{t} ∩ M is maximal in M we get H_{αγ} = H_{α} = H_{γ}, and △(α) ∩ △(γ) = β^{Hα} = △(α) = △(γ), and since Γ is connected, Γ = {α, γ} ∪ △(α). Thus L fixes exactly the vertices α, β and γ. Hence γ = γ^{t} and t ∈ H_{γ} = H_{α}, a contradiction. We have proved

1 H acts transitively on geodesics of length 2, H_{αγ} ∼ 2^{4}Alt(6), |△^{2}(α)| = 77, |△(α)∩△(γ)| = 6 and H_{αβγ} ∼ 2^{4}Alt(5).

By part (c) of Lemma 3.2, H_{αβγ} acts transitively on △(γ)\△(α). Let δ ∈ △(γ)\△(α). Then H_{αγδ} ≅ Alt(6) and H_{αβγδ} ≅ Alt(5). Suppose that δ is at distance 2 from a. Then △^{3}(α) = ∅, thus Γ = {α} ∪ △(a) ∪ △^{2}(α) and |Γ| = 100. It is now easy to see that Γ is isomorphic to the Higman-Sims graph (see for example [19] for a formal proof). As |H| = 100 · |M_{22}| = |HS|, H ≅ HS.

So we may assume from now on that δ is not in distance 2 from α. It follows that

2 δ ∈ △^{3}(α) and H acts transitively on geodesics of length 3.

By Lemma 3.2 part (h), H_{αγδ} has orbits of lengths 1, 6 and 15 on △(δ). Further by (2), H_{αδ} acts transitively on △^{2}(α) ∩ △(δ). Since △(β) ∩ △(δ) ⊂ △^{2}(α) ∩ △(δ), we get | △^{2} (α) ∩ △(δ)| is 7, 16 or 22.

Suppose | △^{2} (α) ∩ △(δ)| = 7. Then |H_{αδ}| = 7 · |H_{αγδ}| and H_{αδ} ≅ Alt(7). Now H_{αγ} ∼ 2^{4}.Alt(6) and H_{αγ} ∩ H_{αδ} = H_{αγδ} ≅ Alt(6). This contradicts Lemma 3.2, part (d).

Suppose that | △^{2} (α) ∩ △(δ)| = 22. Then |H_{αδ}| = 22 · |H_{αγδ}| = 2^{4} · 3^{2} · 5 · 11 and |H_{α}/H_{αδ}| = 56, a contradiction to Lemma 3.2, part (i).

Thus | △^{2} (α) ∩ △(δ)| = 16 and |H_{αδ}| = 16 · |Alt(6)|. Since H_{αδ} acts non-trivially on the six points in △(d) \ △^{2}(α), we conclude that H_{αδ} has a factor group Alt(6) or Sym(6). This implies H_{αδ} ∼ 2^{4}Alt(6) and △(δ) \ △^{2}(α) is a hexad in the H_{δ}−invariant Steiner system △(δ). Since H_{αγδ} has orbits of lengths 1 and 15 on △^{2}(α) ∩ △(δ), the group H_{αβγδ} (≅ Alt(5)) fixes a unique point in △^{2}(α) ∩ △(δ), namely γ. Note that H_{αβδ} ≅ H_{αγδ} ≅ Alt(6) and that H_{αβδ} does not fix γ. So H_{αβδ} fixes no point in △^{2}(α) ∩ △(δ) and by Lemma 3.2, part (g), H_{αβδ} has orbits of lengths 6 and 10 on △^{2}(α) ∩ △(δ). Now parts (e) and (f) of Lemma 3.2 imply that H_{αβγδ} has orbits of lengths 1, 5 and 10 on △^{2}(α) ∩ △(δ) and acts transitively on △(δ) \ △^{2}(a). Let ε ∈ △(d) \ △^{2}(α). Then H_{αδε} ∼ 2^{4}Alt(5). Since H_{αδε} lies in a unique subgroup 2^{4}.Alt(6) of H_{α} and since the stabilizer in H_{α} of points at distance 3 from α are 2^{4}Alt(6)’s, we get that δ is the unique point in △^{3}(α) fixed by H_{αδε}. In particular ε ∉ △^{3}(α). We have proved 3 H_{αδ} ∼ 2^{4}Alt(6), | △^{3} (α)| = 77, ε ∈ △^{4}(α), H_{αδε} ∼ 2^{4}Alt(5), H_{αβγδε} ≅ D_{10} and H acts transitively on geodesics of length 4.

By (1) and part (b) of Lemma 3.2, a subgroup 2^{4}Alt(5) of H_{βγ} which has orbits of lengths 1, 1 and 20 on △(β) has orbits of lengths 1, 5 and 16 on △(γ). Note that △(δ) \ △^{2}(a) is an orbit of length 16 for H_{αδε} on △(δ), and so H_{αδε} has orbits of lengths 1, 1 and 20 on △(ε). Since H_{αε} acts transitively on △^{3}(α) ∩ △(ε) and since △(γ) ∩ △(ε) ⊂ △^{3}(α) ∩ △(ε), we get that | △^{3} (α) ∩ △(ε)| is 21 or 22. Suppose that | △^{3} (α) ∩ △(ε)| = 22, then |H_{αε}| = 22 · |H_{αδε}| = 22 · |2^{4}Alt(5)| and thus |H_{α}/H_{αε}| = 21, a contradiction. Thus | △^{3} (α) ∩ △(ε)| = 21 and △(ε) \ △^{3}(α) = {η} for some η. Thus H_{αε} = H_{αεη} ≅ L_{3}(4). Thus | △^{4} (α)| = 22 and ε is the unique point in △^{4}(α) fixed by H_{αεη} and so η ∉ △^{4}(α). Hence

4 H_{αε} ≅ L_{3}(4), | △^{4} (α)| = 22, η ∈ △^{5}(α), H_{αεη} = H_{αε} and H is transitive on geodesics of length 5.

Since H_{αη} is transitive on △^{4}(α) ∩ △(η) and △(δ) ∩ △(η) ⊂ △^{4}(α) ∩ △(η) we conclude that △^{4}(α) ∩ △(η) = △(η) and H_{αη} = H_{α} ≅ M_{22}. Thus

5 H_{αη} = H_{α}, △^{5}(α) = {η}, Γ = ∑_{0≤i≤5} △^{i}(α) and |Γ| = 200.

Let ϕ be the map that sends a vertex μ in Γ to the unique point at distance 5 from μ. Then ϕ is obviously a bijection and (μ^{ϕ})^{g} = (μ^{g})^{ϕ} for all μ in Γ and g in Aut(Γ). In particular {α^{ϕ}, β^{ϕ}} is the set of fixed-points of H_{αβ} on Γ. So β^{ϕ} is adjacent to α^{ϕ} and therefore ϕ is a graph automorphism of Γ. Put Γ_{0} = {{μ, μ^{ϕ}} | μ ∈ Γ} and let {μ, μ^{ϕ}} be adjacent to {λ, λ^{ϕ}} if μ is adjacent to λ or λ^{ϕ}. Then |Γ_{0}| = 100, M has orbits of lengths 1, 22 and 77 on Γ_{0} and H acts transitively on Γ_{0}. So as above Γ_{0} is the Higman-Sims graph and Aut(Γ_{0}) ≅ Aut(HS). Let N be the kernel of the action of Aut(Γ) on Γ_{0}. We claim that N = {1, ϕ}. Indeed, let n ∈ N. Then α^{n} ∈ {α, α^{ϕ}} and replacing n by nϕ if necessary we may assume that α^{n} = α. Since β^{ϕ} is not adjacent to α, we have β^{n} = β. So n fixes the neighbors of all its fixed-points and since Γ is connected, we conclude that n = 1. Thus N = {1, ϕ} and Aut(Γ) ≅ C_{2} × Aut(HS). Further, |H| = 200·|M| = 2·|HS| and so H ≅ Aut(HS) or C_{2} ×HS and the lemma is proved. QED

Lemma 3.5 (a) G has a subgroup T of order 3 with N_{G}(T) ∼ SU_{6}(5) ∘ SU_{3}(5)).Sym(3).

(b) Put U = N_{G}(T), U_{1} = [V, T] and U_{2} = C_{V} (T). Then V = U_{1} ⊕ U_{2}, moreover, U_{1} and U_{2} are irreducible as U modules, U_{1} ≅ W_{6} ⊗_{GF (25)} W_{3} and GF (25) ⊗_{GF (5)} U_{2} ≅ ⋀^{3} W_{6}, where W_{i} is the natural i−dimensional GF (25)−module for SU_{i}(5), for i=3,6.

(c) N_{G}(U_{1}) = U.

Proof: It is clear from the extended Dynkin diagram of type E_{7} that 2.E_{7}(K) has a subgroup H ≅ SL_{3}(K) ∘ SL_{6}(K). Moreover, the central involution of the Weyl group induces a graph automorphism on both of the factors and so an application of Lang’s theorem yields a subgroup (SU_{6}(5) ∘ SU_{3}(5)).Sym(3) in E_{7}(5). Using the embedding of K^{1+56}2.E_{7}(K) in E_{8}(K), the Steinberg relations and weight theory it is easy to check that the 56−dimensional KE_{7}(K)−module is as an H−module the direct sum of X_{3} ⊗ X_{6}, X_{3}^{∗} ⊗ X_{6}^{∗} and ⋀^{3} X_{6} where X_{i} is a natural module for SL_{i}(K), i = 3, 6. As H is a maximal connected closed subgroup of 2.E_{7}(K) and is of index two in its normalizer, all the statements of the lemma are now readily verified.

Lemma 3.6 There exists an involution f̄ in \overline{U} \\overline{U}′ such that C\overline{U}(f̄) ∼ L_{4}(5).2^{2}×Sym(5)×⟨f̄⟩. Moreover, for any such f̄,

C\overline{G}(f̄) ∼ (2 × L_{8}(5)).2

and C\overline{U}(f̄) acts transitively on the elementary abelian subgroups of order 16 contained in the normal subgroup L_{4}(5) in C\overline{U}(f).

Proof: It is easy to verify that Aut(U_{6}(5)) has three classes of involutions that do not induce a diagonal automorphism on U_{6}(5). The derived groups of the respective centralizers in U_{6}(5) are P Sp_{6}(5), D_{3}(5) and ^{2}D_{3}(5), and the first two of these three classes of involutions lie in the same coset of Inn(U_{6}(5)).

Since Ω_{6}^{−}(5) (≅ 2.^{2}D_{3}(5)) acts absolutely irreducibly on the exterior cube of its natural module, we conclude that the centralizer of Ω_{6}^{−}(5) on the 20−dimensional module lies in the center of the full linear group acting on the 20−space. It follows that \overline{U} \ \overline{U}′ contains no involution centralizing a ^{2}D_{3}(5) in U_{6}(5) and so contains an involution centralizing a D_{3}(5) in U_{6}(5).

Let f̄ be any such involution. Note that the normalizer of D_{3}(5) in U_{6}(5) is P SO_{6}^{−}(5) extended by an involution which multiplies the quadratic form associated to P SO_{6}^{−}(5) by a fourth root of unity. So C_{U_{6}(5)}(f̄) ∼ L_{4}(5).2^{2} and the two classes of elementary abelian subgroups of order 16 in C_{U_{6}(5)}(f̄)′ are fused in C_{U_{6}(5)}(f̄). Moreover, Aut(U_{3}(5)) has exactly one class of involutions outside Inn(U_{3}(5)) and the corresponding centralizers are Sym(5)’s. Therefore C\overline{U}′(f̄) ∼ L_{4}(5).2^{2} × Sym(5).

It remains to determine C\overline{G}(f̄). Note that \overline{G} has three classes of involutions whose centralizers have shapes 2.(L_{2}(5)×D_{6}(5)).2, (C_{2} ×E_{6}(5)).2, and (C_{2} ×L_{8}(5)).2. Under the actions of the derived groups of these centralizers V decomposes into direct sums of irreducible modules of dimensions 24 and 32; 1, 1, 27, and 27; and 28 and 28, respectively. On the other hand, from the action on the 20− and 36−spaces, we know that V is the direct sum of irreducible modules of dimensions 18, 18, 10 and 10 for C_{U}(f)′. It follows that C\overline{G}(f̄) ∼ (C_{2} × L_{8}(5)).2, and the lemma is proved. QED

From now on let T, U and f be as in Lemma 3.6. Note that f^{2} is the central involution of G. Put N = C_{U}(f), F = ⟨f⟩ and R = N_{G}(F). Then \overline{R} = C\overline{G}(f̄) and so by Lemma 3.6 R ∼ C_{4} ∘ 2 · L_{8}(5).2. Let R̂′ be a group with R̂′ ≅ SL_{8}(5) so that R̂′ has R′ as a quotient group. For X in R′, let X̂ be the inverse image of X in R̂′. As Z(N̄′) = 1, it is easy to see that the natural 8−dimensional module for R̂′ is, as a module for N̂′, the tensor product of a 4−dimensional module for SL_{4}(5) and a 2−dimensional module for SL_{2}(5). Since elements in GL_{4}(5) ⊗ GL_{2}(5) ≤ GL_{8}(5) have determinant plus or minus one, \overline{R}′ has two classes of subgroups isomorphic to L_{4}(5) × L_{2}(5). Since R = NR′, R fixes these classes. Moreover, as the element inverting f can be chosen to invert a Cartan subgroup of R′, R does not induce an outer diagonal automorphism on R′.

The character tables of L_{3}(4) and its covers (see [3]) show that there is a group of shape 4.L_{3}(4) which has a faithful irreducible character χ of degree 8. Note that R contains a subgroup L of shape 2.L_{3}(4). We plan to extend L to a subgroup M_{22} and then to a subgroup HS of G. Let S be a Sylow 2−subgroup of L, let B = N_{L}(S) and let A_{1} and A_{2} be the two elementary abelian groups of order 2^{5} in S. Note that L̂ is perfect and that Â_{i} is the central product of a cyclic group of order 4 with an extra-special group of order 2^{5}. Moreover a natural 8−dimensional module for R̂′ is as an Â_{i}−module the direct sum of two isomorphic irreducible modules of dimension 4. Thus C_{R′}(Â_{i}) ≅ GL_{2}(5) and we can choose L, S and A_{1}, so that \overline{A}_{1} is contained in the normal L_{4}(5) in \overline{N}. It follows that N\overline{R}′(\overline{A}_{1}) ≤ N and has shape 2^{4}Sym(6) × Sym(5). Let \overline{B}_{2} be the projection of A_{2} onto the L_{4}(5) in N. Let \overline{B}_{2} be the projection of A_{2} onto the L_{4}(5) in N. Put L_{i} = N_{L}(A_{i}). Our goal is to show that both L_{i}’s can be extended to a subgroup 2^{5}Sym(6) in U. Inside 2^{5}Sym(6) we will choose appropriate subgroups 2^{5}Sym(5) and 2^{5}Alt(6) which will allow us to apply Lemma 3.3 and Lemma 3.4 to find M_{22} and HS in \overline{G}.

Lemma 3.7 In R′, the subgroup A_{2} is conjugate to B_{2} but not to A_{1}.

Proof: Put C_{i} = O^{2}(C_{R′}(Â_{i})). Then C_{i} ≅ SL_{2}(5). Suppose that [S, C_{1}] = 1. Then C_{1} ≤ C_{R′} (Â_{2}) and C_{1} = C_{2}. Since L = ⟨L_{1}, L_{2}⟩, the group L normalizes C_{1}, a contradiction. Hence [S, C_{1}] ≠ 1 and similarly [S, C_{2}] ≠ 1. As B/A_{i} ≅ Alt(4), we have B = O^{2}(B)A_{i}. Since Out(C_{i}) is a 2−group and A_{i} centralizes C_{i} we conclude that S induces inner automorphisms on C_{i}. Hence S normalizes a unique subgroup Q_{8} in C_{i}. Denote this Q_{8} by Q_{i} and note that S induces inner automorphisms on Q_{1}. In particular [Â_{2}, Q_{1}] ≤ Z(Q_{1}) · Z(R̂′). Thus Q_{1} induces inner automorphisms on Â_{2}. Put R_{i} = Q_{1}Â_{i} (where we really mean Q_{1} and not Q_{i}). Then R_{1} and R_{2} are both the central product of an extra-special group of order 2^{7} with a cyclic group of order 4. Put X = C_{R_{2}}(Â_{2}). It follows that R_{2} = XÂ_{2}, that S normalizes X and that X contains a Q_{8} invariant under S. Thus Q_{2} ≤ X, R_{2} = Q_{2}Â_{2} = Q_{2}B_{2} and B̂_{2} = C_{R_{2}}(Q_{1}).

Note that N_{R′}(R_{1})/R_{1} ≅ Sp_{6}(2). Note Ā_{1} and B̄_{2} are both nondegenerate 4−spaces in the 6−dimensional symplectic space R_{1}/Z(R_{1}) (where the symplectic form is given by the commutator map). Hence by Witt’s theorem, A_{1} and B_{2} are conjugate under N_{R′}(R_{1}) proving the first statement of the lemma.

Since C_{R′}(Â_{i}) acts transitively on the Q_{8}’s in C_{R′}(Â_{i}) and since N_{R′}(A_{1})∩N_{_{R}′}(Q_{1}) induces the full automorphism group of Q_{1} on Q_{1}, we conclude that A_{1} and B_{2} are conjugate in R′ if and only if they are conjugate under Y = C_{R′}(Q_{1}). Note that Y is isomorphic to the subgroup of index two in GL_{4}(5). Since B normalizes Y and [A_{1}, B] = A_{1}, A_{1} and hence also B_{2} are contained in Y′. Put S_{0} = A_{1}B_{2}. Since A_{1} and B_{2} normalize each other we conclude that S_{0} is a Sylow 2−subgroup of Y/Z(Y) and that A_{1} and B_{2} are the two maximal elementary abelian subgroups of S_{0}. It follows that A_{1} and B_{2} are not conjugate in Y′. Moreover, Out(Y′) ≅ D_{8}, Out(Y′) acts on {A_{1}^{Y′}, B_{2}^{Y′}} and Out(Y′)′ fixes A_{1}^{Y′} and B_{2}^{Y′}. As the group of outer automorphisms of Y′ induced by Y is Out(Y′)′, A_{1} and B_{2} are not conjugate in Y. So A_{1} and B_{2} are not conjugate in R′ and lemma is established. QED

Lemma 3.8 N\overline{R}′(\overline{A}_{1}) acts transitively on L_{3}(4)’s in \overline{R}′ containing \overline{A}_{1} and on subgroups 2^{4}Alt(5) in N\overline{R}′(\overline{A}_{1}) which can be extended to an L_{3}(4) in \overline{R}′.

Proof: Recall that N\overline{R}′(\overline{A}_{1}) ∼ 2^{4}Sym(6) × Sym(5) and, as [S, C_{i}] ≠ 1 in the notation of the previous lemma, any 2^{4}Alt(5) in N\overline{R}′(\overline{A}_{1}), which can be extended to a L_{3}(4) in \overline{R}′, projects non-trivially on the Sym(5). Hence N\overline{R}′(\overline{A}_{1}) acts transitively on such subgroups. Note that

\overline{L}_{1} is such a subgroup and that L_{1}^{∗} =^{def} N\overline{R}′(\overline{L}_{1}) ∼ 2^{4}Sym(5). Put \overline{W} = N\overline{R}′(\overline{L}_{1}) ∩ N\overline{R}′(\overline{A}_{2}). As Ā_{1} and Ā_{2} are the only maximal elementary abelian subgroups of S̄ = Ā_{1}Ā_{2}, we have N_{L_{1}^{∗}}(A_{2}) = N_{L_{1}^{∗}}(S). Since S/A_{1} is the Sylow 2−subgroup of L_{1}/A_{1}, we have \overline{W} ∼ 2^{4}Sym(4). We claim that W is contained in a 2^{4}Alt(6) × Sym(5) subgroup of N\overline{R}′(\overline{A}_{1}). For this let i be an involution in \overline{W} \ \overline{L}_{1}. Then it is enough to show that the inverse image of [Â_{2}, i] is not elementary abelian. Let C_{2} be the projection of A_{2} onto the normal Sym(5) in \overline{N}. Then [B̂_{2}, i] is elementary abelian, while [Ĉ_{2}, i] is not. So also [Â_{2}, i] is not elementary abelian, and the claim is proved.

It is now easy to see that \overline{W} ∩\overline{L}_{1} is contained in exactly two subgroups 2^{4}Alt(5) of N\overline{R}′(\overline{A}_{2}) and these two subgroups are interchanged by W. Since any L_{3}(4) in \overline{R}′ containing \overline{L}_{1} is generated by \overline{L}_{1} and a 2^{4}Alt(5) in N\overline{R}′(\overline{A}_{2}) containing \overline{W} ∩ \overline{L}_{1} the lemma is proved. QED

Lemma 3.9 Put L_{0} = N_{R}(L). Then |L_{0}/LF | = 2 and A_{1} and A_{2} are conjugate in L_{0}.

Proof: Since Out(R′/Z(R′)) ≅ D_{8}, since R′ has four orbits on A_{1}^{Aut(R′)} and since P GL_{8}(5) acts transitively on those four orbits, the normalizer of any such orbit in Out(R′) is a group of order 2 that is not contained in the center of Out(R′). It follows that R fixes two of these orbits and interchanges the remaining two. Moreover, by Lemma 3.8, L_{8}(5) has four orbits on (A_{1}, L)^{Aut(R′)} and, by Lemma 3.7, each orbit of subgroups L_{3}(4) in \overline{R}′ leads to two orbits of R′ on (A_{1}, L)^{Aut(R′)}. It follows that \overline{R}′ has exactly two classes of subgroups L_{3}(4). Suppose R interchanges those two classes. Then R could not normalize any of the orbits of R′ on A_{1}^{Aut(R′)}, a contradiction. So |L_{0}/LF | = 2. Suppose that A_{1} and A_{2} are not conjugate in L_{0}. Then they are also not conjugate in R. By Lemma 3.7, we conclude that A_{1} and B_{2} are not conjugate in R, and this contradicts Lemma 3.6. So A_{1} and A_{2} must be conjugate in L_{0}. QED

Lemma 3.10 Put X = N_{G}(A_{1}). Then X ≤ U and X ∼ 6.(2^{4}.3^{4}.Sym(6) × U_{3}(5)).Sym(3).

Proof: Let X_{0} = N_{U}(A_{1}). By the action of A_{1} on the natural 6−dimensional GF (25)−module W for SU_{6}(5) we see that, modulo the normal SU_{3}(5) in U, X_{0} is a full monomial subgroup of U_{6}(5).Sym(3). It follows that X_{0} ∼ 6.(2^{4}.3^{4}.Sym(6) × U_{3}(5)).Sym(3). Moreover, X_{0} has two orbits Σ_{1} and Σ_{2} on the hyperplanes in A_{1} which do not contain Z(G). Choose notation so that |Σ_{1}| = 6 and |Σ_{2}| = 10. Recall that V ≅ U_{1} ⊕ U_{2} where U_{1} and U_{2} are defined and described in Lemma 3.5.

Let H_{1}, H_{2}, H_{3} be three different elements of Σ_{1}. Then its is easy to check that H_{1} ∩ H_{2} ≰ H_{3}. As W = ⨁_{H∈Σ_{1}}(C_{W} (H)), this implies that H_{1} acts fixed–point freely on U_{2}. Since U_{1} is a direct sum of copies of W as an A_{1}−module, we get ⨁_{H∈Σ_{1}} C_{V} (H) = U_{1}.

Also C_{V} (H) is either 6−dimensional, if H ∈ Σ_{1}, or 20/10 = 2−dimensional, if H ∈ Σ_{2}. Thus N_{G}(A_{1}) normalizes Σ_{1} and so also U_{1}. Since U is maximal in G, we conclude that X ≤ U. Hence X = X_{0}. QED

Lemma 3.11 There exists t ∈ N_{R}(L) and D ≤ U such that t^{2} ∈ L, A_{1}^{t} = A_{2}, L_{1} ≤ D, D ∼ 2^{5}Sym(6), t normalizes N_{D}(B) and t does not normalize N_{D′}(B).

Proof: Let Y = X/A_{1}. Then Y ∼ 3.(3^{4}Sym(6) × U_{3}(5))Sym(3). Let K be the image of L_{1} in Y. Then K ≅ Alt(5) and we are looking for subgroups Sym(6) of Y containing K. Let Y_{1} be the normal subgroup 3.3^{4}Sym(6) of Y, let Y_{2} be the normal subgroup SU_{3}(5) of Y and let K_{i} be the projection of K onto Y_{i}. Then Y acts transitively on subgroups 3 · Alt(6) (the triple cover of Alt(6)) and transitively on subgroups Alt(5) in Y_{2}. Moreover, the normalizer of an Alt(5) in U_{3}(5).Sym(3) is a C_{2} × Sym(5) and the normalizer of an Alt(6) is Aut(Alt(6)). It follows that K_{2} can be embedded into exactly two subgroups 3·Alt(6) of Y_{2} and f interchanges these two 3 · Alt(6)’s. Furthermore Y has one orbit on subgroups 3 · Alt(6) in Y_{1} and one orbit on subgroups Alt(5) in Y_{1} for which \overline{A}_{1} is a natural SL_{2}(4)−module. (Note here that 3 · Alt(6) exists in Y_{1}, since 3^{4}Alt(6) has 9 classes of Alt(6)’s while 3^{1+4}Alt(6) has only 3 classes.) Now Y/Y_{2} ∼ 3^{4+1}(C_{2} × Sym(6)). The normalizer of the image of K in Y/Y_{2} is a C_{2} × Sym(5) and the normalizer of the image of a 3 · Alt(6) of Y_{1} in Y/Y_{2} is a Sym(6). It follows that K_{1} can be embedded into exactly two subgroups 3 · Alt(6) of Y_{1} and f interchanges these two 3 · Alt(6)’s. Let D_{i} be any of the two subgroups 3 · Alt(6) in Y_{i} with K_{i} ≤ D_{i}. Then there exists precisely one subgroup A in D_{1}D_{2} such that K ≤ A, A = A′ and A/Z(A) ≅ Alt(6). Similarly there exists precisely one subgroup Â in D_{1}^{f}D_{2} such that K ≤ Â, Â = Â′ and Â/Z(Â) ≅ Alt(6). We claim that exactly one of A and Â is an Alt(6), while the other is the triple cover. Indeed, let x_{i}, y_{i} be elements of order 3 in D_{i} such that ⟨x_{1}x_{2}, y_{1}y_{2}⟩Z(Y) is a Sylow 3−subgroup of AZ(Y). Then ⟨x_{1}^{f}x_{2}^{f}, y_{1}y_{2}⟩Z(Y) is a Sylow 3−subgroup of ÂZ(Y). Note that

[x_{1}x_{2}, y_{1}y_{2}] = [x_{1}, y_{1}] [x_{2}, y_{2}] and [x_{1}^{f}x_{2}^{f}, y_{1}y_{2}] = [x_{1}, y_{1}]^{f} [x_{2}, y_{2}].  (∗)

Since [x_{1}, y_{1}] and [x_{2}, y_{2}] are both contained in Z(Y) and unequal to 1 and since f inverts Z(Y), we see that exactly one of the two expressions in (∗) is equal to 1. This proves our claim.

Choose notation so that A ≅ Alt(6). From what we proved so far it follows that A and A^{f} are the only two Alt(6) subgroups in Y which contain K. It is easy to see that N_{Y} (K)/Z(Y) ≅ C_{2} × Sym(5). Furthermore, there exists a subgroup of index 2 in N_{Y} (K) which normalizes A. Since f does not normalize A, this subgroup is, modulo the center of Y, isomorphic to Sym(5). Put E = A(N_{Y} (K) ∪ N_{Y} (A)) and let D̂ be the inverse image of E in X. Then E/Z(Y) ≅ Sym(6) and D̂/T ≅ 2^{5}Sym(6). (Recall that T is the cyclic group of order three with U = N_{G}(T).)

By Lemma 3.9, there exists t ∈ N_{R}(L) with A_{1}^{t} = A_{2}. Choose t so that t normalizes B. Then t^{2} ∈ BF and since t inverts F, we have t^{2} ∈ B. In particular t^{2} ∈ L. Note that A_{1} and A_{2} are the only elementary abelian groups of order 2^{5} in B. Put J = N_{G}(B) ∩ N_{G}(A_{1}). As t ∈ N_{G}(B) \ J, we conclude that J is of index two in N_{G}(B). For Q ⊂ N_{G}(B) let Q^{∗} be the image of Q in N_{G}(B)/B. Since J ≤ X, it is easily verified that J^{∗} is an elementary abelian group of order 9 extended by an elementary abelian group of order 8. Pick a in N_{D′}(B) \ BT and b in N_{D}(B) with [a, B] ≤ A_{1} and such that a^{∗} and b^{∗} are involutions. Since U_{3}(5) contains no Sym(6), b induces an outer automorphism on Y_{2}. The same is true for f and so b and f both invert O_{3}(J^{∗}). Note that a centralizes T and inverts O_{3}(J^{∗})/T. As A_{1} is in the SU_{6}(5) subgroup of U but A_{2} is not, the element t is not in U. Since T ≤ O_{3}(J), we conclude O_{3}(J)^{∗} = T^{∗}T^{∗t} = O_{3}(J^{∗}). It follows that ⟨a^{∗}, t^{∗}⟩ acts as a D_{8} on O_{3}(J^{∗}). Since ⟨a^{∗}, t^{∗}⟩ is a dihedral group, we conclude that ⟨a^{∗}, t^{∗}⟩ ≅ D_{8}. Let x^{∗} = [a^{∗}, t^{∗}].

We claim that x^{∗}T^{∗} = b^{∗}T^{∗}. Indeed, x^{∗} inverts O_{3}(J^{∗}) and |C_{N_{G}(B)^{∗}}(O_{3}(J)^{∗})/O_{3}(J^{∗})| = 2. Hence x^{∗} lies in the same coset of O_{3}(J^{∗}) as f^{∗} or as b^{∗}. Since NR̄′(A_{1}) ∼ 2^{4}.Sym(6)×Sym(5), N_{R′}(B)^{∗} ≅ C_{2} ×C_{2}. Furthermore f and t are in N_{R}(B) and so N_{R}(B)^{∗} is a Sylow 2−subgroup of N_{G}(B)^{∗}. Since f ∉ R′ we conclude that f^{∗} ∉ N_{G}(B)^{∗′}O_{3}(J^{∗}) = ⟨x^{∗}⟩O_{3}(J^{∗}). So x^{∗}O_{3}(J^{∗}) = b^{∗}O_{3}(J^{∗}). Moreover, a^{∗} centralizes b^{∗} and x^{∗} and hence b^{∗}T^{∗} = C_{x^{∗}O_{3}(J^{∗})}(a^{∗}) = x^{∗}T^{∗}.

In particular, we can choose b so that b^{∗} = x^{∗} = [a^{∗}, b^{∗}]. Put D = D̂′⟨b⟩. Then \overline{D} ∼ 2^{4}Sym(6). In addition, N_{D}(B) = B ⟨a, b⟩ and so t normalizes N_{D}(B). On the other hand, N_{D′}(B) = B⟨a⟩ and so N_{D′}(B)^{t} = B⟨ab⟩ ≠ N_{D′}(B). QED

Proof of Theorem 3.1:

We are now able to construct HS and M_{22} in Ḡ. For this choose L, t and D as in Lemma 3.11 and put M_{1} = N_{D}(L_{1}). By Lemma 3.10, N_{D^{′ t}}(B) is contained in D, but not in D′. Since N_{D^{′ t}}(B)/O_{2}(D) ≅ Sym(4), we conclude that N_{D^{′ t}}(B) is contained in M_{1}. Note that L_{1}^{t} = L_{2} and so L_{2} ≤ D^{′ t}. Put M_{2} = D^{′ t}. Then \overline{M}_{1} ∼ 2^{4}Sym(5) and \overline{M}_{2} ∼ 2^{4}Alt(6). Moreover,

M_{1} ∩ L = L_{1}, M_{2} ∩ L = L_{2}, and M_{1} ∩ M_{2} = N_{D^{′ t}}(B).

Put M = ⟨L, M_{1}, M_{2}⟩. Then by Lemma 3.3, \overline{M} ≅ M_{22}. Now D ∩ M = M_{1}, D^{t} ∩ M = M_{2} and D ∩ D^{t} = N_{D}(B). Let H = ⟨D, L, t⟩. Then by Lemma 3.4, H̄ ≅ HS, C_{2} × HS or Aut(HS). Thus in any case H̄′ ≅ HS and Theorem 3.1 is proved. The dedicated reader might check that actually H̄ itself is already the Higman-Sims group.

4 A Computational proof that HS is a Subgroup of E_{7}(5).

In this section we give a computer dependent proof that HS < E_{7}(5). Our strategy is to use a machine calculation to prove that HS acts (absolutely) irreducibly on a 133−dimensional, 5−modular, Lie algebra. We then apply the extensive theory of modular Lie algebras to deduce that the Lie algebra is simple, that it is a classical modular Lie algebra of type E_{7} over GF (5), and thence HS < E_{7}(5).

Computation 4.1 We construct an explicit 133−dimensional (absolutely) irreducible matrix representation of HS over GF (5).

Method: We remark that although it seems natural to construct the 133−dimensional representation as a constituent of a tensor product of smaller representations of HS, all useful tensor products are too large for our implementation of the meataxe. For example, the 133−dimensional, 5−modular representations of HS are constituents in the symmetric cube of the 21−dimensional representation and in the tensor product of a 21−dimensional representation with a 55−dimensional representation, however these representations have degrees 1771 and 1155. In order to avoid such large computations, we shall locate a 133−dimensional representation as a constituent of the symmetric square of a 28−dimensional representation of the double cover of HS: this latter representation can be found in a previously known representation of the Harada–Norton group.

We start with the 133−dimensional matrix representation of HN over GF (5) that is constructed in [15]. We locate matrices, x_{1} and y_{1}, that represent HN elements of classes 40A and 12C (since these classes have small centralizers, we locate such elements by a random search). Let x = x_{1}^{20}. Every dihedral group generated by the 2A−element x and a conjugate of the 2B−element y_{1}^{6} has a central involution. By collecting such involutio ns together with the matrix x_{1}, it is an easy matter to generate enough matrices to represent the centralizer of x in HN (we just keep adding involutions until we observe a group that contains an element of order 11). The group C_{HN}(x) has structure 2.HS.2.

We use the meataxe [10] to decompose the restriction of our HN−module into its irreducible constituents of degrees 1, 21, 28, 28, and 55, under the group 2.HS = (C_{HN}(x))′. Another application of the meataxe to the symmetric square of either of the irreducible 28−dimensional 2.HS−modules yields an (absolutely) irreducible matrix representation of HS of degree 133. QED

Let E denote the 133−dimensional HS−module afforded by the matrix representation of Computation 4.1. In our calculations with E we shall use a fixed basis e_{1}, e_{2}, …, e_{133} of E: moreover for technical reasons, we use a basis on which a particular subgroup

P ∼ 2^{4}.2^{3} < 2^{4}.(2 × Sym(4)) < 2^{4}.Sym(6) < HS

acts monomially. We let e_{1}^{∗}, e_{2}^{∗}, …, e_{133}^{∗} denote the dual basis of the dual module E^{∗} (this module is isomorphic to E, but for computational purposes it is convenient to distinguish E and E^{∗}).

Computation 4.2 We compute an HS−invariant product, ∗: Λ^{2}E → E.

Method: The output from this computation is a list of components, a_{i,j,k} of an invariant rank three tensor such that the map e_{i} ∧ e_{j} ↦ ∑_{k} a_{i,j,k} e_{k} extends to an HS−invariant multilinear map. It is convenient to use duality to observe that the map e_{k}^{∗} ↦ ∑_{i,j} a_{i,j,k} e_{i}^{∗} ∧ e_{j}^{∗} extends to an HS−invariant multilinear map. Moreover, since the group HS acts irreducibly on E^{∗}, it is easy to use the group action to compute these tensor components once we know the image of any single vector of E^{∗} under an HS−invariant map: E^{∗} → Λ^{2}E^{∗}.

The module Λ^{2}E^{∗} has degree 8778 and is too large to decompose directly with our implementation of the meataxe: in order to locate all copies of E^{∗} in Λ^{2}E^{∗} we use the condensation techniques described in [14]. Let π denote the idempotent (∑_{p∈P} p)/|P | of the group algebra GF (5)HS. The condensation programs of [14] compute matrix representations of the Hecke algebra πGF (5)HSπ on the condensed modules E^{∗}π and Λ^{2}E^{∗}π (which have degrees 1 and 50). (It is in the computation of these matrix representations of the Hecke algebra that our programs require the group P to act monomially on E and E^{∗}.) A standard meataxe calculation locates the single copy of E^{∗}π in the πGF (5)HSπ−module Λ^{2}E^{∗}π. Thus there is a single embedding of E^{∗} in the HS−module Λ^{2}E^{∗}. Moreover, the embedding of Hecke algebra modules gives the image of the 1−dimensional space of fixed points of P on E^{∗} under the HS−invariant map: E^{∗} → Λ^{2}E^{∗}. As we remarked above, the action of HS now determines the components of the HS−invariant tensor a_{i,j,k}. QED

For each e ∈ E we write ∗e for the matrix that represents the action of right multiplication by e on our HS−invariant algebra. After completing Computation 4.2, we ran a simple precautionary program to verify HS−invariance of our tensor. For each basis vector e_{i} and for each of generator, h, of HS, we checked that the matrices (∗e_{i})^{h} and ∗(e_{i}^{h}) are identical.

Computation 4.3 The HS−invariant product ∗: Λ^{2}E → E of Computation 2 is a Lie product on E. Moreover the Killing form on (E, ∗) is non-singular.

Method: A straightforward computation shows that for each basis vector e_{i} we have

(∗e_{i})(∗e_{1}) − (∗e_{1})(∗e_{i}) − (∗((e_{i})(∗e_{1}))) = 0.

Hence the Jacobi identity holds for any triple of basis vectors of the form e_{1}, e_{i}, e_{j}. Therefore right multiplication by e_{1} is a derivation of (E, ∗); and, by applying the action of HS, we deduce that ∗(e_{1}^{h}) is a derivation of (E, ∗) for any choice of h ∈ HS. Since HS is irreducible on E, we deduce that ∗ is a Lie product on E.

A random search quickly produces e ∈ E with Tr((∗e)(∗e)) ≠ 0. It follows that the Killing form on E is not identically zero: irreducibility of E as a HS−module now shows that the Killing form is non-singular. QED

We complete the proof that HS < E_{7}(5) by applying standard results to show that the automorphism group of (E, ∗) can only be Aut(E_{7}(5)). The following lemma from [13] shows that (E, ∗) is a simple Lie algebra.

Lemma 4.4 Suppose that (X, ∗) is a finite dimensional (non-associative) algebra and that A ≤ Aut(X, ∗) acts irreducibly on X. Then one of the following holds:

(a) The algebra (X, ∗) is simple.

(b) The A−module X is induced from a module of a proper subgroup of A.

(c) The product ∗ is identically zero.

Proof: Let I be a minimal non-zero ideal of (X, ∗), and let I = {I^{a}|a ∈ A}. We say that a subset of I is independent if it consists of independent (vector) subspaces of X. Let J = {I_{1}, I_{2}, … I_{l}} be a maximal independent subset of I. Let Y = ⊕_{k=1}^{l} I_{k}, then Y is an ideal of (X, ∗).

Let I^{a} be any A−image of I. Maximality of J shows that I^{a} ∩ Y is a non-zero ideal; hence, since I^{a} is a minimal ideal, we have I^{a} ⊂ Y. Thus, Y contains a non-zero A−submodule of X, and, since X is irreducible, we have X = Y.

We now suppose that neither (a) nor (b) holds: thus I ≠ X and hence Stab_{A}I is a proper subgroup of A. Moreover, since X is not induced, there is an a ∈ A with I^{a} ∉ J . Then, I^{a} ∗ I_{k} ⊂ I^{a} ∩ I_{k} = {0}, for each I_{k} ∈ J . Thus I^{a} ∗ X = I^{a} ∗ Y = ∑ I^{a} ∗ I_{k} = {0}. The A−invariance of ∗ now gives I_{k} ∗ X = {0} for each I_{k} ∈ J , and thus X ∗ X = ∑ I_{k} ∗ X = {0}. QED

Let (\overline{E}, ∗) be the Lie algebra obtained from (E, ∗) by extending the scalars to the algebraic closure of GF (5). Since E is absolutely irreducible as a HS−module and it is not induced (since HS has no subgroup of index 133), Lemma 4.4 shows that the Lie algebra (\overline{E}, ∗) is simple. Moreover, by Computation 4.3, the simple Lie algebra (\overline{E}, ∗) has a non-singular Killing form.

Over an algebraically closed field, F say, of characteristic p > 3, the modular Lie algebras with a non-singular trace form are completely classified by a theorem of Block and Zassenhaus [1] (this result is also given in [16], page 49). Block and Zassenhaus show that such a modular Lie algebra is a direct sum of abelian Lie algebras, total matrix algebras M_{n}(F) where p|n, and classical simple Lie algebras of types A_{1}, …, E_{8}. In particular, our algebra (\overline{E}, ∗) must be the classical simple Lie algebra of type E_{7} (since no other simple algebra in the list provided by [1] has dimension 133). Therefore, the HS−invariant algebra (E, ∗) is a GF (5)−form of E_{7}. By Theorem IV.6.1 of [16] there is just one GF (5)−form of E_{7} : thus (E, ∗) is the classical simple Lie algebra of type E_{7} over GF (5). It now follows from [18] that Aut(E, ∗) has structure E_{7}(5).2 and we obtain HS < E_{7}(5).

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