The uniqueness case

G. Stroth Martin-Luther-Universität Halle-Wittenberg Institut für Algebra und Geometrie 06099 Halle Germany

April 9, 1997

This paper is part of the revision of the classification of the finite simple groups. The aim is to give a solution of the so called ”uniqueness case problem”. Originally this problem was solved by M. Aschbacher [Asch1]. We do not follow his proof. The proof given in this paper uses the amalgam method, which has been used successfully in many places for dealing with weak closure. Also our theorem looks different from the one in [Asch1] as definitions have changed over the time. In fact our hypothesis is weaker as in [Asch1]. First of all we need some precise definitions. For any group X let

σ(X) = {p|p odd, m_{p}(H) ≥ 4 for some 2-local H of X with | X : H | odd} .

Furthermore let Q be a p-subgroup of X, k ≤ m(Q), let

Γ_{Q,k}(X) = ⟨N_{X}(R) | R ≤ Q, m(R) ≥ k⟩ .

In this paper we will consider groups G which satisfy the following conditions

(1) (i) If H is a 2-local in G with |G : H| odd. Fix S ∈Syl_{2}(H). Then H = ⟨H̃| O_{2}(H̃) = F^{∗}(H̃), S ≤ H̃⟩ (we call G of even type)

(ii) σ(G) ≠ ∅

(2) A group M is called a uniqueness group provided σ(M) ≠ ∅ and one of the following holds

(α) M is a maximal 2-local of G with F^{∗}(M) = O_{2}(M) .

(β) F^{∗}(M) = O_{2}(M)K, where K is a quasisimple group of Lie type in characteristic 2, Z(K) = O_{2}(K), and for every p ∈ σ(M) we have m_{p}(K) ≥ 3 and m_{p}(C I_{M}(K)) ≤ 1.

(3) Let M be a uniqueness subgroup of G. For every p ∈ σ(M) , and P ∈ Syl_{p}(M) we have Γ_{P,2}(G) ≤ M and one of the following holds

(i) If x ∈ P, o(x) = p, m_{p}(C I_{M}(x)) ≥ 3, then N_{G}(⟨x⟩) ≤ M . (In particular for p = 3, N_{G}(⟨x⟩) ≤ M for any x ∈ M, o(x) = 3)

(ii) F^{∗}(M) = O_{2}(M) and M/O_{p′}(M) = M̄ = X̄Ȳ , X̄ ≤ M̄ where m_{p}(X̄) = 1 and Q̄ = O_{p} (Ȳ) ≤ M̄. Further P̄ = (P̄ ∩ X̄) × Q̄ is elementary abelian of order p^{n}. Furthermore Ȳ is isomorphic to a Borel subgroup of an automorphism group of L_{2}(p^{n}), and Γ_{Q,1}(G) ≤ M, where Q is a preimage of Q̄ in P.

We say G is in the uniqueness case if G is K-simple and satisfies (1) and (3) and the following holds.

(4) (i) For every p ∈ σ(G) there is a uniqueness subgroup M_{p} with p ∈ σ(M_{p}).

(ii) Let M be a uniqueness subgroup of G with p ∈ σ(M). If H is any 2-local subgroup of G such that H ∩ M ≥ E, E ≅ E_{p^{2}}, p ∈ σ(M), Γ_{E,1}(G) ≤ M, then H ≤ M.

(iii) If M is a uniqueness subgroup of G, then N_{G}(S) ≤ M for S ∈ Syl_{2}(M).

If M is as in (3) (ii) we call M exceptional. Now we can state our theorem.

Theorem. Let G be in the uniqueness case. Let M be a uniqueness subgroup. If S ∈ Syl_{2}(M), then M contains every maximal 2-local subgroup of G containing N_{G}(S).

It remains to explain what K-simple means. A group is called K-simple if it is a minimal counterexample to the classification theorem, i.e. all simple nonabelian sections of all proper subgroups are in K. Here K is the set of the groups of Lie type, the alternating groups and the 26 sporadic groups.

For the proof of the theorem we assume there is a 2-local N such that N_{G}(S) ≤ N but N ≰ M. From this we construct a pair of subgroups G_{1}, G_{2} such that G_{2} ≤ N and G_{1} ≤ M. The definition of G_{2} is very simple. We just choose G_{2} minimal with respect to |M : G_{2} ∩ M | odd, F^{∗}(G_{2}) = O_{2}(G_{2}) and G_{2} ≰ M. The construction of G_{1} is much more delicate. The definition of G_{1} is given in chapter 2. The main problem is not to lose the information given from p ∈ σ(M) but have G_{1} not to complicated. Then we use the amalgam method for the pair (G_{1}, G_{2}), to describe their structure and come down with a contradiction. Here action of simple groups on ⟨Ω_{1}(Z(S))^{G_{i}}⟩ and ⟨Ω_{1}(Z(S))^{G_{i}}⟩^{G_{j}} , j, i = 1, 2 , i ≠ j , S ∈ Syl_{2}(G_{1} ∩ G_{2}) plays an important role. In particular results about quadratic modules [MeiStr1] [MeiStr2], [Str] and about SC-modules [Coop], [Asch3] are the key tools. For convenience of the reader we give proofs of almost all the results used from [Asch3], as in this paper are a lot of further references we would like to avoid. These results usually provide us with Lie groups over a field of characteristic two and basic modules V (λ). A careful analysis of centralizers of v ∈ V (λ) in the corresponding Lie group mostly shows that we pick up an E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, which centralize v. Now C I_{G}(v) ≤ M, which then at the end forces G_{2} to be in M.

For what follows we will always assume that G is in the uniqueness case and G is a counterexample to the theorem. Furthermore we just drop the term K-simple. Whenever the letter M occurs it will stand for a uniqueness group we will fix from the beginning. In fact if 3 ∈ σ(G) we choose M with 3 ∈ σ(M). With this restriction we choose M of type (2)(β) if possible, otherwise any M will be fine.

The proof is organized as follows. After the preliminary lemmas we will construct the group G_{2} in §2. Then in §3 we show that for a uniqueness group M we must have F^{∗}(M) = O_{2}(M). In §4 we construct the group G_{1}. To the groups G_{1} and G_{2} we use the amalgam method in §6 - §8 to arrive at a contradiction. Very important for this is a certain class of modules V for the group G_{2}/O_{2}(G_{2}). These are modules such that |V : C I_{V} (A)| ≤ |A|^{2} for some nontrivial elementary abelian 2–subgroup of G_{2}/O_{2}(G_{2}).

§ 1 Preliminaries

(1.1) Lemma. Let X ≅ L_{2}(q) , or Sz(q) , q ≥ 4, q even.

(i) Let t ∈ Ω_{1}(S), S ∈ Syl_{2}(X). Then there are conjugates a, b of t such that X = ⟨a, b, t⟩.

(ii) Let A ≤ Ω_{1}(S) , |A| ≥ 4. Then there is some g ∈ X with X = ⟨A, A^{g}⟩.

Proof. Let ⟨t, a⟩ ≤ A ≤ Ω_{1}(S) , |A| ≥ 4. Let K = N_{X}(S) , |K| = q − 1, and a, b ∈ N_{X}(K^{g}), with N_{X}(K^{g}) = ⟨a, b⟩ , g ∈ X. Then we get ⟨a, b, t⟩ ≥ ⟨Ω_{1}(S), Ω_{1}(S)^{b}⟩. Thus to prove (i) and (ii) it is enough to show ⟨Ω_{1}(S) , Ω_{1}(S)^{b}⟩ = X.

We have that Y = ⟨Ω_{1}(S), Ω_{1}(S)^{g}⟩ contains at least q + 1 conjugates of Ω_{1}(S). Thus we are done if X ≅ L_{2}(q), as ⟨Ω_{1}(S), Ω_{1}(S)^{b}⟩ contains all conjugates.

So let X ≅ Sz(q). The number of conjugates of Ω_{1}(S) in Y is nq + 1. But then nq + 1 | q(q^{2} + 1). Which gives n = q and so Ω_{1}(S)^{X} ≤ Y, hence X = Y.

(1.2) Lemma. Let X ≅ G(q) be a Lie group over a field of characteristic two, q > 2. Let C be the Cartan subgroup and m_{p}(C) ≤ 3 for any prime p. Then X is one of the following: L_{n}(q), n ≤ 4, Sp_{4}(q), Sp_{6}(q), U_{n}(q), n ≤ 7, Ω_{8}^{−}(q), ^{2}F_{4}(q), ^{3}D_{4}(q), G_{2}(q), or Sz(q).

Proof. Let first X be untwisted of Lie rank r. Then

| C | = \frac{1}{d}(q − 1)^{r}

Let r ≥ 4. Then as m_{p}(C) ≤ 3, we have q − 1 = p = d. Furthermore r = 4. But checking the possible values for d gives a contradiction. So we have r ≤ 3 and then X ≅ L_{2}(q), L_{3}(q), L_{4}(q), Sp_{4}(q), Sp_{6}(q) or G_{2}(q).

Assume now that X is twisted. Let X ≅ ^{2}E_{6}(q). Then

| C | = \frac{1}{d}(q − 1)^{2}(q^{2} − 1)^{2} , d = (3, q + 1) .

Let p | q − 1. Then C contains an elementary abelian subgroup of order p^{4}, a contradiction.

Let X ≅ U_{n}(q). If n is even then

| C | = \frac{1}{d}(q − 1)(q^{2} − 1)^{\frac{n}{2} −1}

Thus n ≤ 6.

Let n be odd. Then

| C | = \frac{1}{d}(q^{2} − 1)^{\frac{n−1}{2}}

This implies n ≤ 7.

Finally assume that X ≅ Ω_{2n}^{−}(q). Then

| C | = (q^{2} − 1)(q − 1)^{n−2}

Hence n − 2 ≤ 2. We get n = 4, as Ω_{6}^{−}(q) ≅ U_{4}(q).

(1.3) Lemma. Let X/Z(X) ∈ K , Z(X) a 3-group, then one of the following holds (i) m_{3}(X) = 0 , X ≅ Sz(q) , q even.

(ii) m_{3}(X) = 1 , X ≅ L_{2}(q), L_{3}(q), U_{3}(q), J_{1}.

(iii) m_{3}(X) = 2, X ≅ 3 · A_{6}, 3 · A_{7}, 3 · M_{22}, SL_{3}(q), SU_{3}(q), A_{7}, A_{6}, L_{3}(3), U_{3}(3), L_{3}(q), U_{3}(q), P Sp_{4}(q), G_{2}(q), ^{3}D_{4}(q), ^{2}F_{4}(q), L_{4}(q), U_{4}(q), L_{5}(q), U_{5}(q), M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, HiS, He, Ru, J_{4}.

(iv) m_{3}(X) = 3 and X ≅ 3 · O′N, A_{9}, A_{10}, A_{11}, L_{2}(27), P Sp_{4}(3), Sp_{6}(q), Ω_{8}^{−}(q), L_{4}(q), U_{4}(q), L_{6}(q), U_{6}(q), L_{7}(q), U_{7}(q), J_{3}.

Proof . Just inspection of the groups in K

(1.4) Lemma. Let X/Z(X) ∈ K, X/Z(X) sporadic or alternating, Z(X) a 3–group. Suppose m_{3}(X) ≤ 3, then m_{p}(X) ≤ 2 for any odd prime p ≠ 3.

Proof . By (1.3) we have X ≅ J_{1}, M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, 3 · M_{22}, HiS, He, Ru, J_{4}, 3 · O′N, or J_{3}, or X/Z(X) is alternating, i.e. X ≅ 3 · A_{6}, 3 · A_{7}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}.

Assume now p^{3} | |X| , p odd , p ≠ 3. Then we see X ≅ J_{4} (p = 11), HiS (p = 5), Ru (p = 5), He (p = 7) or 3·O′N (p = 7). But in all these cases we have m_{p}(X) = 2 by [CCNPW].

(1.5) Lemma. Let K be a finite simple group in the list K, m_{p}(K) ≤ 3 for any odd prime p. Then K is one of the following:

(i) L_{2}(q), Sz(q), L_{3}(q), U_{3}(q), P Sp_{4}(q), q some prime power

(ii) Sp_{4}(2^{n}), G_{2}(2^{n}), ^{2}F_{4}(2^{n})′, ^{3}D_{4}(2^{n}), L_{4}(2^{n}), L_{5}(2), L_{6}(2), L_{7}(2), U_{4}(2^{n}), Sp_{6}(2^{n}), Ω_{8}^{−}(2^{n})

(iii) A_{n}, 6 ≤ n ≤ 11

(iv) J_{i}, 1 ≤ i ≤ 4, M_{n}, n ∈ {11, 12, 22, 23, 24}, HiS, Ru, He.

If K ∈ K, U/Z(U) ≅ K, U′ = U, 1 ≠ | Z(U) | odd and m_{p}(U) ≤ 3 for every odd prime p. Then U is isomorphic to 3 · A_{6}, 3 · A_{7}, 3 · M_{22}, SL_{3}(q), SU_{3}(q), q a prime power, or 3 · O′N.

Proof . This is easily established by going over the list in (1.3).

(1.6) Lemma. Let X be a p-group, p odd, cl(X) ≤ 2, X = Ω_{1}(X) and m_{p}(X) ≤ 3. Then X is elementary abelian, extraspecial of width 1, a direct product of a cyclic group of order p with an extraspecial group of width 1, or an extraspecial group of width 2.

Proof . We may assume cl(X) = 2. We have X = {x|x^{p} = 1}. Let m_{p}(X) = 2. Then |Z(X)| = p. Let ω ∈ X \ Z(X). Then [ω, X] = Z(X) and so | X : C I_{X}(ω)| = p. Now m_{p}(C I_{X}(ω)) = 2 and so as | Z(C I_{X}(ω))| ≥ p^{2} we get |C I_{X}(ω)| = p^{2}. This shows |X| = p^{3} and so X is extraspecial of width 1.

Let now m_{p}(X) = 3. Assume | Z(X)| = p^{2}. Again choose ω ∈ X \ Z(X). Then | X : C I_{X}(ω)| ≤ p^{2} and as | C I_{X}(ω)| = p^{3}, we get |X| ≤ p^{5}.

Let |X| = p^{4}. Then X′ = [X, ω] is of order p. Thus X is a direct product of a cyclic group of order p with an extraspecial group of width 1.

Let |X| = p^{5}. Choose C I_{X}(ω) ≤ Y < X, |Y | = p^{4}. Then as just seen Y is a direct product of a cyclic group of order p by an extraspecial group of width 1. Now choose φ ∈ X \ Y. Then [ω, φ] = t ∉ Y′. Let ν ∈ Y, 1 ≠ [ω, ν] = s ∈ Y′. We have

[φ, ν] = s^{i} t^{i} , for some 0 ≤ i, j ≤ p − 1 .

But then

[φ, νω^{j}] = s^{i}

and so we may assume

[φ, ν] = s^{i} .

Hence [X, ν] ≤ ⟨s⟩ and then |X : C I_{X}(ν)| ≤ p, yielding |X| = p^{4}, a contradiction.

So we have |Z(X)| = p. Then Z(X) = Φ(X) = X′ and X is extraspecial. As m_{p}(X) = 3 we have that X is of width 2.

(1.7) Lemma. Let X be some group with O_{2}(X) = 1 and m_{p}(X) ≤ 3 for every odd prime p. Suppose furthermore that for S ∈ Syl_{2}(X) there is exactly one maximal subgroup Y of X containing S. Then either X is solvable, or E(X) is one of the following:

(i) Sz(q), L_{2}(q), (S)L_{3}(q), (S)U_{3}(q), Sp_{4}(q), L_{2}(q)×L_{2}(q), Sz(q)×Sz(q), q even

(ii) L_{2}(p), L_{2}(p^{2}), L_{2}(p^{3}), L_{3}(p), U_{3}(p), P Sp_{4}(p), L_{2}(p) × L_{2}(p), p > 3 some odd prime, or L_{2}(27) or L_{3}(3)

(iii) A_{6}, A_{9}, 3 · A_{6}, 3 · A_{6} ∗ 3 · A_{6}, SL_{3}(4) ∗ SL_{3}(4), SU_{3}(8) ∗ SU_{3}(8).

or F (X) is a p-group with Ω_{1}(F (X)) is elementary abelian of order p^{3} and E(X/F (X)) ≅ L_{2}(p) acts irreducibly on Ω_{1}(F (X)).

If E(X) ≅ (S)L_{3}(q), Sp_{4}(q), q even, 3 · A_{6}, 3 · A_{6} ∗ 3 · A_{6}, or SL_{3}(4) ∗ SL_{3}(4), there is some x ∈ X acting nontrivially on the corresponding Dynkin diagram.

Proof . We may assume that X is nonsolvable. Assume first E(X) = 1. Set F = F (X). Let p | |F (X)| and P ∈ Syl_{p}(F (X)). We may assume that X/C I_{X}(P) is nonsolvable. Let C be a critical subgroup of P and U = Ω_{1}(C). Set X_{1} = C I_{X}(U).

Suppose S ∩ X_{1} ≠ 1. By the Frattini argument we have X = X_{1}N_{X}(S ∩ X_{1}). As O_{2}(X) = 1 we get N_{X}(S ∩X_{1}) ≠ X. But then X_{1}S and N_{X}(S ∩X_{1}) are contained in different maximal subgroups containing S, a contradiction.

So we have S ∩ X_{1} = 1. Let t ∈ Z(S)^{♯} such that C I_{X}(t) covers (X/X_{1}) / O_{p}(X/X_{1}). As O_{2}(X) = 1, we have C I_{X}(t) ≠ X. Let Y be the preimage of O_{p}(X/X_{1}). Then Y S ≠ X and so C I_{X}(t) and Y S are contained in different maximal subgroups, a contradiction.

Now by (1.6) we know the structure of U. The structure of Sp_{4}(p) [Mi2] shows that U has to be elementary abelian of order p^{3} and X/X_{1} is a subgroup of L_{3}(p). Now S X_{1}/X_{1} is contained in exactly one maximal subgroup of X/X_{1}, p | |E(X/X_{1})|, otherwise X splits over P. So with [Mi1] we get that E(X/X_{1}) ≅ L_{2}(p). Hence X acts irreducibly on U.

As above the Frattini argument shows that every Sylow subgroup of X_{1} is normal in X, so X_{1} = F. Suppose F ≠ P. Then choose Q ∈ Syl_{q}(F), q ≠ p. We get X = QC I_{X}(Q). So O^{p′}(X)S ≠ X. Now O^{p′}(X)S and O_{p′}(F)S are contained in different maximal subgroups of X, a contradiction.

Suppose now E(X) ≠ 1. We have X = E(X)S, otherwise N_{X}(S ∩ E(X)), E(X)S are contained in different maximal subgroups. Furthermore this implies that S is transitive on the components of E(X). In particular E(X) contains at most two components.

Let first E(X) be quasisimple. Then E(X) is described in (1.5). If E(X)/Z(E(X)) is a Lie group over GF (q), then S ∩ E(X) ≤ P, P a minimal parabolic. Hence either E(X) is of rank 1 or E(X) is of rank 2 and S induces a diagram automorphism. This is (i). Suppose E(X)/Z(E(X)) is alternating. If n = 11, then S ∩ E(X) ≤ A_{Ω}, Ω = {1, …, 10} and S ∩ E(X) ≤ A_{Ω_{1}}⟨x⟩, Ω_{1} = {1, …, 9}, x = (1, 2)(10, 11). But both groups generate E(X). If n = 10, then S ∩ E(X) is in A_{Ω}⟨x⟩, Ω = {1, …, 8}, x = (1, 2)(9, 10) and N_{E(X)}(⟨(1, 2)(3, 4), (1, 2)(5, 6), (1, 2)(7, 8), (1, 2)(9, 10)⟩). If n = 8, then E(X) ≅ L_{4}(2), a case just done. If n = 7, then S ∩ E(X) is in A_{Ω}, Ω = {1, 2, …, 6} and A_{Ω_{1}}⟨x⟩, Ω_{1} = {1, …, 5}, x = (1, 2)(6, 7). This finishes the case of an alternating group.

Let now K/Z(K) be sporadic. Application of [RoStr] shows that K is generated by minimal parabolics up to K ≅ J_{1} or M_{11}. But the latter contains M_{10} and GL_{2}(3). The group J_{1} contains Z_{2} × A_{5} and the normalizer of a Sylow 2-subgroup. Hence in any such group there are at least two maximal subgroups containing S ∩ E(X).

We are left with K/Z(K) ≅ G(r), r odd, r = p^{f}. As m_{p}(K) ≤ 3, we get from (1.5) K/Z(K) ≅ L_{2}(p), L_{2}(p^{2}), L_{2}(p^{3}), L_{3}(p), U_{3}(p), P Sp_{4}(p). This is (ii).

Let now E(X) = X_{1}X_{2}. Then m_{p}(X_{1}) = 1 for any prime p with p | |̸Z(E(X))| and m_{p}(X_{1}) = 2 for p | |Z(E(X))|. Application of (1.3) shows that X_{1} ≅ Sz(q), L_{2}(q), (S)U_{3}(q), (S)L_{3}(q), J_{1}, 3·A_{6}, 3·A_{7} or 3·M_{22}.

If X_{1} ≅ (S)U_{3}(q) or (S)L_{3}(q) and p | q + 1 or p | q − 1, p ≠ 3, then X_{1} contains an elementary abelian group of order p^{2} intersecting the center trivially. Then we get X_{1} ≅ SL_{3}(4), L_{3}(2) or SU_{3}(8).

Let X_{1} ≅ J_{1}, 3·A_{7}, 3·M_{22}, 3·A_{6}, SL_{3}(4), L_{3}(2). Then there are subgroups A, B of X_{1} such that ⟨A, B⟩ = X_{1}, S ∩ X_{1} ≤ A ∩ B. Let A, B be normal in N_{S}(X_{1}). Choose g ∈ X with X_{1}^{g} = X_{2}. Then ⟨A, A^{g}, S⟩ and ⟨B, B^{g}, S⟩ are both different from X, but ⟨X_{1}, S⟩ = X. This shows X_{1} ≅ 3 · A_{6}, SL_{3}(4) or L_{3}(2) and there is some x ∈ X acting nontrivally on the Dynkin diagram, i.e. A^{x} = B.

(1.8) Lemma. Let O_{2}(X) = 1, S ∈ Syl_{2}(X), S is contained in exactly one maximal subgroup of X. Suppose X is solvable. Let furthermore m_{p}(X) ≤ 3 for all odd prime p. Then X = O_{p}(X)S for some prime p and one of the follwing holds

(i) O_{p}(X) is cyclic and S is cyclic

(ii) O_{p}(X) ≅ Z_{p^{1+2}} or abelian of rank two

(iii) O_{p}(X) ≅ Z_{p^{1+4}}

(iv) There is U ≅ E_{p^{3}}, U ◁ X, m_{2}(S) ≤ 2.

Proof . Let F = F (X). As O_{2}(X) = 1, we have that F is of odd order. Let S_{1} ≤ S be a preimage of O_{2}(X/F). Now N_{X}(S_{1}) and F S are in different maximal subgroups of X, provided F S ≠ X. So we have F S = X. Now O_{p}(F)S and O_{p′}(F)S are in different maximal subgroups or F = O_{p}(F). So we have that F = O_{p}(X).

If m_{p}(F) = 1, we get (i). As S is contained in exactly one maximal subgroup we have that

(∗) S acts irreducibly on F/Φ(F) .

Let now m_{p}(F) = 2, C be a critical subgroup and U = Ω_{1}(C). If m_{p}(C) = 1, then C is cyclic and so X/C is abelian. But then |F/Φ(F)| = p by (∗) and so F is cyclic, i.e. (i) holds.

Let m_{p}(U) = 2. Then U is elementary abelian or of type p^{1+2}. Hence in any case X/C I_{X}(U)U is a subgroup of GL_{2}(p) having a normal Sylow psubgroup. So by (∗) we get C I_{X}(U)U = F. If U ≅ Z_{p^{1+2}} we see U = F by (∗). If U ≅ ZZ_{p} × ZZ_{p}, we get that F is abelian of rank two, this is (ii).

So assume next m_{p}(U) = 3. By (1.6) U is either elementary abelian, ZZ_{p} × Z_{p^{1+2}}, or Z_{p^{1+4}}. Suppose we have U ≅ Z_{p^{1+4}}. Then F = UC I_{F} (U). By (∗) we get C I_{F} (U) ≤ U, so F = U. This is (iii).

Let U ≅ ZZ_{p} × Z_{p^{1+2}}. Set Z = Z(U). Now |F : C I_{F} (Z)| ≤ p. By (∗) we get Z ≤ Z(F). Furthermore there is W ≤ Z, W ∩ U′ = 1 and W ≤ X. Now F = C I_{F} (U/W)U. So by (∗)F = U, a contradiction to (∗) again.

Let finally U be elementary abelian. Then X/C I_{X}(U) is a subgroup of GL_{3}(p). Suppose m_{2}(S) = 3. Then Z(S) contains a fours group ⟨z, t⟩. Now F = ⟨C I_{F} (z), C I_{F} (t), C I_{F} (zt)⟩. We may assume C I_{F} (z) ≠ 1 ≠ C I_{F} (t). Then C I_{F} (z)S and C I_{F} (t)S lie in different maximal subgroups, a contradiction. So m_{2}(S) ≤ 2.

(1.9) Lemma. Let X ≅ A_{n} or Spor. Suppose P ≤ X, P contains a Sylow 2–subgroup of X and P is a {2, p}-group. Then p = 3 or X ≅ J_{1} and p = 7.

Proof . For X ≅ A_{n} this can be found in [Asch2, (6.1)]. For the sporadic groups this is in [RoStr].

(1.10) Lemma. Let K be a Lie group in odd characteristic, K ≅^{2̸}G_{2}(3^{n}). Let ω ∈ Aut(K), o(ω) = p > 3, p prime, S a Sylow 2–subgroup of K, [ω, S] ≤ S. Then either ω induces a field automorphism on K or ω is a diagonal automorphism with [S, ω] = 1.

Proof . This is [Asch2, (6.3)]

We are now going to investigate modules for the groups occuring in the proof of the theorem. Basically there will be two types of modules.

(1) Modules V over GF (2) admitting some elementary abelian 2-group A

such that

[V, A, A] = 0

(2) Modules V over GF (2) for some group G such that |V : C I_{V} (t)| ≤ 2^{m_{2}(G)} for some involution t ∈ G.

First we collect some results concerning modules of the first kind.

(1.11) Lemma. Let a be an involution in G, L be a component of G, O_{2}(G) = 1 and A a four-subgroup of G with a ∈ A, [V, a, A] = 0 for a faithful GF (2) G-module V. If C I_{A}(L) = 1, then either

(a) [L, a] ≤ L, or

(b) ⟨A, L⟩ ≅ O_{4}^{+}(2^{n}) and [L^{A}, V]/C I_{[L^{A},V]}(L^{A}), is a direct sum of natural modules for ⟨L^{A}⟩ ≅ Ω_{4}^{+}(2^{n}).

Proof . Besides of the assertion that we have a direct sum of natural modules this is [GoLyS, (25.14)]. But as shown there an element x ∈ L of order three acts fixed point freely. So we get with [Hi, (8.2)] that V is a direct sum of natural L_{2}(2^{n})-modules. As ⟨a^{G}⟩ ≅ O^{+}(4, 2^{n}), we get the assertion.

(1.12) Lemma. Let G be a group such that F^{∗}(G) is a perfect central extension of a finite simple group. Suppose there is some elementary abelian subgroup A of G, |A| ≥ 4, such that for some irreducible nontrivial faithful module V over GF (2) we have [V, A, A] = 0. Then

(i) If F^{∗}(G)/Z(F^{∗}(G)) is sporadic, then F^{∗}(G)/Z(F^{∗}(G)) ≅ M_{12}, M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2} or Sz. If |A| ≥ 8, then F^{∗}(G) ≅ 3 · M_{22}.

(ii) If F^{∗}(G)/Z(F^{∗}(G)) is a Lie group in odd characteristic which is not a Lie group in even characteristic too, then E(G) ≅ 3 · U_{4}(3). Furthermore V is the 12–dimensional module.

(iii) If F^{∗}(G)/Z(F^{∗}(G)) is alternating, then either V is the natural module or a spin module or F^{∗}(G) ≅ 3·A_{6} and V is the 6-dimensional module. If |A| > 8, then V is natural or G ≅ A_{8} and |V | = 16. If V is the spinmodule and |A| = 4, then A is conjugate to ⟨(12)(34), (13)(24)⟩ or ⟨(12)(34)(56)(78), (13)(24)(57)(68)⟩. If |A| = 8 the A is conjugate to ⟨(12)(34)(56)(78), (13)(24)(57)(68), (14)(26)(37)(48)⟩ under Σ_{n}.

Proof . (i) This is [MeiStr1].

(ii) This is [MeiStr2]. For F^{∗}(G) ≅ L_{2}(q), q odd it was a quotation to [Asch3, (6.4)]. For convenience of the reader we give a proof for the latter. We have F^{∗}(G) ≅ L̸_{3}(2), A_{6}, 3 · A_{6}, A_{5}. If A ≤ F^{∗}(G), set Y = ⟨A^{C I_{G}(a)}⟩, a ∈ A^{♯}. If A ≰ F^{∗}(G) choose a ∈ A \ F^{∗}(G), set Y = C I_{G}(a). Then in all cases G = ⟨Y, Y^{g}⟩ for any g ∈ G \ C I(a) by [Hu, (II.8.27)]. Set W = [V, a]. Then by quadratic action [Y, W] = 0. Now as in the proof of (1.11) we get that a^{G} is a set of odd transpositions. If A ≤ F^{∗}(G), this shows that 8 | |̸F^{∗}(G)| and C I(a) is a 2-group. But then F^{∗}(G) ≅ A_{5}, a contradiction. If a ∉ F^{∗}(G), then C I_{Y} (A ∩ F^{∗}(G)) is a 2-group. Furthermore Sylow 2-subgroups are abelian. This shows that Y ≅ L_{2}(5) and a induces a field automorphism. Hence F^{∗}(G) ≅ L_{2}(25). Now G is generated by three conjugates of a. This shows |C I_{V} (a)/[V, a]| ≤ |[V, a]|. As [Y, V] ≤ C I_{V} (a), [[V, a], Y] = 0. We get either [V, a] ∩ [V, b] ≠ 0 for ⟨a, b⟩ = A, a ∼ b ∈ G or C I_{V} (a) = [V, a] ⨁[V, b].

The former is not possible as [V, a] ∩ [V, b] would be centralized by ⟨Y, Y^{g}⟩ = G. Hence we have C I_{V} (a) = C I_{V} (b) and so C I_{V} (a) is invariant under ⟨Y, Y^{g}⟩, a contradiction.

(iii) This is [MeiStr1]. Suppose |A| ≥ 4. Let a ∈ A^{♯}. Let k be the number of fixed points of a. Then there is K ≤ C I_{G}(a), K ≅ Σ_{k}. Furthermore C I_{C I_{G}(a)}(K′) is an extension of a 2–group by Σ_{m}, m = (n − k)/2. Now choose a ∈ A with m > 2 if possible. Suppose first [A, C I_{C I_{G}(a)}(K′)] ≠ 1. If m ≥ 5, then Σ_{m} is nonsolvable and so C I_{C I_{G}(a)}([V, a]) contains an elementary a belian subgroup of O_{2}(C I_{G}(a)) of order 2^{m−1}. But then this group contains a conjugate of (12)(34) which contradicts [MeiStr1, (4.3)].

Let m = 4. Then a ∼ (12)(34)(56)(78). Furthermore as we may assume that no x ∼ (12)(34) is contained in ⟨A^{C I_{G}(a)}⟩ we see that A is conjugate to a subgroup of ⟨(12)(34)(56)(78), (13)(24)(57)(68), (15)(26)(37)(48)⟩.

Let m = 3. Then C I(K′) ≤ Σ_{6} and a ∼ (12)(34)(56). Then ⟨A^{C I_{G}(a)}⟩ contains some x ∼ (12)(34), contradicting [MeiStr1, (4.3)].

So let [A, O^{2′}(C I_{C I_{G}(a)}(K′))] = 1. If [A, K′] ≠ 1, then [K′, [V, a]] = 0. If k ≥ 4, then K′ contains some x ∼ (12)(34). This again contradicts [MeiStr1, (4.3)]. Let k ≤ 3. As [A, O^{2′}(C I_{C I_{G}(a)}(K′))] = 1 and m > 2, there is x ∼ (12) in A, a contradiction. So we are left with [A, K′] = 1 = [A, C I_{C I_{G}(a)}(K′)]. But this is impossible with m > 2.

So we have m ≤ 2 for all a ∈ A^{♯}. As there is no fours group of transvections we may assume a = (12)(34) ∈ A. Now A ≥ ⟨a, b⟩, b = (13)(24), (13)(24) or (34). Let [b, K′] ≠ 1. Then b = (12)(56) and so K′ contains no involutions by [MeiStr1, (4.3)]. This shows k ≤ 3 and so A ≤ Σ_{7}. But this group A = ⟨(12)(34), (12)(56)⟩ does not act quadratically on the four dimensional module.

Assume now b = (34). Then [a, E(C I_{G}(b))] ≠ 1. Now E(C I_{G}(b)) ≅ Σ_{n−2}, which is nonsolvable. But then ⟨(34), (12)(56)⟩ acts quadratically, a contradiction.

(1.13) Lemma. Let F^{∗}(G) ≅ M_{22} or 3 · M_{22} and V be an irreducible faithful GF (2)G–module. If G contains A with |A| ≥ 4 and [V, A, A] = 0, then one of the following holds

(i) |A| = 4, G ≅ Aut(M_{22}) and V is one of the two 10–dimensional modules.

(ii) 4 ≤ |A| ≤ 16 and ⟨A^{G}⟩ ≅ 3 · M_{22}, V is the 12–dimensional module for G.

Proof . See [MeiStr1]

(1.14) Lemma. Let G = G(q), q = 2^{n}, be a Lie group and V an irreducible GF (2)-module. Let A be a fours group with [V, A, A] = 0. If A intersects some long root group R nontrivially but A ≰ R. Then one of the following holds

(i) G ≅ (S)L_{n}(q), (S)U_{n}(q), Sp_{2n}(q) or F_{4}(q) and V = V (λ) for some fundamental weight λ.

(ii) G = Ω_{2n}^{±}(q) and V is the natural or spin module. (iii) G = E_{6}(q) and V = V (λ_{1}) or V (λ_{6}) (iv) G = E_{7}(q) and V = V (λ_{7})

(v) G = ^{2}E_{6}(q) and V = V (λ_{4}) (vi) G = G_{2}(q) or ^{3}D_{4}(q) and V is the natural module.

Proof . [Str]

The modules from (1.14) will be called strong quadratic in this paper.

(1.15) Lemma. Let G = G(q), q = 2^{n}, be a Lie group, V an irreducible GF (2)-module. Let A be an elementary abelian subgroup of G with [V, A, A] = 0. Then each of the following implies that V is one of the modules of (1.14).

(1) G ≅ (S)L_{n}(q), |A| > q^{\frac{n(n−2)}{4}} or q^{\frac{(n−1)^{2}}{2}} according to n > 2 is even or odd

(2) G ≅ (S)U_{n}(q), n ≥ 4, |A| > q^{[ \frac{n−2}{2} ]^{2}+1}, q^{[ \frac{n−2}{2} ]^{2}} for n > 5.

(3) G ≅ Sp_{2n}(q), n ≥ 3, |A| > q^{\binom{n−1}{2}+2}

(4) G ≅ Sp_{4}(q), q > 2, |A| > q

(5) G ≅ Ω_{2n}^{±}(q), n ≥ 4, |A| > q^{\frac{n^{2}−5n+10}{2}}

(6) G ≅ G_{2}(q), |A| > q

(7) G ≅ ^{3}D_{4}(q), |A| > q

(8) G ≅ ^{2}F_{4}(q), |A| > q (i.e. there is no such module)

(9) G ≅ E_{6}(q), |A| > q^{11}

(10) G ≅ E_{7}(q), |A| > q^{16}

(11) G ≅ E_{8}(q), |A| > q^{28} (i.e. there is no such module)

(12) G ≅ ^{2}E_{6}(q), |A| > q^{10}

(13) G ≅ F_{4}(q), |A| > q^{7}.

Proof . (1): Let P be a parabolic of G with |O_{2}(P)| = q^{n−1} and O^{2′}(P/O_{2}(P)) ≅ SL_{n−1}(q). We may assume A ≤ P. Now the assumption says A ∩ O_{2}(P) ≠ 1. Hence the assertion follows with (1.14).

(2),(5),(9)-(12): Let R be some long root group, P = N_{G}(R). We may assume A ≤ P. Set E = O_{2}(P). Then

|A ∩ E| ≤ q .

Otherwise there is ⟨a, b⟩ ≤ A ∩ E and some g ∈ E with a^{g} = a, b^{g} = br, r ∈ R^{♯}. But then ⟨a, r⟩ also acts quadratically on V and we have the assertion with (1.14).

So we have |A| ≤ q · 2^{m_{2}(P/E)}. But this is exactly the given bound besides the case U_{n}(q) and Ω_{2n}^{±}(q). Here the bound is 2^{m_{2}(P/E)}. Let G ≅ Ω_{2n}^{±}(q), O^{2′}(P/E) ≅ Ω_{2n−4}^{±}(q) × L_{2}(q). If t ∈ A and t corresponds to a nontrivial element of the L_{2}(q), then |[t, E/R]| = q^{2n−4}. If A ∩ E ≠ 1 and there is a ∈ A ∩ E, b ∈ C I_{F} (a) with t^{b} = tc ≠ t, and C I_{E}(a) ≠ C I_{E}(c). Now ⟨a, c⟩ acts quadratically and as C I_{E}(a) ≠ C I_{E}(c) also ⟨a, r⟩, r ∈ R, acts quadratically. The assertion follows with (1.14).

The order of A shows A ∩ E ≠ 1. As |C I_{E}(A ∩ E)| = q^{4n−9}, we see that there is b ∈ C I_{E}(A ∩ E) with t^{b} = tc ≠ t, C I_{E}(c) ≠ C I_{E}(A ∩ E). Let G ≅ U_{n}(q). Then A ∩ E ≠ 1 and A contains an involution x which does not act as an unitary transvection on E/R. But then there is some t ∈ C I_{E}(A ∩ E) with x^{t} = xs, C I_{E}(A ∩ E) ≠ C I_{E}(s). As ⟨a, s⟩, a ∈ (A ∩ E)^{♯} acts quadratically we now get the same contradiction as before.

(3): Let again R be some short root group, P = N_{G}(R). Then O_{2}(P) is a direct product of an elementary abelian group of order q^{2} by a semispecial group of order q^{1+2 (2n−4)}, and O^{2}(P/O_{2}(P)) ≅ Sp_{2n−4}(q) × L_{2}(q). Again we have |(A ∩ O_{2}(P))Z(O_{2}(P))/Z(O_{2}(P))| ≤ q. Furthermore |A ∩ Z(O_{2}(P))| ≤ q^{2}. If |A ∩ Z(O_{2}(P))| > q then A ≤ C I(Z(O_{2}(P))) and so AO_{2}(P)/O_{2}(P) ≤ Sp_{2n−4}(q).Hence we have |A| ≤ q^{2}2^{m_{2}(Sp_{2n−4}(q))}, which is exactly the given bound.

(4): Let S ∈ Syl_{2}(G). Then S = F E, where E and F both are elementary abelian of order q^{3}. We may assume A ≤ E. As A does intersect trivially any root group, we get |A ∩ E ∩ F | ≤ q. So we may assume (maybe after conjugation in N_{G}(E)) that A ∩ F ≠ 1 and |A : A ∩ F | > 2. So let b ∈ A ∩ F and ⟨b_{1}, b_{2}⟩ ≤ A, ⟨b_{1}, b_{2}⟩ ∩ F = 1, |⟨b_{1}, b_{2}⟩| = 4. Then there are c_{1}, c_{2} ∈ F with b_{1}^{c_{1}} = b_{1}bt_{1}, b_{2}^{c_{2}} = b_{2}bt_{2}, t_{1}, t_{2} ∈ Z(O^{2′}(N_{G}(F))). As q > 2, F is an indecomposable O^{2′}(N_{G}(F))-module. This proves t_{1} ≠ 1 or t_{2} ≠ 1. We may assume t_{1} ≠ 1. Then ⟨b, bt_{1}⟩ = ⟨b, t_{1}⟩ acts quadratically on V. But t_{1} is a root element. The assertion follows with (1.14).

(6): Let P be the parabolic with Z(O^{2′}(P)) = 1, Q = O_{2}(P). We have O^{2′}(P)/Q ≅ L_{2}(q) and so A ∩ Q ≠ 1. Let R be a root group in Z(Q) and B = ⟨R^{P} ⟩. Then B is the natural O^{2′}(P)/Q-module. Hence by (1.14) we assume A ∩ B = 1. As G does not contain elementary abelian subgroups of order greater than q^{3}, we get A ≰ Q. Now choose a ∈ A\Q, ã ∈ A∩Q, s ∈ B with a^{s} = at ≠ a. Then ⟨ã, t⟩ acts quadratically and t is a root element. The assertion follows with (1.14).

(7): Let R be a root group, P = O^{2′}(N_{G}(R)). Then P/O_{2}(P) ≅ L_{2}(q^{3}), and O_{2}(P) is semispecial. Hence by (1.14) |A∩O_{2}(P)| ≤ q. If t ∈ A\O_{2}(P), then |[O_{2}(P)/R, t]| = q^{4}. So if A ∩ O_{2}(P) ≠ 1, we find a ∈ A ∩ O_{2}(P), b ∈ C I_{O_{2}(P)}(a), t ∈ A \ O_{2}(P), with t^{b} = tc, C I_{O_{2}(P)}(c) ≠ C I_{O_{2}(P)}(a), and ⟨a, c⟩ acts quadratically.

Suppose A ∩ O_{2}(P) = 1. Then |AO_{2}(P)/O_{2}(P)| > q. This implies there are b, t ∈ A with |[C I_{O_{2}(P)/R}(b), t]| = q^{2}. As C I_{O_{2}(P)/R}(b) = [O_{2}(P)/R, b]. We find ⟨a, b⟩ acts quadratically for some a ∈ C I_{O_{2}(P)/R}(b). But now as before we get some c ∈ O_{2}(P), C I_{O_{2}(P)}(a) ≠ C I_{O_{2}(P)}(c) with ⟨a, c⟩ acts quadratically.

Now as C I_{O_{2}(P)}(a) ≠ C I_{O_{2}(P)}(c) there is x ∈ O_{2}(P), a^{x} = a, c^{x} = cr, r ∈ R^{♯}. Hence ⟨a, r⟩ acts quadratically and the assertion follows with (1.14).

(8): Let R be some long root group, P = O^{2}(N_{G}(R)), Q = O_{2}(N_{G}(R)). We have A ∩ Q ≠ 1. Let a ∈ A \ Q. Then |Z_{2}(Q) : C I_{Z_{2}(Q)}(a)| = q^{2}. Choose b ∈ A ∩ Q. Then |Z_{2}(Q) : C I_{Z_{2}(Q)}(a)| ≤ q. Hence there is t ∈ Z_{2}(Q) with b^{t} = b and a^{b} = ax ≠ a. This shows that ⟨x, b⟩ does act quadratically. Furthermore C I_{Q}(x) ≠ C I_{Q}(b).

If A ≤ Q, a, b ∈ A, with C I_{Z_{2}(Q)}(a) ≠ C I_{Z_{2}(Q)}(b), then there is t ∈ Z_{2}(Q) with b^{t} = b, a^{t} = ax ≠ a and again ⟨b, x⟩ acts quadratically C I_{Q}(x) ≠ C I_{Q}(b).

As |A| > q there are always such pairs a, b. Hence we just have to treat the case of a quadratic fours group ⟨b, x⟩ ≤ Q, x ∈ Z_{2}(Q), C I_{Q}(b) ≠ C I_{Q}(x). By (1.14) ⟨x, b⟩ ∩ R = 1. But then there is some t ∈ Q with b^{t} = br ≠ b, r ∈ R, x^{t} = x. Hence ⟨x, r⟩ does act quadratically and the assertion follows with (1.14).

(13): Let R be a root group, P = O_{2}(N_{G}(R)), Q = O_{2}(P). Then Q is a direct product of an elementary abelian group of order q^{6} by a semispecial group of order q^{1+8}. Further P/Q ≅ Sp_{6}(q). Hence |A : A ∩ Q| ≤ q^{6} and so |A ∩ Q| > q. By (1.14) we have |A ∩ Q : A ∩ Z(Q)| ≤ q. This now implies A ∩ Z(Q) ≠ 1. Let 1 ≠ a ∈ A ∩ Z(Q), b ∈ A ∩ Q \ A ∩ Z(Q). Then there is some t ∈ Q with b^{t} = br, r ∈ R^{♯}, and a^{t} = a. Hence ⟨a, r⟩ acts quadratically and the assertion follows with (1.14).

So we may assume A ∩ Q = A ∩ Z(Q). As |Z(Q)| = q^{7}, we see A ≰ Q. Let now a ∈ A \ Q, t ∈ Q \ Z(Q), b ∈ A ∩ Z(Q) ∖ R. We may choose t with a^{t} = ac, c ∉ Z(Q). Then ⟨c, b⟩ does act quadratically and there is some t_{1} ∈ Q with c^{t_{1}} = cr, r ∈ R^{♯}. Now ⟨r, b⟩ does act quadratically. The assertion follows with (1.14).

(1.16) Lemma. Let G ≅ G(q), q = 2^{n}, be a Lie group. Let G ◁ H and V be a GF (2)H-module. Let A ≤ N_{H}(G), A elementary abelian, C I_{A}(G) = 1, [V, A, A] = 0. Let W ≅ V (λ), λ some fundamental weight, be a G-submodule of V. Let S ∈ Syl_{2}(G). If |A| > |C I_{W} (S)|, then [W, A] ≤ W. (Notice |C I_{W} (S)| = q, q^{2} (for G twisted), q^{3}( for ^{3}D_{4}(q))).

Proof . Let a ∈ A^{♯} with W_{1} = W^{a} ≠ W. Then W_{1} is an irreducible G-module and so W_{1} + W = W_{1} ⨁ W ≅ W ⨁[W, a].

Suppose there is b ∈ A^{♯} with W^{b} = W. Then by quadratic action [[W, b], a] = 0 and so [W, b] ⊆ W ∩ W_{1} = 0. This shows W^{b} ≠ W for all b ∈ A^{♯}. We are going to prove that W_{1} ⨁ W is A-invariant. Suppose there is b ∈ A with W^{b} ⊈ W_{1} ⨁ W. Then W^{b} ⨁ W_{1}^{b} ≠ W ⨁ W_{1}. Furthermore (W_{1} ⨁ W)^{b}∩(W ⨁ W_{1}) is a G-module. As [[W, a], b] = 0 by quadratic action, we see that [W, a] ⊆ (W_{1} ⨁ W)^{b} ∩ (W ⨁ W_{1}). Furthermore |[W, a]| = |W | and so we get (W_{1} ⨁ W)^{b} ∩ (W ⨁ W_{1}) = [W, a]. Hence [W, a] is a G-module. As A is trivial on [W, a], we see that [A, G] ≤ C I_{G}([W, a]) = 1, a contradiction.

This proves W_{1} ⨁ W is A-invariant. As [[W, a], A] = 0, we see [W, a] = [W, A]. Let R = C I_{W} (S), R_{1} = C I_{W_{1}}(S). Then [R × R_{1}, a] = [R × R_{1}, A], and for every a ∈ A^{♯}, C I_{R}(a) = 1. This gives |A| ≤ |R|, the assertion.

(1.17) Lemma. Let X be one of the groups in (1.7)(i) with q > 2 or (1.7)(ii) with E(X) ≅ SL_{3}(4) ∗ SL_{3}(4). Let V be an irreducible GF (2)Xmodule, A an elementary abelian group, A ≤ E(X), [V, A, A] = 0. Then in case of (1.7)(i) |A| ≤ q^{2} while in case of (1.7)(ii) |A| ≤ 16.

Proof . Suppose false. Then (1.7)(i) implies that E(X) ≅ Sp_{4}(q), SL_{3}(4)∗ SL_{3}(4). Suppose first E(X) ≅ Sp_{4}(q). Then by (1.15) and (1.14) V just involves natural or dual natural Sp_{4}(q)-modules. As X induces a graph automorphism on E(X), we see that with each natural module also the dual is involved. But no group of order greater than q^{2} can act quadratically on both modules.

So assume now that we are in (1.7)(ii). Then E(X) = X_{1}X_{2} with X_{1} ≅ X_{2}, X_{1} ≅ SL_{3}(4). Furthermore A ∩ X_{1} ≠ 1 ≠ A ∩ X_{2}. By quadratic action we see that [[V, X_{1}], X_{2}] = 0. This shows that [Z(E(X)), V] = 1. We may assume that |A : C I_{A}(X_{1})| ≥ 8. Now (1.14) implies that [V, X_{1}] just involves natural X_{1}-modules, contradicting [V, Z(X_{1})] = 0.

(1.18) Lemma. Let F^{∗}(G) be a quasisimple group and V be an F -module over GF (2) for G. (i.e. there is some elementary abelian subgroup 1 ≠ A of G with |V : C I_{V} (A)| ≤ |A|). If V is irreducible for F^{∗}(G) one of the following holds:

(i) F^{∗}(G) ≅ A_{n} and V is the natural module.

(ii) F^{∗}(G) ≅ A_{7} and V is the 4-dimensional module, F^{∗}(G) ≅ 3 · A_{6}, |V | = 2^{6}.

(iii) F^{∗}(G) ≅ (S)L_{n}(q), and V ≅ V (λ_{1}), V (λ_{2}), V (λ_{n−2}), V (λ_{n−1}), the natural module, spin module or their duals.

(iv) F^{∗}(G) ≅ (S)U_{n}(q), q even, n > 3 and V is the natural module.

(v) F^{∗}(G) ≅ Ω_{2n}^{±}(q), q even, V is the natural module.

(vi) F^{∗}(G) ≅ Ω_{10}^{+}(q), q even, V is a spin module.

(vii) F^{∗}(G) ≅ Sp_{2n}(q), q even, V the natural module.

(viii) F^{∗}(G) ≅ Sp_{6}(q), q even, V the spin module.

(ix) F^{∗}(G) ≅ G_{2}(q), q even, V the natural 6-dimensional GF (q)-module.

If V is not irreducible and involves an irreducible module as in (i),(ii),(iv) with n = 4, (vi),(viii) or (ix) then V involves exactly one nontrivial irreducible module.

Proof . If F^{∗}(G)/Z(F^{∗}(G)) is a Lie group over a field of characteristic two the assertion follows with [Coop]. If there are transvections on V we get the assertion with [McL]. Now by Thompson replacement [GoLyS, (25.2)] we get a quadratic fours group and we can apply (1.12). If E^{∗}(G)/Z(F^{∗}(G)) is alternating the result follows from (1.12)(iii).

As centralizers of involutions in U_{4}(3) are maximal subgroups and |V : C I_{V} (a)| ≥ 16 for a ∈ A^{♯}; A a quadratic offending subgroup, we get [3U_{4}(3), C I_{V} (A)] ≤ C I_{V} (A), a contradiction. Hence F^{∗}(G) ≅ 3̸U_{4}(3).

Let F^{∗}(X)/Z(F^{∗}(X)) be sporadic. The modules are given in [MeiStr1]. Now we see that |V : C I_{V} (a)| ≥ 8 for any a ∈ F^{∗}(X)^{♯}, a^{2} = 1. Hence by (1.12)(i) we get that F^{∗}(X) ≅ 3 · M_{22} and V is the 12-dimensional module. But now as before we see |A| = 16 and C I_{V} (a) = C I_{V} (A), a ∈ A^{♯}. Hence N_{F^{∗}(X)}(A)/A ≅ 3 · A_{6}. But for this group we know |C I_{V} (A)| = 64, a contradiction.

(1.19) Lemma. Let X ≅ A_{n}, n ≥ 5, V be a GF (2)X-module with V/C I_{V} (X) the natural irreducible permutation module. Assume [V, X] = V. Then |C I_{V} (X)| ≤ 2, and |C I_{V} (X)| = 1 if n is odd. Furthermore V is a submodule of the permutation module.

Proof . For simplicity we prove the dual statement. Let C I_{V} (X) = 1 and [V, X] be the natural module. Then |V : [V, X]| ≤ 2, |V : [V, X]| = 1 for n odd, and V is a factor of the permutation module.

This will be proved by induction on n. For n = 5 this is well known. So let n > 5, K ≅ A_{n−1}, K ≤ X. If n − 1 is odd, then we have that [V, X] = T ⨁ T_{1}, T_{1} the irreducible permutation module for K, [T, K] = 1. By induction V = [V, K] ⨁ T̃, [V, K] the irreducible permutation module. Hence there is v ∈ V \ [V, X], [v, K] = 1, i.e. ⟨v^{X}⟩ = V is a factor of the permutation module.

Let n − 1 be even. Then we have a K-chain. 1 < T < T_{1} < [V, X] < V, with |T | = 2, T_{1}/T the irreducible K-module and |[V, X]/T_{1}| = 2. Now by induction C I_{V/T} (K) ≠ 1. As C I_{V/T} (K) ≰ [V, X]/T, we again get some v ∈ V \ [V, X], [v, K] = 1, and so V is a factor of the permutation module.

For the Lie groups and their modules in (1.18) we also have to know the nonsplit extension. This is basically contained in [JoPa]. But as it is not done for all the groups there, we will give a self contained proof.

(1.20) Lemma. (a) Let L be one of the following groups (S)L_{n}(q), SU_{n}(q), Sp_{2n}(q), Ω_{2n}^{±}(q), q even, and V be a module over GF (2) with [V, L] the natural module and C I_{V} (L) = 0. Then V = [V, L], or one of the following holds:

(i) L ≅ L_{2}(q), and [V : [V, L]] ≤ q

(ii) L ≅ L_{3}(2), and |V | = 16

(iii) L ≅ U_{4}(2), and |V | ≤ 2^{10}

(iv) L ≅ Sp_{2n}(q), and |V : [V, L]| ≤ q

(v) L ≅ Ω_{6}^{+}(2), and |V | = 2^{7} .

(b) Let L ≅ Ω_{8}^{−}(q) or Sp_{6}(q), q even, V be a module over GF (2) with C I_{V} (L) = 0 and [V, L] ≅ V (λ_{3}). Then V ≅ V (λ_{3}).

(c) Let L ≅ L_{n}(q), n ≥ 5, Ω_{10}^{+}(q) or A_{7} and V a GF (2)-module with C I_{V} (L) = 0 and [V, L] ≅ V (λ_{2}), V (λ_{4}) or |[V, L]| = 16, respectively. Then V = [V, L].

(d) If in (a) or (b) we have V = [V, L] and V/C I_{V} (L) is the natural module or V (λ_{3}), respectively, then we either get C I_{V} (L) = 0 or we have the exceptions as in (a) and the bound for |C I_{V} (L)| is the same as the bound for |V : [V, L]| in (a).

Proof . (d) follows by duality. Hence we just have to prove (a) - (c). Let us start with (a).

(1) Let first L ≅ L_{n}(q):

If L ≅ L_{2}(q), then for x ∈ L, o(x) = 2, we have |V : C I_{V} (x)| = q. By (1.1) L is generated by three conjugates of x. Hence |V | ≤ q^{3}.

Let now L ≅ SL_{3}(q). If q > 4, then there are three elements x_{1}, x_{2}, x_{3} in L acting fixed point freely

x_{1} = ( \begin{matrix}ω^{−2} \\ & ω \\ & & ω \end{matrix} ), x_{2} = ( \begin{matrix}ω \\ & ω^{−2} \\ & & ω \end{matrix} ), x_{3} = ( \begin{matrix}ω \\ & ω \\ & & ω^{−2} \end{matrix} ),

o(ω) = q − 1.

We have [x_{i}, V] = [V, L], i = 1, 2, 3, and so as [x_{i}, x_{j}] = 1 for all i, j, we get C I_{V} (x_{1}) = C I_{V} (x_{2}) = C I_{V} (x_{3}). But L = ⟨C I_{L}(x_{i})|i = 1, 2, 3⟩, the assertion.

If q = 4, we have V = [V, Z(L)] = [V, L]. So let q = 2. Then there is x ∈ V \ [V, L], |x^{L}| = 8. Hence V is a factor module of the permutation module, which shows |V | ≤ 16.

Let next L ≅ L_{4}(2). There is ⟨ρ⟩ × A_{5} ≤ L, o(ρ) = 3, [V, L] = [V, ρ] and [V, L] = [V, γ], γ ∈ A_{5}, o(γ) = 3. Hence C I_{V} (ρ) = C I_{V} (γ). Now as ⟨C I_{L}(γ), C I_{L}(ρ)⟩ = L, we get V = [V, L].

Let now n ≥ 4, q > 2 for n = 4. Let P be the parabolic in L with |O_{2}(P)| = q^{n−1} and P′/O_{2}(P) ≅ SL_{n−1}(q). Assume furthermore |C I_{[V,L]}(O_{2}(P))| = q^{n−1}. Then [P′, V] = C I_{[V,L]}(O_{2}(P)), as P′ = P′′. This now shows that [C I_{V} (O_{2}(P)), P′] is the natural module. By induction on n we have that C I_{V} (O_{2}(P)) = C I_{V} (P′) ⨁ C I_{[V,L]}(O_{2}(P)). Now application of [Hu, (I.17.4)] shows C I_{V} (P′) = 0. Hence as |V : C I_{V} (O_{2}(P))| = q, we get V = [V, L].

(2) Let next L ≅ SU_{n}(q):

If L ≅ SU_{3}(q). Then there are x_{1}, x_{2} ∈ L, x_{2} ∉ ⟨x_{1}⟩, o(x_{1}) = q + 1 = o(x_{2}), [x_{1}, x_{2}] = 1 and [V, L] = [V, x_{i}], i = 1, 2. Hence C I_{V} (x_{1}) = C I_{V} (x_{2}). As C I_{V} (x_{1}) ≅ ZZ_{q+1} × L_{2}(q), is a maximal subgroup of L, we get ⟨C I_{L}(x_{1}), C I_{L}(x_{2})⟩ = L, the assertion.

Let now L = U_{4}(q), q > 2. There are x_{1}, x_{2}, x_{3} ∈ L, (o(ω) = q + 1)

x_{1} = ( \begin{matrix}ω^{−3} \\ & ω \\ & & ω \\ & & & ω \end{matrix} ) , x_{2} = ( \begin{matrix}ω \\ & ω^{−3} \\ & & ω \\ & & & ω \end{matrix} ) ,

x_{3} = ( \begin{matrix}ω \\ & ω \\ & & ω \\ & & & ω^{−3} \end{matrix} ) .

As q > 2, we have [V, L] = [V, x_{i}], i = 1, 2, 3. Now C I_{V} (x_{1}) = C I_{V} (x_{2}) = C I_{V} (x_{3}). As C I_{L}(x_{i}) = ⟨x_{i}⟩SU_{3}(q), we get L = ⟨C I_{L}(x_{1}), C I_{L}(x_{2}), C I_{L}(x_{3})⟩. Hence we have the assertion. Let now q = 2. Let P be the parabolic with P/O_{2}(P) ≅ L_{2}(4). Now |C I_{[V,L]}(O_{2}(P))| = 16. As C I_{[V,L]}(O_{2}(P)) ≅ O̸_{2}(P) as P/O_{2}(P)-modules, we get V = [V, L] ⨁ C I_{V} (O_{2}(P)). We have |C I_{V} (O_{2}(P)) : [C I_{V} (O_{2}(P)), P]| ≤ 4. Hence |V : [V, L]| ≤ 4 by [Hu, (I.17.4)].

Let now L ≅ U_{5}(2). In this case we have x_{1}, x_{2}, x_{3}, x_{4},

x_{1} = ( \begin{matrix}ω^{−1} \\ & ω \\ & & ω \\ & & & ω \\ & & & & ω \end{matrix} ) , x_{2} = ( \begin{matrix}ω \\ & ω^{−1} \\ & & ω \\ & & & ω \\ & & & & ω \end{matrix} ) , …

and so on, o(ω) = 3.

Further [x_{i}, x_{j}] = 1. Now [V, L] = [V, x_{j}], i = 1, …, 4 and so C I_{V} (x_{1}) = C I_{V} (x_{2}) = C I_{V} (x_{3}) = C I_{V} (x_{4}). We have C I_{L}(x_{i}) ≅ ⟨x_{i}⟩ × U_{4}(2). Hence L = ⟨C I_{L}(x_{i})|i = 1, 2, 3, 4⟩ and then V = [V, L].

If L ≅ SU_{6}(2), then [Z(L), V] = [V, L] and we get V = [V, L], as |Z(L)| = 3. Let now L ≅ SU_{n}(q), q > 2 for n = 5 or 6.

Let P be the normalizer of a root group R in L. We have |[[V, L], R]| = q^{2}. We have P′/O_{2}(P) ≅ SU_{n−2}(q) and C I_{[V,L]}(R)/[[V, L], R] ≅ O_{2}(P)/R. Now [V, R] = [[V, L], R]. Furthermore as [P′, V] ≤ C I_{[V,L]}(R), we see that V/[[V, L], R] = [V, L]/[[V, L], R]·C I_{V/[[V,L],R]}(O_{2}(P)). Let V_{1} be the preimage of C I_{V/[[V,L],R]}(O_{2}(P)). Then [V_{1}/[[V, L], R], P′] is the natural SU_{n−2}(q)-module. By induction V_{1}/[[V, L], R] = (V_{1}/[[V, L], R])C I_{V_{1}/[[V,L],R]}(P′).

Now as P′ = P′′, we get that [V_{2}, P′] = 0 for a preimage V_{2} of C I_{V_{1}/[[V,L],R]}(P′). Hence V_{2} = 0 and so V = [V, L], the assertion.

(3) Let next L ≅ Ω_{2n}^{±}(q):

Let first L ≅ Ω_{4}^{−}(q). Choose S ∈ Syl_{2}(L). Then S is normalized by ω ∈ L, o(ω) = q + 1. Let V_{1} = [V, S]. Chose x ∈ S^{♯}, then [V, x, x] = 0. Hence [V, x] ≤ [[V, L], x]. This shows [V, S] = [V, L, S]. Furthermore V = [V, L]C I_{V} (x). As [ω, V] ≤ [V, L], we see that V = [V, L]C I_{V} (S). By [Hu, (I.17.4)] we get V = [V, L].

Let L ≅ Ω_{6}^{+}(q), q > 2. Let P be the parabolic with P′/O_{2}(P) ≅ L_{2}(q) × L_{2}(q). Then let V_{1} = C I_{V} (O_{2}(P)). We have |V_{1}| = q, [V, O_{2}(P)]/V_{1} ≅ O_{2}(P) and [P′, V] = [V, O_{2}(P)], as P′ = P′′. This shows |V/V_{1} : C I_{V/V_{1}}(O_{2}(P))| = q. Now |C I_{V/V_{1}}(O_{2}(P)) : C I_{V} (O_{2}(P))/V_{1}| = q^{4}. Hence [C I_{V} (O_{2}(P)), P′] = 0. This gives with [Hu, (I.17.4)] that C I_{V} (O_{2}(P)) ≤ [V, L] and then V = [V, L].

Let L ≅ Ω_{6}^{+}(2). Now A_{8} ≅ Ω_{6}^{+}(2) and [V, L] is the permutation module. Hence the assertion follows with (1.19).

Let L ≅ Ω_{2n}^{±}(q), n ≥ 4 for L ≅ Ω_{2n}^{+}(q). Let P be the parabolic with P′/O_{2}(P) ≅ Ω_{2n−2}^{±}(q). Set V_{1} = C I_{[V/L]}(O_{2}(P)), |V_{1}| = q. We have [P′, V] = [V, O_{2}(P)], as P′ = P′′. Hence we see |V/V_{1} : C I_{V/V_{1}}(O_{2}(P))| = q. Furthermore C I_{V/V_{1}}(O_{2}(P)) = [C I_{V/V_{1}}(O_{2}(P)), P′]C I_{V} (O_{2}(P))/V_{1}. Now we get [C I_{V} (O_{2}(P)), P′] = 0. By [Hu, (I.17.4)] we have C I_{V} (O_{2}(P)) ≤ [V, L] and so [V, L] = V.

(4) Let finally L ≅ Sp_{2n}(q):

If L ≅ Sp_{4}(2)′, the assertion follows with (1.19). Let now L ≅ Sp_{2n}(q), q > 2, for n = 2. Let P be the parabolic with P′/O_{2}(P) ≅ Sp_{2n−2}(q), A_{6} for L ≅ Sp_{6}(2). Now we see V = [V, L]C I_{V} (Z(O_{2}(P))). Set V_{1} = C I_{V} (Z(O_{2}(P))). Now [V_{1}, P′] = [[V, L], P′]. Set V_{2} = [V, Z(O_{2}(P))]. Then |V_{2}| = q. We see that |V_{1}/V_{2} : C I_{V_{1}/V_{2}}(O_{2}(P))| = q. We have V_{1}/V_{2} = ([[V, L], P′]/V_{2})C I_{V_{1}}(O_{2}(P))/V_{2}. Hence as P′ = P′′ we have [C I_{V_{1}}(O_{2}(P)), P′] = 0. This shows C I_{V_{1}}(O_{2}(P)) ≤ [V, L] and so |V : [V, L]| ≤ q by [Hu, (I.17.4)], or L ≅ Sp_{6}(2) and [P, C I_{V_{1}}(O_{2}(P))] ≠ 0. Hence there is some t ∈ P \ P′, o(t) = 2, with [C I_{V_{1}}(O_{2}(P)), t] = V_{2}. We may assume that t induces a transvection on the natural module. As [t, [V, L]/V_{2}] ≠ 0 we now see |[V, t]| ≥ 4. But ⟨t⟩ is conjugate to Z(O_{2}(P)) and [V, Z(O_{2}(P))] = V_{2} is of order 2. This proves (a).

To prove (b) let [V, L] ≅ V (λ_{3}). Let first L ≅ Ω_{8}^{−}(2). Then there are x_{1}, x_{2} ∈ L, o(x_{1}) = o(x_{2}) = 3, x_{1} ∉ ⟨x_{2}⟩, [x_{1}, x_{2}] = 1, [V, L] = [V, x_{i}], i = 1, 2, C I_{L}(x_{i}) = ⟨x⟩ × Ω_{6}^{+}(2). This shows C I_{V} (x_{1}) = C I_{V} (x_{2}) is invariant under L = ⟨C I_{L}(x_{1}), C I_{L}(x_{2})⟩. Hence V = [V, L].

Let now q > 2. Let P be the parabolic with P′/O_{2}(P) ≅ Ω_{6}^{−}(q). Then C I_{[V,L]}(O_{2}(P)) is the natural U_{4}(q)-module for P′ while O_{2}(P) is the natural Ω_{6}^{−}(q)-module. This shows V = [V, L]C I_{V} (O_{2}(P)) and [C I_{V} (O_{2}(P)), P′] = C I_{[V,L]}(O_{2}(P)). As q > 2, we have C I_{V} (O_{2}(P)) = C I_{[V,L]}(O_{2}(P))C I_{V} (P′). Application of [Hu, (I.17.4)] shows C I_{V} (P′) = 0 and then V = [V, L].

Let now L ≅ P Sp_{6}(q). Let L_{1} ≤ L, L_{1} ≅ Ω_{6}^{+}(q) ≅ L_{4}(q). Then [V, L] is an extension of the natural L_{4}(q)-module by the natural module. Hence V = [V, L] ⨁ C I_{V} (L_{1}). Let P_{1} ≤ L_{1} be the parabolic which is the stabilizer of a 2-space in the natural representation of L_{4}(q). Then C I_{V} (P_{1}) = C I_{V} (L_{1}). We have P_{1} ≤ P, P the stabilizer of a 1-space in the natural representation of Sp_{6}(q). Now Z(P′) centralizes P_{1} and so [Z(P′), C I_{V} (P_{1})] = 0. Hence C I_{V} (P_{1}) is centralized by ⟨L_{1}, Z(P′)⟩ = L. This shows C I_{V} (L_{1}) = 0 and then V = [V, L]. This finishes (b).

For (c) let first L ≅ L_{5}(q) and [V, L] ≅ V (λ_{2}). Let P be the parabolic with P′/O_{2}(P) ≅ SL_{n−1}(q), |C I_{[V,L]}(O_{2}(P))| = q^{n−1}. Now C I_{[V,L]}(O_{2}(P)) ≅ O_{2}(P)^{∗}. Furthermore [V, L]/C I_{[V,L]}(O_{2}(P)) ≅ V (λ_{2}) for the group P′/O_{2}(P) ≅ SL_{n−1}(q). As V (λ_{2}) ≅ O̸_{2}(P), we get [V, O_{2}(P)] = C I_{[V,L]}(O_{2}(P)).

If n = 5, q = 2 then let P_{1} ≤ P with O_{2}(P) ≤ P_{1} and P_{1}/O_{2}(P) ≅ A_{7}. Otherwise set P_{1} = P′. Then by (a), induction or (1.19) we see that V/C I_{[V,L]}(O_{2}(P)) = [V, L]/C I_{[V,L]}(O_{2}(P)) · C I_{V/C I_{[V,L]}(O_{2}(P))}(P_{1}). Let V_{1} be the preimage of C I_{V/C I_{[V,L]}(O_{2}(P))}(P_{1}). Then as C I_{[V,L]}(O_{2}(P)) ≅ O̸_{2}(P) as P_{1}/O_{2}(P)- modules, we see [V_{1}, O_{2}(P)] = 0. Hence in any case P′ acts on V_{1}. Now by (a) we have V_{1} = C I_{[V,L]}(O_{2}(P))C I_{V_{1}}(P′). Application of [Hu, (I.17.4)] shows C I_{V} (P′) = 0 and so V = [V, L].

Let now L ≅ Ω_{10}^{+}(q), [V, L] = V (λ_{4}). Let P be the parabolic with P′/O_{2}(P) ≅ Ω_{8}^{+}(q). Then O_{2}(P) is the natural Ω_{8}^{+}(q)-module, while [V, L]/C I_{[V,L]}(O_{2}(P)) and C I_{[V,L]}(O_{2}(P)) both are spin modules for Ω_{8}^{+}(q) and so not isomorphic to O_{2}(P). Hence we get [V, O_{2}(P)] = C I_{[V,L]}(O_{2}(P)) = V_{1}. Now by (a) V/V_{1} = [V, L]/V_{1} · C I_{V/V_{1}}(P′). Let V_{2} be the preimage of C I_{V/V_{1}}(P′). As C I_{[V,L]}(O_{2}(P)) ≅ O̸_{2}(P), we get [V_{2}, O_{2}(P)] = 0. Again by (a) V_{2} = C I_{[V,L]}(O_{2}(P)) · C I_{V_{2}}(P′). By [Hu, (I.17.4)] C I_{V_{2}}(P′) = 0. This shows V = [V, L].

Let finally L ≅ A_{7}, |[V, L]| = 16. Let L_{1} ≅ L_{3}(2), L_{1} ≤ L, with |[[V, L], L_{1}]| = 8. Then by (a)(ii) V = [V, L]C I_{V} (L_{1}). Now by [Hu, (I.17.4)] we see C I_{V} (L_{1}) = 0 and so V = [V, L].

(1.21) Lemma. Let L = F^{∗}(G) be a quasisimple group and V be an F - module over GF (2) for G. Suppose C I_{V} (S) ≤ C I_{V} (F^{∗}(G)) for S ∈ Syl_{2}(G). Then one of the following holds

(i) L ≅ L_{3}(2) and |[V, L]| = 16.

(ii) L ≅ A_{2^{m}} and |C I_{V} (L)| = 2, [V, L]/C I_{[V,L]}(L) is the natural module.

Proof . We may assume G = LS and furthermore V = [V, L]. Set V_{1} = C I_{V} (L). Assume C I_{V} (S) ≤ V_{1}. Let V_{2} be an L-submodule of V, V_{1} ≤ V_{2}, V_{2}/V_{1} irreducible. Let T = S∩N_{G}([V_{2}, L]). If V_{1}∩[V_{2}, L] = 0, then C I_{⟨[V_{2},L]^{S}⟩}(L) = 0, but C I_{⟨[V_{2},L]^{S}⟩}(S) ≠ 0.

Hence by (1.19) and (1.20) we are left with L ≅ L_{2}(q), L_{3}(2), U_{4}(2), Sp_{2n}(q), G_{2}(q) or A_{2x}. Let C be a Cartan subgroup of L. Then C acts on C I_{V_{2}}(S ∩ L), and if C I_{V_{2}}(S ∩ L) ≰ V_{1}, then C I_{V_{2}}(S ∩ L) = V_{1}[C I_{V_{2}}(S ∩ L), C]. As S = (S ∩ L)N_{S}(C). We see that [C I_{V_{2}}(S ∩ L), C] ∩ C I_{V} (S) ≠ 0. So we may assume that either C = 1 or C I_{V_{2}}(S ∩ L) ≤ V_{1}.

Suppose the latter. We may assume V_{2} = [V, L]. If V_{2}/C I_{V_{2}}(L) is the natural L_{2}(q)-module, then as [x, V_{2}]C I_{V_{2}}(L)/C I_{V_{2}}(L) = [L ∩ S, V_{2}]C I_{V_{2}}(L)/C I_{V_{2}}(L) for all x ∈ S ∩ L, x ≠ 1, we see that C I_{V_{2}}(S) ≰ V_{1}.

Let L ≅ L_{3}(2), then |V_{2}| = 16 by (1.20) (a). As V is an F -module, we have V_{2} = V.

Let V ≅ U_{4}(2). By (1.20) (a) |C I_{V_{2}}(L)| ≤ 4. Let P be the parabolic with P/O_{2}(P) ≅ L_{2}(4). We have C I_{V_{2}}(O_{2}(P))/C I_{V_{2}}(L) is the natural L_{2}(4)-module. Hence as seen in the L_{2}(q)-case this implies C I_{V} (S ∩ P) ≰ V_{1}.

Let L ≅ Sp_{2n}(q). By (1.20) (a) |C I_{V_{2}}(L)| ≤ q. Let R be a transvection on V_{2}/C I_{V_{2}}(L). Then [R, V_{2}]C I_{V_{2}}(L)/C I_{V_{2}}(L) is of order q and [R, [R, V_{2}]] = 0. Let P = N_{L}(R), then [P′, [R, V_{2}]] = 0. Hence we have S ∩ L ≰ P′. This shows L ≅ Sp_{6}(2). But in this case |[V_{2}, R]| = 2 and so [S ∩ L, [V_{2}, R]] = 0, too.

Let next L ≅ G_{2}(q). Let R be a root group, r ∈ R^{♯}. Then |[V_{2}, r]| = q^{2} and C I_{L}(r) induces the natural module on [V_{2}, r]. As [V_{2}, r] ∩ C I_{V_{2}}(L) ≠ 0, we have C I_{V_{2}}(S) ≰ V_{1}.

Let finally L ≅ A_{2x}. Then by (1.19) V_{2} is a submodule of the permutation module. As C I_{V_{2}}(S ∩ L) ≤ V_{1}, we have that a Sylow 2-subgroup has to act transitively and so 2x = 2^{m} for some m. So we have

(∗) If C I_{V_{2}}(S ∩ L) ≤ V_{1}, then L ≅ L_{3}(2), |V_{2}| = 16 or L ≅ A_{2^{m}}, V_{2} is a submodule of the permutation module.

Let now C = 1, C I_{V_{2}}(S ∩ L) ≰ V_{1}, but C I_{V_{2}}(L) ≠ 0. Hence we have L ≅ Sp_{2n}(2), G_{2}(2)′ or A_{2x}. Now if L ≅ A̸_{6}, we have |G : L| ≤ 2 and so V = V_{2} + V_{2}^{x} for some x ∈ G \ L. But then x centralizes u + u^{x}, u ∈ C I_{V_{2}}(S ∩ L) \ V_{1}. This leaves us with G ≅ P ΓL_{2}(9). Furthermore V has at least four composition factors which are natural P Sp_{4}(2)-modules. Now for any 1 ≠ A ≤ S, A elementary abelian, |V : C I_{V} (A)| ≥ 16. As |A| ≤ 8, this contradicts the fact that V is an F -module. Hence (∗) holds in general. As V is an F -module we now see that V = V_{2}.

(1.22) Lemma. Let X = G(q) be a Lie group in characteristic two. Let V be an irreducible nontrivial GF (2)-module for X and t ∈ X be an involution. Suppose |V : C I_{V} (t)| ≤ 2^{m_{2}(X)}. Then X ≅ (S)L_{n}(q), (S)U_{n}(q), Sp_{2n}(q), Ω^{±}(2n, q), G_{2}(q), F_{4}(q), E_{6}(q), E_{7}(q), or ^{2}E_{6}(q). Furthermore either V = V (λ) for some fundamental weight λ or

(i) X ≅ L_{n+1}(q) and V ≅ V (λ_{1}) ⨂ V (λ_{1})^{σ} or V (λ_{n}) ⨂ V (λ_{n})^{σ}, σ some field automorphism of X, where n ≥ 6 if o(σ) ≥ 3.

(ii) X ≅ L_{n+1}(q), n ≥ 6, and V ≅ V (λ_{1}) ⨂ V (λ_{n})^{σ}, 1 ≠ σ some field automorphism of X.

(iii) X ≅ L_{n+1}(q), V (λ_{1} + λ_{n}) ≅ V, n ≥ 5 (iv) X ≅ ^{2}A_{n}(q) = SU_{n+1}(q), V ≅ V (λ_{1} + λ_{n}), n ≥ 5, V ≅

V (λ_{1}) ⨂ V (λ_{1})^{σ}, n ≥ 15, V ≅ V (λ_{1}) ⨂ V (λ_{n})^{σ}, n ≥ 7, σ a field automorphism of X, o(σ) ≥ 3.

(v) X ≅ Sp_{2n}(q), V ≅ V (λ_{1}) ⨂ V (λ_{1})^{σ}, n ≥ 7, 1 ≠ σ a field automorphism.

(vi) X ≅ Ω_{2n}^{±}(q), V ≅ V (λ_{1}) ⨂ V (λ_{1})^{σ}, n ≥ 8, σ a field automorphism of X.

Proof . This is [Coop].

(1.23) Lemma. Let X ≅ G(q), q = 2^{n}, be a Lie group. Let V be an irreducible nontrivial GF (2)-module for X and t ∈ X be an involution with |V : C I_{V} (t)| ≤ 2^{m_{2}(X)}. Assume furthermore V = V (λ) for some fundamental weight λ. Then m_{2}(X) and the minimal value of |V : C I_{V} (t)| are given in the following table 1.

Proof . [Coop]

(1.24) Lemma. Let (X, V) be as in (1.22), V ≅ V̸ (λ), λ a basic weight. Then the lower bound for the codimension of C I_{V} (t), t an involution of X is as in table 2.

Proof. [Coop]

(1.25) Lemma. Let F^{∗}(G) be a perfect central extension of a sporadic simple group, V a faithful GF (2)G-module. Then m_{2}(G) and the minimal codimension of C I_{V} (t) in V, for t an involution in G are listed in table 3

Proof . This is [Asch3, (11.1)].

(1.26) Lemma. (i) Let X ≅ (S)L_{3}(q), q even, V an irreducible Xmodule over GF (2), t ∈ X an involution and |V : C I_{V} (t)| ≤ q^{2}. Then V is the natural module or the dual, or q = t^{2} and V is a 9–dimensional GF (t)- module (V ⨂ V ^{σ}, V the natural module, σ a field automorphism of order two).

(ii) Let X ≅ Sp_{4}(q), q even, or 3 · A_{6}, V an irreducible X-module over GF (2), t ∈ X an involution and |V : C I_{V} (t)| ≤ q^{3}. Then V is the natural module or the dual, or [Z(X), V] ≠ 0 and V is a 6-dimensional module.

(iii) Let X ≅ (S)U_{3}(q), q even, q ≥ 4. Let t ∈ X, o(t) = 2, and V be an irreducible GF (2)X-module with |V : C I_{V} (t)| < q^{3}. Then V is the natural module.

X  m_{2}(X)  module  minimal codimension
A_{ℓ}  (\frac{ℓ+1}{2})^{2}n, ℓ odd  V (λ_{1})  1
(\frac{ℓ+2}{4})^{2}n, ℓ even  V (λ_{ℓ})  1
A_{ℓ}, ℓ ≥ 3  V (λ_{2}), V (λ_{ℓ−1})  ℓ − 1
A_{5}  V (λ_{3})  6
A_{6}  V (λ_{3}), V (λ_{4})  10
A_{7}  V (λ_{3}), V (λ_{5})  15
2A_{ℓ}  [\frac{ℓ+1}{2}]^{2}n  V (λ_{1})  2
2A_{3}  V (λ_{2})  2
2A_{ℓ}, ℓ ≥ 4  V (λ_{2})  2(ℓ − 1)
2A_{5}  V (λ_{3})  6
B_{ℓ}  \binom{ℓ+1}{2}n  V (λ_{1})  1
B_{ℓ}  V (λ_{2})  2(ℓ − 1)
B_{ℓ}, 3 ≤ ℓ ≤ 6  V (λ_{ℓ})  2^{ℓ−2}
D_{ℓ}, ℓ ≥ 4  \binom{ℓ}{2}n  V (λ_{1})  2
D_{ℓ}, ℓ ≥ 4  V (λ_{2})  4ℓ − 6
D_{ℓ}, 4 ≤ ℓ ≤ 7  V (λ_{ℓ−1}), V (λ_{ℓ})  2^{ℓ−3}
2D_{ℓ}, ℓ ≥ 4  \binom{ℓ}{2}n  V (λ_{1})  2
2D_{ℓ}, ℓ ≥ 4  V (λ_{2})  4ℓ − 6
2D_{ℓ}, ℓ = 4, 5  V (λ_{ℓ−1})  2^{ℓ−2}
G_{2}  3n  V (λ_{1})  2
F_{4}  11n ≤ m_{2}(X) ≤ 16n  V (λ_{1})  6
E_{6}  16n  V (λ_{1}), V (λ_{6})  6
2E_{6}  14n ≤ m_{2}(X) ≤ 20n  V (λ_{1})  12
E_{7}  27n  V (λ_{1})  7

Table 1

X  V  codimension of C I_{V} (t)
L_{n+1}(q) V (λ_{1}) ⨂ V (λ_{1})^{σ}  2n, n if σ^{2} = 1
V (λ_{n}) ⨂ V (λ_{n})^{σ}  2n, n if σ^{2} = 1
V (λ_{1}) ⨂ V (λ_{n})^{σ}  2n
V (λ_{1} + λ_{n})  n − 1
SU_{n+1}(q) V (λ_{1}) ⨂ V (λ_{n})^{σ}  2n
V (λ_{1}) ⨂ V (λ_{n})^{σ}  n
V (λ_{1} + λ_{n})  n − 1
Sp_{2n}(q) V (λ_{1}) ⨂ V (λ_{1})^{σ} 2(2n − 1), 2n − 1 if σ^{2} = 1
Ω_{2n}^{±}(q) V (λ_{1}) ⨂ V (λ_{1})^{σ}  4(n − 1)

Table 2

Proof . (i),(ii): This follows from (1.22), (1.23) and (1.24). (iii) Let t ∈ Ω_{1}(S), S ∈ Syl_{2}(X). If [[t, V], Ω_{1}(S)] = 0, we get with [Str, Corollary 1] that V is the natural module or the 8-dimensional module over GF (q). In the latter we have |V : C I_{V} (t)| > q^{3}, a contradiction.

So we have [[t, V], Ω_{1}(S)] ≠ 0. Let ω ∈ N(S), o(ω) = (q + 1)/(3, q + 1). Then [[V, t], ω] ≠ 0. Let U ≅ L_{2}(q), [U, ω] = 1. Then [[V, ω], U] ≠ 0. Now t inverts some element γ of order q + 1 in an U. As ω acts freely fixed point freely on [[[V, ω], γ], t] we see that |[V, t] ∩ [V, ω]| ≥ q^{2}. This now implies [U, C I_{V} (ω)] = 1. Hence [[V, t], ω] = [V, t]. If ⟨ω⟩ acts irreducibly on [V, t], we get [[V, t], Ω_{1}(S)] = 0, a contradiction. Hence, we are left with q = 8 and |V : C I_{V} (t)| ≤ 2^{8}. As [S, C I_{[V,t]}(Ω_{1}(S))] ≠ 0, we see that |C I_{[V,t]}(Ω_{1}(S))| ≥ 2^{4}. Now we have that [V, ω] involves at most two irreducible U–modules W_{1}, W_{2}. Let τ ∈ N_{U}(Ω_{1}(S)), o(τ) = 7. Then τ acts irreducibly on C I_{W_{i}}(Ω_{1}(S)) = T_{i}, i = 1, 2. Hence |T_{i}| = 2 or 2^{3}. In both cases W_{1}^{ω} = W_{2}. In the first case [C I_{[V,t]}(Ω_{1}(S)), τ] = 0. As S acts on C I_{[V,t]}(Ω_{1}(S)), we see [S, C I_{[V,t]}(Ω_{1}(S))] = 0. In the second case we see that |C I_{[V,t]}(Ω_{1}(S))| = 2^{6} and ⟨ω⟩ × ⟨τ⟩ acts irreducibly. This again shows [S, C I_{[V,t]}(Ω_{1}(S))] = 0, a contradiction.

(1.27) Lemma. Let X ≅ Sz(q) or L_{2}(q), q even.

(i)If V is some faithful GF (2)-module for X, then |[V, t]| ≥ q^{2}, q, respectively, where t is some involution in X.

(ii)Let X ≅ Sz(q), |V : C I_{V} (t)| ≤ q^{2} for some involution t and V be irreducible. Then V is the natural module.

(iii)Let q > 2, V be some GF (2)-module for X, T ∈ Syl_{2}(X), and S =

F^{∗}(G)/Z(F^{∗}(G)) m_{2}(G) codimension of C I_{V} (t)
M_{11}  2  4
M_{12}  4  4
M_{22}  5  3
M_{23}  4  4
M_{24}  6  4
J_{1}  3  8
J_{2}  4  4
J_{3}  4  6
Mc  4  8
Ly  4  33
HS  5  6
He  6  10
Sz  6  8
Ru  6  12
O′N  4  21
Co_{3}  4  8
Co_{2}  10  6
Co_{1}  11  8
M(22)  10  18
M(23)  11  18
M(24)′  12  18
F_{5}  6  40
F_{3}  5  9
F_{2}  22  54
F_{1}  24  54
J_{4}  11  50

Table 3 Ω_{1}(T). If V = C I_{V} (S) ⨁ C I_{V} (S^{g}) for some g ∈ X, then V is a direct sum of natural modules.

(iv)Let X ≅ L_{2}(q), T ∈ Syl_{2}(X). If V is a faithful irreducible GF (2)- module with [V, T, T] = 0, then V is the natural module.

Proof . (i) We have that t inverts an element of order q + \sqrt{2q} + 1, q + 1, respectively, and so |[V, t]| ≥ q^{2}, q.

(ii) By (1.1) X is generated by three conjugates of t. Hence |V | ≤ q^{6}. Set W = V ⨂_{GF (2)} GF (q). By the tensor product lemma W = ⨁ Mσ where M is a tensor product of algebraic conjugates of the natural module T. Let q = 2^{n}, then dim W ≤ 6n. But this shows W = T.

(iii) We have that for t ∈ S, t inverts some ω ∈ X, o(ω) = 3 for X ≅ L_{2}(q) and o(ω) = 5 for X ≅ Sz(q). As ⟨S, S^{g}⟩ = X by (1.1), we may assume ω = g. Now C I_{V} (ω) = 0. The assertion follows with [Hi, (8.2)] and [Mar].

(iv) By (1.1) there is g ∈ X with ⟨T, T^{g}⟩ = X. Hence V = [V, T] + [V, T^{g}]. Furthermore C I_{V} (T) ∩ C I_{V} (T^{g}) = 0. As [V, T, T] = 0, we get V = C I_{V} (T) ⨁ C I_{V} (T^{g}). The assertion follows with (iii).

(1.28) Lemma. Let F^{∗}(X) = L be a quasisimple Lie group in odd characteristic p, Z(L) a p′-group, L ≅ L̸_{2}(q), ^{2}G_{2}(q), G_{2}(q), ^{3}D_{4}(q) or P Sp_{4}(3). Let V be a faithful GF (2)-module for X and t ∈ X be some involution. Then m([V, t]) ≥ (q − 1)q^{w}d(p)/2εp, where d(p) is the degree of the smallest nontrivial representation of ZZ_{p}, q^{2w+1} is the order of O_{p}(C I_{L}(R)), R a long root group (see table below) and ε = 1 or L ≅ P Sp_{2n}(q), ε = 2, q > p.

\begin{matrix}L & L_{n}(q) & U_{n}(q) & Ω_{n}^{±}(q) & P Sp_{2n}(q) & F_{4}(q) & E_{6}(q) & ^{2}E_{6}(q) & E_{7}(q) & E_{8}(q) \\ w & n − 2 & n − 2 & n − 4 & n − 1 & 7 & 10 & 10 & 16 & 28 \end{matrix}

Proof . This is [Asch3, (10.4)]. We just sketch his proof. Let R be a long root group in X and Q = O_{p}(C I_{L}(R)). Then Q is a special group of order q^{1+2w}, where w is as described above.

By [Asch3, (10.1)] we may assume that t inverts some U, |U| = p, U ≤ Q, U ≰ R, or X = P Sp_{4k}(q) and t induces a field automorphism. In the former we get with [Asch3, (7.2)] the assertion.

So assume the latter. Now E(C I_{L}(t)) ≅ P Sp_{2k}(q^{2}) and t ∼ tz, tz induces a nontrivial inner automorphism on C I_{L}(t).

We can proceed by induction. First of all by (1.12) [[V, t], E(C I_{L}(t))] ≠ 0. Set W = [V, t].

Let k = 1. Then E(C I_{L}(t)) ≅ L_{2}(q^{2}). Let R_{1} be a subgroup of order q^{2} in E(C I_{L}(t)). Then N_{E(C I_{L}(t))}(R_{1}) has at most two orbits on the hyperplanes of R_{1}. Now there is some orbit △ with

[W, R_{1}] = ⨁_{H∈△} C I_{[W,R_{1}]}(H) .

This shows m([W, R_{1}]) ≥ (q^{2} − 1)d(p)/2(p − 1) > (q − 1) q^{w}d(p)/2p, since w = 1.

Let now k > 1. Then by induction

m([V, tz]) ≥ (q^{2} − 1)q^{2u}d(p)/4p ,

where u = k − 1 and w = 2k − 1. Hence w = 2u + 1 and so

m([V, t]) ≥ (q^{2} − 1)q^{w−1}d(p)/4p > (q − 1) q^{w}d(p)/4p .

(1.29) Lemma. Let F^{∗}(X) = L be a quasisimple group such that L/Z(L) is a Lie group over a field of odd characteristic which is not a Lie group over a field of characteristic 2 too. Let t ∈ X be an involution and V be an irreducible faithful GF (2)-module for X. Then one of the following holds:

(1) |V : C I_{V} (t)| ≥ 2^{8}

(2) L ≅ L_{3}(3), |V : C I_{V} (t)| ≥ 2^{4}

(3) L ≅ U_{4}(3), L_{4}(3), G_{2}(3), 3 · G_{2}(3), P Sp_{6}(3) and |V : C I_{V} (t)| ≥ 2^{6}

(4) L ≅ 3 · U_{4}(3)

(5) L ≅ L_{2}(25) or L_{2}(p), p prime, |V : C I_{V} (t)| ≥ 2^{4}

Proof . ([Asch3, (10.5)]). Assume |V : C I_{V} (t)| ≤ 2^{7}. Let first L ≅ L_{2}(q), q odd. Suppose furthermore q = p^{f} > p. Let t ∈ P GL_{2}(q). Then L⟨t⟩ is generated by three conjugates of t and so |V | ≤ 2^{21}. Let P ∈ Syl_{p}(L) and △ be one orbit of hyperplanes under N_{G}(P). Then V = ⨁_{U∈△} C I_{V} (U).

We have |△| = (q − 1)/(p − 1) or G does not induce P GL_{2}(q) on L and |△| = (q − 1)/2(p − 1). Let d (p) = |C I_{V} (U)|. We get d (p)(q − 1) ≤ 42(p − 1). As d (p) ≥ 2, we get a rough bound by p + 1 ≤ 21, and so p ≤ 19. If p > 7, then d (p) ≥ 8 and so p + 1 ≤ 5, a contradiction. Hence p = 3, 5 or 7. Let p = 7, then d (p) = 3 and so q − 1 ≤ 14(p − 1). This shows L = L_{2}(7^{2}) and we have exactly two orbits on the hyperplanes. But now t inverts P, a contradiction.

Let p = 3. Then f ≥ 3. Now L = L_{2}(27). But then there is just one orbit on the hyperplanes and we have 2 · 26 ≤ 21 · 2, a contradiction. Let finally p = 5. Then we get L = L_{2}(25). As |V | ≥ 2^{9}, we get |V : C I_{V} (t)| ≥ 2^{4}.

Suppose next that t induces a field automorphism on L. Let L_{1} = E(C I_{L}(t)) (recall q > 9). Suppose [[V, t], L_{1}] = 0. Now as in (1.11) we see that {a^{L}} is a set of odd transpositions. This shows that C I_{L_{1}}(z) is a 2-group for z ∈ L_{1}, z^{2} = 1, and L_{1} has abelian Sylow 2-subgroup. This shows L_{1} = L_{2}(5) and so L ≅ L_{2}(25).

We are going to prove the same result for [[V, t], L_{1}] ≠ 0. Now L_{1} acts faithfully on [V, t] and |[V, t]| ≤ 2^{7}. In particular L_{1} ≅ L_{2}(p), p | |L_{7}(2)|, or L_{1} = L_{2}(9). But in the latter P GL_{2}(9) acts on [V, t] and so |[V, t]| ≥ 2^{8}. We are left with L_{2}(5), L_{2}(7), L_{2}(31) and L_{2}(127). But in all cases P GL_{2}(p) is involved and so there is some involution s which inverts a Sylow psubgroup. This shows |[V, t]| ≥ 2^{14}(L_{2}(127)), |[V, t]| ≥ 2^{10}(L_{2}(31)). Let C I_{L}(t) ≅ P GL_{2}(7). Then |[V, t]| ≥ 2^{6}. There is z ∈ L_{1}, o(z) = 2, such that t ∼ tz. Hence we see that |[V, P]| ≤ 2^{18} for P ∈ Syl_{7}(L_{1}). As L_{2}(7^{2}) ≰ L_{18}(2), we see that C I_{V} (P) ≠ 1. Let △ be the orbit of P in N_{G}(E), E ∈ Syl_{p}(L), P ≤ E. Then |△| = 4 and V = ⨁_{U∈△} C I_{V} (U). Now there

is some U ∈ △ with [U, t] = U. This shows |C I_{V} (P) ∩ [V, t]| ≥ 8. But then |[V, t]| ≥ 2^{9}, a contradiction.

Let now L ≅ L_{2}(25). We may assume |[V, t]| ≤ 8. Hence [[V, t], L_{1}] = 1. As t inverts some elements of order 13, we see that L⟨t⟩ is generated by three conjugates of t. Now |V | ≤ 2^{9} but 13 | |̸L_{9}(2)|.

So we are left with L ≅ L_{2}(p), p prime, p ≥ 11. Suppose |[V, t]| ≤ 8, then |V | ≤ 2^{9} and so L ≲ L_{9}(2). This shows p = 31, 127, 73, 17. As 37 | |L_{2}(73)|, but 37 | |̸L_{9}(2)|, L ≅ L̸_{2}(73). In L_{9}(2) the normalizer of a Sylow 127-subgroup is of order 2 · 3 · 7 · 127, hence L_{2}(127) ≰ L_{9}(2). On a Sylow 31-subgroup just a group of order 5 is induced, as this is true in L_{5}(2), hence L_{2}(31) ≰ L_{9}(2). As t cannot invert an element of order 17, we get L⟨t⟩ = P GL_{2}(17) if L ≅ L_{2}(17). Now t centralizes a group of order 9. Hence we get some element of order 3 which centralizes [V, t] and V/C I_{V} (t) as well. But |C I_{V} (t) : [V, t]| ≤ 8 and so this element centralizes V, a contradiction.

Let now [[V, t], E(C I_{L}(t))] = 0, then we get the assertion with (1.12). So assume [[V, t], E(C I_{L}(t))] ≠ 0.

If L ≅ ^{2}G_{2}(q), then E(C I_{L}(t)) ≅ L_{2}(q), q ≥ 27. But L_{2}(q) ≰ L_{7}(2).

Let L/Z(L) ≅ G_{2}(q). If t ∈ L, then C I_{L}(t) ≅ (SL_{2}(q) ∗ SL_{2}(q)) · 2. Hence the structure of L_{7}(2) shows q = 3 or 7. If q = 3, then t inverts some element of order 13 and so |V : C I_{V} (t)| ≥ 2^{6}. If q = 7, t inverts some element of order 817.

Let t ∉ L. Suppose t induces a field automorphism. Then E(C I_{L}(t)) ≅ G_{2}(\sqrt{q}) ≰ L_{7}(2). So assume that q = 3^{f} and E(C I_{L}(t)) ≅ ^{2}G_{2}(q). This shows again q = 3 and L_{2}(8) acts on [V, t], i.e. |[V, t]| ≥ 2^{6}.

Let now L/Z(L) ≅ ^{3}D_{4}(q). Then t acts on some G_{2}(q) in L. Hence we may assume q = 3. Now t induces P GL_{2}(27) on a subgroup SL_{2}(27) of L. Hence we see |[V, t]| ≥ 2^{8} as before.

Suppose now that L is none of these groups but p | |̸Z(L)|. Then the conclusion follows from (1.28). If L ≅ L_{3}(3) then either t inverts an element of order 13 or an elementary abelian group E ≅ E_{9}, with N_{L}(E) transitive on E^{♯}. In both cases |[V, t]| ≥ 2^{6}.

So we are left with p | |Z(L)|. This now leaves us with 3 · Ω_{7}(3). Now as 3 · Ω_{7}(3) ≰ L_{14}(2), we get t ∈ C I(Z(L)). Furthermore we may assume [Z(L), V] = V. Now E(C I_{L}(t)) ≤ GL_{3}(4), a contradiction. As t ∈ 3 · Ω_{7}(3), we see E(C I_{L}(t)) ≅ U_{4}(3) or Sp_{4}(3), both not in GL_{3}(4), or we have C I_{L}(t)/Z(L) ≅ Σ_{4} × (SL_{2}(3) ∗ SL_{2}(3)) · 4 [CCNPW]. But now the embedding in GL_{3}(4) gives some kernel which contains a fours group. Now we have a quadratic fours group and can apply (1.12) for a contradiction.

(1.30) Lemma. Let F^{∗}(X) = L be a quasisimple group and V be an irreducible faithful GF (2)-module for X. Let t be some involution in X. Then

(a) If |V : C I_{V} (t)| ≤ 2, then L ≅ SL_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2), A_{n} and V is the natural module. Furthermore if L ≅ A_{n} or Ω_{2n}^{±}(2), then t ∉ L.

(b) If |V : C I_{V} (t)| ≤ 4, then either (L, V) is as in (a) or L ≅ SU_{n}(2), G_{2}(2)′, SL_{n}(4), Sp_{2n}(4), Ω_{2n}^{±}(4) and V is the natural module or one of the following holds:

(i) L ≅ 3 · A_{6}, |V | = 2^{6}

(ii) L ≅ 3 · U_{4}(3), |V | = 2^{12}

(iii) L ≅ A_{7}, |V | = 2^{4}

(iv) L ≅ Sp_{6}(2), |V | = 2^{8}

(c) If |V : C I_{V} (t)| ≤ 8, then either (L, V) is as in (b) or (a) or L ≅ SL_{n}(8), Sp_{2n}(8), Ω_{2n}^{±}(8), V is the natural module, or one of the following holds:

(i) L/Z(L) is a Lie group in characteristic two

(ii) L ≅ M_{22}, |V | = 2^{10} .

Proof . (The proof follows [Asch3, (13.5)]). Let first L/Z(L) = G(q), q even. Then if t ∈ L, we get the result with (1.23), (1.22) and (1.24). So assume t ∉ L. Then there is s ∈ L with t ∼ ts in x. Hence |V : C I_{V} (s)| ≤ 16. Now again we are going to apply (1.23), (1.22) and (1.24). This implies that we either have the assertion or L ≅ L_{5}(2), V = V (λ_{2}) or V (λ_{3}), L ≅ Sp_{8}(2), V = V (λ_{4}), L ≅ Ω_{10}^{+}(2), V = V (λ_{4}) or V (λ_{5}), L ≅ Ω_{8}^{−}(2), V = V (λ_{3}), L ≅ G_{2}(4), V = V (λ_{1}), L ≅ L_{3}(4), |V | = 2^{9}. As t has to act on this module, we get L ≅ Ω_{8}^{−}(2), V = V (λ_{3}), or L ≅ G_{2}(4), V = V (λ_{1}).

Let L ≅ Ω_{8}^{−}(2) and P be the parabolic in L with P/O_{2}(P) ≅ Ω_{6}^{−}(2). Then P/O_{2}(P) induces the natural U_{4}(2)-module on C I_{V} (O_{2}(P)) and V/C I_{V} (O_{2}(P)). Hence t has to induce a transvection on both modules. But as these modules are over GF (4), this is not possible.

Let L ≅ G_{2}(4), then t has to induce a field automorphism. But field automorphisms do act freely on V (λ_{1}), a contradiction.

Let next L/Z(L) ≅ A_{n}. For n ≤ 8, we just check the modules to get the result. Let n ≥ 9. Then C I_{L}(t) induces a subgroup of L_{3}(2) on [V, t]. Hence there is some involution t_{1} ≠ t such that [V, t, t_{1}] = 0. This shows that ⟨t, t_{1}⟩ is a quadratic fours group. Now we apply (1.12), which shows that V is the natural module, or V is the spin module. In the latter ⟨t, t_{1}⟩ corresponds to ⟨(12)(34), (13)(24)⟩ or ⟨(12)(34)(56)(78), (13)(24)(57)(68)⟩ by [MeiStr1, (4.3)] and (1.12). In both cases |V : C I_{V} (t)| ≥ 16, a contradiction.

Let now L/Z(L) be sporadic. Then we quote (1.25) and get L/Z(L) ≅ M_{22}. Furthermore as before we get a quadratic fours group. Application of (1.13) shows that either L ≅ M_{22} and |V | = 2^{10}, or L ≅ 3 · M_{22} and |V | = 2^{12}. In the latter as Z(L) acts fixed point freely we get |[V, t]| ≥ 16 for t ∈ L and |[V, t]| = 64 for t ∉ L.

Let finally L/Z(L) ≅ G(q), q odd, which is not a Lie group over a field of characteristic two. Now the lemma is a consequence of (1.29).

(1.31) Lemma. Let X be as in (1.7) or (1.8) and V an F -module for X over GF (2) which involves a nontrivial irreducible module. Then

(i) E(X) ≅ L_{2}(q), q even, V involves exactly one nontrivial irreducible module which is the natural module, or q = 4 and this module is the orthogonal module. An offending subgroup A is a Sylow 2-subgroup of E(X) or q = 4 and |A| ≤ 4, |A ∩ E(X)| ≤ 2.

(ii) E(X) = X_{1} × X_{2} ≅ L_{2}(q) × L_{2}(q), q even. Denote by W the group [E(X), V]C I_{V} (E(X))/C I_{V} (E(X)). Then W = W_{1} ⊕ W_{2}, [W_{i}, X_{3−i}] = 0, i = 1, 2. Furthermore W_{i} is the natural X_{i}-module, or q = 4 and W_{i} is the orthogonal X_{i}-module. In the first case an offending subgroup is a Sylow 2subgroup of X_{1}, X_{2} or E(X), while in the second case the offending subgroup A normalizes X_{1} and X_{2} and |A ∩ X_{1}| ≤ 2 ≥ |A ∩ X_{2}| , |A| ≤ 16.

(iii) E(X) ≅ A_{9} and there is exactly one nontrivial irreducible module involved which is the natural module.

(iv) F (X) is a 5-group with E(X/F (X)) ≅ L_{2}(5). Furthermore [F (X), V] = 0 and X/F (X) acts as in (i).

(v) X is a solvable {2, 3}-group, |[V, F (X)]| ≤ 16.

Proof . Let first E(X) = X_{1} be quasisimple. Then (1.7) and (1.18) yield X_{1} ≅ L_{2}(q), (S)L_{3}(q), Sp_{4}(q), A_{9}, A_{6} or 3 · A_{6}.

Let X_{1} ≅ (S)L_{3}(q), Sp_{4}(q) or A_{6}, then by (1.18) V involves just trivial and natural modules. By (1.7) there is some x ∈ X acting nontrivially on the Dynkin diagram. Thus with the natural module W also W^{∗}, the dual module, is involved. But then V cannot be an F -module.

If X_{1} ≅ 3·A_{6}, then by (1.30) V involves exactly one nontrivial irreducible module, which is 6-dimensional. But then again by (1.7) both 6-dimensional modules have to be involved, a contradiction.

Let X_{1} ≅ A_{9}, then the assertion follows with (1.30) and (1.18).

Assume now E(X) = X_{1}X_{2}. Let A be some offending subgroup. Let ⟨X_{1}^{A}⟩ = E(X). Then there is some a ∈ A with X_{1}^{a} = X_{2}. Set F = C I_{E(X)}(a). Then F acts on C I_{V} (a). Thus as |A| > 2 (a cannot induce a transvection), we see that C I_{E(X)}(a) induces an F -module in C I_{V} (a), which shows X_{1} ≅ L_{2}(q) as before.

Now a inverts some ω ∈ E(X), o(ω) = q + 1. Hence |V : C I_{V} (a)| ≥ q and so |A| ≥ q. So for b ∈ A ∩ E(X) we have that b inverts an abelian group of order (q + 1)^{2}. Hence |V : C I_{V} (b)| ≥ q^{2} and as |A| ≤ 2q, q = 2, a contradiction.

So we have A ≤ N(X_{1}). Then by the A × B-Lemma C I_{V} (C I_{A}(X_{1})) is an F -module for X_{1}A/C I_{A}(X_{1}). This implies X_{1} ≅ L_{2}(q), q even, as before.

We now determine the offending subgroups. Let E(X) = X_{1} ≅ L_{2}(q), q even. By (1.18) V involves exactly one irreducible module. This is the natural one, or q = 4 and it is the orthogonal module. Furthermore an offending subgroup is a Sylow 2-subgroup or X ≅ Σ_{5} and |A| = 2, 4 with |A : A ∩ X_{1}| = 2.

Let now E(X) = X_{1}X_{2}, X_{1} ≅ L_{2}(q) ≅ X_{2}, q even. Let [A, X_{1}] ≠ 1. Then C I_{V} (C I_{A}(X_{1})) is an F -module for X_{1}. So V involves exactly one irreducible nontrivial X_{1}-module, as |V : C I_{V} (C I_{A}(X_{1}))| ≤ |C I_{A}(X_{1})|. Now [[V, X_{1}], X_{2}] = 0. As there is some x ∈ X with X_{1}^{x} = X_{2}, we get the assertion.

Suppose now E(X) = 1. By (1.7) F (X) = P is a p-group and E(X/F (X)) ≅ L_{2}(p), p odd. So |A| ≤ 4. Suppose [V, F (X)] ≠ 0. Now a ∈ A^{♯} inverts some ω ∈ F (X), o(ω) = p. As |V : C I_{V} (a)| ≤ 4, we get p = 5, |V : C I_{V} (a)| = 4 = |A|. But now as A acts on an elementary abelian group of order 5^{3}, we get U ≅ D_{10} × D_{10} in P A acting faithfully on V. But this is impossible as |V : C I(U)| ≤ 16. So we have [P, V] = 0. Now as just shown L_{2}(p) has to be isomorphic to L_{2}(4) and so p = 5.

Let now X be solvable. By (1.8) |A| ≤ 4. Furthermore X = P S, S ∈ Syl_{2}(X), P = O_{p}(X). So we get p ≤ 5.

Suppose p = 5. Then |A| = 4. As P is noncyclic and S acts irreducibly on P/Φ(P) we get some D_{10} × D_{10} = U involved in AP/Φ(P). But now we get ⟨A, A^{g}⟩ covers U for some g ∈ P. As |V : C I_{V} (⟨A, A^{g}⟩)| ≤ 16, we see that U cannot act faithfully on V and so [P, V] = 0, a contradiction.

So we have p = 3. If P is cyclic, we get |[V, P]| = 4. So let P be noncyclic. Let a ∈ A^{♯} such that |V : C I_{V} (a)| = 2. Then there is some t ∈ Z(S)^{♯} such that |V : C I_{V} (t)| ≤ 4. If |A| = 4, then A∩Z(S) ≠ 1. So in any case |V : C I_{V} (t)| ≤ 4 for t ∈ Z(S). This shows that |[P/Φ(P), t]| ≤ 9. As S acts irreducibly on P/Φ(P), we get |P/Φ(P)| = 9, and P = ⟨ω, ω̃⟩ with |V : C I_{V} (⟨ω, ω̃⟩)| ≤ 16. Hence [Φ(P), V] = 0 and |[V, P]| = 16.

(1.32) Lemma. Let G ≅ L_{2}(q), q even. Let V be a faithful irreducible GF (2)-module for G. Let |V | < q^{4}.

(a) One of the following holds:

(i) V is the natural module

(ii) q = t^{2} and V is the Ω_{4}^{−}(t)-module

(iii) q = t^{3} , |V | = t^{8}.

(b) Suppose |V | = t^{8}, q = t^{3}. Let S ∈ Syl_{2}(G). Then |C I_{V} (S)| = t. If x ∈ S^{♯}, then |[V, x]| = t^{4}.

Proof . (a) Let W = GF (q) ⨁_{GF (2)} V. Then we have that W = ⨁ Mσ, M a tensor product of algebraic conjugates of the natural module T. Let q = 2^{n}, then dim W < 4n. This shows M = T, T ⨂ T τ or T ⨂ T τ ⨂ T τ^{2}. Now in any case just by the dimension formula, we get W = M, and so q = t^{2}, q = t^{3}, as GF (q) possesses an automorphism of order 2 in case of M = T ⨂ T τ, of order 3 in case M = T ⨂ T τ ⨂ T τ^{2}, respectively.

(b) |C I_{V} (S)| = t follows from [Cur]. Now let x ∈ S^{♯}. By (1.27) (i) |[V, X]| ≥ t^{3}. Suppose |C I_{V} (X)| = t^{5}. Let ω ∈ N(S), o(ω) = t^{2} + t + 1. Then [ω, C I_{V} (S)] = 0. Hence |[ω, V]| = t^{3} or t^{6}. As there is y ∈ S with S ≤ ⟨ω, ω^{y}⟩ and |C I_{V} (S)| = t, we see that |[ω, V]| = t^{6}, and so |C I_{V} (ω)| = t^{2}. Let z = X^{g}, ω^{z} = ω^{−1}. Then |[z, [ω, V]]| = t^{3}. Hence [C I_{V} (ω), z] = 0. But C I_{V} (S) ≤ C I_{V} (ω) and so C I_{V} (S) is centralized by ⟨S, y, ω⟩ = G, a contradiction. Hence |C I_{V} (x)| = t^{4}.

(1.33) Lemma. Let G = G(q), q even, and V be an irreducible module for G over GF (2). Let P be some parabolic of G. Then C I_{V} (O_{2}(P)) is an irreducible P/O_{2}(P)-module.

Proof . [Ti]

(1.34) Lemma. Let X ≅ (S)U_{5}(q), q even, t ∈ X be some involution and V be some irreducible nontrivial GF (2)X-module.

(i) If |[V, t]| ≤ q^{4}, then V is the natural module

(ii) Let P = O^{2}(N_{X}(R)), R a transvection group in the natural representation. If |[V, t]| < q^{6}, then C I_{V} (P) ≠ 0.

Proof . (i) follows from (1.22), (1.23) and (1.24)

(ii) Suppose C I_{V} (P) = 0. We may assume t ∈ P \ O_{2}(P). Let r ∈ R^{♯}.

Then [P, [V, r]] ≠ 0. As P/O_{2}(P) ≅ SU_{3}(q), we get with (1.26) (iii) that [V, r] is the natural module and V/C I_{V} (r) is the natural module. Furthermore [P′, C I_{V} (r)] = [V, r]. As C I_{V} (P′) = 0, we get that |C I_{V} (r) : [V, r]| ≤ q^{2}. Now [O_{2}(P), C I_{V} (r)] = [V, r], as C I_{V} (O_{2}(P)) is an irreducible P -module by (1.33). As [V, r] ≅ O_{2}(P)/R as P -modules, we see that C I_{C I_{V} (r)}(x) = [V, r] for every x ∈ O_{2}(P) \ R. Let now r^{g} ∈ O_{2}(P) \ R, g ∈ X. Then we see that [V, r^{g}] = [V, r] as [V, r^{g}] ≤ C I_{V} (r) ∩ C I_{V} (r^{g}). But now [V, r] < ⟨C I_{X}(r), C I_{X}(r^{g})⟩ = X, a contradiction.

(1.35) Lemma. Let X be one of the group in (1.7). Suppose X possesses a nontrivial GF (2)-module V with C I_{V} (F^{∗}(X)) = 0, [V, F^{∗}(X)] = V such that for some elementary abelian subgroup 1 ≠ A of X we have |V : C I_{V} (A)| ≤ |A|^{2}. Then one of the following holds:

(1) V is an F -module for X.

(2) E(X) ≅ SL_{3}(q) and V is a direct sum of the natural and the dual module.

(3) E(X) ≅ Sz(q), V is the natural module, or E(X) = X_{1} × X_{2}, X_{1} ≅ X_{2} ≅ Sz(q) and V = V_{1} ⨁ V_{2}, [V_{i}, X_{3−i}] = 0 and V_{i} is the natural module for X_{i}, i = 1, 2.

(4) E(X) ≅ L_{2}(q), V is a direct sum of two natural modules.

(5) E(X) ≅ L_{2}(q), q = t^{2}, V is the orthogonal module.

(6) E(X) ≅ A_{5} ≅ L_{2}(4), V is a direct sum of two orthogonal modules.

(7) E(X) ≅ SU_{3}(q), V is the natural module.

(8) E(X) ≅ X_{1}X_{2}, X_{1} ≅ X_{2} ≅ L_{2}(q), V = V_{1} ⨁ V_{2}, [V_{i}, X_{3−i}] = 0, [V_{i}, X_{i}] = V_{i}, i = 1, 2, and V_{i} is orthogonal, a direct sum of two natural modules, or q = 4 and V_{i} is a direct sum of two orthogonal modules.

(9) E(X) = X_{1}X_{2}, X_{1} ≅ X_{2} ≅ L_{2}(q) and V is the natural O_{4}^{+}(q)-module.

(10) E(X) = X_{1}X_{2}, X_{1} ≅ X_{2} ≅ L_{3}(2), V = V_{1} ⨁ V_{2}, [V_{i}, X_{3−i}] = 0, [V_{i}, X_{i}] = V_{i}, i = 1, 2, and V_{i} is a direct sum of the natural and the dual module.

(11) E(X) ≅ Sp_{4}(q), V is a direct sum of the natural and the dual module.

(12) E(X) ≅ 3 · A_{6}, [V, Z(E(X))] = 0 and V is a direct sum of the natural 4–dimensional module and its dual.

(13) E(X) = X_{1} ∗ X_{2}, X_{1} ≅ X_{2} ≅ 3 · A_{6}, [V, Z(E(X))] = 0, V = V_{1} ⨁ V_{2}, [V_{i}, X_{3−i}] = 0, [V_{i}, X_{i}] = V_{i}, i = 1, 2 and V_{i} is a direct sum of the 4–dimensional module and its dual.

(14) E(X) ≅ A_{9}

(15) X is a solvable {2, 5}-group.

(16) X is a {2, 3}-group, X = P S, P = O_{3}(X), and one of the following holds

(i) P is cyclic and |V | = 16

(ii) m_{3}(P) = 2, S ≅ ZZ_{4} and |V | = 16

(iii) m_{3}(P) = 2, S ≅ D_{8} and |V | = 2^{8}

(iv) P ≅ 3^{1+2}, |V | = 2^{6}

(v) P ≅ 3^{1+4}, |V | = 2^{8}, [Z(P), V] = 0

(vi) m_{3}(P) = 3, X is as in (1.8) (iv), |V | ≤ 2^{8}.

Proof . Let us go over the list of possibilities for X. Suppose first that X is nonsolvable. Let E(X) be quasisimple.

(i) E(X) ≅ Sz(q): Then for any involution t ∈ E(X) we have |V : C I_{V} (t)| ≥ q^{2}, so we get |A| = q and then by (1.27) V is the natural module. This is (3).

(ii) E(X) ≅ L_{2}(q): Let first A ≤ E(X). If |A| > 2 then by (1.1) E(X)⟨A, A^{g}⟩ for suitable g ∈ E(X). As |A| ≤ q, we get |V | ≤ q^{4}. If |V | = q^{4}, then A ∈ Syl_{2}(E(X)) and by (1.27) V = V_{1} ⨁ V_{2}, V_{i} the natural module for E(X). This is (4).

So assume |V | < q^{4}. Then by (1.32) V is either the natural module, q = t^{2} and V is the orthogonal module, or q = t^{3} and V is the eight dimensional GF (t)-module.

Suppose we have the last case. Then |V : C I_{V} (x)| = t^{4} for x ∈ A^{♯}. Hence |A| ≥ t^{2}. But for B ≤ L_{2}(q), |B| > t, we have |V : C I_{V} (B)| ≥ t^{6}. This shows A ∈ Syl_{2}(E(X)). Then |C I_{V} (A)| = t, a contradiction. So we have (1) or (5).

Suppose now |A| = 2. Then |V : C I_{V} (A)| ≤ 4. As A inverts some element of order q + 1, we get q = 4 and E(X) ≅ L_{2}(4). So we now have (1) or (5) again.

Assume next A ≰ E(X). Then q = t^{2} and |A| ≤ 2t. If |A ∩ E(X)| ≠ 1, then as before |V | ≤ 16t^{4} = 16q^{2}. If |V | < q^{4} we may argue as before. So assume q = 4 and |V | = q^{4}. Then |A| = 4 and so V is a direct sum of two orthogonal modules, which is (6).

Let finally A ∩ E(X) = 1. Then |A| = 2 and |V : C I_{V} (A)| ≤ 4. As A inverts some element of order t + 1, we see t = 2 or 4. If t = 2, then as before we have (1) or (6). Let t = 4, i.e. E(X) ≅ L_{2}(16). As X is generated by four conjugates of A, we get |V | = 2^{8}, and so V is the orthogonal module. This is (5).

(iii) E(X) ≅ (S)L_{3}(q): Suppose first A ≤ E(X). If V is not irreducible, we get some submodule W of V such that |W : C I_{W} (t)| ≤ |A|, for some t ∈ A. By (1.26) W is a natural module or a tensor product of a natural and an algebraically conjugate module. But now by (1.7) also W^{∗} is involved. This shows that W is the natural module and we are in (2).

So assume that V is an irreducible (S)L_{3}(q)-module. Suppose first |A| ≤ q. Then (1.26) implies that V is the natural module or a tensor product of the natural module with an algebraically conjugate module. As X contains some x inducing a Dynkin automorphism on E(X) this is impossible.

So we may assume |A| > q. Let P be the maximal parabolic in E(X) with A ≤ O_{2}(P). Now there is g ∈ G, h ∈ P such that ⟨A^{g}, A^{gh}⟩ = O^{2′}(P) = P_{1}. Furthermore A^{g} ∩ O_{2}(P) ≠ 1. Set V_{1} = C I_{V} (O_{2}(P)). Now |V_{1} : C I_{V_{1}}(A^{g})| ≤ |A^{g} : A^{g} ∩ O_{2}(P)|^{2}. By (1.33) V_{1} is an irreducible P/O_{2}(P)-module. So V_{1} is a natural or orthogonal L_{2}(q)-module, whence |V_{1}| = q^{2}.

Furthermore O_{2}(P) = ⟨A, A^{r}⟩ for some r ∈ P, A∩A^{r} ≠ 1, as |A| > q and |O_{2}(P)| = q^{2}. Now

|V | |V : C I_{V} (A ∩ A^{r})| ≤ |V | |V : C I_{V} (A)C I_{V} (A^{r})| = \frac{|V |^{2}}{|C I_{V} (A)C I_{V} (A^{r})|} =

|V : C I_{V} (A)|^{2}|C I_{V} (A)∩C I_{V} (A^{r})| = q^{2}|V : C I_{V} (A)|^{2} ≤ q^{2}|A|^{4} = q^{2}(|A| |A^{r}|)^{2} =

q^{2} |AA^{r}|^{2} |A ∩ A^{r}|^{2} = q^{6}|A ∩ A^{r}|^{2}.

As we may assume |V : C I_{V} (A ∩ A^{r})| > |A ∩ A^{r}|^{2}, we get |V | < q^{6}. In particular q > 2.

Hence we may assume A = O_{2}(P). Application of (1.10) shows [O_{2}(P), V] ≰ V_{1}. Let V_{3} ≤ V such that V_{3}/V_{1} is a nontrivial irreducible submodule of V/V_{1}. Then [V, O^{2′}(P)] = V_{3}. As [V, O_{2}(P)] ≰ V_{1}, we have V > V_{3} and so V_{3}/V_{1} ≅ O_{2}(P) as L_{2}(q)-module. Choose ω ∈ P, o(ω) = q + 1. There is some t such that t^{g} ∈ A for some g ∈ G with ω^{t} = ω^{−1}. As [t, V] ≤ V_{3} and V_{3} = [V, ω], we get |[V, t]| = q^{2}, contradicting (1.26).

So there is no nontrivial irreducible submodule in V/V_{1}. Let V_{2} be maximal such that [V_{2}, O^{2′}(P)] = V_{1}. We have C I_{V_{2}}(t) = V_{1} for any t ∈ O_{2}(P)^{♯}.

Let now V_{2} ≤ V_{3} ≤ V, V_{3}/V_{2} a nontrivial irreducible L_{2}(q)-module. Then [O_{2}(P), V_{3}] = V_{1}, and so V_{3} ≠ V. Furthermore there is some v ∈ V \ V_{3} such that O^{2′}(P) = O_{2}(P)C I_{O^{2′}(P)}(vV_{1}) and O_{2}(P) ∩ C I_{O^{2′}(P)}(vV_{1}) = 1. But then [O_{2}(P), v] ≅ O_{2}(P) as P/O_{2}(P)-module, and then we have a natural submodule for L_{2}(q) in V/V_{1}, a contradiction.

So assume next A ≰ E(X). Then |A∩E(X)| ≤ q and |A : A∩E(X)| ≤ 4. Let first A ∩ E(X) ≠ 1. Then by (1.9) we may assume |V : C I_{V} (t)| > q^{2} for any t ∈ (A ∩ E(X))^{♯}.

Suppose first |A : A ∩ E(X)| = 4. Then q = t^{2} and |A ∩ E(X)| ≤ t. So |V : C I_{V} (A)| ≤ 16t^{2} = 16q. As |V : C I_{V} (A)| > q^{2}, we get q = 4, |A| = 8. There is some x ∈ A such that C I_{X}(x) contains U ≅ L_{3}(2) · 2. There is t ∈ U′ such that x ∼ xt. As |V : C I_{V} (t)| ≥ 32, we get |V : C I_{V} (x)| ≥ 8. But there is some y ∈ U \ U′, y ∈ A. Hence |C I_{V} (x) : C I_{C I_{V} (x)}(⟨y, t⟩)| ≥ 2^{4}. This implies |V : C I_{V} (A)| ≥ 2^{7}, a contradiction.

So we have |A : A ∩ E(X)| = 2. Suppose A contains a field × diagram automorphism. Then q = t^{2} and |A| ≤ 2t. So |V : C I_{V} (A)| ≤ 4t^{2} = 4q. As |V : C I_{V} (A)| > q^{2}, we get a contradiction. Let p be a primitive prime divisor of q^{3} − 1, p = 9 if q = 4. Then there is some a ∈ A \ E(X) inverting some ω, o(ω) = p. Now we get |V : C I_{V} (a)| ≥ t^{3}, where q = t^{2} or q = t, if q is not a square.

Let a be the diagram automorphism. Then C I_{E(X)}(a) ≅ L_{2}(q). Hence |V : C I_{V} (A)| ≥ qt^{3} = tq^{2}. So 4 ≥ t. This shows E(X) ≅ L_{3}(2), (S)L_{3}(4) or (S)L_{3}(16). In case (S)L_{3}(16) we have |V : C I_{V} (a)| = 4^{3} and so [[V, ω], a] = 1. Hence |C I_{V} (ω) ∩ C I_{V} (A)| = 1. But this implies |C I_{V} (ω)| ≤ 16, i.e. |V | ≤ q^{4}, a contradiction.

Let E(X) ≅ (S)L_{3}(4). Now |C I_{V} (ω) : C I_{C I_{V} (ω)}(a)| ≤ 2. Hence |C I_{V} (ω)| ≤ 2 · 4 = 8. This shows |V | ≤ 2^{9}, again |V : C I_{V} (t)| ≤ 2^{4} for t ∈ E(X), t^{2} = 1, a contradiction.

Let E(X) ≅ L_{3}(2). Then |C I_{V} (ω)| ≤ 4, so |V | ≤ 2^{5}, a contradiction.

So we have q = t^{2} and A contains a, a field automorphism. Hence C I_{X}(a) contains U ≅ (S)L_{3}(t)·2. Again |V : C I_{V} (a)| ≥ t^{3}. As |V : C I_{V} (A)| > q^{2} = t^{4}, we get |A∩U| > t. Now |C I_{V} (a) : C I_{V} (A)| > t^{2}. Hence |V : C I_{V} (A)| > t^{5} = q^{2}t so 4 > t, i.e. q = 4. Now |V : C I_{V} (a)| = 8. So |[V, ω]| = 2^{6} and C I_{C I_{V} (ω)}(A) = 1. This shows |C I_{V} (ω)| ≤ 2^{3}. So |V | ≤ 2^{9}, a contradiction.

So we are left with A ∩ E(X) = 1. Let |A| = 2. Then |V : C I_{V} (A)| ≤ 4 and so for t ∈ E(X), t^{2} = 1, we have |V : C I_{V} (t)| ≤ 16, i.e. E(X) ≅ L_{3}(2) by (1.26). But as a ∈ A^{♯} inverts an element of order 7, |V : C I_{V} (A)| > 4, a contradiction. So we have |A| = 4 and q = t^{2}. There is some a ∈ A^{♯} with |V : C I_{V} (a)| ≥ t^{3}. This shows t = 2. But then for this a we may assume C I_{X}(a) ≥ U ≅ L_{3}(2) · 2 and |C I_{V} (a) : C I_{V} (A)| = 2. But there are now a ∈ (U ∩ A)^{♯} inducing transvections on C I_{V} (a).

(iv) E(X) ≅ (S)U_{3}(q): Then |A| ≤ q. So |V : C I_{V} (a)| ≤ q^{2} for any a ∈ A^{♯}. Now (1.26) yields (7).

(v) E(X) = Sp_{4}(q), including E(X) ≅ A_{6}: Suppose first A ≤ E(X). If V is reducible, then we have some submodule W with |V : C I_{V} (t)| ≤ q^{3} for some t ∈ A^{♯}. Now by (1.26) we may assume that W is the natural module. As X contains some diagram automorphism also the dual module is involved. Now we have (11).

So we may assume that V is irreducible as a Sp_{4}(q)-module and |V : C I_{V} (a)| > q^{3} for any a ∈ A^{♯}. In particular |A| > q. Suppose |A| ≤ q^{2}. Let P be the parabolic with A ≰ O_{2}(P). As [O^{2′}(P), C I_{V} (O_{2}(P))] ≠ 0, we see that |C I_{V} (A ∩ O_{2}(P)) : C I_{V} (A)| ≥ q. Now |V : C I_{V} (A ∩ O_{2}(P))| > q^{3} and so |V : C I_{V} (A)| > q^{4} ≥ |A|^{2}, a contradiction.

So we have |A| > q^{2}. Let now P_{1} be the parabolic with A ≤ O_{2}(P_{1}). Then for some g ∈ P_{1}, O_{2}(P_{1}) = ⟨A, A^{g}⟩. Now choose A with |A| minimal. As in (iii) we see |V | |V : C I_{V} (A ∩ A^{g})| ≤ q^{8}|A ∩ A^{g}|^{2}. Hence |V | < q^{8}. So also |V : C I_{V} (O_{2}(P_{1}))| ≤ |O_{2}(P_{1})|^{2}. Choose a ∈ Z(O_{2}(P_{1}))^{♯}. Then |[V, a]| > q^{3}. As C I_{V} (P_{1}) = 0 we get that [V, a] contains a submodule [[V, a], O^{2′}(P_{1})] = W. Now as [O_{2}(P_{1}), [V, a]] ≠ 0 by (1.10) we have [O_{2}(P_{1}), [V, a]] = W and so W ≅ O_{2}(P_{1})/Z(O^{2′}(P_{1})), i.e. W is the natural module. As W = C I_{[V,a]}(O_{2}(P_{1})), we get |[V, a]| ≤ q^{3}, a contradiction.

So we have A ≰ E(X). Assume first A ∩ E(X) ≠ 1. Recall |V : C I_{V} (t)| > q^{3} for any a ∈ A ∩ E(X), a ≠ 1. Suppose |A ∩ E(X)| ≤ q. Then |V : C I_{V} (A)| ≤ 16q^{2}, 4q^{2}, if q is not a square. Hence q ≤ 4.

Let q = 4, then |A| ≤ 8, so |V : C I_{V} (A)| ≤ 64 = q^{3}, a contradiction. So let q = 2, |A| = 4. But now inspection of the A_{6}-modules just implies (11).

So assume |A ∩ E(X)| > q. Then q = t^{2} and A contains a field automorphism. So |A| ≤ 2t^{3}. Hence |V : C I_{V} (A)| ≤ 4t^{6} = 4q^{3}. Now as there is some x ∈ E(X)^{♯} such that a ∼ ax, a the field automorphism, we get |V : C I_{V} (a)| ≥ 2t^{3}. So |C I_{V} (a) : C I_{V} (A)| ≤ 2t^{3} and by induction just the natural and its dual is involved. As |A ∩ E(X)| = t^{3}, we then get |C I_{V} (a) : C I_{V} (A)| ≥ t^{5}, a contradiction.

So we are left with A∩E(X) = 1. Hence |A| ≤ 4. For u ∈ E(X)^{♯}, u^{2} = 1, we have |V : C I_{V} (u)| ≤ 2^{8}. This implies q ≤ 4. If q = 2, we have E(X) ≅ A_{6}, |A| = 2. Now inspection of the A_{6}-modules gives a contradiction.

So let q = 4, |A| = 4. Now |V : C I_{V} (a)| = 16 for any a ∈ A^{♯}. This yields that E(X)A = ⟨C I_{E(X)}(a)|a ∈ A^{♯}⟩ acts on C I_{V} (A), a contradiction.

(vi) E(X) ≅ L_{2}(r), r ≥ 11, (S)U_{3}(r), (S)L_{3}(r), r odd: Then in any case |A| ≤ 8 and so |V : C I_{V} (A)| ≤ 64. Application of (1.29) shows E(X) ≅ L_{2}(p), p prime, L_{2}(25), or L_{3}(3). Let E(X) ≅ L_{2}(25) or L_{3}(3) and suppose there is some a ∈ A^{♯} inverting some element of order 13. Then |V : C I_{V} (a)| = 64 and so [C I_{V} (a), A] = 0. Hence there is a quadratic fours group in A, contradicting (1.12). So in any case we have |A| = 4. Now E(X) ≅ L̸_{2}(25). As A cannot act quadratically we get |V : C I_{V} (a)| ≤ 8 for a ∈ A^{♯}, which contradicts (1.30). Let finally E(X) ≅ P Sp_{4}(p), p ≥ 5. Then |A| ≤ 16. Hence |V : C I_{V} (A)| ≤ 2^{8}. By (1.29) C I_{V} (A) = C I_{V} (t) for any t ∈ A^{♯}. Hence [V, A, A] = 0. But this contradicts (1.12).

(vii) E(X) ≅ 3·A_{6} and [V, Z(E(X))] ≠ 0: Hence the irreducible modules involved are of dimension 1, 6, 6 or 18. If one 6-dimensional module is involved than also the other 6-dimensional module is involved, as there is a diagram automorphism x in X. Hence |V : C I_{V} (a)| ≥ 16 for any a ∈ A^{♯}, i.e. |A| ≥ 4. But for a fours group we have |V : C I_{V} (A)| ≥ 64. So we have |A| = 8. Now some a ∈ A^{♯} inverts Z(E(X)). Hence |V : C I_{V} (a)| = 64 and |V | = 2^{12} is a direct sum of the two 6-dimensional modules. But now C I_{V} (a) = C I_{V} (A), a contradiction as Z(E(X)) acts on C I_{V} (A ∩ E(X)).

(viii) From now on let E(X) = X_{1}X_{2}. Let X_{1} ≅ L_{2}(q), q even. Suppose first [A, X_{1}] = X_{1}. If C I_{A}(X_{1}) = 1, then we have that as a X_{1}-module V involves just modules which are as in (1), (4), (5) or (6). Suppose [[V, X_{1}], X_{2}] = 0. Then we get V = V_{1} ⨁ V_{2}, [V_{i}, X_{i}] = V_{i} and [V_{i}, X_{3−i}] = 0, i = 1, 2. So we have (8).

Assume now [[V, X_{1}], X_{2}] ≠ 0. Then there are at least two nontrivial irreducible X_{1}-modules involved. This implies that both are natural and so [V, X_{1}] = V_{1} ⨁ V_{2}, V_{i} natural X_{i}-modules, i = 1, 2. Hence V is the O_{4}^{+}(q)- module (9).

So assume now B = C I_{A}(X_{1}) ≠ 1. Then X_{1} acts on C I_{V} (B). By the A×B-Lemma we have [X_{1}, C I_{V} (B)] ≠ 0. If |V : C I_{V} (B)| ≤ |B|^{2}, we may argue as before.

So suppose |V : C I_{V} (B)| > |B|^{2}. Then |C I_{V} (B) : C I_{V} (A)| < |A/B|^{2} and so the action of X_{1} on C I_{V} (B) is as in (1), (4), (5) or (6). If [X_{2}, [V, X_{1}]] = 0, we have the assertion. Now there is just one irreducible nontrivial X_{1}-module in C I_{V} (B). As |V : C I_{V} (B)| < q^{4}, we get that [V, X_{1}] involves at most two nontrivial irreducible X_{1}-modules and so again V is as in (9).

So let now a ∈ A with X_{1}^{a} = X_{2}. This shows |A| ≤ 2q. As a inverts some ω ∈ X_{1}, o(ω) = q + 1, we get |V : C I_{V} (a)| ≥ q. Let |A| = 2, then we get q = 4, and for 1 ≠ y ∈ C I_{E(X)}(a), y^{2} = 1 we get |V : C I_{V} (y)| ≤ 16 = q^{2}.

Assume that [V, X_{1}] involves at least 3 nontrivial irreducible submodules for X_{1}. Then for 1 ≠ y ∈ C I_{E(X)}(a), y^{2} = 1, we get |V : C I_{V} (y)| ≥ q^{3}. Suppose A ∩ E(X) ≠ 1, then |V : C I_{V} (y)| ≤ 4q^{2} and so q = 4. Suppose A ∩ E(X) = 1, then |A| ≤ 4 and so |V : C I_{V} (y)| ≤ 2^{8} and again q = 4.

So we have: If [V, X_{1}] involves at least 3 nontrivial irreducible modules, then q = 4, |A| ≥ 4.

Now suppose [[V_{1}, X_{1}], X_{2}] = 0. Then as [V_{1}, X_{1}]^{a} = [V_{2}, X_{2}], we get that there are at most two X_{1}-modules involved. Hence if there are at least 3 X_{1}-modules involved [[V_{1}, X_{1}], X_{2}] ≠ 0. Suppose these modules are natural modules. Then |A| = 8 and A acts quadratically. But now by (1.11) we have a contradiction. So these modules are all orthogonal modules and then we have four of them. Now again A acts quadratically, contradicting (1.11) again.

So we have that there are at most two nontrivial irreducible X_{1}-modules involved and [[V_{1}, X_{1}], X_{2}] ≠ 0. But then there are exactly two such modules involved and so again we have (9). So let finally [[V_{1}, X_{1}], X_{2}] = 0. Then there is exactly one nontrivial irreducible X_{1}-module involved. As [V, X_{1}]^{a} = [V, X_{2}], we get |V : C I_{V} (a)| ≥ q^{2}.

Furthermore for b ∈ A ∩ N(X_{1}), we have |V : C I_{V} (b)| ≥ t, t^{2} = q or t = q. So we have |V : C I_{V} (b)| ≥ tq^{2} and then t ≤ 4. We get t = q = 4. As C I_{V} (E(X)) = 0, we see |[V, X_{1}]| = 16 and so we have (8).

(ix) Let E(X) = X_{1}X_{2}, X_{i} ≅ Sz(q), i = 1, 2. Let [A, X_{1}] = X_{1}. Then as in (viii) we get that V = V_{1} ⨁ V_{2}, [V_{i}, X_{i}] = V_{i}, [V_{i}, X_{3−i}] = 0. So we have (3). If X_{1}^{a} = X_{2}, for some a ∈ A we get |A| ≤ 2q. So |V : C I_{V} (b)| ≤ 4q^{2} < q^{3} for b ∈ A ∩ E(X). Suppose A ∩ E(X) ≠ 1, then V involves exactly one irreducible X_{1}-module which is nontrivial. So again we have (3).

Let A ∩ E(X) = 1, then |A| = 2 and |V : C I_{V} (a)| ≤ 4. But a inverts some ω, o(ω) = q − 1, a contradiction.

(x) Let E(X) = X_{1}X_{2}, X_{i} ≅ L_{2}(r), r odd, r ≥ 11, i = 1, 2. Suppose [A, X_{1}] = X_{1}. If [[V, X_{1}], X_{2}] = 0 we get the same contradiction as in (vi). So assume [[V, X_{1}], X_{2}] ≠ 0. Now we get that there are at least 5 nontrivial irreducible X_{1}-modules involved. This shows that A has to induce transvections on these modules, contradicting (1.30).

So let a ∈ A, X_{1}^{a} = X_{2}. Then |A| ≤ 16. Now |[V, a]| ≥ 8. Let B = A ∩ N(X_{1}). Then |C I_{V} (a) : C I_{V} (B)| < |B|^{2}, a contradiction.

(xi) Let now E(X) = X_{1}X_{2}, X_{1} ≅ X_{2} ≅ 3 · A_{6}, L_{3}(2), SL_{3}(4) or SU_{3}(8). Suppose first [A, X_{1}] = X_{1}. Let B = C I_{A}(X_{1}) = 1. Then as an X_{1}-module we get that V is as in (2), (7) or (12). In any case [X_{2}, [V, X_{1}]] = 0. So we get that X_{1} ≅ 3 · A_{6} and we have (12) or X_{1} ≅ L_{3}(2) and we have (2). Furthermore we get (13) or (10).

So let B ≠ 1. Then [X_{1}, C I_{V} (B)] ≠ 0. We may assume |V : C I_{V} (B)| > |B|^{2} and so |C I_{V} (B) : C I_{C I_{V} (B)}(A/B)| < |A/B|^{2}. Now it is not possible to apply our results just established as there is no reason why the diagram automorphism should act on C I_{V} (B). We just get X_{1} ≅ S̸U_{3}(8).

Now in any case [[V, X_{1}], X_{2}] ≠ 0. So [V, X_{1}] involves more than one irreducible nontrivial X_{1}-module. Suppose that [V, X_{1}] involves just the modules from (1.26) i.e. natural modules or a 6-dimensional module in case X_{1} ≅ 3 · A_{6}.

Let X_{1} ≅ L_{3}(2), then we have at least 3 natural and 3 dual modules. So |V : C I_{V} (b)| ≥ 2^{6}. This shows |A| = 2^{4}. But then |V : C I_{V} (A)| ≥ 2^{9}, a contradiction.

Let X_{1} ≅ SL_{3}(4). Then again there are six modules involved and so |V : C I_{V} (b)| ≥ 2^{12} and then |A| ≥ 2^{6}. Hence |A ∩ X_{1}| ≥ 4. If |A ∩ X_{1}| > 4, we get |V : C I_{V} (A ∩ X_{1})| ≥ 2^{18}, a contradiction. But now |A ∩ E(X)| ≤ 16 and so we get a contradiction again.

Let X_{1} ≅ 3 · A_{6}. Then we must have two 6-dimension modules and their duals or four natural modules and their duals. In both cases |V : C I_{V} (b)| ≥ 2^{8}. So |A| ≥ 2^{4}. Now |A∩E(X)| ≠ 1 and so |V : C I_{V} (b)| ≥ 2^{16} in the second case, a contradiction. Hence we have the first case. Now [Z(E(X)), V] = V and so there are three 6-dimensional modules and their duals involved. Hence |V : C I_{V} (b)| ≥ 2^{12} and so |A ∩ X_{1}| ≥ 4. But now |V : C I_{V} (A ∩ X_{1})| ≥ 2^{18}, a contradiction.

So we have that [V, X_{1}] also involves modules not from (1.26). In particular for b ∈ X_{2}∩A, we have |V : C I_{V} (b)| > 2^{2m_{2}(X_{2})}. Suppose X_{2}∩A ≠ 1. Then there is some b ∈ X_{2} ∩ A centralized by X_{1} and the diagram automorphism. Now we see that C I_{V} (b) just involves modules from (1.26). But then V just involves modules from (1.26) as X_{1} acts on [V, b] as on V/C I_{V} (b). So we have A ∩ X_{2} = 1. Now the same applies for X_{1}.

Let X_{1} ≅ L_{3}(2). Then |A| ≤ 4. So |A| = 4. But then A contains some a inverting a group of order 7^{2}, i.e. |V : C I_{V} (a)| ≥ 64, a contradiction.

Let X_{1} ≅ 3 · A_{6}, then |A| ≤ 4. Furthermore |A| = 4, otherwise for some b ∈ X_{1}, o(b) = 2, we have |V : C I_{V} (b)| ≤ 16, a contradiction. So there is some a ∈ A^{♯} inverting a group of order 9 in E(X). Hence there is some ζ ∈ X_{1}, o(ζ) = 3, |[V, ζ]| ≤ 16. But now inspection of the A_{6}–modules and 3 · A_{6}–modules gives that there are just X_{1}-modules from (1.26) involved.

Let X_{1} ≅ SL_{3}(4). Now |A| ≤ 16. By (1.26) for b ∈ X_{1}, o(b) = 2, we have |V : C I_{V} (b)| ≥ 2^{15}. This gives |A| = 16 and |V : C I_{V} (a)| = 2^{8} for any a ∈ A^{♯}. In particular C I_{V} (a) = C I_{V} (A) for any a ∈ A^{♯}. But there is some fours group in A acting nontrivially on a group of order 7^{2}, a contradiction.

Now we have some a ∈ A with X_{1}^{a} = X_{2}. Let X_{1} ≅ L_{3}(2), then for some b ∈ E(X), b^{2} = 1, we have |V : C I_{V} (b)| ≤ 2^{6}. This shows that [V, X_{1}] either is irreducible or involves at most three natural and dual modules.

Suppose [[V, X_{1}], X_{2}] = 0. Then |[V, X_{1}]| ≤ 2^{6} and so we have (10). Assume [[V, X_{1}], X_{2}] ≠ 0. Then we have that [V, X_{1}] involves three natural and three dual modules. As N_{A}(X_{2}) induces a fours group on [V, X_{1}], we see that |V : C I_{V} (A ∩ X_{1})| ≥ 2^{9}, a contradiction.

Let X_{1} ≅ 3 · A_{6}. Then for b as above |V : C I_{V} (b)| ≤ 2^{8}. If [V, X_{1}, X_{2}] = 0, then as |V : C I_{V} (a)| ≤ 2^{8} we have (12). Let [[V, X_{1}], X_{2}] ≠ 0. Then inspection of the A_{6}–modules and 3·A_{6}–modules implies that either natural A_{6}–modules and their duals, or 6-dimensional 3·A_{6}-modules and their duals are involved. In any case A ∩ E(X) = 1 and so |A| = 4. But then for b ∈ N(X_{1}) ∩ A, we have |V : C I_{V} (b)| ≥ 2^{6}, a contradiction.

Let X_{1} ≅ SL_{3}(4). Then |V : C I_{V} (b)| ≤ 2^{10}. Let [V, X_{1}, X_{2}] = 0. Then |[V, X_{1}]| ≤ 2^{10}. But now by (1.26) and (1.14) we get a contradiction.

So we have [V, X_{1}, X_{2}] ≠ 0. Now [V, X_{1}] involves more than one nontrivial irreducible (S)L_{3}(4)-module. Suppose that for any b we have |V : C I_{V} (b)| = 2^{10}. Hence |A ∩ N(X_{1})| = 16. But now A ∩ N(X_{1}) acts quadratically on V and so by (1.10) we just get natural modules involved. So in any case we just have the modules from (1.26) involved in V. Now we have three natural and their duals involved. But then |V : C I_{V} (b)| ≥ 2^{12}, a contradiction.

Let finally X_{1} ≅ SU_{3}(8). Now |V : C I_{V} (b)| ≤ 2^{8} for b ∈ E(X), o(b) = 2. So by (1.26) there is just one nontrivial irreducible module involved in [V, X_{1}] which is the natural module. But now [V, X_{1}, X_{2}] = 0 contradicting [Z(E(X)), V] = V.

(xii) Let now F (X) be a p-group and still X be nonsolvable. Now |A| ≤ 4 and then |V : C I_{V} (A)| ≤ 16. As A acts on an elementary abelian group of order p^{3}, we get p = 5. Now |[V, F (X)]| = 2^{12} and so X ≤ GL_{12}(2).

But 5^{4} | |̸GL_{12}(2)|.

(xiii) Now let X be a solvable {2, p}-group. As there is U ≅ D_{2p} × … × D_{2p} in X with A ∈ Syl_{2}(U) by [GoLyS, (24.1)], we see that p ∈ {3, 5}.

(xiv) Let X be a {2, 3}-group. Let O_{3}(X) be cyclic. Then |A| = 2 and so |V : C I_{V} (A)| ≤ 4, hence |[V, O_{3}(X)]| ≤ 16. This is (16) (i). Let next m_{3}(O_{3}(X)) = 2. Then |A| ≤ 4. By (1.8) in any case O_{3}(X) is generated by two elements. So |[V, O_{3}(X)]| ≤ 2^{8}. Suppose |[V, O_{3}(X)]| = 16. Then X/C I_{X}(V) ≲ A_{8}. So either V is an F -module or we have (16) (ii).

Let |[V, O_{3}(X)]| = 2^{6}. Let |A| = 4. Now again X/C I_{X}(V) ≲ GL_{6}(2). Suppose O_{3}(X)/C I_{O_{3}(X)}(V) ≅ ZZ_{3}×ZZ_{3}. Further S ≅ D_{8} or S_{16}. Now we have that for 3 of the 4 subgroups U of O_{3}(X)/C I_{O_{3}(X)}(V) of order 3, |C I_{V} (U)| = 4 while for the forth C I_{V} (U) = 1. But D_{8} induces 2 orbits of length 2 on these subgroups, while S_{16} induces just one orbit of length 4.

Let |A| = 2 and O_{3}(X) be abelian. Now as before we see that we cannot have the proper number of orbits.

So assume |V | = 2^{8}. Let again O_{3}(X)/C I_{O_{3}(X)}(V) ≅ ZZ_{3} × ZZ_{3}. Now we either have |C I_{V} (u)| = 4 for any u which induces a group of order 3 or we have two such subgroups with |C I_{V} (u)| = 16 and two with C I_{V} (u) = 1. Assume the latter. Then S ≅ D_{8} or ZZ_{4} and |A| = 2. But A inverts some fixed point free element of order 3, a contradiction. So we may assume that we have the former. Furthermore |A| = 4 and S ≅ S_{16}. Then for some a ∈ A^{♯} we have C I_{V} (a) = C I_{V} (A). But now A = ⟨a, b⟩ and b inverts some element c of order 8. So [c^{2}, C I_{V} (a)] = 0, a contradiction. So we have (16) (iii).

(xv) Let now O_{3}(X) ≅ 3^{1+4}. Let V = [V, Z(O_{3}(X))]. Then we get |V | ≥ 2^{18}. As m_{2}(X) ≤ 3 (a Sylow 2-subgroup of X is contained in Aut(Sp_{4}(3))), we get that [A, Z(X)] = 1. So |A| ≤ 4. Let a ∈ A, ρ ∈ O_{3}(X) \ Z(X), ρ^{a} = ρ^{−1}. Then |V : C I_{V} (ρ)| ≤ 2^{8}. So |V : C I_{V} (ρ) ∩ C I_{V} (Z(O_{3}(X)))| ≤ 2^{16}, a contradiction.

So we have [V, Z(O_{3}(X))] = 0. Now by irreducible action of S we get |V | = 2^{8} or |V | ≥ 2^{16}. The former is (16) (v). So assume the latter. Let t ∈ A such that C I_{O_{3}(X)}(t) ≤ Z(O_{3}(X)). Then |V : C I_{V} (t)| ≥ 2^{8}. But as |A| ≤ 2^{3} and so |V : C I_{V} (A)| ≤ 64, this is impossible. So we have |A| ≤ 4 and then |V : C I_{V} (A)| ≤ 16. Let a ∈ A^{♯}, then a ∼ at, where C I_{O_{3}(X)}(t) ≤ Z(O_{3}(X)). This shows C I_{V} (a) = C I_{V} (A), for all a ∈ A^{♯}. But as there is some Σ_{3} ×Σ_{3} ≅ U induced on V, with A ∈ Syl_{2}(U), this is impossible.

(xvi) Let m_{3}(O_{3}(X)) = 3. By (1.8) m_{2}(S) ≤ 2. Then for t ∈ Z(S), we get |V : C I_{V} (t)| ≤ 16 and so |V | ≤ 2^{8}.

(1.36) Corollary. Let X, V be as in (1.35). But assume now additionally that |V : C I_{V} (A)| < |A|^{2}. Then one of the following holds:

(1) V is an F -module.

(2) E(X) ≅ (S)L_{3}(q), q even, and V is a direct sum of the natural and the dual module.

(3) E(X) ≅ L_{2}(q), q = t^{2}, q even, V is the orthogonal module.

(4) E(X) ≅ Sp_{4}(q), q even, or 3·A_{6}, V is a direct sum of the natural module and the conjugate 4-dimensional module.

(5) E(X) ≅ A_{9}.

(6) E(X) = X_{1}X_{2} ≅ L_{2}(q) × L_{2}(q), q = t^{2}, q even, V = V_{1} ⨁ V_{2}, [X_{i}, V] = V_{i}, [X_{3−i}, V_{i}] = 0, i = 1, 2, V_{i} is the orthogonal X_{i}-module, i = 1, 2.

(7) E(X) ≅ L_{2}(q) × L_{2}(q), q even, V is the natural O^{+}(4, q)-module.

(8) E(X) = X_{1}X_{2}, X_{i} ≅ L_{3}(2), V = V_{1} ⨁ V_{2}, [V_{i}, X_{i}] = V_{i}, [V_{i}, X_{3−i}] = 0, V_{i} is a direct sum of the natural and the dual X_{i}-module, i = 1, 2.

(9) E(X) = X_{1}X_{2}, X_{i} ≅ 3 · A_{6}, [Z(E(X)), V] = 0, V = V_{1} ⨁ V_{2}, [V_{i}, X_{i}] = V_{i}, [V_{i}, X_{3−i}] = 0, V_{i} is a direct sum of the two 4-dimensional A_{6}- modules, i = 1, 2.

(10) X is solvable, F (X) ≅ 3^{1+4}, |V | = 2^{8}.

Proof . Inspection of the list in (1.35) gives the assertion up to the case in (1.35) (16) (vi). Suppose that there is some A with |A| = 4 and |V : C I_{V} (A)| = 8. Then A ∩ Z(S) ≠ 1. Choose t ∈ A ∩ Z(S), t ≠ 1. Then C I_{O_{3}(X)/Φ(O_{3}(X))}(t) = 1. As |V : C I_{V} (t)| ≤ 8, we get |O_{3}(X) : Φ(O_{3}(X)| ≤ 27. As a 2-group cannot act irreducibly on an elementary abelian group of order 27, we get |O_{3}(x) : Φ(O_{3}(X))| = 9. Now |[V, O_{3}(X)]| ≤ 2^{6} and so O_{3}(X)/C I_{O_{3}(X)}(V) ≅ ZZ_{3} × ZZ_{3} or 3^{1+2}. But in both cases we do not have A with |V : C I_{V} (A)| = 8 and V is not an F -module.

(1.37) Lemma. Let G = G_{2}(q), q even, V be an irreducible faithful F - module, A ≤ G be an offending subgroup. Then |A| = q^{3}.

Proof . By (1.18) V is the natural module, V = V (λ_{1}). By (1.23) the minimal codimension is 2. Assume |A| = q^{2}. Then C I_{V} (a) = C I_{V} (A) for all a ∈ A^{♯}.

Let P be the minimal parabolic with |C I_{V} (O^{2′}(P))| = q. We have that there is no g ∈ P with ⟨A, A^{g}⟩ = O^{2′}(P). Hence by (1.1) A ∩ Z(O_{2}(P)) ≠ 1. Let g ∈ O_{2}(P). Then A ∩ A^{g} ≠ 1. Hence C I_{V} (A^{g}) = C I_{V} (A ∩ A^{g}) = C I_{V} (A). Now either A = Z(O_{2}(P)), or there is some x ∈ ⟨A^{O_{2}(P)}⟩, with x^{2} ≠ 1. Then |C I_{V} (x)| = q^{4} and so C I_{V} (x) = C I_{V} (x^{2}), a contradiction. This now implies that A = Z(O_{2}(P)). But

G = ⟨C I_{G}(a)|a ∈ Z(O_{2}(P))^{♯}⟩

and C I_{G}(a) acts on C I_{V} (a). This implies that C I_{V} (a) is G-invariant, a contradiction.

(1.38) Lemma. Let G = G(q) be a Lie group in even characteristic, V be a faithful GF (2)G-module, and t ∈ G be an involution with |V : C I_{V} (t)| ≤ 2^{m_{2}(G)}. Let ℓ be the Lie rank of G and suppose that V involves all V (λ), λ a fundamental weight, but maybe not V (λ_{1}) and V (λ_{2}). Furthermore assume that there is some irreducible nontrivial module involved in V which is not of type V (λ), λ a fundamental weight. Then G ≅ (S)L_{2}(q) or (S)L_{3}(q).

Proof . By (1.22) G ≅ (S)L_{n}(q), (S)U_{n}(q), n ≥ 6, Sp_{2n}(q), n ≥ 7 or Ω_{2n}^{±}(q), n ≥ 8. Now application of (1.23) shows G ≅ SL_{n}(q), n ≤ 7 or (S)U_{6}(q).

If G ≅ (S)U_{6}(q), then V (λ_{2}) is involved. But by (1.23) |V (λ_{2}) : C I_{V (λ_{2})}(t)| ≥ q^{8}. Hence V just involves V (λ_{2}), as G contains no elementary abelian subgroup of order greater than q^{9} by (1.23).

Let G ≅ (S)L_{7}(q). Then V (λ_{2}), V (λ_{3}), V (λ_{4}), V (λ_{5}) are involved. But by (1.23) |V (λ_{3}) : C I_{V (λ_{3})}(t)| ≥ q^{10} ≤ |V (λ_{4}) : C I_{V (λ_{4})}(t)| and there are no elementary abelian subgroups of order greater than q^{15}.

Let G ≅ (S)L_{6}(q). Then V (λ_{2}), V (λ_{3}), V (λ_{4}) are involved. Now |V (λ_{2}) : C I_{V (λ_{2})}(t)| ≥ q^{4} ≤ |V (λ_{4}) : C I_{V (λ_{4})}(t)| and |V (λ_{3}) : C I_{V (λ_{3})}(t)| ≥ q^{6} by (1.23). But by (1.23) there are no elementary abelian subgroups of order greater than q^{9}.

Let G ≅ (S)L_{5}(q). Then V (λ_{2}) and V (λ_{3}) are involved. By (1.23) |V (λ_{2}) : C I_{V (λ_{2})}(t)| ≥ q^{3} ≤ |V (λ_{3}) : C I_{V (λ_{3})}(t)|. As there are no elementary abelian subgroups of order greater than q^{6} in G by (1.23), we see that just V (λ_{2}) and V (λ_{3}) are involved, a contradiction.

Let finally G ≅ (S)L_{4}(q). Then V (λ_{2}) is involved and |V (λ_{2}) : C I_{V (λ_{2})}(t)| ≥ q^{2}. Furthermore by (1.22) either V (λ_{1}) ⨂ V (λ_{1})^{σ} or V (λ_{3}) ⨂ V (λ_{3})^{σ} is involved. Now by (1.23)

|V (λ_{1}) ⨂ V (λ_{1})^{σ} : C I_{V (λ_{1}) ⨂ V (λ_{1})^{σ}}(t)| ≥ q^{3} .

But (S)L_{4}(q) does not possess elementary abelian subgroups of order q^{5} by (1.23).

(1.39) Lemma. Let F^{∗}(G) ≅ Ω_{6}^{±}(r), r even. Let V be the natural module for G. Suppose there is A ≤ G, A an elementary abelian 2–group with [V, A, A] = 0. If |A| > r^{3} then the following holds

(i) |A| = 2 · r^{3}, A ≰ F^{∗}(G)

(ii) There is some t ∈ A with |[t, V]| = r^{3}

(iii) If F^{∗}(G) ≅ Ω_{6}^{+}(r) ≅ L_{4}(r), then A does not act quadratically on the natural L_{4}(r)-module (iv) If F^{∗}(G) ≅ Ω_{6}^{−}(r) ≅ U_{4}(r), then A does not act quadratically on the

natural U_{4}(r)-module.

Proof . Suppose first A ≤ L = F^{∗}(G). Let P be a subgroup of L, P/O_{2}(P) ≅ Ω_{4}^{±}(r), |O_{2}(P)| = r^{4}. Now O_{2}(P) is the natural module for P/O_{2}(P). As we may assume A ≤ P a moments thought shows A ≤ O_{2}(P). Now we have that for some v ∈ V, [A, ⟨v⟩] = 0, [v^{⊥}, A] ⊆ ⟨v⟩, C I_{v^{⊥}}(A) = ⟨v⟩, [V, A] ⊆ v^{⊥}. By (1.27) (V ≅ V (λ_{2})) we see that [V, A] ⊈ ⟨v⟩. Hence [V, A, A] ≠ 0, a contradiction.

We have A ≰ L. Then we get |A| = 2 · r^{3} and AL ≅ O_{6}^{±}(r). This is (i). Furthermore A contains all types of involutions in O_{6}^{±}(r), which proves (ii). As O_{6}^{±}(r) does not act on the natural L_{4}(r)-module we also have (iii).

Let L ≅ U_{4}(r), x ∈ A \ L. Then x induces a field automorphism on L and on W, the natural U_{4}(r)-module. We have C I_{L}(x) ≅ Sp_{4}(r) and so [[V, x], C I_{L}(x)] ≠ 0. But this contradicts [[V, x], A] = 0. This proves (iv).

(1.40) Lemma. Let X ≅ ^{3}D_{4}(q), ^{2}F_{4}(q), q even, V be a nontrivial irreducible GF (2)-module for X, t ∈ X be an involution. Then |[V, t]| > q^{5}.

Proof . We have |A| ≤ q^{5} for any elementary abelian subgroup of X. By [Coop] |[V, t]| > q^{5}.

(1.41) Lemma. Let F^{∗}(X) ≅ G(r) be a Lie group over a field of odd characteristic which is not a Lie group in characteristic two too. Let V be a faithful irreducible GF (2)-module for X and t be an involution in X with |[V, t]| ≤ 2^{m_{2}(X)+1}. Then F^{∗}(X) ≅ 3 · U_{4}(3), L_{3}(3), U_{4}(3), L_{4}(3), where we have equality in the last three cases and m_{2}(F^{∗}(X)) < m_{2}(X).

Proof . (The proof follows [Asch3, (10.9)]). Let first F^{∗}(X) ≅ L_{2}(r). Then m_{2}(X) ≤ 3 or m_{2}(X) = 2 for r prime . Now the assertion follows with (1.30) and (1.29). If F^{∗}(X) ≅ ^{2}G_{2}(r), (S)L_{3}(r), (S)U_{3}(r), (S)L_{4}(r), (S)U_{4}(r), 3 · U_{4}(3), P Sp_{4}(r), G_{2}(r), 3 · G_{2}(3) or ^{3}D_{4}(r), then m_{2}(X) ≤ 6 and the assertion follows with (1.29), unless F^{∗}(X) ≅ L_{3}(3), U_{4}(3), L_{4}(3) or 3 · U_{4}(3), where we have equality in the first three cases.

Let now T ∈ Syl_{2}(X), △ = Fun(T), k = |△|, Y = ⟨△⟩, and Y T^{∗} = Y T/C I_{Y T} (Y). Then Y ^{∗} is a direct product of k copies of L_{2}(r) permuted by T. Let B^{∗} be an elementary abelian subgroup of T^{∗} of maximal order, K ∈ △, D^{∗} a complement to E^{∗} = N_{B}(K)^{∗} in B^{∗}, t = m_{2}(D^{∗}), and S = ⟨K^{D}⟩ ∩ T. Then S^{∗} is the direct product of 2^{t} copies of L_{2}(r) and ε = m_{2}(C I_{K^{∗}}(E^{∗})) = 1 or 2. Hence m_{2}(C I_{S^{∗}}(E^{∗})) = ε2^{t} and m_{2}(S^{∗} ∩ B^{∗}) = m_{2}(C I_{S^{∗}}(B^{∗})) = ε, so m_{2}(E^{∗}C I_{S^{∗}}(B)) ≥ m_{2}(E^{∗}) + ε2^{t} − ε ≥ m_{2}(E^{∗}) + t = m_{2}(B^{∗}), so that E^{∗}C I_{S^{∗}}(E^{∗}) is also elementary abelian of maximal order. Thus we may choose B to fix each member of △. Let C^{∗} be the subgroup of B^{∗} inducing inner automorphisms in P GL_{2}(r) on each member of △^{∗}. Then m_{2}(B/C) ≤ 1 and m_{2}(C^{∗}) ≤ 2k, so m(B^{∗}) ≤ 2k + 1. Set Z = T ∩ Z(Y). Then m_{2}(Z) ≤ k and m_{2}(C I_{T} (Y)) ≤ m_{2}(Z) + i(F^{∗}(X)), where i(F^{∗}(X)) can be found in [Asch3, (10.8)]. So we have

m_{2}(X) ≤ 3k + 1 + i(F^{∗}(X)) , i(F^{∗}(X)) ≤ 3 .

Let m be the lower bound for [V, t] supplied by (1.28). Then

m ≥ (p − 1)p^{w−1}d(p)/2ε ≥ 2(3^{w−1)}ε,

where r = p^{s}, w is given by |O_{p}(C I_{F ^{∗}(X)}(R))| = r^{1+2w}, R some root group in F^{∗}(X), ε = 1 or F^{∗}(X) = P Sp_{2n}(r), r > p, where ε = 2. Furthermore k is listed in [Asch2, Theorem 2]. In particular k ≤ w unless F^{∗}(X) = P Sp_{2n}(r), where k = n and w = n − 1.

So assume F^{∗}(X) ≠ P Sp_{2n}(r), n ≥ 3. Then

2(3^{w−1}) > 3w + 5 ≥ 3k + 2 + i(F^{∗}(X)) .

Hence the lemma holds.

So let F^{∗}(X) ≅ P Sp_{2n}(r). Then i(F^{∗}(X)) = 0, k = n, w = n − 1. Hence

(p − 1)r^{n−1}d(p)/2εp > 3n + 2 ≥ m_{2}(X) + 1 ,

unless F^{∗}(X) ≅ P Sp_{6}(3). But then |[V, t]| ≥ 2^{6}. As m_{2}(Aut (P Sp_{6}(3))) = 4, the lemma holds.

(1.42) Lemma. Let F^{∗}(X) be a quasisimple group, V be a faithful F_{1}- module for X. Then either

(i) F^{∗}(X) is a Lie group in characteristic two

(ii) F^{∗}(X) ≅ 3 · U_{4}(3)

(iii) F^{∗}(X)/Z(F^{∗}(X)) ≅ M_{22}, M_{23}, M_{24}, M_{12} or J_{2} (in fact M_{12} or J_{2} do not possess F_{1}-modules but we will not need this result)

(iv) F^{∗}(X) ≅ 3 · A_{6} and |[V, X]| = 2^{6}

(v) F^{∗}(X) ≅ A_{n} and V involves just permutation modules, or n = 8, 7, 5 and also 4–dimensional modules are involved, or n = 9 and V has just one nontrivial module involved, a spin module.

Proof . (See also [Asch3]). Let first F^{∗}(X) be a Lie group in odd characteristic. By (1.41) either (i) or (ii) holds, or F^{∗}(X) ≅ L_{3}(3), U_{4}(3) or L_{4}(3).

Now let A be some offending subgroup, i.e. |V : C I_{V} (A)| ≤ 2|A|. Suppose F^{∗}(X) ≅ U_{4}(3), L_{4}(3) or L_{3}(3). Then by (1.41) |V : C I_{V} (A)| = 2|A| = |V : C I_{V} (t)| for any t ∈ A^{♯}. Furthermore |A| > 2. Hence A acts quadratically on V, contradicting (1.12).

Let next F^{∗}(X)/Z(F^{∗}(X)) be sporadic. Application of (1.25) shows that either (iii) holds or F^{∗}(X) ≅ HS, Co_{2} or Co_{1}. In case of HS we get |V : C I_{V} (A)| = |V : C I_{V} (t)| for any t ∈ A^{♯}, and so we get a quadratic fours group, contradicting (1.12).

Let F^{∗}(X) ≅ Co_{1}. For centralizers of involutions in Co_{1} and Co_{2} see [CCNPW]. Let z be a 2-central involution in X. Then C I_{X}(z) ≅ 2^{1+8}Ω_{8}^{+}(2). Set Q = O_{2}(C I_{G}(z)). Suppose the 24-dimensional module is not involved in V. Then by [MeiStr1] there is no quadratic fours group. Hence C I_{C I_{X}(z)}([V, z]) = ⟨z⟩. This shows |[V, z]| ≥ 2^{9}. Let A be an offending subgroup. If z ∈ A, then we have |A| ≥ 2^{8} and so |A : A ∩ Q| ≥ 8. But then |[V, z] : C I_{[V,z]}(A)| ≥ 4. This now shows |A| ≥ 2^{10} and so |A : A∩Q| ≥ 2^{5}. Furthermore |[V, z] : C I_{[V,z]}(A∩Q)| ≥ 2 and so we now see |[V, z] : C I_{[V,z]}(A)| ≥ 16. Hence |A| ≥ 2^{12}, a contradiction. We have z ∉ A. Then we may assume A ∩ Q = 1 as all involutions in Q are conjugated to z. Now |A| ≤ 2^{6}. But this contradicts (1.25).

So we are left with the 24-dimensional module, z ∈ A. Now |[V, z]| = 2^{8}, and A ≰ Q. Hence |C I_{V} (z) : C I_{C I_{V} (z)}(A)| ≥ 16. This gives |A| = 2^{11}. But there is just one elementary abelian subgroup of order 2^{11} in a Sylow 2-subgroup. By [MeiStr1] we have |C I_{V} (A)| = 2, a contradiction.

Let now X ≅ Co_{2}. Let us again assume that there is no quadratic fours group. Let z be 2-central, Q = O_{2}(C I_{G}(z)). As before |[V, z]| ≥ 2^{9}. Suppose z ∈ A. Then A ≰ Q. Furthermore A∩Q ≰ ⟨z⟩. Hence |[V, z] : C I_{[V,z]}(A)| ≥ 8. This would imply |A| ≥ 2^{11}, a contradiction.

Hence z ∉ A. Let a ∈ A with O_{2}(C I_{X}(a)) ≅ 2^{1+6} · 2^{4}, C I_{X}(a)/O_{2}(C I_{X}(a)) ≅ A_{8}. As there is no quadratic fours group we see |[V, a]| ≥ 2^{8}. Furthermore as z ∉ A, |A ∩ O_{2}(C I_{X}(a))| ≤ 16. Now |A : A ∩ O_{2}(C I_{X}(a))| ≥ 8. But |[V, a] : C I_{[V,a]}(A)| ≥ 4, and so |A| ≥ 2^{9}. This shows |A : A∩O_{2}(C I_{X}(a))| ≥ 2^{5}, a contradiction to m_{2}(A_{8}) = 4.

So neither z nor a is in A. Hence A ∩ Q = 1. Now |A| ≤ 2^{6}. Let b ∈ A, then C I_{X}(b) ≅ 2^{10}P ΓL_{2}(9). Now we get |[V, b]| ≥ 2^{8}, a contradiction.

Hence we have quadratic fours groups. By [MeiStr2] this gives |V | = 2^{22}. We show that V is not an F_{1}-module.

We have |C I_{V} (z)| = 2^{16}. Furthermore |C I_{V} (z) : C I_{C I_{V} (z)}(A ∩ Q)| ≥ |A ∩ Q|/2. Hence A ≰ Q. Let z ∈ A. Now for B ≤ A, B ∩ Q = 1, we have |[V, z] : C I_{[V,z]}(B)| ≤ |B|/2^{5}. But this is impossible. Hence z ∉ A. Let a ∈ A^{♯}, a as before. Then |[V, a]| = 2^{8} and |A| ≤ 2^{8} as before. This shows |A : A ∩ O_{2}(C I_{X}(a))| ≥ 8. But then [A, [V, a]] ≠ 0 and so |A : A ∩ O_{2}(C I_{X}(a))| = 16 and A has to induce transvections on [V, a], a contradiction.

So a^{X} ∩ A = ∅. This shows A ⊆ b^{G}. But |[V, b]| ≥ 2^{8}, and |A| ≤ 2^{6}, as A ∩ Q = 1, we have a contradiction. This proves (iii).

(iv) For n ≤ 7 we get the assertion by checking the irreducible modules. Let n = 8 or 9. By (1.30) we have |V : C I_{V} (a)| ≥ 16 for any a ∈ X. Hence |A| ≥ 8. If |A| = 8, then C I_{V} (a) = C I_{V} (A) for any a ∈ A^{♯}. Hence A acts quadratically and the assertion follows with (1.12). So assume |A| = 16. Now |V : C I_{V} (a)| ≤ 32 for any a ∈ A^{♯}. If (12) ∈ A, then C I_{X}(x) ≅ ⟨x⟩ × Σ_{6} or ⟨x⟩ × Σ_{7}. As A ≤ C I_{X}(x) and |[V, x]| ≥ 16, we see that |[V, x] : C I_{[V,x]}(A)| ≤ 2. This shows [E(C I_{X}(x)), [V, a]] = 0. Now there is a quadratic fours group on V. This gives the assertion, or n = 9 and we have the spin module. But there is no quadratic fours group of this type on the spin module by [MeiStr1, (4.3)].

Hence we have A ≤ A_{n}. In particular

A = ⟨(12)(34), (13)(24), (56)(78), (57)(68)⟩.

Let b = (12)(34)(56)(78). Then A ≰ O_{2}(C I_{V} (b)). As |[V, b] : C I_{[V,b]}(A)| ≤ 2, we see again that we have a quadratic fours group containing b. By (1.12) we have the assertion, or n = 9 and we have a spin module. Now A_{1} = ⟨(12)(34), (13)(24)⟩ ≤ A_{5} ≤ A_{9}. As g = (123) acts fixed point freely on V, we see with [Hi, (8.2)] that V is a direct sum of two natural L_{2}(4)- modules. Hence |V : C I_{V} (A_{1})| = 16. We have C I_{X}(A_{1}) ≤ K, K ≅ A_{{5,6,7,8,9}} with A_{2} = ⟨(56)(78), (57)(68)⟩ ≤ K. Now (567) = γ ∈ K and [V, γ] = V. Hence C I_{V} (A_{1}) is the natural K-module. This shows |V : C I_{V} (A)| = 2^{6}, a contradiction.

Let now n > 9. Let Γ∪^{·} Γ′ = Ω, Ω = {1, …, n} be a partition with Γ^{a} = Γ or Γ′ for a ∈ A. Let |Γ| = m, and if A ≰ N(Γ) then n is a power of two. Suppose first Γ^{a} = Γ′ for some a ∈ A. Then 2m = n. Let E = E(X_{Γ′}). Then C I_{E E^{a}}(a) ≅ E. Furthermore C I_{X}(a) = ⟨a⟩ × K, K ≅ Σ_{m}. Let A = ⟨a⟩×B, B ≤ K. We have [C I_{V} (a), E(K)] ≠ 0. Hence as |V : C I_{V} (a)| ≥ 16 by (1.30) we get |C I_{V} (a) : C I_{C I_{V} (a)}(B)| ≤ |B|/4. By induction C I_{V} (a) involves just natural or 4-dimensional modules. But, as can be easily seen, on the natural module W we get |W : C I_{W} (B)| ≥ |B|/2, a contradiction. Hence m ≤ 8 and 4-dimensional modules are involved. Furthermore m ≥ 7. As K ≅ Σ_{m}, we get that both 4-dimensional modules are involved and so again |C I_{V} (a) : C I_{C I_{V} (a)}(B)| ≥ |B|, a contradiction.

So let now A ≤ N(Γ). Choose Γ with m maximal. Then either A has a fixed point and m = n − 1 or m = n − 2.

Let m = n − 1. Then A ≤ X_{Γ′} ≅ A_{n} or Σ_{n−1}. By induction V just involves natural X_{Γ′}-modules, or m = 9. As for a natural module W we have |W : C I_{W} (A)| ≥ |A|/2, we get that just one natural module is involved or |A| ≤ 8.

Suppose the latter. By (1.30) we may assume |V : C I_{V} (a)| ≥ 16 for any a ∈ A. Hence we have |A| = 8 and C I_{V} (a) = C I_{V} (A) for any a ∈ A. In particular A acts quadratically. Now we get the assertion with (1.12), as on the spin module ρ = (1, 2, 3) acts fixed point freely but ρ ∈ A_{m−1}.

Assume V involves exactly one natural X_{Γ′}-module. If C I_{V} (X_{Γ′}) ≠ 1. Then choose v ∈ C I_{V} (X_{Γ′})^{♯}. Now we get that some permutation module for X is involved in V. Then induction on dim V yields the assertion. Hence we may assume C I_{V} (X_{Γ′}) = 1. Now (1.19) shows |V : [V, E(X_{Γ′})]| ≤ 2, and V is a factor module of the permutation module for X_{Γ′}. Hence there is a quadratic group of order eight ⟨(12)(34), (12)(56), (12)(78)⟩. By (1.12) V is a permutation module for X. We are left with m = 9 and V involves a spin module just once. Now |A| = 8 and |V : C I_{V} (a)| = 16 for any a ∈ A^{♯}. Hence A acts quadratically on V and so by (1.12) V is the natural module or a spin module for A_{10}. But |C I_{V} (a)| ≥ 2^{8} for the spin module for A_{10}. We have the assertion.

Assume now m = n − 2. Let Q = O_{2}(X_{Γ})). Then |Q| ≤ 2. Let first A ∩ Q = 1. Now [E(X_{Γ′}), C I_{V} (Q)] ≠ 0. Hence by induction just natural, 4dimensional or a spin module are involved. The last two cases just occur for n = 10, 11, respectively. So assume that just natural modules are involved. If there is more than one natural module then |A| ≤ 8. Now A acts quadratically on V and we get the assertion as above. Hence we may assume that there is just one natural module involved. This shows [[V, Q], E(X_{Γ′})] = 1, as otherwise V involves at least two natural E(X_{Γ′})-modules and we get again |A| ≤ 8.

Now by (1.19) we have V = C I_{V} (E(X_{Γ′}))V_{1}, V_{1}/V_{1}∩C I_{V} (E(X_{Γ′})) is a factor module of the permutation module. As m − 2 ≥ 8, we get a quadratic group of order eight in E(X_{Γ′}). Hence the assertion follows with (1.12).

So let next n = 10, m − 2 = 8. Furthermore there are the natural L_{4}(2)- module and its dual involved in C I_{V} (Q). Now |C I_{V} (Q) : C I_{C I_{V} (Q)}(A)| ≥ |A|. In particular there are no further nontrivial modules involved. This shows with (1.20) that |[E(X_{Γ′}), C I_{V} (Q)]| = 2^{8}. Furthermore [[V, Q], E(X_{Γ′})] = 0. Hence |[E(X_{Γ′}), V]| = 2^{8} and so there is a quadratic group of order 16 (O_{2} of the stabilizer of a 2-space in the L_{4}(2)-representation). By (1.12) V just involves natural modules. Assume now n = 11 and C I_{V} (Q) involves the spin module for A_{9}. As Σ_{9} is involved, both spin modules are in C I_{V} (Q). But then |V : C I_{V} (A)| > 2|A|, as the spin module is not an F -module for A_{9} by (1.18).

So we are left with Q ≠ 1, Q ≤ A. Let B be a complement to Q in A. First of all |V : C I_{V} (Q)| ≥ 16 by (1.30). Hence |C I_{V} (Q) : C I_{C I_{V} (Q)}(B)| ≤ |B|/4. But as C I_{V} (Q) just involves natural modules or n = 10 and C I_{V} (Q) involves both 4-dimensional modules or n = 11 and V involves a spin module we see |C I_{V} (Q) : C I_{C I_{V} (Q)}(B)| ≥ |B|/2, a contradiction.

(1.43) Lemma. Let X be as in (1.7), X nonsolvable, and V be an F_{1}- module. If C I_{V} (X) = 0, V = [V, X], then one of the following holds:

(1) V is an F -module

(2) E(X) ≅ A_{5} or L_{2}(16) and V is the orthogonal module. In case of L_{2}(16) the offending subgroup A is of order eight or two and not contained in E(X).

(3) E(X) ≅ L_{3}(2) and V is a direct sum of the natural module and the

dual module.

(4) E(X) ≅ A_{9}, V is the natural module or a spin module.

(5) E(X) = X_{1}X_{2}, X_{1} ≅ X_{2} ≅ A_{5}, L_{2}(16) or L_{3}(2) and V = V_{1} ⨁ V_{2}, [V_{i}, X_{i}] = V_{i}, [V_{i}, X_{3−i}] = 0, V_{i} is an F_{1}-module for X_{i}, i = 1, 2.

(6) E(X) ≅ A_{5} × A_{5} and V is the O^{+}(4, 4)-module.

(7) E(X) ≅ A_{5} or A_{9} and V = V_{1} ⨁ V_{2}, V_{i} permutation modules for E(X), i = 1, 2, and the offending subgroup A is of order 2.

(8) F (X) is a p-group, p ∈ {5, 7}, E(X/F (X)) ≅ L_{2}(p), [V, F (X)] = 0 and V is an F_{1}-module for X/F (X), as in (2), (3) or (7).

Proof . We have |V : C I_{V} (A)| ≤ 2|A| ≤ |A|^{2}. So all modules will be found in (1.36). Inspection gives the assertion up to the result about A_{9}. Now (4) and (7) follow from (1.42).

(1.44) Lemma. Let X be one of the groups in (1.7) (i), q > 2, or (1.7) (iii) with E(X) ≅ SL_{3}(4) ∗ SL_{3}(4) or E(X) ≅ SU_{3}(8) ∗ SU_{3}(8). Let V be an irreducible X-module over GF (2), t ∈ E(X) be some involution and |[V, t]| ≤ 2q. Then

(i) E(X) ≅ L_{2}(q) or L_{2}(q) × L_{2}(q)

(ii) Let E(X) ≅ L_{2}(q) × L_{2}(q), A = ⟨t|1 = t^{2}, t ∈ E(X), |[V, t]| ≤ 2q⟩. Suppose A ∩ E(X) is not contained in either of the two components, then |V : C I_{V} (A)| ≥ q^{2}.

Proof . By (1.26) and (1.27) we get the assertion (i) as q > 2. We have |[V, t]| < q^{2} and in case of E(X) ≅ (S)L_{3}(q), Sp_{4}(q), SL_{3}(4)∗SL_{3}(4) we have that the natural module and its dual is involved and so q^{2} ≤ |[V, t]|.

To prove (ii) assume |V : C I_{V} (A)| < q^{2}. Let E(X) = X_{1} ×X_{2}, X_{i} ≅ L_{2}(q). Then we see that [V, X_{1}] involves exactly one irreducible nontrivial module T. Hence [[V, X_{1}], X_{2}] = 0. Now A has to be nontrivially on [V, X_{1}] and [V, X_{2}]. This shows |T : C I_{V} (A)| < q. But this contradicts (1.27) (i).

(1.45) Lemma. Let X = (S)U_{n}(q) or Ω_{2n}^{−}(q), q even, and V be the natural module for X. Let A ≤ X be an elementary abelian 2–group with |V : C I_{V} (A)| ≤ q^{2}. Then |A| ≤ q.

Proof . We may assume |A| ≠ 1. Then |V : C I_{V} (A)| = q^{2}. If X ≅ (S)U_{n}(q), then A is a group of transvections to the hyperplane C I_{V} (A) and so |A| ≤ q.

Let X ≅ Ω_{2n}^{−}(q). Now we have some a ∈ A^{♯} such that [a, V]^{⊥} ≥ C I_{V} (A). Let v be some singular vector in [a, V]. Then C I_{X}(v) acts on v^{⊥}. We have O_{2}(C I_{X}(v)) is elementary abelian and C I_{X}(v)/O_{2}(C I_{X}(v)) ≅ Ω_{2n−2}^{−}(q). Furthermore v^{⊥}/v^{⊥⊥} ≅ O_{2}(C I_{X}(v)) as a module for C I_{X}(v)/O_{2}(C I_{X}(v)). As there are no transvections on the natural module in Ω_{2n−2}^{−}(q) and |v^{⊥} : C I_{V} (A)| = q, we get A ≤ B ≤ O_{2}(C I_{X}(v)) where B ≅ C I_{V} (A)^{⊥}/v^{⊥⊥}. So |B| = q and we are done.

(1.46) Lemma. Let X ≅ Sp_{6}(r), r = 2^{n}, V be the natural module. Let A be an elementary abelian 2–subgroup of X. If |V : C I_{V} (A)| ≤ r^{2}, then |A| ≤ r^{3}.

Proof . Let R be a transvection group. We may assume A ≤ C I_{X}(R). Let V_{1} = C I_{V} (R) and V_{2} = [V, R]. Then for B = O_{2}(C I(R)), we have [V_{1}, B] = V_{2} and [V/V_{2}, B] = V_{1}/V_{2}. Hence for B_{1} ≤ B, B_{1}∩R = 1, we get |V : C I_{V} (B_{1})| ≥ |B_{1}|r. This now implies |B ∩ A : B ∩ R| ≤ r. Hence |B ∩ A| ≤ r^{2}. Let a ∈ A \ B. Then |[V_{1}/V_{2}, a]| ≥ r. Suppose |V_{1}/V_{2} : C I_{V_{1}/V_{2}}(A)| ≥ r^{2}. Then B ∩ A = 1. As Sp_{4}(r) does not contain elementary abelian subgroups of order greater than r^{3}, we have the assertion.

So let R ∩ A ≠ 1. Then |V_{1}/V_{2} : C I_{V_{1}/V_{2}}(A)| ≤ r and so |A : A ∩ B| ≤ r, whence |A| ≤ r^{3}.

(1.47) Lemma. Let X ≅ (S)L_{n}(r), r = 2^{m}, V an irreducible faithful Xmodule over GF (2). Any one of the following implies that V is the natural module.

(a) n = 6, 7, |V : C I_{V} (t)| ≤ r^{3}, t some involution in X

(b) n = 5, |V : C I_{V} (t)| ≤ r^{2}, t some involution in X

(c) n = 4, |V : C I_{V} (t)| < r^{2}, t some involution in X.

Proof . We have that V is an SC-module. By (1.22) V ≅ V (λ), λ a fundamental weight or V ≅ V (λ_{1}) ⨂ V (λ_{1})^{σ}, V (λ_{n−1}) ⨂ V (λ_{n−1})^{σ}, V (λ_{1}) ⨂ V (λ_{n−1})^{σ}, σ a field automorphism, or we have (a) and V ≅ V (λ_{1} + λ_{n−1}).

Suppose V ≅ V̸ (λ). Then application of (1.24) shows |[V, t]| ≥ r^{(n−1)} or V ≅ V (λ + λ_{n−1}) and |[V, t]| ≥ r^{4}, a contradiction

So we have V ≅ V (λ), λ a fundamental weight. Now (1.23) yields the assertion.

(1.48) Lemma. Let X ≅ G (r) be a Lie group over a field of characteristic 2, X ≅ ^{3̸}D_{4}(r). Let V be some irreducible GF (2)-module for X and t be an involution in X. Suppose |V : C I_{V} (t)| ≤ r^{2} and |V : C I_{V} (t)| ≤ r^{4} if X is twisted. Then X ≅ (S)L_{n}(r), Sp_{2n}(r), (S)U_{n}(r), Ω_{2n}^{±}(r), G_{2}(r) or Sz(r).

Proof . Suppose first that X contains elementary abelian subgroups of order r^{2}, r^{4}, respectively. Then V is an SC-module. By (1.22) X ≅ SL_{n}(r), Sp_{2n}(r), SU_{n}(r), Ω_{2n}^{±}(r), G_{2}(r), F_{4}(r), E_{6}(r), ^{2}E_{6}(r), E_{7}(r).

In the last four cases V = V (λ), λ a fundamental weight, by (1.22). Now the assertion follows with (1.23).

If X is untwisted and does not contain elementary abelian subgroups of order r^{2}, we see X ≅ L_{2}(r).

If X is untwisted and does not contain elementary abelian subgroups of order r^{4}, we see X ≅ (S)U_{3}(r) or Sz(r).

(1.49) Lemma. Let X = Sp_{4}(q), q even, V a GF (2)X-module such that V/C I_{V} (X) is the natural module. Let S ∈ Syl_{2}(X) and Z = Z(S). If [Z, V, Z] = 0, then there is some A ≤ V, |A| = q, A ∩ C I_{V} (X) = 0 and |Z : C I_{Z}(A)| = q.

Proof . Let U ≤ Z, |U| = q, be the group of transvections on V/C I_{V} (X). Let V_{1} be the preimage of C I_{V/C I_{V} (X)}(U). Let A ≤ V_{1}, |A| = q, with C I_{V/C I_{V} (X)}(Z)AC I_{V} (X)/C I_{V} (X) = V_{1}/C I_{V} (X). We are going to show that [A, U] = 0. Let P = C I_{X}(U). Then P/O_{2}(P) ≅ L_{2}(q) and P/O_{2}(P) acts irreducibly on V_{2} = (V_{1}/C I_{V} (X))/([U, V]C I_{V} (X)/C I_{V} (X)). As [Z, V, Z] = 0, we have [[U, V]C I_{V} (X), U] = 0. Hence [V_{1}, U] is trivial or isomorphic to V_{2} as a P/O_{2}(P)-module. As [V_{1}, U] ≤ C I_{V} (X), we get [V_{1}, U] = 0 and so [A, U] = 0.

(1.50) Lemma. (a) Let X ≅ U_{4}(r), r = 2^{n}, and V be the natural module. If v ∈ V ^{♯}, A ≤ X, A an elementary abelian 2–group, [V, A, A] = 0, then |[v, A]| ≠ r^{4}.

(b)Let X ≅ Sp_{6}(r), r = 2^{n}, and V be the spin module. If v ∈ V ^{♯}, A ≤ X, A elementary abelian, [V, A, A] = 0, then |[v, A]| ≠ r^{4}.

(c) Let X_{1}, X_{2} ≤ X, T_{i} < X_{i}, X_{i}/C I_{X_{i}}(T_{i}) ≅ Sp_{6}(r), T_{i} the spin module for X_{i}/C I_{X_{i}}(T_{i}), i = 1, 2. Let ⟨T_{1}, T_{2}⟩ ≤ X_{1} ∩ X_{2}. Then [T_{1}, T_{2}] = 1.

Proof . (a) and (b): Suppose |[v, A]| = r^{4} for some v ∈ V ^{♯}. Then as [V, A, A] = 0, [v, A] ≤ C I_{V} (A). Let W ≤ V be a 4-dimensional subspace such that N_{X}(W) induces L_{2}(r^{2}) or Sp_{4}(r) on W. Then N_{X}(W) also induces L_{2}(r^{2}) and Sp_{4}(r) on V/W. Hence all cosets of W in V are conjugated under N_{X}(W). We may assume A ≤ N_{X}(W). As |[v, A]| = r^{4}, we have v ∉ W. Now we may assume v^{A} = v + W. This implies that X acts transitively on V ^{♯}, a contradiction.

(c): Suppose [T_{1}, T_{2}] ≠ 1. We have [T_{1}, T_{2}, T_{2}] = 1 = [T_{2}, T_{1}, T_{1}]. By (b) there is no t ∈ T_{2} with |[T_{1}, t]| = r^{4}. Let W ≤ T_{1}, |W | = r^{4}, such that N_{X_{1}}(W) induces Sp_{4}(r) on W and T_{1}/W. Set Z(O^{2′}(N_{X_{1}}(W))/C I_{X_{1}}(T_{1})) = Z. Then [Z, T_{1}] = W. We may assume that [T_{2}C I_{X_{1}}(T_{1})/C I_{X_{1}}(T_{1}), Z] = 1. Hence T_{2} induces transvections on either W or T_{1}/W. Set U_{2} = T_{2}C I_{X_{1}}(T_{1})/C I_{X_{1}}(T_{1}) ∩ O_{2}(N_{X_{1}}(W)/C I_{X_{1}}(T_{1})). Then |U_{2}| ≤ r^{3}. Hence |T_{2}C I_{X_{1}}(T_{1})/C I_{X_{1}}(T_{1})| ≤ r^{4}. Furthermore the same holds for T_{1}, i.e. |T_{1}C I_{X_{2}}(T_{2})/C I_{X_{2}}(T_{2})| ≤ r^{4}. By (1.23) we see that |T_{2} : C I_{X_{2}}(T_{1})| ≥ r^{3}. Hence |U_{2}| ≥ r^{2}.

Suppose u ∈ U_{2}, u is an element of type c_{2} in the Suzuki notation. Then there is some g ∈ X_{1} with u^{g} ∈ N_{X_{1}}(W) \ C I_{X_{1}}(W) and u^{g} is an element of type c_{2} in Sp_{4}(r). Hence |[W, u^{g}]| = r^{2} = |[T_{1}/W, u^{g}]|, a contradiction.

So we have that all elements in U_{2} are of type a_{2}. Hence |U_{2}| = r^{2} and |T_{1} : C I_{T_{1}}(U_{2})| = r^{2}. But now C I_{T_{1}}(u) = C I_{T_{1}}(U_{2}) for all u ∈ U_{2}^{♯}. Let R be a root group of type a_{2}, then N_{X_{1}}(R) is a maximal subgroup of X_{1} and so we have ⟨C I_{X_{1}/C I_{X_{1}(T_{1})}}(u)|u ∈ U^{♯}⟩ = X_{1}/C I_{X_{1}}(T_{1}), a contradiction.

(1.51) Lemma. Let O_{r}(X) be elementary abelian of order r^{3}, and X/O_{r}(X) ≅ L_{2}(r), r an odd prime, r > 3. Let C I_{X}(O_{r}(X)) = O_{r}(X) and V be a GF (2)X-module. If for some involution t ∈ X we have |[V, t]| ≤ 32, then [O_{r}(X), V] = 0.

Proof . Suppose [O_{r}(X), V] ≠ 0. Now |[O_{r}(X), t]| ≥ r^{2}. This implies r = 5 and there is some ω ∈ O_{5}(X), |[V, ω]| = 16. As |O_{5}(X)| = 5^{3}, we get |[V, O_{5}(X)]| ≤ 2^{12}. But 5^{4} | |̸GL_{12}(2)|, a contradiction.

(1.52) Lemma. Let X = O_{p} (X)F, O_{p} (X) elementary abelian of order p^{n}, n ≥ 3 and X a subgroup contained in the Borel subgroup of Aut (L_{2}(p^{n})), containing the Borel subgroup of L_{2}(p^{n}). Let V be some GF (2)X-module with C I_{V} (O_{p} (X)) = 0. If ω ∈ O_{p} (X), o(ω) = p, then |[ω, V]| > 4. In particular for x ∈ F, o(x) = 2, then |V : C I_{V} (x)| > 2.

Proof . Suppose |V : C I_{V} (x)| = 2. Then p = 3. Let ω ∈ O_{p} (X) with ω^{x} = ω^{−1}. Then |[V, ω]| = 4. So we get |[V, O_{p} (X)]| ≤ 2^{2n}. Now F contains a group of order (3^{n} − 1)/2 acting fixed point freely on O_{p} (X). This shows that n ≥ (3^{n} − 1)/4, which gives n = 2, a contradiction.

(1.53) Lemma. Let X = G (r) be a Lie group, r = 2^{n}, X ≅ Sz(r), L_{2}(r), U_{3}(r). Let S ∈ Syl_{2}(X), A◁S, A elementary abelian. Then there is some parabolic P of X, O^{2′}(P/O_{2}(P)) ≅ L_{2}(r), L_{2}(r^{2}) or U_{3}(r), such that A ≤ O_{2}(P). If X ≅ ^{2}F_{4}(r), A ≤ O_{2}(P) for both minimal parabolics. If X ≅ 3D_{4}(r), O^{2′}(P/O_{2}(P)) ≅ L_{2}(r^{3}).

Proof . By way of induction it is enough to prove the assertion for Lie groups of rank two, i.e. X ≅ L_{3}(r), Sp_{4}(r), U_{4}(r), U_{5}(r), G_{2}(r), ^{3}D_{4}(r), 2F_{4}(r).

Let P_{1}, P_{2} be the two minimal parabolics. In case of X ≅ U_{4}(r), U_{5}(r), 3D_{4}(r), ^{2}F_{4}(r) choose notation such that O^{2′}(P_{1}/O_{2}(P_{1})) ≅ L_{2}(r), SU_{3}(r), L_{2}(r^{3}), Sz(r), respectively.

If X ≅ L_{3}(r) or Sp_{4}(r), O_{2}(P_{1}) and O_{2}(P_{2}) are the only maximal elementary abelian subgroups of X. Hence A ≤ O_{2}(P_{1}) or O_{2}(P_{2}).

Let X ≅ U_{4}(r). Let A ≰ O_{2}(P_{1}). We have O_{2}(P_{1})/Z(O_{2}(P_{1})) is elementary abelian and for a ∈ A \ O_{2}(P_{1}), |[O_{2}(P_{1})/Z(O_{2}(P_{1})), a]| = r^{2}. This implies |⟨Z(O_{2}(P_{1})), A⟩| > r^{3}.

Let now A ≰ O_{2}(P_{2}). We have Ω_{1}(O_{2}(P_{2})) is elementary abelian of order r^{4} and O^{2′}(P_{2}/O_{2}(P_{2})) ≅ L_{2}(r^{2}). Furthermore Ω_{1}(O_{2}(P_{2})) is the Ω^{−}(4, r)-module for L_{2}(r^{2}). As A acts quadratically on Ω_{1}(O_{2}(P_{2})), we see |A : A ∩ O_{2}(P_{2})| ≤ r. Now |⟨Z(O_{2}(P_{1})), A⟩ ∩ Ω_{1}(O_{2}(P_{2}))| > r^{2}. But this contradicts (1.27).

Let X ≅ G_{2}(r). Let O_{2}(P_{1})/Z(O_{2}(P_{1})) be elementary abelian of order r^{4}. If a ∈ A \ O_{2}(P_{1}), then |[a, O_{2}(P_{1})/Z(O_{2}(P_{1}))]| = r^{2}. Hence |⟨A, Z(O_{2}(P_{1}))⟩| > r^{3}, contradicting the fact that A contains no elementary abelian subgroup of order greater than r^{3} (see (1.23)).

Let X ≅ ^{3}D_{4}(r). Then V = O_{2}(P_{1})/Z(O_{2}(P_{1})) is the 8-dimensional GF (r)-module for L_{2}(r^{3}). Suppose that A ≰ O_{2}(P_{1}). Now by (1.32) |[a, O_{2}(P_{1})/Z(O_{2}(P_{1}))]| = r^{4}. Then [a, O_{2}(P_{1})] = C I_{O_{2}(P_{1})}(a). Let γ ∈ N(S), o(γ) = r^{2} + r + 1. Then [γ, C I_{V} (S)] = 0 by (1.32). Hence C I_{O_{2}(P_{1})}(γ) is a Sylow 2-subgroup of L_{3}(r). But we may assume there is a^{g} = b ∈ L_{2}(r^{3}) with γ^{b} = γ^{−1}. But there is no automorphism b of a Sylow 2-subgroup T of L_{3}(r) such that [T, b] is elementary abelian of order r^{2}.

Let X ≅ ^{2}F_{4}(r). We have Z_{2}(O_{2}(P_{1})) = Ω_{1}(O_{2}(P_{1})). Now [a, O_{2}(P_{1})] ≤ Z_{2}(O_{2}(P_{1})) for a ∈ A, and so A ≤ O_{2}(P_{1}). Hence A ≤ Z_{2}(O_{2}(P_{1})) ≤ C I(Z(O_{2}(P_{1}))) ≤ O_{2}(P_{2}).

(1.54) Lemma. Let X ≅ U_{5}(q), q even, A ≤ X be an elementary abelian 2–group, |A| > q^{2}. Then A contains some t which induces a transvection on the natural module.

Proof . Let P be the parabolic with Z(O_{2}(P)) = 1. Then |Ω_{1}(O_{2}(P))| = q^{4} and O^{2′}(P/O_{2}(P)) ≅ L_{2}(q^{2}), Ω_{1}(O_{2}(P)) is the natural Ω^{−}(4, q)-module.

Let R = Z(S), S ∈ Syl_{2}(P). We may assume A ≤ P and R ∩ A ≠ 1. Hence |A R| > q^{3}. If a ∈ S \ O_{2}(P), then |C I_{Ω_{1}(O_{2}(P))}(a)| = q^{2}. Hence A R ≤ Ω_{1}(O_{2}(P)). Now A ≤ Ω_{1}(O_{2}(P)). As the conjugates of t correspond to the singular vectors and the largest subgroup of the orthogonal 4-dimensional module containing no singular vector is of order q^{2}, we get the assertion.

(1.55) Lemma. Let E(X) ≅ 3 · U_{4}(3) and u ∈ X be an involution. Then there is an elementary abelian subgroup E, |E| = 27 of E(X) such that u ∈ N(E), |C I_{E}(u)| ≥ 9. If u ∈ C I(Z(E(X))), even [E, u] = 1.

Proof . According to [CCNPW] any involution in Aut (U_{4}(3)) centralizes an elementary abelian subgroup F of order 9 in U_{4}(3). An easy inspection shows that this group has an elementary abelian preimage E.

(1.56) Lemma. Let G = G(q) be a group of Lie type over GF (q), q = 2^{n}, G ≅ L̸_{2}(q), L_{3}(q), U_{3}(q), Sz(q), G_{2}(q) or ^{2}F_{4}(q). Let R be a long root group, Q = O_{2}(N_{G}(R)/R) and L be a Levi complement in N_{G}(R). Then Q has the following L–module structure

(i) G ≅ L_{n}(q), O^{2′}(L) ≅ L_{n−2}(q), Q = V_{1} ⊕V_{2}, V_{1} is the natural L–module and V_{2} its dual.

(ii) G ≅ Sp_{2n}(q), O^{2′}(L) = Sp_{2n−4}(q) × L_{2}(q) = L_{1} × L_{2}, Q = V_{1} ⊕ V_{2}, [V_{2}, L_{1}] = 1, V_{1} is the natural L_{2}–module, V_{1} = V_{1}^{(1)} ⊕ V_{1}^{(2)}, V_{1}^{(i)}, i = 1, 2, are natural L_{1}–modules and [L_{2}, V_{1}] = V_{1}.

(iii) G ≅ Ω_{2n}^{±}(q), O^{2′}(L) = Ω_{2n−4}^{±}(q) × L_{2}(q) = L_{1} × L_{2}, Q = V_{1} ⊕ V_{2}, V_{i}, i = 1, 2, are natural L_{1}–modules and [Q, L_{2}] = Q.

(iv) G ≅ U_{n}(q), O^{2′}(L) = U_{n−2}(q), Q is the natural module.

(v) G ≅ E_{6}(q), O^{2′}(L) ≅ L_{6}(q), Q ≅ V (λ_{3}).

(vi) G ≅ ^{2}E_{6}(q), O^{2′}(L) ≅ U_{6}(q), Q ≅ V (λ_{3}).

(vii) G ≅ E_{7}(q), O^{2′}(L) ≅ Ω_{12}^{+}(q), Q ≅ V (λ_{6}).

(viii) G ≅ E_{8}(q), O^{2′}(L) ≅ E_{7}(q), Q ≅ V (λ_{1}).

(ix) G ≅ F_{4}(q), O^{2′}(L) ≅ Sp_{6}(q), Q is an extension of the natural module by a spin module.

(x) G ≅ ^{3}D_{4}(q), O^{2′}(L) ≅ L_{2}(q^{3}), Q is the 8–dimensional GF (q)–module for L.

Proof . This can easily be checked using the Chevalley commutator formula (see also [AschSe]).

(1.57) Lemma. Let G = G(q) ≅ ^{2̸}F_{4}(q), q = 2^{m}, be a group of Lie type and R be a long root group. Set Q = O_{2}(C I_{G}(R)). Let A ≤ Q be elementary abelian with [A, Q] ≠ 1. Then there is U ≤ Q, |U| = q, with |A : C I_{A}(U)| ≤ q.

Proof . We have R = Q′ is of order q. Hence |[A, Q]| ≤ q. Furthermore Q is generated by subgroups R_{i}, |R_{i}| = q, and R_{i} is a TI–set in Q. Let x ∈ A with [x, r] = 1 for some r ∈ R_{i}, r ≠ 1. Then [x, R_{i}] = 1, as R_{i} is a TI–set. This now implies that R_{1} ∩ C I(A) = 1 for some R_{1} and |A : C I_{A}(R_{1})| = |A : C I_{A}(r)| ≤ q, for r ∈ R_{1}, r ≠ 1.

(1.58) Lemma. Let G = F_{4}(q), q = 2^{m}, R be a root group, A be an elementary abelian subgroup of N_{G}(R). Let Z = Z(O_{2}(N_{G}(R))) and S ∈ Syl_{2}(N_{G}(R)). If A ≰ O_{2}(N_{G}(R)) but A ≤ O_{2}(C I_{N_{G}(R)}(Z(S))), then |Z : C I_{Z}(A)| ≥ |A : A ∩ O_{2}(N_{G}(R))|.

Proof . We have that Z/R is the natural Sp_{6}(q)–module by (1.56). Furthermore by assumption about A we have that there is a subgroup Z_{1} of Z, |Z : Z_{1}| = q, with Z_{1} ≥ [Z, A] and [Z_{1}, A] ≤ Z(S). Now obviously |Z_{1} : C I_{Z_{1}}(A)| ≥ |A : A ∩ O_{2}(N_{G}(R))|.

(1.59) Lemma. Let G be a group of Lie type over GF (q), q = 2^{n}, G ≅ 2F_{4}(q). Let R be a long root group, Q = O_{2}(N_{G}(R)/R). If t is an involution in C I(R)\O_{2}(N_{G}(R)) such that |[Q, t]| ≤ q^{2}, then G ≅ L_{n}(q), U_{n}(q) or Sp_{2n}(q) and |[Q, t]| = q^{2}, or G ≅ G_{2}(q).

Proof . We have that Q is a GF (q)–module (see (1.56)) and so |[Q, t]| = q or q^{2}. In the former t induces a GF (q)–transvection and so by (1.22), (1.23) and (1.24) we get a contradiction to (1.56).

So let |[Q, t]| = q^{2}. If G ≅ Ω_{2n}^{±}(q), then by (1.56) Q is a sum of two modules. On neither of them t can induce a transvection. If G ≅ F_{4}(q), then again by (1.56) there are two modules in Q and t cannot induce a transvection on the spin module by (1.23).

If G ≅ E_{6}(q), E_{7}(q), E_{8}(q), or ^{2}E_{6}(q), then by (1.23) and (1.56) the minimal codimension for centralizers of involutions in Q is greater than 2.

So we are left with G ≅ ^{3}D_{4}(q). But in this case the minimal codimension is four by (1.32).

(1.60) Lemma. Let G = G(2) be a group of Lie type over GF (2).

(a) Assume m_{3}(G) ≥ 4. Let x be a long root element, t ∈ O_{2}(C I_{G}(x)) be an involution, S ∈ Syl_{2}(C I_{G}(x)) and [t, S] ≤ ⟨x⟩. Then 3||C I_{G}(t)|.

(b) Assume m_{3}(G) ≥ 3. If S ∈ Syl_{2}(G) then either G ≅ Sp_{6}(2) or m_{3}(C I_{G}(Z(S))) ≥ 2.

Proof . (a) Set H = C I_{G}(x). If G ≅ L̸_{n}(2), then O_{2}(H)/Z(O_{2}(H)) is an irreducible module for H/O_{2}(H) and so tZ(O_{2}(H)) is centralized by some parabolic P in H/O_{2}(H) by (1.56). Hence 3||P | or H/O_{2}(H) is solvable. The latter just occurs for G ≅ Ω_{8}^{+}(2). But in this group any involution is centralized by a 3–element.

Assume now G ≅ L_{n}(2). Then by (1.56) O_{2}(H)/Z(O_{2}(H)) = H_{1} ⊕ H_{2}, where H_{1} is the natural L_{n−2}(2)–module and H_{2} its dual. So tZ(O_{2}(H)) is centralized by a 3–element as n ≥ 8.

Hence we may assume that tZ(O_{2}(H)) is centralized by a 3–element in H. If ⟨x⟩ = Z(O_{2}(H)) the same applies to t. So assume ⟨x⟩ < Z(O_{2}(H)). This means G ≅ Sp_{2n}(2) or F_{4}(2).

Let G ≅ Sp_{2n}(2) and set H̃ = C I_{H}(Z(O_{2}(H))). Then O_{2}(H)/Z(O_{2}(H)) ≅ H_{1} ⊕ H_{2}, where H_{i}, i = 1, 2, are natural modules for H̃/O_{2}(H) ≅ Sp_{2n−4}(2) by (1.56). Now as 2n − 4 ≥ 4, we have that tZ(O_{2}(H)) is centralized by a 3–element in H̃ and then also t is centralized by a 3–element.

So we hare left with G ≅ F_{4}(2). Now tZ(O_{2}(H)) is centralized by a subgroup U ≅ 2^{6}L_{3}(2) in H/O_{2}(H). As [O_{2}(U), t] ≤ ⟨x⟩, we see that |t^{U}⟨x⟩| = 1 or 7. In both cases t is centralized by a 3–element.

(b) Let H be as in (a). If |Ω_{1}(Z(S))| = 2 this is the same as saying m_{3}(H/O_{2}(H)) ≥ 2. Hence the assumption holds or H/O_{2}(H) ≅ Σ_{3} or L_{3}(2) by (1.3). But as m_{3}(G) ≥ 3, an easy inspection shows that this is not possible. So we have |Ω_{1}(Z(S))| = 4 and G ≅ Sp_{2n}(2) or F_{4}(2). In the latter Ω_{1}(Z(S)) is centralized by Sp_{4}(2) and in the former by Sp_{2n−4}(2) (see (1.56)). Hence m_{3}(C I(Z(S))) ≥ 2 or G ≅ Sp_{6}(2).

(1.61) Lemma. Let G ≅ U_{n}(q), Ω_{2n}^{−}(q) or ^{2}E_{6}(q), q = 2^{n}. Suppose no parabolic subgroup contains an elementary abelian subgroup of order 27. Then G ≅ U_{n}(q), n ≤ 7, or Ω_{8}^{−}(q).

Proof . In ^{2}E_{6}(q) we have a parabolic which involves L_{2}(q) × L_{3}(q^{2}) and so it has 3–rank at least 3.

In Ω_{2n}^{−}(q) we have a parabolic which involves L_{2}(q) × Ω_{2n−4}^{−}(q). As m_{3}(Ω_{6}^{−}(q)) ≥ 2, we have the assertion (remember Ω_{6}^{−}(q) ≅ U_{4}(q)).

In U_{n}(q) we have a parabolic which involves L_{2}(q) × U_{n−4}(q). As m_{3}(U_{4}(q)) ≥ 2, we have the assertion.

(1.62) Lemma. Let G be a group with F^{∗}(G) = O_{2}(G) ≠ 1 and A ≤ G, A elementary abelian, A ≰ O_{2}(G), but A ≤ S ∈ Syl_{2}(G). Then there is some g ∈ G such that for X = ⟨A, A^{g}⟩ the following hold

(1) X/O_{2}(X) ≅ D_{2u}, u odd, L_{2}(q), Sz(q) or P Sp(4, q), q even.

(2) Y = (A ∩ O_{2}(X))(A^{g} ∩ O_{2}(X)) ≤ X

(3) Y ≠ A ∩ O_{2}(X)

(4) |A : C I_{A}(Y/C I_{Y} (A))| ≤ |Y : C I_{Y} (A)|^{2},

or there is some B ≤ S, B elementary abelian, with [B, A] ≠ 1 and |A : C I_{A}(B)| ≤ |B : C I_{B}(A)|.

Proof . Set Ḡ = G/O_{2}(G). Let [O_{r}(Ḡ), Ā] ≠ 1 for some odd prime r. Then we get some g ∈ O_{r}(Ḡ) with X/O_{2}(X) ≅ D_{2r}, X = ⟨A, A^{g}⟩.

Let now L be some component of Ḡ with 1 ≠ [L, Ā] ≤ L. If |Ā : C IĀ(L)| = 2 we again get some g ∈ L with X/O_{2}(X) ≅ D_{2u}, u odd. As O_{2}(X) ≤ AO_{2}(G) ≤ S in both cases we see ⟨A ∩ O_{2}(X), A^{g} ∩ O_{2}(X)⟩ = Y ≤ X. As [X, O_{2}(G)] ≤ Y and C I_{G}(O_{2}(G)) ≤ O_{2}(G), we see A ∩ O_{2}(G) ≠ Y. We have A ∩ A^{g} = C I_{A}(Y) and A ∩ Y = C I_{Y} (A). Now

|A : C I_{A}(Y)| = 2|A ∩ Y : A ∩ A^{g}| ≤ |A^{g} ∩ Y : A ∩ A^{g}||A ∩ Y : A ∩ A^{g}|

= |Y : A ∩ A^{g}| = |Y : Y ∩ A||A ∩ Y : A ∩ A^{g}| = |Y : A ∩ Y |^{2}.

We have that A acts quadratically on O_{2}(G). From now on assume 1 ≠ [Ā, L] ≤ L, but |Ā : C IĀ(L)| ≥ 4. If L is a group of Lie type in odd characteristic which is not a group of Lie type in even characteristic too. Then by (1.12) L/Z(L) ≅ U_{4}(3). Let i be a 2–central involution in S̄ ∩ L. If Ā ≰ O_{2}(C I_{L}(i)) we get for suitable g ∈ C I_{L}(i) that X/O_{2}(X) ≅ Σ_{3}, where X = ⟨A, A^{g}⟩. Now the assertion follows as above. So let Ā ≤ O_{2}(C I_{L}(i)). As C I_{L}(i) does not contain an elementary abelian normal subgroup of order at least four, we get some g ∈ C I_{L}(i) with Ā ≠ Ā^{g}. Now 1 ≠ [A, A^{g}] ≤ A ∩ A^{g}. Hence the second alternative holds.

Let now L ≅ A_{n}, n ≥ 5. Let m_{1}, …, m_{r} be the dyadic decomposition of n. Then S ∩A_{n} is in Σ_{m_{1}} ×…×Σ_{m_{r}}. We may assume m_{i} ≥ 4 for all i. If m_{i} ≥ 8 for all i, we have induction. So we may assume m_{r} = 4. Then we may suppose Ā/C IĀ(L) ≤ A_{4} ≤ A_{5}. Hence for g ∈ L either ⟨A, A^{g}⟩/O_{2}(⟨A, A^{g}⟩) ≅ D_{2u}, u odd, or A_{5}. In the first case we get the assertion. If X/O_{2}(X) ≅ A_{5}, we get |Y : A ∩ A^{g}| ≥ 16, as A_{5} has to act nontrivially on Y/A ∩ A^{g}. But now again |A : C I_{A}(Y)| ≤ |Y : C I_{Y} (A)|^{2}.

Assume that L is sporadic. With (1.12) we get L/Z(L) ≅ M_{12}, M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2}, or Suz. Again choose i ∈ Z(S ∩L)^{♯}. If L∩Ā ≰ O_{2}(C I_{L}(i)), then we have induction. So assume L ∩ Ā ≤ O_{2}(C I_{L}(i)). If there is some g ∈ C I_{L}(i) with [Ā, Ā^{g}] ≠ 1, then as before we get the second alternative of the statement. So assume that ⟨Ā^{C I_{L}(i)}⟩ is abelian. Then we have L/Z(L) ≅ M_{12}, M_{22}, or M_{24}. In the first case Ā ∩ L ≤ P ΓL_{2}(9) and we have the assertion by induction. If L ≅ M_{24}, we get Ā ∩ L ≤ M_{23}, a contradiction to (1.12). If L/Z(L) ≅ M_{22}, then Ā ∩ L ≤ (S)L_{3}(4), a case we will handle next.

Let now L ≅ G(r) be a group of Lie type in even characteristic. Let R be a root group in Z(S̄ ∩ L). If Ā ≰ O_{2}(N_{L}(R)), we have induction again. Hence assume Ā ≤ O_{2}(N_{L}(R)). Furthermore we may assume that ⟨A^{N_{L}(R)}⟩ is abelian. If Ā ≤ R, then Ā ≤ L̃ ≤ L, where L̃ ≅ L_{2}(r) or Sz(r), and we have induction again. So we are left with L ≅ L_{n}(r), Sp_{2n}(r), F_{4}(r), ^{2}F_{4}(r), Sz(r), or U_{3}(r).

Let L ≅ L_{2}(r). Then we may assume X/O_{2}(X) ≅ L_{2}(r). So |A : C I_{A}(Y)| ≤ |Y : C I_{Y} (A)|^{2}, as |Y : A ∩ A^{g}| ≥ r^{2}.

Let L ≅ L_{n}(r), n ≥ 3. Then we have two parabolics P_{1} and P_{2} which involve L_{n−1}(r). If Ā ≰ O_{2}(P_{i}), for some i, we have induction again. This shows Ā ≤ O_{2}(P_{1}) ∩ O_{2}(P_{2}) = R. Now Ā ≤ L_{2}(r) and we are done.

Let L ≅ Sp_{2n}(r). We have Ā∩L ≤ Z(O_{2}(N_{L}(R))). Hence Ā∩L ≤ Sp_{4}(r). So it is enough to prove the assertion for L ≅ Sp_{4}(r). Then we have two parabolics P_{1} and P_{2}, which involve L_{2}(r). If Ā ≰ O_{2}(P_{1}) ∩ O_{2}(P_{2}), we have induction. So assume Ā ≤ O_{2}(P_{1}) ∩ O_{2}(P_{2}). Then either X/O_{2}(X) ≅ L_{2}(r) or P Sp_{4}(r). In the latter |Y : A∩A^{g}| ≥ r^{4}. As |A : A∩Y | ≤ r^{2}, we again get |A : C I_{A}(Y)| ≤ |Y : C I_{Y} (A)|^{2}. In the former we get the assertion by induction.

Let now L ≅ F_{4}(r). We have two root subgroups R_{1} and R_{2} and Ā ∩ L ≤ Z(O_{2}(N_{L}(R_{1}))) ∩ Z(O_{2}(N_{L}(R_{2}))) = Z(S̄ ∩ L). But this group is contained in some P Sp_{4}(r), and we may argue by induction.

Let L ≅^{2}F_{4}(r). As A acts quadratically, we get with (1.14) that Ā∩L ≤ R and so X/O_{2}(X) ≤ L_{2}(r), and induction applies. The same is true for L ≅ U_{3}(r).

Let L ≅ Sz(r). Then |Y : A∩A^{g}| ≥ r^{4} and so |A : C I_{A}(Y)| ≤ |Y : C I_{Y} (A)|^{2}. Let now finally [A, L] ≰ L. As Ā ≤ S̄, we have that L possesses abelian

Sylow 2–subgroups. Furthermore we have |A : C I_{A}(L)| ≥ 4. By (1.11) we get L ≅ L_{2}(r), r even. Now N(S̄ ∩ L)/S̄ ∩ L ≅ ZZ_{r−1} × ZZ_{r−1} and so some ā ∈ Ā^{♯} acts nontrivially on this group. Hence g can be chosen in such a way that X/O_{2}(X) is dihedral.

(1.63) Lemma. Let G be in the uniqueness case. Let R ≤ G, E ≤ R ∩ M, E elementary abelian p-subgroup, M not exceptional. Suppose one of the following holds:

(a) p ∈ σ(M), |E| ≥ p^{2} and Γ_{E,1}(G) ≤ M

(b) p ∉ σ(M), F^{∗}(M) = O_{2}(M), |E| = p^{3}, Γ_{E,1}(G) ≤ M and there is H ⊴ M such that EO_{2}(H) ◁ H, H/O_{2,p}(H) ≅ L_{2}(p), (H/O_{2,p}(H))′ = H/O_{2,p}(H).

Then R ≤ M or (a) holds and E(R/O_{p′}(R)) = L is simple and we have one of the following (P ∈ Syl_{p}(R ∩ M)):

(i) L ≅ L_{2}(p^{n}), n > 1, U_{3}(p^{n}) or ^{2}G_{2}(3^{n}), n > 1; M ∩ L = N_{L}(P ∩ L)

(ii) p = 3 , L ≅ L_{2}(8) , L_{3}(4), M_{11}, A_{6}; M ∩ L = N_{L}(P ∩ L)

(iii) p = 5 , L ≅ Sz(32), McL, ^{2}F_{4}(2)′; M ∩ L = N_{L}(P ∩ L)

(iv) p = 5 or 7 and L ≅ HS, Ru, He, O′N, or M(24)′ ; m_{p}(L) = 2 and Γ_{P ∩L,1}(G) ≰ M

(v) p = 5 , L ≅ A_{10}

(vi) p = 11 , L ≅ J_{4}; M ∩ L = N_{L}(P ∩ L).

(vii) p = 5, L ≅ M(22) and E(M) = F^{∗}(M).

In any case O_{p′}(R) = O_{2′}(R) and O_{2}(R ∩ M) = 1 or L ≅ L_{3}(4) and some graph × field automorphism is involved.

Proof . Let R ≰ M. Set K = O_{p′}(R). As K = ⟨C I_{K}(x)|x ∈ E^{♯}⟩ we get K ≤ M. Set U = O_{p′,p}(R). Let P ∈ Syl_{p}(R), E ≤ P and 1 ≠ x ∈ Z(P) ∩ U. Then [x, E] = 1 and so x ∈ M. If x ∈ E, then P ≤ M. Assume x ∉ E. Then as m_{p}(M) = 3 in case (b) we have (a). Now m_{p}(⟨x, E⟩) ≥ 3 and so P ≤ N_{G}(⟨x⟩) ≤ M.

So U ≤ M in any case. Suppose first U = K. Let now W be the preimage of E(R/U). Then W > K and p | |W |. We first show that W/K is simple. Let E ≤ P ∈ Syl_{p}(R), then as before P ≤ M. Let W_{1}/K · W_{2}/K ⋯ W_{r}/K = W/K. Suppose there is ω ∈ E with (W_{1}/K)^{ω} ≠ W_{1}/K. Then we see that r ≥ p. As p ≥ 3, we get in case (a) that W_{2}/K ⋯ W_{r}/K ≤ N_{W/K}((P ∩ W_{1})K/K) ≤ M and W_{1}/K ⋯ W_{r−1}/K ≤ N_{W/K}((P ∩ W_{r})K/K) ≤ M.

Hence W ≤ M and as R = W N_{R}(W ∩ P) we see R ≤ M. Suppose (b), then r = 3. Now |N_{E}(W_{1}/K)| = 3^{2}. As m_{3}(M) = 3, we get N_{E}(W_{1}/K) ≤ W. But now C I_{Ω_{1}(P ∩W)}(ω) ≥ N_{E}(W_{1}/K) which contradicts (W_{1}/K)^{ω} = W_{2}/K, (W_{2}/K)^{ω} = W_{3}/K.

So we have E ≤ N(W_{1}/K). There is x ∈ Z(P ∩ W_{1}) with N_{G}(⟨x⟩) ≤ M, so W_{2}W_{3} ⋯ W_{r} ≤ M. But the same is true for W_{r}, i.e. W_{1}W_{2} ⋯ W_{r−1} ≤ M. Hence we have W ≤ M. If we are in case (a), we get R ≤ M. So assume (b). As m_{p}(M) = 3, we get r ≤ 3. We have E ∩ W_{i} ≠ 1, i = 1, …, r. So at least one W_{i} has a cyclic Sylow p-subgroup, say W_{1}. Now if W_{1} ◁ R, we get R = W N_{R}(E ∩ W_{1}) ≤ M. Hence we have that R/N_{R}(W_{1}) ≲ Σ_{3}, and N_{R}(W_{1}) ≤ M. Furthermore R = N_{R}(W_{1}) N_{R}(E ∩ W). Where N_{R}(E ∩ W) = C I_{R}(E ∩ W)⟨x|C I_{E∩W} (x) ≠ 1⟩ ≤ M.

So we have that W/K is simple. Now we may apply [GoLy, (24-9)] if p ∈ σ(M), [GoLy, (24-12)] in case (b). We get a list of possibilities for W/K.

Suppose first case (a): As W ≰ Γ_{P,2}(G) we get W/K ≅ P̸ Sp_{4}(p), L_{3}(p) or A_{3p}. Let W/K not be as in (i). Suppose furthermore p = 3. Then Γ_{P,1}(G) ≤ M. Now we get W/K ≅ L_{2}(8), L_{3}(4) or M_{11}. In any case M ∩ W/K = N_{M∩W} (P)K/K.

So let p ≥ 5. If W/K ≅ M(22), then M ∩ R involves D_{4}(2). If F^{∗}(M) = O_{2}(M) then 3 ∈ σ(M) and so m_{3}(M ∩ R) ≥ 4. But then R ≤ M.

Let W/K ≅ ^{2}F_{4}(32). Then m_{p}(P) ≥ 3 and so also W satisfies the assumption for some Ẽ ≤ P ∩ W. But this contradicts [GoLy, (24-9)].

Let W/K ≅ A_{2p}, then A_{p} × A_{p} is in M ∩ R. If p > 5, then A_{p} contains a 2–group which is normalized by an elementary abelian group of order 9, a contradiction as before.

So assume (b). As |E| = p^{3} and m_{p}(P) = 3, we get W/K ≅ A_{3p},L_{2}(p^{3}), G_{2}(p), P Sp_{4}(p) or p = 3 and W/K ≅ Sp_{6}(2), A_{10}, A_{11}, J_{3}, Ly, F_{3}, F_{5}, Co_{1} (p = 5), or F_{1} (p = 7).

As E ∩ W ◁ M ∩ W/O_{2}(M ∩ W) we see by inspection of [CCNPW] that W/K ≅ L_{2}(p^{3}). Now there is some H ≤ M ∩W with |H/H ∩K| = (p^{3}−1)/2 and H acts irreducibly on E. Hence M induces L_{3}(p) on E, a contradiction.

So we just have to show that U = K. Suppose P ∩ U ≠ 1. Then R = UN_{R}(Z(U ∩ P)). So in case (a) we have N_{R}(Z(U ∩ P)) ≤ M. Thus we are in (b). If [E, Ω_{1}(Z(U ∩ P))] ≠ 1, then there is some ω ∈ Z(U ∩ P) inducing a transvection on E. But now M induces L_{3}(p) on E, a contradiction. So Ω_{1}(Z(U ∩ P)) ≤ E. So all we have to show is:

(∗) Let 1 ≠ F ≤ E, then N_{G}(F) ≤ M .

As EO_{2}(M) ◁ M, we get some Ẽ, |Ẽ| = r^{2}, r ∈ σ(M) with Γ_{E,1}(G) ≤ M and Ẽ ≤ N_{G}(F). Now the just proved part (a) tells us that either N_{G}(F) ≤ M or O_{r′}(N_{G}(F)) ≤ M and let X be a preimage of E(N_{G}(F)/O_{p′}(N_{G}(F))) then N_{G}(F) = N_{M}(F)X. As Γ_{F,1}(G) ≤ M, we see that C I_{X}(F) ≤ O_{p′}(N_{G}(F)). This gives X/O_{p′}(N_{G}(F)) ≤ L_{3}(p). But none of the groups in the assertion has this property. So we have N_{G}(F) ≤ M.

In case (a) we just have to prove the additional assertion about M ∩ R. Let 2 | |K|. Then let T ∈ Syl_{2}(K). So R = K N_{R}(T). As we may assume

E ≤ N_{R}(T), we get N_{R}(T) ≤ M, a contradiction. So 2 | |̸K|.

We see that O_{2}(N_{L}(P ∩ L)) = 1. So assume there is some ω ∈ Aut (L) with [ω, P ∩ L] = 1, o(ω) = 2. Then we just have L ≅ L_{3}(4).

(1.64) Lemma. Let X = G(q) be a Lie group, q even, q > 2. Let r = 2^{n} and x be a primitive prime divisor of r − 1, x = 9 in case r = 64. Suppose r > q. Let ω ∈ Aut(X), o(ω) = x, ω normalizes a Borel subgroup B of X. Then one of the following holds

(i) r = q^{2} and X ≅ U_{n}(q), Ω_{2n}^{−}(q) or ^{2}E_{6}(q).

(ii) r = q^{3} or r^{2} = q^{3} and X ≅ ^{3}D_{4}(q).

(iii) x = 3 or 9 and X ≅ ^{3}D_{4}(q), D_{4}(q), q ≤ 32.

Proof . Suppose first that ω induces a graph automorphism on G(q). Then X ≅ ^{3}D_{4}(q) or D_{4}(q) and x = 3 or 9. If x = 3, we get r = 4 and so by assumption q = 2, a contradiction. Let x = 9, then r = 64. So q ≤ 32 and (iii) holds.

So assume now that ω induces a field automorphism. Then q = t^{x} or x = 9 and q = t^{3}. Suppose the former. As x | t^{x−1} − 1 < q − 1, we see that r ≤ q, which contradicts the choice of x. So we have q = t^{3} and x = 9. Hence r = 64 and so q = 8. This shows r = q^{2}. Now ω^{3} induces an inner × diagonal automorphism. As 3 | q̸ − 1, we see that X is a 2-fold twisted group and so we have (i).

So assume finally that ω induces an inner × diagonal automorphism. Let x = p, prime. Then x | q̸ − 1. This now shows x | q^{2} − 1 or X ≅ ^{3}D_{4}(q) and

x | q^{3} − 1.

In the former by the choice of x we have r = q^{2} and this is (i). In the latter we get r = q^{3} or r^{2} = q^{3} and so we have (ii).

So we are left with x = 9, r = 64, q ≤ 32. As the torus of B is an abelian group, we see that 9 | q − 1, 9 | q^{2} − 1 or 9 | q^{3} − 1. This gives q = 8, q = 4 and X ≅ ^{3}D_{4}(q), or q = 16 and X ≅ ^{3}D_{4}(q). Hence we have (i) or (ii).

(1.65) Lemma. Let X = G(q) be a Lie group, q even. Let r = 2^{n} and x be a primitive prime divisor of r − 1, x = 9 for r = 64. Suppose r > q. Let ω ∈ Aut(X), o(ω) = x. Let S ∈ Syl_{2}(X). Suppose that ⟨S, ω⟩ is a {2, x}-group. Then one of the following holds:

(i) q > 2, ω normalizes a Borel subgroup of X

(ii) q = 2, o(ω) = 3 or 9 and ω ∈ X if ⟨S, ω⟩ = S⟨ω⟩, S ◁̸⟨S, ω⟩

(iii) X ≅ U_{n}(2), Ω_{2n}^{−}(2), ^{2}E_{6}(2), ^{3}D_{4}(2), o(ω) = 3 or 9 and ω normalizes a Borel subgroup

(iv) X ≅ ^{2}F_{4}(2)′, o(ω) = 5, ω ∈ X, if ⟨S, ω⟩ = S⟨ω⟩, S ◁̸ S⟨ω⟩.

(v) X ≅ ^{3}D_{4}(2), o(ω) = 7, ω normalizes a Borel subgroup

(vi) X ≅ D_{4}(2), o(ω) = 3 or 9, ω induces a diagram automorphism on X.

Proof . As ⟨S, ω⟩ is a {2, x}-group we either have that ω induces an automorphism which normalizes a Borel subgroup or X possesses a solvable parabolic.

Suppose the former. Then if q > 2 we have (i). Let q = 2. Then ω induces a graph or diagonal automorphism. In the first case we have (vi). So assume that ω induces a diagonal automorphism. Then X ≅ U_{n}(2), Ω_{2n}^{−}(2), 2E_{6}(2) or ^{3}D_{4}(2). This is (iii) or (v).

So assume that ω does not normalize a Borel subgroup of X. We then have a solvable minimal parabolic P. This first shows q = 2. Furthermore ⟨S, ω⟩∩ X ≥ P. We have P/O_{2}(P) ≅ Σ_{3}, SU_{3}(2) or F_{20}. Which shows x = 3 or 9, or X ≅ ^{2}F_{4}(2) and o(ω) = 5.

Suppose now that we have ⟨S, ω⟩ = S⟨ω⟩. Then ω induces an inner automorphism. As S ◁̸ S⟨ω⟩, we see that ω ∈ ⟨S, ω⟩′ ≤ X.

(1.66) Lemma. Let X be one of the groups of (1.7) or (1.8). Let M be the maximal subgroup of X containing S. Suppose there is some ω ∈ M, o(ω) = p, p > 2, a prime such that N_{X}(⟨ω⟩) ≤ M, then X is one of the following: L_{2}(r), L_{3}(r), U_{3}(r), r odd, P Sp_{4}(r), L_{2}(r) × L_{2}(r), r prime, (S)L_{3}(q), Sp_{4}(q), q even , or F (X) is elementary abelian of order r^{3} and E(X/F (X)) ≅ L_{2}(r), L_{2}(r) ≅ A̸_{5} or X is solvable and (1.8) (i) or (iv) holds. If E(X) ≅(S)L_{3}(q) or Sp_{4}(q), q even, then there is E ≤ M, E elementary abelian of order p^{2} such that N_{X}(E) ≰ M.

Proof . Suppose first that E(X) is as in (1.7) (i). Then M ∩ E(X) = N_{E(X)}(S ∩ E(X)). In any case there is some x ∈ E(X) with ω^{x} = ω^{−1}, x ∉ M.

Suppose now that E(X) is as in (1.7) (iii). Then we have E(X) ≅ A_{9}. Now obviously p = 5 or 7 as M ≅ A_{8}. But in both cases N_{A_{9}}(⟨ω⟩) ≰ A_{8}.

So we have F (X) ≠ 1. Then X is solvable. By (1.8) we have (1.8) (i) or (iv).

(1.67) Lemma. Let X be quasisimple 1 ≤ m_{p}(X) ≤ 2 for some odd prime p, m_{r}(X) ≤ 3 for all odd prime. Suppose X admits an outer automorphism ω of order p such that m_{p}(X⟨ω⟩) > m_{p}(X), then X ≅ L_{2}(r^{3}), SL_{3}(r), SU_{3}(r), 3 | |Z(X)|, r prime and p = 3, or X ≅ G(q), a Lie group over a field of characteristic 2.

Proof . As no sporadic or alternating group does admit an outer automophism of order p, we have that X is a Lie group, we may assume X = G(q), q odd.

Let q = r^{t}, r prime. Now m_{r}(X) ≤ 3 ≥ m_{3}(X). This shows X/Z(X) ∈ {L_{2}(r), L_{2}(r^{2}), L_{2}(r^{3}), L_{3}(r), U_{3}(r), Sp_{4}(r), G_{2}(r)}.

As X has to admit an outer automorphism of order p we get X/Z(X) = L_{2}(r^{3}), p = 3, L_{3}(r), p = 3, p | r − 1, U_{3}(r), p = 3, p | r + 1. Let p | |̸Z(X)|. In case L_{3}(r) and U_{3}(r), we have m_{3}(X) = 2. Furthermore ω is a diagonal automorphism, i.e. C I_{X}(ω) ≅ SL_{2}(r). So m_{3}(C I_{X}(ω)) = 1, hence m_{3}(X⟨ω⟩) = 2, a contradiction. So we have p | |Z(X)|.

(1.68) Lemma. Let G be in the uniqueness case. There is no normal subgroup M_{1} of M with |M : M_{1}| = p, p ∈ σ(M).

Proof . Let P ∈ Syl_{p}(M). By Alperin’s lemma [Go, chapter 7, Theorem (2.6)] we have P ∩ G′ = ⟨N_{G}(X)′|X ≤ P, X = O_{p}(N_{G}(X))⟩. Now choose X ≤ P, X = O_{p}(N_{G}(X)). As m_{p}(P) ≥ 2, we see that m_{p}(X) ≥ 2. Hence N_{G}(X) ≤ M. This shows P ∩ G′ ≤ P ∩ M′ ≤ M_{1}, contradicting P ≤ G′, as G is simple.

§2 The group G_{2}

For the proof of the main theorem we need two groups G_{1} and G_{2} on which we are going to apply the amalgam method. In this chapter we will to give the construction of the group G_{2}. This group is dependent of the uniqueness group M and a Sylow 2-subgroup S of M.

We fix some uniqueness group M. All we assume is that 3 ∈ σ(M) if 3 ∈ σ(G). Furthermore we fix some Sylow 2-subgroup S of M.

Let

N_{1} = {Ñ | Ñ is a maximal 2-local subgroup of G, N_{G}(S) ≤ Ñ, Ñ ≰ M}

and

N = {N | S ≤ N, F^{∗}(N) = O_{2}(N), N ≤ Ñ ∈ N_{1}, N ≰ M}

and

N_{min} = {N | N ∈ N , S is in exactly one maximal subgroup of N}

By assumption N_{min} ≠ ∅. Now choose G_{2} ∈ N_{min} such that the maximal subgroup containing S is contained in M. If possible choose G_{2} to be solvable.

(2.1) Lemma. (a) If σ(G_{2}) = ∅, then G_{2}/O_{2}(G_{2}) is as in (1.7) and (1.8)

(b) If σ(G_{2}) ≠ ∅, then 3 ∉ σ(G_{2}) and either G_{2}/O_{2}(G_{2}) = ES, E extraspecial p - group or elementary abelian, or G_{2}/O_{2}(G_{2}) = E(G_{2}/O_{2}(G_{2}))S, E(G_{2}/O_{2}(G_{2})) ≅ L_{2}(r) × L_{2}(r), or L_{2}(r), r odd.

(c) If G_{2} is one of the groups in (b), then G_{2} does not possess an F -module V with O_{2}(G_{2}/C I_{G_{2}}(V)) = 1.

Proof. (a) If σ(G_{2}) = ∅, then m_{p}(G_{2}) ≤ 3 for any odd prime p. So the assertion follows with (1.7) and (1.8).

(b) Let p ∈ σ(G_{2}). Then by the choice of G_{2} and M, we have p ≠ 3. In particular m_{3}(G_{2}) ≤ 3. Let X = G_{2}/O_{2}(G_{2}). Assume first E(X) = 1. Then C I_{X}(F (X)) ⊆ F (X). Let R ∈ Syl_{r}(F (X)). Suppose m_{r}(R) ≤ 3. Then as in (1.7) we get a contradiction. So we have m_{r}(R) > 3, r ∈ σ(G_{2}). Now R ≰ M, and so X = RS, in particular X is solvable.

Let C be a critical subgroup of R and C_{1} = Ω_{1}(C). Suppose first m_{r}(C_{1}) ≥ 3. Then C_{1}S = X, hence m_{r}(C_{1}) ≥ 4. Let D be an elementary abelian subgroup of order r^{2}, D ≤ X. Then D ≤ M. So C I_{C_{1}}(D) = D, but this is impossible. So we have C_{1} is elementary abelian or extraspecial.

So suppose now m_{r}(C_{1}) ≤ 2. Then C_{1}S ≤ M. But Γ_{C_{1},1}(G) ≤ Ñ, where N ≤ Ñ, Ñ a uniqueness group for the prime r, a contradiction.

If O_{2}(M) = F^{∗}(M) we get a contradiction by definition of the uniqueness group. So assume E(M) = K is quasisimple. Application of (1.63) shows M ⊆ Ñ. Now another application of (1.63) gives Ñ ⊆ M and so G_{2} ⊆ Ñ = M, a contradiction.

So assume now E(X) ≠ 1. Then X = E(X)S. Suppose that E(X) contains at least four components. Then these components are of type Sz(q). But then N_{X}(E(X)∩S) is covered by (G_{2} ∩M)/O_{2}(G_{2}) and m_{r}(N_{X}(E(X)∩ S)) ≥ 4 for some odd r, a contradiction.

So we have that E(X) contains at most two components. Suppose that E(X) is quasisimple. Then m_{3}(E(X)) ≤ 3 and so (1.3) applies. As m_{p}(E(X)) ≥ 4, we get by (1.4) that E(X)/Z(E(X)) is not sporadic or alternating. If E(X)/Z(E(X)) is a Lie group in characteristic two we have by the minimal choice of G_{2} that the Lie rank is at most two. But then m_{p}(E(X)) ≤ 3. So E(X)/Z(E(X)) is a Lie group in odd characteristic. As M ∩ G_{2} cannot contain an E ≅ E_{p^{2}} with m_{p}(C I_{G_{2}}(E)) ≥ 3, we see that E(X) ≅ L_{2}(p^{n}).

So assume E(X) = E_{1} ∗ E_{2}. Then m_{3}(E_{1}) = 1 if 3 |̸ |Z(E_{1}) | and m_{3}(E_{1}) = 2 if 3 || Z(E_{1}) |. Again with (1.3) and (1.4) we get that E_{1} ≅ L_{2}(r), (S)L_{3}(r) or (S)U_{3}(r). Suppose r to be odd. In the last two cases for r ∈ σ(G_{2}) we get SL_{2}(r)∗SL_{2}(r) is involved in G_{2}∩M. But this is impossible as before. So E(X) ≅ L_{2}(r) × L_{2}(r), r odd. If r is even, then E_{1} ≅ (S)L_{3}(r) or (S)U_{3}(r) and for p ∈ σ(G_{2}) we get p | r − 1 or r + 1, respectively. But again G_{2} ∩ M contains E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ Ñ, where Ñ is a uniqueness group containing G_{2}, a contradiction as before.

(c) If G_{2} is nonsolvable the assertion follows with (1.18). So let G_{2} be solvable. Let A be some offending subgroup on V. By [GoLyS, (24.1)] G_{2}/C I_{G_{2}}(V) contains some U ≅ D_{2p} × … D_{2p}, | U |= 2^{n}p^{n}, where | A |= 2^{n}. Now | V : C I_{V} (A) |≤| A | and so | V : C I_{V} (U) |≤ 2^{2n}. But as p > 3 and GL_{2n}(2) does not contain an elementary abelian subgroup of order p^{n}, p > 3, we have a contradiction.

(2.2) Lemma. Let E(L_{2}/Q_{2}) ≅ L_{2}(r) × L_{2}(r), Sz(r) × Sz(r), Sp_{4}(r), (S)L_{3}(r), r even, or SU_{3}(8) ∗ SU_{3}(8), SL_{3}(4) ∗ SL_{3}(4). Let ℤ_{p} × ℤ_{p} ≅ E ≤ L_{2} ∩ M. Then p ∉ σ(M).

Proof. The assertion follows from the fact that L_{2} = ⟨L_{2} ∩ M, N_{L_{2}}(E)⟩.

§3 The nonconstrained case

The purpose of this chapter is to prove that a uniqueness group M has to satisfy F^{∗}(M) = O_{2}(M). For this we assume that F^{∗}(M) = KO_{2}(M),where O_{2}(M) might be trivial.

(3.1) Lemma. If 3 ∈ σ(G), then 3 ∈ σ(M).

Proof. Let M̃ be a uniqueness group for the prime 3 such that M̃ ∩ M contains a Sylow 3-subgroup of M. By (1.3) m_{3}(M) > 1. Now application of (1.63) gives the assertion.

Next we choose a long root group R in K/Z(K). Let R̃ be the preimage in K and G_{1} = N_{M}(R̃). Then G_{1} ∩ K is 2-constrained. (in case of K ≅ F_{4}(q) extended by a field automorphism we choose G_{1} such that G_{1} ∩ K is the parabolic with Sp_{4}(q) on top.)

(3.2) Lemma. There ist E ⊆ G_{1}, E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M.

Proof. Let first K ≅ G(q), q > 2. Then application of (1.2) shows that we have the assertion or K/Z(K) ≅ L_{4}(q), Sp_{6}(q), U_{n}(q), n ≤ 7, Ω_{8}^{−}(q). If m_{p}(K) ≤ 2, then there is a diagonal automorphism of K of order p contradicting (1.61).

So in case K/Z(K) ≅ L_{4}(q), Sp_{6}(q), U_{4}(q), Ω_{8}^{−}(q), we have m_{p}(K) ≥ 3. Hence p | q − 1, q^{2} − 1, q + 1, q^{2} − 1 respectively, and we see again that the assertion holds. We are left with K/Z(K) ≅ U_{n}(q), 5 ≤ n ≤ 7, and p = 5, 3, 7, respectively. Now in any case p | q + 1 and then again the assertion holds.

Let now K = G(2) be defined over GF (2). If 3 ∉ σ(M) then by (3.1) K/Z(K) ≅ L_{n}(2), 4 ≤ n ≤ 7, Sp_{6}(2), U_{4}(2), Ω_{8}^{−}(2), G_{2}(2). But m_{p}(K) ≥ 3, and so we get a contradiction. Hence 3 ∈ σ(G). Now it is enough to find Z_{3} × Z_{3} in G_{1} ∩ K. Inspection shows that we may assume K ≅ L_{4}(2), L_{5}(2), U_{4}(2), G_{2}(2). But now m_{3}(K) ≤ 3 and so m_{3}(C I(K)) = 1. Hence it is enough to have 3 | | G_{1} ∩ K | which is the case.

Now we choose G_{2} as in chapter 2 as a 2−local with S ⊆ G_{2}, where S ∈ Syl_{2}(G_{1} ∩ K) and G_{2} is minimal with respect to S ⊆ G_{2} and G_{2} ⊈ M.

(3.3) Lemma. O_{2}(⟨G_{1}, G_{2}⟩) = 1

Proof. This follows from (3.2) and the definition of the uniqueness case.

Set Z_{2} = ⟨Ω_{1}(Z(S))^{G_{2}}⟩. We have Z_{2} ⊆ Q_{2} = O_{2}(G_{2}). Set Q_{1} = O_{2}(G_{1}).

(3.4) Lemma. [Z_{2}, O_{2}(G_{1} ∩ K)] ≠ 1.

Proof. Suppose false

(α) C I_{G}(Z_{2}) ⊆ M.

Suppose C I_{G}(Z_{2}) ⊈ M. As m_{p}(K) ≥ 2, we get Z_{2} ∩ C I(K) = 1. In particular O_{2}(M) = 1. Now (3.2) shows that K = G(q) and p | q − 1. Furthermore m_{p}(G_{1}) = 2. Hence m_{p}(K) ≥ 4. Application of (1.2) shows K ≅ L_{4}(q), Sp_{6}(q), U_{n}(q), 5 ≤ n ≤ 7, or Ω^{−}(8, q). As m_{p}(K) ≥ 4 and p | q − 1, we get a contradiction.

(β) If P ∈ Syl_{p}(C I_{M}(Z_{2})), then N_{G}(P) ⊈ M, in particular

m_{p}(P) ≤ 1.

Otherwise N_{G}(Z_{2}) = C I_{G}(Z_{2})N_{G}(P) ⊆ M by (α) and the Frattiniargument.

Let Z(O_{2}(G_{1} ∩ K))/Z(K) be a root group R. Then by (β) we have m_{p}(G_{1}) = 2 and p | q − 1. Application of (1.2) and (1.61) yield K ≅ Ω_{8}^{−}(q). But then C I_{K}(R) involves L_{2}(q^{2}) × L_{2}(q), a contradiction to (β).

So let now K ≅ Sp_{2n}(q) or F_{4}(q) and Z_{2} ∩ K/Z(K) not be contained in R/Z(K). If K ≅ Sp_{2n}(q), then Z_{2} is centralized by Sp_{2n−4}(q). If n ≥ 4, then p | q^{2} − 1, a contradiction. Hence we have K ≅ Sp_{6}(q). But in this case there is a p-element ω ∈ C I(Z_{2}) with m_{p}(C I_{M}(ω)) ≥ 3, hence N_{G}(⟨ω⟩) ⊆ M, contradicting (β).

Now Z_{2} ⊆ Z(N(R)). But then any x ∈ Z_{2}^{♯} is centralized by Sp_{4}(q) in K. Hence C I_{G}(x) ⊆ M for every x ∈ Z_{2}^{♯}. This shows N_{G}(Z_{2}) ⊆ M, a contradiction.

(3.5) Lemma. We have [Z_{2}, K ∩ G_{1}] ⊈ O_{2}(G_{1}).

Proof. Suppose false. By (3.4) we get [Z_{2}, O_{2}(G_{1} ∩ K)] ≠ 1. Suppose that in case of K ≅ F_{4}(q) we do not have a diagram automorphism in M. Now there is a group U of order q in O_{2}(G_{1} ∩ K) with U ∩ C I(Z_{2}) = 1 and | Z_{2} : C I_{Z_{2}}(U) |≤ q (see (1.57)). Hence Z_{2} is an F - module. If K ≅ F_{4}(q) and [Z_{2}, N_{K}(R)] ⊈ O_{2}(N_{K}(R))), we see with (1.58) that | Z_{2} : C I_{Z_{2}}(Z(O_{2}(N_{K}(R)))) | ≤| Z(O_{2}(N_{K}(R))) : C I_{Z(O_{2}(N_{K}(R)))}(Z_{2}) |. Hence also in this case Z_{2} is an F - module. Now by (2.1) and (1.31) we have that G_{2} is solvable or E((G_{2}/Q_{2})/F (G_{2}/Q_{2})) ≅ L_{2}(r), L_{2}(r) × L_{2}(r), r even, or A_{9}.

Assume first that in case of F_{4}(q) there is no diagram automorphism involved.

Let first K = G(q), q > 2. Then for every u ∈ U^{♯} we have C I_{Z_{2}}(u) = C I_{Z_{2}}(U) and so we get E(G_{2}/Q_{2}) ≅ L_{2}(r) or L_{2}(r) × L_{2}(r), and | U |= r. Now we see q = r. Let ω ∈ M ∩ G_{2}, o(ω) = p | q − 1. Then N_{G}(⟨ω⟩) ≰ M. By (1.2) we see K ≅ L_{n}(q), n ≤ 4, Sp_{6}(q), or U_{n}(q), n ≤ 7.

If K ≅ L_{n}(q) or Sp_{6}(q), then every t ∈ σ(M) divides q−1, a contradiction. Let K ≅ U_{n}(q), n ≤ 7. Now there is t ∈ σ(M), t | q + 1. Then there is ν ∈ K, o(ν) = t, N_{G}(⟨ν⟩) ≤ M, [ν, a] = 1, a ∈ Z_{2}, a ∼ x ∈ R^{♯}. But C I_{G}(x) ⊆ M and so C I_{G}(a) ⊆ M^{g} for suitable g ∈ G_{2} \ M. Hence ν ∈ M^{g}. But then M = M^{g}, a contradiction.

So we are left with K = G(2). Now U induces transvections and so G_{2} is solvable or E((G_{2}/Q_{2})/F (G_{2}/Q_{2})) ≅ L_{2}(4) or A_{9}. For some prime p we have m_{p}(K) ≥ 3. This shows with (1.3) p = 3 and so G_{2}/Q_{2} ≅ Σ_{3}. As before we see that t ∈ Z_{2} with [t, O_{2}(G_{1})] ≠ 1 is not centralized by a 3−element in K. By (1.60) we see that m_{3}(K) = 3 and so O_{2}(K) ≠ 1. Hence Ω_{1}(Z(S)) ∩ O_{2}(K) ≠ 1. This gives Z(G_{2}) ≠ 1 and so m_{3}(C I_{K}(Z(S) ∩ K)) ≤ 1. Application of (1.60) again implies that no 2-local in K contains an elementary abelian group of order 27, contradicting the definition of an uniqueness group.

So we are left with K ≅ F_{4}(q) and some diagram automorphism is involved. Furthermore we may assume that Z_{2} ⊈ O_{2}(N_{K}(R)), R a root group. Now choose x ∈ Z(O_{2}(N_{K}(R))), [x, Z_{2}] ≠ 1. Then | [x, Z_{2}] |≥ q. Let y ∈ O_{2}(N_{K}(R))\Z(O_{2}(N_{K}(R))) with [y, Z_{2}] ≠ 1, then | [y, Z_{2}] |≥ q^{2}.

Suppose first q > 2. Then we get | [y, Z_{2}] |≥ 16 and so we see that Z_{2}/C I_{Z_{2}}(G_{2}) is the natural module and E(G_{2}/C I_{G_{2}}(Z_{2})) ≅ L_{2}(r), r ≥ q^{2}. As [x, Z_{2}] ≠ [y, Z_{2}], we see that one of x or y has to induce a field automorphism on L_{2}(r). As ⟨x, y⟩ acts quadratically on Z_{2} this is impossible.

So we have q = 2. Now 3 ∈ σ(G) and O_{2}(M) ≠ 1. Hence again Z_{2}/C I_{Z_{2}}(G_{2}) is the natural module. As | [y, Z_{2}] |≥ 4, we get E(G_{2}/C I_{G_{2}}(Z_{2})) ≅ L_{2}(r), r ≥ 4. As there is a field automorphism involved, we get that 3 | r −1. But then for some ω ∈ G_{2} ∩ M, o(ω) = 3, we have N_{G}(⟨ω⟩) ⊆ M, a contradiction.

Now we have Z_{2} ⊈ O_{2}(G_{1}). By (1.62) we get that there is some g ∈ G_{1} such that for X = ⟨Z_{2}, Z_{2}^{g}⟩ we have

(1) X/O_{2}(X) ≅ D_{2u} (u odd), L_{2}(q_{1}) or P Sp_{4}(q_{1}), q_{1} even.

(2) Y = (Z_{2} ∩ O_{2}(X))(Z_{2}^{g} ∩ O_{2}(X) ≤ X (3) Y ≠ Z_{2} ∩ O_{2}(X)

(4) | Z_{2} : C I_{Z_{2}}(Y/C I_{Y} (Z_{2})) |≤| Y : Y ∩ Q_{2} |^{2}

Suppose equality in (4). Then we have | Y : Z_{2} ∩ Z_{2}^{g} |= 4, q_{1}^{2}, q_{1}^{4}, respectively. In all cases X acts transitively on (Y/Z_{2} ∩ Z_{2}^{g})^{♯}. Hence Y′ = 1. But then | Z_{2} : C I_{Z_{2}}(Y) |= 2, q_{1}, q_{1}^{2}, a contradiction. So we have

(3.6) Lemma. | Z_{2} : C I_{Z_{2}}(Y) |<| Y : Y ∩ Q_{2} |^{2}.

Now application of (1.36) implies that we have one of the following possibilities for the structure of G_{2} and Z̄_{2} = Z_{2}/C I_{Z_{2}}(E(G_{2}/Q_{2})).

(1) Z_{2} is an F-module

(2) E(G_{2}/Q_{2}) ≅ (S)L_{3}(r) and Z̄_{2} is the natural module extended by its dual

(3) E(G_{2}/Q_{2}) ≅ L_{2}(r), r = t^{2}, Z̄_{2} is the orthogonal module.

(4) E(G_{2}/Q_{2}) ≅ P Sp_{4}(q) or 3 · A_{6}, Z̄_{2} is a direct sum of the natural and the dual module.

(5) E(G_{2}/Q_{2}) ≅ A_{9}

(6) E(G_{2}/Q_{2}) = X_{1} × X_{2} ≅ L_{2}(t^{2}) × L_{2}(t^{2}), Z̄_{2} = Z̄_{2}^{(1)} ⨁ Z̄_{2}^{(2)}, Z̄_{2}^{(i)} is the orthogonal X_{i}-module and [X_{i}, Z̄_{2}^{(j)}] = 1 for i ≠ j, i, j = 1, 2.

(7) E(G_{2}/Q_{2}) ≅ L_{2}(r) × L_{2}(r) and Z_{2} ist the natural O_{4}^{+}(r)-module.

(8) E(G_{2}/Q_{2}) = X_{1} × X_{2} ≅ L_{3}(2) × L_{3}(2), Z̄_{2} = Z̄_{2}^{(1)} ⨁ Z̄_{2}^{(2)}, Z̄_{2}^{(i)} is the direct sum of the natural module and its dual for X_{i} and [X_{i}, Z̄_{2}^{(j)}] = 1 for i ≠ j, i, j = 1, 2.

(9) E(G_{2}/Q_{2}) = X_{1}X_{2} ≅ 3A_{6} ∗ 3A_{6}, Z̄_{2} = Z̄_{2}^{(1)} ⨁ Z̄^{(2)}, Z̄_{2}^{(i)} is the direct sum of the natural and dual module for X_{i} and [X_{i}, Z̄_{2}^{(j)}] = 1 for i ≠ j, i, j = 1, 2.

(10) G_{2} is solvable, F (G_{2}/Q_{2}) ≅ 3^{1+4}, | Z̄_{2} |= 2^{8}.

We now fix the following notation. Let t be a Zsigmondy prime dividing r − 1, t = 9 in case of r = 64, and ω ∈ G_{2} ∩ K, o(ω) = t.

(3.7) Lemma. [K, ω] ≠ 1.

Proof. Otherwise by (1.63) we have N_{G}(⟨ω⟩) ⊆ M. But this contradicts the structure of G_{2}.

(3.8) Lemma. ω normalizes a Borel subgroup of K.

Proof. Suppose false. As ⟨S, ω⟩ is a {2, t} - group, we have that K has to have a solvable minimal parabolic. This now implies K = G(2) and t = 3 or 5. In particular r = 4, 64 or 16 and t = 3, while t = 5 implies K ≅ ^{2}F_{4}(2), a contradiction. As m_{p}(K) ≥ 3 for p ∈ σ(M), we get p = 3, contradicting o(ω) = 3 or 9 and N_{G}(⟨ω⟩) ⊈ M.

(3.9) Lemma. G_{2} is solvable or E(G_{2}/Q_{2}) ≅ L_{3}(2), A_{6}, 3·A_{6}, A_{9}, L_{3}(2)× L_{3}(2) or 3A_{6} ∗ 3A_{6}.

Proof. Let r ≤ q. By (3.8) ω normalizes a Borel subgroup of K. This implies O_{2}(G_{1} ∩ K)Q_{2}/Q_{2} ⊆ E(G_{2}/Q_{2}). Now we see | O_{2}(G_{1} ∩ K)/R : C I_{O_{2}(G_{1}∩K)/R}(t) |≤ r^{2} ≤ q^{2}, for t ∈ Z_{2}, and | O_{2}(N_{K}(R)/R) : C I_{O_{2}(N_{K}(R)/R)}(t) |≤^{2}≤ q^{2} for K ≅ F_{4}(q) and M involves a diagram automorphism.

This now implies with (1.59) that q = r and K ≅ (S)L_{n}(q), (S)U_{n}(q), Sp_{2n}(q). Furthermore | O_{2}(G_{1} ∩ K)/R : C I_{O_{2}(G_{1}∩K)/R}(t) |= q^{2}. Inspection of the groups in (1) - (7) shows that we have E(G_{2}/Q_{2}) ≅ (S)L_{3}(q), Sp_{4}(q) or L_{2}(q) × L_{2}(q), where Z̄_{2} is the O^{+}(4, q) - module in the latter.

Let E(G_{2}/Q_{2}) ≅ (S)L_{3}(q) or Sp_{4}(q). Then we may assume that Z_{2}^{(1)} ⊈ O_{2}(G_{1} ∩ K), where Z̄_{2}^{(1)} is the natural module. Hence R ∩ Z_{2}^{(1)} ≠ 1, R a root group. But C I_{G}(u) ⊆ M, for u ∈ R^{♯}, a contradiction. We are left with the O^{+}(4, q) - module. Now p |̸ q − 1 for p ∈ σ(M). This implies K ≅ Sp_{6}(q) or (S)U_{n}(q), n ≤ 7, by (1.2). As M ∩ G_{2} has a factorgroup isomorphic to ℤ_{q−1} ≀ ℤ_{2}, we see that in case of K ≅ Sp_{6}(q) there is some ω_{1} ∈ M ∩ K, o(ω_{1}) = t, [ω_{1}, K] = 1. Now we get with (1.63) N_{G}(⟨ω_{1}⟩) ⊆ M, a contradiction.

Hence we have K ≅ (S)U_{n}(q). Then there is p ∈ σ(M) such that Ω_{1}(Z(S)) is centralized by an elementary abelian group of order p^{2} in K. This implies Z(G_{2}) = 1 and so | Z_{2} |= q^{4}. Let R be the root group in K, then we see R ⊆ [Z_{2}, O_{2}(G_{1} ∩K)] and so Ω_{1}(Z(S)) ⊆ R. Hence F^{∗}(M) = K.

Now we see | O_{2}(G_{1}) : C I_{O_{2}(G_{1})}(Z_{2}) |= q^{2}. Hence C I_{O_{2}(G_{1})Z_{2}}(Z_{2}) = Z_{2} · T, where T is a special group of order q^{1+2(n−2)}. In particular Z_{2} contains a conjugate R^{g}, g ∈ K, R^{g} ≠ R. As N_{G}(R) ⊆ M, we have that N_{G}(R^{g}) ⊆ M. But ⟨N_{G_{2}}(R), N_{G_{2}}(R^{g})⟩ = G_{2}, a contradiction.

So assume now r > q. By (1.64) K ≅ (S)U_{n}(q),^{2}E_{6}(q), Ω_{2n}^{−}(q), r = q^{2}. Now 3 | r − 1 and so in any case 3 ∉ σ(M). This shows K ≅ (S)U_{n}(q), n ≤ 7, Ω_{8}^{−}(q) by (1.61). Now in any case there is p | q^{2} − 1, p ∈ σ(M). This implies that there is some ω_{1} ∈ G_{2} ∩ M, o(ω_{1}) = p. If m_{p}(C I(K)) ≠ 1, we see m_{p}(C I_{M}(ω_{1})) ≥ 3. Hence N_{G}(⟨ω_{1}⟩) ⊆ M, a contradiction.

So we have m_{p}(Aut(K)) ≥ 4. As there is no field automorphism of order p involved in M by (1.68), we see K ≅ (S)U_{n}(q), 5 ≤ n ≤ 7, p | q + 1. Now ω_{1} normalizes a parabolic P in Aut(K) with P/O_{2}(P) contains (S)U_{3}(q) × ℤ_{q+1}, L_{2}(q) × L_{2}(q^{2}) × ℤ_{q+1}, or L_{2}(q) × (S)U_{3}(q), respectively. Hence in all cases m_{p}(C I_{M}(ω_{1})) ≥ 3, the same contradiction as before.

(3.10) Proposition. If M is a uniqueness group in G then F^{∗}(M) = O_{2}(M).

Proof. Suppose first E(G_{2}/Q_{2}) ≅ A̸_{9}. Then (1) - (10) shows that there is some t ∈ Z_{2}\O_{2}(G_{1} ∩ K) such that | O_{2}(G_{1} ∩ K) : C I_{O_{2}(G_{1}∩K)}(t) |≤ 4. Now (1.59) yields K ≅ L_{n}(2), (S)U_{n}(2) or Sp_{2n}(2). As m_{p}(Aut(K)) ≥ 3, we see 3 ∈ σ(M). Hence we get that E(G_{2}/Q_{2}) ≅ L_{3}(2) or G_{2}/C I_{G_{2}}(Z_{2}) ≅ Σ_{3}. In the latter | O_{2}(G_{1} ∩ K) : C I_{O_{2}(G_{1}∩K)}(t) |= 2, a contradiction.

So we are left with E(G_{2}/Q_{2}) ≅ L_{3}(2). It is O_{2}(G_{1} ∩ K)Q_{2}/Q_{2} an elementary abelian normal subgroup in S/Q_{2}. But then | O_{2}(G_{1} ∩ K) : O_{2}(G_{1} ∩ K) ∩ Q_{2} |= 2 and so | O_{2}(G_{1} ∩ K) : C I_{O_{2}(G_{1}∩K)}(t) |≤ 2, a contradiction.

So we are left with E(G_{2}/Q_{2}) ≅ A_{9}. We have 3 ∉ σ(M). Further M∩G_{2}/Q_{2} ≥ A_{8}. If (M∩G_{2})^{(∞)} ⊆ C I(K), then m_{3}(K) = 1. But then we have a contradiction to (1.3) (i),(ii). Hence we have that K possesses a parabolic P with O^{2′}(P/O_{2}(P)) ≅ A_{8}. Then K ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{+}(2), E_{n}(2), or 2E_{6}(2). As 3 ∉ σ(M), we get K ≅ L_{n}(2), n ≤ 7, Sp_{8}(2), or Ω_{8}^{+}(2). As m_{p}(K) ≥ 3 for some prime p ≠ 3, we have a contradiction.

§ 4 The group G_{1}

In this chapter we construct the group G_{1}. To the pair G_{1}, G_{2} we are going to apply the amalgam method.

By (3.10) we have F^{∗}(M) = O_{2}(M) for the uniqueness group M. If 3 ∈ σ(G), we choose M such that m_{3}(M) ≥ 4.

(4.1) Lemma. Let S ∈ Syl_{2}(M) and N be any 2-local of G with S ≤ M ∩ N. If m_{3}(N) ≥ 4, then N ≤ M. Furthermore if m_{3}(M) ≥ 4, and m_{3}(N) ≥ 2, then N ≤ M.

Proof . In both cases by choice of M we have 3 ∈ σ(G) and m_{3}(M) ≥ 4. Let P ∈ Syl_{3}(N). Then P contains some E, E elementary abelian of order 9. Let P ≤ Q ∈ Syl_{3}(G). Then there is some g ∈ G such that Q ≤ M^{g}. As Γ_{E,1}(G) ≤ M^{g}, we now get N ≤ M^{g}. So S ≤ M^{g}. But then we may assume that g ∈ N_{G}(S) ≤ M, i.e. M = M^{g} ≥ N.

(4.2) Lemma. Let p ∈ σ(M) and E be an elementary ablian group of order p^{3} in M such that E ∩ M^{g} ≠ 1 for some g ∈ G, where E ≤ O_{p′,p}(M) if M is exceptional. Then M = M^{g}.

Proof . Let ν ∈ E ∩ M^{g}, o(ν) = p. Then N_{G}(⟨ν⟩) ≤ M. Let P ∈ Syl_{p}(C I_{M^{g}}(⟨ν⟩)). Then m_{p}(P) ≥ 2. Now N_{G}(P) ≤ M^{g}. Using this argument several times, we get that M and M^{g} share a Sylow p-subgroup. But then M = M^{g}, as M^{g} contains E and Γ_{E,1}(G) ≤ M.

(4.3) Lemma. Denote by ¯ the natural homomorphism onto M/O_{2}(M). Let \overline{K} < \overline{M}, m_{r}(\overline{K}) ≤ 3, for all odd primes r. Let ω ∈ K, o(ω) = r, r an odd prime. Then N_{G}(ω) ≤ M or M is exceptional or L = O^{p′}(M/O_{p′}(M)) is a Sylow p-subgroup of L_{2}(p^{n}), U_{3}(p^{n}) or ^{2}G_{2}(3^{n}) and M induces a Borel subgroup on L. Furthermore in the latter C I_{O_{2}(M)}(ω) = 1.

Proof . Let ω ∈ R ∈ Syl_{r}(\overline{K}). Then \overline{M} = \overline{K}N\overline{M}(R). Set Z = Ω_{1}(Z(R)). Then m_{p}(N_{G}(Z)) ≥ 4 and so application of (1.63) shows that E(N_{G}(Z)/O_{p′}(N_{G}(Z))) ≅ L_{2}(p^{n}), U_{3}(p^{n}), ^{2}G_{2}(3^{n}), or M is exceptional, or N_{G}(Z) ≤ M.

Suppose that N_{G}(Z) ≤ M. Let H = N_{G}(⟨ω⟩). Then Z ≤ H. Suppose furthermore H ≰ M. Suppose finally that N_{M}(⟨ω⟩) contains an elementary abelian p - group E such that Γ_{E,1}(G) ≤ M. Then the structure of H is given in (1.63). We have R ∩ H < H ∩ M/H ∩ O_{2}(M). Now we get that R ∩ H ≤ O_{p′}(H). Let R_{1} ∈ Syl_{r}(O_{p′}(H)), R ∩ H ≤ R_{1}. Then H = O_{p′}(H)N_{H}(R_{1}). In particular N_{H}(R_{1}) ≰ M. As Z ≤ R_{1}, we get that C I_{H}(R_{1}) ≤ O_{p′}(H). As O_{p′}(H) = ⟨C I_{O_{p′}(H)}(x)|x ∈ E^{♯}⟩ ≤ M, we have R_{1} ≤ M. If m_{r}(R_{1}) ≥ 4, then r is an uniqueness prime, contradicting H ≰ M. So we have m_{r}(R_{1}) ≤ 3. Let C be a critical subgroup of R_{1}, C_{1} = Ω_{1}(C). Then C_{1} ≅ ZZ_{r}, ZZ_{r} × ZZ_{r}, Z_{r^{1+2}}, ZZ_{r} × Z_{r^{1+2}}, ZZ_{r} × ZZ_{r} × ZZ_{r} or Z_{r^{1+4}}. In any case C_{H}(C_{1}) ≤ O_{p′}(H). So E(H/O_{p′}(H)) ≲ L_{3}(r) or Sp_{4}(r). Application of (1.63) and [Mi1],[Mi2] shows E(H/O_{p′}(H)) ≅ A_{6}. But either 2 · A_{6} is induced, contradicting 2 | |̸O_{p′}(H)| by (1.63) or 3 · A_{6} is induced, but p = 3,

as m_{p}(H/O_{p′}(H)) = 2, and now 3 | |O_{p′}(H)|, a contradiction.

So we have N_{G}(⟨ω⟩) ≤ M if N_{G}(Z) ≤ M and N_{G}(⟨ω⟩) contains E with Γ_{E,1}(G) ≤ M.

So let now F ≤ M, F elementary abelian of order p^{4}, F ≤ N(R). Suppose furthermore |C I_{F} (ω)| ≤ p.

Let D be a critical subgroup of R and D_{1} = Ω_{1}(D). Then |F : C I_{F} (D_{1})| ≤ p. This immediately shows that D_{1} ≅ ZZ_{r} × ZZ_{r} × ZZ_{r} , or D_{1} ≅ Z_{p^{1+4}} and p | r − 1.

Suppose the latter. Then we get that R ≅ D_{1} ∗ D_{2}, D_{2} cyclic. As ω ∈ R, we get ω ∈ D_{1} and so ⟨ω⟩ is normalized by an elementary abelian group of order p^{2} in F.

So we have D_{1} is elementary abelian of order r^{3}. But now F induces an elementary abelian group of order p^{3} in GL_{3}(r). This now shows that M possesses a normal subgroup of index p contradicting (1.68). So we have that N_{G}(⟨ω⟩) does contain some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. Hence N_{G}(⟨ω⟩) ≤ M, if N_{G}(Z) ≤ M.

Suppose finally N_{G}(Z) ≰ M. Then a Borel subgroup Y of L_{2}(p^{n}), U_{3}(p^{n}) or ^{2}G_{2}(3^{n}) acts on R. Now let again D_{1} = Ω_{1}(D), D a critical subgroup of R. As m_{r}(D_{1}) ≤ 3, we see that O_{p}(Y) acts trivially on D_{1} and so it centralizes R. Hence by (1.63) we have C I_{O_{2}(M)}(ω) = 1.

The choice of G_{1} is a little bit complicated. The aim is to have G_{2} ∩ M ≤ G_{1} and m_{p}(G_{1}) large enough for some p ∈ σ(M). Furthermore G_{1} should have a more or less simple structure.

Denote by \overline{M} = M/O_{2}(M) = K_{0}. Fix some p_{1} ∈ σ(M). Let K_{1} = O_{p_{1}}′(\overline{M}). Suppose there is some p_{2} ∈ σ(M) with m_{p_{2}}(K_{1}) ≥ 4. Then set K_{2} = O_{p_{2}}′(K_{1}). After finitely many steps we end with some K = K_{i} such that m_{r}(K) ≤ 3 for any prime r, r | |K|. Furthermore m_{p}(K_{i−1}) ≥ 4 for some p ∈ σ(M). Set H = K_{i−1}(\overline{G_{2} ∩ M}). Now F (H/K) is a p-group. We construct G_{1} in a preimage of H.

(4.4) Lemma. One of the following holds

(a) For suitable choice of G_{1} we have that G_{2} ∩ M ≤ G_{1} and one of the following holds for Z_{1} = ⟨Ω_{1}(Z(S))^{G_{1}}⟩

(i) G_{1} = C I_{G_{1}}(Z_{1})(G_{2} ∩ M), m_{p}(C I_{G_{1}}(Z_{1})) ≥ 3. If G_{2}/O_{2}(G_{2}) ≅ Sz(q) (including q = 2, i.e. F_{20}), ⟨Ω_{1}(Z(S))^{G_{2}}⟩ = Z_{2} is the natural module and for Z = C I_{Z_{2}}(Ω_{1}(S/O_{2}(G_{2}))) we have T_{1} = ⟨Z^{G_{1}}⟩ ≠ Z then m_{p}(C I_{G_{1}}(T_{1})) ≤ 1.

(ii) G_{1} ≠ C I_{G_{1}}(Z_{1})(G_{2}∩M), m_{p}(G_{1}) ≥ 2. If m_{p}(G_{1}) ≤ 4 set m_{p}(G_{1}) = 4−r. If r ≠ 0, then we have m_{p}(C I_{M}(Z_{1})) ≥ 1 and m_{p}(C I_{M}(Z_{1})) ≥ 2 if not r = 1 and m_{p}(C I_{G_{1}}(Z_{1})) = 0. Furthermore if x ∈ C I_{G_{1}}(Z_{1}) is a 2–element then x ∈ G_{2} ∩ M. If r = 1, any E ≅ E_{p^{2}}, E ≤ G_{1} is contained in some F ≅ E_{p^{3}} with |F ∩ C I_{M}(Z_{1})| ≥ p. If r = 2 any x ∈ Z_{1} is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M.

(b) G_{2}/O_{2}(G_{2}) ≅ Sz(q), ⟨Ω_{1}(Z(S))^{G_{2}}⟩ = Z_{2} is the natural module and for Z = C I_{Z_{2}}(Ω_{1}(S/O_{2}(G_{2}))) we have T_{1} = ⟨Z^{G_{1}}⟩ ≠ Z and m_{p}(C I_{M}(T_{1})) ≥ 2. If y ∈ G_{1}, y a 2–element with [T_{1}, y] = 1, then y ∈ O_{2}(G_{1}). If x ∈ T_{1} and x ≠ 1, then M is the unique maximal 2–local of G containing C I_{G}(x).

Furthermore in (a) and (b) if F (H/K) is a Sylow p–subgroup of L_{2}(p^{n}), U_{3}(p^{n}), or ^{2}G_{2}(3^{n}) with m_{p}(F (H/K)) ≥ 3 and M induces a Borel subgroup B of the corresponding simple group on F (H/K), then G_{1}K/K ≥ B.

Proof . We will prove (4.4) in several steps according to the structure of H/K.

(4.4.1) If m_{p}(F (H/K)) ≥ 4 the lemma holds.

Let H_{1} ≤ H be minimal with respect to G_{2} ∩ M ≤ H_{1} and H_{1}K/K ≥ F (H/K). Set Z_{1} = ⟨Ω_{1}(Z(S))^{H_{1}}⟩ and T_{1} = ⟨Z^{H_{1}}⟩ if G_{2}/O_{2}(G_{2}) ≅ Sz(q) and Z_{2} is the natural module.

Let first m_{p}(C I_{H_{1}}(Z_{1})) ≥ 3. Let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{H_{1}}(Z_{1})K/K. Hence (a) holds or G_{2}/O_{2}(G_{2}) ≅ Sz(q), Z_{2} is the natural module, T_{1} ≠ Z and m_{p}(C I_{G_{1}}(T_{1})) ≥ 2. Suppose the latter. If m_{p}(C I_{H_{1}}(T_{1})) ≥ 3 set G_{1} minimal with respect to G_{1}K/K ≥ C I_{H_{1}}(T_{1})K/K and G_{2} ∩ M ≤ G_{1} hence again (a)(i) holds.

So assume next that m_{p}(C I_{H_{1}}(T_{1})) = 2. Suppose C I_{H_{1}}(T_{1})K/K ≤ F (H/K). Then set G_{1} = C I_{H_{1}}(Z_{1})(G_{2} ∩ M). Hence either (a)(i) or (b) holds. In fact we have that any 2–element y ∈ C I(T_{1}) is contained in the preimage of F (H/K) and so is contained in O_{2}(G_{1}). Finally any x ∈ T_{1} is centralized by some E ≅ E_{p^{2}}, E ≤ H_{1}, with Γ_{E,1}(G) ≤ M, as m_{p}(F (H/K)) ≥ 4.

So assume now that C I_{H_{1}}(T_{1}) ≰ F (H/K). Set E = F (H/K) and U = [C I_{H_{1}}(T_{1}), E/Φ(E)]. As O_{p}((H/K)/F (H/K)) = 1 we see that (H_{1}K/K)/E is a subgroup of GL_{2}(p). Set S̃ = O_{2}(C I_{H_{1}}(T_{1})). Then S̃K/K is nontrivial. Set Ũ = C I_{E}(S̃). As m_{p}([S̃, E]) ≤ 2, we see that m_{p}(Ũ) ≥ 2. Let m_{p}(C I_{E}(T_{1})) ≥ 2, then let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and Ũ ≤ G_{1}K/K. Then we have (b). If y is a 2–element centralizing T_{1}, then y ∈ S̃ and so [y, Ũ] = 1, i.e. y ∈ O_{2}(G_{1}).

So let m_{p}(C I_{E}(T_{1})) = 1 and then also m_{p}([S̃, E]) = 1. In this case set G_{1} = (G_{2} ∩ M)C I_{H_{1}}(Z_{1}) and so (a)(i) holds.

So assume now m_{p}(C I_{H_{1}}(Z_{1})) ≤ 2. If C I_{H_{1}}(Z_{1})K/K ≤ F (H/K), then set G_{1} = H_{1} and we have (a)(ii).

So assume C I_{H_{1}}(Z_{1})K/K ≰ F (H/K). Set again E = F (H/K) and let U = [C I_{H_{1}}(Z_{1}), E/Φ(E)]. Let S̃ = O_{2}((C I_{H_{1}}(Z_{1})K/K)/E). Then again S̃ is nontrivial and (H_{1}K/K)/E is isomorphic to a subgroup of GL_{2}(p). Set Ũ = C I_{E}(S̃). Suppose first [C I_{G_{2}∩M}(Z_{1}), Ũ] = 1. As m_{p}([S̃, E]) ≤ 2, we see m_{p}(Ũ) ≥ 2. Now let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ Ũ. Then we have (a)(ii). Obviously any 2–element in G_{1} centralizing Z_{1} is contained in G_{2} ∩ M. Let m_{p}(Ũ) = 3 and let Û ≤ Ũ with Û ≅ E_{p^{2}}. Set U_{1} = [S̃, E]. As m_{p}(U_{1}) ≤ 2 a critical subgroup V of U_{1} is of rank at most two. As S̃ acts on V we get some W ≤ Ω_{1}(V), |W | = p, W ∩ Û = 1 and [W, Û] = 1. So ⟨Û, W ⟩ ≅ E_{p^{3}}.

Suppose now m_{p}(Ũ) = 2, then m_{p}([S̃, E]) = 2 and so m_{p}(C I_{H_{1}}(Z_{1})) ≥ 2. Furthermore any x ∈ Z_{1} is centralized by some F ≅ E_{p^{2}}, F ≤ [S̃, E] with Γ_{F,1}(G) ≤ M, as m_{p}(F (H/K)) ≥ 4 and so m_{p}(C I_{F (H/K)}(F)) ≥ 3.

So we are left with [C I_{G_{2}∩M}(Z_{1}), Ũ] ≠ 1. But then the argument above yield some S̃^{(1)}, a nontrivial 2–subgroup with [S̃^{(1)}, Ũ] ≠ 1 and S̃^{(1)} ≤ C I_{H_{1}}(Z_{1})K/K, a contradiction to the choice of S̃.

(4.4.2) If m_{p}(F^{∗}(H/K)) ≥ 4 the lemma holds.

By (4.4.1) we may assume m_{p}(F (H/K)) ≤ 3. As before choose H_{1} minimal with respect to G_{2} ∩ M ≤ H_{1} and H_{1}K/K ≥ F^{∗}(H/K). Define Z_{1} and T_{1} as before.

If m_{p}(C I_{H_{1}}(Z_{1})) ≥ 3 set G_{1} = C I_{H_{1}}(Z_{1})(G_{2} ∩ M) or if m_{p}(C I_{H_{1}}(T_{1})) ≥ 3 set G_{1} = C I_{H_{1}}(T_{1})(G_{2} ∩ M). Now as in (4.4.1) we get (a)(i).

So we may assume m_{p}(C I_{H_{1}}(Z_{1})) ≤ 2 or m_{p}(C I_{H_{1}}(Z_{1})) ≥ 3 and m_{p}(C I_{H_{1}}(T_{1})) ≤ 2. Set U_{1} = Z_{1} in the first case and U_{1} = T_{1} in the second case. We have m_{p}(C I_{H_{1}}(U_{1})) ≤ 2. Let E_{1} be some component with [E_{1}, U_{1}] = 1. Then m_{p}(E_{1}) ≤ 2. If m_{p}(E_{1}) = 2, then choose G_{1} minimal with respect to G_{2} ∩M ≤ G_{1} and G_{1}K/K ≥ C I_{H/K}(E_{1}). Let t be a 2-element with [Z_{1}, t] = 1 but t ∉ G_{2} ∩ M. Then Z(E_{1}) ≠ 1 and [Z(E_{1}), t] ≠ 1. We then let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{C I_{H/K}(E_{1})}(t) ≤ G_{1}. Now any 2-element centralizing Z_{1} is in G_{2} ∩ M. So we have (a)(ii) as we have m_{p}(G_{1}) ≥ 2.

So let next m_{p}(E_{1}) = 1. If there is a second component E_{2} with [U_{1}, E_{2}] = 1, m_{p}(E_{2}) = 1, then choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{H/K}(E_{1}E_{2}) and we get (a)(ii) as before.

So we have

(∗) There is at most one component E_{1} of F^{∗}(H/K) with [E_{1}, U_{1}] = 1. Furthermore m_{p}(E_{1}) = 1 if it exists.

We may assume m_{p}(C I_{H_{1}}(Z_{1})) ≥ 1, otherwise choose H_{1} = G_{1}. So assume first m_{p}(C I_{H_{1}}(Z_{1})) ≥ 3, m_{p}(C I_{H_{1}}(T_{1})) ≤ 2. Further we assume there is some 2– element x ∈ C I_{H_{1}}(T_{1}) with [x, F (H/K)] ≠ 1. As in (4.4.1) there is a nontrivial 2–group S̃ ≤ C I_{H_{1}}(T_{1}) with S̃ ≤ (H/K)/F^{∗}(H/K). Now m_{p}([F (H/K), S̃]) ≤ 2. Let again Ũ = C I_{F (H/K)}(S̃). If m_{p}([F (H/K), S̃]) = 2, we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and

G_{1}K/K ≥ ⟨C I_{E(H/K)}(C I_{E(H/K)}(T_{1})), Ũ⟩ .

Then we have (b).

Suppose m_{p}([F (H/K), S̃]) = 1. If C I_{E(H/K)}(T_{1}) ≠ 1, we just choose G_{1} as before. Hence assume that C I_{E(H/K)}(T_{1}) = 1. Now choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{H_{1}}(Z_{1}) ≤ G_{1}K/K, and we have (a)(i).

So we may assume that [C I_{H_{1}}(T_{1}), F (H/K)] = 1. Let E_{1} be some component with [E_{1}, T_{1}] = 1. If there is no such component we may choose H_{1} = G_{1} and so any 2-element centralizing T_{1} is contained in O_{2}(G_{1}) and we have (b).

Now m_{p}(E_{1}) = 1 by (∗) or there is a component E_{1} with [T_{1}, E_{1}] = 1. We choose G_{1}K/K ≥ C I_{F^{∗}(H_{1}/K)}(E_{1}) and G_{2} ∩ M ≤ G_{1}. Then we have (b).

Assume now m_{p}(C I_{H_{1}}(Z_{1})) ≤ 2. Suppose there is some 2-element x ∈ G_{2}∩ M, [x, Z_{1}] = 1 and [x, F (H/K)] ≠ 1. As before there is some S̃ ≤ C I(Z_{1}), S̃ < (H/K)/F^{∗}(H/K). Hence m_{p}([F (H/K), S̃]) ≤ 2. If m_{p}([F (H/K), S̃]) < m_{p}(F (H/K)) we may choose Ũ as before. Then choose G_{1} minimal with G_{2} ∩ M ≤ G_{1}.

G_{1}K/K ≥ ⟨E(H/K), Ũ⟩ if C I_{E(H/K)}(Z_{1}) = 1
G_{1}K/K ≥ ⟨C I_{E(H/K)}(E_{1}), Ũ⟩ if E_{1} = C I_{E(H/K)}(Z_{1}) .

Hence we have (a)(ii).

So assume m_{p}([F (H/K), S̃]) = m_{p}(F (H/K)). In particular we have m_{p}(F (H/K)) ≤ 2. If C I_{E(H/K)}(Z_{1}) = 1, we let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ E(H/K). In the other case we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{E(H/K)}(E_{1}), E_{1} = C I_{E(H/K)}(Z_{1}). We get (a)(ii) again.

So we may assume [C I_{H_{1}}(Z_{1}), F (H/K)] = 1. If C I_{E(H/K)}(Z_{1}) ≤ F (H/K), then choose G_{1} = H_{1} and we have (a)(ii). So assume there is a component E_{1} with [E_{1}, Z_{1}] = 1. Then by (∗) m_{p}(E_{1}) = 1 and E_{1} is the only component with this property. Now choose G_{1} with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{F^{∗}(H/K)}(E_{1}). Again we have (a)(ii).

Assume now that we have the exceptional case or F (H/K) is a Sylow p-subgroup of L_{2}(p^{n}), U_{3}(p^{n}) or ^{2}G_{2}(3^{n}), where H induces a Borel subgroup of the corresponding group on F (H/K).

Then choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K covers the Borel subgroup of L_{2}(p^{n}), U_{3}(p^{n}) or ^{2}G_{2}(3^{n}). In these cases we always have m_{p}(G_{1}) ≥ 3, where m_{p}(G_{1}) = 3 just happens if we have the situation of L_{2}(p^{3}). Hence we get the assertion.

So assume from now on m_{p}(F^{∗}(H/K)) ≤ 3.

(4.4.3) If there is a component E_{1} of H/K with m_{p}(E_{1}) ≥ 3, then the lemma holds.

By (4.4.2) we may assume that E_{1} = F^{∗}(H/K). Let H_{1} = E_{1}. Define Z_{1} and T_{1} as before. Assume [E_{1}, Z_{1}] ≠ 1. Choose U > E_{1}, m_{p}(U) = 4, (G_{2} ∩ M)K/K ≤ U. Set H_{1} = U. Now C I_{U}(Z_{1}) ≤ G_{2} ∩ M and so we may choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ U. So assume now [E_{1}, Z_{1}] = 1. Let G_{1} be minimal with G_{2} ∩M ≤ G_{1} and G_{1}K/K ≥ E_{1}. Then we have (a)(i) or we have the Suzuki case and [E_{1}, T_{1}] ≠ 1.

(4.4.4) If there is some component E_{1} of H/K with m_{p}(E_{1}) = 2, then the lemma holds.

First of all E_{1} ≠ F^{∗}(H/K). Otherwise E_{1} admits an outer automorphism group E ≅ E_{p^{2}}. By (1.67) E_{1} is a Lie group and E contains some field automorphism. By construction K < \overline{M} and E(H/K) < \overline{M}/K besides possibly for A_{8} ≤ \overline{G_{2} ∩ M}, A_{8} a component of H/K. As A_{8} does not possess an outer automorphism group of order p^{2}, we have that E_{1} < \overline{M}/K. In particular M possesses a normal subgroup of index p. This contradicts (1.68).

Now we have F^{∗}(H/K) = E_{1}E_{2}, m_{p}(E_{2}) ≤ 2 and m_{p}(E_{1}E_{2}) = 3. Again there is some U > E_{1}E_{2}, m_{p}(U) = 4, \overline{G_{2} ∩ M} ≤ U. Set H_{1} = U. Let U_{1} = Z_{1} or T_{1}. If [E_{1}E_{2}, U_{1}] = 1, let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and E_{1}E_{2} ≤ G_{1}K/K. Assume [E_{1}E_{2}, U_{1}] ≠ 1. Let [E_{1}, U_{1}] = 1. If m_{p}(C I_{U}(U_{1})) = 3 let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{U}(U_{1}) ≤ G_{1}K/K.

So assume m_{p}(C I_{U}(U_{1})) = 2. As m_{p}(U) = 4 we see that there is a p– element ω ∈ U \E_{1}E_{2} such that m_{p}(E_{1}E_{2}⟨ω⟩) = 4. Suppose first [ω, E_{1}] = 1. Then let G_{1} be minimal with G_{1} ≥ G_{2} ∩ M and G_{1}K/K ≥ E_{2}⟨ω⟩. If E_{2} is nonsolvable or E_{2} is cyclic we have that C I_{E_{2}}(U_{1}) = 1 and so we get (a)(ii) or (b). So let E_{2} be solvable of rank two. Then p||Z(E_{1})|. By (1.5) we see p = 3. Now E_{2} = F (C I_{H}(E_{1})) and so as ω ∉ E_{2}, we get C I_{H}(E_{1})/E_{2} ≅ SL_{2}(3) or GL_{2}(3). Hence for a critical subgroup C of E_{2} we have Ω_{1}(C) ≅ Z_{3^{1+2}}. This shows that any 2–element in C I_{G_{1}}(U_{1}) is contained in C I(E_{2}) and so we have (a)(ii) or (b).

Assume now [ω, E_{1}] ≠ 1. If p | |̸Z(E_{1})| we get m_{p}(E_{2}) = 1 and m_{p}(E_{2}⟨ω⟩) = 2. Furtermore C I_{E_{2}⟨ω⟩}(U_{1}) = 1 and so let G_{1} be as before we get (a)(ii) or (b) provided ω normalizes \overline{G_{2} ∩ M} ∩ E_{1}. As m_{p}(E_{1}⟨ω⟩) = 3 we get with (1.67) that E_{1} ≅ SL_{3}(r), SU_{3}(r) or a Lie group in even characteristic. If we have a Lie group in even characteristic \overline{G_{2} ∩ M} ∩ E_{1} is a parabolic in E_{1} and so ω normalizes \overline{G_{2} ∩ M} ∩ E_{1}. If E_{1} ≅ SL_{3}(r) or SU_{3}(r), r odd, then r has to be a prime and ω induces a diagram automorphism on E_{1}. But then we get a contradiction with (1.68).

So assume now that p||Z(E_{1})|. Then again p = 3 and E_{1} ≅ SL_{3}(2^{n}) or SU_{3}(2^{n}). Hence ω normalizes \overline{G_{2} ∩ M} ∩ E_{1}. If E_{2} is nonsolvable we let G_{1} be as before. So let E_{2} be solvable. Then m_{3}(E_{2}) = 2. Now suppose there is x ∈ U, o(x) = 2 with [x, U_{1}] = 1 and [x, E_{2}] ≠ 1. Let C be a critical subgroup of E_{2} and C_{1} = Ω_{1}(C). Then we see that C_{1} ≅ E_{9} or ZZ_{3}. As C_{1} ≥ Z(E_{1}), we get that U induces ZZ_{2} or Σ_{3} on C_{1}. In the latter we have that U = E_{1}[x, U]C I_{E_{2}}(x)⟨x⟩. But C I_{E_{2}}(x) is cyclic and so m_{p}(⟨E_{1}, [x, U]⟩) ≥ 3. But [⟨E_{1}, [x, U]⟩, U_{1}] = 1, a contradiction. So we have the former and then we may assume that ω normalizes C I_{E_{2}}(x) and then let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ ⟨C I_{E_{2}}(x), ω⟩K/K. Hence we get (a)(ii) or (b) again.

So we may assume [E_{1}, U_{1}] ≠ 1. Suppose [E_{2}, U_{1}] = 1. Then either E_{2} is solvable or m_{p}(E_{2}) = 1. Choose ω as before. If [ω, U_{1}] = 1 choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ E_{1}. Now we have (a)(ii) or (b).

So let [ω, U_{1}] ≠ 1. Suppose [E_{1}, Z_{1}] = 1. Then m_{p}(C I_{E_{1}E_{2}}(Z_{1})) = 3. If m_{p}(E_{2}) = 2, we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1} ≤ G_{1}K/K and (b) holds. If m_{p}(E_{2}) = 1 we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1}E_{2} ≤ G_{1}K/K and we have (a)(i).

So assume [E_{1}, Z_{1}] ≠ 1. If [ω, Z_{1}] = 1, then [ω, E_{1}] = 1 and so for G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1} ≤ G_{1}K/K we get (a)(ii). So assume [ω, Z_{1}] ≠ 1. If m_{p}(E_{2}) = 2 then choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1} ≤ G_{1}K/K. Hence we have (a)(ii). If m_{p}(E_{2}) = 1 and [ω, E_{2}] = 1, then choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and E_{1}⟨ω⟩ ≤ G_{1}K/K. hence we have (a)(ii) again.

So let [E_{2}, ⟨ω⟩] ≠ 1. If E_{2} is nonsolvable then by (1.67) we get (a)(ii) for G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1}⟨ω⟩ ≤ G_{1}, or E_{2} ≅ L_{2}(r^{3}) and p = 3. Furthermore ω induces a field automorphism on E_{2}. But now we get a normal subgroup of index p in M, contradicting (1.68). So assume that E_{2} is a cyclic p–group. Let x ∈ U be a 2–element with [x, U_{1}] = 1 but [x, E_{1}] ≠ 1. Then we may assume that [ω, x] = 1. Let ⟨x⟩C I_{U}(E_{2})/C I_{U}(E_{2}) ∈ Syl_{2}(U/C I_{U}(E_{2})). Then let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and E_{1}C I_{U}(x) ≤ G_{1}K/K. Again we have (a)(ii).

So assume now also [E_{2}, U_{1}] ≠ 1. If E_{2} is nonsolvable or any 2-element of U in C I(U_{1}) is contained in G_{2} ∩ M, let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and G_{1}K/K = U and we have (a)(ii).

So let x be a 2-element, [x, U_{1}] = 1, x ∉ G_{2} ∩ M. Again we may assume [x, U] ≤ E_{2}. As [E_{2}, U_{1}] ≠ 1, we have m_{p}(C I_{E_{2}}(x)) ≥ 1 and also m_{p}(C I_{E_{2}}(U_{1})) ≥ 1, as [[x, E_{2}], U_{1}] = 1. We have m_{p}(E_{2}) = 2 and as m_{p}(E_{1}E_{2}) = 3, we see p | |Z(E_{1})|, [E_{2}, x] is cyclic and [E_{2}, x] ∩ E_{1} = 1, as [E_{1}, x] = 1. Furthermore C I_{E_{2}}(x) ∩ Z(E_{1}) ≠ 1. If [Z_{1}, E_{1}] = 1, we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and ⟨E_{1}, [E_{2}, x]⟩ ≤ G_{1}K/K and we have (a)(i).

So assume [Z_{1}, E_{1}] ≠ 1. Let [Z_{1}, E_{2}] = 1. Then we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and E_{1} ≤ G_{1}K/K and we have (a)(ii).

So we may assume [Z_{1}, E_{2}] ≠ 1. If m_{p}(⟨E_{1}, C I_{E_{2}}(x)⟩) ≥ 3, we let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and ⟨E_{1}, C I_{E_{2}}(x)⟩ ≤ G_{1}K/K. Now any E ≤ G_{1}, E ≅ E_{p^{2}} centralizes some ν ∈ [E_{2}, x]^{♯}, and we have (a)(ii).

It remains the case that C I_{E_{2}}(x) is cyclic and m_{p}(E_{1}C I_{E_{2}}(x)) = 2. Now let ω be as before. We may assume [ω, x] = 1. Hence ω acts on E_{1}C I_{E_{2}}(x). If [ω, U_{1}] ≠ 1, we let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and ⟨E_{1}, C I_{E_{2}}(x), ω⟩ ≤ G_{1}K/K. As [ω, Ω_{1}([E_{2}, x])] = 1, we have (a)(ii).

Let [ω, U_{1}] = 1. In this case [ω, E_{1}] = 1. As E_{2} = O_{p}(U) and [x, U] ≤ E_{2}, we get m_{p}([x, E_{2}]) ≥ 2, a contradiction, as [x, E_{2}]∩E_{1} = 1 and m_{p}(E_{1}E_{2}) = 3. This proves (4.4.4).

So we have m_{p}(E_{1}) = 1 for any component of H/K.

(4.4.5) If E_{1}, E_{2} are two components of H/K. Then the lemma holds.

We have F^{∗}(H/K) = E_{1} × E_{2} × E_{3}, E_{3} cyclic or nonsolvable of p-rank 1. Let E_{1}E_{2}E_{3} < U, m_{p}(U) = 4. Now there is some ω with [E_{1}, ω] ≠ 1, m_{p}(E_{1}⟨ω⟩) = 2. By (1.67) ω induces a field automorphism on E_{1}. Now (1.68) implies that there is some t ∈ S with E_{1}^{t} = E_{2}, ω^{t} = ω^{−1}. Now we may consider ⟨E_{1}, E_{2}⟩ as a rank 2 component and we get the assertion as in (4.4.4).

(4.4.6) If m_{p}(E(H/K)) = 1, the lemma holds.

Now we have F^{∗}(H/K) = E_{1}E_{2}, m_{p}(E_{1}) = 1, E_{2} is solvable with 1 ≤ m_{p}(E_{2}) ≤ 2. Suppose first m_{p}(E_{2}) = 1. Then we get some ⟨ω_{1}, ω_{2}⟩ ≤ Out (E_{1}), ⟨ω_{1}, ω_{2}⟩ an elementary abelian p-subgroup. By (1.67) we may assume that ω_{1} is a field automorphism. But this contradicts (1.68).

So we have m_{p}(E_{2}) = 2. Now choose again U > E_{1}E_{2}, m_{p}(U) = 4, \overline{G_{2} ∩ M} ≤ U. Again by (1.67) and (1.68) for some ω ∈ U \ E_{1}E_{2}, o(ω) = p, we have [ω, E_{1}] = 1. Hence U = E_{1}C I_{U}(E_{1})(\overline{G_{2} ∩ M}).

If C I_{E_{1}E_{2}}(Z_{1}) = 1 then choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K = U. We have (a)(ii).

Assume next [E_{1}, Z_{1}] = 1. If C I_{E_{2}}(Z_{1}) = 1. The structure of U tells us [ω, Z_{1}] ≠ 1. Hence choose G_{1} minimal with respect to E_{2}⟨ω⟩ ≤ G_{1}K/K and G_{2} ∩ M ≤ G_{1}. Then as any E ≅ E_{p^{2}} in G_{1} centralizes some p-element in E_{1}, we have (a)(ii) again.

We may assume C I_{E_{2}}(Z_{1}) ≠ 1. Let m_{p}(C I_{E_{2}}(Z_{1})) = 1. Then choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ ⟨E_{2}, ω⟩. As O_{p}(U) = E_{2}, we have that SL_{2}(p) is induced on E_{2} by ⟨ω, \overline{G_{2} ∩ M}⟩. If x is a 2–element in G_{1} centralizing Z_{1} not contained in G_{2} ∩ M then we see that there is also a 2–element y with m_{p}([E_{2}, y]) = 2 and [y, Z_{1}] = 1, a contradiction. Hence we have (a)(ii).

Suppose m_{p}(C I_{E_{2}}(Z_{1})) = 2. Then we choose G_{1} minimal with G_{2}∩M ≤ G_{1} and G_{1}K/K ≥ C I_{U}(Z_{1}). We have (a)(i) as long as we do not have the Sz(q)- case with C I_{E_{2}}(T_{1}) = 1. So let assume that we have the Sz(q)–case. Let [T_{1}, E_{1}] = 1. If m_{p}(C I_{E_{2}}(T_{1})) = 2, we choose G_{1} minimal with G_{2} ∩ M ≤ G_{1} and C I_{U}(T_{1}) ≤ G_{1}K/K and we have (a)(i). So assume next m_{p}(C I_{E_{2}}(T_{1})) = 1. As in the case of m_{p}(C I_{E_{2}}(Z_{1})) = 1, we choose G_{1} minimal with G_{2} ∩M ≤ G_{1} and G_{1}K/K ≥ ⟨E_{2}, ω⟩ and we have (b).

So assume now [T_{1}, E_{1}] ≠ 1. If ω ∈ C I_{U}(T_{1}) ≰ E_{2}, then m_{p}(C I_{E_{2}}(T_{1})) = 2. Now we just choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{U}(T_{1}) ≤ G_{1}K/K. So let ω ∉ C I_{U}(T_{1}). Again we have m_{p}(C I_{E_{2}}(T_{1})) = 2, otherwise we have (a)(ii). If x is a 2–element with [x, T_{1}] = 1 and x ∉ G_{2} ∩ M, then we see that m_{p}(C I_{U}(x)) ≥ 2 and so we choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{U}(x) ≤ G_{1}K/K. Hence we have (b) in any case.

So assume now [E_{1}, Z_{1}] ≠ 1. If ω ∈ C I_{U}(Z_{1}), then m_{p}(C I_{E_{2}}(Z_{1})) = 2 and we choose G_{1} minimal with respect to G_{2}∩M ≤ G_{1} and C I_{U}(Z_{1})K/K ≤ G_{1}K/K. So we have (a)(i). So assume ω ∉ C I_{U}(Z_{1}). Then we are going to prove that we have (a)(ii). For this we may assume that there is some 2-element x ∉ G_{2} ∩ M, [x, Z_{1}] = 1. Then m_{p}(C I_{E_{2}}(Z_{1})) = 2 = m_{p}(C I_{U}(Z_{1})). Now let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{U}(x) ≤ G_{1}K/K. As m_{p}(C I(Z_{1})) = 2, we have (a)(ii).

To prove the lemma we just have to show

(4.4.7) If F^{∗}(H/K) = F (H/K) and m_{p}(F (H/K)) ≤ 3, then the lemma holds.

Let C be a critical subgroup of E = F (H/K) and C_{1} = Ω_{1}(C). As H/F (H/K) acts faithfully on C_{1}, and m_{p}(Out(C_{1})) ≤ 1 for m_{p}(C_{1}) ≤ 2, we get m_{p}(E) = m_{p}(C_{1}) = 3. As E ≥ C I_{H/K}(C_{1}/Φ(C_{1})), we see that H/E is involved in Aut (C_{1}). So H/E is involved in GL_{2}(p), GL_{3}(p) or Sp_{4}(p). Now choose X minimal with respect to G_{2}∩M ≤ X, m_{p}(X) ≥ 4 and XK/K ≥ C_{1}. Set Z_{1} = ⟨Ω_{1}(Z(S))^{X}⟩, T_{1} = ⟨Z^{X}⟩.

Suppose m_{p}(C I_{X}(Z_{1})) ≥ 3. Then choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I(Z_{1}). If we do not have the Sz(q)-case then we get (a)(i). The same holds for m_{p}(C I_{X}(T_{1})) ≤ 1. So assume we have the Sz(q)-case. If m_{p}(C I_{X}(T_{1})) ≥ 3, we choose G_{1} minimal with respect to G_{2} ∩ M ≤ G_{1} and C I_{X}(T_{1}) ≤ G_{1}K/K. Again we have (a)(i). So for the remainder set U_{1} = Z_{1} or T_{1}. We just have m_{p}(C I_{X}(U_{1})) ≤ 2. If any 2– element centralizing U_{1} is in E we may choose G_{1} = X and so we have (a)(ii) or (b).

Suppose next |C I_{X}(U_{1})/O_{2}(X)| is even. Set U = C_{1}/Φ(C_{1}). Now [C I\overline{X}(U_{1}), U] ≤ C I_{U}(U_{1}). Let |[C I_{X}(U_{1}), U]| = p. Then C I\overline{X}(U_{1})/O_{p}(C I\overline{X}(U_{1})) is cyclic. Let T = S ∩ C I(U_{1}), X_{1} = ⟨N_{X}(T/O_{2}(X)), G_{2} ∩ M⟩. As m_{p}(C I_{E}(T/O_{2}(X))) ≥ 2, m_{p}(X) = 4 and X_{1}KE/KE = XKE/KE, we get m_{p}(X_{1}) ≥ 2 and m_{p}(X_{1}) ≥ 3 if m_{p}(C I_{C_{1}}(U_{1})) = 1. Furthermore any 2-element in X_{1} centralizing U_{1} is in G_{2} ∩ M. We may choose G_{1} minimal in X_{1} such that G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ N_{X}(T/O_{2}(X)). Now we have (a)(ii) or (b) provided we can prove that any Ẽ ≅ E_{p^{2}}, Ẽ ≤ G_{1} is contained in some F ≅ E_{p^{3}}, |F ∩ C I(U_{1})| ≥ p. But |C I_{U}(Ẽ)| ≥ p^{2} and so C I_{C_{1}}(Ẽ) contains some F_{1} ≅ E_{p^{2}} and then F = F_{1}Ẽ does the job or Ẽ ≤ C_{1} and C I_{C_{1}}(Ẽ) = Ẽ. But then C_{1} ≅ Z_{p^{1+2}} and so X induces SL_{2}(p) on C_{1} and all elementary abelian subgroups of order p^{2} in C_{1} are conjugate, hence they are contained in some F ≅ E_{p^{3}}.

Suppose now |[U, C I(U_{1})]| = p^{2}. Then m_{p}(C I_{C_{1}}(U_{1})) = 2. Now W = O^{p}(C I(U_{1})/O_{p}(C I(U_{1}))) ≲ GL_{2}(p). Let T = S ∩ O_{2}(W). Let C_{1} be elementary abelian of order p^{3}. Then (H/K)/E ≲ GL_{3}(p). As O_{p}((H/K)/E) = 1 and m_{p}(E) = 3, we now get p^{2}GL_{2}(p) ≲ (H/K)/C I_{H/K}(C_{1}). But C I_{H/K}(C_{1}) ≤ E and so |[E, C_{1}]| = p, and C I_{E}(U_{1})C I_{E}(C_{1}) = E. This finally gives [C I(U_{1}), C_{1}] = C_{1}, a contradiction.

Assume next C_{1} ≅ Z_{p^{1+4}}. Then C I_{C_{1}}(T) ≅ Z_{p^{1+2}}. Set X_{1} = C I_{H/K}(T). Then m_{p}(X_{1}) ≥ 2. So let G_{1} be minimal with respect to G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{C_{1}}(T). Then any 2–element of G_{1} centralizing U_{1} is contained in G_{2} ∩ M. So we have (a)(ii) or (b).

Suppose now C_{1} ≅ ZZ_{p} × Z_{p^{1+2}} or C_{1} ≅ Z_{p^{1+2}}. Let F ≤ (H/K)/E, F ≅ SL_{2}(p). Suppose that C I(U_{1}) does not cover F. Then let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and G_{1}K/K ≥ C I_{H/K}(T). Now m_{p}(G_{1}) ≥ 2 and any 2-element centralizing U_{1} is in G_{2} ∩ M. Again we have (a)(ii) or (b) provided we can prove that Ẽ ≅ E_{p^{2}}, Ẽ ≤ G_{1} is contained in some F ≅ E_{p^{3}}. This is clear as long C_{1} ≅ Z̸_{p^{1+2}} as before. In the latter we see that C ≅ ZZ_{p^{n}} ∗ Z_{p^{1+2}} and E ≅ C_{1}C I_{E}(C_{1}). Either there is a normal Z_{p^{1+2}} in C I_{E}(C_{1}) and we are done, or F covers C I_{X}(C I_{E}(C_{1})) as we may assume p > 3. But now we have no Ẽ ≤ G_{1}, Ẽ ≅ E_{p^{2}} and Ẽ = Ω_{1}(C I_{G_{1}}(Ẽ)).

So assume that C I(U_{1}) does cover F. Now if p > 3, then C I(U_{1}) does contain a subgroup F_{1} ≅ SL_{2}(p) and C I(U_{1}) does contain C_{2} ≤ C_{1}, C_{2} ≅ Z_{p^{1+2}}. Now m_{p}(C_{2}F_{1}) = 3, a case handled before.

So we have p = 3 and (H/K)/E ≅ SL_{2}(3) or GL_{2}(3). If H/K ≠ EC I_{H/K}(U_{1}), then choose G_{1} as before. So assume first C I_{H/K}(U_{1})/C I_{E}(U_{1}) ≅ GL_{2}(3). Let P ∈ Syl_{3}(H/K), T_{1} = T ∩ N(P). Then T = ⟨x_{1}, x_{2}⟩ is a fours group. Furthermore [x_{1}, P] = C_{2} ≅ Z_{3^{1+2}} and [C I_{P} (x_{1}), x_{2}] is cyclic. Hence [P, T_{1}] ≅ ZZ_{3^{n}} ∗ Z_{3^{1+2}}. As m_{3}(P) ≥ 4, we get m_{3}(C I_{P} (T_{1})) ≥ 2. But C I_{P} (T_{1}) ≤ E and so as [C I_{P} (T_{1}), C_{2}] = 1 and C I_{P} (T_{1}) ∩ C_{2} = 1, we see m_{3}(E) ≥ 4, a contradiction.

So we have C I_{H/K}(Z_{1})/C I_{E}(Z_{1}) ≅ SL_{2}(3). Let Q be the quaternion group in SL_{2}(3). Then we see that [Q, E] = C_{2} and E = C_{2}C I_{E}(Q) = C_{2} ∗ C I_{E}(Q). Now as we may assume that C I(U_{1}) does not contain SL_{2}(3), otherwise we get a contradiction as before, we get m_{3}(C I_{E}(Q)) ≥ 2. Now let G_{1} be minimal with G_{2} ∩ M ≤ G_{1} and N_{H/K}(Q) ≤ G_{1}K/K. As Z_{1} = ⟨Ω_{1}(Z(S))^{G_{1}}⟩, we see that any 2-element centralizing U_{1} is in O_{2}(G_{1}). Furthermore U_{1} is centralized by E_{9} and we have (a)(ii) or (b).

So assume finally C_{1} ≅ ZZ_{p} × ZZ_{p} and [C_{1}, U_{1}] = 1. Now (H/K)/E acts as a subgroup of GL_{2}(p) on C_{1}. Let Ẽ ≤ H/K, Ẽ be elementary abelian of order p^{4}. Then |Ẽ : Ẽ ∩ E| = p. But |C_{1} : C I_{C_{1}}(Ẽ)| = p. This implies |(Ẽ ∩ E)C_{1}| = p^{4}, a contradiction. This proves the lemma.

In (5.8) we will see that (4.4)(b) does not exist.

If G_{2} is solvable or E(G_{2}/O_{2}(G_{2})) ≅ L_{2}(r), L_{2}(r) × L_{2}(r), r odd, r ≥ 11 or P Sp_{4}(r), r ≥ 7, then we change the choice of G_{1} in (4.4) a little bit. We are just interested in having S in G_{1} ∩ G_{2} not the whole of G_{2} ∩ M.

(4.5) Lemma. Let G_{2} solvable or E(G_{2}/O_{2}(G_{2})) ≅ L_{2}(r) or L_{2}(r) × L_{2}(r), r odd, r ≥ 11 or P Sp_{4}(r), r ≥ 7, then choose G_{1} minimal such that S ≤ G_{1} and G_{1}/K satisfies (4.4)(a).

(4.6) Lemma. We may choose G_{1} such that (4.4) holds and \overline{G_{2} ∩ G_{1}} ∩ K ≤ O_{2}(\overline{G}_{1}), or F (H/K) is a Sylow p–subgroup of L_{2}(p^{n}), U_{3}(p^{n}) or ^{2}G_{2}(p^{n}) with m_{p}(F (H/K)) ≥ 3 and [Ω_{1}(Z(S)), G_{1} ∩ G_{2}] ≠ 1.

Proof . Suppose \overline{G_{2} ∩ G_{1}} ∩ K is a 2-group T. Then \overline{G}_{1}K = KN_{\overline{G}_{1}}(T). So we may assume T < \overline{G}_{1}.

Suppose \overline{G_{2} ∩ G_{1}} ∩ K is not a 2–group and furthermore N_{G}(⟨ν⟩) ≤ M, for all ν ∈ \overline{G_{2} ∩ G_{1}} ∩ K, o(ν) odd. Then by (1.66) G_{2} is solvable, E(G_{2}/O_{2}(G_{2})) ≅ L_{2}(r), L_{2}(r) × L_{2}(r), (S)L_{3}(r), (S)U_{3}(r), P Sp_{4}(r) or F (G_{2}/O_{2}(G_{2})) is elementary abelian of order r^{3} and E((G_{2}/O_{2}(G_{2}))/F (G_{2}/O_{2}(G_{2}))) ≅ L_{2}(r). Let T ∈ Syl_{2}(K). Then \overline{G}_{1}K = KN_{\overline{G}_{1}}(T). So by choice of G_{1} in (4.5), we get that E(G_{2}/O_{2}(G_{2})) ≅ (S)L_{3}(r) or (S)U_{3}(r), r odd, or |F (G_{2}/O_{2}(G_{2}))| = r^{3} and G_{2} is nonsolvable.

In all cases r | |K ∩ \overline{G_{2} ∩ G_{1}}| and so even r^{2} | |K ∩ \overline{G_{2} ∩ G_{1}}|. This implies that we have |F (G_{2}/O_{2}(G_{2}))| = r^{3}, as in the other cases ((G_{1} ∩ G_{2})/O_{2}(G_{2})) ∩ E(G_{2}/O_{2}(G_{2})) ≅ GL_{2}(r). Now r^{3} | |K ∩ \overline{G_{2} ∩ G_{1}}|. But then G_{2} ≤ M, a contradiction.

So we have that there is some ν ∈ \overline{G_{1} ∩ G_{2}} ∩ K, o(ν) odd, such that N_{G}(⟨ν⟩) ≰ M.

The assertion follows with (4.3). As C I_{O_{2}(M)}(ν) = 1, we also see that [ν, Ω_{1}(Z(S))] ≠ 1.

If we have the exceptional case of (4.6) we also change the definition of G_{1} a little bit.

If Ω_{1}(Z(S)) ≮G_{2} and G_{1} is as in the exceptional case in (4.6) then replace G_{1} by some group minimal with respect to containing the normalizer of the corresponding Sylow p–subgroup..

Now in any case we have

(4.7) Lemma. \overline{G_{2} ∩ G_{1}} ∩ K ≤ O_{2}(\overline{G}_{1}).

(4.8) Lemma. The group G_{1} has the following structure

G_{1}/O_{2}(G_{1}) = (X_{1} ∗ X_{2} ∗ … ∗ X_{t} ∗ Y_{1} ∗ Y_{2} ∗ … ∗ Y_{f})U,

where X_{1}, …, X_{t} are the quasisimple groups, Y_{i} ′ = Y_{i}, Y_{i}/O_{r_{i}}(Y_{i}) ≅ L_{2}(r_{i}) or L_{3}(r_{i}), and U is p-constrained. If ν is a p-element with [ν, Y_{i}] ≤ O_{r_{i}}(Y_{i}) then ν ∈ Y_{i}C I(Y_{i}).

Proof . Let XK/K be a component of G_{1}K/K. Let F = F (K) and R ∈ Syl_{r}(K). We have m_{r}(R) ≤ 3. Suppose that C I_{XK}(R) ≤ K. Then XK/K ≲ L_{3}(r) or Sp_{4}(r). As 2 | |̸K|, we get XK/K ≲ L_{3}(r). So by [Mi1] we get XK/K ≅ L_{2}(r), A_{5}, L_{3}(r), L_{2}(7) or A_{6}. In case of A_{6} the group 3 · A_{6} will be induced. So we get G_{1}/O_{2}(G_{1}) = (X_{1} × … × X_{t} ∗ H_{1})F, X_{1}, …, X_{t} quasisimple groups, E(H_{1}K/K) has just components L_{3}(r), L_{2}(r), A_{5}, L_{2}(7) or A_{6}, F is an automorphism group of X_{1} × … × X_{t}.

Let now Y_{1}K/K ≅ L_{3}(r_{1}), L_{2}(r_{1}), Y_{2}K/K ≅ L_{3}(r_{2}), L_{2}(r_{2}), A_{5}, L_{2}(7), or 3 · A_{6} and Y_{1}Y_{2}K/K ≤ H_{1}K/K.

Set U_{1} = (Y_{1}K)^{∞}, U_{2} = (Y_{2}K)^{∞}. Then [U_{1}, U_{2}] ≤ U_{1} ∩ U_{2} ≤ K. Let U_{1} = A_{0} > A_{1} > A_{2} > … > A_{f} = 1, A_{i} characteristic in U_{1}. A_{0}/A_{1} ≅ L_{2}(r_{1}), L_{3}(r_{1}), A_{5}, L_{2}(7), or 3 · A_{6} and A_{i}/A_{i+1} groups of prime power order s_{i}^{a_{i}}, s_{i} ≠ s_{i+1}. Let [U_{2}, A_{i}] ≰ A_{i+1} for some i. Then Aut(A_{i}) either involves U_{1}K/K × U_{2}K/K or U_{1} = A_{1}C I_{U_{1}}(A_{i}/A_{i+1}).

Suppose the former. As |A_{i}/A_{i+1}| = s_{i}^{a_{i}} for some prime s_{i}, we get a s_{i}- subgroup W of A_{i} such that Aut(W) involves U_{1}K/K×U_{2}K/K. But we have m_{s_{i}}(W) ≤ 3, so Aut (Ω_{1}(C)), C a critical group in W, has a subgroup which is an extension of a group of odd order by a direct product of two groups in {L_{3}(r_{1}), L_{2}(r_{2}), L_{2}(r_{1}), L_{2}(r_{2}), A_{5}, A_{6}, L_{2}(7)}. But we have that this group is involved in L_{3}(s_{i}) or Sp_{4}(s_{i}), a contradiction to [Mi1] and [Mi2].

So we have the latter. Choose i minimal with respect to C I_{U_{1}}(A_{i}/A_{i+1}) ≰ A_{1}. Now A_{i}/A_{i+1} ≤ Z(U_{1}/A_{i+1}) as U_{1} = U′_{1}. Let i > 1. Then A_{i−1}/A_{i+1} ≅ A_{i}/A_{i+1} × A_{i−1}/A_{i}. So we produce a new series A_{0} > A_{1} = B_{1} > B_{2} > … > B_{f}, where C I_{A_{0}}(B_{j}/B_{j+1}) ≰ A_{1} for j < i .

This shows that we may assume A_{1}/A_{2} ≤ Z(A_{0}/A_{2}). Now the Schur multipliers yield A_{0}/A_{1} ≅ L_{3}(r_{1}) and |A_{1}/A_{2}| = 3 so A_{0}/A_{2} ≅ SL_{3}(r_{1}).

Now there is an elementary abelian subgroup C of A_{2}, |C| = r_{1}^{3}, such that A_{0}/A_{2} induces SL_{3}(r_{1}) on C. As C I_{C}(Z(A_{0}/A_{1})) = 1, we see that A_{0} contains a subgroup A ≅ SL_{3}(r_{1}). Now m_{r_{1}}(C A) = 4, a contradiction.

So we have that C I_{A_{0}}(A_{i}/A_{i+1}) ≤ A_{1} for all i. In particular [U_{1}, U_{2}] = 1. Furthermore the same argument implies that A_{1}/A_{3} ≅ A_{1}/A_{2} × A_{2}/A_{3} and so A_{1} is nilpotent.

Let now ν be a p-element with [Y_{1}, ν] ≤ A_{1}. Suppose [ν, A_{1}] ≠ 1. Then ν acts on some E ≅ E_{r^{3}}. As Y_{1}/A_{1} acts irreducibly on E, we see that ν acts as an element in Z(GL_{3}(r)). We have \overline{M} = KN\overline{M}(\overline{E}). Hence by (1.68) we know that ν ∈ Y_{1} ′. Now as seen before we have Y_{1}/A_{1} ≅ 3 · A_{6}, o(ν) = 3, and so ν ∈ Y_{1}C I(Y_{1}).

Suppose gcd(|Y_{1}K/K|, |K|) = 1. Then by Zassenhaus’ lemma [Hu, (I.18.1),(I.18.2)] we get a subgroup isomorphic to Y_{1}/A_{1} which is normalized by (\overline{G_{2} ∩ G_{1}}). By the minimal choice of G_{1}, (4.7) and the fact that any p-element normalizing Y_{1} is in Y_{1}C I(Y_{1}), we have A_{1} = 1.

If Y_{1}/A_{1} ≅ L_{3}(r_{1}), L_{2}(r_{1}), r_{1} | |A_{1}|. Then by the same argument, minimal choice and (4.7) again, we have that A_{1} is an r_{1}-group.

Now we make the following choice for L_{1}, L_{2}. Set L_{2} = G_{2}. Suppose f ≥ 1, Y_{1}/O(Y_{1}) ≅ L_{2}(r). Let Ŷ_{1} be the preimage of Y_{1}. Set L_{1} = Ŷ_{1}(G_{1} ∩ G_{2}).

So we have the following:

(4.9) Lemma. L_{1} is of one of the following types

(i) L_{1}/O_{2}(L_{1}) = (X_{1} ∗ … ∗ X_{t} ∗ Y_{1} ∗ … ∗ Y_{f} ∗ U)A, X_{1}, …, X_{t} quasisimple groups, Y_{i}/O_{r_{i}}(Y_{i}) ≅ L_{3}(r_{i}), r_{i} prime, U is p-constrained, A is an automorphism group of (X_{1} ∗ … ∗ X_{t} ∗ Y_{1} ∗ … ∗ Y_{f} ∗ U). Furthermore if m_{p}(X_{1} ∗ … ∗ X_{t} ∗ Y_{1} ∗ … ∗ Y_{f} ∗ U) ≥ 4, A ≤ G_{2}.

(ii) L_{1}/O_{2}(L_{1}) ≅ A · B, B/O_{r}(B) ≅ L_{2}(r), A = G_{2} ∩ L_{1}/O_{2}(L_{1}), r > 3.

(4.10) Lemma. Let L_{1} be as in (4.9)(ii). Then M is the only maximal subgroup of G containing an elementary abelian subgroup of order r^{3} from O_{r}(B).

Proof . By (4.3) we have N_{G}(⟨ω⟩) ≤ M for any ω ∈ O_{r}(B), o(ω) = r. Now the assertion follows with (1.63).

If L_{2} is solvable with σ(L_{2}) ≠ ∅ we are going to change notation a little bit in order to get more symmetry. This will become important in (6.35).

Let N be the uniqueness group with L_{2} ≤ N. If we go over the construction of L_{2} in (2.1) we see that either L_{2}/O_{2}(N) ≤ F (N/O_{2}(N))S or L_{2}/O_{2}(N) ≤ ⟨E, S/O_{2}(N)⟩, E ≤ E(N/O_{2}(N)), E ≅ Sz(q). Hence in any case we get that ⟨L_{1} ∩N, L_{2}⟩ is of the same shape, i.e. ⟨L_{1} ∩N, L_{2}⟩/O_{2}(⟨L_{1} ∩ N, L_{2}⟩) is an extension of a p-group by G_{1} ∩ N/O_{2}(⟨G_{1} ∩ N, L_{2}⟩).

Now we are going to choose H_{1} ≤ L_{1} minimal with respect to H_{1} ≰ N and S ≤ H_{1}. If σ(H_{1}) = ∅, we replace L_{1} by ⟨H_{1} ∩ N, L_{2}⟩ and L_{2} by H_{1}. Hence we just interchange the roles of L_{1} and L_{2}.

So from now on assume that we are not able to choose H_{1} with σ(H_{1}) = ∅.

(4.11) Lemma. Let L_{2} be solvable with σ(L_{2}) ≠ ∅. Suppose there is no H_{1} ≤ G_{1}, H_{1} ≰ N, S ≤ H_{1}, σ(H_{1}) = ∅. Then if H_{1} is minimal with respect to H_{1} ≰ N, S ≤ H_{1} we have either

(i) H_{1} is a {2, p}-group, m_{p}(H_{1}) ≥ 4, or

(ii) E(H_{1}/O_{2}(H_{1})) = K_{1} × … × K_{t}, K_{i} are Lie groups over fields of odd characteristic, not U_{3}(3), P Sp_{4}(3), ^{2}G_{2}(3) or 3 · U_{4}(3). Furthermore S acts transitively on {K_{1}, …, K_{t}}.

Proof . Choose notation as in (4.8). Let X_{1} be a quasisimple component of G_{1}/O_{2}(G_{1}). Suppose X_{1} ≰ (G_{1} ∩ N)/O_{2}(G_{1}).

Let first X_{1} ≅ A_{n} or 3A_{7}, n ≥ 7. As m_{p}(⟨X_{1}^{G_{1}}⟩) ≥ 4 we have p = 3. There is a subgroup U_{1} of X_{1}, U_{1}/O_{2}(U_{1}) ≅ Σ_{⌊ \frac{n}{2} ⌋}. We may assume U_{1} ≤ N, otherwise we argue by induction. Hence m_{3}(Σ_{⌊ \frac{n}{2} ⌋}) = 1. In particular ⌊\frac{n}{2}⌋ ≤ 5 and so n ≤ 11. If n is odd there is A_{n−1} ≤ X_{1}. Hence we may assume n to be even. Now n = 8 or 10. As Σ_{10} contains ZZ_{2} × Σ_{8}, which has the same Sylow 2-subgroup, it is enough to treat X_{1} ≅ A_{8}. But Σ_{8} is generated by S-invariant {2, 3}-groups, the assertion.

Let X_{1} be a sporadic group. By (1.4) p = 3. Let X_{1} ≅ M̸_{11}. Then X_{1} is generated by minimal parabolics P and in case X_{1} ≅ J_{1} we have X_{1} is generated by N_{X_{1}}(S ∩ X_{1}) and C I_{X_{1}}(t) for some involution t ∈ S ∩ X_{1}.

Suppose X_{1} is normalized by S. Then we see from [RoStr] that 1 ≤ m_{3}(P) ≤ 3, for any such P, a contradiction. As m_{3}(M_{11}) = 2, we get X_{1} is not normalized by S if X_{1} ≅ M_{11}.

So assume that [X_{1}, S] ≰ X_{1}. Now as in the A_{n}-case we may assume that there is no solvable parabolic. As GL_{2}(3) ≤ M_{11}, [RoStr] provides us with a contradiction.

So let X_{1} be a Lie group G(q), q = 2^{n}. As X_{1} is generated by minimal parabolics we may assume Ĥ_{1} ≤ G_{1}, Ĥ_{1} ≰ N, S ≤ Ĥ_{1} with:

F^{∗}(Ĥ_{1}/O_{2}(Ĥ_{1})) ≅ K_{1} × … × K_{t}, K_{1} ≅ K_{2} ≅ … ≅ K_{t}, m_{p}(Ĥ_{1}) ≥ 4 .

K_{i} ≅ L_{2}(r), U_{3}(r), Sz(r), L_{3}(r), Sp_{4}(r), r even.

If K_{i} ≅ L_{2}(q), Sz(q), then t ≥ 4, as m_{p}(Ĥ_{1}) ≥ 4. In the remaining cases t ≥ 2. Let r > 2. Then we get that m_{p}(C) ≥ 2, where C is a Cartan subgroup of E(Ĥ_{1}/O_{2}(Ĥ_{1})). Hence C ≰ N and as C is solvable we get the assertion.

If r = 2, then either Ĥ_{1} is solvable or K_{1} ≅ L_{3}(2) ≅ L_{2}(7) or K_{1} ≅ Sp_{4}(2)′ ≅ L_{2}(9), which are Lie groups in odd characteristic, the assertion.

As U_{3}(3) ≅ G_{2}(2)′, P Sp_{4}(3) ≅ Ω_{6}^{−}(2) and ^{2}G_{2}(3)′ ≅ L_{2}(8), it just remains to handle X_{1} ≅ 3 · U_{4}(3). Hence p = 3. But 3 · U_{4}(3) is generated by subgroups T with T/O_{2}(T) ≅ Σ_{3}. Now we get that H_{1} is solvable as before. So we have shown that:

The assertion holds or any quasisimple component of G_{1}/O_{2}(G_{1}) is covered by N.

Now we consider Y_{1}. Suppose G_{1} ∩ N/O_{2}(G_{1}) ≱ Y_{1}. In any case S normalizes Y_{1}. But m_{p}(Y_{1}) ≤ 3 for any prime p, a contradiction. So we are left with U. But U is p-constrained and so we are done if O_{p}(U)S ≰ N. Suppose O_{p}(U)S ≤ N. Then m_{p}(O_{p}(U)) ≤ 1. But m_{p}(X_{1} ∗ … ∗ X_{t} ∗ Y_{1} ∗ … ∗ Y_{f}) ≤ 1, as X_{1}∗…∗X_{t}∗Y_{1}∗…∗Y_{f} is covered by N. This is impossible as m_{p}(G_{1}) ≥ 4. The lemma is proved.

Now we choose H_{1} according to (4.11) and set L_{1} = H_{1}. In particular we have symmetry between L_{1} and L_{2}.

(4.12) Lemma. Let L_{1} be as in (4.9). Suppose there is some ω ∈ L_{1}, o(ω) = p^{a}, ω induces a field automorphism on some of the X_{i} or 1 ≠ [ω, Y_{i}] ≤ O_{r}(Y_{i}) for some i. Then one of the following holds

(1) There is some s ∈ S, X_{1}^{s} = X_{2}, ω induces a field automorphism on both of X_{1} and X_{2}, ω^{s} = ω^{−1}, and X_{1} ≅ X_{2} ≅ L_{2}(q) or Sz(q), q even

(2) X ≅ X_{1} ×X_{2} or X_{1} ×X_{2} ×X_{3}, X < L_{1}/O_{2}(L_{1}), X_{1} ≅ X_{2} ≅ X_{3} ≅ L_{2}(q) or Sz(q), C I_{L_{1}/O_{2}(L_{1})}(X) ≤ (L_{1} ∩ L_{2})/O_{2}(L_{1}).

Proof . Suppose first that ω induces a field automorphism on X_{1} (notations are as in (4.9)). We recall the construction given in (4.4). This shows that X_{1} is a component of (H/K)/F (H/K). Furthermore by (1.68) we see that there is some g ∈ M with X_{1}^{g} ≠ X_{1} and ω induces a field automorphism on X_{1}^{g}. If m_{p}(F^{∗}(H/K)) ≥ 4, the construction given in (4.4) shows that we may assume ω ∈ G_{1} ∩ G_{2}. But as m_{p}(C I_{M}(ω)) ≥ 3, we see with (1.66) and the construction of L_{1} in (4.5) that E(G_{2}/O_{2}(G_{2})) ≅ L_{3}(r) or U_{3}(r) or E((G_{2}/O_{2}(G_{2}))/F (G_{2}/O_{2}(G_{2}))) ≅ L_{2}(r) and F (G_{2}/O_{2}(G_{2})) is an elementary abelian group of order r^{3}. As N_{G}(⟨ω⟩) ≤ M and m_{p}(G_{1}∩G_{2}) ≤ 1, we see that E(G_{2}/O_{2}(G_{2})) ≅ L_{3}(r) or U_{3}(r) and ωO_{2}(G_{2}) ∈ (G_{1} ∩ G_{2}/O_{2}(G_{2}))^{(∞)}. But this is impossible as ω has to induce a field automorphism on X_{1}. Hence m_{p}(F^{∗}(H/K)) ≤ 3. If there are just two conjugates of X_{1} in E(H/K) we have (1), as m_{x}(X_{1}) = 1 for any odd prime x and (1.3). If |X_{1}^{M}| = 3, we have F^{∗}(H/K) = X_{1} × X_{2} × X_{3}, and we get (2).

That 1 ≠ [Y_{i}, ω] ≤ O_{r}(Y_{i}) is not possible has been shown in (4.8).

Let Z_{1} = ⟨Ω_{1}(Z(S))^{L_{1}}⟩ and Z_{2} = ⟨Ω_{1}(Z(S))^{L_{2}}⟩, Q_{i} = O_{2}(Li), i = 1, 2. Because of C_{L_{i}}(Q_{i}) ≤ Q_{i}, we have Z_{i} ≤ Q_{i}, i = 1, 2.

There are a few special cases to be treated in (7.3) and chapter 8, which we will investigate now.

(4.13) Lemma. Suppose Z_{1} ≤ Z_{2}, L_{1} = C I_{L_{1}}(Z_{1})(L_{1} ∩ L_{2}) and L_{1} is not of type (ii) in (4.9). Let Z̃_{2} = C I_{Z_{2}/Z_{1}}(S) mod Z_{1}. Assume Z̃_{2} ≤ O_{2}(L_{1}). Set V_{1} = ⟨Z̃_{2}^{L_{1}}⟩ and suppose O^{2′}(C I_{L_{1}}(V_{1})) ≰ O_{2}(L_{1}). Let (L_{2}, Z_{2}) be one of the following:

(i) E(L_{2}/Q_{2}) ≅ L_{2}(q), q even, or E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ L_{2}(4) with [F (L_{2}/Q_{2}), Z_{2}] = 1. Further assume that Z_{2} is the natural module, a direct sum of two natural modules or the natural O^{−}(4, t)-module t = 2, or q = t^{2}.

(ii) E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q even, Z_{2} the natural O^{+}(4, q)-module. (iii) E(L_{2}/Q_{2}) ≅ SU_{3}(q), q even, Z_{2} the natural module. (iv) E(L_{2}/Q_{2}) ≅ Sz(q), Z_{2} the natural module, Z̃_{2} ≮L_{1}, or Z̃_{2} < L_{1} and we

replace Z̃_{2} by C I_{Z_{2}/Z̃_{2}}(S) mod Z̃_{2}, m_{p}(C I_{L_{1}}(Z̃_{2})) ≤ 1.

(v) L_{2} is solvable and one of the following holds (α) |Z_{2}| = 4, L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} (β) |Z_{2}| = 16 and one of the following holds

(a) L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{10} or F_{20}, m_{p}(C I_{L_{1}}(Z̃_{2})) ≤ 1.

(b) L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3}
(c) L_{2}/C I_{L_{2}}(Z_{2}) ≅ E_{9}ZZ_{4}
(d) L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}

(γ) |Z_{2}| = 64 and O_{3}(L_{2}/C I_{L_{2}}(Z_{2})) ≅ Z_{3^{1+2}} (vi) E(L_{2}/Q_{2}) ≅ A_{9}, Z_{2} is the natural module. Then there is L̂_{1} ≤ L_{1}, L_{2} ∩ L̂_{1} = L_{2} ∩ L_{1} with

(1) m_{p}(L̂_{1}) ≥ 2

(2) If V̂_{1} = ⟨Z̃ L̂_{1} ; 2⟩, then O^{2′}(C IL̂_{1}(V̂_{1})) ≤ O_{2}(L̂_{1})

(3) L̂_{1} is of the same general shape as G_{1} in (4.9)

(4) Suppose m_{p}(L̂_{1}) = 2. If ω ∈ L̂_{1}, o(ω) = p, there is ν ∈ M, o(ν) = p, [ν, V̂_{1}] = 1 and for E = ⟨ω, ν⟩ we have Γ_{E,1}(G) ≤ M.

Proof . We choose L_{1}^{(1)} ≤ L_{1} minimal with respect to (1) L_{1}^{(1)} is of the same general shape as L_{1} in (4.9).

(2) m_{p}(L_{1}^{(1)}) ≥ 3.

(3) L_{1} ∩ L_{2} ≤ L_{1}^{(1)}. Hence we may assume L_{1}^{(1)} = L_{1}. Set U_{1} = X_{1} × … × X_{t}, W_{1} =

Y_{1} × … × Y_{f} (notation as in (4.9)). Let I ⊆ {1, …, t}, [X_{i}, V_{1}] ≠ 1 for i ∈ I, J ⊆ {1, …, f}, [Y_{j}, V_{1}] ≠ 1, j ∈ J. If m_{p}(∏_{i∈I} X_{i} × ∏_{j∈J} Y_{j}) ≥ 2, then let L̂_{1}^{(1)} be the preimage of (∏_{i∈I} X_{i} × ∏_{j∈J} Y_{j})(L_{1} ∩ L_{2}). Now O^{2′}(C I_{L_{1}}(V_{1})O_{2}(L_{1})/O_{2}(L_{1})) ∩ (∏_{i∈I} X_{i} × ∏_{j∈J} Y_{j}) = 1. If O^{2′}(C IL̂_{1}(V̂_{1})) ≰ O_{2}(L̂_{1}), then there is E ≅ E_{p^{2}} in L_{1}, Γ_{E,1}(G) ≤ M such that [Z̃_{2}, E] = 1. But in all cases except L_{2}/Q_{2} ≅ Sz(q), we have L_{2} = ⟨C I_{L_{2}}(x)|x ∈ Z̃_{2}^{♯}⟩. So we are left with L_{2}/Q_{2} ≅ Sz(q). Now by assumption we have Z̃_{2} ≮L_{1} and m_{p}(C I_{L_{2}}(Z̃_{2})) ≤ 1, a contradiction. Obviously (4) is satisfied, so we have found L̂_{1}.

Let us now assume m_{p}(U_{1} × W_{1}) ≤ 1. If U_{1} × W_{1} = 1, then m_{p}(X_{1} × … × X_{t} × Y_{1} × … × Y_{f}) ≤ 2. In case of equality we have some E ≅ E_{p^{2}} ≤ X_{1} × … × X_{t} × Y_{1} × … × Y_{f} with Γ_{E,1}(G) ≤ M, a contradiction as before. So we have m_{p}(X_{1} × … X_{t} × Y_{1} × … × Y_{f}) ≤ 1. By (4.12) we see that [O^{p′}(U), X_{1} × … X_{t} × Y_{1} × … × Y_{f}] = 1, and so m_{p}(O_{p}(U)) ≥ 2.

Suppose m_{p}(U_{1} × W_{1}) = 1. Then m_{p}(X_{1} × … X_{t} × Y_{1} × … × Y_{f}) ≤ 2. Suppose that U contains some ω which induces a field automorphism on one of the components. Then by (4.12) C I_{L_{1}/Q_{1}}(X_{1} × … × X_{t} × Y_{1} × … × Y_{f}) ≤ X_{1} × … X_{t} × Y_{1} × … × Y_{f}. Now we may choose X_{1} = U_{1} × W_{1} and X_{2} ≠ X_{1}. Let L̂_{1} = ⟨(X_{2} mod Q_{1}), ω, L_{1} ∩ L_{2}⟩. Then (1) - (4) holds. Hence in any case we may assume [O_{p}(U), X_{1} × … X_{t} × Y_{1} × … × Y_{f}] = 1.

Suppose O_{p}(U) contains an elementary abelian characteristic subgroup C of order ≥ p^{3}. Then by choice of L_{1} we have L_{1}/O_{2}(L_{1}) = C · (L_{1} ∩ L_{2})/O_{2}(L_{1}). Now as before |C I_{C}(Z̃_{2})| ≤ p. We have [C, Q_{2}] ≤ C I_{C}(Z̃_{2}). Let D = C I_{C}(Q_{2}). Suppose D ≠ C. Let L̂_{1} be the preimage of D · (L_{1} ∩ L_{2})/O_{2}(L_{1}). Obviously L̂_{1} satisfies (1) - (4).

Assume D = C. Then Q_{2} ≤ Q_{1}. Hence Z_{2} is an F -module with offending subgroup in Q_{1}. As Q_{2} ≠ Q_{1}, (vi) is not possible. Hence with (1.31) we see that Z_{2} is the natural L_{2}(q)-module, or L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}. In the former we get a contradiction with (4.2). So assume the latter. Let now D be an S-invariant complement of C I_{C}(Z̃_{2}). Let L̂_{1} be the preimage of D · S/O_{2}(L_{1}). Then L̂_{1} satisfies (1) - (4).

Let now C be a characteristic subgroup of O_{p}(U), C ≅ E_{p^{2}}. Then m_{p}(C I_{L_{1}}(C)) ≥ 3. If C I_{C}(Z̃_{2}) = 1, we let L̂_{1} be the preimage of C(L_{1} ∩ L_{2})/O_{2}(L_{1}). Then (1) - (4) holds. (Recall Q_{2} ≤ Q̂_{1} and so Z_{2} is an F - module, hence p | |̸L_{1} ∩ L_{2}|).

So assume |C I_{C}(Z̃_{2})| = p. If U_{1} × W_{1} ≠ 1, set D = C I_{C}(Q_{2}). Suppose D ≠ C. Let L̂_{1} be a preimage of (U_{1} × W_{1} × D) (L_{1} ∩ L_{2})/O_{2}(L_{1}). Then (1)

- (4) holds.

Assume [C, Q_{2}] = 1. Now C ≤ N_{M}(J(Q_{2}Q_{1})) and so we have that Z_{2} is an F -module with offending subgroup in Q_{2}Q_{1}. Now as before L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}. Let D be an S-invariant complement of C I_{C}(Z̃_{2}) in C, and L̂_{1} be a preimage of (U_{1} × W_{1} × D)S/O_{2}(L_{1}). Again (1) - (4) holds.

We may assume U = L_{1}/O_{2}(L_{1}). Suppose Ω_{1}(O_{p}(U)) = C. Now C I_{U}(C) ≤ O_{p}(U) and we see U = O_{p}(U)(L_{1} ∩ L_{2})/O_{2}(L_{1}), as any p-element stabilizes 1 < C I_{C}(Z̃_{2}) < C. But now there is ω ∈ L_{1} ∩ L_{2}, o(ω) = p with [ω, C] = 1. This contradicts N_{G}(⟨ω⟩) ≰ M by (1.66).

We have m_{p}(O_{p}(U)) ≥ 3 and so U = O_{p}(U)(L_{1} ∩ L_{2})/O_{2}(L_{1}) again. Let [Q_{2}, O_{p}(U)] ≠ 1. Then m_{p}([Q_{2}, O_{p}(U)]) = 1. Hence m_{p}(C I_{O_{p}(U)}(Q_{2})) ≥ 2. The same holds if [Q_{2}, O_{p}(U)] = 1. Hence in any case N_{G}(J(Q_{1}Q_{2})) contains E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M. This shows that Z_{2} is an F -module with offending subgroup in Q_{1}Q_{2}. Now as before this implies L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}. Then L_{1} ∩ L_{2} = S. Let T = C I_{S}(Z̃_{2}) and let L̂_{1} be the preimage of C I_{O_{p}(U)}(T)S. As m_{p}(C I_{O_{p}(U)}(T)) ≥ 2, we have (1) - (4).

So we are left with O_{p}(U) is extraspecial, Ω_{1}(O_{p}(U)) = O_{p}(U). Furthermore we may assume C I_{O_{p}(U)}(Z̃_{2}) ≰ Φ(O_{p}(U)). As C I_{O_{p}(U)}(V_{1})< L_{1}/O_{2}(L_{1}), we now have m_{p}(C I_{O_{p}(U)}(V_{1})) = 2. We know that for E ≤ C I_{O_{p}(U)}(V_{1}), m_{p}(E) = 2, Γ_{E,1}(G) ≰ M. This implies U_{1} × W_{1} = 1 and |O_{p}(U)| = p^{3} and U/O_{p}(U) ≲ GL_{2}(p). As m_{p}(L_{1}) = 3, we see that p | |U : O_{p}(U)|. But now again there is some E ≤ O_{p}(U), E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, as E ≤ F, F ≅ E_{p^{3}}. This is a contradiction, proving the lemma.

Hence if we are in the situation of (4.13) we replace L_{1} by L̂_{1}.

(4.14) Lemma. O_{2}(⟨L_{1}, L_{2}⟩) = 1.

Proof . If L_{1} is as in (4.9)(ii) the assertion follows with (4.10) and L_{2} ≰ M. Otherwise m_{p}(L_{1}) ≥ 2 and L_{1} contains an elementary abelian group E of order p^{2} with Γ_{E,1}(G) ≤ M. Hence ⟨L_{1}, L_{2}⟩ ≤ M if O_{2}(⟨L_{1}, L_{2}⟩) ≠ 1, a contradiction.

Now for the remainder of this paper let us fix some notations:

Let Γ = Γ(L_{1}, L_{2}) be the coset graph. Furthermore let

b_{i} = min{d(i, α)|α ∈ Γ such that Z_{i} ≤ L_{α} but Z_{i} ≰ L_{β} for some β ∈ △(α)}, i = 1, 2, where △(α) denotes the set of neighbours of α in Γ. Let

b = min{b_{1}, b_{2}}.

A critical pair (i, α) is a pair of vertices in Γ such that b = b_{i}, d(i, α) = b and Z_{i} ≰ L_{β} for some β ∈ △(α).

We recall some basic facts about Γ, which might be found in [Gol].

(1) Γ is connected.

(2) L = ⟨L_{1}, L_{2}⟩ acts edge transitively on Γ.

(3) Let α ∈ Γ, then L_{α} is either conjugate to L_{1} or to L_{2} (this will be denoted by α ∼ 1 or α ∼ 2).

(4) Let B = L_{1} ∩ L_{2}, U ≤ B such that L_{i} = N_{L_{i}}(U)B, i = 1, 2, then U = 1.

We fix the following notation:

(a) There are vertices 1, 2 ∈ Γ whose stabilizers in L which are L_{1} and L_{2}, respectively.

(b) Let N_{i} ≤ G_{i}, i = 1, 2, α ∈ Γ and g ∈ L with i^{g} = α. Then set N_{α} = N_{i}^{g}. As g is uniquely determined up to h ∈ G_{i}, we have that N_{α} is well defined.

There is some unsymmetry between L_{1} and L_{2}. While the structure of L_{2} is restricted, all we know about L_{1} is some information about p-elements. The proof of the theorem is devided into several steps.

(i) We show that the exceptional Sz(q)–situation from (4.4) does not occur (§5)

(ii) We prove that Ω_{1}(Z(S)) < L_{1}. ((5.1),§6)

(iii) We show that Ω_{1}(Z(S)) < L_{2}. (§5,7,8)

Obviously (i) and (ii) yield a contradiction to (4).

In both cases (i) and (ii) there is a basic subdivision b > 2 and b ≤ 2, where arguments are different. The so called amalgam method, i.e. investigation of Γ, just works for b > 2.

§ 5 Action of L_{1}/Q_{1} and L_{2}/Q_{2} on some modules

(5.1) Lemma Let V ≤ Z_{1} be some module for L_{1}/Q_{1}. Let K be a quasisimple component of L_{1}/Q_{1}. Suppose V is an F -module for K. Then there is a submodule U of V, [U, K] = U, such that any x ∈ U^{♯} is centralized by some E_{x} ≅ E_{p^{2}} with Γ_{E_{x},1}(G) ≤ M. In particular C I_{G}(x) ≤ M, or

(i) K ≅ L_{2}(q), q even, U = [V, K] and U/C I_{U}(K) is the natural module. Furthermore C I_{U}(K) ≠ 1 and U is centralized by ⟨ρ⟩ with o(ρ) = p and N_{G}(⟨ρ⟩) ≤ M.

(ii) K ≅ U_{4}(2), U = [V, K] and U/C I_{U}(K) is the natural module. Furthermore |C I_{U}(K)| ≤ 4 and every x ∈ U^{♯} is centralized by some ρ ∈ L_{1} with o(ρ) = 3 = p.

Proof. By (1.18) K/Z(K) ≅ A_{n}, L_{n}(q), Sp_{2n}(q), Ω_{2n}^{±}(q), U_{n}(q) or G_{2}(q), q even. Further by (4.4) we may assume that m_{p}(G_{1}) ≥ 4.

(1) K/Z(K) ≅ A_{n}

Let U be a submodule with [U, K] = U. Suppose that U/C I_{U}(K) is the permutation module. If p = 3 and n ≥ 8, then every x ∈ U is centralized by E_{x} ≅ E_{9} in K.

So assume next p ≠ 3. Then n ≤ 11. Furthermore U = [V, K]. We have m_{p}(K) ≤ 2. Hence by (4.4) some E ≅ E_{p^{2}} centralizes U, the assertion. This also applies for n ≤ 7, p = 3.

Hence by (1.18) we may assume that n = 8 or 7 and |U/C I_{U}(K)| = 16, n = 6 and |U/C I_{U}(K)| = 2^{6}, or n = 5 and U/C I_{U}(K) is the L_{2}(4)-module. If n ≠ 8, then U = [V, K] and we can argue as above.

So we are left with n = 8, U ≠ [V, K]. As V is an F -module, we have that [V, K] has at least three nontrivial composition factors. Hence for p > 3, we get with (4.4) some E ≅ E_{p^{2}} with [E, [V, K]] = 1, the assertion. Let p = 3. Then by (4.4) there is some 3-element ρ with [ρ, [V, K]] = 1. But every x ∈ U^{♯} is centralized by some 3-element in K, the assertion.

(2) K/Z(K) ≅ L_{n}(q), q even

Suppose U/C I_{U}(K) is the natural module. Then every x ∈ U^{♯} is centralized by L_{n−1}(q). So we may assume that there is no E ≅ E_{p^{2}}, E ≤ L_{n−1}(q) with Γ_{E,1}(G) ≤ M. This shows K/Z(K) ≅ L_{2}(q), L_{3}(q), L_{4}(q), L_{5}(2), L_{6}(2) or L_{7}(2).

Suppose next p | q^{2} − 1. Then we have K/Z(K) ≅ L_{n}(q), n ≤ 4. Furthermore m_{p}(K) ≤ 2. If m_{p}(C I(K)) ≥ 3, we get U with [E, U] = 1 for some E ≅ E_{p^{2}} with m_{p}(C I(E)) ≥ 3.

So we may assume m_{p}(C I(K)) ≤ 2. Hence by (4.12) either K ≅ L_{2}(q) and field automorphisms are involved, or m_{p}(K) = 2. In the latter U is centralized by ρ with o(ρ) = p. Now for every x ∈ U^{♯}, p | |C I_{U}(x)| as p | q^{2} − 1 and so we get E_{x} ≅ E_{p^{2}}, [E_{x}, x] = 1 and Γ_{E_{x},1}(G) ≤ M.

Suppose K ≅ L_{2}(q). Then just one module is involved. Now [U, ρ] = 1, for some ρ ∈ C I(U). This is (i).

So assume now p | q̸^{2} − 1. Let m_{p}(K) ≥ 2. Then we get K ≅ L_{6}(2) or L_{7}(2) and p = 7. Now p | |L_{n−1}(2)|. This shows K ≅ L_{6}(2). By the construction given in § 4 there is E_{7^{2}} ≅ E ≤ C I(K). Now we may choose U such that U is centralized by ρ ∈ E, o(ρ) = 7, and now every x ∈ U^{♯} is centralized by E_{x} ≅ E_{7^{2}} with Γ_{E_{x},1}(G) ≤ M.

We may assume that U/C I_{U}(K) is not natural. By (1.18) U/C I_{U}(K) ≅ V (λ_{2}), or K ≅ L_{2}(r^{2}), r^{2} = q, and U/C I_{U}(K) is the orthogonal Ω_{4}^{−}(q)- module. In the latter by (1.20) C I_{U}(K) = 1 and so every x ∈ U^{♯} is centralized by E_{x} ≅ E_{p^{2}} with Γ_{E_{x},1}(G) ≤ M.

Suppose now U/C I_{U}(K) ≅ V (λ_{2}). Then V is the set of forms on V^{∗}. Hence for any x ∈ U, we get C I_{K}(x) picks up L_{2}(q) × L_{n−2}(q) or Sp_{4}(q). So we may assume that p | q̸^{2} − 1. In this case n ≤ 7. Furthermore q = 2 for n > 4. Finally we have m_{p}(K) ≥ 2, otherwise we may choose U with [E, U] = 1 for some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. This shows K ≅ L_{6}(2) or L_{7}(2) and p = 7. By (1.23) we see in both cases that there are at most two composition factors which are isomorphic to V (λ_{2}). Hence some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M centralizes U.

(3) K/Z(K) ≅ Sp_{2n}(q), q even

Suppose first U/C I_{U}(K) is the natural module. Then every x ∈ U^{♯} is centralized by Sp_{2n−2}(q). Hence we have the assertion or K/Z(K) ≅ Sp_{4}(q) or Sp_{6}(q). In the latter we may assume m_{p}(K) = 1. Anyway, if m_{p}(K) = 1 by (4.12) and (4.4) there is some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and [E, U] = 1.

So let K/Z(K) ≅ Sp_{4}(q), m_{p}(K) = 2. Now by (4.12) and (4.4) there is ρ ∈ C I(K), o(ρ) = p, and we may assume [ρ, U] = 1. As any x ∈ U^{♯} is centralized by L_{2}(q) in K and p | q^{2} −1 (as m_{p}(K) = 2), we get the assertion.

Hence by (1.18) we may assume K ≅ Sp_{6}(q) and U/C I_{U}(K) is the spin module. Furthermore we may assume that m_{p}(K) ≥ 2. Hence p | q^{2} − 1.

As elements in U are centralized by L_{3}(q) or G_{2}(q), we see that p | q̸ − 1.

Hence p | q +1. As m_{p}(C I(K)) ≥ 1 by (4.12) we get that there are at least two composition factors which are spin modules. Hence |A| ≥ q^{4} by (1.23), A an offending subgroup. Furthermore |A| ≠ q^{6}, as |[U, t]| ≥ q^{4} for a transvection t (in the natural representation). Now |U : C I_{U}(A)| = q^{2} and so by (1.23) again C I_{U}(A) = C I_{U}(a) for every a ∈ A^{♯}. But ⟨C I_{K}(a) | a ∈ A^{♯}⟩ = K, a contradiction.

(4) K/Z(K) ≅ Ω_{2n}^{±}(q), n ≥ 4, q even

Let U/C I_{U}(K) be the natural module. Then every x ∈ U^{♯} is centralized by Ω_{2n−2}^{±}(q). Hence we may assume m_{p}(Ω_{2n−2}^{±}(q)) ≤ 1. This shows K ≅ Ω_{8}^{−}(q) and m_{p}(K) = 1. Now by (4.12) we may assume that there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, and [E, U] = 1. Now by (1.18) we may assume that K ≅ Ω_{10}^{+}(q) and U/C I_{U}(K) is the spinmodule. In this module every element is centralized by L_{4}(q) or Sp_{6}(q). As p | q^{2} − 1, we get the assertion.

(5) K/Z(K) ≅ U_{n}(q), n ≥ 4, q even

If U/C I_{U}(K) is the natural module, then as every x ∈ U^{♯} is centralized by U_{n−1}(q) or U_{n−2}(q), we get n ≤ 5. Furthermore in case n = 5 we have q = 4. Let first n = 5. Then p ≠ 5 as m_{5}(U_{3}(4)) = 2 = m_{5}(U_{4}(4)). Hence m_{p}(K) ≤ 2. If m_{p}(K) = 2, then p = 3. Now there is ρ ∈ C I(K), [⟨ρ⟩, U] = 1. As 3 | |U_{4}(4)| and |U_{3}(4)|, we have the assertion.

So let m_{p}(K) = 1. Then we may assume that there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and [U, E] = 1.

Assume n = 4. If m_{p}(K) = 1, we may argue as before. Hence m_{p}(K) ≥ 2. Now p | |U_{3}(q)| and p | |U_{2}(q)|. The same argument as before shows m_{p}(K) ≥

3, i.e. p | q + 1. By (1.20) C I_{U}(K) = 1 or q = 2 and |C I_{U}(K)| ≤ 4. The latter just is (ii). In the former by (1.45), we see that U is invariant under L_{1}. But now every x ∈ U^{♯} is centralized by E_{x} ≅ E_{p^{2}} with Γ_{E_{x},1}(G) ≤ M.

So assume that n = 4, U_{n}(q) ≅ Ω_{6}^{−}(q) and U/C I_{U}(K) is the natural orthogonal module. Again by (1.20) C I_{U}(K) = 1. Furthermore by (1.45) U ◁ G_{1}. Hence p | |̸L_{2}(q^{2})| or p | |̸Sp_{4}(q)|. In particular m_{p}(K) = 1. But now by (4.12) we may assume that there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and [U, E] = 1.

(6) K/Z(K) ≅ G_{2}(q), q even.

By (1.18) U/C I_{U}(K) is the natural 6-dimensional module. As before we may assume m_{p}(K) = 2, i.e. p | q^{2} − 1. Now every x ∈ U^{♯} is centralized by some p-element in K. By (4.12) we may assume there is some ρ ∈ C I(K), o(ρ) = p and [ρ, U] = 1, the assertion.

(5.2) Hypothesis. (a) Z_{1} ≤ Z_{2}, m_{p}(C I_{L_{1}}(Z_{1})) ≥ 3.

(b) b = b_{2} > 2.

Set V_{1} = ⟨Z_{2}^{G_{1}}⟩. We have V_{1} ≤ Q_{1} if b > 2.

(5.3) Lemma. Assume (5.2). Let J(S) ≰ Q_{1} and J(S) ≤ Q_{2}. There is U_{1} ≤ L_{1}, S ≤ U_{1} such that for Q̃_{1} = O_{2}(U_{1}), we either have that E(U_{1}/Q̃_{1}), is a direct product L^{(1)}×…×L^{(n)}, L^{(1)} ≅ L^{(2)} ≅ … ≅ Ł^{(n)}, L^{(1)} ≅ L_{2}(r), U_{3}(r), Sz(r), ^{2}F_{4}(r), r even, sporadic, a group of Lie type in odd characteristic, a solvable {2, r}-group, or L_{1} is isomorphic to r^{3}L_{2}(r), r an odd prime. Furthermore J(S) ≰ Q̃_{1} and Z_{2} ⋪ U_{1}.

Proof. Let L/Q_{1} be a component of L_{1}/Q_{1} with [J(S), L] ⊈ Q_{1}. Suppose L/Q_{1} is alternating or a group of Lie type G(r) ≅ ^{2̸}F_{4}(r) in even characteristic. Then there is some minimal parabolic P such that J(S) ≰ O_{2}(⟨P^{J(S)}⟩J(S)). If P ≰ N_{L_{1}}(Z_{2}), we choose P as L^{(1)}. So assume P ≤ N_{L_{1}}(Z_{2}) for every P with J(S) ≰ O_{2}(⟨P^{J(S)}⟩J(S)). Then [O^{2}(P), Z_{2}] = 1. Let p ∈ σ(M) with p | |P |. By the Frattini argument we have m_{p}(C I_{G}(Z_{2})) ≤ 1. Hence we see that for some x ∈ P with o(x) = p we get N_{G}(Z_{2}) = C I_{G}(Z_{2})N_{G}(⟨x⟩). As N_{G}(Z_{2}) ≰ M, we have N_{G}(⟨x⟩) ≰ M and so m_{p}(C I_{M}(x)) ≤ 2 for all x ∈ P with o(x) = p. By the construction in chapter 4 we may assume m_{p}(L) ≥ 4. But then we easily see m_{p}(C I_{M}(x)) > 2 for some x ∈ P, with o(x) = p ∈ σ(M). So we have p | |̸P |. In particular

p ≠ 3 and p | r̸^{2} − 1 if L/Q_{1} ≅ G(r). This implies m_{p}(L) ≤ 2.

Suppose first m_{p}(L) = 2. Then we see L ≤ L_{1}/Q_{1} and L/Q_{1} ≅ A_{10}, A_{11}, L_{6}(2), L_{7}(2). In all cases p | |C I_{L_{1}/Q_{1}}(L/Q_{1})|. We first suppose [J(S)Q_{1}/Q_{1}, C I_{L_{1}/Q_{1}}(L/Q_{1})] ≠ 1. As m_{3}(L) = 3, p ≠ 3, we have that C I_{L_{1}/Q_{1}}(L/Q_{1}) is solvable or has components of type Sz(r). Now choose L^{(1)} as such a component. Suppose now [J(S)Q_{1}/Q_{1}, C I_{L_{1}/Q_{1}}(L/Q_{1})] = 1. Let L/Q_{1} ≅ A_{10}, A_{11}. Then p = 5 and there is some parabolic P in L with P/O_{2}(P) ≅ S_{5} and J(S) ≤ O_{2}(⟨P^{J(S)}⟩J(S)). Now J(S) is normalized by E ≅ E_{25} with Γ_{E,1}(G) ≤ M, a contradiction to L_{2} ≰ M.

Let next L/Q_{1} ≅ L_{6}(2) or L_{7}(2). Then p = 7 . Now there is more than one parabolic P with J(S) ≰ O_{2}(⟨P^{J(S)}⟩J(S)) and so we get Σ_{3} × Σ_{3} which is involved in C I(Z_{2}). Call this group U_{2}. We are going to prove a similar result for m_{p}(L) = 1.

Let now m_{p}(L) = 1. If m_{r}(L) ≤ 1 for all primes r, we may choose L as L^{(1)}. So we may assume that there is some prime r with m_{r}(L) > 1 and so L ≤ L_{1}/Q_{1}. Then [J(S)Q_{1}/Q_{1}, C I_{L_{1}/Q_{1}}(L/Q_{1})] ≠ 1. Now we may assume that there is a subgroup L̃/Q_{1} of C I_{L_{1}/Q_{1}}(L/Q_{1}) with [J(S), L̃] ≰ Q_{1} too. We may assume that L̃ is alternating or of type G(r̃). Finally as we may interchange the rôles of L and L̃ we have m_{p}(L̃) = 1 too. As 3 ∉ σ(M), we may assume m_{3}(L) = 1. Hence L ≅ (S)L_{3}(r). Now m_{3}(L̃) ≤ 2. Then L̃ ≅ L_{3}(r̃), Sp_{4}(r̃), G_{2}(r̃), ^{3}D_{4}(r̃), L_{4}(r̃), U_{4}(r̃), U_{5}(r̃), L_{5}(r̃), or A_{7} by (1.4). Now we get L_{2}(r)×L_{2}(r̃) or L_{2}(r)×U_{3}(r̃) ≲ C I(Z_{2}). Hence we have that there is U_{2} ≤ L_{1}, [U_{2}, Z_{2}] = 1, U_{2}/O_{2}(U_{2}) ≅ K_{1} × K_{2}, K_{1} ≅ L_{2}(r), K_{2} ≅ L_{2}(r̃) or SU_{3}(r̃) including r̃ = 2, r = 2. For each ω ∈ K_{i}, o(ω) prime, o(ω) > 2, we have that ω is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. As p > 3 and C I_{O_{2}(M)}(ω) ≠ 1, we get with (1.63) N_{G}(⟨ω⟩) ≤ M.

But we have U_{2} ≤ M_{3}, M_{3} ≠ M_{1}. Now by construction L/O_{p′}(L) was a component of M/O_{p′}(M). Let L̂ be the corresponding group in M_{3}. So assume first 1 ≠ [L̂O_{p′}(M_{3})/O_{p′}(M_{3}), U_{2}] ≤ L̂O_{p′}(M_{3})/O_{p′}(M_{3}). Then as L ≅ L_{6}(2), L_{7}(2) or L_{3}(r), we see C I_{U_{2}}(L̂O_{p′}(M_{3})/O_{p′}(M_{3})) ≰ O_{2}(U_{2}) or L ≅ L_{6}(2), L_{7}(2) and (L̂O_{p′}(M_{3}) ∩ U_{2})O_{2}(U_{2}) = U_{2}. But in the latter 3 | |̸O_{p′}(M_{3})| as 3 ∉ σ(M) and so ω ∈ K_{1} is centralized by ν ∈ M_{3}, o(ν) = 7, ν ∈ E ≅ E_{p^{3}}, E ≤ M_{3}. As ν ∈ N_{G}(⟨ω⟩) ≤ M, we have a contradiction to (4.2).

So we have that C I_{U_{2}}(L̂O_{p′}(M_{3})/O_{p′}(M_{3}))O_{2}(U_{2}) ≥ K_{1} or K_{2}. Now choose ω ∈ C I_{U_{2}}(L̂O_{p′}(M_{3})/O_{p′}(M_{3})), o(ω) = 3. As m_{3}(O_{p′}(M_{3})) ≤ 3, and L/Q_{1} ≴ L_{3}(3) or Sp_{4}(3), we see that for U_{3} ∈ Syl_{3}(C I_{M_{3}}(L̂O_{p′}(M_{3})/O_{p′}(M_{3}))) we have C I_{M_{3}}(U_{3})O_{p′}(M_{3})/O_{p′}(M_{3}) ≥ L̂O_{p′}(M_{3})/O_{p′}(M_{3}). Hence there is some ν ∈ M_{3}, o(ν) = p with [ν, ω] = 1 and ν ∈ E ≅ E_{p^{3}}, contradicting (4.2).

(5.4) Lemma Assume (5.2)(a).

(a) Let J(S) ≰ Q_{1} and [J(S), Z_{1}] = 1. Then one of the following holds

(i) [⟨J(S)^{L_{1}}⟩, Z_{2} ∩ Z(O_{2}(⟨J(S)^{L_{1}}⟩))] = 1.

(ii) There is a component L_{3}(2) of L_{1}/Q_{1} which induces an indecomposable submodule U of dimension 4 in Ω_{1}(Z(O_{2}(⟨J(S)^{L_{1}}⟩))) .

(iii) There is a component A_{2^{m}} of L_{1}/Q_{1} which induces a submodule U in Ω_{1}(Z(O_{2}(⟨J(S)^{L_{1}}⟩))) which is an extension of the trivial module by the permutation module.

(b) In (a)(ii) and (a)(iii) for every 1 ≠ x ∈ Z_{2} ∩ U we have C I_{G}(x) ≤ M.

Proof. Set W = ⟨J(S)^{L_{1}}⟩ and \overline{W} = W/O_{2}(W). As [J(S), Z_{1}] = 1, we have [W, Ω_{1}(Z(S))] = 1. Hence [O(\overline{W}), Z(O_{2}(W))] = 1. Assume now that we do not have (i). Then by construction in chapter 4 we have that there are components of L_{1}/Q_{1} which act nontrivially on Ω_{1}(Z(O_{2}(W))) = V. Then V is an F -module. Let L̃ be a component of \overline{W}, such that L̃ induces an F - module on V, which exists by [GoLyS, (26.24)] Then by (1.18), L̃ ≅ SL_{n}(r), SU_{n}(r), Sp_{2n}(r), Ω_{2n}^{±}(r),G_{2}(r), r even, or A_{n}. Now we have some submodule U such that U is a nonsplit extension of a trivial module by an irreducible F -module for L̃. Furthermore C I_{U}(S ∩ N_{L_{1}}(L̃)) = C I_{U}(L̃). Application of (1.21) shows L̃ ≅ L_{3}(2) or A_{2^{m}}.

In case L̃ = L_{3}(2) we have that U is the indecomposable 4-dimensional module.

If m_{p}(L̃) = 1, there is E ≅ E_{p^{2}} with [E, L̃] = 1 and Γ_{E,1}(G) ≤ M. Now [E, U] = 1, and we have the assertion with (1.63).

So assume L̃ ≅ A_{2^{m}} and p = 3, m_{p}(L̃) ≥ 2. As every v ∈ U is centralized by E_{9} in L̃, we again get the assertion with (1.63).

(5.5) Lemma. Assume (5.2). Then one of the following holds

(i)Z_{2} is an F –module

(ii) J(S) ≤ Q_{2} and there is U_{1} such that U_{1} is as in (5.3) or U_{1}/O_{2}(U_{1}) is a component of L_{1}/Q_{1} with m_{p}(U_{1}) ≥ 2, p ∈ σ(M). Furthermore ⟨Z_{2}^{U_{1}}⟩/Z_{1} involves at least two nontrivial irreducible U_{1}-modules

(iii) J(S) ≤ Q_{2} and there is U_{1} such that U_{1}/O_{2}(U_{1}) ≅ A_{2^{m}} is a component of L_{1}/Q_{1}. Furthermore there is U ≤ Z(Q_{1}), U is an extension of the natural U_{1}–module by a trivial module, |Z_{2} ∩ U| ≥ 8 and for every x ∈ (Z_{2} ∩ U)^{♯}, C I_{G}(x) ≤ M.

Proof. Suppose Z_{2} is not an F –module. Then J(S) ≤ Q_{2} and J(S) ≰ Q_{1}. By (5.3) we have that there is U_{1} as in (ii) or (iii) which does not normalize Z_{2}. Set W_{1} = ⟨Z_{2}^{U_{1}}⟩. Let W̃_{1} be the largest submodule in W_{1} which possesses just trivial U_{1}-composition factors. Let W ≥ W̃_{1} such that W/W̃_{1} is a nontrivial irreducible U_{1}-module. Suppose W_{1} = W Z_{2}. Set X = C I_{O_{2}(U_{1})}(W̃_{1}). Suppose first [X, W_{1}]W̃_{1} ≥ W. As X ≤ L_{2} we have W_{1} = W̃_{1}Z_{2}. Now as J(S) ≤ Q_{2}, [Z_{2}, J(S)] = 1. This implies J(S) ≤ O_{2}(U_{1}), a contradiction.

So we have [W_{1}, X] ≤ W̃_{1}. Suppose next [W, X] ≠ 1. Let W̃ be a hyperplane in W̃_{1} such that [W, X] ≰ W̃. We have C I_{W_{1}/W̃} (U_{1}) = W̃_{1}/W̃. Now C I_{W_{1}/W̃} (X) = W̃_{1}/W̃. Hence X induces an offending subgroup on Z_{2}, a contradiction.

So we have [W, X] = 1. As [O_{2}(U_{1}), U_{1}] ≤ X, we get W ≤ W̃_{1}Z(O_{2}(U_{1})). Now we apply (5.4). As [U_{1}, Z(O_{2}(U_{1}))] ≠ 1, we do not have (5.4)(i). Hence we get some component L̃ ≅ L_{3}(2) or A_{2^{m}}. But we know that Z_{2} is not an F - module. Let first L̃ ≅ L_{3}(2). As [J(S), Z_{2}] = 1, we see that |C I_{U}(J(S))| = 4, where U is as in (5.4). Now there is x ∈ S̃ \ Q_{1} with |Z_{2} : C I_{Z_{2}}(x)| = 2, a contradiction.

Let L̃ ≅ A_{2^{m}}. Then |Z_{2} ∩ U| ≥ 8, otherwise again Z_{2} is an F -module. Now the assertion follows with (5.4)

(5.6) Proposition. Assume (5.2). Then one of the following holds

(i) Z_{2} is an F -module.

(ii) E(L_{2}/Q_{2}) ≅ L_{2}(t^{2}), t even and Z_{2} is the natural O_{4}^{−}(t)-module, or F (L_{2}/Q_{2}) ≅ E_{r^{3}} and E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ L_{2}(4) and we have the orthogonal module.

(iii) E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q even, Z_{2} is the natural O_{4}^{+}(q)-module.

(iv) E(L_{2}/Q_{2}) ≅ L_{2}(q), q even, Z_{2} is a direct sum of two natural modules.

(v) E(L_{2}/Q_{2}) ≅ SU_{3}(q), q even, Z_{2} is the natural module.

(vi) E(L_{2}/Q_{2}) ≅ Sz(q), q even, Z_{2} is the natural module.

(vii) L_{2} is solvable and one of the following holds

(α) L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{10}, F_{20}, |Z_{2}| = 16.

(β) L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3}, |Z_{2}| = 16.

(γ) L_{2}/C I_{L_{2}}(Z_{2}) ≅ E_{9}ZZ_{4}, |Z_{2}| = 16.

(δ) O_{3}(L_{2}/C I_{L_{2}}(Z_{2})) ≅ 3^{1+2}, |Z_{2}| = 64.

(Notice that (α) is the solvable version of (vi), (β) the solvable version of (iv), (γ) the solvable version of (iv) and (δ) the solvable version of (v))

Proof. Suppose Z_{2} is not an F -module. Then J(S) ≤ Q_{2}. Now J(S) ≰ Q_{1}. Hence we may apply (5.3). Let the notation be as in (5.3). Let A ∈ A(S), A ≰ Q̃_{1}. We have A ∩ Q̃_{1} = (A ∩ C IQ̃_{1}(V_{1}))C with C ∩ C IQ̃_{1}(V_{1}) = 1.

Assume first C = 1. Then |A : A ∩ Q̃_{1}| ≥ |V_{1} : C I_{V_{1}}(A)|. Hence V_{1} is an F -module for U_{1}. This now implies with (1.18) and [GoLyS, (26.24)] that L^{(1)} ≅ L_{2}(r) or solvable and |A : A ∩ Q̃_{1}| = |V_{1} : C I_{V_{1}}(A)|. But now by (1.27) there is just one nontrivial irreducible module involved which contradicts (5.5).

So we have C ≠ 1. Then there is some g ∈ U_{1} such that [Z_{2}^{g}, C] ≠ 1. Set C_{1} = C I_{C}(Z_{2}^{g}). Suppose there is no elementary abelian subgroup A_{1} of L_{2} with A_{1} ≰ Q_{2} such that

|Z_{2} : C I_{Z_{2}}(A_{1})| ≤ |A_{1} : A_{1} ∩ Q_{2}|^{2}

Then we get

|Z_{2}^{g} : C I_{Z_{2}^{g}}(C)| > |C : C_{1}|^{2}

Now repeating this argument implies

|V_{1} : C I_{V_{1}}(C)| > |C|^{2}

As |V_{1} : C I_{V_{1}}(A)| ≤ |C||A : A ∩ Q̃_{1}| we get

|C| < |A : A ∩ Q̃_{1}|

and then

|V_{1} : C I_{V_{1}}(A)| < |A : A ∩ Q̃_{1}|^{2}

.

Now with (5.5) we get that V_{1} involves an F -module for U_{1}. Hence we get L^{(1)} ≅ L_{2}(r) or solvable again. Now

|A : A ∩ Q̃_{1}|^{2} ≤ |V_{1} : C I_{V_{1}}(A)|

by (1.27). This gives a contradiction. So we have shown: There is some A_{1} ≤ L_{2} with A_{1} ≰ Q_{2} and

|Z_{2} : C I_{Z_{2}}(A_{1})| ≤ |A_{1} : A_{1} ∩ Q_{2}|^{2}

Now we may apply (1.35). By (5.2)(a) we have that G_{2} ≠ ⟨C I_{G_{2}}(x) | x ∈ Z_{1}^{♯}⟩. Hence we see that E(L_{2}/Q_{2}) ≅ L_{3}(2) in (1.35)(2), E(L_{2}/Q_{2}) ≅ Sz(q) in (1.35)(3), q = 4 and the V_{i} are direct sums of two orthogonal modules in (1.35)(8) and E(L_{2}/Q_{2}) ≅ A_{6} in (1.35)(11).

Suppose first that L_{2} is nonsolvable. Let E(L_{2}/Q_{2}) be as in (1.35)(2), (6), (8), (10), (11), (12), (13). Then in all cases |Z_{1}| ≤ 4. Furthermore in (1.35)(6) we get that V_{1} involves exactly two F -modules for L_{2}(r). In all other cases even |Z_{2} : C I_{Z_{2}}(A_{1})| ≤ |A_{1} : A_{1} ∩ Q_{2}|^{\frac{3}{2}} .

The same argument as before now shows that

|V_{1} : C I_{V_{1}}(A)| ≤ |A : A ∩ Q̃_{1}|^{3}.

Hence in any case there are at most three modules involved.

Set L̂_{1} = ⟨J(S)^{L_{1}}⟩. Then m_{p}(L̂_{1}) ≥ 2. As U_{1} ≤ L̂_{1}, we have that also L̂_{1} induces at most three nontrivial irreducible modules and at most two of them in case (1.35)(6). Let Q̂_{1} = O_{2}(L̂_{1}).

Set Z̃_{2} = C I_{Z_{2}/Z_{1}}(Q̂_{1})(mod Z_{1}). Then by (5.4)(b) [Z̃_{2} ∩ Z(Q̂_{1}), L̂_{1}] = 1 and so Ẑ_{2} ∩ Z(Q̂_{1}) = Z_{1} or of order 4 in case (6). Furthermore for V_{1}^{(1)} = ⟨Z̃L̂_{1} ; 2⟩ we have V_{1}^{(1)} > Z_{1} and V_{1}^{(1)}/Z_{1} is a nontrivial module. As [V_{1}^{(1)}, Q̂_{1}] = Z_{1} we get that Q̃_{1} acts nontrivially on the set of irreducible E(L_{2}/Q_{2})-submodules or E(L_{2}/Q_{2}) ≅ L_{2}(4). Set Z_{2}^{(1)} = C I_{Z_{2}/Z̃_{2}}(Q̂_{1}). Then Z_{2}^{(1)} ≰ V_{1}^{(1)} and as

[Q̂_{1}, Z_{2}^{(1)}] ≤ Z̃_{2}, we see that ⟨Z_{2}^{(1)}L̂_{1}⟩ = V_{1}^{(2)} > V_{1}^{(1)} and V_{1}^{(2)}/V_{1}^{(1)} involves a nontrivial module. But even in (1.35)(6) we have Z_{2} ≠ Z_{2}^{(1)}. So ⟨ZL̂_{1} ; 2⟩ > V_{1}^{(2)}, a contradiction. Now in all other cases set Z_{2}^{(2)} = C I_{Z_{2}/Z_{2}^{(1)}}(Q̂_{1}). We have Z_{2}^{(2)} ≠ Z_{2} and so V_{1}^{(3)} = ⟨Z_{2}^{(2)L̂_{1}}⟩ > V_{1}^{(2)} and V_{1}^{(3)}/V_{1}^{(2)} and V_{1}/V_{1}^{(3)} are nontrivial modules, again a contradiction.

Suppose now E(L_{2}/Q_{2}) ≅ A_{9}. Then Q_{1} ≤ Q_{2} and so Z_{2} ≤ Z(Q̂_{1}). But this contradicts (5.4).

So it remains L_{2} to be solvable. Let L_{2} be a {2, 5}-group. Then again there are exactly two modules involved in V_{1}. Suppose we do not have (vii)(α). By (1.8) O_{5}(L_{2}/Q_{2}) ≅ 5^{1+2} or 5^{1+4}. Suppose [Φ(O_{5}(L_{2}/Q_{2})), Z_{2}] ≠ 1 and [A_{1}, Z(O_{5}(L_{2}/Q_{2}))] ≠ 1. As |Z_{2} : C I_{Z_{2}}(A_{1})| ≤ 2^{6}, we get that |[Z(O_{5}(L_{2}/Q_{2})), Z_{2}]| ≤ 2^{12}. But Sylow 5-subgroups of GL_{12}(2) are abelian. Hence [A_{1}, Z(O_{5}(L_{2}/Q_{2}))] = 1. Then we have that |A_{1} : A_{1} ∩ Q_{1}| = 2 in the first case and |A_{2} : A_{1} ∩ Q_{1}| ≤ 4 in the second. Now x ∈ A_{1}^{♯} \ Q_{1} inverts a group of order 25 in O_{5}(L_{2}/Q_{2})/Φ(O_{5}(L_{2}/Q_{2})). But then |Z_{2} : C I_{Z_{2}}(A_{1})| > 4, 16, respectively.

So [Φ(O_{5}(L_{2}/Q_{2})), Z_{2}]) = 1. But now |Z_{2}| = 2^{8}, |Z_{2}| = 2^{16} respectively and Z_{2} = Z_{2}^{(1)} ⊕Z_{2}^{(1)}, Z_{2} = Z_{2}^{(1)} ⊕Z_{2}^{(2)} ⊕Z_{2}^{(3)} ⊕Z_{2}^{(4)}, Z_{2}^{(i)} O_{5}(L_{2}/Q_{2})-modules. Now the same argument as above shows that we have more than two modules involved in V_{1}.

So let L_{2} be a {2, 3}-group. By (1.35) we have to handle O_{3}(L_{2}/Q_{2}) ≅ ZZ_{3} × ZZ_{3} and S ≅ D_{8}, |Z_{2}| = 2^{8}, or O_{3}(L_{2}/Q_{2}) ≅ 3^{1+4} and |Z_{2}| = 2^{8}, or m_{3}(O_{3}(L_{2}/Q_{2})) = 3 and |Z_{2}| ≤ 2^{8}.

If O_{3}(L_{2}/Q_{2}) ≅ ZZ_{3} × ZZ_{3}, then as O_{3}(L_{2}/Q_{2})-module we have Z_{2} = Z_{2}^{(1)} ⊕ Z_{2}^{(2)}, |Z_{2}^{(i)}| = 16 and so Q_{1}Q_{2} = S. Now we see that |Z_{1}| = 4, |C I_{Z_{2}/Z_{1}}(S)(mod Z_{1})| = 16, |[S, Z_{2}]| = 64 and so again three modules are in V_{1}, a contradiction.

Let O_{3}(L_{2}/Q_{2}) ≅ 3^{1+4}. Then |Z_{1}| = 2. We have as O_{3}(L_{2}/Q_{2})-module that Z_{2} = Z_{2}^{(1)} ⊕ Z_{2}^{(2)} ⊕ Z_{2}^{(3)} ⊕ Z_{2}^{(4)} and S induces a transitiv subgroup of S_{4} on these modules. Hence Q_{1} contains either ⟨(12)(34), (13)(24)⟩ or ⟨(1234)⟩. Now we easily see that there are at least four modules involved in V_{1}.

So let finally m_{3}(O_{3}(L_{2}/Q_{2})) = 3 and m_{2}(S/Q_{2}) = 2. Then |Z_{2}| ≤ 2^{8} and so O_{3}(L_{2}/C I_{L_{2}}(Z_{2})) ≅ 3^{1+2}, 3^{2} or 3^{4}. But now we may argue as before.

(5.7) Lemma. Assume C I_{G}(x) ≤ M for every x ∈ Z_{1}^{♯}. If E(L_{2}/Q_{2}) ≅ Sz(q) or L_{2}/C I_{L_{2}}(Z_{2}) ≅ F̸_{20} then assume additionally Z_{1} ≤ Z_{2}. Let Z_{2} be one of the modules in (5.6) or E(L_{2}/Q_{2}) ≅ L_{2}(4) × L_{2}(4) = X_{1} × X_{2}, Z_{2} = Z_{2}^{(1)} ⊕ Z_{2}^{(2)}, [Z_{2}^{(i)}, X_{j}] = 1 for i ≠ j, Z_{2}^{(i)} the orthogonal X_{i}-module, i = 1, 2. Then Z_{2} ≤ O_{2}(M).

Proof. Assume Z_{2} ≰ O_{2}(M). Then O_{2}(M) ≰ Q_{2}. Suppose first E(L_{2}/Q_{2}) ≅ U_{3}(q) or Sz(q) including the solvable case, i.e. O_{3}(L_{2}/Q_{2}) ≅ 3^{1+2}, or L_{2}/Q_{2} ≅ F_{20}.

(1) Ω_{1}(O_{2}(M)) ≰ Q_{2}.

Otherwise [Ω_{1}(O_{2}(M)), Z_{2}] = 1. As [O_{2}(M), Z_{2}] ≤ O_{2}(M) ∩ Z_{2} ≤ Ω_{1}(O_{2}(M)), we get x ∈ C I_{M}(O_{2}(M)) with o(x) odd and x^{t}O_{2}(M) = x^{−1}O_{2}(M) for some t ∈ Z_{2}. Hence C I_{M}(O_{2}(M)) ≰ O_{2}(M), a contradiction.

Now in any case |Ω_{1}(O_{2}(M)) : Ω_{1}(O_{2}(M)) ∩ Q_{2}| = q.

(2) Ω_{1}(O_{2}(M))′ = 1.

Suppose Z(Ω_{1}(O_{2}(M))) ≠ Ω_{1}(O_{2}(M)). We have Z_{1} ≤ Z(Ω_{1}(O_{2}(M))). Now Z_{1} ≤ Ω_{1}(O_{2}(M))′, as N_{M∩G_{2}}(Ω_{1}(Z(S))) acts transitively on Ω_{1}(Z(S))^{♯}. Choose t ∈ Z_{2} \ C I_{Z_{2}}(Ω_{1}(O_{2}(M))) with [t, Ω_{1}(O_{2}(M))] ≤ Z_{1}. Then t ∈ Ω_{1}(O_{2}(M)). If E(L_{2}/Q_{2}) ≅ U_{3}(q), then [Ω_{1}(O_{2}(M)), Z_{2}] = Z_{1}, a contradiction. If E(L_{2}/Q_{2}) ≅ Sz(q), then Z(Ω_{1}(O_{2}(M))) ≤ Q_{2} and so [Z_{2}, Z(Ω_{1}(O_{2}(M)))] = 1. But [Z_{2}, Ω_{1}(O_{2}(M))] ≤ Z(Ω_{1}(O_{2}(M))) and [Z_{2}, O_{2}(M)] ≤ Ω_{1}(O_{2}(M)). This contradicts C I_{M}(O_{2}(M)) ≤ O_{2}(M).

(3) If E(L_{2}/Q_{2}) ≅ Sz(q), then O_{2}(M)′ = 1.

As [Z_{2}, O_{2}(M)] ≤ Ω_{1}(O_{2}(M)), we have [Z_{2}, Ω_{1}(Z(O_{2}(M)))] ≠ 1. By (2) |Z_{2} ∩ O_{2}(M)| = q^{2}. Now as [O_{2}(M), Z_{2}] ≤ Z_{2} ∩ O_{2}(M) we have |O_{2}(M) : O_{2}(M) ∩ Q_{2}| = q. But then there is t ∈ Z_{2} \ O_{2}(M) with [t, O_{2}(M)] ≤ Z_{1} and so O_{2}(M)′ = 1, as otherwise Z_{1} ≤ O_{2}(M)′.

(4) If E(L_{2}/Q_{2}) ≅ SU_{3}(q), then |Z_{2} ∩ O_{2}(M)| = q^{4}.

As [Z_{2}, O_{2}(M)] ≤ Ω_{1}(O_{2}(M)), we have C I_{Z_{2}}(Ω_{1}(O_{2}(M))) ≤ O_{2}(M), and so |Z_{2} ∩ O_{2}(M)| = q^{4}.

(5) If E(L_{2}/Q_{2}) ≅ SU_{3}(q), then O_{2}(M)′ = 1.

We have Z_{1} ≤ O_{2}(M)′. Now there is some t ∈ Z_{2} \ O_{2}(M) with [Ω_{1}(O_{2}(M))O_{2}(M)′/O_{2}(M)′, t] = 1. But [t, O_{2}(M)] ≤ Ω_{1}(O_{2}(M)), contradicting C I_{M}(O_{2}(M)) ≤ O_{2}(M).

So we have shown :

(*) If E(L_{2}/Q_{2}) ≅ SU_{3}(q) or Sz(q) (including q = 2), then O_{2}(M)′ = 1.

Assume next E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q > 2, Z_{2} the orthogonal O^{+}(4, q)-module, q = 4, Z_{2} = Z_{2}^{(1)} ⊕ Z_{2}^{(2)}, Z_{2}^{(i)} orthogonal L_{2}(4)-modules, or |Z_{2}| = 16, F (L_{2}/C I_{L_{2}}(Z_{2})) ≅ ZZ_{3} × ZZ_{3}, S/Q_{2} ≅ D_{8} or ZZ_{4}. As C I_{G}(x) ≤ M for any x ∈ Z_{1}^{♯}, we see that for q > 2 just the orthogonal module is possible.

Suppose q > 2. Then |O_{2}(M) : O_{2}(M) ∩ Q_{2}| = q^{2}. Hence in any case |Z_{2} ∩ O_{2}(M)| = q^{3} and |Z_{2} : Z_{2} ∩ O_{2}(M)| = q.

Set W = ⟨(Z_{2} ∩ O_{2}(M))^{M}⟩. Then as C I_{M}(O_{2}(M)) ≤ O_{2}(M), we have [W, Z_{2}] ≠ 1, and so |W : W ∩ Q_{2}| = q^{2}. As O_{2}(M)′ ≠ 1, we get W ≰ O_{2}(M)′. As [O_{2}(M)′, Z_{2}] = 1, we see that we have A ≤ Z_{2}, |A| = q, |[W/W ∩ O_{2}(M)′, A]| = q^{2}, if Z_{2} is the O_{4}^{+}(q)-module and |A| = t, |[W/W ∩ O_{2}(M)′, A]| = t^{2}, t^{2} = q, if Z_{2} is a sum of two O_{4}^{−}(t)-modules. Furthermore in that case [W ∩ O_{2}(M)′, Z_{2}] = 1. So we have

(6) If E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q > 2, then for W = ⟨(Z_{2} ∩ O_{2}(M))^{M}⟩ we have [W ∩ O_{2}(M)′, Z_{2}] = 1. There is A ≤ Z_{2}, A ∩ O_{2}(M) = 1, |A| = q, t, respectively and |[W/W ∩ O_{2}(M)′, A]| = |A|^{2}.

Suppose now that L_{2} is solvable, i.e. |Z_{2}| = 16. If |Ω_{1}(O_{2}(M)) : Q_{2} ∩ Ω_{1}(O_{2}(M))| = 2, we get as before that O_{2}(M) is abelian and |Z_{2} : Z_{2} ∩ O_{2}(M)| = 4, |O_{2}(M) : O_{2}(M) ∩ Q_{2}| = 2.

Let now |Ω_{1}(O_{2}(M)) : Ω_{1}(O_{2}(M)) ∩ Q_{2}| > 2, then S/Q_{2} ≅ D_{8}. Let [Ω_{1}(O_{2}(M)), Z_{2}, Ω_{1}(O_{2}(M))] = 1. Then we see that O_{2}(M)′ = 1. Furthermore there is x ∈ Z_{2} \ O_{2}(M) with |[x, O_{2}(M)]| = 2.

If Ω_{1}(O_{2}(M)) does not act quadratically. We see |O_{2}(M)∩Z_{2}| = 8. Define W as before. Then O_{2}(M)′ ≠ 1 and |[Z_{2}, W/W ∩O_{2}(M)′]| = 4. Furthermore [W ∩ O_{2}(M)′, Z_{2}] = 1.

Let now E(L_{2}/Q_{2}) ≅ L_{2}(q), Z_{2} the natural or orthogonal module, or the direct sum of two natural modules, including q = 2, or L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{10}.

If we have the natural module or a sum of two natural modules or F_{10}, then [O_{2}(M), Z_{2}] ≤ Z(O_{2}(M)). We have Z(O_{2}(M))∩Z_{2} ≤ O_{2}(M)′, if O_{2}(M)′ ≠ 1, a contradiction to C I_{M}(O_{2}(M)) ≤ O_{2}(M). So we have O_{2}(M)′ = 1.

(7) If E(L_{2}/Q_{2}) ≅ L_{2}(q), or L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{10}, then O_{2}(M)′ = 1, or E(L_{2}/Q_{2}) ≅ L_{2}(q) and Z_{2} is the orthogonal module.

Let Z_{2} be the orthogonal module. Then |Z_{2} ∩ O_{2}(M)| = t^{3}, t^{2} = q. Now set W = ⟨(Z_{2} ∩ O_{2}(M))^{M}⟩. Again [W, Z_{2}] ≠ 1 and W ∩ O_{2}(M)′ ≠ 1. Hence there is A ≤ Z_{2}, |A| = t with A ∩ O_{2}(M) = 1 and |[W/W ∩ O_{2}(M)′, A]| = q.

Now we have all possible cases from (5.6) besides E(L_{2}/Q_{2}) ≅ A_{9}. But in that case O_{2}(M) ≤ Q_{2}, contradicting Z_{2} ≰ O_{2}(M). We collect

(8) One of the following holds:

(α) E(L_{2}/Q_{2}) ≅ L_{2}(q), Sz(q), including q = 2, or L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{10}, and O_{2}(M)′ = 1.

(β) E(L_{2}/Q_{2}) ≅ SU_{3}(q), O_{2}(M)′ = 1. |Z_{2} : Z_{2} ∩ O_{2}(M)| = q^{2} and |[Z_{2}, O_{2}(M)]| = q^{2}, C I_{O_{2}(M)}(z) = C I_{O_{2}(M)}(Z_{2}) for every z ∈ Z_{2} \ O_{2}(M).

(γ) E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q) or L_{2}/C I_{L_{2}}(Z_{2}) is a solvable {2, 3}–group with m_{3}(L_{2}/C I_{L_{2}}(Z_{2})) = 2. In the latter set q = 2. For W = ⟨(Z_{2} ∩ O_{2}(M))^{M}⟩ we have [W ∩ O_{2}(M)′, Z_{2}] = 1 and there is some A ≤ Z_{2}, A ∩ O_{2}(M) = 1 with |A| = q or q = t^{2} and orthogonal L_{2}(t^{2})-modules are involved in which case |A| = t, such that |[W/W ∩ O_{2}(M)′, A]| = |A|^{2} and C I_{W/W ∩O_{2}(M)′}(a) = C I_{W/W ∩O_{2}(M)′}(A) for every a ∈ A^{♯} we have

Furthermore for u ∈ [W, Z_{2}]^{♯}, we have C I_{G}(u) ≤ M iff u ∈ Z_{1}.

(δ) L_{2} is a solvable {2, 3} - group with m_{3}(L_{2}) > 1, O_{2}(M)′ = 1. There is x ∈ Z_{2}, |[O_{2}(M), x]| = 2, C I_{G}(x) ≰ M.

Now we show

(9) (8)(α) holds.

Suppose we have (β), (γ) or (δ). Set A = Z_{2}/Z_{2} ∩ O_{2}(M) in case (8)(β) and A = ⟨x⟩ in case (δ). Set V = O_{2}(M) in case (8)(β), or (δ), V = W/W ∩ O_{2}(M)′ in case (8)(γ).

Let K be a component of E(M̄), M̄ = M/O_{2}(M) with 1 ≠ [Ā, K] ≤ K. As C I_{V} (A) = C I_{V} (a) for all a ∈ A^{♯} we have C I_{A}(K) = 1. Furthermore |[V, A]| = |A| or |A|^{2}. In case (γ) [K, O_{2}(M)′ ∩W] = 1, as [Z_{2}, O_{2}(M)′ ∩W] = 1. Hence we have

(9.1) If (γ) holds, then [K, Z_{1}] = 1.

If |A| > 2, then by quadratic action (1.12) we have K/Z(K) ≅ A_{n}, M_{12}, M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2}, Sz, U_{4}(3) or K/Z(K) ≅ G(r), a group of Lie type in even characteristic. If |A| = 2, then |V : C I_{V} (A)| ≤ 4 and by (1.30) the same holds.

We first are going to show that |A| = 2. So until further notice we assume |A| > 2. Then there is some ω ∈ L_{1}∩L_{2}, o(ω) is odd and ⟨ω⟩ acts transitively on A^{♯}.

Suppose that K is not a group of Lie type in even characteristic. Then by (1.12) and (1.25) o(ω) = 3 and so 3 ∉ σ(M). This shows m_{p}(K) ≤ 2 and there is E ≅ E_{p^{2}} with [E, K] = 1. Furthermore by (1.25) V involves exactly one nontrivial irreducible K–module, i.e. [[K, V], E] = 1. This implies that there is some ρ ∈ K, o(ρ) odd, with ρ^{a} = ρ^{−1} for some a ∈ A^{♯} and [[V, ρ], E] = 1. As [[V, ρ], a] ≤ Z_{2} but [V, ρ] ∩ Z_{1} = 1, we see that there is some x ∈ Z_{2}∩O_{2}(M)\Z_{1} with C I_{G}(x) ≤ M. But L_{2} = ⟨L_{2}∩L_{1}, C I_{L_{2}}(x)⟩, a contradiction. So we have

(9.2) If |A| > 2 then K is a group of Lie type in even characteristic

Next we show

(9.3) [K, ω] ≤ K.

Suppose [K, ω] ≰ K. Then [K^{ω}, K] = 1 and K^{⟨ω⟩} contains at least three components K, K^{ω}, K^{ω^{2}}. Assume first [[K, V], K^{ω}] = 1. Then we may assume [[K, V], K^{ω} ∗K^{ω^{2}}] = 1 too. Now there is x ∈ Z_{2}∩M \Z_{1} with [x, K^{ω} ∗ K^{ω^{2}}] = 1. This shows that C I_{G}(x) ≤ M and so L_{2} = ⟨L_{1} ∩ L_{2}, C I_{L_{2}}(x)⟩ ≤ M, a contradiction. So we have [K^{ω}, [K, V]] ≠ 1 and so [K, V] involves at least two modules which then are F –modules with A offending.

We have that m_{p}(C I_{W} (S ∩ K)) ≤ 1 for any prime p | |K|, where W is some nontrivial irreducible K–submodule in V. Hence we see with (1.18) that K induces on V the group A_{7}, 3 · A_{6}, L_{4}(2), L_{3}(r), L_{2}(r), SU_{4}(r), L_{2}(r^{2}), Sp_{4}(r), or G_{2}(r). Furthermore we have some irreducible module W invariant under A. Now the action of K^{ω} ∗ K^{ω^{2}} on [V, K] shows that there are at least 8,8,8,6,4,16,8,8, or 12 such modules involved, respectively. Hence we have |A| ≥ r^{4}, r^{4}, r^{4}, r^{3}, r^{2}, r^{8}, r^{4}, r^{4}, r^{6}, respectively. This shows that |A| > 2^{m_{2}(Aut(K))}, a contradiction. So we have (9.3).

(9.4) [K, ω] ≠ 1.

This is clear, as A = [A, ω].

By (9.2) we have that K is a group of Lie type in even characteristic. Let first K ≅ G(r) with |A| ≤ r. Then either K ≅ Sz(r), U_{3}(r) or L_{2}(r) or V is an SC-module for K with offending subgroup A and so by (1.22) and (1.23) K ≅ (S)L_{n}(r), (S)U_{n}(r), Sp_{2n}(r), Ω_{2n}^{±}(r) or G_{2}(r). Hence |A| = r or K ≅ (S)L_{n}(r) or Sp_{2n}(r) and |A| = \sqrt{r}.

As |A| ≤ r, we do not have case (β). Suppose first some p | r − 1 for some p ∈ σ(M). If there is some x ∈ Z_{2}∩M \Z_{1} centralized by Sp_{4}(r), L_{3}(r), U_{4}(r) or Ω^{+}(4, r), then x is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and so C I_{G}(x) ≤ M. But L_{2} = ⟨L_{2} ∩ L_{1}, C I_{L_{2}}(x)⟩. Hence we get K ≅ SL_{3}(r), L_{2}(r), U_{3}(r), U_{4}(r), U_{5}(r), G_{2}(r) or Sz(r). Now m_{p}(K) ≤ 2 and so m_{p}(C I(K)) ≥ 1. As |A| = r, there is ν ∈ ⟨ω⟩, o(ν) = p and so m_{p}(C I_{M}(ν)) ≥ 3. Hence N_{G}(⟨ν⟩) ≤ M, but L_{2} = ⟨L_{1} ∩ L_{2}, N_{L_{2}}(⟨ν⟩)⟩. So we have |A| = \sqrt{r} and K ≅ SL_{3}(r) or L_{2}(r). Furthermore V involves just one nontrivial irreducible module, the natural one. Finally A ≤ K. Hence there is some ρ ∈ K, o(ρ) odd, and some E ≅ E_{p^{3}}, with [E, ⟨ρ, a⟩] = 1. But then some F ≤ E, |F | = p^{2} centralizes some z ∈ Z_{2} ∩ O_{2}(M) \ Z_{1}, a contradiction.

So we have if p ∈ σ(M) then p | r̸ − 1. Let p | r^{2} − 1 and K ≅ (S)U_{n}(r). Again no element x ∈ (V/Z_{1})^{♯} is centralized by U_{3}(r) or U_{4}(r) in K. This shows K ≅ SU_{4}(r) or SU_{3}(r) and m_{p}(K) ≤ 3. If K ≅ SU_{3}(r) then there is ρ ∈ K, o(ρ) odd, with ρ^{a} = ρ^{−1} for some a ∈ A^{♯} and [E, ⟨a, ρ⟩] = 1 for some E ≅ E_{p^{3}}. We get the same contradiction as above. So we have K ≅ SU_{4}(r). In any case E(L_{2}/Q_{2}) ≅ L_{2}(|A|^{2}) or L_{2}(|A|) × L_{2}(|A|). In the former there is some ν ∈ L_{1} ∩ L_{2}, o(ν) = p. Now m_{p}(C I_{M}(ν)) ≥ 3, a contradiction. So we have the latter. Then T = ZZ_{r−1} × ZZ_{r−1} ≤ L_{1} ∩ L_{2}. This implies T ∩ K ≠ 1, otherwise any prime u with u | r − 1 is in σ(M). As [T ∩ K, Z_{1}] = 1, we have [T ∩ K, A] = 1. Now m_{p}(N_{M}(T ∩ K)) ≥ 2 and there is E ≤ N_{M}(T ∩ K), E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. This implies with (1.63) that N_{G}(T ∩ K) ≤ M, a contradiction.

Suppose now first that [K, V] involves exactly one irreducible nontrivial module. As for p ∈ σ(M) we have p | r̸ − 1, and p |̸ r + 1 in case of K ≅ (S)U_{n}(r), we get that any p-element μ, with [K, μ] = 1 centralizes [K, V]. Hence m_{p}(C IM̄(K)) ≤ 1. Now m_{p}(Aut(K)) ≥ 3. This implies m_{p}(K) ≥ 2 and so K ≅ L_{4}(r), Sp_{4}(r), Sp_{6}(r), Ω_{8}^{−}(r), G_{2}(r). In any case p | r^{2} − 1. But there is some a ∈ A^{♯} such that [a, V] is centralized by L_{2}(r) in K. Hence m_{p}(C IM̄(K)) = 0 and so m_{p}(K) ≥ 3. So we are left with Sp_{6}(r), Ω_{8}^{−}(r). Now [a, V] is centralized by Sp_{4}(r), Ω_{6}^{−}(r), and so we again see that C I_{G}([a, V]) ≤ M, a contradiction.

Hence we have K ≅ SL_{n}(r) or Sp_{2n}(r), |A| = r and [K, V] involves exactly two natural modules. Again p | r̸ − 1 for p ∈ σ(M). So we have K ≅

L_{2}(r), SL_{3}(r), SL_{4}(r), Sp_{4}(r) or Sp_{6}(r). If K ≅ Sp_{6}(r), then even p | r̸^{2}−1, as otherwise some x ∈ Z_{2}∩M \Z_{1} is centralized by Sp_{4}(r) and so C I_{G}(x) ≤ M, a contradiction. But now in the case of Sp_{6}(r) we have m_{p}(K) = 1 and so m_{p}(C IM̄(K)) ≥ 2. But [[K, V], C IM̄(K)] = 1, a contradiction. So in any case m_{p}(K) ≤ 2, m_{p}(C IM̄(K)) = 1 and [C IM̄(K), [V, K]] ≠ 1. Hence we have a subgroup K_{1} of C IM̄(K), K_{1} isomorphic to a subgroup of L_{2}(r). Furthermore A cannot centralize this group otherwise we may argue as before. Hence K_{1} ≅ L_{2}(r) and [V, K_{1}] is a sum of two natural modules, i.e. |[V, K_{1}]| = |A|^{4} and so |[V, K]| = |A|^{4}. This shows K ≅ L_{2}(r) too, [V, K×K_{1}] is the O^{−}(4, r)- module. As above T = ZZ_{r−1} × ZZ_{r−1} ≤ L_{1} ∩ L_{2}. As L_{2}(r) × L_{2}(r) does not contain ZZ_{r−1} × ZZ_{r−1} normalizing A, we get that T ∩ C I(K × K_{1}) ≠ 1, a contradiction to (1.63).

So we are left with |A| > r. By (1.64) and (1.65) we get K ≅ SU_{n}(r), Ω_{2n}^{−}(r), ^{2}E_{6}(r), ^{3}D_{4}(r), or ^{2}F_{4}(2)′, or r = 2, o(ω) = 3, |A| = 4. In the latter we get a contradiction as in the sporadic and alternating case. As |A| acts quadratically K ≅ ^{3̸}D_{4}(r) nor ^{2}F_{4}(2)′ by (1.15) and (1.14). Hence we have |A| = r^{2} and K ≅ SU_{n}(r) or Ω_{2n}^{−}(r) by (1.23). In any case by (1.45) |V : C I_{V} (A)| > |A| and so (8)(β) is not possible. Now A ∩ Z(S/O_{2}(M)) ≠ 1. As always ω normalizes a Borel subgroup we get |Z(S ∩ K)| ≥ |A|, a contradiction.

So we have shown

(9.5) |A| = 2.

Now we have (8)(γ) or (δ) and |V : C I_{V} (A)| ≤ 4, and so by (1.30) K/Z(K) ≅ L_{n}(2), SL_{n}(4), SU_{n}(2), Sp_{2n}(2), Sp_{2n}(4), Ω_{2n}^{±}(2), Ω_{2n}^{±}(4), G_{2}(2), A_{n} or U_{4}(3).

In any case either 3 | |L_{1} ∩ L_{2}| or m_{3}(L_{2}) ≥ 2 as we assume that (α) does not hold. Hence we have 3 ∉ σ(M). This now implies K/Z(K) ≅ L_{n}(2), n ≤ 7, SL_{n}(4), n ≤ 4, SU_{4}(2), Sp_{4}(2), Sp_{6}(2), Sp_{4}(4), Sp_{6}(4), Ω_{8}^{−}(2), Ω_{8}^{−}(4), G_{2}(2), A_{n}, n ≤ 11. Now any p-element μ ∈ C IM̄(K) centralizes [V, K]. This implies m_{p}(Aut(K)) ≥ 3, and so m_{p}(K) ≥ 3. This implies K ≅ Sp_{6}(4) or Ω_{8}^{−}(4), p = 5. Now [V, A] is centralized by Sp_{4}(4), or Sp_{6}(4), and so by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

So we have shown

(9.6) If [K, A] ≠ 1 and [K, V] ≠ 1, then [K, A] ≰ K.

Next we show

(9.7) [[A, E(M̄)], V] = 1.

Let [K, A] ≠ 1 and [K, V] ≠ 1. If |A| > 2 then by (1.11) K ≅ L_{2}(r), r even, and [V, K^{A}] just involves O_{4}^{−}(r)-modules. But now there is T ≅ D_{2(r+1)} × D_{2(r+1)} ≤ ⟨K^{A}⟩, A ∩ T ∈ Syl_{2}(T). As C I_{V} (a) = C I_{V} (A) for every a ∈ A^{♯} this is impossible.

Hence |A| = 2. As 3 ∉ σ(M), K possesses cyclic Sylow 3-subgroups. Hence K ≅ L_{2}(r), U_{3}(r), L_{3}(r), Sz(r) or J_{1}. Furthermore as |[V, A]| ≤ 4, we have that |K| = 2^{x}3^{y}5^{z}, with cyclic Sylow 5 subgroups as A cannot invert an elementary abelian group of order 5^{2}. Hence K ≅ L_{2}(4). Now we see that for P ∈ Syl_{p}(K × K^{a}), |[V, P]| = 2^{8}. Then there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and [[V, P], E] = 1. This shows that for some v ∈ O_{2}(M) ∩ Z_{2} \ Z_{1}, we have C I_{G}(v) ≤ M, a contradiction.

(9.8) Let P̄ ≤ M̄ be a u-group, u an odd prime, and [P̄, A] ≤ P̄. If |A| > 2, then [A, P̄] ≤ C IP̄(V).

Let [A, P̄] ≠ 1. Then as C I_{V} (a) = C I_{V} (A) for a ∈ A^{♯}, we have C I_{A}([A, P̄]) = 1. Now by [GoLyS, (24.1)] there is U_{1} ≤ P̄A, U_{1} ≅ D_{2u} × … × D_{2u}, A ∈ Syl_{2}(U_{1}). But then [O_{u}(U_{1}), V] = 1. As [A, O_{2}(M)] ≤ V, we see [O_{u}(U_{1}), O_{2}(M)] = 1, a contradiction.

Now we come to the final contradiction. Let P̄ ∈ Syl_{u}(F (M̄)), u an odd prime with [A, P̄] ≠ 1 and [V, P̄] ≠ 1. Then by (9.8) |A| = 2. Furthermore u ∈ {3, 5} . Suppose u ∉ σ(M). Then m_{u}(P̄) ≤ 3. Hence m_{p}(Aut(P̄)) ≤ 1. This gives some E ≅ E_{p^{3}} with Γ_{E,1}(G) ≤ M and [E, [V, A]] = 1, a contradiction.

So we have u = 5 ∈ σ(M). If m_{5}(P̄) ≥ 3, then by [GoLyS, (11.18)] A normalizes some E ≅ E_{5^{3}} and so some F ≤ E, |F | = 25, centralizes [V, A], a contradiction again. So we have m_{5}(P̄) ≤ 2 and σ(M) = {5}. Now there is a critical subgroup P̄_{1} of P̄ with Ω_{1}(P̄_{1}) ≅ ZZ_{5}, ZZ_{5} × ZZ_{5} or 5^{1+2}. In the last two cases C I_{G}(x) ≤ M for all x ∈ Ω_{1}(P̄_{1})^{♯}. Hence [V, A] is centralized by some E ≅ E_{25} with Γ_{E,1}(G) ≤ M, a contradiction. This shows that P̄_{1} is cyclic with [P̄, A] ≤ P̄_{1} and [P̄_{1}, A] = P̄_{1}. Furthermore |[P̄_{1}, V]| = 16 and so C I_{M}([P̄_{1}, A]) contains an elementary abelian group of order 5^{3}, a contradiction.

So we have shown that [F^{∗}(M̄), A] ≤ C I_{F^{∗}(M̄)}(V). This now implies [O_{2}(M), [F^{∗}(M̄), A]] ≤ V and so [[F^{∗}(M̄), A], O_{2}(M)] = 1, a contradiction to C I_{M}(O_{2}(M)) ≤ O_{2}(M). This proves (9).

By (8) and (9) we know that O_{2}(M)′ = 1. Set L̃_{2} = ⟨O_{2}(M_{1}), O_{2}(M_{3})⟩. Then L̃_{2}/O_{2}(L̃_{2}) ≅ L_{2}(q), Sz(q) or F_{10}.

Now [O_{2}(M_{1}) ∩ O_{2}(M_{3}), L̃_{2}] = 1. As [L̃_{2}, Q_{2}] ≤ O_{2}(L̃_{2}), and Z(L_{2}) = 1, we get O_{2}(M_{1}) ∩ O_{2}(M_{3}) = 1 or L̃_{2}/C IL̃_{2}(Z_{2}) ≅ F_{10} and L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{20}. In the latter C I_{Q_{2}}(L̃_{2}Q_{2}) ≠ 1 and so also C I_{Q_{2}}(L_{2}) ≠ 1, a contradiction. So we have

(10) O_{2}(M_{1}) ∩ O_{2}(M_{3}) = 1 and W̃ = ⟨O_{2}(M_{1}) ∩ Q_{2}, O_{2}(M_{3}) ∩ Q_{2}⟩ is a direct sum of natural modules.

The latter follows from (1.27).

So we have that O_{2}(M) is an F -module for M with offending subgroup O_{2}(M_{3}) ∩ Q_{2}. As |O_{2}(M)| > 16, we have |O_{2}(M_{3}) ∩ Q_{2}| > 4. Furthermore C I_{O_{2}(M)}(x) = C I_{O_{2}(M)}(O_{2}(M_{3}) ∩ Q_{2}) for every 1 ≠ x ∈ O_{2}(M_{3}) ∩ Q_{2}. Hence there is some component K of M̄ with 1 ≠ [K, O_{2}(M) ∩ Q_{2}] ≤ K. Finally C I_{O_{2}(M_{3})∩Q_{2}}(K) = 1. If |O_{2}(M) : O_{2}(M) ∩ Q_{2}| = 2, then O_{2}(M_{3}) ∩ Q_{2} induces the full group of transvections on O_{2}(M) i.e. K ≅ L_{n}(2), |O_{2}(M)| = 2^{n}.

Let q > 2. We know that O_{2}(M) involves exactly one nontrivial irreducible module. Hence let ω ∈ L_{1} ∩ L_{2}, o(ω) = q − 1. Then [ω, K] ≤ K. As [ω, O_{2}(M_{3}) ∩ Q_{2}] = O_{2}(M_{3}) ∩ Q_{2}, we have [ω, K] ≠ 1. By (1.18) we know the structure of K. By (1.65) and (1.64) we see that K is a Lie group over GF (q) or K ≅ Ω_{2n}^{−}(\sqrt{q}) or U_{n}(\sqrt{q}). But as |O_{2}(M_{3})∩Q_{2}| > q, we get with (1.45) that K is a Lie group over GF (q) and then |O_{2}(M) : C I_{O_{2}(M)}(O_{2}(M) ∩ Q_{2})| = q. Hence by (1.23) we see that K ≅ SL_{n}(q) and |O_{2}(M)| = q^{n}. Now in any case M/O_{2}(M) ≲ ΓL_{n}(q). Hence (O_{2}(M_{3}) ∩ Q_{2})O_{2}(M) is normalized by SL_{n−1}(q) in M. This implies that there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and E ≤ N_{M}((O_{2}(M_{3}) ∩ Q_{2})O_{2}(M)). But then also E ≤ N_{M}(W̃), and so L_{2} ≤ N_{G}(W̃) ≤ M, a contradiction.

(5.8) Lemma. The configuration in (4.4)(b) does not occur.

Proof. We consider the graph Γ(G_{1}, G_{2}) = Γ. Let α ∈ Γ be of minimal distance from 1 such that T_{1} ≤ G_{α} but there is some β ∈ Δ(α) with T_{1} ≰ G_{β}.

(1) α ≁ 1

Suppose α ∼ 1. Then T_{α} ≤ G_{1} and so [T_{1}, T_{α}] ≤ T_{1} ∩T_{α}. As T_{1} ≰ O_{2}(G_{α}) we have [T_{1}, T_{α}] ≠ 1 by (4.4). Furthermore M is the unique maximal 2–local containing C I_{G}(x) for x ∈ [T_{1}, T_{α}]^{♯}. Hence M = M_{α}. By (5.7) Z_{2} ≤ O_{2}(M) but T_{1} ≰ O_{2}(M_{α}), a contradiction.

(2) Z_{α} ≰ G_{1}

Suppose Z_{α} ≤ G_{1}. Then C I_{G}(x) ≤ M for every x ∈ [T_{1}, Z_{α}]^{♯}. We have |[T_{1}, Z_{α}]| = q^{2} and there is some x ∈ [T_{1}, Z_{α}]^{♯} such that M_{α−1} is the unique maximal 2–local containing C I_{G}(x). This again proves M = M_{α−1}. By (5.7) Z_{α} ≤ O_{2}(M_{α−1}) = O_{2}(M). We have [T_{1}, Ω_{1}(O_{2}(M))] = 1 and so [Z_{α}, T_{1}] = 1, a contradiction.

Now we have Z_{α} = ⟨T_{β} | β ∈ Δ(α)⟩, and so Z_{α} ≤ G_{2}. As Z_{α} ≰ G_{1} we have [Z_{2}, Z_{α}] ≠ 1. Furthermore Z_{2} = ⟨T_{γ} | γ ∈ Δ(2)⟩ and so Z_{2} ≤ G_{α}. But this implies that Z_{2} is an F –module, a contradiction to (1.18).

§ 6 The case b = b_{1}

In this chapter we are going to investigate the case b = b_{1}. The notation is as introduced in chapter 4.

(6.1) Lemma. Let α ∈ Γ with Z_{1} ≤ L_{α} and Z_{α} ≤ L_{1}. If [Z_{1}, Z_{α}] ≠ 1, then α ∼ 2.

Proof. Suppose false, i.e. α ∼ 1. We have

[Z_{1}, Z_{α}] ≤ Z_{1} ∩ Z_{α}

Hence Z_{1} is an F-module with offending subgroup Z_{α}/Z_{α} ∩ C I(Z_{1}). Let L_{1} not be in the exceptional case. Suppose [E(L_{1}/Q_{1}), Z_{α}] ≠ 1. By (1.11) Z_{α} induces orbits of length at most two on the set of components of L_{1}/Q_{1}. Furthermore either for a component K we have [Z_{α}, K] ≤ K, or |Z_{α} : N_{Z_{α}}(K)| = 2 and K ≅ L_{2}(2^{n}), or |Z_{α} : C I_{Z_{α}}(K)| = 2 and N_{Z_{α}}(K) = C I_{Z_{α}}(K). Hence in the latter two cases by (1.11) Z_{1} is not an F-module for ⟨K^{Z_{α}}⟩Z_{α} with offending group Z_{α}/C I_{Z_{α}}(K). Hence there is a component K of L_{1}/Q_{1} with [Z_{α}, K] ≤ K such that Z_{1} is an F-module for KZ_{α} with offending subgroup Z_{α}/C I_{Z_{α}}(K).

Now we may apply (5.1). Let U be the K-module given by (5.1). Let K^{(α)} be the corresponding component to K in L_{α}/Q_{α}.

We will assume that we do not have case (5.1) (i) or (ii). Choose u ∈ U^{♯}. We have x ∈ [Z_{α}, u]^{♯} is centralized by E_{x} ≅ E_{p^{2}} with Γ_{E_{x},1}(G) ≤ M.

Assume [u, K^{(α)}] = 1. Let [[u, Z_{α}], K^{(α)}] = 1 . Then K^{(α)} is covered by M ∩ M_{α}. Now (4.2) implies M = M_{α}. But Z_{1} ≤ Z(O_{2}(M)) = Z(O_{2}(M_{α})) ≤ Q_{α}, a contradiction. So [[u, Z_{α}], K^{(α)}] ≠ 1. Moreover we have that m_{p}(K^{(α)}) = 1. But in this case we have by (1.18) that U is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. Now again K^{(α)} is covered by M ∩ M_{α}, a contradiction.

So we have [u, K^{(α)}] ≠ 1. Now there is z ∈ Z_{α} such that [u, z] is centralized by E ≅ E_{p^{2}} in M and M_{α} as well. This gives M = M_{α}, a contradiction.

Assume now (5.1) (i). In this case U = [Z_{1}, K] and [Z_{α}, U] is normalized by E ≅ E_{p^{3}}. For u ∈ U, we have |[u, Z_{α}]| ≤ q. Hence for [u, K^{(α)}] = 1, we have [[u, Z_{α}], K^{(α)}] = 1. By (5.1) (i) we see that we may choose u with [u, Z_{α}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. Hence for such u we have [u, K^{(α)}] ≠ 1. This now shows that Z_{α} is an F-module for K^{(α)} with offending subgroup U. But then [U, Z_{α}] is normalized by E ≅ E_{p^{3}} in M and M_{α}, contradicting M ≠ M_{α}.

So we are left with (5.1) (ii). Then U = [Z_{1}, K] and [Z_{α}, U] is normalized by E ≅ E_{9}. But [Z_{α}, U] is normalized by E ≅ E_{9} in M_{α} as well, which is clear for [U, K_{α}] = 1 and in case [U, K_{α}] ≠ 1 we get this as U offends on U_{α}. Hence we have M = M_{α}, a contradiction.

So we have [E(L_{1}/Q_{1}), Z_{α}] = 1. By (4.7) O_{p′}(L_{1}) ∩ L_{2} ≤ Q_{1}. So we get a faithfull {2, p}-group K_{1} ≤ L_{1}/Q_{1} on Z_{1} with O_{2}(K_{1}) = 1 and having Z_{α}Q_{1}/Q_{1} as a Sylow 2–subgroup.. This implies p = 3, and by symmetry we may assume that there are x ∈ Z_{1}, y ∈ Z_{α} with [x, y] = [x, Z_{α}] = [y, Z_{1}] ≤ [Z_{1}, ρ], ρ of order 3, |[Z_{1}, ρ]| = 4. By (4.4) we get some E ≅ E_{9}, with [[x, y], E] = 1 and Γ_{E,1}(G) ≤ M. As the same holds in M_{α}, we get a contradiction.

So we are left with the exceptional case: L_{1}/O_{2}(L_{1}) = BA. A = L_{1} ∩ L_{2}/O_{2}(L_{1}) and B is an extension of an r-group by L_{2}(r), r odd, r > 3. Then [Z_{1}, O_{r}(B)] = 1. Hence we get a contradiction with (4.10).

(6.2) Lemma. Let b = b_{1} and [Z_{1}, Z_{α}] ≠ 1. Then [Z_{1}, Z_{2}] ≠ 1.

Proof . Suppose Z_{1} ≤ Q_{2}. By (6.1) we have that α ∼ 2. So either Z_{2} or Z_{1} is an F-module. Suppose that Z_{2} is an F-module. Then by (1.31) and (2.1) either E(L_{2}/Q_{2}) ≅ L_{2}(q), L_{2}(q) × L_{2}(q), q even, A_{9}, or E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ A_{5}, or L_{2} is a solvable {2, 3}-group. So in any case by inspection we get that Z_{1} is an F-module too. So we may assume that Z_{1} is an F-module wright from the begining. To prove the lemma we now are going to argue as in (6.1).

Let K be some component of L_{1} such that Z_{α}/C I_{Z_{α}}(K) offends on Z_{1}. Then (5.1) provides us with some module U. Suppose C I_{G}(u) ≤ M for any u ∈ U^{♯}. Let now α + 1 be some neighbor of α such that U ≰ M_{α+1}. Let U_{α+1} be the corresponding module in L_{α+1}. Then U_{α+1} ≤ G_{2}.

Suppose [U_{α+1}, Z_{2}] ≠ 1. Then Z_{1} ≤ C I_{G}([U_{α+1}, Z_{2}]) ≤ M_{α+1}, by (5.1), a contradiction. So we have [U_{α+1}, Z_{2}] = 1, i.e. U_{α+1} ≤ L_{1}. Now C I_{U_{α+1}}(Z_{1}) = 1, i.e. U_{α+1} ∩ Q_{1} = 1 otherwise Z_{1} ≤ M_{α+1}. As b > 1, we see that ⟨U_{α+1}^{L_{α}}⟩ is abelian. hence U_{α+1} acts quadratically on Z_{1}. This together with (1.11) shows that [U_{α+1}, K] ≤ K. As [U_{α+1}, Z_{α}] = 1, we see that [U_{α+1}, U] ≤ U, and so for u ∈ U \ M_{α+1}, we get [u, U_{α+1}] ≅ U, a contradiction.

Now by (5.1) K ≅ L_{2}(q) or U_{4}(2). Then Z_{α} is an F-module too. In the first case we get E(L_{2}/K_{2}) ≅ L_{2}(q) or L_{2}(q) × L_{2}(q). Furthermore p | q − 1. But now there is ω ∈ L_{1} ∩ L_{2} with o(ω) = p. This shows m_{p}(C I_{L_{1}}(ω)) ≥ 3. But then N_{G}(⟨ω⟩) ≤ M, a contradiction to (1.66).

If K ≅ U_{4}(2), then |Z_{1} : Z_{1} ∩ Q_{α}| = 16. Furthermore 3 ∈ σ(M). Now by (1.31) E(L_{2}/Q_{2}) ≅ L_{2}(r). As |Z_{1} : Z_{1} ∩ Q_{α}| = 16, we get r = 16, but then 3 | |L_{1} ∩ L_{2}|, contradicting (1.66).

So we have [E(L_{1}/Q_{1}), Z_{α}] = 1. Let K be a t–group on which Z_{α} acts nontrivially. Suppose first t = p. Then we get p = 3. Furthermore, as Z_{1} is an F-module, there is some t ∈ Z_{α} inducing a transvection on Z_{1}. By (1.52) we do not have the exceptional case. By quadratic action there is also some u ∈ Z_{1} inducing a transvection on Z_{α}, [GoLyS, (25.12)], [GoLyS, (24.1)]. So by (1.31) E(L_{2}/Q_{2}) ≅ A_{5}, A_{5} × A_{5}, A_{9}, or E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ A_{5}, or L_{2} is a solvable {2, 3}-group. By (2.1) m_{3}(L_{2}) = 1.

Furthermore by (1.66) m_{3}(L_{1} ∩ L_{2}) = 0. This implies that L_{2}/Q_{2} ≅ Σ_{3} and Z_{2} = Z(L_{2}) × Z̃_{2}, |Z̃_{2}| = 4. Now let Z_{1} ≰ M_{α+1}, for some α + 1 ∈ △(α). We have |Z_{α+1} : Z_{α+1} ∩ L_{1}| ≤ 2. As ⟨Z_{α+1}^{L_{α}}⟩ = V_{α} is abelian by (6.1) we see that V_{α} ∩ L_{1} acts quadratically on Z_{1}. Now there is a {2, 3}-group P (V_{α}∩L_{1})Q_{1}/Q_{1} where P is normal in this group. Let A = C I_{V_{α}∩L_{1}}(P) and C be a complement of A in (V_{α} ∩L_{1})Q_{1}/Q_{1}. We have that P C acts nontrivially on C I_{Z_{1}}(A) by the A × B-Lemma. Now by [GoLyS, (24.1)] and [GoLyS, (25.12)] there is some u ∈ C I_{Z_{1}}(A) \ Q_{2} such that |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(u)| = 2. Hence |V_{α} : C I_{V_{α}}(u)| ≤ 4, i.e. |Z_{α+1} : C I_{Z_{α+1}}(u)| ≤ 4.

By (4.4) every subgroup of index 4 in Z_{α+1} contains some v ≠ 1, such that v is centralized by an elementary abelian group E_{α+1} ≤ L_{α+1} such that Γ_{E_{α+1},1}(G) ≤ M_{α+1}. So C I_{G}(v) ≤ M_{α+1} and then u ∈ M_{α+1}. But u ∉ Q_{α}, a contradiction.

So let t = p. Then we have the exceptional case L_{1}/Q_{1} = BA, B/O_{r}(B) ≅ L_{2}(r). Again [O_{r}(B), Z_{1}] = 1.

Let α + 1 ∈ △(α) with Z_{1} ≰ M_{α+1}. Then by (4.10) we get that C I_{G}(v) ≤ M_{α+1} for every v ∈ Z_{α+1}^{♯}. As Z_{1} ≤ C I([Z_{2}, Z_{α}]), we see that [Z_{α+1}, Z_{2}] = 1 and so Z_{α+1} ≤ L_{1}. Now Z_{α+1} ∩ Q_{1} = 1. Choose u ∈ Z_{1}, u ∉ M_{α+1}. Then |[Z_{α+1}, u]| = |Z_{α+1}| = |Z_{1}|, a contradiction as |u^{L_{1}}| ≤ |Z_{1}| − 1.

Suppose b = b_{1} and [Z_{1}, Z_{α}] ≠ 1. Then by (6.2) α = 2. Furthermore as in (6.2) we see that Z_{1} is an F-module with Z_{2}/Z_{2} ∩ Q_{1} offending.

(6.3) Hypothesis. b = b_{1} and [Z_{1}, Z_{α}] ≠ 1. Suppose furthermore there is some component K such that Z_{1} is an F-module for K with Z_{2}/C I_{Z_{2}}(K) offending.

Under (6.3) we have

(6.4) Hypothesis. L_{2} is nonsolvable with E(L_{2}/Q_{2}) ≅ L_{2}(r), (S)L_{3}(r), (S)U_{3}(r), Sz(r), Sp_{4}(r), L_{2}(r) × L_{2}(r), Sz(r) × Sz(r), r > 2 even, SL_{3}(r) ∗ SL_{3}(r), r = 4, SU_{3}(r) ∗ SU_{3}(r), r = 8, or F (L_{2}/Q_{2}) ≠ 1 and E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ L_{2}(r), r = 4.

As by construction L_{2} ∩ M = L_{2} ∩ L_{1} we see that |Z_{1} : Z_{1} ∩ Q_{2}| = r or E(L_{2}/Q_{2}) ≅ Sp_{4}(r), L_{2}(r) × L_{2}(r), Sz(r) × Sz(r), SL_{3}(r) ∗ SL_{3}(r) or SU_{3}(r) ∗ SU_{3}(r) in which cases |Z_{1} : Z_{1} ∩ Q_{2}| = r^{2}.

Now let x be a primitive prime divisor of r − 1, which exists by [Zy], or x = 9 in case r = 64. We fix the following notation

ω ∈ L_{1} ∩ L_{2}, o(ω) = x .

By (1.66) there is some ω with N_{G}(⟨ω⟩) ≰ M. Furthermore x ∉ σ(M) for E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r), r even.

(6.5) Lemma. Suppose (6.3) and (6.4). If N_{G}(⟨ω⟩) ≰ M, then we have [K, ω] ≤ K.

Proof . Suppose false. Then let K_{1}K_{2} … K_{y} be some orbit of ⟨ω⟩, i.e. y = x or y = 3 and x = 9.

Let x = y. Then C I_{M}(ω) involves some group isomorphic to K. Application of (1.63) yields that m_{p}(K) = 1 for every prime p | |K|. So we get with (1.5) that K ≅ L_{2}(q), Sz(q) or J_{1}. As K has to admit an F-module, we get with (1.18) K ≅ L_{2}(q), q even or K ≅ L_{2}(7).

If y = 3 and x = 9, then 3 ∉ σ(M) and so m_{3}(K) = 1. Hence by (1.5) we see K ≅ L_{2}(q), Sz(q), J_{1}, (S)L_{3}(q) or (S)U_{3}(q). Again (1.18) implies K ≅ L_{2}(q) or (S)L_{3}(q), q even.

So in any case we have K ≅ L_{2}(q) or (S)L_{3}(q), q even. Now we see |Z_{1} : Z_{1} ∩ Q_{2}| ≥ q^{y}. This shows r ≥ q^{y} or r^{2} ≥ q^{y}.

Suppose y = x. Then x|2^{y−1} − 1. Hence 2^{y−1} ≥ r. This shows q = 2. So K ≅ L_{3}(2) and |Z_{1} : Z_{1} ∩ Q_{2}| = r^{2}.

Suppose first 3 ∈ σ(M). Then by (4.1) E(L_{2}/Q_{2}) ≅ Sz(r) × Sz(r). Now 4^{y} ≥ |Z_{2} : C I_{Z_{2}}(⟨K^{⟨ω⟩}⟩)| ≥ r^{4}. Hence as r^{2} ≥ 2^{y}, we see that r^{2} = 2^{y}. But then y is even, a contradiction. So we have 3 ∉ σ(M). So y = 3 = x and then r = 4. But we have [K^{⟨ω⟩}, Z_{1}] is a direct sum of three natural modules for the three components. Hence we see that |[Z_{1}, K^{⟨ω⟩}]/C I_{[Z_{1},K^{⟨ω⟩}]}(Z_{2})| = 8, a contradiction as L_{1} ∩ L_{2} acts on this group and so the order has to be power of r.

As C I_{G}(u) ≤ M for any u ∈ C I_{Z_{2}}(⟨K^{⟨ω⟩}⟩)^{♯}, we see |Z_{2} : C I_{Z_{2}}(⟨K^{⟨ω⟩}⟩)| ≥ r^{4}.

We are left with x ≠ y, i.e. y = 3, x = 9. Then r = 64. Furthermore q ≤ 16. Now choose ν ∈ L_{1} ∩ L_{2}, o(ν) = 7. Then [K, ν] ≤ K as 3 ∉ σ(M). As N_{G}(⟨ν⟩) ≰ M, we get m_{7}(K) ≤ 1. This shows that K ≅ L_{2}(q) or K ≅ (S)L_{3}(q) and [ν, K] = 1. In the latter we get m_{p}(K) = 1 for p ∈ σ(M) by (1.63). Hence K ≅ L_{2}(7).

So in any case K ≅ L_{2}(q), q ≤ 16, or L_{2}(7). Suppose 7 | |̸K|, then [ν, K] = 1 and so [ν, K^{⟨ω⟩}] = 1. Again by (1.63) we get a contradiction. So we have K ≅ L_{2}(8) or L_{2}(7). So in both cases 7 ∈ σ(M) and ν ∈ E, E elementary abelian of order 7^{3}, so N_{G}(⟨ν⟩) ≤ M, a contradiction. This proves (6.5).

(6.6) Lemma. Assume (6.3) and (6.4). There is some ω ∈ L_{1} ∩ L_{2} with N_{L_{2}}(⟨ω⟩) ≰ M and [K, ω] ≠ 1, [K, ω^{3}] ≠ 1 for o(ω) = 9.

Proof . Suppose [K, ω] = 1 or [K, ω^{3}] = 1 for all ω ∈ L_{1} ∩ L_{2}, o(ω) = x, with N_{L_{2}}(⟨ω⟩) ≰ M. Then by (1.63) m_{p}(K) = 1 for p ∈ σ(M). By (1.18)

K ≅ (S)L_{n}(q), n ≤ 4 for q > 2, n ≤ 7 for q = 2, Sp_{4}(q), Sp_{6}(q), Ω_{8}^{−}(q), U_{4}(q), G_{2}(q), A_{n}, n ≤ 11.

Set Ω = ⟨ω⟩ if (r − 1)^{2} |̸ |L_{1} ∩ L_{2}| and Ω = ⟨ω, ω_{1}⟩, where |Ω| = x^{2} if (r −1)^{2} | |L_{1} ∩L_{2}|.Furthermore we choose notation such that N_{G}(⟨ω_{1}⟩) ≰ M, too. So we have [Ω, K] = 1 or [Ω^{3}, K] = 1, where Ω^{3} = ⟨ω^{3}, ω_{1}^{3}⟩. We have that |C I_{S}(Ω) : C I_{Q_{1}Q_{2}}(Ω)| ≤ 2. Hence we get Z_{2}/Z_{2} ∩ Q_{1} projects into Z(S/Q_{1} ∩ K) or q = 2 and |Z_{2}/C I_{Z_{2}}(K)| ≤ 4, or 8 if K ≅ Sp_{4}(2) or Sp_{6}(2).

In any case we may choose ω such that [Z_{1}, ω] ≰ Q_{2} and then K acts nontrivially on V_{1} = [Z_{1}, ω] or V_{1} = [Z_{1}, ω^{3}], respectively, which is a module over GF (r) or GF (4).

Suppose first |Z_{2}/C I_{Z_{2}}(K)| ≤ q, then we have K ≅ (S)L_{n}(q), Sp_{4}(q) or Sp_{6}(q), K ≅ A_{11}, and in V_{1} there is just one natural module involved (1.18). This shows [Z_{2}, V_{1}] ≤ V_{1}. We have |V_{1} : V_{1} ∩ Q_{2}| = r or r^{2} or 4^{t}. On the other hand |V_{1} : C I_{V_{1}}(Z_{2})| = q. So we have q = r or q = r^{2}, or q is a power of 4. So K ≅ SL_{n}(q), q > 2, Sp_{4}(q) or Sp_{6}(q). As N_{G}(⟨ω⟩) ≰ M, we get m_{x}(K) ≤ 2, m_{3}(K) ≤ 2 for x = 9. This shows K ≅ L_{2}(q), SL_{3}(q) or Sp_{4}(q). Now also Z_{2} is an F-module.

Suppose next that K ≅ Sp_{4}(q) or Sp_{6}(q) and |Z_{2}/C I_{Z_{2}}(K)| > q. Now |Z_{1} : Z_{1} ∩ Q_{2}| = q^{2}. Hence again V_{1} is irreducible and we get r = q again. Again we get that Z_{2} is an F-module.

So assume |Z_{2}/C I_{Z_{2}}(K)| ≤ 4, 8 respectively. As |[V_{1}, u]| ≥ 4 for u ∈ Z_{2} with [u, K] ≠ 1 we see that V_{1} involves just one nontrivial irreducible module again. So either Z_{2} is an F –module or |Z_{2}/C I_{Z_{2}}(K)| = 8 and |V_{1} : C I_{V_{1}}(Z_{2})| = 4. Now m_{3}(K) ≤ 2 and so K ≅ Sp_{4}(2)′ and the natural module is involved, which is not a module over GF (4).

So in any case Z_{2} is an F –module with Z_{1} offending. Now as [Ω, Z_{2}/C I_{Z_{2}}(K)] = 1, we get a contradiction to (1.31) and the action of an offending group on an F-module for L_{2}. Now the assertion follows with (6.5).

We now fix the ω from (6.6) for the remainder.

(6.7) Lemma. Assume (6.3) and (6.4). If K ≅ G(q), a group of Lie type in even characteristic, then r ≠ q.

Proof . So suppose r = q. Let ⟨ν⟩ ≤ L_{1} ∩ L_{2}, o(ν) = q − 1, ω ∈ ⟨ν⟩. Let ⟨ν_{1}⟩ = N_{⟨ν⟩}(K). By (6.5) ω ∈ ⟨ν_{1}⟩. We are going to show

(1) p | q̸ − 1 .

Suppose false. As r = q, we see that by (6.6) ω induces an inner-diagonal automorphism on K. As N_{L_{2}}(⟨ω⟩) ≰ M we get m_{x}(K) ≤ 3 and so that K ≅ L_{n}(q), n ≤ 4, Sp_{2n}(q), n ≤ 3, U_{n}(q), n ≤ 7, Ω_{8}^{−}(q) or G_{2}(q).

On the other hand let μ ∈ ⟨ν⟩, o(μ) = p. As N_{L_{2}}(⟨ω⟩) ≰ M also N_{L_{2}}(⟨μ⟩) ≰ M . If μ ∈ ⟨ν_{1}⟩, then m_{p}(C I_{M}(μ)) ≤ 2. If furthermore p | |C I_{M/O_{p′}(M)}(KO_{p′}(M)/O_{p′}(M))|, then m_{p}(C I_{M}(μ)) ≥ 3. Hence we have m_{p}(K) ≥ 4. As p | q − 1, this is impossible.

So we have ⟨ν⟩ ≠ ⟨ν_{1}⟩. As ω normalizes K^{ν^{i}} for any i and C I_{K^{ν^{i}}}(ω) contains an element of order q − 1 for all i, we get that there are at most three components K^{ν^{i}}, otherwise N_{G}(⟨ω⟩) ≤ M. So |⟨ν⟩ : ⟨ν_{1}⟩| = 3. As μ ∉ ⟨ν_{1}⟩, we get p = 3. But now 3 | |⟨ν⟩| and so by (1.66), we get again L_{2} ≤ M, a contradiction.

By (1) we get

(2) K ≅ L_{n}(q), n ≤ 4, Sp_{2n}(q), n ≤ 3, U_{n}(q), n ≤ 7, Ω_{8}^{−}(q) or G_{2}(q).

(3) E(L_{2}/Q_{2}) ≅ S̸p_{4}(q)

Suppose E(L_{2}/Q_{2}) ≅ Sp_{4}(q). Then |Z_{1} : Z_{1} ∩ Q_{2}| = q^{2}. Furthermore by quadratic action and (1.14) Z_{2} just involves natural, dual of natural and trivial modules.

As L_{2} contains some element inducing the diagram automorphism on E(L_{2}/Q_{2}), we see that with any natural module also its dual is involved. So |Z_{2} : Z_{2} ∩ Q_{1}| ≥ q^{4}. As before we have that K^{L_{1}∩L_{2}} contains at most three components. As p |̸ q − 1 by (1), we get K ≅ L_{2}(q) or U_{3}(q). But then |Z_{2}/C I_{Z_{2}}(⟨K^{L_{1}∩L_{2}}⟩)| ≤ q^{2}, a contradiction. So we have that L_{1} ∩ L_{2} normalizes K. As |Z_{1} : C I_{Z_{1}}(Z_{2})| = q^{2}, we see with (1.45) and (1.18) that K ≅ L_{4}(q), Sp_{6}(q). Now L_{1}∩L_{2} contains (ZZ_{q−1}×ZZ_{q−1})·2. Hence m_{p}(K(L_{1}∩ L_{2})) ≥ 4 for p | q−1, if K ≅ Sp_{6}(q). This implies K ≅ L_{4}(q). Furthermore Z_{2} acts as the uniquely determined subgroup of order q^{4} on Z_{1}. This shows with (1.18) that [K, Z_{1}] = U_{1} is the natural module. But now for t ∈ U_{1} we have |Z_{2} : C I_{Z_{2}}(t)| ≤ q^{2}. But this is impossible as Z_{2} involves the natural module and its dual and no element can induce transvections on both modules. This contradiction proves (3).

By (3) in any case we now have

(4) Z_{2} = ⟨t | |Z_{1} : C I_{Z_{1}}(t)| ≤ q⟩ .

Hence Z_{2} induces transvections on Z_{1}. So we have

(5) Z_{1} involves exactly one irreducible nontrivial module.

Now by (5.1) and (1) and q > 2 we get that for [K, Z_{1}] = U_{1}, every x ∈ U_{1} is centralized by some elementary abelian group E of order p^{2} with Γ_{E,1}(G) ≤ M. So let g ∈ L_{2} with U_{1}^{g} ≰ M. The structure of E(L_{2}/Q_{2}) tells us that for X = ⟨U_{1}, U_{1}^{g}⟩ we may assume X/O_{2}(X) ≅ L_{2}(q) or Sz(q) and O_{2}(X) ≤ L_{1}. Now U_{1} ∩ U_{1}^{g} ≤ Z(X). So U_{1} ∩ U_{1}^{g} = 1.

As U_{1}∩O_{2}(X)≤ O_{2}(X), we get ⟨U_{1}∩O_{2}(X), U_{1}^{g}∩O_{2}(X)⟩ = (U_{1}∩O_{2}(X))× (U_{1}^{g} ∩ O_{2}(X)). So U_{1}^{g} ∩ O_{2}(X) induces the full group of transvections of U_{1} to the hyperplane U_{1} ∩ O_{2}(X). This shows that

(6) K ≅ L_{n}(q), n ≤ 4, U_{1} = [Z_{1}, K] is the natural module.

(7) Z_{1} = Q_{1} = O_{2}(M)

By (6) we have Z_{1} = U_{1} ⨁ C I_{Z_{1}}(K). Suppose U_{1} ≠ Z_{1}. We are going to show that there is some V_{1} with V_{1} ≤ Z_{1} and U_{1} < V_{1}, such that for any u ∈ V_{1}^{♯} we have C I_{G}(u) ≤ M. By (4.4) we may assume m_{p}(L_{1}) ≥ 3 and m_{p}(C I_{L_{1}}(Z_{1})) = 0 for m_{p}(L_{1}) = 3. We have m_{p}(K) ≤ 2, and m_{p}(K) = 1 for n ≤ 3. So we may assume m_{p}(C I(K)) ≥ 1, and m_{p}(C I(K)) ≥ 2 for n ≤ 3. If m_{p}(C I(K)) ≥ 3, there is some subgroup U of order p^{2} in C I(K) such that C I_{Z_{1}}(U) > V_{1} and we are done.

Let m_{p}(C I(K)) = 2. And assume there is no such V_{1}. There is some ρ ∈ C I(K), o(ρ) = p and some Z_{1} > Ṽ_{1} > U_{1}, such that [ρ, Ṽ_{1}] = 1. If every z ∈ Z_{1} is centralized by some p-element in K, we get that any element in V_{1} = Ṽ_{1} is centralized by some elementary abelian p-group E of order p^{2} with Γ_{E,1}(G) ≤ M.

Assume now that there is some z ∈ U_{1} which is not centralized by some p-element in K. Then we see m_{p}(K) = 1. If m_{p}(L_{1}) > 3 there is some x ∈ L_{1}, o(x) = p, x induces a field automorphism on K. Furthermore by (4.12) we may assume [x, ⟨ρ, ρ_{1}⟩] = 1, ⟨ρ, ρ_{1}⟩ ≤ C I(K), elementary abelian of order p^{2}. Now we may replace x by xρ_{1}^{i}, for suitable i, such that V_{1} = U_{1}C IṼ_{1}(xρ_{1}^{i}) > U_{1}. But now as every z ∈ U_{1} is centralized by a conjugate of xρ_{1}^{i}, we see that every z ∈ V_{1} is centralized by some E, E elementary abelian, |E| = p^{2}, Γ_{E,1}(G) ≤ M.

So let m_{p}(L_{1}) = 3. Then m_{p}(C I_{L_{1}}(Z_{1})) = 0 and so by (4.4) we have some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, such that ρ ∈ E and C I_{E}(Z_{1}) ≠ 1. Set V_{1} = Ṽ_{1}.

Let finally m_{p}(C I(K)) = 1. Now m_{p}(K) = 2. By (4.12) no field automorphisms are involved. So we have m_{p}(L_{1}) = 3. Now any u ∈ Z_{1}^{♯} is centralized by some p–element in K and so by (6.4) any u ∈ Z_{1}^{♯} is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M.

So we have found our V_{1}. But now we replace U_{1} by V_{1} and see that V_{1}^{g}∩L_{1} has to induce all transvections on the hyperplane V_{1} ∩ L_{1}^{g}, a contradiction, as this hyperplane is of order greater than q^{n−1}.

So we have Z_{1} = U_{1}. If |Z_{2} : Z_{2} ∩ Q_{1}| = q, then we see that Z_{2} is an F-module, so E(L_{2}/Q_{2}) ≅ L_{2}(q) or L_{2}(q) × L_{2}(q) by (1.31). Hence Z_{1} ≰ Q′_{1}.

So assume |Z_{2} : Z_{2} ∩ Q_{1}| > q. Then Q_{1}Q_{2} does not contain a Sylow 2subgroup of K. This again shows that Q_{1}Q_{2} = Z_{1}Q_{2}, so Z_{1} ≰ Q′_{1}. As Z_{1} is irreducible, we get (7).

By (1) we have m_{p}(K) ≤ 2. As C I_{M}(O_{2}(M)) ≤ O_{2}(M) we get with (1) and (6.4) that m_{p}(L_{1}) ≥ 4. Hence we have that m_{p}(C I(K)) = 0. But as p | q̸ − 1 this would force some p-element in M to centralize O_{2}(M) = Z_{1}, a contradiction again.

(6.8) Lemma. Assume (6.3) and (6.4). If K ≅ G(q) is a group of Lie type, then r = 4, q = 2, ⟨S, ω⟩ ∩ K/O_{2}(⟨S, ω⟩ ∩ K) ≅ Σ_{3}, ω ∈ K.

Proof . By (6.7) we know q ≠ r. First we show

(1) r > q.

Suppose false. By quadratic action, (1.14) and (1.49) in case E(L_{2}/Q_{2}) ≅ Sp_{4}(r), we get some t ∈ Z_{2} with |Z_{1} : C I_{Z_{1}}(t)| ≤ r, a contradiction to (1.18).

Suppose first q = 2. By (1.18) we have K ≅ L_{n}(2), SU_{n}(2), Ω_{2n}^{±}(2), Sp_{2n}(2) or G_{2}(2)′. Now (1.65) tells us that o(ω) = 3 or 9. Hence we have 3 ∉ σ(M). This shows that K ≅ L_{n}(2), n ≤ 7, SU_{n}(2), n ≤ 4, Sp_{2n}(2), n ≤ 3, Ω_{8}^{−}(2) or G_{2}(2)′. As these groups do not posses automorphisms of order 9 interchangeable with a Sylow 2-subgroup, we see that r = 4.

If K ≅ Sp_{2n}(2), G_{2}(2)′ or L_{n}(2), we have the assertion as long as ⟨S, ω⟩ = S⟨ω⟩. So suppose false. Then we have 9 | |L_{1} ∩ L_{2}| and so E(L_{2}/Q_{2}) ≅ (S)L_{3}(4), Sp_{4}(4), L_{2}(4) × L_{2}(4) or SL_{3}(4) ∗ SL_{3}(4). But in all cases any element of order 3 in E(L_{2}/Q_{2}) has its normalizer not contained in M. So as ⟨S, ω⟩ ∩ K/O_{2}(⟨S, ω⟩ ∩ K) ≅ Σ_{3}, we see that there is some ω_{1} ∈ L_{1} ∩ L_{2}, o(ω_{1}) = 3, which is in K. Hence we may assume ω ∈ K.

So assume now q > 2. Then by (1.65) and (1.64) we get r = q^{2} and K ≅ U_{n}(q) or Ω_{2n}^{−}(q), or o(ω) = 3 or 9 and K ≅ O_{8}^{+}(q), q ≤ 32. But in the latter 3 ∈ σ(M), a contradiction. So we have

(2) r = q^{2} and K ≅ U_{n}(q) or Ω_{2n}^{−}(q).

By (5.1) there is U_{1} ≤ Z_{1}, such that every x ∈ U_{1} is centralized by an elementary abelian subgroup E of order p^{2} in M, such that Γ_{E,1}(G) ≤ M. Recall that (5.1) (ii) cannot occur as in this case 3 ∈ σ(M) and r = 4, i.e. o(ω) = 3. Let g ∈ L_{2} such that ⟨U_{1}, U_{1}^{g}⟩ = X ≰ M. Suppose |U_{1} : U_{1} ∩ O_{2}(X)| = r, then U_{1}^{g} ∩ O_{2}(X) induces the full group of transvections to the subgroup U_{1} ∩ O_{2}(X). As |U_{1} : U_{1} ∩ O_{2}(X)| = q^{2}, we get with (1.45) that |U_{1}^{g} ∩ O_{2}(X)| ≤ q. But then |U_{1}| ≤ q^{3}, a contradiction.

So we have |U_{1} : U_{1} ∩ O_{2}(X)| > r. This shows E(L_{2}/Q_{2}) ≅ Sp_{4}(r). But now again by (1.49) there is some Z̃_{2} ≤ Z_{2}, |Z̃_{2}| = r, Z̃_{2} ∩ Q_{1} = 1 but |Z_{1} : C I_{Z_{1}}(Z̃_{2})| = r = q^{2}. But this contradicts (1.45).

(6.9) Lemma. Assume (6.3) and (6.4). Then ⟨S, ω⟩/O_{2}(⟨S, ω⟩) ≅ Σ_{3}, ω ∈ K and r = 4.

Proof . By (6.8) and (1.18) it is enough to treat the case K ≅ A_{n}, n ≥ 6. Now by (1.9) we get o(ω) = 3, so r = 4. Furthermore ⟨S, ω⟩/O_{2}(⟨S, ω⟩) ≅ Σ_{3} as in (6.8).

By (6.9) we have 3 ∉ σ(M). Hence K ≅ L_{n}(2), n ≤ 7, Sp_{6}(2), U_{4}(2), Ω_{8}^{−}(2), G_{2}(2)′ or A_{n}, n ≤ 11.

If m_{p}(C I_{L_{1}/Q_{1}}(K)) ≥ 2, we have that ω is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. Suppose m_{p}(C I_{L_{1}/Q_{1}}(K)) ≤ 1. If K ≤ E(M/O_{p′}(M)), then we see that m_{p}(C I_{M}(KO_{p′}(M)/O_{p′}(M)) ≥ 2, as m_{p}(K) ≤ 2 and K does not allow outer p–automorphisms. Hence in this case too ω is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. If K ≰ E(M/O_{p′}(M)), then by the construction in (4.4) we have that [E(M/O_{p′}(M)), K] = 1, so we may assume m_{p}(E(M/O_{p′}(M)) ≤ 1 and so we have m_{p}(F (M/O_{p′}(M)) ≤ 3. Hence as K acts faithfully on F (M/O_{p′}(M)), we see K ≅ L_{2}(7) or K ≅ A_{6} and Z_{1} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M.

Suppose we do not have the latter. Then by (1.63) C I_{Z_{1}}(ω) = 1. We have that ω is in some minimal parabolic. If K ≅ L_{n}(2) then ω centralizes some nontrivial vector in the natural module. Hence the same is true for the module V (λ_{2}) as the maximal parabolic with L_{n−1}(2) on top induces the natural module. Hence we have that K ≅ L̸_{n}(2). If K ≅ Sp_{6}(2) then by (1.18) Z_{1} involves the natural or spin module. Again in the natural module ω has a fixed point. So we have that Z_{1} is the spin module. If K ≅ U_{4}(2) we must have the natural O_{6}^{−}(2)–module. If K ≅ Ω_{8}^{−}(2) then Z_{1} is the natural module by (1.18) and so ω has a fixed point. If K ≅ A_{n} and Z_{1} is the permutation module, we get a fixed point for ω or n = 6.

So we summarize. We have K ≅ L_{3}(2), A_{6}, A_{7}, U_{4}(2) (and Z_{1} is the O_{6}^{−}(2)–module), G_{2}(2)′, or Sp_{6}(2) (and Z_{1} is the spin module). In any case for all u ∈ Z_{1}^{♯} we have C I_{G}(u) ≤ M.

Now choose g ∈ Z_{2} with Z_{1}^{g} ≰ M. Set X = ⟨Z_{1}, Z_{1}^{g}⟩. Suppose first |Z_{1} : Z_{1}^{g} ∩ O_{2}(X)| = 4. As we have Z_{1}^{g} ∩ O_{2}(X) ∩ Z_{1} = 1, we see that Z_{1}^{g} ∩O_{2}(X)∩Q_{1} = 1. Hence |Z_{1}^{g} ∩O_{2}(X)| ≤ 2^{m_{2}(Aut(K))}. Further Z_{1}^{g} ∩O_{2}(X) centralizes a subgroup of index 4 in Z_{1}. As |Z_{1}^{g} ∩ O_{2}(X)| ≤ 8 for K ≅ G_{2}(2)′ we see that |Z_{1}| ≤ 2^{5}, a contradiction. If K ≅ Sp_{6}(2), we get |Z_{1}^{g} ∩ O_{2}(X)| = 2^{6} and so Z_{1}^{g} ∩ O_{2}(X) is the uniquely determined subgroup of order 64 in a Sylow 2-subgroup of Sp_{6}(2). But the centralizer of this group in the spin module is of order two. Let K ≅ U_{4}(2). Then |Z_{1}^{g} ∩ O_{2}(X)| = 2^{4}. But a subgroup of Aut(K) centralizing a subgroup of order 4 in the natural O_{6}^{−}(2)– module is never of order 16.

So we are left with K ≅ L_{3}(2), A_{6} or A_{7} and |Z_{1}| = 2^{3} or 2^{4}. Now there is t ∈ Z_{1} \ Q_{2} with |Z_{2} : C I_{Z_{2}}(t)| ≤ 4. Application of (1.43) shows E(L_{2}/Q_{2}) ≅ A_{5} or A_{5} × A_{5} and so Z_{1} ≰ Q′_{1}. If Z_{1} is irreducible then Q′_{1} = 1. We show that Q_{1} is abelian in general. So we may assume |Z_{1}| = 16 and K ≅ L_{3}(2). but as |Z_{1} : C I_{Z_{1}}(u)| = 4 for any u ∈ Z_{2} \ Q_{1}, we see that Z_{1} is an indecomposabe module having the natural module U_{1} as a submodule. Now |U_{1} : U_{1} ∩ Q_{2}| = 4 and so U_{1} ≰ Q′_{1}. This again shows Q′_{1} = 1.

Suppose now that |Z_{1}^{g} : Z_{1}^{g} ∩ O_{2}(X)| = 16 in which case E(L_{2}/Q_{2}) ≅ Sp_{4}(4). Then X/O_{2}(X) ≅ Sp_{4}(4) and again Z_{1} ∩ O_{2}(X) ∩ Z_{1}^{g} = 1. Furthermore |Z_{1} ∩ O_{2}(X)| ≥ 16. This shows that K ≅ U_{4}(2) or Sp_{6}(2). Furthermore |Z_{1}| ≥ 2^{8} and so K ≅ Sp_{6}(2) and Z_{1} is the spin module. Hence |Z_{1} ∩ O_{2}(X)| = 16. As Z_{2} centralizes a subgroup of index 16 in Z_{1} we see that |Z_{2} : C I_{Z_{2}}(Z_{1})| ≤ 2^{5} and so Z_{1} offends as an F_{1}–module on Z_{2}. But this contradicts (1.43).

So we have Q′_{1} = 1 and so Q_{1} = O_{2}(M) is elementary abelian. Furthermore |Z_{1}| ≤ 16 and K ≅ L_{3}(2), A_{6} or A_{7} and E(L_{2}/Q_{2}) ≅ L_{2}(4). Hence O_{2}(M) is an F –module for K with Z_{2}/C I_{Z_{2}}(O_{2}(M)) offending. So we see that [K, O_{2}(M)] ≤ Z_{1}. As Ω_{1}(Z(S)) ≤ Z_{1}, we get with (1.20) that |O_{2}(M)| ≤ 2^{5}. But this contradicts m_{p}(M) ≥ 4 and C I_{M}(O_{2}(M)) ≤ O_{2}(M).

So we have with (2.1)

(6.10) Lemma. Assume (6.3). Then L_{2} is solvable or E(L_{2}/Q_{2}) ≅ L_{2}(r), L_{3}(r), U_{3}(r), P Sp_{4}(r), L_{2}(r)×L_{2}(r), r odd, A_{6}, A_{9}, 3·A_{6}, 3A_{6} ∗3A_{6}.

As Z_{1} ≰ Q_{2}, we have E(L_{2}/Q_{2}) ≅ A̸_{9}. Next we show

(6.11) Lemma. Assume (4.3). Then |Z_{1} : Z_{1} ∩ Q_{2}| > 2.

Proof . For this assume |Z_{1} : Z_{1} ∩ Q_{2}| = 2. Then Z_{2} induces transvections on Z_{1} and so K ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{−}(2) or A_{n}. By (5.1) there is U_{1} ≤ Z_{1}, such that any z ∈ U_{1} is centralized by some elementary abelian group E of order p^{2} in M with Γ_{E,1}(G) ≤ M. So let g ∈ L_{2} such that ⟨U_{1}, U_{1}^{g}⟩/O_{2}(⟨U_{1}, U_{1}^{g}⟩) is dihedral of order 2t, t odd. Then we see that U_{1}^{g}∩Q_{2} induces the full group of transvection on U_{1} to the hyperplane U_{1} ∩ Q_{2}. This implies K ≅ L_{n}(2) and U_{1} is the natural module. If Z_{1} > U_{1}, then we see as in (6.7) (3) that there is some V_{1} > U_{1} such that every z ∈ V_{1} is centralized by some elementary abelian E of order p^{2} with Γ_{E,1}(G) ≤ M, a contradiction. So we have Z_{1} is the natural module. In particular

(1) |Ω_{1}(Z(S))| = 2.

Now let ν ∈ ⟨U_{1}, U_{1}^{g}⟩, o(ν) = t. Then (U_{1} ∩ Q_{2}) × (U_{1}^{g} ∩ Q_{2}) = [Q_{2}, ν] and Ω_{1}(Z(S)) ≤ [Q_{2}, ν]. As Φ(Q_{2}) ≤ C I_{Q_{2}}(ν) we see that Φ(Q_{2}) = 1. As in a linear group the full transvection group is its own centralizer we see that

(2) S/Q_{1}Q_{2} involves a Sylow 2-subgroup of L_{n−1}(2).

Next suppose that z ∈ Z_{1} \ Q_{2} inverts an elementary abelian group of order t^{2}, t a prime. Then there is some ω ∈ L_{2}, ω^{z} = ω^{−1}, C I_{[Q_{2},z]}(ω) ≠ 1. But [Q_{2}, z] ≤ Z_{1} and every x ∈ Z_{1} is centralized by some elementary abelian group E of order p^{2} with Γ_{E,1}(G) ≤ M. So we have ω ∈ M, a contradiction as z ∈ O_{2}(M). As z ∈ Z(S/Q_{2}), this now implies together with (2.1)

(3) E(L_{2}/Q_{2}) ≅ L_{2}(r), L_{3}(r), U_{3}(r), 3 · A_{6}, P Sp_{4}(r), or L_{2} is a solvable {2, t}-group and σ(L_{2}) = ∅.

Suppose now n ≥ 5 and n ≥ 6 in case of E(L_{2}/Q_{2}) ≅ P Sp_{4}(r). Then (3) implies that L_{2} is solvable. Let C be a critical subgroup of O_{t}(L_{2}/Q_{2}), C_{1} = Ω_{1}(C). As Aut(C_{1}) involves a Sylow 2-subgroup of type A_{8}, and m_{t}(C_{1}) ≤ 3, we get a contradiction with (1.6). So (3) and (2) imply

(4) n ≤ 4 and n ≤ 5 in case of E(L_{2}/Q_{2}) ≅ P Sp_{4}(r).

Suppose now that L_{2} is solvable and O_{t}(L_{2}/Q_{2}) is not cyclic or E(L_{2}/Q_{2}) ≅ 3·A_{6}. Then by (1.6) and (2) z ∈ Z_{1} \Q_{2} acts on some extraspecial group T of order t^{3} such that C I_{T} (z) = Z(T). As |[z, Q_{2}]| ≤ 8, we see t = 3. Let u ∈ T \ Z(T). We may assume [u, Q_{2}] = (Z_{1} ∩ Q_{2}) × (Z_{1}^{g} ∩ Q_{2}). Now Z(T) acts on [u, Q_{2}]. As T ∩M ≤ Z(T) and every element in Z_{1} ∩Q_{2} is centralized by an elementary abelian group E of order p^{2} with Γ_{E,1}(G) ≤ M, we see C I_{Q_{2}}(Z(T)) ∩ ((Z_{1} ∩ Q_{2}) × (Z_{1}^{g} ∩ Q_{2})) ≠ 1. But then T acts as an elementary abelian group of order 9 on C I_{Q_{2}}(Z(T)), which gives that some ũ ∈ T \ Z(T) centralizes some x ∈ Z_{1} ∩ Q_{2}, x ≠ 1, a contradiction. So we have

(5) (i) If L_{2} is solvable, then O_{t}(L_{2}/Q_{2}) is cyclic.

(ii) E(L_{2}/Q_{2}) ≅ 3̸ · A_{6}.

Let L_{2} be nonsolvable. Then |Q_{2} : C I_{Q_{2}}(Z_{1})| ≤ 8 and |Q_{2} : C I_{Q_{2}}(Z_{1})| ≤ 16 in case of E(L_{2}/Q_{2}) ≅ P Sp_{4}(r). With (1.30) we see E(L_{2}/Q_{2}) ≅ L_{2}(7) ≅ L_{3}(2).

Now E(L_{2}/Q_{2}) is generated by four conjugates of Z_{1}. This shows that |Q_{2} : C I_{Q_{2}}(E(L_{2}/Q_{2}))| ≤ |[Z_{1}, Q_{2}]|^{4} ≤ 2^{12}. As Ω_{1}(Z(S)) ≤ Z_{1} ∩ Q_{2}, we get with (1) C I_{Q_{2}}(E(L_{2}/Q_{2})) = 1. So we now see |Q_{1}| ≤ 2^{9}.

If O_{t}(L_{2}/Q_{2}) ≠ 1 we get n = 3 and |Q_{2} : C I_{Q_{2}}(O_{t}(L_{2}/Q_{2}))| ≤ 16. Hence |Q_{2}| ≤ 16 and so t = 3 or 5, and |Q_{1}| ≤ 2^{3}. So we have

(6) |Q_{1}| ≤ 2^{9} and |Q_{1}| ≤ 2^{3} for L_{2} solvable.

As O_{2}(M) ≤ Q_{1} and some elementary abelian subgroup of order p^{4}, p odd, acts faithfully on O_{2}(M)/Φ(O_{2}(M)), we see |O_{2}(M)/Φ(O_{2}(M))| ≥ 2^{8}.

This shows 2^{8} ≤ |Q_{1}| ≤ 2^{9}, n = 4, and p = 3. As |Z(Q_{1})| ≥ 2^{4}, we also get that Q_{1} is elementary abelian. But as |Ω_{1}(Z(S))| = 2 by (1), we now have C I_{Q_{1}}(K) = 1, a contradiction to |[Q_{1}, K]| = 2^{4}. This proves (6.11).

(6.12) Lemma. Assume (6.3). Set U_{1} = [K, Z_{1}]. Suppose there is some x ∈ Z_{2} with [x, K] ≠ 1 such that |U_{1} : C I_{U_{1}}(x)| = 2. Then for z ∈ [Z_{2}, U_{1}] with z ≠ 1, we have C I_{L_{2}}(z) ≤ M

Proof . By assumption K ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2) or A_{n} and U_{1}/C I_{U_{1}}(K) is the natural module. Now by (5.1) every z ∈ U_{1} is centralized by some elementary abelian subgroup E of order p^{2} in M with Γ_{E,1}(G) ≤ M. Hence C I_{G}(z) ≤ M, if z ≠ 1.

(6.13) Lemma. Assume (6.3). Then L_{2} is solvable.

Proof . Let E(L_{2}/Q_{2}) = X_{1} or X_{1}X_{2}, X_{1} quasisimple. If |Z_{1} : C I_{Z_{1}}(X_{1})| ≤ 2, we see that [E(L_{2}/Q_{2}), Z_{2}] is a sum of two modules Z_{2}^{(1)}, Z_{2}^{(2)} with [Z_{2}^{(i)}, X_{3−i}] = 1. Furthermore there is some x ∈ Z_{2}^{(1)} with [x, U_{1}] ≠ 1 and [x, K] ≠ 1, where U_{1} = [K, Z_{1}]. Now |U_{1} : C I_{U_{1}}(x)| = 2 and [U_{1}, Z_{2}^{(1)}] is centralized by X_{2}. But this contradicts (6.12).

So we have |Z_{1} : C I_{Z_{1}}(X_{1})| > 2. Application of (1.12) shows X_{1} ≅ L_{3}(2), A_{6} or 3A_{6}. As Z_{1} ≤ S and there is no normal fours group in a Sylow 2-subgroup of Aut(L_{3}(2)), we get X_{1} ≅ A_{6} or 3A_{6} and N_{L_{2}/Q_{2}}(X_{1})/C I_{L_{2}/Q_{2}}(X_{1}) ≅ P ΓL_{2}(9). Hence |Z_{1} : C I_{Z_{1}}(X_{1})| = 4 and Z_{1}/C I_{Z_{1}}(X_{1}) corresponds to the center of a Sylow 2-subgroup of Σ_{6}. By (1.14) Z_{2} involves natural and dual Sp_{4}(2)-modules. But now again there is some x ∈ Z_{2} (in some natural module) such that |U_{1} : C I_{U_{1}}(x)| = 2. If [x, K] = 1, we would get [x, U_{1}] ≤ C I_{U_{1}}(K) and so [x, U_{1}] = 1. Thus we have [x, K] ≠ 1. But in the natural module every element is centralized by some 3-element in Σ_{6}. So [U_{1}, Z_{2}]^{♯} contains an element z with C I_{L_{2}}(z) ≰ M, contradicting (6.12).

(6.14) Lemma. We have that (6.3) does not hold.

Proof . Suppose false. Then by (6.13) L_{2} is solvable. Now by [GoLyS, (24.1)] there is some subgroup L̂_{2} of L_{2}/C I_{L_{2}}(Z_{2}) such that L̂_{2} = D_{1}×…×D_{n} with D_{i} ≅ D_{2t} where |Z_{1} : Z_{1} ∩ C I_{L_{2}}(Z_{2})| = 2^{n} and Z_{1}C I_{L_{2}}(Z_{2})/C I_{L_{2}}(Z_{2}) is a Sylow 2-subgroup of L̂_{2}.

Set U_{1} = [K, Z_{1}]. By [GoLyS, (25.12)] there is some z ∈ Z_{2} with |[U_{1}, z]| = 2 and [K, z] ≠ 1. By (6.11) n > 2. But now by quadratic action there is some 1 ≠ x ∈ [U_{1}, Z_{2}] which is centralized by some t-element in L̂_{2}. As Z_{1} ≤ O_{2}(M), we have L̂_{2} ∩ (L_{2} ∩ M)/C I_{L_{2}}(Z_{2}) = Z_{1}C I_{L_{2}}(Z_{2})/C I_{L_{2}}(Z_{2}), a contradiction to (6.12).

(6.15) Proposition. If b = b_{1}, then [Z_{1}, Z_{α}] = 1.

Proof . Suppose b = b_{1} and [Z_{1}, Z_{α}] ≠ 1. Let first K be p-constraint such that Z_{1} is an F –module for KZ_{2} with Z_{2}/C I_{Z_{2}}(K) offending. Then we get p = 3. Furthermore by quadratic action there is some z ∈ Z_{1} inducing a transvection on Z_{2}. This implies that L_{2} is a solvable {2, 3}- group, E(L_{2}/Q_{2}) ≅ A_{5}, A_{5} × A_{5} or A_{9}, or E(L_{2}/Q_{2})/F (L_{2}/Q_{2}) ≅ L_{2}(4). As p = 3, we get m_{p}(L_{2} ∩ M) = 0. This implies that L_{2}/Q_{2} ≅ Σ_{3}. Hence Z_{2} induces transvections on Z_{1} to the hyperplane Z_{1} ∩ Q_{2}. We get that [Z_{2}, O_{3}(K/C I_{K}(Z_{1}))] = N is of order three and |[N, Z_{1}]| = 4. As Z_{2}Q_{1}/Q_{1} < S/Q_{1}, we see that [N, Z_{1}] ≤ S. Now |[[N, Z_{1}], Q_{2}]| = 2. So let ω ∈ L_{2}, o(ω) = 3. Then |Q_{2} : C I_{Q_{2}}(ω)| = 4. Hence ω centralizes a subgroup of index 8 in Z_{1}. Let F be an elementary abelian 3-subgroup of maximal rank in L_{1}. Then |F : C I_{F} (Z_{1})| ≥ 9. So every subgroup of Z_{1} of index 8 contains some 1 ≠ z such that |C I_{F} (z)| ≥ 9. As Γ_{C I_{F} (z),1}(G) ≤ M, we get C I_{G}(z) ≤ M. But then ω ∈ M, a contradiction.

So by (6.14) we may assume that we have the exceptional case K_{1}/Q_{1} = BA, B/O_{r}(B) ≅ L_{2}(r). Now again [O_{r}(B), Z_{1}] = 1. As Z_{1} is an F-module we have r = 7 or r = 5. By (4.10) C I_{G}(z) ≤ M for any z ∈ Z_{1}^{♯}. Hence

(1) C I_{Z_{2}}(L_{2}) = 1. If ω ∈ L_{2}, C I_{Z_{1}∩Z_{2}}(ω) ≠ 1, then ω ∈ M.

(2) If z ∈ Z_{2} and [B, z] ≠ 1, then [Q_{1}, z] ≤ Z_{1}.

Otherwise [Q_{1}, z] ≤ Z_{1} and so [O_{r}(B), Q_{1}] ≤ Z_{1}. But then [O_{r}(B), Q_{1}] = 1, a contradiction.

(3) |Z_{1} : Z_{1} ∩ Q_{2}| = 2

Otherwise Z_{2} is an F-module with Z_{1} offending. As [Z_{1}, Z_{2}] ≤ Z_{1}, we see with (1.31) that L_{2} is solvable or L_{2}/C I_{L_{2}}(Z_{2}) ≅ A_{5} or S_{5} and Z_{2} is the natural L_{2}(4)-module.

In the latter Z_{1}∩Z_{2} = C I_{Z_{2}}(Z_{1}). So let z ∈ Z_{2}\Z_{1}∩Z_{2} with [z, S] ≤ Z_{1}∩Z_{2}. Then [z, Q_{1}] = [Z_{1}, z], a contradicting (2).

So we have that L_{2} is solvable. Now by [GoLyS, (24.1)] there is a subgroup L̂_{2} in L_{2}/C I_{L_{2}}(Z_{2}) such that L̂_{2} = D_{1} × … × D_{n} with D_{i} ≅ Σ_{3} and Z_{1}C I_{L_{2}}(Z_{2})/C I_{L_{2}}(Z_{2}) ∈ Syl_{2}(L̂_{2}). Furthermore n ≥ 2 by assumption. By quadratic action [D_{2}, [Z_{1} ∩ D_{1}, Z_{2}]] = 1. But this contradicts (1) and Z_{1} ≤ O_{2}(M).

By (2) we have that Z_{2} induces transvections and so r = 7. Hence p = 3, Sylow 3-subgroups of L_{2} are cyclic and m_{3}(L_{2} ∩ M) = 0 by (4.1) and (1.66). This together with (1.23) and (1.43) shows that E(L_{2}/Q_{2}) ≅ L_{3}(2), L_{2}/Q_{2} ≅ Σ_{3} or L_{2} is a solvable {2, 5}-group.

In the first case we have by (1.43) a submodule in Z_{2} which is the natural one. But then every element in this submodule is centralized by a 3-element in L_{2}, contradicting (1).

So L_{2} is solvable. Let 5 | |L_{2}|. As Z_{1}Q_{2}/Q_{2} ≤ Z(S/Q_{2}) we have that Z_{1}Q_{2}/Q_{2} acts fixed point freely on O_{5}(L_{2}/Q_{2})/Φ(O_{5}(L_{2}/Q_{2})). But |Z_{2} : C I_{Z_{2}}(Z_{1})| ≤ 4 and so we have that O_{5}(L_{2}/Q_{2}) is cyclic. As Z_{1} ≤ O_{2}(M) we have |O_{5}(L_{2}/Q_{2})| = 5 and |Z_{2}| = 16. So |Z_{2} : Z_{2} ∩ Q_{1}| = 4. As B/O_{7}(B) ≅ L_{2}(7) we see that Q_{1}Q_{2} ≠ S and so L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{20}.

But now in both cases L_{2}/C I_{L_{2}}(Z_{2}) ≅ F_{20} and L_{2}/Q_{2} ≅ Σ_{3} there is some z ∈ Z_{2} \ Z_{1} ∩ Z_{2} with [S, z] ≤ Z_{1} ∩ Z_{2}. Hence [S, z] ≤ Z_{1}, contradicting (2).

(6.16) Proposition. Suppose b = b_{1} > 1. Set V_{2} = ⟨Z_{1}^{L_{2}}⟩. Then V_{2} involves at least two nontrivial irreducible L_{2}/Q_{2}-modules.

Proof . Suppose b = b_{1} > 1. By (6.15) [Z_{1}, Z_{α}] = 1 hence α ∼ 2. This shows Ω_{1}(Z(S)) < L_{2}. Let (1, α) be a critical pair and (1, 2 …, α) be a path of length b. Then V_{2} ≤ L_{α} and V_{α} ≤ L_{2}. We have

[V_{2}, V_{α}] ≤ V_{2} ∩ V_{α}, V′_{2} = 1 and [Z_{1}, V_{α}] ≠ 1 .

This shows

[Z_{1}, Z_{α+1}] ≠ 1 for some α + 1 ∈ △(α) .

Choose V̂_{2} maximal in V_{2} such that V̂_{2} ⊴ L_{2} and [V̂_{2}, O^{2′}(L_{2})] = 1. Let V̂_{2} ≤ A ≤ V_{2} such that A/V̂_{2} is an irreducible L_{2}/Q_{2}-module. Suppose [V_{2}, O^{2′}(L_{2})] ≤ A. We show

V_{2} ≠ AZ_{1}

Suppose false. Assume V_{2} = V̂_{2}Z_{1}. As Z_{1} ≰L_{2}, we get Q_{1} ≤ Q_{2}. But this contradicts Z_{1} ≤ Q_{α−1}, Z_{1} ≰ Q_{α}. So we have V̂_{2}Z_{1} ≠ V_{2}. We have [Q_{2}, Z_{1}] ≤ Z_{1}, and so [V_{2}/V̂_{2}, Q_{2}] ≤ Z_{1}V̂_{2}/V̂_{2}, as [A, Q_{2}] ≤ V̂_{2}. As V̂_{2}Z_{1} ≠ V_{2}, we have V̂_{2}Z_{1} ≱ A. This now shows that [V_{2}, Q_{2}] ∩ A ≤ V̂_{2}. Hence we have [V_{2}, Q_{2}] ≤ V̂_{2}.

Set X = C I_{Q_{2}}(V̂_{2}). Assume [X, A] ≠ 1. Choose 1 ≠ v ∈ V̂_{2} ∩ [A, X]. Let F be a hyperplane of V̂_{2} not containing v. Then [A/F, X] = V̂_{2}/F.

Now we get C I_{V_{2}/F} (X) = V̂_{2}/F as [V_{2}, O^{2′}(L_{2})] ≤ A. Hence |X : C I_{X}(Z_{1})| ≥ |Z_{1} : Z_{1} ∩ V̂_{2}| and so Z_{1} is an F-module with offending subgroup X/C I_{X}(Z_{1}). We have [X, Z_{1}] ≤ V̂_{2}. Hence [Z_{1}, X, X] = 1. This shows that X acts quadratically. By (1.11) X/C I_{X}(Z_{1}) induces orbits of lenght ≤ 2 on the components of L_{1}/Q_{1}. Suppose [X, E(L_{1}/Q_{1})] ≠ 1. Hence again by (1.11) and as X ≤ S there is some component K normalized by X such that X/C I_{X}(K) offends on Z_{1}. Now (5.1) provides us with some submodule U_{1} of Z_{1} such that every u ∈ U_{1}^{♯} is centralized by some E ≤ M, E elementary abelian of order p^{2} and Γ_{E,1}(G) ≤ M, or L_{1} possesses a component K ≅ L_{2}(q), or U_{4}(2) and U_{1} the natural module.

As C I_{U_{1}}(X) ≠ 1, we get U_{1} ∩ V̂_{2} ≠ 1, and then O_{2′}(L_{2}) ≤ C I(U_{1} ∩ V̂_{2}) ≤ M in the first case, a contradiction. Hence we just have the latter. By (5.1) we have C I_{U_{1}}(K) ≠ 1. But C I_{U_{1}}(K) = 1, as C I_{U_{1}}(K) ∩ V̂_{2} ≠ 1, we have a contradiction.

So we have [E(L_{1}/Q_{1}), X] = 1. We get p = 3. By (4.10) we have p ∈ σ(M). As C I_{Z_{1}}(X) ≤ V̂_{2}, we get with [GoLyS, (24.1)] that |X/C I_{X}(K)| ≤ 4. Now there is a p–group K ∈ Syl_{2}(F (L_{1}/Q_{1})) with [K, X] ≠ 1. There is some element ρ of order 3 in K with |[Z_{1}, ρ]| = 4 and |[Z_{1}, ρ] ∩ V̂_{2} ≠ 1. Then there is some E elementary abelian of order 9 with Γ_{E,1}(G) ≤ M and [[Z_{1}, ρ] ∩ V̂_{2}, E] = 1, contradicting L_{2} ≰ M.

We have shown

X = C I_{Q_{2}}(A) .

As [O^{2′}(L_{2}/Q_{2}), Q_{2}] ≤ X, we see that A/V̂_{2} ≤ Z(Q_{2}) ∩ V_{2}/V̂_{2}. In particular Z(Q_{2}) is a nontrivial O^{2′}(L_{2}/Q_{2})-module. But now we may argue as in (6.15).

So we have V_{2} ≠ AZ_{1}. As V_{2} = ⟨Z_{1}^{L_{2}}⟩, there is some nontrivial irreducible module in V_{2}/A. This proves the lemma.

Let [V_{α} ∩ Q_{2}, Z_{β}] = 1 for all β ∈ △(2) with Z_{β} ≰ Q_{α}. Suppose furthermore [V_{2} ∩ Q_{α}, Z_{δ}] = 1 for all δ ∈ △(α) with Z_{δ} ≰ Q_{2}. Let γ ∈ △(2) with [V_{α} ∩ Q_{2}, Z_{γ}] ≠ 1. Then Z_{γ} ≤ Q_{α} and so there is some ε ∈ △(α) with [Z_{γ}, Z_{ε}] ≠ 1. Now Z_{γ} ≤ Q_{α} ≤ L_{ε} and Z_{ε} ≤ Q_{2} ≤ L_{γ}, a contradiction to (6.1). So we have [V_{2}, V_{α} ∩ Q_{2}] = 1 = [V_{2} ∩ Q_{α}, V_{α}]. Hence V_{2} is an F-module for L_{2}/Q_{2} with offending subgroup V_{α}/V_{α}∩Q_{2}. By (1.31) V_{2} involves exactly one irreducible nontrivial O^{2′}(L_{2}/Q_{2})-module, contradicting (6.16). So we have

(6.17) Lemma. Let b = b_{1} > 1. Then we may choose notation such that [V_{α} ∩ Q_{2}, Z_{1}] ≠ 1.

For what follows it is very important to show V_{α} ≤ M. For this it is enough to show that there is some 1 ≠ W ≤ V_{α} ∩ L_{1} such that W is normalized by E, E elementary abelian of order p^{2} with Γ_{E,1}(G) ≤ M. In this case we get V_{α} ≤ C I_{G}(W) ≤ M.

If (6.4) holds then as before we choose ω ∈ L_{1} ∩ L_{2}, o(ω) = x, where x is minimal with x | r − 1, i.e. it does not divide any smaller power of two minus one, and x = 9 for r = 64. According to (1.66) we may choose ω with N_{G}(⟨ω⟩) ≰ M. Fix α + 1 ∈ Δ(α) with [Z_{1}, Z_{α+1}] ≠ 1, Z_{1} ≰ M_{α+1}.

(6.18) Lemma. Assume (6.4), b = b_{1} > 1. Let K be a component of L_{1}/Q_{1}. Then [ω, K] ≤ K or [(V_{α} ∩ L_{1})Q_{1}/Q_{1}, K^{⟨ω⟩}] = 1.

Proof . Suppose false. Let K_{1}, …, K_{y} be some orbit under ⟨ω⟩, i.e. y = x or y = 3 and x = 9.

Let y = x. Then by (1.63) we get m_{p}(K) = 1 for every prime p | |K|. So with (1.5) we have K ≅ L_{2}(q), Sz(q) or J_{1}.

If y = 3 and x = 9, then 3 ∉ σ(M), so m_{3}(K) = 1. Hence by (1.3) we see K ≅ L_{2}(q), Sz(q), (S)L_{3}(q) or (S)U_{3}(q). Let K̂_{1} = C I_{K_{1}}(ω^{3}). If p | |K̂_{1}|, then p ∉ σ(M), p prime.

Suppose K_{1} = (S)L_{3}(q) or (S)U_{3}(q). If K_{1} ≅ (S)U_{3}(q), then ω^{3} centralizes some γ ∈ K_{1}, o(γ) | q + 1/3. This implies q = 8, a contradiction. Suppose K_{1} ≅ (S)L_{3}(q). Then ω^{3} is not inner × diagonal, so ω^{3} induces a field automorphism. Now it centralizes (S)L_{3})( ^{3} \sqrt{q}) and so ^{3} \sqrt{q} = 2, i.e. q = 8. But m_{7}((S)L_{3}(8)) ≥ 2 where 7 | |SL_{3}(2)|, a contradiction. So in any case we have K ≅ L_{2}(q), Sz(q) or J_{1}. Further K^{⟨ω⟩} ≤ L_{1}/Q_{1}.

Let x ∈ V_{α} ∩ L_{1}, [x, K^{⟨ω⟩}] = 1 and [x, Z_{1}] ≠ 1. But now C I_{L_{1}/Q_{1}}(K^{⟨ω⟩}) ≠ 1 and as [(V_{α} ∩L_{1})Q_{1}/Q_{1}, K^{⟨ω⟩}] ≠ 1 we get by quadratic action C I_{Z_{1}}(K^{⟨ω⟩}) ≠ 1. As C I_{Z_{1}}(K^{⟨ω⟩}) ≤ S there is some 1 ≠ t ∈ Ω_{1}(Z(S)) such that t is centralized by some elementary abelian subgroup E of order p^{2} in L_{1} with Γ_{E,1}(G) ≤ M, a contradiction to L_{2} ⊈ M. This gives C I_{(V_{α}∩L_{1})Q_{1}/Q_{1}}(K^{⟨ω⟩}) = 1.

Let y > 3. Then every p with p | |K| is in σ(M). Let first 3 ∈ σ(M). By

(6.4) we get Z_{1} = ⟨t | |V_{α} : C I_{V_{α}}(t)| ≤ sr^{3}⟩, where s = q, if q is even, s = 2 for K ≅ J_{1} and s = 4 for q odd . Now we get some E with E elementary abelian of order p^{2} and Γ_{E,1}(G) ≤ M_{α+1} such that |C I_{Z_{α+1}}(E)| ≥ q^{2(y−3)}. As Z_{1} ≰ M_{α+1}, we see q^{2(y−3)} ≤ qr^{3} and so q^{2y−5} ≤ r^{3}. As r ≤ 2^{y−1} we get q^{2y−5} ≤ 2^{3(y−1)}. As q ≥ 4 we have 4y − 10 ≤ 3y − 3, i.e. y ≤ 7. Hence y = 5 or y = 7. Furthermore we have that r^{3} divides |E(L_{2}/Q_{2})| which shows E(L_{2}/Q_{2}) ≅ Sp_{4}(r). Then by (1.3) 3 ∉ σ(M). Hence K ≅ Sz(q) and so q ≥ 8. This now shows even 6y − 15 ≤ 3y − 3 and so y ≤ 4, a contradiction.

So we have y = 3 and then 3 ∉ σ(M). This implies x = 9 or x = 3, so r = 64 or r = 4.

We have K^{L_{1}∩L_{2}} = K^{⟨ω⟩} otherwise we may argue as before. Let ZZ_{9}×ZZ_{9} ≅ F ≤ L_{1} ∩L_{2}. As 3 ∉ σ(M), we see m_{3}(⟨K, (L_{1} ∩L_{2})/Q_{1}⟩) ≤ 3. Hence F acts faithfully on K^{⟨ω⟩}. But then C I_{K_{1}^{⟨ω⟩}}(ω) contains an elementary abelian group of order 9 and we get the contradiction m_{3}(K^{⟨ω⟩}) ≥ 6. . The same argument shows that E(L_{2}/Q_{2}) ≅ S̸L_{3}(4) ∗ SL_{3}(4) as L_{1} ∩ L_{2} does not contain an elementary abelian group of order 27.

So we have |V_{2} : V_{2}∩Q_{α}| ≤ r^{2} = 16 or r = 64 and |V_{2} : V_{2}∩Q_{α}| ≤ r = 64.

Let [Z_{α+1} ∩ L_{1}, Z_{1}] = 1. There is some E ≤ M_{α+1}, E elementary abelian of order p^{2} with Γ_{E,1}(G) ≤ M_{α+1}, and |C I_{Z_{α+1}}(E)| ≥ 64, as p ≥ 5. This shows p = 7 ,r = 64 and |Z_{α+1} : Z_{α+1} ∩ L_{1}| = 64. But now there is ν ∈ L_{1} ∩ L_{2}, o(ν) = 7, [K, ν] ≤ K. So m_{7}(K) = 1 and ν centralizes some elementary abelian group of order 7^{3} in L_{1}, which shows N_{G}(⟨ν⟩) ≤ M. But as L_{2} = ⟨L_{2} ∩ M, N_{L_{2}}(⟨ν⟩)⟩ ≤ M, we have a contradiction.

Thus [Z_{α+1} ∩ L_{1}, Z_{1}] ≠ 1. Suppose next that [[K_{1}, Z_{1}], K_{2} × K_{3}] = 1. Then as we may assume [Z_{α+1} ∩ L_{1}, [K, Z_{1}]] ≠ 1, we get V_{α} ≤ L_{1}. Now Z_{1} = ⟨t | |Z_{α+1} : C I_{Z_{α+1}}(t)| ≤ 2^{m_{2}(K)}⟩. Hence |[K, Z_{1}]| ≤ 2^{m_{2}(K)}, which is not possible.

So we have that [[K_{1}, Z_{1}], K_{2} × K_{3}] ≠ 1. In particular there are at least 4 nontrivial irreducible K_{1}-modules involved. By symmetry we have some s ∈ Z_{α+1} ∩ L_{1}, [s, K_{1}] ≠ 1 with 1 ≠ |Z_{1} : C I_{Z_{1}}(s)| ≤ 2^{m_{2}(K_{1})} · 64. Hence there is some irreducible module V for K_{1} such that for some involution s ∈ Aut (K_{1}) we have |V : C I_{V} (s)| ≤ 2^{m_{2}(K_{1})/4} · ^{4} \sqrt{64}.

Let K_{1} ≅ L_{2}(p) or J_{1}, p odd. Then |V : C I_{V} (s)| ≤ 4. This implies with (1.30)(a) K_{1} ≅ L_{2}(5) or L_{2}(7). Suppose K_{1} ≅ L_{2}(q), q even. Then |V : C I_{V} (s)| ≥ q. So q^{3} ≤ 64, i.e. q ≤ 4. If K_{1} ≅ Sz(q), then |V : C I_{V} (S)| ≥ q^{2} and so q^{7} ≤ 64, a contradiction.

Let K_{1} ≅ L_{2}(5). Then r = 64. But now some ν ∈ L_{1} ∩ L_{2}, o(ν) = 7, centralizes K_{1}^{⟨ω⟩}, and so N_{G}(⟨ν⟩) ≤ M by (1.63) as O_{2}(M) ∩ N_{G}(⟨ν⟩) ≠ 1, a contradiction.

So we have K_{1} ≅ L_{2}(7). But in this case at least 6 nontrivial modules are involved. Hence also r = 64 and then again N_{G}(⟨ν⟩) ≤ M.

(6.19) Lemma. Assume (6.4), b = b_{1} > 1. Let K/Z(K) ≅ A_{n}, with m_{p}(K) ≥ 2, be a component of L_{1}/Q_{1}. Then [(V_{α} ∩ L_{1})Q_{1}/Q_{1}, K] = 1.

Proof . Suppose false. By (6.18) [ω, K] ≤ K. By (1.63) we have [ω, K] ≠ 1. Hence by (1.9) o(ω) = 3 and so r = 4. Furthermore 3 ∉ σ(M) and then n ≤ 11. As m_{p}(K) ≥ 2, we get n = 10 or 11 and p = 5. Now we see that KO_{3′}(M)/O_{3′}(M) ≤ E(M/O_{3′}(M)). The construction of G_{1} in §4 tells us that KO_{3′}(M)/O_{3′}(M) = E(M/O_{3′}(M)). Hence we have that a Sylow 3–subgroup of M is centralized by some E ≅ E_{5^{2}} with Γ_{E,1}(G) ≤ M. As [ω, Z_{2}] = 1, we see with (1.63) that N_{G}(⟨ω⟩) ≤ M, a contradiction.

(6.20) Lemma. Assume (6.4) with b = b_{1} > 1. Let K/Z(K) be a component of L_{1}/Q_{1}. If K/Z(K) is a sporadic simple group with m_{p}(K) ≥ 2, then [(V_{α} ∩ L_{1})Q_{1}/Q_{1}, K] = 1.

Proof . Suppose false. By (6.18) [ω, K] ≤ K and then [ω, K] ≠ 1 by (1.63). By (1.9) o(ω) = 3 or o(ω) = 7 and K ≅ J_{1}. The latter is not possible as m_{p}(K) ≥ 2. So we have r = 4 and 3 ∉ σ(M).

As m_{3}(K) ≤ 3 we get with (1.3) K/Z(K) ≅ J_{1}, M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, HiS, He, Ru, J_{4}, O′N, J_{3}. As m_{p}(K) ≥ 2, we get K/Z(K) ≅ J_{2}, J_{3}, J_{4}, HiS, He, Ru, or O′N. Finally p = 5, 7 or 11.

As K cannot centralize any element in C I_{Z_{1}}(S)^{♯} we get that (V_{α}∩L_{1})Q_{1}/Q_{1} acts faithfully on K. Let first |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ 2. As there is E ≤ M_{α+1} and |C I_{Z_{α+1}}(E)| ≥ 2^{6} we get |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≥ 2^{6}. But this contradicts (1.17).

So we have |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≥ 4. As C I_{Z_{α+1}∩L_{1}/Z_{α+1}∩Q_{1}}(K) = 1, we get with (1.12)

K ≅ J_{2} and p = 5 .

Now |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 2^{6} by (1.12)(i) and so every t ∈ Z_{1} centralizes some x ∈ Z_{α+1} such that x is centralized by some elementary abelian E, E ≤ M_{α+1}, |E| = 25 and Γ_{E,1}(G) ≤ M. Hence t ∈ C I_{G}(x) ≤ M_{α+1}. This implies the contradiction Z_{1} ≤ M_{α+1}.

(6.21) Lemma. Assume (6.4) with b = b_{1} > 1. Let K/Z(K) be a component of L_{1}/Q_{1}, K/Z(K) = G(q) be a group of Lie type over a field GF (q) of odd characteristic. If m_{p}(K) ≥ 2, then [(V_{α} ∩ L_{1})Q_{1}/Q_{1}, K] = 1.

Proof . Suppose false. Let |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 4. Then by (1.12) K/Z(K) ≅ U_{4}(3), A_{6} or U_{3}(3), and so 3 ∈ σ(M). As [ω, K] = 1, we get a contradiction. So we have |V_{α} ∩L_{1} : C I_{V_{α}∩L_{1}}(K)| = 2. Let K_{1} ∗⋯∗K_{u} = K^{S}. Then as L_{2} ⊈ M, we have V_{α} ∩ L_{1}/V_{α} ∩ Q_{1} acts fatihfully on K^{S}. Hence we get that Z_{1} = ⟨t | |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(t)| ≤ 2⟩. If V_{α} ≤ L_{1} then even

Z_{1} = ⟨t | |[V_{α}, t]| ≤ 2⟩. By (6.16) then Z_{1} ≤ Q_{α}, a contradiction. So we have V_{α} ⊈ L_{1} and C I_{V_{α}∩L_{1}}(K) = V_{α} ∩ Q_{1}. Hence |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| = 2. Now |V_{α} :

C I_{V_{α}}(Z_{1})| ≤ 2r^{2} by (1.17). By (6.16) and (1.44) E(L_{2}/Q_{2}) ≅ L_{2}(r) × L_{2}(r) or L_{2}(r) and |Z_{1} : Z_{1} ∩ Q_{α}| ≤ r. But in the latter by (6.16) we would get |V_{α} : C I_{V_{α}}(Z_{1})| ≥ r^{2}, which contradicts |V_{α} : V_{α} ∩ L_{1}| ≤ r. As V_{2} ≰ M_{α=1} we see that |V_{2} : V_{2} ∩ Q_{α+1}| ≤ 2 too and so |V_{2} : C I_{V_{2}}(Z_{α+1})| ≤ 2r^{2}, we get |V_{2} : V_{2} ∩ Q_{α}| ≥ \frac{1}{2}r^{2} ≤ |V_{α} : V_{α} ∩ Q_{2}|. Hence there is β ∈ △(2) such that V_{α} ≤ G_{β}, and Z_{β}Q_{α} ≠ Z_{1}Q_{α}, Z_{β} ≰ Q_{α}. Now we get |V_{α} : C I_{V_{α}}(Z_{1}Z_{β})| ≤ 4r^{2}. By (6.16) and (1.44) we may choose Z_{β} such that |V_{α} : C I_{V_{α}}(Z_{1}Z_{β})| ≥ r^{4}, a contradiction.

(6.22) Lemma. Assume (6.4) with b = b_{1} > 1. Let K be a component of L_{1}/Q_{1} with K/Z(K) = G(q) a group of Lie type over a field of characteristic two, q > 2, If m_{p}(K) ≥ 2 then [(V_{α} ∩ L_{1})Q_{1}/Q_{1}, K] = 1.

Proof . Suppose first that ω induces a field automorphism on K. Then q = t^{x} and ω centralizes K̃ ≅ G(t). By (1.63) m_{p}(K̃) ≤ 1. Then K ≅ L_{n}(q), n ≤ 7, where t = 2 for n > 4 , Sp_{4}(q) or Sp_{6}(q), G_{2}(q), U_{n}(q), n ≤ 7, Ω_{8}^{−}(q), 2F_{4}(q) or ^{3}D_{4}(q).

Suppose that ω induces an inner × diagonal automorphism. Then ω centralizes some Cartan subgroup C. Hence m_{x}(C) ≤ 3 and m_{x}(C) ≤ 2 for m_{x}(K) ≥ 4. This shows with (1.2) that K is one of the groups above.

Suppose that K ≰L_{1}/Q_{1}. Then in the first case m_{ℓ}(K̃) ≤ 1 for all prime ℓ, so K̃ ≅ L_{3}(2) and K ≅ L_{3}(2^{x}). Replace K by ⟨K^{L_{1}}⟩. We have that ⟨K^{L_{1}}⟩ contains at most three components.

In the second case we get m_{s}(C) ≤ 1 for any prime s. But this implies that K/Z(K) ≅ U_{3}(q) or L_{3}(4). If K/Z(K) ≅ L_{3}(4) we have 3 ∈ σ(M), a contradiction as o(ω) = 3. So we have the former. But now x centralizes some E, E ≅ E_{p^{2}}, p | q + 1, a contradiction too.

So we have

(1) K < L_{1}/Q_{1} or K ≅ L_{3}(2^{x}) and there are at most three conjugates.

Next we show

(2) [Z_{1} ∩ Q_{α}, V_{α}] ≠ 1 .

Otherwise V_{α} centralizes a subgroup of index r^{2} in Z_{1} by (1.17). Application of (1.22) shows that Z_{1} has a submodule T ≅ V (λ), λ some weight, or r > q and K ≅ U_{3}(q).

Suppose K ≤ L_{1}/Q_{1}. Let first K ≅ U̸_{3}(q), if r > q and T ≅ V̸ (λ). Suppose furthermore T ≤ L_{1}. Then Ω_{1}(Z(S)) ∩ T ≠ 1. Hence there is no E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, such that C I_{C I_{T} (S)}(E) ≠ 1 as L_{2} ≰ M.

As p | q^{2} − 1, we see that K ≅ L_{4}(q), L_{3}(q), Sp_{4}(q), Sp_{6}(q), G_{2}(q), U_{n}(q), n ≤ 5 or Ω_{8}^{−}(q).

In any case there is some μ ∈ K, o(μ) = p, such that [C I_{T} (S), μ] = 1. By (4.4) we get m_{p}(L_{1}) ≥ 4. Further we see that on C I_{T} (S ∩ K) there is just a cyclic group induced. But this now implies that we have K ≅ U_{5}(q), p | q +1. Hence T is not the natural module. As r^{2} ≤ q^{4} this contradicts (1.34)(i).

So we have T_{1} ⨁ T_{2}, T_{i} ≅ T or T^{⋆} as GF (2)-modules, involved in Z_{1}. Now V_{α} ∩ L_{1} induces GF (q)–transvections on T_{i}, i.e. T is the natural module for L_{n}(q), Sp_{2n}(q) and q = r, Ω_{6}^{−}(q), Ω_{8}^{−}(q), or U_{n}(q) and q = r^{2}.

Now as before we get K ≅ L_{3}(q), L_{4}(q), Sp_{4}(q) or U_{4}(q). Furthermore as m_{p}(L_{1}) ≥ 4 we get that either K ≅ U_{4}(q) and p | q + 1, K ≅ L_{4}(q) or L_{3}(q), T_{1} ⨁ T_{2} ≅ T ⨁ T^{⋆}. In the latter we also get that p | q − 1 = r − 1. Now ω centralizes some F ≅ E_{p^{2}} with Γ_{F,1}(G) ≤ M. By (1.63) we have N_{G}(⟨ω⟩) ≤ M.

So we are left with K ≅ U_{4}(q), r = q^{2}, p | q + 1. Furthermore T is the natural U_{4}(q)– or Ω_{6}^{−}(q)–module. Now L_{1}∩L_{2} contains Ẽ = ⟨ω̃⟩×⟨ω̃_{1}⟩ ≅ E_{p^{2}}, contradicting (2.2).

Assume now K ≅ U_{3}(q), r > q. Then ω induces a diagonal automorphism and so r = q^{2}. We get p | q + 1 and L_{1} ∩ L_{2} contains an elementary abelian group of order p^{2}, a contradiction as before.

So we are left with K ≰L_{1}/Q_{1} and K ≅ L_{3}(2^{x}). Now r < q and so by (1.26)(i) [K, Z_{1}] is the natural module. In particular [[K, Z_{1}], K^{g}] = 1 for K^{g} ≠ K, g ∈ L_{1}/Q_{1}. But then [K^{L_{1}/Q_{1}}, Z_{1}] ∩ Ω_{1}(Z(S)) is centralized by E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. This proves (2).

Now choose β ∈ △(α) such that [Z_{1} ∩ Q_{α}, Z_{β}] ≠ 1.

(3) Let K ≤ L_{1}/Q_{1} and K_{β} be the component in L_{β} corresponding to K in L_{1}. If there is t ∈ Z_{1} ∩ Q_{α} such that [t, K_{β}] = 1, then [t, Z_{β}] = 1.

Suppose false. We have [K_{β}, [t, Z_{β}]] ≤ [t, Z_{β}] and so V_{2} ≤ L_{β}. If [K_{β}, [t, Z_{β}]] = 1, then K centralizes some x ∈ Ω_{1}(Z(S))^{♯}, a contradiction. So we have

[K_{β}, [t, Z_{β}]] ≠ 1 and [K_{β}, Z_{1}] = 1.

As [V_{α}∩L_{1}, K] ≠ 1, we get C I_{V_{α}∩L_{1}/V_{α}∩Q_{1}}(K) = 1. Let |V_{α}∩L_{1} : V_{α}∩Q_{1}| = q^{x}. Then by (1.17) |[Z_{β}, t]| ≤ q^{x}r^{2}. If K ≅ ^{3}D_{4}(q), then x ≤ 5 and by (1.64) and (1.65) r ≤ q^{3}. This gives |[Z_{β}, t]| ≤ q^{11}, contradicting (1.23). If K ≅ 2F_{4}(q), then r ≤ q^{2} and so we have |[Z_{β}, t]| ≤ q^{10} a contradiction to (1.23) again. If K ≅ G_{2}(q), we have r ≤ q and so [Z_{β}, t]| ≤ q^{5}, a contradiction.

So we are left with K/Z(K) ≅ L_{n}(q), Sp_{4}(q), Sp_{6}(q), U_{n}(q) or Ω_{8}^{−}(q). Let K/Z(K) ≅ Ω_{8}^{−}(q), then r ≤ q^{2} and so |[Z_{β}, t]| ≤ q^{x}q^{4}. We have |[Z_{β}, t]| ≥ q^{8} and so x ≥ 4. By (1.15) and (1.14) there is some natural or spin module involved in [Z_{β}, t]. But for the latter the dimension is 16 and x < 12, a contradiction. So just one natural module is involved. But now C I_{K}(C I_{T} (S)) contains some E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

If K/Z(K) ≅ U_{6}(q) or U_{7}(q) then x ≤ 9. Then we see as before that x = 8 and K/Z(K) ≅ U_{6}(q). But now again with (1.14) and (1.15) we get a contradiction.

By the same argument we get T is the spin module for K ≅ Sp_{6}(q). But then |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| = q^{6}. The spin module does not admit a quadratic group of order q^{6}. We see that K/Z(K) ≅ U̸_{5}(q).

If K ≅ U_{4}(q), and T is the natural module, then r = q^{2} and so L_{1} ∩ L_{2} contains an elementary abelian group of order p^{2}, a contradiction to (2.2). If T is the orthogonal module, then |V_{α}∩L_{1} : V_{α}∩Q_{1}| ≥ q^{2} and so there is some v ∈ V_{α} ∩L_{1} such that [Z_{1}, v] contains some 1 ≠ x with C I_{K}(x) involves Sp_{4}(q). So x is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, which gives V_{α} ≤ C I_{G}(x) ≤ M. But this contradicts (1.44) and (6.16), as |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q^{3} < r^{2}.

Let K ≅ Sp_{4}(q). Then T is the natural module and |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ q^{2}. By (1.16) V_{α} ∩ L_{1} normalizes T^{g} for every g ∈ L_{1}. As |T | = q^{4}, we see that V_{α} ≰ M. Hence we have with (4.4) that m_{p}(L_{1}) ≥ 4. Let U be the kernel of the action of L_{1} on {T^{g} | g ∈ L_{1}}. Then we see that m_{p}(U) = 2 as V_{α} ∩ L_{1} ≤ U. Now we also see that C I_{U}(T) is solvable. Let U_{1} be the largest solvable normal subgroup of U. Then [O^{2′}(U_{1}), T] = 1 and so O^{2′}(U_{1}) centralizes T^{g} for all g in question. As p | |̸U_{1}| this implies that there is some t ∈ V_{α} ∩ L_{1} \ Q_{1} such that tO^{2′}(U_{1}) is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, and so [⟨T^{g}| g ∈ L_{1}⟩, t] is normalized by E, i.e. V_{α} ≤ L_{1}, a contradiction.

So we have K ≅ L_{n}(q). If p | q − 1, we see that n ≤ 4, otherwise for some 1 ≠ x ∈ Ω_{1}(Z(S)) we have C I_{G}(x) ≤ M. Furthermore for n ≥ 5, we have p | q − 1, so we have n ≤ 4 anyway.

Let K ≅ L_{4}(q), then T is the natural or dual or the orthogonal module. In the latter some 1 ≠ x ∈ C I_{T} (S) is centralized by E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and E ≤ K, a contradiction. So we have that T is the natural module. Furthermore as V_{α} ≰ L_{1} by (6.1), we get p | q + 1, i.e. m_{p}(K) = 2. Now by (1.16) we see that (V_{α} ∩ L_{1})Q_{1}/Q_{1} normalizes some conjugate of T in L_{1} as |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ q^{2}. But now we may argue as in the case of Sp_{4}(q).

So we have K ≅ L_{3}(q), p | q − 1 and T is the natural module. If |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q we may argue as before. So assume |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q. Thus we must have r = q. Now L_{1} ∩L_{2} contains an elementary abelian group E of order p^{2}. As m_{p}(L_{1}) = 4, we see that Γ_{E,1}(G) ≤ M, a contradiction.

(4) Let K < L_{1}/Q_{1}. Then Z_{1} is either a strong quadratic module or an

SC-module, or Z_{1} is the natural module for K ≅ U_{3}(q).

Suppose false. Let t ∈ Z_{1} ∩ Q_{α} with [t, Z_{β}] ≠ 1. By (3) [t, K_{β}] ≠ 1. If

K ≅ L_{n}(q), then we get |[Z_{β}, t]| > q^{( \frac{n}{2} )^{2}} or q^{\frac{(n^{2}−1)}{4}} , according to n is even or odd. This shows |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q^{\frac{n^{2}−8}{4}} or q^{\frac{n^{2}−9}{4}} . Hence (1.15) implies n = 3 and |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q. As r ≤ q we see that |Z_{β} : Z_{β} ∩ Q_{2}| > r. Now we have ⟨ω⟩ × ⟨ω_{1}⟩ involved in L_{1} ∩ L_{2}. Hence we may assume that ω_{1} induces an inner × diagonal automorphism on K. As p| q − 1, we get N_{G}(⟨ω_{1}⟩) ≤ M. Application of (1.66) shows E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r). We have |V_{2} : C I_{V_{2}}(Z_{β})| ≥ r^{6} by (6.16). In particular |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ r^{4}. As r^{2}q > q^{2}, i.e. r > q, we see that (V_{2} ∩ L_{β})Q_{β}/Q_{β} ∩ C I(K_{β}) ≠ 1. Hence C I_{Z_{β}}(K_{β}) ≠ 1 and so as K ≤ L_{1}/Q_{1} we have that C I_{Ω_{1}(Z(S))}(K) ≠ 1, contradicting L_{2} ≰ M.

Let K ≅ U_{n}(q), n ≤ 7. Now |[Z_{β}, t]| > q^{(⌊ \frac{n}{2} ⌋)^{2}}. Hence |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q^{(⌊ \frac{n}{2} ⌋)^{2}−4}. By (1.15) we get n ≤ 5. Suppose n > 3. Then by (1.15) |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q^{2}. This now implies r > q. Hence r = q^{2} and ω induces some diagonal automorphism. We have that p| r − 1. So there is ℤ_{p} × ℤ_{p} ≅ E ≤ L_{1} ∩ L_{2}, which contradicts (2.2). So |V_{α} : V_{α} ∩ L_{1}| ≤ r = q^{2}, a contradiction too.

Let K ≅ U_{3}(q). If r > q, then ω induces a diagonal automorphism. So we have r = q^{2}. As p | q+1, we have some ν ∈ L_{1}∩L_{2}, o(ν) = p. But ν centralizes some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and E ≤ K. This gives N_{G}(⟨ν⟩) ≤ M, a contradiction. So r ≤ q. As before we see that |V_{α} : V_{α} ∩ Q_{2}| ≤ r and now |[Z_{β}, t]| ≤ q^{2}. By (1.26)(iii) Z_{β} is the natural module.

Let K ≅ Sp_{4}(q) or Sp_{6}(q). Then |[Z_{β}, t]| > q^{3}, q^{6} respectively. Hence |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q^{4} in the latter, contradicting (1.15). So we have K ≅ Sp_{4}(q) and |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q, which contradicts (1.15) too.

Let K ≅ Ω^{−}(8, q), then |[Z_{β}, t]| > q^{6}. Hence |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q^{2}. By (1.15) we have |V_{α}∩L_{1} : V_{α}∩Q_{1}| ≤ q^{3}. So we get r = q^{2} and |V_{α} : V_{α}∩L_{1}| > r. But as ω induces a diagonal automorphism and p| r−1 we get a contradiction.

Let K ≅ G_{2}(q). Then |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q, in which case Z_{β} is strong quadratic by (1.15).

Let K ≅ ^{3}D_{4}(q). As Z_{β} is not strong quadratic we get with (1.15) that |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q. Then as |[Z_{β}, t]| > q^{5}, we get r = q^{3} and ⟨ω⟩ × ⟨ω_{1}⟩ ≤ L_{1} ∩ L_{2}, o(ω) = o(ω_{1}). So we again get N_{G}(⟨ω_{1}⟩) ≤ M and then |V_{2} : C I_{V_{2}}(Z_{β})| ≥ r^{6}. Hence |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ r^{4} = q^{12}. This implies that (V_{2} ∩ L_{β})Q_{β}/Q_{β} ∩ C I(K_{β}) ≠ 1. Hence we have C I_{Z_{β}}(K_{β}) ≠ 1, a contradiction.

Let K ≅ ^{2}F_{4}(q). By (1.15) |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q^{2}. So |[Z_{β}, t]| ≤ q^{4}, a contradiction. This proves (4).

As no element 1 ≠ x ∈ Ω_{1}(Z(S)) is centralized by E ≅ E_{p^{2}}, we now get one of the following for K ≤ L_{1}/Q_{1}

(a) K ≅ L_{n}(q), n ≤ 4, Z_{1} contains the natural module or the dual or V (λ) ⨁ V (λ^{σ})

(b) K ≅ Sp_{4}(q), Z_{1} just involves the natural module

(c) K ≅ Sp_{6}(q), Z_{1} just involves spin modules

(d) K ≅ U_{n}(q), n ≤ 5. In case n = 5 there is no natural submodule involved

(e) K ≅ G_{2}(q) and the natural module is involved

(f) K ≅ ^{3}D_{4}(q) and the natural module is involved

(g) K ≅ Ω_{8}^{−}(q) and just the spin module is involved.

(5) Let K < L_{1}/Q_{1}. If |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > q, q^{2} in case (d) and (g) and q^{3} in case (f), then V_{α} ≤ L_{1}. This is also true if α is replaced by 2 and 1 is replaced by β.

By (1.16) V_{α} ∩ L_{1} normalizes every K-submodule T. In any case above we see that there is some x ∈ V_{α} ∩ L_{1} such that some 1 ≠ U ≤ [x, Z_{1}] is normalized by some μ ∈ K, o(μ) = p. If there is some μ_{1} ∈ M with [T, μ_{1}] = 1, then V_{α} ≤ L_{1}.

So we may assume there is no such μ_{1}. In particular ⟨L_{1} ∩V_{α}, K⟩Q_{1}/Q_{1} ∩ C I(K) = 1. Now let again U_{1} be the kernel of the action of L_{1} on {T^{g}| g ∈ L_{1}}. Then V_{α} ∩ L_{1} ≤ U_{1}. Furthermore let U_{2} = C I_{U_{1}}(T). Then p |̸ |U_{2}| and m_{p}(Aut(KO^{2′}(U_{2})/O^{2′}(U_{2}))) ≥ 4.

If m_{p}(K) = 2, this implies that U is normalized by E_{p^{2}}, a contradiction. So m_{p}(K) ≥ 3. By (4.12) no field automorphism is induced.

This now implies K ≅ U_{5}(q), p | q + 1. By (1.54) V_{α} ∩ L_{1}/V_{α} ∩ Q_{1} contains some u ∈ K, u a transvection in the natural representation. But [u, Z_{1}] is normalized by U_{3}(q) and so N_{G}([u, Z_{1}]) ≤ M, i.e. V_{α} ≤ L_{1}.

(6) K ≰L_{1}/Q_{1}

Suppose false. By (6.1) either Z_{1} ⊈ L_{β} or Z_{β} ⊈ L_{1}. As in then proof of (6) we will never use V_{2} ⊈ Q_{α} we now by (5) have symmetry between 1 and β. So we may assume Z_{β} ⊈ L_{1}. So we may assume that |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q, q^{2} in case (d) and q^{3} in case (f). Hence for t ∈ Z_{1} ∩ L_{β} we have

|[Z_{β}, t]| ≤ qr^{2}, q^{2}r^{2} or q^{3}r^{2}

Suppose first that Z_{β} and also Z_{1} involve exactly one irreducible nontrivial module T_{1}. Let first K ≅ ^{3}D_{4}(q). As |[t, Z_{β}]| ≥ q^{6}, we see that r > q. Suppose first that W = ⟨ω⟩ × ⟨ω_{1}⟩ ≤ L_{1} ∩ L_{2}, o(ω) = o(ω_{1}). As r > q, we see that C I_{W} (K) ≠ 1 if all elements in W induce inner × diagonal automorphism on K. But then for some 1 ≠ ω_{2} ∈ W we have N_{G}(⟨ω_{2}⟩) ≤ M, by (1.63). This implies with (1.66) that E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r) and so by (6.16) |V_{2} : V_{2} ∩ Q_{β}| ≥ r^{6}. Hence |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ r^{4}. As r^{2} ≥ q^{3}, we have |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ q^{6} and then (V_{2} ∩ L_{β})Q_{β}/Q_{β} ∩ C I(K_{β}) ≠ 1. This shows C I_{Z_{1}}(K) ≠ 1, contradicting L_{2} ≰ M.

So we may assume that ω induces a field automorphism on K. By (1.64) and (1.65) this implies x = 3 or 9, and ω_{1} induces a diagonal × inner automorphism. Let U_{1} be the parabolic with L_{2}(q^{3}) on top. Then C I_{T_{1}}(O_{2}(U_{1})) is the natural L_{2}(q^{3})–module and U_{1}/O_{2}(U_{1}) ≅ L_{2}(q^{3}) × Z_{q−1}. As x |̸ q − 1 and as r > q, we see that x ≠ 3. If 3 |̸ q^{3} − 1, we have that C I_{⟨K,ω_{1}⟩}(T_{1}) ≠ 1. Hence we have [ω_{1}, K] = 1 and so N_{G}(⟨ω_{1}⟩) ≤ M. Now we may argue as above. Hence 3 | q^{3} − 1, i.e. 3 | q − 1. Hence as r^{2} ≥ q^{3}, we get q = 4 or q = 16. But then m_{3}(⟨K, W ⟩) = 4, a contradiction.

So we may assume |[Z_{β}, t]| ≤ q^{3}r. Hence r ≥ q^{3}. Now by (1.64) and (1.65) either r = q^{3} and ω induce a diagonal × inner automorphism or x = 9 and q = 4, and so r = q^{3} too. If p |̸ q^{3} − 1, then any p–element μ ∈ C I(K) centralizes [Z_{1}, K]. As by (4.12) no field automorphism of order p is involved we see that [Z_{1}, K] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, or p = 3. The former implies L_{2} ≤ M, a contradiction. So we have p = 3 and m_{3}(K) = 3. But then [Z_{1}, K] is centralized by a cyclic group of order 3. As C I_{[Z_{1},K]}(S) is centralized by a 3–element in K, we get that N_{G}(C I_{[Z_{1},K]}(S)) ≤ M, a contradiction. So p| q^{3} − 1. Now choose ν ∈ L_{1} ∩ L_{2}, o(ν) = p. By (4.4)(ii) we get m_{p}(L_{1}) ≥ 3. By (4.12) no field automorphisms of order p are involved. Hence p | |C I_{L_{1}/Q_{1}}(K)| and so m_{p}(C I_{L_{1}}(ν)) ≥ 3, which gives N_{G}(⟨ν⟩) ≤ M, a contradiction to (1.66) and (2.2).

So we may assume K ≅ ^{3̸}D_{4}(q). In this case there is some μ ∈ K, o(μ) = p such that [μ, C I_{T_{1}}(S)] = 1. Hence we see that m_{p}(K) ≥ 4. This implies K ≅ U_{5}(q), p | q + 1. By (1.34) (i) we have r > q. Hence by (1.64) and (1.65) r = q^{2}. Further L_{1} ∩ L_{2} contains a subgroup of order (r − 1)^{2}. As p | r − 1 we get a contradiction with (2.2).

So assume next that T_{1} ⨁ T_{2} is involved. Then we have K ≅ L_{n}(q), n ≤ 4, T_{1} is the natural module, K ≅ U_{4}(q) or Ω_{6}^{−}(q) and T_{1} is the natural module. Suppose K ≅ U_{4}(q) ≅ Ω_{6}^{−}(q). If r = q^{2} then p | |L_{1} ∩ L_{2}| and so again by (4.4) and (4.12) N_{G}(⟨μ⟩) ≤ M for μ ∈ L_{1} ∩ L_{2}, o(μ) = p, a contradiction. So we have r = q by (1.64) and (1.65). Now ω induces an inner × diagonal automorphism and p | q +1. But now ⟨ω⟩×⟨ω_{1}⟩ ≤ L_{1} ∩L_{2} and so C I_{⟨ω⟩×⟨ω_{1}⟩}(K) ≠ 1, as [⟨ω⟩ × ⟨ω_{1}⟩, Ω_{1}(Z(S))] = 1. So there is ω_{2} ∈ ⟨ω⟩ × ⟨ω_{1}⟩ with N_{G}(ω_{2}⟩) ≤ M. By (1.66) we get E(L_{2}/Q_{2}) ≅ (S)L_{3}(q) or Sp_{4}(q) and so by (6.16) |V_{2} : V_{2} ∩ Q_{β}| ≥ r^{6}. Hence |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ r^{4} = q^{4}. If (V_{2} ∩ L_{β})Q_{β}/Q_{β} ∩ C I(K_{β}) ≠ 1 we get C I_{Z_{1}}(K) ≠ 1, contradicting L_{2} ≰ M. So we have (V_{2} ∩L_{β})Q_{β}/Q_{β} ∩C I(K_{β}) = 1 and then |V_{2} ∩L_{β} : V_{2} ∩Q_{β}| = q^{4},i.e. T_{1} and T_{2} are natural U_{4}(q)–modules as V_{2} ∩ L_{β} acts quadratically on Z_{β}. But now as |Z_{β} ∩ L_{1} : Z_{β} ∩ Q_{1}| = q^{2}, we see that |Z_{1} : C I_{Z_{1}}(Z_{β} ∩ L_{1})| ≥ q^{8} = r^{8} and then |Z_{1} ∩ L_{β} : Z_{1} ∩ Q_{β}| ≥ r^{6}, a contradiction.

So we have K ≅ (S)L_{3}(q) or L_{4}(q). Suppose first that p | q + 1. Then K ≅ L_{4}(q) and m_{p}(K) = 2. Further by (4.4) and (4.12) we have that m_{p}(C I_{L_{1}/Q_{1}}(K)) ≥ 2. As Z_{1} involves at most three natural modules, we get that C I_{L_{1}/Q_{1}}(K) induces on ⟨T_{1}^{g} | g ∈ L_{1}⟩ a subgroup of SL_{3}(q) and so as p | q + 1 we have that p | |C I_{L_{1}}(⟨T_{1}^{g} | g ∈ L_{1}⟩)|. But then as p | |C I_{K}(s)| for s ∈ V_{α} ∩ T_{1}, we see that V_{α} ≤ L_{1}, a contradiction. So we have that p | q − 1. As V_{α} ≰ L_{1}, no element in T_{1}^{♯} is centralized by an elementary abelian subgroup of order p^{2} in K. Hence n = 3. As ω cannot centralize a subgroup of order (q − 1)^{2} in K, we see that ω induces a field automorphism on K. This implies 2r = q or x = 9, r = 64 and q = 2^{9}. Hence in any case we get that (r − 1)^{2} | |L_{1} ∩ L_{2}| and so there is some ω_{2} ∈ L_{1} ∩ L_{2}, o(ω_{2}) = x and C I_{G}(ω_{2}) ≤ M. Now by (1.66) E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r) and so by (6.16) |V_{2} : V_{2} ∩ Q_{β}| ≥ r^{6}. Hence |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≥ r^{4}. On the other hand we have |V_{2} ∩ L_{β} : V_{2} ∩ Q_{β}| ≤ q^{2}, so r^{2} ≤ q. As r ≥ 4 this is impossible.

For proving the lemma we just have to treat the case K ≅ L_{3}(2^{x}) and K ≰L_{1}/Q_{1}.

Let t ∈ Z_{1} ∩ Q_{α} with [t, Z_{β}] ≠ 1. Assume first [K_{β}, t] = 1. Then V_{2} ⊆ L_{β} and so V_{α} ⊈ L_{1} by (6.1). Hence by quadratic action |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ q^{2} and so |[Z_{β}, t]| ≤ r^{2}q^{2}. As q = 2^{x} we see that 2r = q or q = 2^{9} and r = 64, i.e. x = 9. If [[Z_{β}, t], K_{β}] ≠ 1, then we have (r − 1)^{2} | |L_{1} ∩ L_{2}|. But then there is some ω_{1} ∈ L_{1} ∩ L_{2}, o(ω_{1}) = 3 or x with [ω_{1}, K] = 1. This implies N_{G}(⟨ω_{1}⟩) ≤ M and so by (1.66) E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r). Now |V_{α} : C I_{V_{α}}(Z_{1})| ≥ r^{6} by (6.16) and then |V_{α} ∩ L_{1} : V_{α} ∩ L_{2}| ≥ r^{4}. As (V_{α}∩L_{1})Q_{1}/Q_{1}∩C I(K) = 1, we see that |V_{α}∩L_{1} : V_{α}∩Q_{1}| ≤ q^{2}. But as r ≥ 4 this is not possible. Hence we have [[Z_{β}, t], K_{β}] = 1. If [⟨K_{β}^{L_{β}}⟩, t] = 1, we get [[Z_{β}, t], ⟨K_{β}^{L_{β}}⟩] = 1 and so there is 1 ≠ u ∈ Ω_{1}(Z(S)) with [u, ⟨K^{L_{1}}⟩] = 1, i.e. C I_{G}(u) ≤ M, contradicting L_{2} ≰ M. So we get a conjugate K_{β}^{g} with [t, K_{β}^{g}] ≠ 1.

So we may assume [t, K_{β}] ≠ 1 from the beginning. Assume V_{α} ≰ L_{1} again and so |[Z_{β}, t]| ≤ r^{2}q^{2}. Let g ∈ L_{1}, K^{g} ≠ K. Then K^{g} acts nontrivially on the set of irreducible K–modules involved in Z_{1}. In particular |[Z_{β}, t]| ≥ q^{3}. But now we have (r − 1)^{2} | |L_{1} ∩ L_{2}| and we may argue as above to get a contradiction. So we have V_{α} ≤ L_{1}. By (6.1) V_{2} ≰ L_{β}. Now by quadratic action for s ∈ Z_{β} we have |[Z_{1}, s]| ≤ q^{2}r^{2}. As before |[Z_{1}, s]| ≥ q^{3} and so (r − 1)^{2} | |L_{1} ∩ L_{2}| and |V_{α} : C I_{V_{α}}(Z_{1})| ≥ r^{6}. If V_{α} ∩ C I(K) ≠ 1 then we have there is some u ∈ [t, Z_{β}]^{♯} which is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{β} and so V_{2} ≤ L_{β}, a contradiction. Hence we have |V_{α} : V_{α} ∩ Q_{1}| ≤ q^{2}. So r^{6} ≤ q^{2}. But q ≤ r^{2} and then r^{6} ≤ r^{4}, a contradiction.

(6.23) Lemma. Assume (6.4) with b = b_{1} > 1. Let K be a component of L_{1}/Q_{1} with K/Z(K) ≅ G(2) and m_{p}(K) ≥ 2. Then [V_{α} ∩ L_{1}, K] ≤ Q_{1}.

Proof . Suppose false. As [ω, K] ≠ 1, we see with (1.65) that o(ω) = 3 or o(ω) = 5 and K ≅ ^{2}F_{4}(2) or o(ω) = 7 and K ≅ ^{3}D_{4}(2).

Suppose E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r). Then |V_{α} : C I_{V_{α}}(Z_{1})| ≥ r^{6}. Hence |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ r^{4} ≥ 2^{8}. As K ≤ L_{1}/Q_{1} we see that C I_{(V_{α}∩L_{1})Q_{1}/Q_{1}}(K) = 1. Hence K ≅ L_{n}(2), o(ω) = 3. But now m_{3}(K) = 3 and so ω ∈ K. As m_{p}(K) = 2, we see that ω is centralized by E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and so N_{G}(⟨ω⟩) ≤ M by (1.63), a contradiction. So we have by (1.66) that N_{G}(⟨ω_{1}⟩) ≰ M for any ω_{1} ∈ L_{1} ∩L_{2}, o(ω_{1}) = y. If ω ∈ K then we argue as before. So let ω ∉ K. Let first y ≠ 7. then ⟨K, ω⟩ ≅ K × Z_{y} and ω_{1} ∈ C I_{⟨K,ω⟩}(K) ≤ L_{1} ∩ L_{2}, as elements of order y in parabolics are inverted. But then N_{G}(⟨ω_{1}⟩) ≤ M by (1.63), a contradiction.

So we are left with o(ω) = 7 and K ≅ ^{3}D_{4}(2). Further p = 3. Finally 49 |̸ |L_{1} ∩ L_{2}|. So |V_{α} : V_{α} ∩ L_{1}| ≤ 8. This shows that [Z_{1} ∩ Q_{α}, V_{α}] ≠ 1 and so we may assume [Z_{1} ∩ Q_{α}, Z_{β}] ≠ 1 for some β ∈ Δ(α). Choose t ∈ Z_{1} ∩ Q_{α} with [t, Z_{β}] ≠ 1. Suppose [[t, Z_{β}], K_{β}] = 1. Then C I_{Z_{β}}(K_{β}) ≠ 1 and so C I_{Ω_{1}(Z(S))}(K) ≠ 1, a contradiction. Hence as |[t, Z_{β}]| ≤ 2^{8}, we see [t, K_{β}] ≠ 1 and by (1.40) |[Z_{β}, t]| ≥ 2^{6}. This gives that |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ 2^{3}, as |V_{α} : V_{α} ∩ Q_{1}| ≤ r = 8. Hence by (1.15) Z_{β} is a strong quadratic module, i.e. it involves just one nontrivial irreducible module T, the natural one. But as 3 | |C I_{M}(T)|, we see that every x ∈ T^{♯} is centralized by an elementary abelian group of order 9, which gives C I_{G}(Ω_{1}(Z(S))) ≤ M, a contradiction.

(6.24) Lemma. Assume (6.4) with b = b_{1} > 1. Let K be p–constrained with m_{p}(O_{p}(K)) ≥ 2 or O_{t}(K) ≠ 1 and K/O_{t}(K) ≅ L_{3}(t) or L_{2}(t), then [V_{α} ∩ L_{1}, K] ≤ Q_{1}. (In particular the exceptional case in (4.9) does not occur.)

Proof . Assume first V_{α} ≤ L_{1}. Then there is some t ∈ Z_{1} with |V_{α} : C I_{V_{α}}(t)| ≤ 4. By (6.16) t ∈ Q_{α}. Let β ∈ △(α) with [Z_{β}, t] ≠ 1. Then Z_{1} ≤ L_{β}. But this contradicts (6.1).

So we have V_{α} ≰ L_{1}. In particular either C I_{V_{α}∩L_{1}/V_{α}∩Q_{1}}(K) = 1 or C I_{V_{α}∩L_{1}/V_{α}∩Q_{1}}(O_{p}(K)) = 1 . Furthermore by [GoLyS, (24.1)] and [GoLyS, (25.12)] we see that |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≤ 4. Let t ∈ Z_{1} \ Q_{α}. Then |V_{α} : C I_{V_{α}}(t)| ≤ 4|V_{α} : V_{α}∩L_{1}|. By (6.16) and (1.44) we get E(L_{2}/Q_{2}) ≅ L_{2}(4) or L_{2}(r)×L_{2}(r). Suppose the latter with r > 4. Then as |V_{α} : C I_{V_{α}}(Z_{1})| ≤ 4r^{2} we get with (6.16) that |Z_{1} : Z_{1} ∩ Q_{α}| ≤ r. Choose α + 1 ∈ △(α) such that Z_{1} ≰ L_{α+1}. Then also Z_{α+1} ≰ L_{1}.

Hence |V_{2} : C I_{V_{2}}(Z_{α+1})| ≥ r^{2}. This shows V_{2}Q_{α} > Z_{1}Q_{α}. Let λ ∈ △(2) with Z_{λ}Q_{α} ≰ Z_{1}Q_{α}. Then also |V_{α} : C I_{V_{α}}(Z_{λ})| ≤ 4r^{2} and so |V_{α} : C I_{V_{α}}(Z_{1}Z_{λ})| ≤ 16r^{2}. On the other hand by (1.44) and (6.16) |V_{α} : C I_{V_{α}}(Z_{1}Z_{λ})| ≥ r^{4}, a contradiction. This shows r = 4.

Choose α + 1 ∈ △(α) as before. Now |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 64. This implies p = 7 and |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| = 64, hence E(L_{2}/Q_{2}) ≅ L_{2}(4) × L_{2}(4) and |Z_{α+1} : Z_{α+1} ∩ L_{1}| = 16 = |Z_{1} : Z_{1} ∩ Q_{α}|. But then by (6.16) |V_{α} : C I_{V_{α}}(Z_{1})| ≥ 2^{8}, a contradiction.

(6.25) Lemma. Assume (6.4) with b = b_{1} > 1. Suppose we have components K_{1}, K_{2}, K_{3} of L_{1}/Q_{1}, m_{p}(K_{i}) = 1, where the K_{i} might be cyclic p-groups. Then there is some i with [V_{α} ∩ L_{1}, K_{i}] ≤ Q_{1}.

Proof . Suppose false. As in the proof of (6.24) we may assume that |V_{α}∩L_{1} : C I_{V_{α}∩L_{1}}(K_{i})| ≥ 8 for every i, in particular the K_{i} are all nonsolvable.

We have K_{i} ≅ L_{2}(q_{i}), (S)L_{3}(q_{i}), (S)U_{3}(q_{i}) or Sz(q_{i}), i = 1, 2, where we may assume q_{1} ≤ q_{2}. Suppose first r ≤ q_{1}, r ≤ q_{1}^{2} in case of K_{1} ≅ (S)U_{3}(q_{1}).

(1) If Z_{1} ≰ L_{β} then [Z_{β}, Z_{1} ∩ Q_{α}] = 1.

Suppose false. Then let [K_{1β}, Z_{1} ∩ Q_{α}] ≠ 1.

Never [[K_{i}, Z_{1}], K_{j} × K_{k}] = 1 for {i, j, k} = {1, 2, 3}. Otherwise we have [K_{iβ}, Z_{1} ∩ L_{β}] = 1. But then even [K_{iβ}^{L_{β}}, Z_{1} ∩ L_{β}] = 1. This implies K_{iβ} ≤ L_{β}/Q_{β} and so there is some x ∈ Ω_{1}(Z(S)) ∩ [Z_{1}, K_{i}] which is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, contradicting L_{2} ≰ M.

If now C I_{V_{α}∩L_{1}}(K_{k}) ≰ Q_{1} for some k ∈ {1, 2, 3} then [[K_{k}, Z_{1}], K_{j}] = 1 for some j ∈ {1, 2, 3} \ {k}. As V_{2} ≰ L_{β} we have [[K_{k}, Z_{1}], K_{i}] ≠ 1 for {i, j, k} = {1, 2, 3}. In particular [K_{i}, C I_{V_{α}∩L_{1}}(K_{k})] = 1. But then [[K_{j}, Z_{1}], K_{i}] = 1 = [[K_{j}, Z_{1}], K_{k}], a contradiction. So we have C I_{V_{α}∩L_{1}}(K_{i}) ≤ Q_{1} for all i.

There is X ≤ Z_{β}, |X| ≥ q_{1}^{2} (q_{1}^{4} for K_{1} ≅ (S)U_{3}(q_{1})) with X ∩ C I(Z_{1}) = 1. Suppose X ∩L_{1} ≤ Q_{1}. Then q_{1} = r (q_{1}^{2} = r respectively) and |V_{α} : V_{α} ∩L_{1}| = r^{2}. As |V_{2} : C I_{V_{2}}(X)| ≥ r^{4} by (6.16) we get K_{1} ≅ (S)L_{3}(r) and [Z_{1}, K_{1}] involves exactly two natural K_{1}–modules. As L_{1} ∩ L_{2} centralizes Ω_{1}(Z(S)) we get m_{x}(K_{1}(L_{1} ∩ L_{2})) ≥ 4, a contradiction. So we have [X ∩ L_{1}, Z_{1}] ≠ 1. Now let t ∈ X ∩ L_{1}, then |[t, Z_{1}]| ≤ q_{1}^{3} (q_{1}^{5} in case of K_{1} ≅ (S)U_{3}(q_{1})) or q_{1}^{4} in case of K_{1} ≅ (S)L_{3}(q_{1}) and |V_{2} ∩ Q_{α} : C I_{V_{2}∩Q_{α}}(K_{1β})| = q_{1}^{2}. As C I_{V_{α}∩L_{1}}(K_{1}) ≤ Q_{1} we may assume [t, K_{1}] ≠ 1. If there are four nontrivial irreducible modules involved in [Z_{1}, K_{1}] then we again see q_{1} = r and as before we get a contradiction. Suppose there are exactly three nontrivial irreducible modules involved. Then K_{1} ≅ (S)U_{3}(q_{1}) and [K_{1}, Z_{1}] is centralized by K_{2} or K_{3}. In particular |[t, [K_{1}, Z_{1}]]| ≥ q_{1}^{3}. Suppose [K_{2}, [K_{1}, Z_{1}]] = 1. We have |[t, [K_{2}, Z_{1}]]| ≥ q_{2}^{2} ≥ q_{1}^{2} as C I_{V_{α}∩L_{1}}(K_{1}) ≤ Q_{1}, a contradiction. So [K_{2}, [K_{1}, Z_{1}]] ≠ 1. As [t, [K_{3}, Z_{1}]] ≠ 1, we even have |[t, Z_{1}]| > q_{1}^{3} and so K_{1} ≅ SL_{3}(q_{1}). We have |[K_{2}, [K_{1}, Z_{1}], t]| ≥ q_{2}^{3}. Now K_{2} is isomorphic to a subgroup of SL_{3}(q_{1}). As q_{2} ≥ q_{1}, we have K_{2} ≅ L_{2}(q_{2}). We have |[[K_{2}, [Z_{1}, K_{1}]], t]| ≥ q_{2}^{3}. As [K_{3}, [Z_{1}, K_{1}]] = 1, we see that [K_{2}, [K_{3}, Z_{1}]] ≠ 1 , so |[K_{2}, [K_{1}, Z_{1}], t]| ≥ q_{2}, and so |[Z_{1}, t]| ≥ q_{2}^{4}. But then q_{1} = q_{2} = r, a contradiction as (r − 1)^{2} | |L_{1} ∩ L_{2}|.

So there are exactly two nontrivial irreducible modules involved. Again either [[K_{1}, Z_{1}], K_{2}] = 1 or [[K_{1}, Z_{1}], K_{3}] = 1. In particular |[Z_{1}, t]| > q_{1}^{2} (q_{1}^{4} for K_{1} ≅ (S)U_{3}(q_{1})). Let [K_{2}, [K_{1}, Z_{1}]] = 1. Then we have |[Z_{1}, t]| ≥ q_{1}^{2}q_{2}^{2} ≥ q_{1}^{4} (or for K_{1} ≅ (S)U_{3}(q_{1}) we have |[Z_{1}, t]| ≥ q_{1}^{4}q_{2}^{2} ≥ q_{1}^{6}). So K_{1} ≅ (S)U_{3}(q_{1}) and we have q_{1} = q_{2} = r, a contradiction again.

Hence [K_{2}, [K_{1}, Z_{1}]] ≠ 1 and so K_{2} is isomorphic to a subgroup of L_{2}(q_{1}) or L_{2}(q_{1}^{2}) in case of K_{1} ≅ (S)U_{3}(q_{1}). Let first K_{1} ≅ (S)U_{3}(q_{1}). As q_{2} ≥ q_{1}, we have K_{2} ≅ L_{2}(q_{1}) or L_{2}(q_{1}^{2}). We have further that there are at least six nontrivial irreducible K_{2}–modules involved if K_{2} ≅ L_{2}(q_{1}) and at least three nontrivial irreducible modules involved if K_{2} ≅ L_{2}(q_{1}^{2}) , so |[t, Z_{1}]| ≥ q_{1}^{6}, a contradiction. Hence we have K_{1} ≅ (S)U_{3}(q_{1}). This gives q_{2} = q_{1} and K_{2} ≅ L_{2}(q_{1}). Suppose K_{1} ≅ Sz(q_{1}). Then we have [Z_{1}, t]| ≤ q_{1}^{3}. But on the other hand |[t, Z_{1}]| ≥ q_{1}^{4}, a contradiction. So we have K_{1} ≅ L_{2}(q_{1}) or (S)L_{3}(q_{1}) and then K_{3} ≅ L_{2}(q_{3}), (S)L_{3}(q_{3}), (S)U_{3}(q_{3}) or Sz(q_{3}). Hence q_{3} < q_{1}. If K_{1} ≅ L_{2}(q_{1}), then |[Z_{1}, t]| ≤ q_{1}^{3} and so there are at most three nontrivial irreducible L_{2}(q_{1})–modules involved. As [K_{1}, [K_{2}, Z_{1}]] ≠ 1, we see [K_{3}, [K_{2}, Z_{1}]] = 1. But then [K_{3}, [K_{1} × K_{2}, Z_{1}]] = 1 and so [K_{1} × K_{2}, [K_{3}, Z_{1}]] = 1, a contradiction. So we have K_{1} ≅ (S)L_{3}(q_{1}) and |[Z_{1}, t]| > q_{1}^{3}. Furthermore [[K_{2}, Z_{1}], K_{3}] ≠ 1. But then [K_{2}, Z_{1}] has to involve at least five nontrivial irreducible modules, a contradiction.

So we have [Z_{1} ∩ Q_{α}, K_{1β}] = 1, and we may assume [Z_{1} ∩ Q_{α}, K_{1β}^{L_{β}}] = 1. Furthermore we see K_{1β} ≤ L_{β}/Q_{β} otherwise V_{2} ≤ L_{β}, a contradiction. Now we have [Z_{1} ∩ Q_{α}, K_{2β}] ≠ 1 ≠ [Z_{1} ∩ Q_{α}, K_{3β}] as otherwise again V_{2} ≤ L_{β}. In particular q_{1} < q_{2}.

Let K_{1} ≅ (S)U_{3}(q_{1}). Then either [K_{1}, [Z_{1}, K_{2}]] ≠ 1 or [K_{1}, [Z_{1}, K_{3}]] ≠ 1. In any case there is X ≤ Z_{β}, |X| ≥ q_{1}^{6} and C I_{X}(Z_{1}) = 1. As |X ∩ L_{1} : X ∩ Q_{1}| ≤ q_{1}, we get r > q_{1}^{2}, a contradiction.

Now we have r ≤ q_{1}. There is X ≤ Z_{β}, |X| ≥ q_{2}^{2} with C I_{X}(Z_{1}) = 1. If X ∩ L_{1} ≤ Q_{1} then r ≥ q_{2} but r ≤ q_{1}, a contradiction. So we have X ∩ L_{1} ≰ Q_{1}. Now for t ∈ X ∩ L_{1} \ Q_{1} we have |Z_{1} : C I_{Z_{1}}(t)| ≤ q_{2}^{2}r^{2} or q_{2}r^{2} (for K_{2} ≅ (S)L_{3}(q_{2})). As r < q_{2} there are at most three nontrivial irreducible modules involved (if K_{2} ≅ (S)L_{3}(q_{2}) there are exactly two). So we have [K_{2}, Z_{1}] is centralized by K_{1} or K_{3}. If [K_{1}, [K_{2}, Z_{1}]] = 1, then as r ≤ q_{1} and r < q_{2}, and there are at least two nontrivial irreducible modules involved, we see that |Z_{1} : C I_{Z_{1}}(t)| ≥ q_{2}^{2}q_{1}^{2} ≥ q_{2}^{2}r^{2}. Hence we have K_{2} ≅ L_{3}(q_{2}) and r = q_{1}. Finally both [K_{2}, Z_{1}] and [K_{1}, Z_{1}] involve exactly two nontrivial irreducible modules. This shows K_{3} is isomorphic to a subgroup of L_{2}(q_{1}) and L_{2}(q_{2}) as well. So K_{3} ≅ L_{2}(q_{3}) for q_{3} ≤ q_{1}. We have (r − 1)^{2} | |L_{1} ∩ L_{2}|. So let Z_{x} × Z_{x} ≅ T = ⟨ω_{1}⟩ × ⟨ω⟩ ≤ L_{1} ∩ L_{2}. As we may assume q_{3} < r, we have [T, K_{3}] = 1. Furthermore if C I_{T} (K_{1}) = 1, we have K_{1} ≅ (S)L_{3}(r). Now as m_{x}(K_{1}T) ≤ 3, we see that T ∩ K_{1} ≠ 1. Hence in any case we have some ω_{2} ∈ T^{♯}, with N_{G}(⟨ω_{2}⟩) ≤ M. Application of (1.66) shows E(L_{2}/Q_{2}) ≅ (S)L_{3}(r) or Sp_{4}(r). Let now t ∈ Z_{1}. then |V_{α} : C I_{V_{α}}(s)| ≤ q_{3}r^{2} < r^{3}. But this contradicts (6.16).

So we have [K_{1}, [K_{2}, Z_{1}]] ≠ 1. Now [K_{3}, [K_{2}, Z_{1}]] = 1. Furthermore K_{1} is isomorphic to a subgroup of SL_{3}(q_{2}) and so K_{1} ≅ L_{2}(q_{1}) or (S)L_{3}(q_{1}). In particular 3 | |K_{1}|. If K_{2} ≅ Sz(q_{2}), then |[t, Z_{1}]| ≥ q_{2}^{4}, a contradiction. So we have 3 | |K_{2}| and then K_{3} ≅ L_{2}(q_{3}), (S)L_{3}(q_{3}), (S)U_{3}(q_{3}) or Sz(q_{3}). So by symmetry we have q_{3} < r ≤ q_{1} < q_{2}. Then we have |V_{α} : C I_{V_{α}}(Z_{1})| ≤ q_{3}^{2}r^{2}. If |V_{α} : V_{α} ∩ L_{1}| ≤ r, we even get |V_{α} : C I_{V_{α}}(Z_{1})| ≤ q_{3}^{2}r and on the other hand by (6.16) we may assume that there is s ∈ Z_{1} with |V_{α} : C I_{V_{α}}(s)| ≥ r^{2}. This shows K_{3} ≅ (S)L_{3}(q_{3}). If |V_{α} : V_{α} ∩ L_{1}| > r and |Z_{1} : Z_{1} ∩ Q_{α}| > r, we have with (6.16) that |V_{α} : C I_{V_{α}}(Z_{1})| ≥ r^{4}, a contradiction. So we have |Z_{1} : Z_{1} ∩ Q_{α}| ≤ r and so |Z_{1} : C I_{Z_{1}}(t)| ≤ q_{2}^{2}r < q_{2}^{3}. Now K_{2} ≅ (S)L_{3}(q_{2}) and there is exactly one nontrivial irreducible K_{1}-module in C I_{Z_{1}}(K_{2}). But this would imply [[K_{1}, C I_{Z_{1}}(K_{2})], K_{3}] = 1 and [[K_{1}, Z_{1}], K_{3}] = 1, a contradiction as [K_{3}, [K_{2}, Z_{1}]] = 1 . So we have K_{3} ≅ (S)L_{3}(q_{3}) and |V_{α} : V_{α} ∩ L_{1}| ≤ r. As |Z_{1} : C I_{Z_{1}}(t)| ≤ q_{2}^{2}r^{2}, we get r = q_{1} and [C I_{Z_{1}}(K_{2}), K_{1}] involves at most two nontrivial irreducible modules. Hence [K_{3}, [C I_{Z_{1}}(K_{2}), K_{1}]] = 1, a contradiction again. This proves (1).

(2) We have q_{1} < r and q_{1}^{2} < r for K_{1} ≅ (S)U_{3}(q_{1}).

Suppose false. Choose β ∈ Δ(α) with Z_{1} ≰ L_{β}. Then we have [Z_{β}, Z_{1} ∩ Q_{α}] = 1 by (1). Obviously [Z_{β}, Z_{1}] ≠ 1. Let t ∈ Z_{β}∩L_{1} with [t, Z_{1}] ≠ 1. Then |Z_{1} : C I_{Z_{1}}(t)| ≤ r^{2} ≤ q_{1}^{2} (or q_{1}^{4} in case of K_{1} ≅ (S)U_{3}(q_{1})). So there are exactly two nontrivial irreducible K_{1}–modules involved and q_{1} = r (or q_{1}^{2} = r). But as C I_{K_{2}×K_{3}}([Z_{1}, K_{1}]) ≠ 1, we have a contradiction. So we have Z_{β} ∩ L_{1} ≤ Q_{1}. Now ⟨Q_{1}, Z_{β}⟩ centralizes Z_{1} ∩ Q_{α}. In particular L_{2} = SC I_{L_{2}}(Z_{1} ∩ Q_{α}). But [V_{α}, Z_{3}] = 1 and so for t ∈ V_{α} ∩ L_{1} we have |[t, Z_{1}]| ≤ r^{2}. Again we get V_{α} ∩ L_{1} ≤ Q_{1}, contradicting (6.17). This proves (2).

We will show that K_{3} ≅ L_{2}(q_{3}), (S)L_{3}(q_{3}), (S)U_{3}(q_{3}) or Sz(q_{3}). If not either K_{1} ≅ Sz(q_{1}) or K_{2} ≅ Sz(q_{2}). Let K_{1} ≅ Sz(q_{1}). Then |V_{α} : C I_{V_{α}}(Z_{1})| ≤ q_{1}r^{i} < r^{i+1}, i = 1, 2. By (6.16) we get i = 2 and so Z_{x}×Z_{x} ≅ E ≤ L_{1}∩L_{2}. By (1.66) N_{G}(⟨ω̃⟩) ≰ M for all ω̃ ∈ E. So we have that C I_{E}(K_{2}) = 1. This gives K_{2} ≅ (S)L_{3}(q_{2}). But then either m_{x}(K_{2}E) ≥ 4 or there is some 1 ≠ ω_{2} ∈ E with ω_{2} ∈ K_{2}. In both cases N_{G}(⟨ω_{2}⟩) ≤ M, a contradiction. Assume now K_{2} ≅ Sz(q_{2}). Then |V_{α} : C I_{V_{α}}(Z_{1})| < r^{4} and so by (1.66) and (6.16) we have that N_{G}(⟨ω̃⟩) ≰ M for all ω̃ ∈ L_{1} ∩ L_{2} with o(ω̃) = x. Hence we have (r − 1)^{2} |̸ |L_{1} ∩ L_{2}| as any ω̃ ∈ L_{1} ∩ L_{2} with o(ω̃) = x has to centralize K_{1} by (1.64) and (1.65).

Now we get that K_{1} ≅ (S)L_{3}(q_{1}). Let β ∈ Δ(α) with Z_{1} ≰ L_{β}. Assume [Z_{1} ∩ Q_{α}, Z_{β}] = 1. Then as before Z_{β} ≰ Q_{2} and so ω centralizes a subgroup of index r in Z_{1}. Now we have [[ω, Z_{1}], K_{1}] = 1 and so [[K_{1}, Z_{1}], ω] = 1. But then [K_{2} × K_{3}, [K_{1}, Z_{1}]] = 1, a contradiction.

Let [Z_{1} ∩ Q_{α}, K_{2β}] ≠ 1. Then we have X ≤ Z_{β}, |X| ≥ q_{2}^{4} with |Z_{1} : C I_{Z_{1}}(X)| ≤ rq_{2}, a contradiction. So we have [Z_{1} ∩ Q_{α}, K_{2β}] = 1. Hence [Z_{1} ∩ Q_{α}, K_{1β}] ≠ 1, and there is X ≤ Z_{β} with |X| = q_{1}^{2} and |Z_{1} : C I_{Z_{1}}(X)| ≤ q_{1}^{2}r ≤ q_{1}^{4} as r ≤ q_{1}^{2}. If X ∩ L_{1} ≰ Q_{1} then X cannot act on K_{2} as for s ∈ K_{2} we have |Z_{1} : C I_{Z_{1}}(s)| ≥ q_{2}^{2} > q_{1}^{4}. Now we have that X ∩ L_{1} ≤ Q_{1}. Hence as q_{1}^{2} ≥ r, we get q_{1}^{2} = r and there are exactly two nontrivial irreducible K_{1}–modules involved. Now [[K_{1}, Z_{1}], K_{2}] = 1. So we have [K_{3}, [K_{1}, Z_{1}]] ≠ 1 and then K_{3} ≅ L_{2}(q_{3}).

So in any case we may assume that K_{3} ≅ L_{2}(q_{3}), (S)L_{3}(q_{3}), (S)U_{3}(q_{3}) or Sz(q_{3}). Let q_{i} ≤ q_{j}, {i, j} = {2, 3}. As before we get q_{i} < r (q_{i}^{2} < r in case of K_{i} ≅ (S)U_{3}(q_{i})). Hence [ω, K_{1}K_{i}] = 1 and so N_{G}(⟨ω⟩) ≤ M, a contradiction.

(6.26) Lemma. Suppose there is exactly one component K of L_{1}/Q_{1} with [V_{α} ∩ L_{1}, K] ≠ 1. Then K ≰L_{1}/Q_{1}.

Proof . Suppose false. We have m_{p}(K) = 1. By (6.16) we have |V_{α} : V_{α} ∩ Q_{1}| ≥ 16. As V_{α} acts quadratically on Z_{1}, we have that K is a group of Lie type in even characteristic or U_{4}(3), 3 · M_{22} or A_{n}. As m_{3}(K) ≤ 3, we get K ≅ 3 · M_{22} or A_{n}, 9 ≤ n ≤ 11 in the latter. Furthermore just natural modules are involved. So assume first that K is not a group of Lie type in even characteristic. As [Z_{1}, K_{1}] is not centralized by E ≅ E_{p^{2}} we may assume with (4.4) that m_{p}(L_{1}) ≥ 4. Furthermore there is β ∈ Δ(α) with Z_{1} ⊈ L_{β}. But there is some t ∈ Z_{1} with |Z_{β} : C I_{Z_{β}}(t)| ≤ 16. As p ≥ 5, we get that there is X ≤ Z_{β} with |X| ≥ 64 which is centralized by some E ≅ E_{p^{2}} and Γ_{E,1}(G) ≤ M_{β}, a contradiction.

So we have that K is a group of Lie type in characteristic two. This now shows that K is isomorphic to L_{n}(q), n ≤ 4, L_{5}(2), L_{6}(2), L_{7}(2), U_{n}(q), n ≤ 4, P Sp_{4}(q), P Sp_{6}(q), Ω_{8}^{−}(q), G_{2}(q), ^{2}F_{4}(q), ^{3}D_{4}(q), or Sz(q).

If K ≅ L_{n}(2), n ≥ 5, then as before K ≅ L_{7}(2), p = 5 or 31 and |V_{α} : V_{α} ∩ Q_{1}| ≥ 2^{8}. Furthermore there are at least 8 irreducible nontrivial K - modules involved. By (6.16) |Z_{1} : Z_{1} ∩ Q_{α}| ≤ 2^{6}. So there is t ∈ Z_{1} ∩ Q_{α} with [t, Z_{β}] ≠ 1. If [K_{β}, t] = 1, we see that [[Z_{β}, K_{β}], t] contains the natural module. If [K_{β}, t] ≠ 1, then as 8 modules are involved and |[Z_{β}, t]| ≤ 2^{12}, we also see that there is a natural module involved. So now a 2-space is centralized by a p-element in K and then there are at least 12 modules involved. But now for t ∈ Z_{1} we have |V_{α} : C I_{V_{α}}(t)| ≤ 2^{6}, a contradiction.

So let K ≅ ^{2}F_{4}(q). Then |V_{α} : V_{α} ∩ Q_{1}| ≤ q and so again Z_{1} ≤ L_{β}. Let K ≅ ^{3}D_{4}(q), then |V_{α} : V_{α} ∩ Q_{1}| ≤ q^{5}. We have |V_{α} : V_{α} ∩ Q_{1}| > q and so Z_{1} just involves natural modules. The same applies for K ≅ G_{2}(q). In the former we get [Z_{1} ∩ Q_{α}, Z_{β}] = 1. But this contradicts (6.16). So we have the latter and |V_{α} : V_{α} ∩ Q_{1}| ≥ q^{2}. But then [Z_{1}, K] involves exactly one nontrivial irreducible module, a contradiction again.

Let K ≅ Ω_{8}^{−}(q), Sp_{6}(q), or Sp_{4}(q). Let [Z_{1} ∩ Q_{α}, Z_{β}] ≠ 1. If [K_{β}, t] = 1, then K ≅ Sp_{6}(q) and |V_{α} : V_{α} ∩ Q_{1}| = q^{6}. Furthermore the natural module is involved. But now there is a direct sum of two such modules Z_{1}^{(i)}, i = 1, 2 where |Z_{1}^{(i)} : C I_{Z_{1}^{(1)}}(V_{α})| = q^{6} and [Z_{1}^{(i)} ∩ Q_{α}, Z_{β}] = 1, contradicting (6.16). Hence there is t with [K_{β}, t] ≠ 1. As there is no field automorphism of order p, we get that p^{3} | |C I(K_{β})|. Further p |q̸ − 1. So we see that there are at least 6 nontrivial irreducible K-modules involved. Hence we have K ≅ Ω_{8}^{−}(q) or Sp_{6}(q), |V_{α} : V_{α} ∩ Q_{1}| = q^{6} and t induces transvections on the natural module. As V_{α} has to act quadratically we get K ≅ Ω_{8}^{−}(q). But now |Z_{1} ∩ Q_{α} : Z_{1} ∩ C I(K_{β})| = 2. So [Z_{β}, K_{β}] centralizes a subgroup of index 2 · |Z_{1} : Z_{1} ∩ Q_{α}| in Z_{1}, which is at least q^{12}. Hence by (6.16) and (6.4) we have that |V_{α} : V_{α} ∩ Q_{1}| ≥ q^{11}, a contradiction.

So we have that [Z_{1} ∩ Q_{α}, V_{α}] = 1. Now we see by (6.16) again that |Z_{1} : Z_{1} ∩ Q_{α}| ≤ q^{3} (q^{2} in case of K ≅ Sp_{4}(q)). In any case we see that there is at most one nontrivial irreducible K - module involved, and so L_{2} ≤ M, a contradiction.

So we have K ≅ L_{2}(q), L_{3}(q), L_{4}(q), U_{3}(q), U_{4}(q), or Sz(q). Suppose [Z_{1} ∩Q_{α}, V_{α}] ≠ 1 and [K_{β}, t] = 1. Then we get K ≅ L_{4}(q), |V_{α} : V_{α} ∩Q_{1}| = q^{4} and natural modules are involved. Now Z_{1} is generated by elements s with |V_{α} : C I_{V_{α}}(s)| ≤ q^{2}. As C I_{L_{1}/Q_{1}}(K) contains an elementary abelian group of order p^{3}, there is some U ≤ Z_{β}, |U| = q^{4} such that U is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{β}, a contradiction. So we have [K_{β}, t] ≠ 1. Hence K ≅ U̸_{3}(q) or Sz(q). In case of K ≅ L_{4}(q) or U_{4}(q), we have that p | |̸L_{2}(q)| and so there are at least 6 nontrivial irreducible K-modules involved, a contradiction.

So we have K ≅ L_{3}(q) or L_{2}(q). In case of L_{2}(q) we just have one nontrivial irreducible involved, a contradiction. So we have K ≅ L_{3}(q) and exactly two nontrivial irreducible K-modules are involved. As p | q̸ − 1,

we get p| q + 1 and so p^{3}| |C I_{L_{1}/Q_{1}}(K)|. But then there is some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M which centralizes [Z_{1}, K], a contradiction. So we have [Z_{1} ∩ Q_{α}, V_{α}] = 1. By (6.16) |Z_{1} : Z_{1} ∩ Q_{α}| ≤ q^{2}. So there are at most two nontrivial irreducible K-modules involved and V_{α} induces transvections. But then |V_{α} : V_{α} ∩ Q_{1}| = q^{4}, a contradiction.

(6.27) Lemma. There are at least two components K_{1}, K_{2} of L_{1}/Q_{1}, where one of the K_{i} might be a cyclic p-group normal in L_{1}/Q_{1}, with [V_{α} ∩ L_{1}, K_{1}] ≠ 1 ≠ [V_{α} ∩ L_{1}, K_{2}].

Proof. Suppose false. Then by (6.26) K_{1} ≰L_{1}/Q_{1}. In particular m_{3}(K_{1}) ≤ 1. So we have K_{1} ≅ L_{2}(q), Sz(q) or J_{1}. As |V_{α} : V_{α} ∩ Q_{1}| ≥ 16, as in the proof of (6.25), we get K ≅ L_{2}(q) or Sz(q), q even. So in L_{β}/Q_{β} we have K_{1β} × K_{2β} and K_{1β} ≅ K_{1} ≅ K_{2β}. Suppose there is t ∈ Z_{1} ∩ Q_{α} with [t, K_{1β}] ≠ 1 or [t, K_{2β}] ≠ 1. As |Z_{β} : C I_{Z_{β}}(t)| ≤ q, we see K ≅ L_{2}(q) and there is exactly one nontrivial irreducible module involved, which then is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M , a contradiction. So we have [Z_{1} ∩ Q_{3}, V_{α}] = 1. Now we get |Z_{1} : Z_{1} ∩ Q_{α}| < q, a contradiction.

(6.28) Lemma. Assume (6.4) with b = b_{1} > 1. Then V_{α} ≤ L_{1}.

Proof . Suppose false. By (6.16) - (6.27) we get that there are exactly two components K_{1}, K_{2} of L_{1}/Q_{1} such that [V_{α} ∩ L_{1}, K_{i}] ≠ 1, i = 1, 2. We have m_{p}(K_{1}) = m_{p}(K_{2}) = 1 for p ∈ σ(M). Furthermore [V_{α} ∩L_{1}, O^{p′}(C I_{L_{1}/Q_{1}}(K_{1} × K_{2}))] = 1. If [[Z_{1}, K_{1}], K_{2}] = 1 we have that p |̸ |C I_{L_{1}/Q_{1}}(K_{1} × K_{2})|. Further by (4.4) we have m_{p}(L_{1}) ≥ 4. Now there is E ≅ E_{p^{2}} with E ∩ K_{1} × K_{2} = 1 and E induces field automorphism on K_{1} and K_{2} as well. This contradicts (4.12). So we have

(1) [[Z_{1}, K_{1}], K_{2}] ≠ 1.

As before we may assume

(2) |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K_{i})| ≥ 8, i = 1, 2 .

Further by (4.4) we have

(3) m_{p}(L_{1}) ≥ 3 and p | |C I_{L_{1}}(Z_{1})| if m_{p}(L_{1}) = 3.

Next we choose notation such that [K_{1}, ω] ≠ 1. This is possible by (1.63) as [ω, Ω_{1}(Z(S))] = 1. As m_{3}(K_{1}) ≤ 3 we get with (1.5) and (1.12) that K_{1} ≅ A_{n}, n ≤ 11, M_{12}, M_{22}, M_{24}, 3·M_{22}, J_{2}, L_{n}(q), n ≤ 4, L_{n}(2), 5 ≤ n ≤ 7, Sp_{2n}(q), n ≤ 3, U_{n}(q), n ≤ 4, U_{5}(4), Ω_{8}^{−}(q), G_{2}(q), ^{3}D_{4}(q), Sz(q), ^{2}F_{4}(q), q even. Set q = 2 for K sporadic or alternating. As V_{α} ≰ L_{1} we have m_{p}(C I_{L_{1}/Q_{1}}(K_{1} × K_{2})) ≤ 1. If m_{p}(L_{1}) ≥ 4, we get that there are outer automorphism of order p in L_{1}. these have to be field automorphism. By (4.12) we get K_{1} ≅ K_{2} ≅ L_{2}(q) or Sz(q). If m_{p}(L_{1}) = 3 we get the same result with (3) as m_{p}(C I(Z_{1})) ≥ 1. Now any involution in K_{1}K_{2} is centralized by an element of order p and so we get for t ∈ V_{α} ∩ L_{1} \ Q_{1} that [t, Z_{1}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. So we have shown V_{α} ≤ L_{1}.

(6.29) Lemma. Let b = b_{1} > 1. If L_{2} is nonsolvable, then E(L_{2}/Q_{2}) ≅ L_{2}(r), L_{3}(r), U_{3}(r), P Sp_{4}(r), L_{2}(r) × L_{2}(r), r odd, (in case of L_{2}(r) or L_{2}(r) × L_{2}(r) we assume r > 5) A_{6}, A_{9}, 3 · A_{6}, 3A_{6} ∗ 3A_{6} or E(L_{2}/F (L_{2}/Q_{2})) ≅ L_{2}(r), r odd.

Proof . Suppose false. Then (6.4) holds. By (6.28) V_{α} ≤ L_{1}. Furthermore by (6.27) there are two components K_{1}, K_{2} of L_{1}/Q_{1}, where one of the K_{i} might be a cylcic normal p-group, with [V_{α}, K_{i}] ≠ 1. We may assume that m_{3}(K_{1}) ≤ 1 and so K_{1} is solvable or K_{1} ≅ L_{2}(q), L_{3}(q), U_{3}(q), Sz(q) or J_{1}. Set Z̃_{1} = [Z_{1}, K_{1}]. Let [Z̃_{1} ∩ Q_{α}, Z_{β}] ≠ 1 for some β ∈ Δ(α) with Z_{1} ≰ L_{β}. Then [Z̃_{1} ∩ Q_{α}, K_{1β} × K_{2β}] ≠ 1 and C I_{Z̃_{1}∩Q_{α}}(K_{1β} × K_{2β}) = 1. Let t ∈ Z̃_{1} ∩ Q_{α} with [t, K_{1β}] ≠ 1. Then either [K_{1}, Z_{1}] is irreducible or K_{1} ≅ SL_{3}(q) and [K_{1}, Z_{1}] involves exactly two natural modules. In the former, as by (4.4) we may assume that m_{p}(L_{1}) ≥ 4, we get that [K_{1}, Z_{1}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. So we have the latter. Now C I_{L_{1}/Q_{1}}(K_{1}) contains some F ≅ E_{p^{3}}, and so as p |̸ q − 1, we get some E ≤ F, |E| = p^{2} with [E, [K_{1}, Z_{1}]] = 1, a contradiction.

So we have [Z̃_{1} ∩ Q_{α}, K_{1β}] = 1. This implies [Z̃_{1} ∩ Q_{α}, K_{2β}] ≠ 1. Now for t ∈ Z̃_{1} ∩ Q_{α} we get [[t, Z_{β}], K_{1β}] = 1 and so [[K_{1}, Z_{1}], K_{2}] = 1. By (4.12) m_{p}(C I_{L_{1}/Q_{1}}(K_{1} × K_{2})) ≥ 1. So there is some component K_{3} of L_{1}/Q_{1} or cyclic normal subgroup with [K_{1} × K_{2}, K_{3}] = 1. We have [t, K_{3β}] ≠ 1. So [K_{2} × K_{3}, [K_{1}, Z_{1}]] = 1. But as [t, K_{1β}^{L_{β}}] = 1, we have K_{1} ≤ L_{1}/Q_{1} and so [K_{1}, Z_{1}] ∩ Z_{2} ≠ 1. So we get L_{2} ≤ M, a contradiction.

This shows [Z̃_{1}∩Q_{α}, Z_{β}] = 1 for β ∈ Δ(α) with Z_{1} ≰ L_{β}. By (6.1) we have [Z̃_{1} ∩ Q_{α}, V_{α}] = 1. Hence |Z̃_{1} : C IZ̃_{1}(V_{α})| ≤ r^{2}. If K_{1} is solvable or isomorphic to J_{1} then |V_{α} : C I_{V_{α}}(Z̃_{1})| ≤ 2 contradicting (6.16). Let K_{1} not isomorphic to L_{3}(q). Then |V_{α} : C I_{V_{α}}(Z̃_{1})| ≤ q. By (6.16) |V_{α} : C I_{V_{α}}(Z̃_{1})| ≥ r^{2}, so r^{2} ≤ q and then we get r^{2} = q. But now by (6.16) again |V_{α} : C I_{V_{α}}(Z̃_{1})| ≥ r^{4}, a contradiction. So we have K_{1} ≅ (S)L_{3}(q) and so r = q. But in this case Z̃_{1} is irreducible. As K_{1} ≤ L_{1}/Q_{1}, we get L_{2} ≤ M, a contradiction.

(6.30) Hypothesis. Either L_{2} is solvable or one of the groups in (6.29). If L_{2} is solvable then σ(L_{2}) = ∅.

As Z_{1} ≰ Q_{α}, we see that E(L_{2}/Q_{2}) ≅ A̸_{9}. Now we have |V_{α} : V_{α}∩Q_{2}| ≤ 8 or E(L_{2}/Q_{2}) ≅ L_{2}(r) × L_{2}(r), or 3 · A_{6} ∗ 3 · A_{6} and |V_{α} : V_{α} ∩ Q_{2}| ≤ 64. But V_{α} has to act quadratically which shows |V_{α} : V_{α} ∩ Q_{2}| ≤ 16. So we have

(6.31) Lemma. Assume (6.30). Then |V_{α} : V_{α} ∩ Q_{2}| ≤ 16. If E(L_{2}/Q_{2}) ≅ 3̸ · A_{6} ∗ 3 · A_{6} or L_{2}(r) × L_{2}(r) then even |V_{α} : V_{α} ∩ Q_{2}| ≤ 8 and |V_{α} : V_{α} ∩ Q_{2}| ≤ 4 if L_{2} is nonsolvable.

(6.32) Lemma. Suppose (6.30) with b = b_{1} > 1. Then L_{1} is not as in (4.9) (ii) and also not the special case.

Proof . Suppose false. We see that |V_{α} : C I_{V_{α}}(Z_{1})| ≤ 2^{2} · |V_{α} : V_{α} ∩ Q_{2}|. Hence with (1.31) and (6.16), (6.28) we get that |Z_{1} : Z_{1} ∩ Q_{α}| ≤ 4 and |Z_{1} : Z_{1} ∩ Q_{α}| = 2 for p = 3.

Let first L_{1} to be nonsolvable. Suppose there is some t ∈ Z_{1} ∩ L_{β} with [Z_{β}, t] ≠ 1 for β ∈ △(α). Let [O_{r}(L_{β}), Z_{β}] ≠ 1. We have that t inverts some subgroup R of order r^{2} in O_{r}(L_{β}). As |[Z_{β}, t]| ≤ 2^{6}, this implies |[R, Z_{β}]| ≤ 2^{12} and |[O_{r}(L_{β}), Z_{β}]| ≤ 2^{18}. Now Sylow r-subgroups of L_{18}(2) have to be non abelian. This shows r = 7 and |[Z_{β}, t]| = 2^{6}. But then L_{β} does not involve A_{5} and so |[Z_{β}, t]| ≤ 2|V_{α} : V_{α} ∩ Q_{2}| ≤ 2^{5}, a contradiction.

This shows [Z_{1}, O_{r}(L_{1})] = 1. But then [O_{r}(L_{1}), Ω_{1}(Z(S))] = 1 and so by (4.10) L_{2} ≤ M, a contradiction.

This now proves [Z_{1} ∩ Q_{α}, V_{α}] = 1. But then V_{α} centralizes a subgroup of index four in Z_{1}, a contradiction as before as [V_{α} ∩ L_{1}, Z_{1}] ≠ 1 by (6.16).

So we have that L_{1} is solvable. If Z_{α+1} ⊆ L_{1}, then Z_{1} = ⟨t | |Z_{α+1}(t)| ≤ 2⟩. But then any such t centralizes some x ∈ Z_{α+1} such that C I_{G}(x) ≤ M_{α+1}, a contradiction. So Z_{α+1} ≰ L_{1}. Now we get |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ 4, otherwise there is some 1 ≠ x ∈ [Z_{α+1} ∩ L_{1}, Z_{1}] such that C I_{G}(x) ≤ M and so Z_{α+1} ≤ L_{1}, a contradiction. Hence we have |Z_{α+1} : C I_{Z_{α+}}(Z_{1})| ≤ 16. As C I_{Z_{α+1}}(Z_{1}) does not contain some x with C I_{G}(x) ≤ M_{α+1} we get m_{p}(L_{1}) = 3 or p = 3 and m_{p}(L_{1}) = 4. Suppose m_{p}(L_{1}) = 4. Then |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 8, a contradiction again. So in any case m_{p}(L_{1}) = 3. Now Sylow 2-subgroup of L_{1}/Q_{1} are cyclic. Let t ∈ Z_{α+1} ∩ L_{1}, [t, Z_{1}] ≠ 1. Then t acts fixed point freely on O_{p}(L_{1}/Q_{1}). As |[t, Z_{1}]| ≤ 8, this implies p = 3 and |[ω, Z_{1}]| = 4 for some ω ∈ O_{3}(L_{1}/Q_{1})^{♯}. But this contradicts (1.52).

We now have that [Z_{α+1} ∩ Q_{2}, Z_{1}] = 1 and |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 4. But then p = 3 as before. Now |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 2 and then as before Z_{1} ≤ M_{α+1}, a contradiction.

(6.33) Lemma. Suppose (6.30). Then [Z_{1} ∩ Q_{α}, Z_{α+1}] ≠ 1.

Proof . Suppose false. Then |Z_{1} : C I_{Z_{1}}(Z_{α+1})| ≤ 16 and |Z_{1} : C I_{Z_{1}}(Z_{α+1})| ≤ 8 for 3 ∈ σ(M). Hence as Z_{1} ≰ M_{α+1}, we have Z_{α+1} ∩ L_{1} ≰ Q_{1}. So let K be a component of L_{1}/Q_{1} with [K, Z_{α+1} ∩ L_{1}] ≠ 1.

(1) [K, Z_{α+1} ∩ L_{1}] ≤ K

Let t ∈ Z_{α+1} ∩ L_{1} with K^{t} ≠ K. Let first |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| ≥ 4. Then by (1.11) we have K ≅ L_{2}(q), q even. As |Z_{1} : C I_{Z_{1}}(t)| ≤ 16 we then get q ≤ 4. Hence K ≅ A_{5} and [K × K^{t}, Z_{1}] is the O_{4}^{+}(4)-module. Finally 3 ∉ σ(M). Now for s ∈ L_{1} ∩ Z_{α+1}, with [s, K] = 1, we see that [s, [K ×K^{t}, Z_{1}]] = 1. But as C I_{Z_{1}}(u) = C I_{Z_{1}}(Z_{α+1} ∩L_{1}) for u ∈ Z_{α+1} ∩K ×K^{t}, we get that [s, Z_{1}] = 1. Hence |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ 8. As 5 ∈ σ(M) we now see that Z_{1} ≤ M_{α+1}, a contradiction.

So we have |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| = 2. There is U ≤ [K × K^{t}, Z_{1}] with U ∩ C I(Z_{α+1}) = 1 and |U| ≥ 4. Now by (6.16) |V_{α} : C I_{V_{α}}(U)| ≥ 2^{6}. This shows that |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{2}. In particular K ≅ A_{5} again. Further |V_{α} : V_{α} ∩ L_{1}| ≥ 8. If 3 ∉ σ(M) then 5 ∈ σ(M). As [Z_{1}, K × K^{t}] is generated by elements x which centralize a subgroup of index 2^{7} in V_{α}, we see that [Z_{1}, K × K^{t}] ≤ ⋂_{β∈Δ(α)} M_{β}. So [Z_{1}, K × K^{t}] ≤ Q_{α}, a contradiction. Hence we have 3 ∈ σ(M). Now we get that L_{2} is solvable with m_{r}(L_{2}) = 3. Furthermore L_{2} possesses an F_{1}-module with offending foursgroup. This contradicts (1.36).

(2) K/Z(K) ≅ A̸_{n}, n ≠ 8

Suppose K/Z(K) ≅ A_{n}. Assume furthermore 3 ∈ σ(M). Then |[t, Z_{1}]| ≤ 8 for t ∈ Z_{α+1} ∩L_{1}. By (1.30) [K, Z_{1}] just involves natural modules or n ≤ 7. So assume n ≥ 9. Then any element in the natural module is centralized by E ≅ E_{9} with Γ_{E,1}(G) ≤ M. Hence we get L_{2} ≤ M, a contradiction.

So we have n ≤ 7. This shows |V_{α} ∩ L_{1} : V_{α} ∩ C I(K)| ≤ 8. Now as before there is no foursgroup U with C I_{U}(Z_{α+1}) = 1. So we get that t induces a transvection and then [K, Z_{1}] is the natural module. This again shows K ≅ A_{6} or A_{5}. Furthermore K ≰L_{1}/Q_{1}. But then again L_{2} ≤ M.

So we have 3 ∉ σ(M). Hence n ≤ 11. As |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≤ 2^{5}, we see that p = 5 or 7. In the former |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{4} and so n ≥ 9. As 3 ∉ σ(M), we see that K ≤ L_{1}/Q_{1} and so as m_{p}(K) ≤ 2, we see that some x ∈ Ω_{1}(Z(S)) ∩ [K, Z_{1}] is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, contradicting L_{2} ⊈ M. So we have p = 7 and |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{2}. Now n ≥ 7 and K ≤ L_{1}/Q_{1}. But then as before we get L_{2} ≤ M, a contradiction.

(3) K/Z(K) is not sporadic.

As there are no transvections we get |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 4. So K is one of the groups in (1.12). By (1.30) |Z_{1} : C I_{Z_{1}}(t)| ≥ 8 and so we even get |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 16. Hence by (1.12) we have K ≅ 3 · M_{22}. Now by (1.30) we have |Z_{1} : C I_{Z_{1}}(t)| = 16 and so 3 ∉ σ(M). But then we even get |V_{α} : C I_{V_{α}}([Z_{1}, K])| ≥ 2^{10}, a contradiction.

(4) If K/Z(K) is of Lie type, then the characteristic is two.

Suppose K is of of odd characteristic. As there are no transvections we get |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 4 and so by (1.12) K/Z(K) ≅ U_{4}(3). So 3 ∈ σ(M). Now |V_{α} : C I_{V_{α}}(K)| ≤ 2^{8}. As [Z_{1}, K] ≰ Q_{α}, we get that m_{3}(L_{1}) ≤ 6 and so K ≤ L_{1}/Q_{1}. By the construction of L_{1} we see C I_{L_{1}/Q_{1}}(K) = Z(K). Now we have that |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| = 2. In particular |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 16. As m_{3}(L_{1}) = 5, we see that Z_{1} ≤ M_{α+1}, a contradiction.

(5) [E(L_{1}/Q_{1}), Z_{α+1} ∩ L_{1}] = 1.

Otherwise by (1) - (4) there is K ≅ G(q), q = 2^{m}, with 1 ≠ [K, Z_{α+1} ∩ L_{1}] ≤ K.

Suppose K ≅ L_{n}(q). Then there are at most four nontrivial irreducible modules involved. Assume there are exactly four. Then q = 2 and these modules are natural ones. Further 3 ∉ σ(M) and so n ≤ 7. If n ≥ 4, we have K ≤ L_{1}/Q_{1} and so E ≅ E_{p^{2}} centralizes [K, Z_{1}], a contradiction or p = 7 and n ≥ 6. Then some 7-element centralizes [K, Z_{1}]. But some 1 ≠ x ∈ Ω_{1}(Z(S)) ∩ [Z_{1}, K] is centralized by a 7-element in K, so we have the contradiction L_{2} ≤ M. So let n = 3, K ≅ L_{3}(2). Then p = 7. But now [Z_{1}, K] centralizes in V_{α} a subgroup of index 2^{6}, a contradiction.

Assume now there are exactly three modules involved. Again q = 2 and these modules are natural modules or n = 4 and one of them is the orthogonal module. If 3 ∉ σ(M), we get a contradiction as before. So let 3 ∈ σ(M). If n ≥ 7 any x ∈ Ω_{1}(Z(S)) is centralized by E ≅ E_{9}, a contradiction. So we have n ≤ 6. As |V_{α} ∩ L_{1} : V_{α} ∩ C I(K)| ≥ 2^{5}, we get n ≥ 5. If n = 5, then m_{3}(K) = 2, and as any x ∈ Ω_{1}(Z(S)) ∩ [Z_{1}, K] is centralized by some 3-element in K, we get L_{2} ≤ M. So we have n = 6. But then the natural module and its dual is involved. Hence in one of these modules we get some x ∈ Ω_{1}(Z(S)) which is centralized by some E ≅ E_{p^{2}} in K, a contradiction.

So let now be exactly two modules involved. Then q ≤ 4. Suppose q = 4. then we have two natural modules or duals. Furthermore 3 ∉ σ(M). Let n ≥ 3. Then we have K ≤ L_{1}/Q_{1}. In particular we have m_{p}(K) ≥ 2 if p = 5 and m_{p}(K) ≥ 3 else. As 3 ∉ σ(M) we have K ≅ L_{4}(4). This shows p = 5. But any x ∈ Ω_{1}(Z(S)) ∩ [K, Z_{1}] is centralized by a 5-element in K and so L_{2} ≤ M, a contradiction. So K ≅ L_{2}(4). Now x ∈ [Z_{1}, K] centralizes in V_{α} a subgroup of index 2^{7}, which yields x ∈ Q_{α}, as p = 5, a contradiction.

So we have q = 2 again. By (1.30) we just have two nontrivial modules involved and we get a contradiction as before, or n = 4, q = 2 and we have two orthogonal modules. But this case we will handle lateron together with the orthogonal groups.

So we are left with [K, Z_{1}] involves exactly one irreducible nontrivial module and so q ≤ 16. Let [t, [K, Z_{1}]] ≤ 8, then q ≤ 8 and by (1.30) the module is the natural one, or n = 2, q = 4 and we have the orthogonal module. Suppose 3 ∈ σ(M). Then we see that n ≤ 4. Let n = 4, then K ≅ SL_{4}(8) or L_{4}(2). Let n ≥ 3, then x ∈ Ω_{1}(Z(S)) is centralized by E ≅ E_{9}, a contradiction. So we have K ≅ L_{2}(8) or L_{2}(4). So for x ∈ [Z_{1}, K] we have |V_{α} : C I_{V_{α}}(x)| ≤ 2^{6}. This shows K ≅ L_{2}(4). But then |V_{α} : C I_{V_{α}}(x)| ≤ 2^{5}, a contradiction. So we have 3 ∉ σ(M). Let m_{3}(K) ≥ 2. Then K ≤ L_{1}/Q_{1}. Hence we get m_{p}(K) ≥ 2 for p| q − 1 and m_{p}(K) ≥ 3 for p | q̸ − 1. This shows p = 7 and K ≅ L_{n}(8), 4 ≤ n ≤ 7. As no element in Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{7^{2}}, we get a contradiction. So let m_{3}(K) = 1, i.e. K ≅ L_{2}(4), L_{2}(8), L_{3}(2) or L_{3}(8). We have K ≅ L̸_{2}(4). Otherwise p = 5 and as any x ∈ [Z_{1}, K] centralizes a subgroup of index 2^{7} in V_{α}, we get a contradiction. So we have p = 7 and K has at most two conjugates under S. As no x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{7^{2}}, we get K ≅ L_{2}(8) or L_{3}(8). If K ≅ L_{3}(8), then there is some ω with o(ω) = 7 and [K, ω] = 1. Now as any x ∈ Ω_{1}(Z(S)) is centralized by some 7-element in K we see that there is x ∈ Ω_{1}(Z(S))^{♯} which is centralized by E ≅ E_{7^{2}}, a contradiction. So we have K ≅ L_{2}(8) and K is not normalized by S. As before there is U ≤ [K, Z_{1}], |U| = 8 with U ∩ C I(Z_{α+1}) = 1. By (6.16) |V_{α} : C I_{V_{α}}(U)| ≥ 2^{8}. But as V_{α} normalizes K, we get |V_{α} : C I_{V_{α}}(U)| ≤ 2^{7}, a contradiction.

So we have |[t, [Z_{1}, K]]| = 16. In particular 3 ∉ σ(M). Now n ≤ 7. Suppose K ≤ L_{1}/Q_{1}, further q > 2. If n ≥ 5, then p| q − 1. Now by (1.33) we have E ≅ E_{p^{2}} which centralizes Ω_{1}(Z(S)), a contradiction. So we have n ≤ 4. As E(L_{2}/Q_{2}) ≅ L_{2}(r) × L_{2}(r), or 3 · A_{6} ∗ 3 · A_{6}, we see with (6.16) that for some U ≤ [Z_{1}, K], |U| = 16, we have |V_{α} : C I_{V_{α}}(U)| ≥ 2^{12}. Hence |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{8}. This shows K ≅ L_{3}(16), or L_{4}(q). In the former p = 5 and [K, Z_{1}] involves the natural module just once. But now we get some x ∈ Ω_{1}(Z(S))^{♯} which is centralized by E ≅ E_{25} in L_{1}, a contradiction. So we have K ≅ L_{4}(q). If p|q − 1, then m_{p}(K) = 3 and E ≅ E_{p^{2}} centralizes Ω_{1}(Z(S)). So we have p |̸ q − 1. Now m_{p}(K) ≤ 2 and so some E ≅ E_{p^{2}} centralizing K centralizes [K, Z_{1}] too, a contradiction.

So we are left with K ≰L_{1}/Q_{1} and then m_{3}(K) = 1, i.e. K ≅ L_{2}(q) or L_{3}(q), or 3||Z(K)| and so K ≅ SL_{3}(q). Let K ≅ L_{3}(q), q > 2 or SL_{3}(q), q > 4. Then p|q − 1 and so some x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{p^{2}}, a contradiction. So we have K ≅ L_{2}(4), L_{2}(8), L_{2}(16), L_{3}(2) or SL_{3}(4). As |V_{α} : V_{α} ∩ C I(K)| ≥ 2^{12}, we get a contradiction.

Let next K ≅ Sp_{2n}(q), n ≥ 2 or Ω_{2n}^{±}(q), n ≥ 3. If [K, Z_{1}] involves more than one nontrivial irreducible K-module then by (1.30) these are natural modules or q = 2 , n = 3 and the spin module is involved. If four modules are involved we have q = 2 and 3 ∉ σ(M). So n = 2, 3 or K ≅ Ω_{8}^{−}(2). Now m_{p}(C I(K)) ≥ 3, contradicting L_{2} ≰ M. Let exactly three nontrivial irreducible modules be involved. Again q = 2 and these modules are natural modules or one of them is the spin module for Sp_{6}(2) or Ω_{6}^{−}(2). In the latter 3 ∉ σ(M) and |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{8}, a contradiction. So we have three natural modules. If 3 ∉ σ(M) we get a contradiction as before. So assume that 3 ∈ σ(M). Then we get K ≅ Sp_{4}(2) or Ω_{6}^{−}(2). Furthermore we have K ≤ L_{1}/Q_{1}. This shows K ≅ Ω_{6}^{−}(2). But now |V_{α} : C I_{V_{α}}(K)| ≥ 2^{8}, a contradiction.

Let now exactly two nontrivial irreducible modules be involved. Again by (1.30) these are natural ones q ≤ 4 or K ≅ Sp_{6}(2) or Ω_{6}^{−}(2) and there is a spin module. If q = 4 we have 3 ∉ σ(M). So we have K ≅ Sp_{4}(4), Sp_{6}(4), Ω_{6}^{±}(4). But K ≅ Sp_{4}(4) is impossible. We have that K ≤ L_{1}/Q_{1} and so either m_{p}(K) ≥ 3 or p = 5 and m_{5}(K) = 2. This in any case shows p = 5. As there is no E_{p^{2}} centralizing some x ∈ Ω_{1}(Z(S))^{♯}, we get K ≅ Ω_{6}^{−}(4). But now we see |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 2^{5} and so Z_{1} ≤ M_{α+1}, a contradiction.

So we have q = 2. Suppose 3 ∉ σ(M), then K ≅ Sp_{6}(2), Ω_{6}^{−}(2) or Ω_{8}^{−}(2) and K ≤ L_{1}/Q_{1}. Further m_{p}(K) = 1 and so some E ≅ E_{p^{2}} centralizes some x ∈ Ω_{1}(Z(S))^{♯}, a contradiction.

So we have 3 ∈ σ(M). As no x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{9}, we see K ≅ Ω_{6}^{−}(2). As m_{3}(K) = 3, we must have two natural modules and |Z_{α} ∩ L_{1} : C I_{Z_{α}∩L_{1}}(K)| = 2. This shows |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≤ 16. Hence we have U ≤ Z_{1}, |U| = 4 with U ∩ C I(Z_{α+1}) = 1. If L_{2} is solvable then |V_{α} : C I_{V_{α}}(U)| ≥ 2^{8} by (6.16), a contradiction. So L_{2} is nonsolvable and |[Z_{1}, t]| = 4, In particular |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| = 8, contradicting Z_{1} ≰ M_{α+1}.

So we have that [K, Z_{1}] contains exactly one nontrivial irreducible Kmodule. Let first |[t, [Z_{1}, K]]| ≤ 8. Then we have the natural module or the spin module in case of K ≅ Sp_{6}(2) or Ω_{6}^{−}(2). Suppose p | q − 1, then we get K ≅ Sp_{4}(q) or Ω_{6}^{−}(q) or K ≅ Sp_{6}(q) and we have the spin module. In the latter q = 4 but then Ω_{1}(Z(S))^{♯} is centralized by an elementary abelian group of order 9, a contradiction. So we have K ≅ Sp_{4}(q) or Ω_{6}^{−}(q) and the natural module. In both cases m_{p}(K) = 2. But then there is a p-element centralizing [K, Z_{1}], a contradiction. So we have p |̸ q−1. Then K ≅ Sp_{4}(q), Sp_{6}(q), Ω_{6}^{−}(q), Ω_{8}^{−}(q), Ω_{6}^{+}(q). Further m_{p}(K) ≥ 3. But this is impossible.

So we have |[t, [Z_{1}, K]]| = 16 and 3 ∉ σ(M). This again shows K ≅ Sp_{4}(q), Sp_{6}(q), Ω_{6}^{−}(q), Ω_{8}^{−}(q), Ω_{6}^{+}(q). Further K ≤ L_{1}/Q_{1}. As before p | q−1. This shows q = 8 and p = 7 or q = 16 and p = 5. Finally m_{p}(K) ≥ 3. So K ≅ Sp_{6}(q), Ω_{8}^{−}(q) or Ω_{6}^{+}(q). By (1.22) in any case we have the natural module. But now some x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{p^{2}} in K with Γ_{1,E}(G) ≤ M, a contradiction.

Let now K ≅ U_{n}(q) or G_{2}(q). Then q ≤ 4 and there are at most two nontrivial irreducible modules involved. Assume there are exactly two. Then q = 2 and these modules are natural ones by (1.30). Further 3 ∉ σ(M). So K ≅ U_{4}(2) or G_{2}(2)′. But U_{4}(2) ≅ Ω_{6}^{−}(2), a case just handled. So K ≅ G_{2}(2)′ and K ≤ L_{1}/Q_{1}. Now there is U ≤ Z_{1}, |U| = 16 with U ∩ C I(Z_{α+1}) = 1 and |V_{α} : C I_{V_{α}}(U)| ≤ 16 · 8, a contradiction.

Hence [K, Z_{1}] involves exactly one nontrivial irreducible K-module. If K ≅ U_{n}(q), n ≥ 5, then p | q + 1 and so some x ∈ Ω_{1}(Z(S))^{♯} is centralized by some E ≅ E_{p^{2}} with Γ_{1,E}(G) ≤ M. So we have K ≅ U_{4}(2), U_{4}(4), U_{3}(4), G_{2}(2)′ or G_{2}(4). Suppose q = 4. Then we have |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{8} and just K ≅ U_{4}(4) is possible. Now we get m_{p}(K) ≥ 3 and so p = 5. But now |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| = 4 and we get |Z_{α+1} : C I_{Z_{α+1}}(Z_{1})| ≤ 2^{6}, contradicting Z_{1} ≰ M_{α+1}.

So we have q = 2. Now we see |[[K, Z_{1}], t]| = 4 and so just the natural module is involved in [K, Z_{1}]. If 3 ∉ σ(M), then K ≤ L_{1}/Q_{1} and we are done as m_{p}(K) = 1. So we have 3 ∈ σ(M). Again K ≤ L_{1}/Q_{1}. This shows K ≅ U_{4}(2) and m_{3}(C I_{L_{1}/Q_{1}}(K)) = 1. Now let U be as before. We see that |V_{α} : C I_{V_{α}}(U)| ≤ 2^{7}, and so L_{2} is nonsolvable and |Z_{1} : Z_{1} ∩ Q_{α}| = 4. Hence |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| = 2. So Z_{1} centralizes a subgroup of index 8 in Z_{α+1}, a contradiction.

By (1.22) we are done.

(6) If K ≤ L_{1}/Q_{1}, K a p-group, p ∈ σ(M), then [K, Z_{α+1} ∩ L_{1}] = 1.

Suppose false. Choose t ∈ Z_{α+1} ∩ L_{1}, 1 ≠ ω ∈ K with ω^{t} = ω^{−1}. Then as |[Z_{1}, t]| ≤ 16, we get o(ω) = 3, 5, 7, 17. Suppose p = 5 or 17. Then |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≥ 2^{4} and so there is X ≤ K((Z_{α+1} ∩ L_{1})Q_{1}/Q_{1}) X ≅ D_{2p} × D_{2p} × D_{2p} × D_{2p} with (Z_{α+1} ∩ L_{1})Q_{1}/Q_{1} ∈ Syl_{2}(X). But then there is E ≅ E_{p^{2}} centralizing Z_{1} , a contradiction.

So we have p = 7 or 3. If p = 7 the same argument as before shows that |V_{α} : V_{α} ∩ L_{1}| = 16. Now |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ 2^{8}. But then we see that there is some subgroup U of Z_{1} of order 8 centralizing a subgroup of index 2^{5} in V_{α}, a contradiction.

So p = 3 and |[Z_{1}, t]| ≤ 8. If |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≥ 4, then we find U ≤ Z_{1}, |U| = 4 with C I_{U}(Z_{α+1}) = 1 and |V_{α} : C I_{V_{α}}(U)| ≤ 2^{5}, a contradiction. So we have |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| = 2. Now we get m_{3}(L_{1}) = 4. Further |Z_{α+1} : Z_{α+1} ∩ L_{1}| = 8, and so L_{2} is solvable. Now we see that |[Z_{1}, t]| = 2. Let s ∈ Z_{1}, [s, t] ≠ 1 with |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(s)| = 2. Then |V_{α} : C I_{V_{α}}(s)| ≤ 16 and so by (6.16) we have equality and L_{2} is a {2, 5}-group. Let ω be as before. Now |[Z_{1}, ω]| = 4. Hence ⟨ω⟩ ≰L_{1}/Q_{1}. But then there is E ≅ E_{9} which centralizes a subgroup of index 16 in Z_{1}. As Z_{1} ≰ M_{α+1}, we see that |Z_{1}| ≤ 2^{8}. Furthermore 3 |̸ |C I(Z_{1})|. So K is isomorphic to a subgroup of ℤ_{3} ≀ ℤ_{3} × ℤ_{3}. Now we have |V_{2} : C I_{V_{2}}(Z_{α+1})| ≥ 2^{12}, which shows that |V_{2} ∩ L_{α+1}/V_{2} ∩ Q_{α+1}| ≥ 2^{9}. As L_{1} is solvable this is impossible.

To prove the assertion it is enough to treat the case |O_{r}(L_{1}/Q_{1})| = r^{3} and L_{1}/O_{2,r}(L_{1}) ≅ L_{2}(r), r odd. By (1.51) we have [O_{r}(L_{1}), Z_{1}] = 1. But then by (4.10) L_{2} ≤ M, a contradiction.

(6.34) Lemma. If b = b_{1} > 1 then L_{2} is solvable with σ(M) ≠ ∅.

Proof. We choose notation as before.

(1) [Z_{α+1} ∩ L_{1}, Z_{1}] ≠ 1

Otherwise Z_{1} centralizes a subgroup of index 2^{4} in Z_{α+1}. This shows p = 3, |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 16 and Sylow 3-subgroup of L_{2} are cyclic, a contradiction to m_{2}(L_{2}/Q_{2}) = 1.

Let K be a component of L_{1}/Q_{1} with 1 ≠ [K, Z_{α+1} ∩ L_{1}]. Until further notice assume m_{p}(K) ≥ 2.

(2) [K, Z_{α+1} ∩ L_{1}] ≤ K

Otherwise |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| = 2. We have |[[K × K^{t}, Z_{1}], t]| ≥ 4, for some t ∈ Z_{α+1} ∩ L_{1}. Suppose there is U ≤ [K × K^{t}, Z_{1}] , |U| = 4 with U ∩ M_{α+1} = 1. Then |V_{α} : C I_{V_{α}}(U)| ≥ 64, a contradiction. So there is some s ∈ [K × K^{t}, Z_{1}]^{♯} with s ∈ L_{α+1} and [Z_{α+1}, s] ≠ 1. Hence we have |[Z_{α+1}, s]| ≤ 2^{5}. As [Z_{α+1}, K_{α+1}] has to involve at least three nontrivial irreducible modules, we get that s induces a transvection on one of them. So K ≅ A_{n}, L_{n}(2), Sp_{2n}(2) or Ω_{2n}^{±}(2). Further K is isomorphic to a subgroup of L_{5}(2). This shows K ≅ L_{4}(2), L_{5}(2) or A_{6}. In any case p = 3 and so even |[Z_{α+1}, s]| ≤ 2^{4}. Hence K ≅ L_{4}(2) or A_{6}. But in both cases [s, Z_{1}] is normalized by E ≅ E_{9}, a contradiction.

(3) K ≅ A̸_{n}

Suppose K ≅ A_{n}. As m_{p}(K) ≥ 2, we have n ≥ 6. If n ≥ 12, then p = 3. Let s ∈ Z_{1} ∩ L_{α+1}. If n ≥ 13, then s is centralized by E ≅ E_{9} in K, a contradiction.

So we have n ≤ 12. Suppose next that |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| = 2 and |[t, [Z_{1}, K]]| ≥ 16, for t ∈ Z_{α+1} ∩ L_{1}. Then there is U ≤ [Z_{1}, K], |U| ≥ 8 with |Z_{α+1} : C I_{Z_{α+1}}(U)| ≤ 32. Further U acts quadratically. So in any case we get with (1.30) or (1.12)(iii) that [Z_{1}, K] just involves natural modules T, spin modules T or n ≤ 8 and the four dimensional module T is involved.

Suppose first p = 3 and natural modules are involved. If n > 8, then any element in the natural module is centralized by E ≅ E_{9}, a contradiction. So n ≤ 8. By (4.4)(ii) we may assume that C I_{L_{1}/Q_{1}}(K) contains an elementary abelian group of order 3^{2}. We have |[Z_{1}, t]| ≤ 2^{7}. As there are at least four conjugates of T under C I_{L_{1}/Q_{1}}(K) we see that t has to induce a transvection on some of these modules. So we get U ≅ E_{16} in Z_{1} with |Z_{α+1} : C I_{Z_{α+1}}(U)| ≤ 16.. This shows |U : U ∩ Q_{α}| ≥ 8. Now |V_{α} : C I_{V_{α}}(U)| ≥ 2^{8}. Hence V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{5}, a contradiction.

Now we have U as before |U| = 16 and so U ∩ Q_{α} induces a quadratic fours group on K_{α+1}. So by (1.12) s ∈ U corresponds to (1, 2)(3, 4) or (1, 2)(3, 4)(5, 6)(7, 8). As C I_{K_{α+1}}(s) does not contain E ≅ E_{9}, we get n ≤ 10. By (1.12) we also see that |[Z_{α+1}, s]| ≤ 2^{6}. So there is exactly one spin module involved. As K is centralized by a 3-element in L_{1}, we see that the spin module is centralized by a 3-element in L_{1}. Now [T, s] is normalized by E ≅ E_{9}, a contradiction.

So assume n = 7, 8 and we have the four dimensional module involved. The case K ≅ A_{8} ≅ L_{4}(2) will be treated among the groups of Lie type. So we have n = 7. Then |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| ≤ 4 and so |[s, Z_{α+1}]| ≤ 2^{5}. Now we get a contradiction as before.

So we have p = 5, n = 10, 11. Again by (4.4)(ii) we may assume that C I_{L_{1}/Q_{1}}(K) contains E ≅ E_{25}. Now we have at least 8 modules involved. Hence we get some U ≤ L_{α+1} ∩ Z_{1}, U ∩ Q_{α+1} = 1, |U| ≥ 16 and |Z_{α+1} : C I_{Z_{α+1}}(U)| ≤ 2^{9}, a contradiction.

(4) K is not sporadic.

Suppose that |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| = 2 = |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}|. Then for t ∈ Z_{α+1} ∩ L_{1} we have |Z_{1} : C I_{Z_{1}}(t)| ≤ 2^{5}. This shows with (1.25) that K ≅ M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, or J_{2}. Further |Z_{1} : C I_{Z_{1}}(t)| ≥ 8. Hence we get U ≤ Z_{1}, |U| = 4, U ≤ L_{α+1} with U ∩ Q_{α+1} = 1, a contradiction.

So we have that K admits a quadratic fours group. Hence by (1.12) K ≅ M_{12}, M_{22}, 3 · M_{22}, M_{24}, Co_{1}, Co_{2}, J_{2} or 3 · Suz. Let p = 3. Then for M_{12}, M_{22}, 3 · M_{22} or M_{24} we have by (4.4)(ii) that 3 | |C I_{L_{1}/Q_{1}}(K)| and so there are at least two quadratic modules involved. So |[Z_{1}, t]| ≥ 2^{8}, 2^{6}, 2^{8}, 2^{8}. But this contradicts the fact that besides for 3 · M_{22} we always have |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ 4 by (1.12). In case of 3 · M_{22} we get |[Z_{1}, t]| ≤ 2^{7}, a contradiction again. Let K ≅ 3Suz, or Co_{1}. Then |[Z_{1}, t]| ≥ 2^{8}. But |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ 4, a contradiction.

Also for K ≅ J_{2}, or Co_{2}, we get |[Z_{1}, t]| ≤ 2^{6}, (2^{5} if p = 3). As |[Z_{1}, t]| ≥ 2^{6} and p = 3 for K ≅ Co_{2}, we see K ≅ J_{2} and Z_{1} involves exactly one irreducible module. But m_{p}(K) = 2, a contradiction as before.

(5) K is not a group of Lie type over a field of odd characteristic, besides K is a group of Lie type over a field of characteristic two, too.

Suppose K ≅ 3̸ · U_{4}(3). Then by (1.30) |[[Z_{1}, K], t]| ≥ 16 and so we see that K admits a quadratic fours group. So by (1.12) K ≅ 3 · U_{4}(3). Hence p = 3. By (1.55) s ∈ Z_{α+1} ∩ L_{1} is centralized by E ≅ E_{9} in K_{α+1} and so L_{2} ≤ M, a contradiction.

(6) There is no component K of L_{1}/Q_{1} with [K, Z_{α+1} ∩L_{1}] ≠ 1, such that m_{p}(K) > 1.

Assume that K ≅ G(q), q even. We will establish (6) in a series of steps.

(6.1) K induces some SC-module or strong quadratic module on Z_{1}.

We have |[Z_{1}, t]| ≤ 16 · |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}|. Then the assertion follows with (1.15) or K ≅ L_{3}(q), U_{3}(q), L_{n}(2), n ≤ 7, Sp_{6}(2), U_{4}(2), U_{5}(2), G_{2}(2)′, Ω_{8}^{±}(2), ^{2}F_{4}(2)′. Now in all these cases [Z_{1}, K] involves exactly one nontrivial irreducible module. So if m_{p}(K) = 2, we see that [s, Z_{α+1}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α+1}, a contradiction.

Hence we have K ≅ L_{6}(2), L_{7}(2), Sp_{6}(2), U_{4}(2), U_{5}(2) or Ω_{8}^{±}(2) and p = 3. Now |[Z_{1}, t]| ≤ 8|Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}|. So by (1.15) we have that K ≅ U_{4}(2) or U_{5}(2) and |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≥ 4. By (1.14) we have equality and |Z_{1} : Z_{1} ∩ Q_{α}| = 8. But then |[Z_{1}, K] : [Z_{1}, K] ∩ Q_{α}| ≥ 8 and so |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{5}, a contradiction.

(6.2) We have K ≅ L_{n}(q), n ≤ 7, (n ≤ 4 for q > 2), Sp(2n, q), n ≤ 3, U_{n}(q), n ≤ 5, Ω_{8}^{−}(q), Ω_{8}^{+}(2), G_{2}(q), ^{3}D_{4}(q).

By (6.1), (1.14) and (1.22) we know one irreducible module T in [Z_{1}, K]. If m_{p}(C I_{K}(C I_{T} (S ∩ K))) ≥ 2, we see that there is 1 ≠ x ∈ Ω_{1}(Z(S)) ∩ T^{L_{1}} with m_{p}(C I_{K}(x)) ≥ 2 or T is the natural module for L_{n}(q) or Ω^{+}(2n, q) and the graph automorphism is induced by L_{1}. The former is not possible as L_{2} ≰ M. Now inspection gives (6.2).

We will check the groups in (6.2) to arrive at a contradiction in any case. Let K ≅ L_{n}(q). Let further first q = 2. If m_{p}(K) = 2 then E ≅ E_{p^{2}} centralizes K. As Z_{1} ∩ L_{α+1} acts trivially on {T_{α+1}^{L_{α+1}}}. Then [⟨T_{α+1}^{L_{α+1}}⟩, Z_{1} ∩ L_{α+1}] is centralized by E ≅ E_{p^{2}}, a contradiction. So we have p = 3 and K ≅ L_{6}(2) or L_{7}(2). As 7^{2} | |K|, we have K ≤ L_{1}/Q_{1}. But for T we then have E ≤ C I_{K}(C I_{T} (S ∩ K)), E ≅ E_{9}, a contradiction.

So we have q > 2 and K ≅ L_{4}(q) or L_{3}(q). Now T is the natural module or in case of L_{4}(q) the orthogonal module. Let |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| > q, then we see that m_{p}(K) = 2, otherwise some 1 ≠ x ∈ [s, Z_{α+1}] is centralized by E ≅ E_{p^{2}}. But now by (1.16) Z_{1} ∩ L_{α+1} acts trivially on {T_{α+1}^{L_{α+1}}} and so we get a contradiction as before . So we have |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≤ q. In particular |[Z_{1}, t]| ≤ 16q ≤ q^{3}. Hence there are at most three nontrivial irreducible modules involved. Further p ≠ 3. This shows K ≅ L_{4}(4), p = 5. Now m_{p}(C I(K)) ≥ 2. Then [K, Z_{1}] is centralized by some 5-element. As there is a 5-element in K which centralizes [s, Z_{α+1}], we have a contradiction.

Assume there are exactly two modules involved. If p |̸ q −1, there is a pelement centralizing [K, Z_{1}]. Furthermore both modules are normalized by S. But then some 1 ≠ x ∈ [S, T]^{♯} is centralized by E ≅ E_{p^{2}}, a contradiction. So we have p | q − 1. The same applies if [Z_{1}, K] is irreducible. Now K ≤ L_{1}/Q_{1} and so we see m_{p}(K) = 2, otherwise some x ∈ Ω_{1}(Z(S))^{♯} is centrailzed by E ≅ E_{p^{2}}. So we have K ≅ SL_{3}(q). If there is exactly one nontrivial irreducible module involved we have some p-element centralizing [K, Z_{1}]. But x ∈ [s, Z_{1}] is centralized by a p-element in K too, a contradiction. So we have exactly two nontrivial irreducible modules involved. Then q ≤ 8, p = 3 or 7. As K ≤ L_{1}/Q_{1} it is now easy to see that there is some x ∈ Ω_{1}(Z(S))^{♯} centralized by E ≅ E_{p^{2}}, a contradiction.

Let K ≅ Sp_{2n}(q). In case n = 3 we have that T is the spin module. We may assume p | |C I_{L_{1}}(K)| by (4.4). So by (1.16) again |Z_{α+1} ∩ L_{1} : Z_{α+1} ∩ Q_{1}| ≤ q. Hence q^{2} ≤ |[Z_{1}, t]| ≤ 16q. So q ≤ 16. If there are at least two modules involved we get q = 2 and so p = 3. Now |[Z_{1}, t]| = 16 and so |Z_{1} : Z_{1} ∩ Q_{α}| ≥ 8. This shows |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{9}, a contradiction. So we just have one nontrivial irreducible module involved. Now there is x ∈ [s, Z_{α+1}]^{♯} centralized by E ≅ E_{p^{2}}, as p | |C I_{L_{α+1}}([K, Z_{1}])|.

So let now n = 2. We may assume that there is E ≅ E_{p^{2}} centralizing K. In particular there are at least two nontrivial modules and again |Z_{α+1} ∩ L_{1} : Z_{α+1}∩Q_{1}| ≤ q. Suppose p | q+1. Then either there is a p-element centralizing [K, Z_{1}] or there are at least four nontrivial irreducible modules involved in [K, Z_{1}] . In the former we get that [s, Z_{α+1}] is centralized by E ≅ E_{p^{2}}, a contradiction. In the latter we have q = 2. But then |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ 2^{5}, a contradiction. So we have that p | q − 1 and then q > 2. Suppose there are three nontrivial irreducible modules involved. Then q = 4 and p = 3. But now |[Z_{1}, t]| ≤ 8 · 4, a contradiction. So there are exactly two nontrivial irreducible modules involved and q ≤ 16. Further K ≤ L_{1}/Q_{1}. Now |Ω_{1}(Z(S)) ∩ [K, Z_{1}]| ≤ q^{2} and there is F ≅ E_{p^{4}} acting on a group of order q^{2} containing Ω_{1}(Z(S)) ∩ [K, Z_{1}]. In any case there is E ≅ E_{p^{2}} in F centralizing some x ∈ Ω_{1}(Z(S)^{♯}, a contradiction.

Let now K ≅ U_{n}(q). Let n = 5. Then T is not the natural module as otherwise some 1 ≠ x ∈ [s, Z_{α+1}] is centralized by some E ≅ E_{p^{2}}, p | q+1 with Γ_{E,1}(G) ≤ M_{α+1}. Now from (1.34) (ii) we get |[Z_{1}, t]| ≥ q^{6}. Furthermore Z_{1} ∩ L_{α+1} does not contain transvections s on the natural module, otherwise [s, Z_{α+1}] is normalized by E ≅ E_{p^{2}}. So we have |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≤ q^{2}. Now q = 2 . Then p = 3 and |[Z_{1}, t]| ≤ 8q^{2}, a contradiction.

Let now n = 4. Then p | q̸−1, as otherwise for both modules, the natural U_{4}(q)-module and the natural Ω_{6}^{−}(q)-module, [T, s] contains a subgroup 1 ≠ U normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. So we must have p | q + 1. Furthermore there are at least two nontrivial irreducible modules involved. Suppose both are natural modules. If there is t ∈ Z_{α+1} with |[[Z_{1}, K], t]| ≥ q^{8}, we get q = 2. But then 3 ∈ σ(M), a contradiction. So we have that any t ∈ Z_{α+1} just induces transvections. In particular |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| ≤ q. Now we have |U| = q^{4}, U ≤ Z_{1}, [U, Z_{α+1}] ≠ 1. If q > 2 we get U_{1} ≤ U, |U_{1}| = q^{2}, U_{1} ≤ L_{α+1}. As q^{8} ≤ |Z_{α+1} : C I_{Z_{α+1}}(U_{1})| ≤ q · 16, we have a contradiction. So we get q = 2 and then |U| = 16. Now |U ∩ L_{α+1}| = 2. But then |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| ≥ 2^{5}, a contradiction.

If s ∈ N(K), then either s centralizes a subgroup E ≅ E_{p^{2}} in K or s corresponds to an element of type c_{2} (in Suzuki - notation) on the natural module. But in the latter [s, Z_{1}] contains some 1 ≠ x with C I_{K}(x) involves Sp_{4}(q). So in any case V_{2} ≤ M_{α+1} a contradiction.

Let now n = 3. Then as before |[Z_{1}, t]| ≥ q^{5}. But |[Z_{1}, t]| ≤ 16q, a contradiction.

Let K ≅ Ω_{8}^{−}(q). Then T is the spin module and p | q + 1. Furthermore p | |C I_{L_{1}/Q_{1}}(K)|. Hence there are at least two such modules involved. Now |[Z_{1}, t]| ≥ q^{8}. But |[Z_{1}, t]| ≤ q^{6} · 16 and so q ≤ 4. As elements of type c_{2} or a_{2} are centralized by E ≅ E_{p^{2}} in K, we get |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≤ q^{4} and so we get q = 2. But in this case p = 3 and so |[Z_{1}, t]| ≤ 2^{7}, a contradiction.

Let K ≅ Ω_{8}^{+}(2). Then both natural modules are involved and p = 3. But in the natural module every element is centralized by E ≅ E_{9} in K, a contradiction.

Let K ≅ G_{2}(q). Then |T | = q^{6}. Now m_{p}(C I_{L_{1}/Q_{1}}(K)) ≥ 2. So there are at least two nontrivial irreducible modules involved. Now |[Z_{1}, t]| ≤ 16q^{3}. As |[Z_{1}, t]| ≥ q^{4}, we have q ≤ 16. If there are three such modules involved, we get q = 2 and as usual |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K)| > 2^{5}, a contradiction. So we have exactly two natural modules involved. Let |Z_{1} ∩L_{α+1} : Z_{1} ∩Q_{α+1}| > q. Then with (1.16) we see that there is E ≅ E_{p^{2}} in L_{α+1} acting on [Z_{1} ∩ L_{α+1}, Z_{α+1}], a contradiction to Z_{1} ≰ M_{α+1}. So |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≤ q. Now |[Z_{1}, t]| ≤ 16 · q and so q = 2 again. Hence p = 3. But then there is ν ∈ L_{1}, o(ν) = 3 and [ν, [Z_{1}, K]] = 1. As K ≤ L_{1}/Q_{1}, we get some x ∈ Ω_{1}(Z(S))^{♯} which is centralized by E ≅ E_{9}, a contradiction.

Let finally K ≅ ^{3}D_{4}(q). Then T is the natural module. Furthermore |[t, T]| ≥ q^{6}. In particular just one nontrivial irreducible module is involved and |Z_{α+1} ∩ L_{1} : C I_{Z_{α+1}∩L_{1}}(K)| ≤ q^{5}. Now K ≤ L_{1}/Q_{1} and so T ≅ [Z_{1}, K]. This implies with (4.4) m_{p}(K) ≥ 3, a contradiction.

So we have (6).

(7) If K ≤ L_{1}/Q_{1}, K a p–group with [Z_{α+1} ∩L_{1}, K] ≠ 1, then m_{p}(K) = 1.

Let m_{p}(K) ≥ 2. Obviously |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| ≤ 4. Let m_{p}(K) ≥ 3. By [GoLyS, (11.18)] s normalizes some E ≅ E_{p^{2}}. Now [Z_{1}, E] = ⨁_{|E:E_{i}|=p} C I_{Z_{1}}(E_{i}). We have that [C I_{Z_{1}}(E_{i}), s] = 1 for [E_{i}, s] ≤ E_{i}. As [s, E] ≠ 1 there is E_{i} with [E_{i}, s] ≰ E_{i}. Now C I_{Z_{1}}(E_{i}) ∩ C I_{Z_{1}}(E_{i}^{s}) = 1 and C I_{Z_{1}}(E_{i}) ⊕ C I_{Z_{1}}(E_{i}^{s}) is centralized by E_{i} ∩ E_{i}^{s}. Let E_{i} = ⟨E_{i} ∩ E_{i}^{s}, ρ⟩. Then ρρ^{s} is centralized by s and so ρρ^{s} centralizes [C I_{Z_{1}}(E_{i}) ⊕ C I_{Z_{1}}(E_{i}^{s}), s], a contradiction.

So we have m_{p}(K) = 2. Now |V_{2} : C I_{V_{2}}(K_{α+1})| ≤ 2^{6}. Let β ∈ Δ(α). Application of (6.16) shows |[Z_{β}, K_{β}] : [Z_{β}, K_{β}] ∩ Q_{2}| ≤ 4. Hence [Z_{1}, K] is generated by elements t with |[Z_{β}, K_{β}] : C I_{[Z_{β},K_{β}]}(t)| ≤ 8. This shows t ∈ M_{β}. Hence [Z_{1}, K] ≤ ⋂_{β∈Δ(α)} M_{β} ≤ L_{α+1}. Now there is some ρ ∈ K, ρ^{s} = ρ^{−1} such that ρ ∈ E ≅ E_{p^{3}}. Suppose p > 3. Then |[ρ, Z_{1}]| ≤ 2^{6}. As x ∈ [[ρ, Z_{1}], s]^{♯} is not centralized by a hyperplane in E, we get p = 7 and |[ρ, Z_{1}]| = 2^{6}. Now |C I_{E}([ρ, Z_{1}])| = 7. Hence we get that the orbits of E on [ρ, Z_{1}] are 49, 7,7. We have that C I_{L_{1}/Q_{1}}(ρ) induces a subgroup of GL_{2}(8) on [ρ, Z_{1}], whose order is divisble by 7^{2}. We see that there is a normal Sylow 7subgroup and so s acts on this group. As s cannot invert a Sylow 7-subgroup of GL_{2}(8) we see that there is some 7-element ν centralized by s. But then [s, [ρ, Z_{1}]] is centralized by Ẽ ≅ E_{7^{2}} in L_{1}, a contradiction.

So let p = 3. Then we get [V_{2} : V_{2}∩C I_{L_{α+1}/Q_{α+1}}(K_{α+1})| ≤ 2^{5} and |[Z_{β}, K_{β}] : [Z_{β}, K_{β}] ∩ Q_{2}| ≤ 2 for β ∈ Δ(α). Let ρ be as before. We get |[ρ, Z_{1}]| ≤ 2^{4}. But now we may argue as before.

(8) If K_{1} × K_{2} × K_{3} ≤ L_{1}/Q_{1}, m_{p}(K_{i}) = 1 and the K_{i} are components, or a cyclic p–group, i = 1, 2, 3, then [Z_{α+1} ∩ L_{1}, K_{i}] = 1 for suitable i.

Suppose false. Assume first K_{1α+1}^{s} ≠ K_{1α+1}, where K_{1α+1} is the component in L_{α+1} corresponding to K_{1}. If |V_{2} ∩ L_{α+1} : C I_{V_{2}∩L_{α+1}}(K_{1})| = 2, we get that Z_{α+1} centralizes a subgroup of index at most 32 in V_{2}. In particular |Z_{α+1} : Z_{α+1} ∩Q_{2}| ≤ 2. Now there are at least two nontrivial irreducible K_{1} - modules in Z_{1}, which tells us that for any t ∈ Z_{α+1} ∩L_{1} we have that [K_{1}, Z_{1}] involves some nontrivial irreducible module W with |[W, t]| ≤ 4. This gives with (1.30) as all Sylow r-subgroups , r odd, have to be cyclic and then that K_{1} ≅ L_{2}(4) or L_{3}(2). In both cases there is a subgroup U of order 16 in [Z_{1}, K_{1}] which centralizes a subgroup of index 64 in V_{α}. Hence |U ∩ Q_{α}| ≥ 4. But U ∩ C I(Z_{α+1}) = 1, a contradiction.

So we have that |V_{2}∩L_{α+1} : C I_{V_{2}∩L_{α+1}}(K_{1α+1})| > 2 and then by (1.11) K_{1} ≅ L_{2}(q) and [Z_{α+1}, ⟨K_{1α+1}, s⟩] involves the orthogonal O^{+}(4, q)-module. Now |V_{2} ∩L_{α+1} : V_{2} ∩Q_{α+1}| ≤ 2q and |Z_{1} : C I_{Z_{1}}(Z_{α+1})| ≤ 2^{5}q. Furthermore at least two such irreducible modules are involved, which shows |[Z_{1}, t]| ≥ q^{2} and so q ≤ 32 and K_{1}^{t} ≠ K_{1}. Now for s ∈ [⟨K_{1}, t⟩, Z_{1}] we have |Z_{α+1} : C I_{Z_{α+1}}(t)| ≤ 2^{5}. If q > 4, we see that [[⟨K_{1}, t⟩, Z_{1}] ∩ Q_{α}, Z_{α+1}] = 1. So |[[⟨K_{1}, t⟩, Z_{1}], t]| < 2^{4}. So we have q = 4. As |[[⟨K_{1}, t⟩, Z_{1}], t]| < q^{4}, we have at most three natural modules involved. This shows p = 3. We have |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 8. So |V_{2} : V_{2} ∩ C I(Z_{α+1})| ≥ 2^{12}. This shows |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 2^{9}, a contradiction.

So we have [K_{iα+1}, s] ≤ K_{iα+1}, for i = 1, 2, 3. In particular there are at least two i with [K_{iα+1}, s] ≠ 1.

Suppose [[K_{1}, Z_{1}], K_{2}K_{3}] = 1. Then [V_{2} ∩ L_{α+1}, K_{1α+1}] = 1 and [[K_{1α+1}, Z_{α+1}], V_{2} ∩ L_{α+1}] = 1. Now [K_{1α+1}, Z_{α+1}] ≤ L_{1}. Further [K_{1α+1}, Z_{α+1}] ∩ Q_{1} = 1. We may assume m_{3}(K_{2}) ≤ 1. Assume furthermore [[K_{1α+1}, Z_{α+1}], K_{2}] ≤ K_{2}. As |[K_{1α+1}, Z_{α+1}]| > 2, we get with (1.3) and (1.12) that K_{2} ≅ L_{2}(q), Sz(q), U_{3}(q), or L_{3}(q), q even. As [Z_{1}, K_{2}] involves at least two nontrivial irreducible modules we see that K_{2} ≅ L_{2}(4), L_{3}(4), or L_{3}(2). As Z_{1} ≰ M_{α+1} we get K_{2} ≅ SL_{3}(4) and |[K_{1}, Z_{1}]| ≤ 16. Furthermore there are exactly two nontrivial irreducible K_{2}-modules involved which shows that K_{3} is a subgroup of L_{2}(4) and so p = 5. But then [K_{2}, Z_{1}] is centralized by E ≅ E_{25}, a contradiction.

So let K_{2}^{t} ≠ K_{2} for some t ∈ [K_{1α+1}, Z_{α+1}]. If [K_{1α+1}, Z_{α+1}] induces a quadratic fours group on K_{2}^{t}K_{2}, we get K_{2} ≅ L_{2}(q) and q = 4. Now p = 5 and we get a contradiction as before. So let |[K_{1α+1}, Z_{α+1}] : C I_{[K_{1α+1},Z_{α+1}]}(K_{2}^{t}K_{2})| = 2. Then we get |[K_{1α+1}, Z_{α+1}]| = 2, a contradiction.

Now for any i there is some j ≠ i with [[K_{i}, Z_{1}], K_{j}] ≠ 1. Let [[K_{1}, Z_{1}], K_{3}] ≠ 1 and [[K_{1}, Z_{1}], K_{2}] = 1. Then we get [[K_{2}, Z_{1}], K_{1}] = 1 by the three subgroup lemma and so [[K_{2}, Z_{1}], K_{3}] ≠ 1. Then also [[K_{3}, Z_{1}], K_{1}] ≠ 1 ≠ [[K_{3}, Z_{1}], K_{2}]. So we may choose notation such that [[K_{1}, Z_{1}], K_{2}] ≠ 1 ≠ [[K_{1}, Z_{1}], K_{3}].

Assume now [K_{i}, Z_{α+1} ∩ L_{1}] ≤ K_{i} for i = 1, 2, 3. Then V_{α} ∩ L_{1} ∩ C I(K_{1}) ≤ C I(K_{1}K_{2}K_{3}). Let K_{i} be some component with m_{3}(K_{i}) = 0. We have that there are at least three nontrivial irreducible modules involved. Let now 1 ≠ t ∈ C I_{Z_{α+1}∩L_{1}}(K_{i}) \ C I(K_{1}K_{2}K_{3}). Then [t, K_{1}] ≠ 1 and for s ∈ Z_{α+1} ∩ L_{1} with [K_{i}, s] ≠ 1, we have [t, Z_{1}, s] = 1, so [[t, Z_{1}], K_{i}] = 1, but then [[K_{1}, Z_{1}], K_{i}] = 1, a contradiction.

So we have C I_{Z_{α+1}∩L_{1}}(K_{i}) ≤ C I(K_{1}K_{2}K_{3}). Let first |V_{α} ∩L_{1} : C I_{V_{α}∩L_{1}}(K)| = 2. Then for s ∈ [K_{1}, Z_{1}] we have |V_{α} : C I_{V_{α}}(s)| ≤ 32. Suppose p > 3, then [K_{1}, Z_{1}] ≤ L_{α+1}. Let s with [K_{1α+1}, s] ≠ 1. Then s induces a transvection on [K_{1α+1}, Z_{α+1}]. As now K_{2} × K_{3} is a nonsolvable subgroup of L_{5}(2), we get K_{2} ≅ ℤ_{3} and K_{3} ≅ A_{5} or L_{3}(2), contradicting p > 3. So we have [[K_{1}, Z_{1}], K_{1α+1}] = 1. Then K_{1α+1} acts nontrivially on [s, Z_{α+1}]. In particular K_{1} is a subgroup of L_{5}(2). So K_{1} ≅ L_{2}(4), L_{2}(9), L_{2}(7), A_{7}, or A_{8}, and p = 5 or 7. But now we have [[K_{2}, Z_{1}], K_{3}] ≠ 1 and then [K_{2}, Z_{1}] involves at least four nontrivial irreducible modules. Now we may interchange roles. So we have p = 3 and |V_{α} : C I_{V_{α}}(s)| ≤ 16. Now there is U ≤ Z_{1}, |U| = 16 with U ∩ C I(Z_{α+1}) = 1. Hence there is s ∈ U ∩ L_{α+1} with [s, Z_{α+1}] ≠ 1. As before we get that one of the irreducible modules admits transvections and so we get K_{1}K_{2}K_{3} ≅ ℤ_{3} × A_{5} × A_{5} or ℤ_{3} × A_{5} × L_{3}(2). Now as [Z_{1}, K_{1}] centralizes a subgroup of index 16 in V_{α}, we again see that [Z_{1}, K_{1}] ≤ L_{α+1}. But |[Z_{1}, K_{1}] : [Z_{1}, K_{1}] ∩ C I(Z_{α+1})| ≥ 16, which shows that [K_{1}, Z_{1}, K_{2}] = 1, a contradiction.

So we have |V_{α}∩L_{1} : V_{α}∩L_{1}∩C I(K_{1}K_{2}K_{3})| ≥ 4 and so K_{i} ≅ Sz(q). For s ∈ [K_{1}K_{2}K_{3}, Z_{1}] we have |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 16q. Furthermore there is some s ∈ L_{α+1} \ Q_{α+1}. This now shows [s, K_{iα+1}] = 1 and even [[s, Z_{α+1}], K_{iα+1}] = 1. But then [K_{1α+1}, s] = 1, which shows K_{i} = K_{1}. Now [s, K_{2α+1}] ≠ 1 and so [[K_{2α+1}, Z_{α+1}], K_{1α+1}] = 1, a contradiction.

Now by (1.12) and (1.3) we have K_{1} ≅ (S)U_{3}(q), L_{2}(q) or (S)L_{3}(q), q even. We get |[s, Z_{α+1}]| ≤ 16q (or 16q^{2} in case of K_{1} ≅ (S)L_{3}(q)), where s ∈ Z_{1}. Suppose s ∈ L_{α+1} If [s, K_{1α+1}] ≠ 1, we get K_{1} ≅ SL_{3}(4) or L_{3}(2). As m_{3}(K_{i}) ≥ 1 for all i, we see K_{1} ≤ L_{3}(2). Now K_{2}K_{3} is a subgroup of GL_{6}(2) and so K_{2} ≅ K_{3} ≅ L_{3}(2) and there are at least six nontrivial irreducible K_{1}-modules involved. But now we get U ≤ Z_{1}, |U| = 2^{6} with C I_{U}(Z_{α+1}) = 1. Then |U ∩ L_{α+1}| ≥ 2^{4}, a contradiction. So we have [s, K_{1α+1}] = 1. Then [K_{1α+1}, [Z_{α+1}, s]] ≠ 1. This shows K_{1} ≅ L_{2}(q), q ≤ 16 or K_{1} ≅ L_{3}(2). If [K_{2}, Z_{1}, K_{3}] ≠ 1 we may argue as before. So we may assume [K_{2}, Z_{1}, K_{3}] = 1. Now we either have [[[K_{2α+1}, Z_{α+1}], s], K_{1α+1}] = 1 or [[[K_{3α+1}, Z_{α+1}], s], K_{1α+1}] = 1, both yields a contradiction.

This shows [[Z_{1}, K_{1}] ∩ L_{α+1}, Z_{α+1}] = 1. Then |[Z_{1}, K_{1}] : C I_{[Z_{1},K_{1}]}(Z_{α+1})| ≤ 16. As there are at least four nontrivial irreducible modules involved we have K_{1} ≅ L_{3}(2) or A_{5} and there are exactly four nontrivial irreducible modules involved. Further p > 3. But then K_{2}K_{3} is a subgroup of L_{4}(2), a contradiction.

So we may assume K_{i}^{t} ≠ K_{i} for some t ∈ Z_{α+1} ∩ L_{1}. We may furthermore assume that K_{i}^{t} = K_{j}. Let |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K_{i}K_{j})| = 2. Let further s ∈ [⟨K_{i}, t⟩, Z_{1}]. Then |V_{α} : C I_{V_{α}}(s)| ≤ 2^{5}. Now by (6.16) we get s ∈ L_{α+1}. If [K_{1α+1}, s] ≠ 1, we have K_{1} ≅ L_{3}(2), A_{5}, or K_{1} is a 3-group. Now K_{2}K_{3} is a subgroup of L_{5}(2) and so we get K_{1}K_{2}K_{3} ≅ A_{5}×A_{5}×ℤ_{3} or L_{3}(2)×L_{3}(2)×ℤ_{3}. In any case p = 3 and so we have K_{1}K_{2}K_{3} ≅ A_{5} × A_{5} × ℤ_{3}. Now s induces a transvection and so s centralizes 3-elements in K_{1α+1}K_{2α+1}K_{3α+1}. As there is a 3-element centralizing [K_{1}K_{2}K_{3}, Z_{1}] by (4.4), we get a contradiction.

So we have [K_{1α+1}, s] = 1. Again K_{1} is a subgroup of L_{5}(2) and m_{3}(K_{1}) = 1. This shows K_{1} ≅ L_{3}(2) or A_{5} again. Now we may interchange the roles of 1 and 2 and get a contradiction as before.

So we have |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}(K_{i} × K_{j})| > 2 and then by (1.11) K_{i}K_{j} ≅ L_{2}(q)×L_{2}(q) and O_{4}^{+}(q)-modules are involved. Furthermore there are at least two of them. Now for s ∈ Z_{1} ∩ L_{α+1} we have that |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 32q. On the other hand we have |Z_{α+1} : C I_{Z_{α+1}}(s)| ≥ q^{4}, which shows q = 2, a contradiction.

(9) If K_{1} × K_{2} ≤ L_{1}/K_{1} where K_{i} are as in (8). Then [Z_{α+1} ∩ L_{1}, K_{i}] = 1 for suitable i ∈ {1, 2}.

Suppose false. First of all [Z_{α+1} ∩ L_{1}, C I_{L_{1}/Q_{1}}(K_{1} × K_{2})] = 1 by (8). Let first |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| = 2. Then |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 2^{5} for s ∈ Z_{1}. If p > 3, then Z_{1} ≤ M_{α+1} a contradiction. So p = 3 and we have |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 16. Furthermore there is some s with equality. We may assume [K_{1α+1}, s] ≠ 1. If [K_{1α+1}, s] ≰ K_{1α+1} we may define K_{3α+1} = K_{1α+1}^{s}. Now C I_{V_{2}}(K_{1α+1}K_{3α+1}) = 1 and so either we have orthogonal O_{4}^{+}(q) - modules involved or |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| = 2 too. In the former there is some s_{1} ∈ Z_{1} ∩ L_{α+1} with |Z_{α+1} : C I_{Z_{α+1}}(s_{1})| ≥ q^{2} and so q = 4 and there is exactly one nontrivial irreducible K_{1}K_{3} - module involved in Z_{1}. But this gives some E ≅ E_{9} in M_{α+1} centralizing [Z_{α+1}, s], a contradiction. So we have |Z_{1} ∩ L_{α+1} : Z_{1} ∩ Q_{α+1}| = 2. But we have |Z_{α+1} : Z_{α+1} ∩ L_{1}| = 2 and so |[Z_{α+1}, s]| ≤ 4. This gives K_{1} ≅ A_{5} and again there is just one nontrivial irreducible K_{1}K_{3} - module involved in Z_{1}, a contradiction as before. So we have [K_{1α+1}, s] ≤ K_{1α+1}. As we may assume that m_{p}(L_{1}) ≥ 4, we have that [K_{1}, Z_{1}] involves at least two nontrivial irreducible modules and so for some of them we have that s centralizes a subspace of codimension two. As p = 3 we have K_{1} ≅ L_{3}(2) or L_{2}(4). Now in any case there is x ∈ [Z_{α+1}, s]^{♯} centralized by E ≅ E_{9} in M_{α+1}, a contradiction. So we have |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ 4.

Suppose K_{1} ≅ K_{2} ≅ L_{2}(q) and Ω_{4}^{+}(q)-modules are involved in Z_{1}. Then |[Z_{α+1}, s]| ≤ 2^{5}q. Suppose [K_{1}, s] ≠ 1. If [K_{1α+1}, s] ≤ K_{1α+1}, then we get q ≤ 2^{5} and there is just one Ω_{4}^{+}(q)-module involved. Further K_{1}K_{2} ≤ L_{1}/Q_{1} and so we get some E ≅ E_{p^{2}} centralizing some element in Ω_{1}(Z(S))^{♯}, a contradiction. So we have K_{1α+1}^{s} ≠ K_{1α+1}. Suppose K_{1α+1}^{s} = K_{2α+1}. Again we may assume that there are at least two nontrivial irreducible modules involved. Now q ≤ 32 again. Furthermore there is no p-element centralizing [Z_{1}, K_{1}K_{2}]. This shows that there is some field automorphism of order p, i.e. q = 8, p = 3. But now for t ∈ Z_{α+1} ∩ L_{1} we have |[Z_{1}, t]| ≤ 2^{4}, a contradiction. So K_{1α+1}^{s} ≠ K_{2α+1}. Then [K_{2α+1}] = 1 and so s centralizes E ≅ E_{p^{2}}, a contradiction.

So we now have [Z_{α+1} ∩ L_{1}, K_{i}] ≤ K_{i}, i = 1, 2. Let m_{3}(K_{1}) ≤ 1. Then K_{1} ≅ Sz(q), L_{2}(q), L_{3}(q), U_{3}(q), J_{1} or ℤ_{p}. If q is odd or K_{1} ≅ J_{1} or ℤ_{p} set r = 2 else r = q or r = q^{2} in case of K_{1} ≅ L_{3}(q). Then we have that |V_{α} : C I_{V_{α}}([Z_{1}, K_{1}])| ≤ 16r. Let s ∈ [Z_{1}, K_{1}]^{♯} ∩ L_{α+1} with [Z_{α+1}, s] ≠ 1.

Assume further [K_{1α+1}, s] ≠ 1. Suppose [K_{1α+1}, s] ≰ K_{1α+1}. If |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| = 2, we have for t ∈ Z_{α+1} that |[Z_{1}, t]| ≤ 32. As [t, K_{1}] ≠ 1, we get K_{1} ≅ L_{2}(2^{f}), L_{3}(2^{f}) f ≤ 32, U_{3}(4), L_{2}(25), L_{2}(x), x prime. As |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| > 2, and [K_{1}, Z_{1}, K_{2}] = 1 in the last two cases, we see that K_{1α+1}^{s} = K_{2α+1}. Now K_{1} ≅ L_{2}(x) and |[s, Z_{α+1}]| ≤ 2^{5}. This shows K_{1} is a subgroup of L_{5}(2), so K_{1} ≅ L_{2}(4) or L_{3}(2). So we may assume that K_{1} ≅ L_{2}(q), q even, or L_{3}(2). If |V_{2}∩L_{α+1} : V_{2}∩Q_{α+1}| > 2 we get K_{1} ≅ L_{2}(q) too. Now |[V_{α}, s]| ≤ 16q or 64 in case of L_{3}(2). As we may assume that there are at least two nontrivial irreducible modules involved, we get |[V_{α}, s]| ≥ q^{2}, 2^{6}, respectively. Now in any case K_{1}K_{2} ≤ L_{1}/Q_{1}. Assume that we have a group of order q^{2}, in case of L_{3}(2) of order 16, centralized by E ≅ E_{p^{2}}. In the latter we have p = 3 and so there is E ≅ E_{9} centralizing some element in [Ω_{1}(Z(S)), K_{1}K_{2}]^{♯}. So we have K_{1} ≅ L_{2}(q). If |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| > 2 there are at least four nontrivial irreducible K_{1}- modules involved, so q^{4} ≤ 32q, a contradiction. Hence we have |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| = 2, and then q^{2} ≤ 32, i.e. q = 4. We have K_{1} ≅ L_{2}(4). Now we have [K_{1}, Z_{1}, K_{3}] = 1 and [K_{1α+1}, Z_{α+1}]^{s} = [K_{3α+1}, Z_{α+1}]. So |[K_{1}, Z_{1}]| ≥ q^{4}, a contradiction.

So we have [K_{1α+1}, s] ≤ K_{1α+1}. Further we know that [K_{1}, Z_{1}] involves at least two nontrivial irreducible modules. We see |[s, Z_{α+1}]| ≤ 16r. Now by (1.29) we have K_{1} ≅ L_{2}(q), L_{3}(q), q ≤ 16, ℤ_{3}, or ℤ_{5}. Suppose there are exactly two nontrivial irreducible modules involved. If there is no field automorphism of order p in L_{1}, we get E ≅ E_{p^{2}} centralizing [K_{1}, Z_{1}], a contradiction. So there is a field automorphism and so K_{1} ≅ L_{2}(8), p = 3. Now there is a 3-element centralizing [Z_{1}, K_{1}]. Furthermore s centralizes some 3-element in K_{1α+1}⟨ν⟩, where ν induces the field automorphism. But the [Z_{α+1}, s] is normalized by E ≅ E_{9} in L_{α+1}, a contradiction. So we have at least three modules involved. This shows K_{1} ≅ L_{2}(4), L_{3}(q), q ≤ 16 or ℤ_{3}. Let K_{1} ≅ L_{3}(q), then p |̸ q − 1. So we see that in all this cases there are at least four modules involved and then K_{1} ≅ L_{3}(4) or L_{3}(2).

Let K_{1} ≅ L_{3}(2). Then if p = 7, we get E ≅ E_{p^{2}} in L_{1} centralizing [Z_{1}, K_{1}]. If p = 3, we see ℤ_{3} centralizing [Z_{1}, K_{1}] and there is some 1 ≠ x ∈ [s, Z_{α+1}] centralized by E ≅ E_{9} in L_{α+1}, a contradiction.

Let K_{1} ≅ L_{3}(4). Then there are at most four nontrivial irreducible modules involved, and p = 5 or p = 7. As there is no E ≅ E_{p^{2}} in M centralizing [Z_{1}, K], we get p = 5 and there are exactly four modules involved. Now |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 16 and so |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 2^{8}, which shows that there is a 5-element centralizing [K_{1α+1}, Z_{α+1}], and so there is 1 ≠ x ∈ [Z_{α+1}, s] centralized by E ≅ E_{p^{2}} in M_{α+1}, a contradiction.

Let K_{1} ≅ L_{2}(4). Then we have three nontrivial irreducible modules and p = 5. But then [K_{1}, Z_{1}] is centralized by E ≅ E_{p^{2}} in M a contradcition.

So we have K_{1} ≅ ℤ_{3} and |[K_{1}, Z_{1}]| ≤ 2^{8}. We have [K_{2}, [K_{1}, Z_{1}]] = 1. Let U ≤ [Z_{α+1}, K_{1α+1}], |U| = 8, U ∩ C I(Z_{1}) = 1. Then we see that |U ∩ C I([K_{1}, Z_{1}])| ≤ 2. So |U : U ∩ Q_{2}| ≥ 4. Hence |V_{2} : C I_{V_{2}}(U)| ≥ 64. This gives |V_{2} ∩ L_{α+1} : C I_{V_{2}∩L_{α+1}}(U)| ≥ 8, a contradiction.

So we finally have [K_{1α+1}, s] = 1. Suppose [K_{1α+1}, [Z_{α+1}, s]] = 1. We must have [K_{2α+1}, s] ≠ 1 and there is K_{3} with [K_{3α+1}, s] ≠ 1. Hence [Z_{1}, K_{1}] is centralized by E ≅ E_{p^{2}} in M. So C I_{[Z_{α+1},K_{1α+1}]}(Z_{1}) = 1. If u ∈ [Z_{α+1}, K_{1α+1}] then |V_{2} : C I_{V_{2}}(u)| ≤ 16. So we get [Z_{α+1}, K_{1α+1}] ≤ Q_{2}. So [[Z_{α+1}, K_{1α+1}], K_{2}] ≠ 1. Now we have that [K_{2}, Z_{1}] involves at least two nontrivial irreducible modules. So there is u ∈ [Z_{α+1}, K_{1α+1}] which centralizes codimension two in such a modules, and this has to be an F-modules. This shows that K_{2} ≅ L_{n}(2), n ≤ 7, L_{n}(4), n ≤ 3, Sp_{2n}(2), n ≤ 3, Sp_{2n}(4), n ≤ 3, Ω_{2n}^{+}(2), n ≤ 4, G_{2}(2)′, A_{n}, n ≤ 11. In any case there are at most four nontrivial irreducible modules involved . As there is no E ≅ E_{p^{2}} centralizing [K_{1}, Z_{1}] in M, we get p = 3. But now there are at most three nontrivial irreducible modules involved, and we have E ≅ E_{9} centralizing [K_{2}, Z_{1}], a contradiction.

So we have [K_{1α+1}, [Z_{α+1}, s]] ≠ 1. In particular [K_{1}, Z_{1}, K_{2}] ≠ 1. So we see that K_{1} ≅ ℤ_{p}, L_{2}(q) or L_{3}(q), q even, where q ≤ 16. Suppose [K_{1α+1}, [Z_{α+1}, s]] contains exactly one nontrivial irreducible module. Then either K_{1} or K_{2} has to admit a nontrivial field automorphism of order p. Suppose this is true for K_{1} then K_{1} ≅ L_{2}(8) and p = 3, |[Z_{α+1}, s]| = 2^{6}. But now there is E ≅ E_{9} centralizing s modulo Q_{α+1}, a contradiction. So we have that this is true for K_{2} and then K_{2} ≅ L_{2}(r^{p}). As there are at least four nontrivial irreducible K_{2}-modules involved we get r^{p} ≤ 8 and so p = 3 and K_{2} ≅ L_{2}(8). But now we must have K_{1} ≅ L_{3}(16), a contradiction.

So we have at least two nontrivial irreducible K_{1}-modules involved. This shows K_{1} ≅ L_{3}(2) or ℤ_{3}. Now we have that K_{2} admits transvections. So K_{2} ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2), A_{n}. As m_{3}(K_{2}) ≤ 2, we see that K_{2} ≅ L_{3}(2), L_{4}(2), L_{5}(2), A_{5}, A_{6}, A_{7}. Let K_{1} ≅ L_{3}(2). Then |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 16, so |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 2^{8}, a contradiction. So we have K_{1} ≅ ℤ_{3}. Now |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 8 and so |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 2^{5}. As m_{3}(K_{2}) = 1, we have a contradiction again.

So it remains the case that [[Z_{1}, K_{1}] ∩ L_{α+1}, Z_{α+1}] = 1. Then there is t ∈ Z_{α+1}^{♯} with |[Z_{1}, K_{1}], t]| ≤ 16. We have K_{1} ≅ L_{2}(q), L_{3}(q), q ≤ 16, U_{3}(4), L_{2}(r), r odd, or ℤ_{p}. If |[[Z_{1}, K_{1}], t]| = 16, we have |V_{α} ∩ L_{1} : C I_{V_{α}∩L_{1}}([Z_{1}, K_{1}])| ≥ 2^{8}, a contradiction. So we have |[[Z_{1}, K_{1}], t]| ≤ 8 and so K_{1} ≅ L_{2}(q), L_{3}(q) q ≤ 8, or ℤ_{p}. Now we get |V_{α} ∩L_{1} : C I_{V_{α}∩L_{1}}([Z_{1}, K_{1}])| ≥ 2^{5}, which gives K_{1} ≅ L_{3}(8) and [Z_{1}, K_{1}] involves exactly one nontrivial irreducible module. If |[Z_{1}, K_{1}] : [Z_{1}, K_{1}] ∩ Q_{α}| ≤ 4, we get K_{1} ≅ L_{2}(4) or L_{3}(2). In any case there is just one nontrivial irreducible module involved and so [K_{1}, Z_{1}] is centralized by E ≅ E_{p^{2}} in M. This shows [V_{2} ∩ L_{α+1}, K_{1α+1}] = 1. We have [K_{1α+1}, Z_{α+1}] ≤ L_{1} and [K_{1α+1}, Z_{α+1}] ∩ Z_{1} = 1. Furthermore [[K_{1α+1}, Z_{α+1}], [K_{1}, Z_{1}]] = 1. Now we see that |[[K_{2}, Z_{1}], t]| ≤ 8 for t ∈ [K_{1α+1}, Z_{α+1}]. As [K_{2α+1}, Z_{α+1}] involves at least three nontrivial irreducible modules, we see that t has to induce transvections. But then K_{2} ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2) or A_{n} and we have exactly three modules involved. This implies p = 3 or 7 and there is E ≅ E_{p^{2}} in M centralizing [K_{2}, Z_{1}], a contradiction.

(10) We have |O_{r}(L_{1}/Q_{1})| = r^{3} for some r ∉ σ(M).

Suppose false. Then by (9) there is exactly one component K with [Z_{1} ∩ L_{α+1}, K] ≠ 1. Further m_{p}(K) = 1. Let K_{α+1} the corresponding component, U_{1} = [Z_{1}, K] and suppose there is s ∈ U_{1} with 1 ≠ [K_{α+1}, s] ≤ K_{α+1}.

Let first m_{3}(K) ≤ 1. Then K ≅ L_{2}(q), L_{3}(q), U_{3}(q), Sz(q), J_{1} or ℤ_{p}. Now |[Z_{α+1}, s]| ≤ 16r, where r = q or r = q^{2} if q is even, and r = 2 else. As [[s, Z_{α+1}]^{♯} contains no element which is centralized by E ≅ E_{p^{2}}, we see that there are at least four nontrivial irreducible K-modules involved in Z_{1}. Hence we get K ≅ L_{3}(2) or ℤ_{3}. In both cases p = 3 and so |Z_{α+1} : Z_{α+1} ∩ Q_{2}| ≥ 4. Now |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 8. So there is some 3-element centralizing [K, Z_{1}], which shows K ≅ ℤ_{3}. Now for t ∈ Z_{α+1} ∩ L_{1}, we have |Z_{1} : C I_{Z_{1}}(t)| ≥ 8, which gives |V_{α} : C I_{V_{α}}(t)| ≥ 64 and |V_{α} ∩ L_{1} : V_{α} ∩ Q_{1}| ≥ 8, a contradiction.

So we have m_{3}(K) ≥ 2 and so p ≠ 3. Now there must be some component or normal p-subgroup K_{1} or of L_{1}/Q_{1}, K_{1} ≠ K with [K_{1α+1}, s] ≠ 1. Suppose m_{p}(K_{1}) ≥ 2. Then [K_{1}, Z_{1}, K] ≠ 1. Now let r be as before and t ∈ [K_{1α+1}, Z_{α+1}]. Then |[Z_{1}, t]| ≤ 16r. On the other hand [t, K_{1}] = 1. So K_{1} acts faithfully on [t, Z_{1}]. We see that K_{1} is solvable or K_{1} ≅ L_{3}(q), q ≤ 16. Suppose the latter. As m_{3}(K_{1}) ≤ 1, we get K_{1} ≅ L_{3}(q), p = 7. Now we see that [KK_{1}, Z_{1}] involves exactly one nontrivial irreducible KK_{1}- module. As we may assume that m_{7}(L_{1}) ≥ 4, we see that there is some 1 ≠ x ∈ Ω_{1}(Z(S)) centralized by E ≅ E_{49} in M with Γ_{E,1}(G) ≤ M, a contradiction. So K_{1} is a p-group. Then for t as before we get |V_{2} : C I_{V_{2}}(t)| ≤ 32. As p > 3, we get [K_{1α+1}, Z_{α+1}] ≤ L_{1} and so |[t, Z_{1}]| ≤ 32. On the other hand there has to be E ≅ E_{p^{2}} acting faithfully on [Z_{1}, t], a contradiction. So we are left with [K_{1α+1}, Z_{α+1}] ∩ Q_{2} ≤ Q_{1}. Then |[Z_{α+1}, s]| ≤ 16. This shows K_{1} ≅ L_{3}(8), U_{3}(4), L_{2}(x), x prime, L_{2}(25). Besides of the first case we get |V_{2} : C I_{V_{2}}(t)| ≤ 64 and so [K_{1α+1}, Z_{α+1}] ≤ L_{1}, a contradiction. So we have K_{1} ≅ L_{3}(8), p = 7. But we must have that there are at least two nontrivial irreducible modules in Z_{1}, a contradiction.

So we have m_{p}(K_{1}) = 1 and there is a third component K_{2} such that [K_{2α+1}, s] ≠ 1. Also m_{p}(K_{2}) = 1 and m_{3}(K_{2}) ≤ 1. We even may assume m_{3}(K_{2}) = 0 and so K_{2} ≅ Sz(q) or ℤ_{p}. Let t ∈ [K_{2α+1}, Z_{α+1}] ∩ L_{1} with [t, Z_{1}] ≠ 1. Suppose further [[t, Z_{1}], K_{2}] = 1. In this case [K_{1}, [K, Z_{1}]] ≠ 1 as [[K, Z_{1}], K_{2}] = 1. Now choose t_{1} ∈ [K_{1α+1}, Z_{α+1}] ∩ L_{1} with [t, Z_{1}] ≠ 1. Then we get [K_{1}, [t, Z_{1}]] ≠ 1. Further we have K_{1} ≅ L_{2}(q), L_{3}(q), or K_{1} is a p-group. If [Z_{1}, t] involves exactly one nontrivial irreducible K_{1}-module, we get some x ∈ Ω_{1}(Z(S))^{♯} which is centralized by E ≅ E_{p^{2}} or K_{1} is a p-group. In the latter |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| = 2 and so Z_{α+1} ≤ L_{1}.

Now also |[Z_{1}, t_{1}]| ≤ 32. As now [K_{1}, Z_{1}] involves at least four nontrivial irreducible K-modules we get that it involves an F-module for K in which t_{1} centralizes a subgroup of index 4. This now shows that |C I_{[Z_{1},K_{1}]}(S∩K)| ≤ 32. Now on this centralizer acts an elementary abelian group of order p^{3} and so some x ∈ Ω_{1}(Z(S))^{♯} is centralized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

So we may assume [K_{1α+1}, Z_{α+1}]∩L_{1} ≤ Q_{1}. Then s centralizes a subgroup of index 16 in [K_{1α+1}, Z_{α+1}]. As [K_{1α+1}, Z_{α+1}] involves at least four nontrivial irreducible modules we have that s induces transvections. This shows K_{1} ≅ L_{3}(2) or L_{2}(4). As [K_{2}, [K_{1}, Z_{1}]] ≠ 1 we see that K_{2} ≅ ℤ_{p}. Now |Z_{α+1} : Z_{α+1} ∩ Q_{2}| = 16 and then |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| ≥ 2^{8}, a contradiction.

So we have [[Z_{1}, t], K_{2}] ≠ 1. Now K_{2} ≅ ℤ_{p}. Then Z_{α+1} ≤ L_{1} and |[t, Z_{1}]| ≤ 32. We have that K_{1} is nonsolvable and there are more than one nontrivial irreducible K_{1}-modules involved. So again we see |C I_{[Z_{1},K_{1}]}(S ∩ K)| ≤ 32, which give the same contradiction as before.

So we are left with [Z_{α+1}, K_{2α+1}]∩L_{1} ≤ Q_{1}. Now |[[Z_{α+1}, K_{2α+1}], s]| ≤ 16 and so K_{2} ≅ ℤ_{p}. Then again Z_{α+1} ≤ L_{1}, a contradiction.

We now have that K_{α+1}^{s} ≠ K_{α+1}. Then K ≅ L_{2}(q), L_{3}(2) or J_{1}. Let |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| = 2, then Z_{α+1} ≤ L_{1} and so we see that K ≅ L_{2}(q) and |[t, Z_{α+1}]| = q. Then [K × K^{s}, Z_{1}] is irreducible and so there is some x ∈ Ω_{1}(Z(S))^{♯} centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. So we have |V_{2} ∩ L_{α+1} : V_{2} ∩ Q_{α+1}| > 2 and then again K ≅ L_{2}(q) and natural O_{4}^{+}(q)

- modules are involved. Now for t ∈ [K_{1α+1} × K_{1α+1}^{s}, Z_{α+1}] we have |Z_{1} : C I_{Z_{1}}(t)| ≤ 32q and as there are at least two orthogonal modules involved we see |Z_{1} : C I_{Z_{1}}(t)| ≥ q^{4} or [K_{1α+1} × K_{1α+1}, Z_{α+1}] ∩ L_{1} ≤ Q_{1}. The former provides us with a contradiction. In the latter we have two such modules involved and then q = 4. Now |Z_{α+1} : Z_{α+1} ∩Q_{2}| = 16, a contradiction again.

So we have shown that [[Z_{1}, K] ∩ L_{α+1}, K_{α+1}] = 1. Let [K_{α+1}, [Z_{α+1}, s]] = 1. We have K_{1} with [K_{1α+1}, s] ≠ 1 again. If m_{p}(K_{1}) = 1, we must have another K_{2} with [K_{2α+1}, s] ≠ 1 too. In any case [K, Z_{1}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. Now we have K ≰L_{1}/Q_{1} and so we get [s, KK^{x}] = 1, for some x ∉ N_{L_{1}/Q_{1}}(K), a contradiction. So we have [K_{α+1}, [Z_{α+1}, s]] ≠ 1. Let m_{3}(K) ≤ 1, then we see that K ≅ ℤ_{p}, L_{2}(q), L_{3}(q), q even, q ≤ 16. We have [Z_{α+1}, K_{α+1}] ≤ L_{1}. In any case there is some E ≅ E_{p^{3}} centralizing K_{α+1}. Suppose there is some nontrivial irreducible module in [Z_{α+1}, K_{α+1}] centralized by F ≅ E_{p^{2}}. As this module intersects Q_{1} nontrivially, we see Z_{1} ≤ L_{α+1}, a contradiction. So we have that K acts trivially on C I_{Z_{1}}(F) for any F ≤ E, |F | = p^{2}. But Z_{α+1} = ⟨C I_{Z_{α+1}}(F) | |F | = p^{2}⟩, a contradiction.

So we have m_{3}(K) ≥ 2 and 3 ∉ σ(M). As before there is an irreducible module in Z_{1} whose dimension is less than the 2-rank of K. With (1.5) we get that K is a group of Lie type in characteristic two. This shows K ≅ L_{4}(q), Sp_{6}(q), L_{5}(2), L_{6}(2) or L_{7}(2). We have p |̸ q − 1. As there is E ≅ E_{p^{3}} in C I(K) there are at least two nontrivial modules in Z_{1} which are centralized by some elementary abelian group of order p^{2}, or [[E, Z_{α+1}], K_{α+1}] = 1. Assume the former. Then we get U ≤ Z_{α+1} ∩ L_{1}, U ∩ Q_{1} = 1 with |U| ≥ q^{8}, q^{12}, 2^{10}, 2^{12} and 2^{14} , respectively. This is impossible. So we have the latter. Now [E, [Z_{α+1}, K_{α+1}]] = 1 . Further [Z_{1}, K] is the natural module. But then [K_{1α+1}, [K_{α+1}, Z_{α+1}]] = 1, a contradiction.

So it remains [[K, Z_{1}] ∩ L_{α+1}, Z_{α+1}] = 1. Then for t ∈ Z_{α+1} we have |[[K, Z_{1}], t]| ≤ 16 or 8. Now we see that some E ≅ E_{p^{2}} centralizes [K, Z_{1}]. So [V_{2} ∩ L_{α+1}, K_{α+1}] = 1. This gives that [K_{α+1}, Z_{α+1}] ≤ L_{1}. Now we get the same groups for K as before. Further [K, Z_{1}] involves exactly one nontrivial irreducible module, which is the natural one. We have [K_{α+1}, Z_{α+1}]Q_{1}/Q_{1} ∩ K ≠ 1. So C I_{Z_{1}}(K) ≤ M_{α+1}. Now Z_{α+1}/C I_{Z_{α+1}}(K_{α+1}) has to be a nonsplit extension of the natural module by a trivial module. Application of (1.20) shows K ≅ Sp_{6}(q). Now |Z_{1} : Z_{1} ∩ M_{α+1}| ≤ q. But [K_{α+1}, Z_{α+1}] acts quadratically on Z_{1}, so some x ∈ [K_{α+1}, Z_{α+1}]^{♯} centralizes some y ∈ Z_{1} \ [K, Z_{1}]C I_{Z_{1}}(K), a contradiction. So we have Z_{1} ≤ M_{α+1}, a contradiction again.

(11) Z_{α+1} ∩ L_{1} ≤ Q_{1}.

Suppose false. By (10) we have (L_{1}/Q_{1})/O_{r}(L_{1}/Q_{1}) ≅ L_{2}(r) and |O_{r}(L_{1}/Q_{1})| = r^{3}. Now |Z_{α+1} ∩ L_{1}/Z_{α+1} ∩ Q_{1}| ≤ 4. And so for s ∈ Z_{1} we have |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 64. By (6.16) we see that |Z_{α+1} : Z_{α+1} ∩ Q_{2}| ≤ 2. So |Z_{α+1} : C I_{Z_{α+1}}(s)| ≤ 8. Application of (1.51) shows [O_{r}(L_{1}/Q_{1}), Z_{1}] = 1. But this contradicts (4.10).

Now (11) contradicts (1) and we are done.

(6.35) Proposition. If b = b_{1}, then b = 1.

Proof . By (6.34) it remains to treat the case L_{2}/Q_{2} is solvable with σ(L_{2}) ≠ ∅. By construction in (4.11) we may assume that either L_{1} is solvable too with σ(L_{1}) ≠ ∅ or E(L_{1}/Q_{1}) ≅ K_{1} ×…×K_{t}, K_{i} ≅ G(q), q odd, G(q) a group of Lie type different from U_{3}(3), P Sp_{4}(3), ^{2}G_{2}(3) and 3U_{4}(3).

(1) [F^{∗}(L_{1}/Q_{1}), Z_{1}] = Z_{1}, C I_{Z_{1}}(F^{∗}(L_{1}/Q_{1})) = 1 .

Otherwise some 1 ≠ x ∈ Ω_{1}(Z(S)) is centralized by F^{∗}(L_{1}/Q_{1}) and so L_{2} ≤ C I_{G}(x) ≤ M, a contradiction.

Application of [GoLyS, (24.1)], [GoLyS, (25.12)] and (1.12) (ii) shows that Z_{1} is generated by elements z such that

(2) |V_{α} : C I_{V_{α}}(z)| ≤ 4|V_{α} : V_{α} ∩ L_{1}| and |V_{α} : C I_{V_{α}}(z)| ≤ 2|V_{α} : V_{α} ∩ L_{1}| for K_{1} ≅ L̸_{2}(7) or L_{2}(9)

Suppose V_{α} ≤ M. Then |V_{α} : V_{α} ∩ L_{1}| ≤ 2 and so by (2) Z_{1} = ⟨z| |Z_{α+1} : C I_{Z_{α+1}}(z)| ≤ 8⟩ for α + 1 ∈ Δ(α) with Z_{1} ≰ M_{α+1}. But this is a contradiction. So we have V_{α} ≰ M and then we have symmetry. So we may assume that |V_{α} : V_{α} ∩ Q_{2}| = 2^{s} ≥ |V_{2} : V_{2} ∩ Q_{α}|. Set V_{α} = (V_{α} ∩ Q_{2})A, A = ⟨z_{β_{1}}, …, z_{β_{s}}⟩, z_{β_{i}} ∈ Z_{β_{i}}, β_{i} ∈ △(α). As Z_{β_{1}} is generated by elements z with |V_{2} ∩ L_{β_{1}} : C I_{V_{2}∩L_{β_{1}}}(z)| ≤ 4, we may assume that |V_{2} ∩ Q_{α} : C I_{V_{2}∩Q_{α}}(z_{β_{i}})| ≤ 4. Hence |V_{2} : C I_{V_{2}}(A)| ≤ 2^{3s}. On the other hand by (2.1) L_{2} is a {2, r}-group with r > 3. Now |T : C I_{T} (A)| ≥ 2^{2s} for any nontrivial irreducible L_{2}/Q_{2}-module T with O_{2}(C I_{L_{2}/Q_{2}}(T)) = 1. By (6.16) we get |V_{2} : C I_{V_{2}}(A)| ≥ 2^{4s}, a contradiction.

(6.36) Lemma. Let b = b_{1} = 1. Then there is 3 ∈ △(2) such that T = ⟨Z_{3}, Z_{1}⟩/(Q_{2} ∩ ⟨Z_{1}, Z_{3}⟩) ≅ L_{2}(q), Sz(q) or D_{2t}, t prime. Furthermore

Y = (Z_{1} ∩ Q_{2}) (Z_{3} ∩ Q_{2}) < ⟨Z_{1}, Z_{3}⟩

and Y/Z_{1} ∩ Z_{3} is a direct sum of natural T-modules. Furthermore there is some ω ∈ T with o(ω) = x | q − 1 and N_{G}(⟨ω⟩) ≰ M.

Proof . We have Z_{1} ≰ Q_{2} and Z_{1}Q_{2} < S. Furthermore Z_{1} acts quadratically on Q_{2}. If |Z_{1} : Z_{1} ∩ Q_{2}| = 2, we get some g such that ⟨Z_{1}, Z_{1}^{g}⟩/(⟨Z_{1}, Z_{1}^{g}⟩∩Q_{2}) ≅ D_{2t}, t odd. So assume |Z_{1} : Z_{1}∩Q_{2}| > 2. Then by (1.12) E(L_{2}/Q_{2}) is a product of groups of Lie type in characteristic two or 3 · A_{6}, or there is some g ∈ L_{2} such that ⟨Z_{1}, Z_{1}^{g}⟩/(⟨Z_{1}, Z_{1}^{g}⟩ ∩ Q_{2}) ≅ D_{2t}, t odd. Now with (1.1) it is easy to see that we get g ∈ L_{2} such that ⟨Z_{3}, Z_{1}⟩/Q_{2} ∩ ⟨Z_{1}, Z_{3}⟩ ≅ L_{2}(q), Sz(q) or D_{2t} or E(L_{2}/Q_{2}) ≅ Sp_{4}(q) or 3A_{6} ∗ 3A_{6} and then |Z_{1} : Z_{1} ∩ Q_{2}| = q^{2}, 2^{4} respectively. If we have L_{2}(q) or Sz(q), then as Z_{1} ≤ L_{1} ∩ L_{2}, we see that Z_{1} cannot contain a root group and so ⟨Z_{1}, Z_{3}⟩ contains some ω, o(ω) = x | q − 1, x a Zsigmondy prime or x = 9 in case of q − 1 = 63, such that ω ∈ L_{1} ∩ L_{2} and N_{G}(⟨ω⟩) ≰ M.

If E(L_{2}/Q_{2}) ≅ Sp_{4}(q) or 3A_{6} ∗ 3A_{6} there is some h ∈ L_{2} such that ⟨Z_{1}, Z_{1}^{h}⟩Q_{2}/Q_{2} = E(L_{2}/Q_{2}). By quadratic action we see that Y/Z_{1} ∩ Z_{3} just involves natural modules (1.15) and (1.14).

Now choose Z̃_{1} ≤ Z_{1} such that |Z̃_{1} : Z̃_{1} ∩ Q_{2}| = q and Z̃_{1}Q_{2}/Q_{2} corresponds to a transvection group on the natural module. There are two cases

(i) ⟨Z̃_{1}, Z̃_{1}^{h}⟩/(⟨Z̃_{1}, Z̃_{1}^{h}⟩ ∩ Q_{2}) ≅ L_{2}(q) or

(ii) ⟨Z̃_{1}, Z̃_{1}^{h}⟩ ≅ E_{q^{2}}.

Suppose (i). Then Y/Z_{1} ∩ Z_{3} is a direct sum of natural L_{2}(q)-modules by (1.27)(iv). Which gives that |[Z̃_{1}, Y/(Z_{1} ∩ Z_{3})]| = |C I_{Y/(Z_{1}∩Z_{3})}(Z̃_{1})|. But this contradicts the fact that Y/Z_{1} ∩ Z_{3} involves some natural module and Z̃_{1} induces the transvection group on this module.

So we have (ii). Then ⟨Z̃_{1}, Z̃_{1}^{h}⟩Q_{2}/Q_{2} ≤ E, E ≅ E_{q^{3}} and there is some g ∈ N(E) with Z̃_{1}Q_{2} = Z̃_{1}^{hg}Q_{2}, hg ∈ N(Z̃_{1}Q_{2}). As Z_{1} ≤ O_{2}(N_{L_{2}}(Z̃_{1}Q_{2})), we see that Z_{1}^{hg} ≤ N(E) and so Z_{1}^{h} ≤ N(E). Now ⟨Z_{1}, Z_{1}^{h}⟩ ≤ N(E), a contradiction. This proves the first part of the assertion.

As C I_{L_{2}}(Q_{2}) ≤ Q_{2}, we get Y ≠ Z_{1} ∩ Z_{3}. Now the second part follows with (1.27)(iii).

(6.37) Lemma. Assume b = b_{1} = 1. Let K be some component of L_{1}/Q_{1}, or normal p-group , m_{p}(K) ≥ 2. Then [K, Y] = 1.

Proof . Suppose [K, Y] ≠ 1. Set q = 2 for ⟨Z_{1}, Z_{3}⟩/O_{2}(⟨Z_{1}, Z_{3}⟩) ≅ D_{2t}.

(1) [K, Y] ≤ K. Furthermore [K, ω] ≤ K if T ≅ L_{2}(q) or Sz(q).

Suppose first that (6.4) holds. Then [Y, ω]Q_{1} = Y Q_{1}. As m_{p}(K) ≥ 2, we get by (1.63) that [ω, K] ≤ K or q = 64 and ω^{3} ∈ N_{L_{1}/Q_{1}}(K). If [K, ω] ≤ K we see that [Y, K] ≤ K.

Assume q = 64 But then as [Y, ω^{3}] = Y we get [Y, K] ≤ K too. Now 3 ∉ σ(M). Hence m_{3}(K) = 1, K^{⟨ω⟩} = K_{1} × K_{2} × K_{3}. Let ν ∈ L_{1} ∩ L_{2}, o(ν) = 7. Then 1 ≠ [ν, K_{i}] ≤ K_{i}. As m_{p}(K) ≥ 2, and there is some subgroup isomorphic to E_{64}ℤ_{21} normalizing K, we get with (1.5) that K ≅ L_{3}(r) or U_{3}(r), r even. Suppose that ν induces a field automorphism on K_{1} × K_{2} × K_{3}. Then m_{p}(C I_{K}(ν)) ≤ 1 for any odd prime p. So K ≅ L_{3}(r) and then E ≅ E_{7^{3}} is contained in C I_{K_{1}×K_{2}×K_{3}}(ν). This shows m_{7}(M) ≥ 4 and so 7 ∈ σ(M) a contradiction as N_{G}(⟨ν⟩) ≰ M. So we have that ν induces an inner automorphism. Then we see that ν is in a 2-local of K, i.e. o(ν) | r^{2}−1. As m_{7}(K) = 1, we get that K ≅ L_{3}(r) and 7 | r + 1, p | r − 1 or K ≅ U_{3}(r), 7 | r − 1 and p | r + 1. In the last case there is E ≅ E_{p^{3}} which centralizes ν contradicting (1.63). So we have K ≅ L_{3}(r) and 7 | r + 1, p | r − 1. As ν is in a 2-local it centralizes an element of order r − 1 in K_{1} and so it centralizes E ≅ E_{p^{3}} in K_{1} × K_{2} × K_{3}, a contradiction.

So we have ⟨Z_{1}, Z_{3}⟩/⟨Z_{1}, Z_{3}⟩ ∩ Q_{2} ≅ D_{2t}. Hence Z_{1} is an F_{1}-module with offending subgroup Y. Now let y ∈ Y with K^{y} ≠ K. Then y acts on some E ≅ E_{p^{4}}. Hence there is some F ≤ E, |F | = p^{2}, |C I_{[y,Z_{1}]}(F)| ≥ 4. But |[y, Z_{1}] : [y, Z_{1}] ∩ Z_{3}| = 2 and so some x ∈ (Z_{1} ∩ Z_{3})^{♯} is centralized by F. As Γ_{F,1}(G) ≤ M, we get ⟨Z_{1}, Z_{3}⟩ ≤ M, a contradiction.

(2) C I_{Y} (K) = 1.

Let y ∈ C I_{Y} (K). Then K acts on [y, Z_{1}]. So suppose first [K, [y, Z_{1}]] = 1. Then [y, Z_{1}] ∩ Z_{3} = 1. Hence |[y, Z_{1}]| = q. As C I_{Z_{1}}(K^{S}) = 1, we get some K_{1} ∼ K in L_{1} such that 1 ≠ [K_{1}, y] ≤ K_{1}. But then we see [[K, Z_{1}], K_{1}] = 1 = [[K_{1}, Z_{1}], K]. So there is some y_{1} ∈ Y with [y_{1}, K_{1}] = 1 and [y_{1}, K] ≠ 1.

Hence also |[y_{1}, Z_{1}]| = q. But then by (6.36) Z_{1} ∩ Y = C I_{Z_{1}}(y) = C I_{Z_{1}}(y_{1}) = C I_{Z_{1}}(y y_{1}) which is impossible, as [y, [K_{1}, Z_{1}]] = 1 and [y_{1}, [K_{1}, Z_{1}]] ≠ 1.

So we have [[y, Z_{1}], K] ≠ 1. Set W = [y, Z_{1}]. Then |W : C I_{W} (Y)| ≤ q by (6.36). Furthermore W ≤ Y and so [Y, W] ≤ Z_{1}∩Z_{3} i.e. Y acts quadratically on W. As |Y : C I_{Y} (K)| ≥ q, by the action of ω, we see that W is an F-module for K. Hence by (1.18) we get K ≅ A_{n}, L_{n}(r), Sp_{2n}(r), U_{n}(r), Ω_{2n}^{±}(r) or G_{2}(r), or K is solvable. Furthermore we have that Z_{1}/C I_{Z_{1}}(y) ≅ W as Kmodule and |[Y, Z_{1}/C I_{Z_{1}}(y)]| = |Z_{1}/C I_{Z_{1}}(y)|/q.

Let K be solvable. As some A ≤ Y, |A| = q with A⟨ω⟩ ≤ N(K) and A⟨ω⟩ ∩ C I(K) = 1 centralizes a subgroup of index q in W, we see that |Y : C I_{Y} (K)| = 2. This implies q = 2 and m_{3}(K) = 2. Finally |W | = 4 as Y induces the full transvection group on W. Let y_{1} ∈ C I_{Y} (K)^{♯}, y_{1} ≠ y. If |Z_{1} : C I_{Z_{1}}(⟨y, y_{1}⟩)| = 8 there is some u ∈ Z_{1}\C I_{Z_{1}}(⟨y, y_{1}⟩), [u, E] ≤ C I_{Z_{1}}(⟨y, y_{1}⟩) with E ≅ E_{9}, E ≤ K. But C I_{Z_{1}}(E) ≤ C I_{Z_{1}}(y), as [E, W] = W. So 1 ≠ [C I_{Z_{1}}(E), y_{1}] ≤ Z_{1} ∩ Z_{3}, a contradiction.

This shows |Z_{1} : C I_{Z_{1}}(C I_{Y} (K))| = 4. Furthermore |Z_{1} : Z_{1} ∩ Z_{3}| > 8 and so |C I_{Y} (K)| ≥ 4. Hence Z_{1} is an F-module for C I(K) with offending subgroup C I_{Y} (K). But then the action of K shows that we have two submodules T_{1}, T_{2} of Z_{1} such that C I_{Y} (K) induces the full group of transvections on T_{i} and T_{1} ∼_{K} T_{2}. This gives K_{1} ≤ C I(K), K_{1} ≅ L_{n}(2), |Y : Y ∩ Q_{1}| = 2^{n}. But now every x ∈ T_{1} is centralized by E ≅ E_{9}, E ≤ L_{1}. Hence [T_{1}, C I_{Y} (K)] ∩ Z_{3} = 1. The same is true for T_{2}. In particular T_{1}, T_{2} ≤ Y and so |⟨T_{1}, T_{2}⟩ : C I_{⟨T_{1},T_{2}⟩}(y)| = 2 for y ∈ C I_{Y} (K)^{♯}, a contradiction. So we have that K is nonsolvable.

Suppose first q = 2, then Y induces the full transvection group on W i.e. K ≅ L_{n}(2). Now [Y, W] ≤ Z_{3} ∩ Z_{1} and is normalized by L_{n−1}(2). This implies m_{p}(L_{n−1}(2)) ≤ 1. Hence K ≅ L_{4}(2), p = 3 or K ≅ L_{6}(2), p = 7. But in both cases [W, Y] contains a subgroup U of dimension \frac{n}{2} normalized by E ≅ E_{p^{2}}, a contradiction.

So we have q > 2. Now Y ∩K ≠ 1. Let K ≅ U_{n}(r) or Ω_{2n}^{±}(r), then q ≥ r^{2} by (1.18). As ω ∈ L_{1} ∩ L_{2}, we have that ⟨S, ω⟩ is a {2, x} - group, and so by (1.65) and (1.64) we have q = r^{2}. But then |[Y, W]| ≠ |W |/q.

Let K ≅ L_{n}(r), Sp_{2n}(r) or G_{2}(r), then q ≥ r. If q = r, we get as before K ≅ L_{n}(r) and W is the natural module. Now as above (K ≅ L_{4}(2)) we get a contradiction. So let q > r. By (1.65) and (1.64) q = 4, r = 2 and ω ∈ K. Now m_{p}(K) ≥ 3, p ≠ 3 and 3 ∉ σ(M). But this is impossible.

So we are left with A_{n}. Again by (1.9) ω ∈ K and so m_{p}(K) ≥ 3, p ≠ 3, 3 ∉ σ(M). The same contradiction as before.

(3) If y ∈ Y then |Z_{1} : C I_{Z_{1}}(y)| ≤ |Y |. Furthermore K ≅ G(r), a group of Lie type in characteristic two.

The first assertion is clear. Now we go over all possibilities for K.

Let K ≅ A_{n}. By (1.9) q ≤ 4 or q = 64. If o(ω) = 3, or 9 then ω ∈ K and so m_{p}(K) ≥ 3, p ≠ 3, 3 ∉ σ(M), a contradiction.

Hence q = 2 and so Z_{1} is an F-module. By (1.42) Z_{1} is the natural module, or n = 7 or 8 and Z_{1} involves a four-dimensional module, or n = 9 and Z_{1} involves a spin module. Suppose that Z_{1} is natural. Let 3 ∈ σ(M). Then some x ∈ Ω_{1}(Z(S))^{♯} is centralized by an elementary abelian group E of order 9 and so L_{2} ≤ M, a contradiction. So let p = 5, K ≅ A_{10} or A_{11}. Then by (4.4) we may assume that [K, Z_{1}] is centralized by E_{25}, a contradiction. Let next K ≅ A_{9} and Z_{1} involve a spin module. Now we have that just one such module is involved. Hence as m_{3}(K) = 3, we see that [Z_{1}, K] is centralized by an element of order 3. As every element in the spin module is centralized by a 3-element, we get some E ≅ E_{9}, Γ_{E,1}(G) ≤ M, with C I_{Ω_{1}(Z(S))}(E) ≠ 1. Hence L_{2} ≤ M, a contradiction.

So we have K ≅ A_{7} (K ≅ A_{8} ≅ GL_{4}(2) is a group of Lie type), p = 3. In that case by (4.4) any z ∈ [K, Z_{1}] is centralized by an elementary abelian group of order 9, a contradiction.

Let K be sporadic. By (1.9) q ≤ 4 or q = 64, or q = 8 and K ≅ J_{1}. But m_{p}(K) ≥ 2, so K ≅ J_{1} is not possible.

Let q = 4 or 64. Then p ≠ 3. But m_{p}(K) ≥ 3, a contradiction to (1.4). So assume q = 2. Then Z_{1} is an F_{1} - module and so K ≅ M_{12}, M_{22}, M_{23}, M_{24} or J_{2} by (1.42). Now p = 3 or p = 5 in case J_{2} and m_{p}(K) = 2. Then Ω_{1}(Z(S))^{♯} contains x, which is centralized by E ≅ E_{9} in M, and so L_{2} ≤ M, a contradiction.

Let K ≅ G(r) be a group of Lie type in odd characteristic. Suppose first q > 2. As ω acts fixed point freely on Y, we see that Y corresponds to a subgroup of K. Now (1.41) yields a contradiction. So we have q = 2 and Z_{1} is an F_{1} - module. By (1.42) K ≅ 3U_{4}(3), p = 3. But now Y ∩ K = 1, as every involution in K is centralized by E ≅ E_{27} by (1.55) and so some 1 ≠ z ∈ Z_{1} ∩ Z_{3} ∩ [W, Y ∩ K] would be centralized by F ≅ E_{9}. So we have |Y | = 2 by (2). Then Y induces a transvection, a contradiction.

It remains K to be solvable. Let |Y : Y ∩ Q_{1}| = 2^{n}. Now by [GoLyS, (24.1)] there is E ≅ E_{p^{n}} such that [E, y] = E for some y ∈ Y. As |Z_{1} : C I_{Z_{1}}(y)| ≤ |Y | there is some F ≅ E_{p^{2}} in E such that |Z_{1} : C I_{Z_{1}}(F)| ≤ 16. Now C I_{Z_{1}}(F) ∩ Z_{3} = 1 and so |C I_{Z_{1}}(Y) : C I_{C I_{Z_{1}}(Y)}(F)| ≤ 4. Hence we have

|Z_{1} ∩ Z_{3}| ≤ 4 .

As |[y, Z_{1}]| = |Y |, we get |Z_{1}| ≤ |Z_{1} ∩ Z_{3}|^{2} · q^{2}. Suppose first |Y | ≥ q^{2}. Then we see q ≤ 4 and so |Z_{1}| ≤ 2^{8}. If q = 4, then |Y | = 16 and so |Z_{1}| = 2^{8}. But then the action of K on Z_{1} is uniquely determined i.e. there is y ∈ Y ^{♯} with |[Z_{1}, y]| = 2, a contradiction. This shows q = 2 and so |Z_{1}| ≤ 2^{6}. This now implies |Z_{1}| = 64 and |Y | = 8. But now we have ⟨y_{1}, y_{2}⟩ ≤ Y with |⟨y_{1}, y_{2}⟩| = 4 and |Z_{1} : C I_{Z_{1}}(y)| = 2 = |Z_{1} : C I_{Z_{1}}(y_{1})|. This would imply that |Z_{1} : C I_{Z_{1}}(⟨y_{1}, y_{2}⟩)| = 2 by the action of Y on Z_{1}. But this contradicts [GoLyS, (26.11)].

So we have |Y | = q and then C I_{Z_{1}}(Y) = C I_{Z_{1}}(y) for every y ∈ Y \ Q_{1}. Now

|Z_{1}| = |Z_{1} ∩ Z_{3}|q^{2}

and Y acts quadratically on Z_{1}. So by [GoLyS, (24.1)] and [GoLyS, (25.12)] there is t ∈ Z_{1}^{♯} with |Y : C I_{Y} (t)| = 2. This shows q = 2 and so |Z_{1}| = 16. But then there is E ≅ E_{9} with [Z_{1}, E] = 1, whence L_{2} ≤ M, a contradiction.

Now we are able to prove the lemma. Let T ≤ Z_{1} be an irreducible Aut(K)-submodule. Then C I_{K}(C I_{T} (S ∩ K)) does not contain an elementary abelian group of order p^{2}. Now (3), (1.22) and (1.24) imply K ≅ L_{3}(r), L_{4}(r), L_{n}(2), n ≤ 6, U_{n}(r), n ≤ 6, Sp_{4}(r), Sp_{6}(r), Ω_{8}^{−}(r), G_{2}(r).

If K ≅ Ω_{8}^{−}(r), then T is the spin module and p | r + 1. Now in the spin module the codimension of C I(v) is at least four, v an involution. Hence T = [K, Z_{1}] < L_{1} and so E ≅ E_{p^{2}} centralizes some 1 ≠ x ∈ Ω_{1}(Z(S)) ∩ T by (4.4), as m_{p}(K) ≤ 3.

Let K ≅ G_{2}(r), then also T = [K, Z_{1}] < L_{1} and also some 1 ≠ x ∈ Ω_{1}(Z(S)) is centralized by E ≅ E_{p^{2}} again.

Let K ≅ Sp_{6}(r), then T is the spin module, m_{p}(K) = 3. If T ≰L_{1}, then |[T, y]| = r^{2} for every y ∈ Y \ Q_{1}. Furthermore |Y/Y ∩ Q_{1}| ≥ r^{4}. Hence we see |Y/Y ∩ Q_{1}| = r^{4} and |C I_{T} (Y)| > r^{5}. But this is impossible. So we have T < L_{1} and the same contradiction as before.

Let K ≅ Sp_{4}(r), T the natural module. As |Y : Y ∩ Q_{1}| ≤ r^{3}, we get that T < L_{1} and so we have the same contradiction as before.

Let K ≅ U_{6}(r). Then T ≅ V (λ_{1} + λ_{l}) in the notation of (1.22). As T is not an F_{1}-module we see q > 2. If 3 | o(ω) then 3 ∉ σ(M) and so r > 2. Hence by (1.65) and (1.64) we either have q ≤ r or q = r^{2}. But p | r + 1 and so p | q − 1, i.e. there is ν ∈ L_{1} ∩ L_{2}, o(ν) = p and N_{G}(⟨ν⟩) ≰ M. So we have q ≤ r. Let R be a transvection group in the natural representation with [Y, R] ≤ Q_{1}. We have |[T, R]| ≥ r^{6}, as C I_{K}(R)′ acts faithfully on [T, R]. If R∩Y ≠ 1, then |Y : Y ∩Q_{1}| ≥ r^{6} by (3). Now we have |[T, R] : C I_{[T,R]}(Y)| ≤ q ≤ r. As Y Q_{1}/Q_{1} ≰ O_{2}(C I_{K}(R)), there is some module V for U_{4}(r) such that |V : C I_{V} (u)| ≤ r for some 2-element u ∈ U_{4}(r), contradicting (1.22) - (1.24). So we have R ∩ Y = 1. As [Y, Q_{2}] ≤ Y, we see that Y ∩O_{2}(N_{K}(R)) is abelian and |O_{2}(N_{K}(R))/O_{2}(N_{K}(R))∩Q_{2}| ≥ r^{4} ≥ q^{4}. Hence S/Q_{2} involves an elementary abelian subgroup of order q^{4}. By (1.7) we get either q = r = 8 and E(L_{2}/Q_{2}) ≅ SU_{3}(8) ∗ SU_{3}(8) or q = r = 4 and E(L_{2}/Q_{2}) ≅ SL_{3}(4) ∗ SL_{3}(4). In the former 3 ∈ σ(M), contradicting (2.2). So we have the latter with 3 ∉ σ(M). Now 5 ∈ σ(M). We have [R, Y] ≤ Y ∩ Q_{1} ≤ Z_{1}, As [R, ⟨Z_{1}, Z_{3}⟩] ≤ Y, we see [R, Y] ≤ Z_{1} ∩ Z_{3}. So |[Z_{1}, R] : Z_{1} ∩ Z_{3} ∩ [Z_{1}, R]| ≤ q = 4. Furthermore [Z_{1}, R] ∩ Z_{3} ≠ 1. As R is centralized by F ≅ E_{5^{3}}, there is E ≅ E_{5^{2}} with Γ_{E,1}(G) ≤ M such that C I_{Z_{1}∩Z_{3}}(E) ≠ 1. But then ⟨Z_{1}, Z_{3}⟩ ≤ M, a contradiction.

Let now K ≅ U_{n}(r), n = 4 or 5. Then |Y : Y ∩ Q_{1}| ≤ r^{4}. If n = 5, then as before R ≰ Y and q = r = 4. As |[T, R] : C I_{[T,R]}(Y)| ≤ r we see that Y ≤ O_{2}(N_{K}(R)) and so |Y : Y ∩ Q_{1}| ≤ r^{2}. But then we have a contradiction to (1.24).

So we have n = 4. Let T be the natural U_{4}(r)-module. Then we get |Y : Y ∩ Q_{1}| ≥ r^{2} and then even |Y : Y ∩ Q_{1}| = r^{4}, as any such subgroup of order r^{2} in U_{4}(r) contains an element u with |[T, u]| = r^{4}.

Now Y Q_{1}/Q_{1} is uniquely determined and so the action of Y on the natural module tells us [Y, T, Y] = 1, i.e Y acts quadratically and T < L_{1}. But m_{p}(K) ≤ 3 and so by (4.4) some 1 ≠ x ∈ Ω_{1}(Z(S)) is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

So we have T is the natural Ω^{−}(6, r)-module. As before we get T ≰L_{1}. Then as |[y, T]| = r^{2} for every y ∈ (Y ∩K)^{♯}, we see that |Y | = r^{4} and again Y is uniquely determined. Now let T^{g} ≠ T. Then |T^{g} × T : C I_{T ^{g}×T} (Y)| = r^{10}. On the other hand |Z_{1} : C I_{Z_{1}}(Y)| ≤ q·r^{4}. This implies r^{6} ≤ q, a contradiction.

Let K ≅ L_{3}(r). We may assume that T ≰L_{1}, as m_{p}(K) = 2. Then T is the natural module and |Y : Y ∩ Q_{1}| = r^{2}. Suppose [T, Y, Y] ≠ 1. Then |[T, Y, Y]| = q and [T, Y, Y] ≤ Z_{1} ∩ Z_{3}. Then Z_{1} ∩ Z_{3} contains a nontrivial subgroup which is normalized by E ≤ K, E ≅ E_{p^{2}}, a contradiction. But now [Z_{1}, Y, Y] = 1 and then |Z_{1} : C I_{Z_{1}}(Y)| = q. As T ≰L_{1}, we have q ≥ r^{2}. But this contradicts (1.65) and (1.64).

So assume next K ≅ L_{4}(r). Again T ≰L_{1}. This shows that T is the natural module. Suppose there is some y ∈ Y ^{♯} with |[T, y]| = r^{2}. Then |Y | = r^{4} and [T, Y, Y] = 1. So Y acts quadratically and |Z_{1} : C I_{Z_{1}}(Y)| = q. Now q = r^{4}, a contradiction as before. So we have that |Y | ≤ r^{3} and Y is a transvection group. But every transvection group acts quadratically and so |Z_{1} : Z_{1} ∩ Y | = q ≥ r^{2}, contradicting (1.64) and (1.65).

Let finally K ≅ L_{n}(2), 5 ≤ n ≤ 7. Suppose first p = 3. Then by (1.65) we have q = 2 and so Z_{1} is an F_{1}-module. By (1.22) - (1.24) we see that K ≅ L_{5}(2), otherwise some x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{9}, E ≤ K, a contradiction to L_{2} ≰ M. So K ≅ L_{5}(2). Now by (1.22) - (1.24) natural modules, duals or V (λ_{1} + λ_{4}) are involved. Let T be one of these modules. As m_{3}(K) = 2, T ≰L_{1}. If T ≅ V (λ_{1} + λ_{4}) we see that |Y : Y ∩ Q_{1}| = 2^{6}. As Y does not act quadratically, we see that 1 ≠ [Y, Z_{1}, Y] ≤ Z_{1} ∩ Z_{3} is normalized by N_{K}(Y) ≅ Y Q_{1}/Q_{1}(Σ_{3} × L_{3}(2)) and so it is normalized by E ≤ K, E ≅ E_{9}, a contradiction. So T is the natural module and as no x ∈ Ω_{1}(Z(S))^{♯} is centralized by E ≅ E_{9}, E ≤ K, we get that T^{∗}, the dual module is also involved. Now there are at least two conjugates of each T and T^{⋆}. Hence Y is a transvection group. In particular |Y | = 16 and |Z_{1} : C I_{Z_{1}}(Y)| ≥ 2^{10}. But |Z_{1} : C I_{Z_{1}}(Y)| ≤ 32, a contradiction.

Let K ≅ L_{6}(2) or L_{7}(2), p = 7. As m_{p}(K) = 2 we first get that there are at least three conjugates of T. Furthermore T is the natural module. As 7 |̸ |C I_{K}(C I_{T} (S ∩ K))| we get with (4.4) that at least 6 nontrivial irreducible modules are involved. Now K ≅ L_{7}(2) and |Y | = 2^{6}, Y is a transvection group. Then |Z_{1} : C I_{Z_{1}}(Y)| ≥ 2^{12}. But |Z_{1} : C I_{Z_{1}}(Y)| ≤ 2 · 2^{6}, a contradiction.

(6.38) Lemma. Assume b = b_{1} = 1. Let K_{1} × K_{2} be a direct product of two components of L_{1}/Q_{1} of p-rank 1 or a component with a cyclic normal p-group. Then we may assume [K_{2}, Y] = 1.

Proof . Suppose [K_{1}, Y] ≠ 1 ≠ [K_{2}, Y].

(1) [K_{i}, ω] ≤ K_{i}, i = 1, 2,, or q = 64 and for ν = ω^{3}ν_{1}, o(ν_{1}) = 7, ⟨ν, ω⟩ ≅ ZZ_{q−1} we have [K_{i}, ν] ≤ K_{i}, or q = 4.

Suppose [K_{1}, ω] ≰ K_{1}. Then we get m_{t}(K_{1}) = 1 for every odd prime t | |K_{1}|. Hence K_{1} ≅ L_{2}(r), Sz(r) or J_{1} by (1.5). Let first o(ω) = x > 3. Then

K_{1}^{⟨ω⟩} = K_{1} × … × K_{y}, y = x or x = 9, y = 3 .

Let ω ∈ ⟨ν⟩, o(ν) = q − 1. Suppose first y > 3. If 1 ≠ μ ∈ ⟨ν⟩ with [K_{1}, μ] ≤ K_{1}. Then μ induces an outer automorphism on K_{1} with C I_{K_{1}}(μ) a 2-group. But K_{1} does not possess such an automorphism. This implies

K_{1}^{⟨ν⟩} = K_{1} × … × K_{q−1} .

Let t ∈ Y, [K_{1}, t] ≠ 1. Then t acts on an elementary abelian group E ≅ E_{p^{q−1}}. We have |[Z_{1}, t] : [Z_{1}, t] ∩ Z_{3}| ≤ q. If [Z_{1}, t] ∩ Z_{3} ≠ 1. Then we see that there is F ≅ E_{p^{2}} ≤ E, C I_{Z_{3}}(F) ≠ 1, a contradiction. So we have |[t, Z_{1}]| ≤ q. In particular t ∈ Z((Z_{1} ∩ Q_{2})(Z_{3} ∩ Q_{2})). Then C I_{E}(t) ≠ 1, otherwise |[t, Z_{1}]| ≥ \sqrt{|[E, Z_{1}]|} > q. But then we may assume that [t, K_{i}] = 1 for some i. As [Y, [t, Z_{1}]] ≤ [Z_{1}, t] ∩ Z_{3} = 1 and [Y, K_{i}] ≠ 1, we see that [[t, Z_{1}], K_{i}] = 1. But now [[Z_{1}, K_{i}], t] = 1 and so [Z_{1}, K_{i}] ≤ Q_{2} and [s, [Z_{1}, K_{i}]] = 1, for s ∈ Y, [K_{2}, s] ≠ 1 by the same argument as for t. But this is not possible.

So we are left with q = 64, ω^{3} ∈ N(K_{1}), C I_{K_{1}}(ω) is not a 2-group. This implies [K_{1}, ν_{1}] ≤ K_{1}.

(2) We may choose K_{1}, K_{2} such that [K_{1} × K_{2}, Y] ≤ K_{1} × K_{2}.

If q > 4 this follows from [Y, ω] = Y, [Y, ν] = Y, respectively by (1). If q = 4, ω induces at most one orbit of length 3 on K_{1}^{Y ⟨ω⟩}. Thus either (2) holds or Y ⟨ω⟩ induces A_{4} on K_{1}^{Y ⟨ω⟩}. Suppose the latter. Then as 3 ∉ σ(M), we have K_{1} ≅ Sz(r) and ω induces a field automorphism on K_{1}. Now there is ν ∈ K_{1}, o(ν) = p, [ω, ν] = 1. In particular C I_{K_{1}^{Y ⟨ω⟩}}(ω) ≥ ⟨ν⟩ × K̃, K̃ ≅ K_{1}. But then N_{G}(⟨ω⟩) ≤ M by (1.63), a contradiction.

So let q = 2 and without loss K_{2}^{y} = K_{3} ≠ K_{2}, K_{1} for some y ∈ Y ^{♯}. Now K_{2} ≅ L_{2}(r), Sz(r), or J_{1} again. Let y_{1} ∈ Y ^{♯} with K_{2}^{y_{1}} ≠ K_{3}, K_{2}. Then K_{2}^{⟨y,y_{1}⟩} = K_{2} × K_{3} × K_{4} × K_{5} and so y y_{1} acts on E ≅ E_{p^{4}}, which shows that there is 1 ≠ x ∈ Z_{1} ∩ Z_{3} centralized by F ≅ E_{p^{2}}, F ≤ E, a contradiction. So we have [K_{2} × K_{3}, Y] ≤ K_{2} × K_{3}.

(3) We may choose K_{1}, K_{2} with (1), (2) and C I_{Y} (K_{1} × K_{2}) = 1.

Suppose false. Choose 1 ≠ y ∈ C I_{Y} (K_{1} × K_{2}). Set W = [y, Z_{1}]. Then K_{1} × K_{2} acts on W. Let first [K_{1} × K_{2}, W] = 1. Then W ∩ Z_{3} = 1 and so |W | = q. Hence K_{1} × K_{2} ≰L_{1}/Q_{1}. So we may assume that there is some s ∈ S with K_{2}^{s} ≠ K_{2} and [y, K_{2}^{s}] ≠ 1. We now have

[K_{1} × K_{2}, Z_{1}, K_{2}^{s}] = 1 .

Furthermore [[K_{1} × K_{2}, Z_{1}], Y] = 1 and so [K_{1} × K_{2}, Z_{1}] ≤ Y. So for ỹ ∈ Y we have [ỹ, [K_{i}, Z_{1}]] ≤ Z_{1} ∩ Z_{3}. Now

[K_{2}, Z_{1}, K_{1}^{s^{−1}} × K_{2}^{s^{−1}}] = 1 .

Let ỹ ∈ Y with [K_{2}, ỹ] ≠ 1. Then [[K_{2}, Z_{1}], ỹ] is centralized by a group of order p^{2} in K_{1} × K_{2} × K_{1}^{s^{−1}} × K_{2}^{s^{−1}}, a contradiction. So we have

[W, K_{1} × K_{2}] ≠ 1

and then

1 ≠ |W : C I_{W} (Y)| ≤ q .

Suppose [W, K_{i}] ≠ 1, i = 1, 2 . By (1) we have |Y : C I_{Y} (K_{i})| ≥ q, i = 1, 2, or q = 4 and |Y : C I_{Y} (K_{1})| = 2. In the former W is an F-module for both K_{1} and K_{2}. In the latter K_{1} ≅ L_{2}(r), Sz(r) or J_{1} and |W : C I_{W} (Y)| = 4. This implies K_{1} ≅ L_{2}(4) or L_{3}(2) with (1.30). Hence also in this case just F-modules are involved in W.

If W is an F-module, we also have K_{1} ≅ L_{2}(r) or L_{3}(r), or a 3-group, if we choose notation such that m_{3}(K_{1}) ≤ 1. By (1) we get q = 4 or r = q, as r ≤ q anyway. Again for q = 4 we may have K_{1} ≅ L_{3}(2). Hence in any case

K_{1} ≅ L_{2}(q), L_{3}(q) or q = 4 and K_{1} ≅ L_{3}(2), or K_{1} is a 3-group.

Suppose first [W, K_{1}, K_{2}] = 1. There is ỹ ∈ Y ^{♯} with [K_{1}, ỹ] ≠ 1 ≠ [K_{2}, ỹ]. As |[W, ỹ]| ≤ q, we see that K_{1} ≅ L_{3}(2), q = 4, and [W, K_{1}] is the natural module. Furthermore also K_{2} ≅ L_{3}(2), [W, K_{2}] is the natural module and p = 7. Finally K_{1}^{ω} = K_{2}, K_{2}^{ω} = K_{3}. Now we have C I_{Y} (K_{1} × K_{2}) = C I_{Y} (K_{1} × K_{2} ×K_{3}). But we have C I_{W} (Y) = C I_{W} (ỹ) which implies [W, K_{1} ×K_{2}, K_{3}] ≠ 1, a contradiction. So we have

[W, K_{1}, K_{2}] ≠ 1 .

In particular there are at least two nontrivial K_{1}-modules involved. This shows q = 4 and K_{1} ≅ L_{3}(2). Now K_{2} ≅ L_{3}(2). But then [W, K_{1}, K_{2}] = 1, as there are exactly two K_{1}-modules involved. So we may assume

[W, K_{1}] = 1 or [W, K_{2}] = 1 .

Choose notation such that [K_{1}, W] = 1. Hence [K_{1}, Z_{1}] ≤ C I_{Z_{1}}(Y). Let K_{3} ≤ C I(K_{1} × K_{2}) with [K_{3}, Y] ≠ 1. Then [[K_{1}, Z_{1}], K_{3}] = 1. Hence we may assume [K_{3}, Y] ≤ K_{3}, otherwise there is K_{3} × K_{4} ≤ C I(K_{1}) with [K_{3} × K_{4}, [K_{1}, Z_{1}]] = 1, a contradiction.

As [K_{2}, W] ≠ 1, we have [K_{1}, Y] ≤ K_{1} by (2) and so [K_{1} × K_{3}, Y] ≤ K_{1} × K_{3}. Now as [[K_{1}, Z_{1}], K_{3}] = 1 we have C I_{Y} (K_{1} × K_{3}) = 1, as shown before. Replace {K_{1}, K_{2}} by {K_{1}, K_{3}}.

We now choose notation such that m_{3}(K_{1}) ≤ 1; i.e. K_{1} ≅ L_{2}(r), L_{3}(r), U_{3}(r) Sz(r), J_{1} or solvable.

(4) C I_{Y} (K_{1}) = 1 .

Suppose Y_{1} = Y ∩ C I(K_{1}) ≠ 1. Let y ∈ C I_{Y} (K_{1}) and W = [y, Z_{1}]. Assume [W, K_{1}] = 1. Then [Z_{1}, K_{1}, K_{2}] = 1 by (3). Let t ∈ [Z_{1}, K_{1}] with t ∉ Y. Then [Y, t](Z_{1} ∩ Z_{3}) = Y ∩ Z_{1}. This shows [[Z_{1}, K_{2}], Y] = 1, a contradiction. So we have [Z_{1}, K_{1}] ≤ Y ≥ [Z_{1}, K_{2}]. Let K_{3} be a component in C I_{L_{1}/Q_{1}}(K_{1} × K_{2}) or a normal p-group. Then K_{3} acts on [Z_{1}, K_{1}] and [Z_{1}, K_{2}]. If [y, K_{3}] = 1, we see that K_{3} acts on 1 ≠ [y, [Z_{1}, K_{2}]] ≤ Z_{1} ∩ Z_{3}. Let [y, K_{3}] ≠ 1. Then K_{3} acts trivially on [Z_{1}, K_{1}]. Hence in both cases there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, such that E normalizes 1 ≠ U ≤ Z_{1} ∩ Z_{3}, a contradiction to L_{2} ≰ M.

Hence C I_{L_{1}/Q_{1}}(K_{1} × K_{2}) = 1. By (4.4) we see m_{p}(L_{1}) ≥ 4. But then we have some μ ∈ L_{1}, o(μ) = p, such that μ induces a field automorphism on K_{1} and so μ acts on Ω_{1}(Z(S)) ∩ [Z_{1}, K_{1}] and then Ω_{1}(Z(S)) ∩ [Z_{1}, K_{1}] ≤ Z_{1} ∩ Z_{3} is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. So we have

[W, K_{1}] ≠ 1 .

Now for ỹ ∈ Y we get |W : C I_{W} (ỹ)| ≤ q. If ω or ν act nontrivially on K_{1}, we get r ≥ q or K_{1} ≅ U_{3}(r), r^{2} = q. Hence we have K_{1} ≅ L_{2}(q), L_{3}(q) or U_{3}(\sqrt{q}) and W involves the natural module just once. Suppose q = 4 and [K_{1}, ω] ≰ K_{1}. Now |W : C I_{W} (ỹ)| ≤ 4. By (1.30) we get K_{1} ≅ L_{2}(4) or L_{3}(2), and there are at most two nontrivial modules involved. Furthermore for y_{1} ∈ C I_{Y} (K_{1}) we get [W, y_{1}] = 1. Now as Z_{1}/C I_{Z_{1}}(y) ≅ W as K_{1}-module, we see that Y has to induce the full group of transvection. Hence K_{1} ≅ L_{2}(q), SL_{3}(q) or L_{3}(2). In particular |W | = q^{2}, q^{3}. In case K_{1} ≅ L_{3}(2), we have |Y : C I_{Y} (K_{1})| ≤ 4. But Y has to induce a group of order 16, a contradiction.

So we have [ω, K_{1}] ≤ K_{1} and |W | = q^{2}, q^{3}. Because of the action of K_{1} we get that there are K_{2}-submodules T_{1}, T_{2} for K_{1} ≅ L_{2}(q), T_{1}, T_{2}, T_{3} for K_{1} ≅ SL_{3}(q) with |T_{i} : C I_{T_{i}}(C I_{Y} (K_{1}))| ≤ q, i = 1, 2, 3. Furthermore [T_{i}, C I_{Y} (K_{1})] = C I_{T_{i}}(C I_{Y} (K_{1})).

Let first q = 2. Then K_{2} ≅ L_{n}(2). As m_{p}(K_{2}) ≤ 3 for any odd prime p, we get n ≤ 7. As 3 | |K_{2}|, we get 3 ∉ σ(M) or m_{3}(K_{2}) = 1. Hence n ≤ 5. But now there is E ≅ E_{p^{2}} such that [Z_{1}, K_{1} × K_{2}] is centralized by E and Γ_{E,1}(G) ≤ M, a contradiction.

So we have q ≥ 4. We have ρ ∈ K_{1}, o(ρ) = q − 1, such that [ρ, T_{1}] = T_{1}. This shows K_{2} ≅ L_{n}(q) and so n ≤ 5 again. Let n ≥ 3, then we get K_{1} ≅ L_{2}(q), otherwise m_{p}(K_{1}) = 2. Now p | q^{2} − 1 and so we have n = 3. As every p-element ν ∈ C I(K_{1} × K_{2}) has to centralize T_{1} ⨁ T_{2}, we see C I(K_{1} × K_{2}) = 1. So by (4.4) there is some μ, o(μ) = p, μ induces a field automorphism on K_{1} or K_{2}. But this contradicts (4.12).

So let n = 2. By symmetry we may assume K_{1} ≅ L_{2}(q) too. If p | q + 1,

then we get a contradiction as before. Hence assume that p | q − 1. Now if [ω, K_{1}] ≠ 1, [ω, K_{1}] ≤ K_{1}, then ω induces a diagonal automorphism and so ω centralizes E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, contradicting N_{G}(⟨ω⟩) ≰ M by (1.63). Hence q = 4 by (1). But now K_{1} ≅ L_{2}(4), q − 1 = 3. But 3 ∉ σ(M) if q = 4. This proves (4).

So we have C I_{Y} (K_{1}) = 1. Then [K_{1}, Z_{1}] is an SC-module. This implies K_{1} ≅ L_{2}(r), L_{3}(r) or a {2, 3}-group, r even by (1.25) - (1.27) and (1.29).

Suppose that [K_{1}, Z_{1}] involves exactly one irreducible nontrivial module. If K_{1} < L_{1}/Q_{1}, then [K_{1}, Z_{1}] ∩ Ω_{1}(Z(S)) is normalized by E ≅ E_{p^{2}}, a contradiction. So we have K_{1} ≰L_{1}/Q_{1}. As m_{p}(K_{1}) = 1 this now gives K_{1} ≅ L_{2}(r). If K_{2} is a p-group then (4) implies |Y : Y ∩ Q_{1}| = 2. Hence |Z_{1} : C I_{Z_{1}}(Y)| = 2 and so K_{1} ≅ L_{2}(2) is solvable too, a contradiction. So K_{2} is nonsolvable. Now [[K_{1}, Z_{1}], K_{2}] = 1. As [Y, [K_{1}, Z_{1}]] ≠ 1, and [K_{2}, Y] ≠ 1, we see that [K_{1}, Z_{1}] ≤ Y ≥ [K_{2}, Z_{1}] again. By (4.4) we may assume that m_{p}(L_{1}) ≥ 4. But now there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, such that C I_{K_{1}}(E) contains some s which acts as some t ∈ Y \ Q_{1} on [Z_{1}, K_{1}]. But then 1 ≠ [s, [Z_{1}, K_{1}]] ≤ Z_{1} ∩ Z_{3}, and so E normalizes a nontrivial subgroup of Z_{1} ∩ Z_{3}, a contradiction.

So we have that [K_{1}, Z_{1}] involves two natural modules and K_{1} ≅ SL_{3}(r). If K_{1} ≤ L_{1}/Q_{1}, then as p | r̸ − 1 we see that there is 1 ≠ U ≤ [K_{1}, Z_{1}] ∩ Ω_{1}(Z(S)) normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. So we see K_{1} ≰L_{1}/Q_{1} and then K_{1} ≅ L_{3}(2). Now [K_{2}, [K_{1}, Z_{1}]] = 1, as K_{2} is nonsolvable as above. But now for y ∈ Y \ Q_{1}, we have |Z_{1} : C I_{Z_{1}}(y)| ≥ 8 > |Y : Y ∩ Q_{1}| = 4 or 2, a contradiction.

(6.39) Proposition. b ≠ b_{1}.

Proof . Suppose false. By (6.35) we know b = b_{1} = 1. Choose Y as before. Let L_{1} not be exceptional. Then by (6.38) either [E(L_{1}/Q_{1}), Y] ≠ 1 and [Y, K] ≠ 1 for exactly one component K of L_{1}/Q_{1} or K = F (L_{1}/Q_{1}) and [Y, K] ≠ 1. In any case m_{p}(K) = 1.

Let K̃ = C I_{L_{1}/Q_{1}}(K). Then [Y, K̃] = 1. Choose y ∈ Y. Set W = [y, Z_{1}]. Suppose [W, K̃] = 1. Then we get m_{p}(C I_{M}(W)) ≥ 2, as m_{p}( Aut (K)) ≤ 2. But then |W | = q and W ∩ Z_{3} = 1. We furthermore have [[Z_{1}, K], K̃] = 1. If K ≤ L_{1}/Q_{1}, then we see that Ω_{1}(Z(S)) ∩ [Z_{1}, K] is centralized by E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, a contradiction. So K ≰L_{1}/Q_{1} and then K ≅ L_{2}(r), Sz(r) or J_{1}. As ω acts on K, we get K ≅ L_{2}(q) and [Z_{1}, K] is the natural module, |Z_{1} : Z_{1} ∩ Z_{3}| = q^{2}. Hence |Z_{1} : C I_{Z_{1}}(K × K^{s})| ≤ q^{4}, s ∈ S. As |C I_{Z_{1}}(K) : C I_{Z_{1}}(K) ∩ Z_{3}| ≤ q, we see that |Z_{1}| ≤ q^{5} and so K × K^{s} ≤ L_{1}/Q_{1}, for some s ∈ S. Now C I_{Z_{1}}(K × K^{s}) = 1 and so |Z_{1}| = q^{4}. Furthermore as ω induces a diagonal automorphism we see that p | q̸ − 1. But now p | q + 1. Then every x ∈ [K, Z_{1}] is centralized by E_{p^{3}} and so Z_{3} ∩ [K, Z_{1}] = 1, i.e. C I_{[K,Z_{1}]}(ω) = 1. The same is true for [K^{s}, Z_{1}] and so C I_{Z_{1}}(ω) = 1, contradicting [Z_{1} ∩ Z_{3}, ω] = 1.

We may assume [W, K̃] ≠ 1 for every y ∈ Y ^{♯}. But this implies [W, Y] = 1 otherwise [Y, W] ≤ Z_{1} ∩ Z_{3} is normalized by E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, a contradiction. This now implies that Y acts quadratically and so |Z_{1} : C I_{Z_{1}}(Y)| = q. But now [K̃, W] = 1 as |W | = q and [W, ω] = W.

It remains the exceptional case. Here |Y : Y ∩ Q_{1}| ≤ 4 and so |Z_{1} : Z_{1} ∩ Z_{3}| ≤ 8. Inparticular |Ω_{1}(Z(M)) : Ω_{1}(Z(M)) ∩ Z_{1} ∩ Z_{3}| ≤ 8. As Ω_{1}(Z(S)) ≤ Ω_{1}(Z(M)), we have Ω_{1}(Z(M)) ∩ Z_{1} ∩ Z_{3} ≠ 1. But then there is 1 ≠ z ∈ Z_{1} ∩ Z_{3} centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

§ 7 The case [Z_{2}, Z_{α}] ≠ 1

In this chapter we assume b = b_{2}. By (6.39) we know α ∼ 2. Let (1, 2, 3, …, α) be a path of length b + 1.

(7.1) Lemma. Z_{2} is an F - module with offending subgroup Z_{α}Q_{2}/Q_{2}. Furthermore one of the following holds :

(1) E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ L_{2}(q), q even, Z_{2}/C I_{Z_{2}}(E(L_{2}/Q_{2})) is the natural module

(2) E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q even, Z_{2}/C I_{Z_{2}}(E(L_{2}/Q_{2})) is a direct sum of two natural modules.

(3) L_{2} is solvable and one of the following holds

(a) |Z_{2}/C I_{Z_{2}}(L_{2})| = 4, L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3}

(b) |Z_{2}/C I_{Z_{2}}(L_{2})| = 16 and L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}

(Notice that (3) are special cases of (1) and (2).)

Proof. As [Z_{2}, Z_{α}] ≤ Z_{2} ∩ Z_{α}, we see that Z_{2} is an F - module with offending subgroup Z_{α}Q_{2}/Q_{2}. Now application of (2.1) and (1.31) gives the structure of L_{2}/Q_{2}. Suppose E(L_{2}/Q_{2}) ≅ A_{9}. Then L_{2} ∩ M ≤ L_{1}. So we see that Q_{3} ≤ Q_{2} which is not possible as [Z_{α}, Z_{2}] ≠ 1.

We have to show that orthogonal modules are not possible in case (1) or (2). In the case of the orthogonal module an offending group is not contained in E((L_{2}/Q_{2})/F (L_{2}/Q_{2})). But this contradicts L_{2} ∩ M ≤ L_{1} and so Q_{3}Q_{2}/Q_{2} ≤ E((L_{2}/Q_{2})/F (L_{2}/Q_{2})).

(7.2) Lemma. Z_{1} ≤ Z_{2}, L_{1} = C I_{L_{1}}(Z_{1})(L_{1} ∩ L_{2})

Proof. Suppose false. As b_{1} > b, we get Z_{1} ≤ L_{α} and there is some t ∈ Z_{α} \ Q_{2} with 1 ≠ [Z_{2}, t] = [Z_{1}Z_{2}, t] ≤ Z_{2} by (7.1). This shows Z_{1}Z_{2} ≤ ⟨L_{1} ∩ L_{2}, t⟩ = L_{2}.

Suppose first L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} or L_{2} is nonsolvable. We may assume that Q_{1} = C I_{S}(Z_{1}) and Q_{1} ∩ Q_{2} = C I_{S}(Z_{1}Z_{2}). This shows Q_{1} ∩ Q_{2} ≤ L_{2}. Now Q_{1} ∈ Syl_{2}(⟨Q_{1}^{L_{2}}⟩). So we get some L̃_{2} ≤ L_{2}, L̃_{2} ≰ M, L̃_{2}/O_{2}(L̃_{2}) ≅ L_{2}(q) and Q_{1} ∈ Syl_{2}(L̃_{2}).

We are going to prove the same result for L_{2}/C I_{L_{2}}(Z_{2}) ≅ Σ_{3} ≀ ZZ_{2}. There is some ω ∈ L_{2} \ M, o(ω) = 3, with |Z_{1} : C I_{Z_{1}}(ω)| ≤ 2. If m_{p}(L_{1}) ≥ 3, then ω centralizes some x ∈ Z_{1}, C I_{G}(x) ≤ M, a contradiction.

According to (4.9) assume m_{p}(L_{1}) = 1. Let P ∈ Syl_{3}(L_{2}), then [P, Z_{1}Z_{2}] = [P, Z_{2}] ≅ E_{16}. Let x ∈ Z_{1}Z_{2} with [Q_{2}, x] ≤ Z_{2}. As [x, P] = 1 and Z_{2} ≰ Z_{1}, we have [x, Q_{2}] ≤ C I_{Z_{2}}(P), so |Z_{2}| > 16. Hence there is 1 ≠ x ∈ C I_{Ω_{1}(Z(S))}(P). Set Z̃_{1} = ⟨x^{L_{1}}⟩. If Z̃_{1} is an F - module for L_{1}, we see with [GoLyS, (25.12)] that [O_{r}(L_{1}/Q_{1}), Z̃_{1}] = 1. But this contradicts (4.10). Hence J(S) = J(Q_{1}) and |J(S) : J(S) ∩ Q_{2}| = 4. But now [Q_{1}, [J(S), Z_{2}]] = 1, and so |Q_{1} : Q_{1} ∩ Q_{2}| = 4. In particular ⟨Q_{1}^{L_{2}}⟩Q_{2}/Q_{2} ≅ Σ_{3} × Σ_{3}, and then we get the same result as above.

Application of [Ste, Theorem3] provides us with a subgroup 1 ≠ C ≤ Q_{1} which is normalized by automorphism of Q_{1} of odd order and which is normal in L̃_{2}. Hence as Q_{1} is normalized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, or by O_{r}(L_{1}), we get the contradiction L̃_{2} ≤ M.

This shows Z_{1} ≤ Z_{2} and furthermore Q_{1} < C I_{S}(Z_{1}). But then by construction L_{1} = C I_{L_{1}}(Z_{1})(L_{1} ∩ L_{2}) and Z_{1} ≤ Z_{2}.

(7.3) Lemma. C I_{Z_{2}}(L_{2}) = 1 and (7.1)(2) does not occure.

Proof. By (7.2) C I_{Z_{2}}(L_{2}) ≤ Z_{1} and so N_{G}(C I_{Z_{2}}(L_{2})) ≥ ⟨L_{1}, L_{2}⟩. By (4.14) O_{2}(⟨L_{1}, L_{2}⟩) = 1 and so C I_{Z_{2}}(L_{2}) = 1.

Let L_{2} be as in (7.1)(2). Then as L_{2} ∩M ≤ L_{1}, we see that |Z_{1}| = q^{2}. But now there is 1 ≠ t ∈ Z_{1} with C I_{L_{2}}(t)/O_{2}(C I_{L_{2}}(t)) ≅ L_{2}(q). By (7.2) C I_{G}(t) ≤ M and then L_{2} = ⟨S, C I_{L_{2}}(t)⟩ ≤ M, a contradiction.

In the next series of lemmas we are going to prove b = 2. Let O^{2}(L_{2}/C I_{L_{2}}(Z_{2})) ≅ L_{2}(q). Then Z_{2} is the natural module by (7.1) and |Z_{1}| = q. Set V_{1} = ⟨Z_{2}^{L_{1}}⟩. We get [Q_{1}, V_{1}] = Z_{1}, V′_{1} = 1 iff b > 2.

Let L_{2} be as in (7.1)(3)(b). Then |Z_{1}| = 2. Let Z̃_{2} = C I_{Z_{2}/Z_{1}}(S) (mod Z_{1}) and V_{1} = ⟨Z̃_{2}^{L_{1}}⟩. Also V′_{1} = 1 if b > 2.

Suppose first that V_{1} = ⟨Z_{2}^{L_{1}}⟩. Then we have [Z_{2}, Z_{α}] = Z_{3} = Z_{α−1}. If b > 2, then we have [V_{1}, Z_{2}] = 1 = [V_{1}, Z_{3}] = [V_{1}, Z_{α−1}]. As V_{1} ≤ L_{α−2}, we get V_{1} ≤ L_{α−2} ∩M_{α−1} ≤ L_{α−1}. Hence [V_{1}, V_{α−1}] ≤ V_{α−1}. Furthermore V_{α−1} ≤ G_{2} but Z_{α} ≰ M. Finally C I_{L_{1}}(V_{1}/Z_{1}) ≤ Q_{1}. Otherwise some 1 ≠ ω ∈ L_{1}, ω ∈ E ≅ E_{p^{3}}, centralizes V_{1}. Then ω centralizes Z_{3} and so ω ∈ M_{3}. By (4.2) M = M_{3}. The same applies for m_{p}(L_{1}) = 1 as then [O_{r}(L_{1}), Z_{3}] = 1, which contradicts (4.10). But ⟨L_{1} ∩ L_{2}, L_{2} ∩ L_{3}⟩ = L_{2}, a contradiction.

We are going to prove the same result for L_{2} as in (7.1)(3)(b). We have Z_{α−2} ≤ Q_{1}. As [V_{1}, Q_{1}] ≤ Z_{1}, we get [Z̃_{β}, Z_{α−2}] ≤ Z_{1}, for every β ∈ Δ(1). Suppose [Z̃_{β}, Z_{α−2}] ≠ 1 for some β ∈ Δ(1). Then we may assume that [Z̃_{2}^{g}, Z_{α}] = Z_{3}, for M^{g} = M_{3}. But then [Z_{2}, Z_{α}] = Z̃_{2}^{g} and so [Z̃_{2}^{g}, Z_{α}] = 1, a contradiction.

So we get [Z_{α−2}, V_{1}] = 1 and again V_{1} ≤ L_{α−1}. This shows [V_{1}, V_{α−1}] ≤ V_{α−1}. Furthermore Z_{α} ≰ M.

Suppose next [V_{1}, V_{α−1}] ≤ Z_{α−1}. Then either V_{1} ≤ Q_{α−1} or we have (7.1)(3)(b) with C I_{S}(V_{1}) > Q_{1}. By (4.10) in the latter m_{p}(L_{1}) > 1. Further p | |C I_{L_{1}}(V_{1})|. As L_{2} = ⟨C I_{L_{2}}(Z̃_{2}), S⟩ ≰ M, we get that m_{p}(C I_{L_{1}}(V_{1})) = 1. But now the construction in (4.13) gives a contradiction. So we may assume V_{1} ≤ Q_{α−1}. Then we see V_{1} ≤ L_{α} and so for some t ∈ Z_{α} \Q_{2} we get [V_{1}, t] = [Z_{2}, t]. Then V_{1}Z_{2} ≤ ⟨S, t⟩ = L_{2}. This shows that we are in (7.1)(3)(b). Let u ∈ V_{1}, u ∉ Z_{2}, [u, S] ≤ Z_{2}. Then [u, Q_{2}] ≤ L_{2}. As [u, Q_{2}] ≤ V_{1}, we see Z_{2} ≤ V_{1} and so V_{1} ≤ ⟨L_{1}, L_{2}⟩, a contradiction.

(7.4) Lemma. Let b > 2. Then V_{1} ≤ L_{α−1} and [V_{1}, V_{α−1}] ≰ Z_{α−1}.

(7.5) Lemma. Let b > 2. Then [V_{1}, V_{α−1}, V_{1}] = 1.

Proof. Let W_{2} = ⟨V_{1}^{L_{2}}⟩. Let s ∈ L_{2}. Then ⟨V_{1}, V_{1}^{s}⟩ ≤ Q_{1} ∩ Q_{1}^{s} as b > 2. Hence [V_{1}, V_{1}^{s}] ≤ Z_{1} ∩ Z_{1}^{s} = 1, as M ≠ M^{s}. This shows that W_{2} is abelian. As [V_{1}, V_{α−1}] ≤ W_{2}, we are done.

(7.6) Lemma. Let L_{2} be as in (7.1)(3)(b). Then 3 ∉ σ(M), [V_{1}, V_{α−1}] ≤ V_{1} ∩ V_{α−1} and [V_{α−1} ∩ Q_{1}, V_{1}] = 1.

Proof. We have [Z_{2}, Z_{α} ∩ V_{α−1}] = 1. This implies [Z_{2}, V_{α−1}] = 1. In particular V_{α−1} ≤ L_{1} and so

[V_{1}, V_{α−1}] ≤ V_{1} ∩ V_{α−1} (1)

Hence V_{1} and V_{α−1} act quadratically on each other. Furthermore as m_{3}(L_{2}) ≥ 2 we have

3 ∉ σ(M) (2)

[V_{α−1} ∩ Q_{1}, V_{1}] = 1 (3)

We have [V_{α−1} ∩ Q_{1}, V_{1}] ≤ Z_{1}. If [V_{α−1} ∩ Q_{1}, V_{1}] ≠ 1, then Z_{α} centralizes ⟨Z_{2} ∩ V_{3}, z⟩ with 1 ≠ z ∈ [V_{α−1} ∩ Q_{1}, V_{1}]. But Q_{2} ∈Syl_{2}(C I_{L_{2}}(⟨Z_{2} ∩ V_{3}, z⟩)), and so Z_{α} ≤ L_{1}, a contradiction.

Now we fix the following notation : K is a component or normal psubgroup of L_{1}/Q_{1} with 1 ≠ [V_{α−1}, K] ≤ K and K_{α−1} is the corresponding component in L_{α−1}/Q_{α−1}. Furthermore we may assume that Z_{α} ≰ M. We set Ṽ_{1} = [V_{1}, K].

(7.7) Lemma. Let Ṽ_{1} ≰ Q_{α−1} and L_{2}/Q_{2} be as in (7.1)(3)(b). Let [K_{α−1}, t] = 1, for some 1 ≠ t ∈ Ṽ_{1}. Let K_{α−1}^{(1)} be a component of L_{α−1}/Q_{α−1} with [t, K_{α−1}^{(1)}] ≠ 1. If [[t, V_{α−1}], K_{α−1}] = 1 and K^{(1)} is the group corresponding K_{α−1}^{(1)} in L_{1}/Q_{1}, then [Ṽ_{1}, K^{(1)}] = 1 = [[K^{(1)}, V_{1}], K]. Furthermore there is no E ≤ L_{1}, E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M with [Ṽ_{1}, E] = 1 or [[K^{(1)}, V_{1}], E] = 1. In particular m_{p}(K) = m_{p}(K^{(1)}) = 1.

Proof. The first assertion is obvious by the three subgroup lemma. Assume [E, Ṽ_{1}] = 1. Then there is some 1 ≠ x ∈ [Ṽ_{1}, V_{α−1}] ≤ V_{α−1} with E ≤ C I_{G}(x) and so C I_{G}(x) ≤ M. But [V_{α−1}, Z_{α}] = 1 and so Z_{α} ≤ M, a contradiction. Hence m_{p}(K^{(1)}) = 1.

Let now [E, [Ṽ_{1}, K^{(1)}]] = 1. Then by the same reason there is 1 ≠ x ∈ [t, V_{α−1}] with C I_{G}(x) ≤ M_{α−1}. But [[Ṽ_{1}, V_{α−1}], K^{(1)}] = 1 and so K^{(1)} ≤ M_{α−1}. But this shows M = M_{α−1} by (4.2), which contradicts Z_{α} ≰ M.

(7.8) Hypothesis. L_{2}/Q_{2} is as in (7 .1)(3)(b) and Ṽ_{1} ≰ Q_{α−1}, K is a quasisimple component of L_{1}/Q_{1} with 1 ≠ [K, V_{α−1}] ≤ K.

(7.9) Lemma. Assume (7.8). Let |V_{α−1} : C I_{V_{α−1}}(K)| ≤ 4. Then K/Z(K) ≅ A_{7}, A_{9}, A_{10}, A_{11}, or K/Z(K) is a group of Lie type in even characteristic.

Proof. By (7.6) 3 ∉ σ(M). Let t ∈ Ṽ_{1}. Then |V_{α−1} : C I_{V_{α−1}}(t)| ≤ 4 by (7.6). If [t, K_{α−1}] ≠ 1, we have the assertion with (1.30). So assume [Ṽ_{1}, K_{α−1}] = 1. Then [K_{α−1}, [Ṽ_{1}, V_{α−1}]] = 1. Let K_{α−1}^{(1)} be a component or normal p-subgroup in L_{α−1}/Q_{α−1} with [t, K_{α−1}] ≠ 1 and K_{1}^{(1)} the corresponding group in L_{1}/Q_{1}. Then [K^{(1)}, V_{1}] involves at most two nontrivial irreducible K^{(1)} - modules. By (7.7) m_{p}(K^{(1)}) = 1. Hence by (4.4) m_{p}(C I_{L_{1}}(K^{(1)})) ≥ 2. As p > 3 this shows that there is some E ≤ M, E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, and [[V_{1}, K^{(1)}], E] = 1. But this contradicts (7.7).

(7.10) Lemma. Assume (7.8). If t ∈ V_{α−1} induces some 1 ≠ s ∈ K on Ṽ_{1}, then m_{p}(N_{L_{1}}(K)/C I_{L_{1}}(K)) > 1. If we have the assumptions of (7.7) then m_{p}(N_{L_{1}}(K^{(1)})/C I_{L_{1}}(K^{(1)})) > 1.

Proof. Suppose m_{p}(N_{L_{1}}(K)/C I_{L_{1}}(K)) = 1. Then there is some E ≅ E_{p^{2}}, E ≤ C I(K) with Γ_{E,1}(G) ≤ M and some ν ∈ K, o(ν) = r, r prime, ν^{s} = ν^{−1}. Hence E acts on [[ν, Ṽ_{1}], t] ≤ V_{α−1}, and so Z_{α} ≤ M, a contradiction. The same argument works for K^{(1)}.

(7.11) Lemma. Assume (7.8). Then K ≅ G(r), a group of Lie type ine characteristic two.

Proof. Suppose false. By (7.6), (7.9) and (1.12) K ≅ A_{7}, A_{9}, A_{10}, A_{11} or 3 · M_{22}. Now |V_{α−1} : C I_{V_{α−1}}(K)| ≤ 2^{5}. Hence for t ∈ Ṽ_{1} we have |V_{α−1} : C I_{V_{α−1}}(t)| ≤ 32, where |V_{α−1} : C I_{V_{α−1}}(t)| ≤ 4 for K ≅ A_{7} by (1.12). Suppose [t, K_{α−1}] = 1. Then we may apply (7.7) and get |V_{α−1} : C I_{V_{α−1}}(t)| ≥ 8. Let K ≅ A_{n}, then K_{α−1}^{(1)} is solvable or K_{α−1}^{(1)} ≅ Sz(q). But the latter is impossible as q ≥ 8 and (1.27) implies |[V_{α−1}, t]| ≥ q^{2}. So K_{α−1}^{(1)} is solvable, contradicting (7.10). Hence as in case of K ≅ A_{n} we have C IṼ_{1}(K_{α−1}) = 1.

If K ≅ 3 · M_{22}, then K_{α−1}^{(1)} ≅ L_{2}(q), Sz(q), L_{3}(q), U_{3}(q), or J_{1}. As |[V_{α−1}, t]| ≤ 2^{4}, we get with (1.25), (1.26) and (1.27) that K^{(1)} ≅ L_{2}(q), q ≤ 16, q even, or L_{3}(q), q = 2, 8. By (7.7) m_{p}(K^{(1)}) = 1. But this now contradicts (7.10) as p > 3.

So in any case Ṽ_{1} acts faithfully on K_{α−1}. Suppose first m_{p}(K) = 1. Then by (7.10) Ṽ_{1} involves at least two nontrivial irreducible modules for K. By (1.25) K ≅ 3̸ · M_{22}. Suppose there are at least three modules involved. As |V_{α−1} : C I_{V_{α−1}}(Ṽ_{1})| ≤ 2^{5} we have |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = 2. But then for z ∈ V_{α−1} we have |Ṽ_{1} : C IṼ_{1}(z)| ≤ 4, a contradiction. Hence we have that there are at most two nontrivial irreducible modules involved, which are spin modules or natural modules. But now there is some E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M, [E, Ṽ_{1}] = 1, a contradiction.

So we have p = 5, K ≅ A_{10} or A_{11}. Furthermore for some ω ∈ M, o(ω) = p, we have [Ṽ_{1}, ω] = 1. As in the natural module every element is centralized by a 5 - element in K, we see that we have the spin module as a submodule. Now |V_{α−1} : C I_{V_{α−1}}(Ṽ_{1})| ≤ 8 by (1.12). But this contradicts (1.30).

As 3 ∉ σ(M) we now have :

(7.12) Lemma. Assume (7.8). If K ≅ G(r), r even, then K is one of the following

(i) (S)L_{n}(r), n ≤ 7

(ii) (S)U_{n}(r), n ≤ 7

(iii) Sp(2n, r), n ≤ 3

(iv) Ω^{−}(8, r)

(v) G_{2}(r), ^{3}D_{4}(r), ^{2}F_{4}(r), Sz(r).

(7.13) Lemma. Assume (7.8). Then Ṽ_{1} acts faithfully on K_{α−1}

Proof. Suppose first [K^{(1)}, Ṽ_{1}] = 1. Then by (7.7) m_{p}(K) = m_{p}(K^{(1)}) = 1. Furthermore by (7.10) m_{p}(N_{L_{1}}(K^{(1)})/C I_{L_{1}}(K^{(1)})) > 1. Application of (1.67) shows that K^{(1)} is one of the groups of (7.12). Then we see that there is some ω ∈ L_{α−1}, o(ω) = p, ω induces a field automorphism on K_{α−1}. Hence we may assume t ∈ C I_{K_{α−1}^{(1)}}(ω), for some t ∈ Ṽ_{1}. But now there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α−1} and E acts on [t, V_{α−1}] ≤ Ṽ_{1}. So K^{(1)} ≤ C I_{G}([t, V_{α−1}]) ≤ M_{α−1}. But this contradicts (4.2).

So we have [K_{α−1}, [t, V_{α−1}]] ≠ 1 for some t ∈ Ṽ_{1}, [t, K_{α−1}] = 1. As |V_{α−1} : C I_{V_{α−1}}(t)| ≤ 2^{m_{2}(K)}, we see that K ≅ L_{n}(r), 4 ≤ n ≤ 7, or Sp_{6}(r). Furthermore as [K_{α−1}, [t, V_{α−1}]] ≠ 1, we see that [Ṽ_{1}, K_{α−1}] = 1 and so [K_{α−1}, [t, V_{α−1}]] ≠ 1 for every t ∈ Ṽ_{1} \ Q_{α−1}. We have m_{3}(K^{(1)}) ≤ 1 and so K^{(1)} ≅ L_{2}(q), (S)L_{3}(q), (S)U_{3}(q), Sz(q), J_{1}, or solvable. If q is odd, K^{(1)} is solvable or K^{(1)} ≅ J_{1}, we have t ∈ V_{α−1}, [t, Ṽ_{1}] ≠ 1 and |Ṽ_{1} : C IṼ_{1}(t)| ≤ 4 by (1.12), [GoLyS, (24.1)] and [GoLyS, (25.12)]. Hence Ṽ_{1} involves at most two nontrivial irreducible modules. As p ≠ 3, this would imply [K^{(1)}, Ṽ_{1}] = 1, a contradiction. So we have K^{(1)} ≅ L_{2}(q), (S)L_{3}(q), (S)U_{3}(q), or Sz(q), q even. Now |Ṽ_{1} : C IṼ_{1}(t)| ≤ 2q, 2q^{2} in case of K^{(1)} ≅ (S)L_{3}(q). As there are at least four irreducible nontrivial K^{(1)} - modules involved in V_{1}, we get that [t, K^{(1)}] = 1. Now [[t, Ṽ_{1}], K^{(1)}] ≠ 1 and so we get K^{(1)} ≅ L_{3}(2), |[Ṽ_{1}, t]| = 8 and Ṽ_{1} involves at least three irreducible nontrivial modules on which t induces transvections. As m_{3}(K) ≤ 2, we get K ≅ L_{4}(2) or L_{5}(2), p = 7. But now m_{p}(K × K^{(1)}) = 2 and so there is some ω ∈ M, o(ω) = 7, [Ṽ_{1}, ω] = 1. As every element in the natural module is centralized by some ν ∈ K, o(ν) = 7, we see that some 1 ≠ x ∈ [Ṽ_{1}, t] is centralized by E ≅ E_{49}, Γ_{E,1}(G) ≤ M, a contradiction to Z_{α} ≰ M.

(7.14) Lemma. Let L_{2}/Q_{2} be as in (7.1)(3)(b). There is no component K of L_{1}/Q_{1} with 1 ≠ [K, V_{α−1}] ≤ K.

Proof.

K ≅ (S)L_{n}(r) (1)

By (7.13) we know that Ṽ_{1} is an SC - module. Suppose furthermore every module involved in Ṽ_{1} is of type V (λ), λ some fundamental weight. In what follows we will use (1.23) freely.

Let n = 7. Then by (1.23) we may assume that V (λ_{4}) is not involved. Let V (λ_{3}) be involved. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{10}. Now V_{α−1} contains some z, z a root element in K. But now ⟨V_{α−1}^{C I_{K}(z)}⟩ ≥ O^{2}(C I_{K}(z)). This implies [Ṽ_{1}, z, O^{2}(C I_{K}(z))] = 1, and so V (λ_{3}) cannot be involved. Hence we have that only V (λ_{1}), V (λ_{2}) V (λ_{5}), or V (λ_{6}) may be involved. If both V (λ_{2}) and V (λ_{5}) are involved, then again |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{10} and we get a contradiction as before. So we may assume that just V (λ_{1}), V (λ_{6}) and V (λ_{2}) are involved. Then C IṼ_{1}(S) is centralized by L_{4}(r) in K. As now E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, centralizes Z̃_{2}, we see r = 2. Furthermore there is a p - element ω in L_{1}, o(ω) ≥ 5, such that [Ṽ_{1}, ω] ≠ 1 and [ω, K_{1}] = 1. Let V̂_{1} = [Ṽ_{1}, ω]. Then V̂_{1} cannot involve V (λ_{2}). So we have natural submodules.

As 7^{2} | |L_{6}(2)|, we see that m_{p}(K) = 1. This now implies that we have at

least 7 natural modules involved (p = 127) or if p | |L_{6}(2)| there is a faithful E_{p^{2}} and there are even 8 modules involved.

We have |[V_{α−1}, t]| ≤ 2^{12} for t ∈ Ṽ_{1}. This shows that Ṽ_{1} induces transvections on the natural module and so |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2^{6}. Hence |Ṽ_{1} : C IṼ_{1}(z)| ≤ 2^{7} for every z ∈ V_{α−1}. Then p = 127, and there are exactly 7 natural modules involved. Now some ω, o(ω) = p, centralizes Ṽ_{1}. But we have Z_{α−1} ≤ Ṽ_{1}, and so ω ∈ M_{α−1}. By (4.2) M = M_{α−1}, a contradiction.

Let n = 6. If |V_{α−1} : C I_{V_{α−1}}(K)| > r^{6}, we may argue as before to see that just natural or dual modules are involved. So let first |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r^{6}. Then by (1.23) V (λ_{3}) is not involved. Furthermore either V (λ_{2}) or V (λ_{4}) is not involved. So in any case C IṼ_{1}(S) is centralized by L_{3}(r). This implies r = 2, and there is ω, o(ω) = p, [ω, K] = 1 and [Ṽ_{1}, ω] ≠ 1. As o(ω) > 3, we get again that V̂_{1} = [Ṽ_{1}, ω] just involves natural or dual modules. As p | |L_{5}(2)|, we get p = 7 and there are at least three natural modules involved, or p ≠ 7 and there are at least 8 natural modules involved. Now as in the case n = 7, we get p = 7. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ 8. By (1.16) some t ∈ V_{α−1} induces some s ∈ K on V̂_{1}. Hence there is some ν ∈ K, o(ν) > 2 prime, and ν^{s} = ν^{−1}. Then we may assume [ω, ν] = 1, and so ω acts on [[V̂_{1}, ν], s]. As there is some ω_{1} ∈ K, o(ω_{1}) = 7, C I_{[[V̂_{1},ν],s]}(ω_{1}) ≠ 1 and [ω_{1}, ω] = 1, we get some 1 ≠ U ≤ [V_{1}, V_{α−1}] which is normalized by E ≅ E_{49} with Γ_{E,1}(G) ≤ M, a contradiction.

Let next n = 5. As before not both V (λ_{2}) and V (λ_{3}) are involved. So assume neither V (λ_{2}) nor V (λ_{3}) are involved. Then C IṼ_{1}(S) is centralized by SL_{3}(r) and so r = 2. Hence m_{p}(K) = 1 and so either p | |L_{4}(2)| and there are at least 6 natural modules involved or p = 31 and there are at least 5 natural modules involved. But now as in the case K ≅ L_{7}(2) we get a contradiction.

So we have V (λ_{2}) is involved and |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{3}. As V (λ_{2}) is involved just once, we see that there has to be involved V (λ_{1}) or V (λ_{4}) too. This now gives |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}. As before we see |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r^{4} and so we have |V_{α−1} : C I_{V_{α−1}}(K)| = r^{4}. Now either L_{2}(r)×L_{2}(r) or SL_{3}(r) centralizes C IṼ_{1}(S). This implies r = 2. But then m_{p}(K) = 1 and E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M centralizes Ṽ_{1}, a contradiction.

Let now n = 4. Suppose V (λ_{2}) is involved. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r^{3} and so there is at most one further nontrivial irreducible module involved which then has to be the natural module. Suppose V (λ_{2}) is a submodule. Every v ∈ V (λ_{2}) is centralized by E ≅ E_{p^{2}}. If p | r^{2} − 1, we get m_{p}(K) = 1. But now E ≅ E_{p^{2}} centralizes V (λ_{2}), a contradiction.

So in any case by (1.20) we have a natural submodule and so p | r̸ − 1. Now m_{p}(K) ≤ 2 and so there is some ω ∈ C I(K), o(ω) = p with [Ṽ_{1}, ω] ≠ 1. Hence there are at least two natural modules for r > 2 and at least three natural modules for r = 2 involved. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{2} and so there is some t ∈ V_{α−1} inducing some 1 ≠ s ∈ K on Ṽ_{1} by (1.16). If p | |SL_{3}(r)| we get as before some 1 ≠ U ≤ [Ṽ_{1}, V_{α−1}] normalized by some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

So we may assume that there are four natural modules involved. Now |V_{α−1} : C I_{V_{α−1}}(K)| = r^{4}, and |Ṽ_{1} : Ṽ_{1} ∩ Q_{α}| = r^{3}. Hence Z_{α−1} ≤ [V_{α−1}, Ṽ_{1}]. But as m_{p}(K) = 1, we see that Z_{α−1} is centralized by ω, o(ω) = p, ω ∈ E, E ≅ E_{p^{3}}. This contradicts (4.2) and M ≠ M_{α−1}.

Let n = 3. Now we have at most two nontrivial modules involved. If m_{p}(K) = 1, we get some ω ∈ C I(K), [Ṽ_{1}, ω] ≠ 1, so there are exactly two nontrivial irreducible modules involved and then p | r + 1. But now |V_{α−1} : C I_{V_{α−1}}(K)| = r^{2} and some t ∈ V_{α−1} induces s ∈ K on Ṽ_{1} by (1.16). Hence again some 1 ≠ U ≤ [V_{1}, V_{α−1}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. This shows m_{p}(K) = 2 and so p | r − 1, i.e. r > 2. By (1.20) the natural module is a submodule of Ṽ_{1}/Z_{1}. Now in the natural module every element is centralized by L_{2}(r) and so there is some ω, o(ω) = p, which does not normalize some natural module in Ṽ_{1}. But we may assume [ω, s] = 1 for some 1 ≠ s ∈ K where s acts as t ∈ V_{α−1} on Ṽ_{1}. Now as before we get 1 ≠ U ≤ [V_{1}, V_{α−1}] normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M.

Let finally n = 2. Now |V_{α−1} : C I_{V_{α−1}}(K)| = r and Ṽ_{1}/C IṼ_{1}(K) is the natural module. This implies that some 1 ≠ x ∈ [V_{α−1}, Ṽ_{1}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M.

So we have seen that there is some module involved which is not a fundamental module. These modules are described in (1.22).

Let T be such a module. If Ṽ_{1}/C IṼ_{1}(K) ≅ T, then we see immediately that C IṼ_{1}(S) is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

Now application of (1.24) shows that r is a square and either T ≅ V (λ_{1}) ⨂ V (λ_{1})^{σ} or V (λ_{n}) ⨂ V (λ_{n})^{σ}. As 3 ∉ σ(M) and (7.6), we get 3 ≤ n ≤ 4. But now we see that |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{n−1}, contradicting (1.15).

K ≅ (S)U_{n}(r) (2)

By (1.22) every irreducible module involved in Ṽ_{1} is a fundamental module or V (λ_{1} + λ_{l}). But the latter contradicts (1.24) and (1.15). So we just have fundamental modules involved.

Let n = 6 or 7. Then p | r + 1. If V (λ_{3}) is not involved then C IṼ_{1}(S) is centralized by SU_{3}(r), and so by some E ≅ E_{p^{2}}, E ≤ K with Γ_{E,1}(G) ≤ M, a contradiction.

So we have K ≅ (S)U_{6}(r) and V (λ_{3}) is involved. As C I_{V (λ_{3})}(S) is centralized by (S)L_{3}(r^{2}), we see that there has to be some additional module involved. This has to be the natural module. Now |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{8}. But then Ṽ_{1} just contains elements which induce transvections on the natural K_{α−1} - module. This implies |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ r. Then |Ṽ_{1} : C IṼ_{1}(z)| ≤ 2r for every z ∈ V_{α−1}, a contradiction.

If n = 5 then we just have natural modules involved, a contradiction as p | r + 1.

So we are left with n = 4. Let first Ṽ_{1}/C IṼ_{1}(K) = T be irreducible. If T ≅ V (λ_{2}), then [t, Ṽ_{1}] contains elements centralized by Sp_{4}(r). Hence p | r̸^{2} − 1. This shows m_{p}(K) = 1. But then here is some 1 ≠ x ∈ [Ṽ_{1}, V_{α−1}] with E ≤ C I_{G}(x), E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction. So T ≅ V (λ_{1}), the natural module.

If there are two modules involved, then |V_{α−1} : C I_{V_{α−1}}(K)| = r^{4}. If the natural Ω_{6}^{−}(r) - module is involved we get K ≅ Ω_{6}^{−}(2) by (1.39) and both modules are natural Ω_{6}^{−}(2) - modules. Hence either both modules are natural Ω_{6}^{−}(2) - modules or U_{4}(r) - modules. Now by (1.39) we get some z ∈ Ṽ_{α−1} with |Ṽ_{1} : C IṼ_{1}(z)| ≥ r^{4}, 2^{6} in case of Ω_{6}^{−}(2) - modules. Hence |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≥ r^{4}/2, 2^{5}, respectively. The latter is not possible. In the former we get that T is the natural U_{4}(r) - module. But then we get some x ∈ Ṽ_{1} with |[V_{α−1}, x]C IṼ_{1}(K)/C IṼ_{1}(K)| = r^{4}, contradicting (1.50).

K ≅ S̸p_{2n}(r), n ≤ 3 (3)

By (1.22) Ṽ_{1} just involves V (λ), λ fundamental weight. Let first n = 3. If |V_{α−1} : C I_{V_{α−1}}(K)| > r^{5}, then there is some root element z ∈ V_{α−1} with ⟨V_{α−1}^{C I_{K}(z)}⟩ ≥ O^{2}(C I_{K}(z)). This implies that [[Ṽ_{1}, z], O^{2}(C I_{K}(z))] = 1 and so there are just natural modules involved.

Suppose T is the natural module, T a submodule of Ṽ_{1}. As every x ∈ T is centralized by Sp_{4}(r), we see that p | r̸^{2} − 1. Hence m_{p}(K) = 1 and there is some ω, o(ω) = p, which acts nontrivially on the set of natural modules involved in Ṽ_{1}. So there are at least three of them. Now Ṽ_{1} centralizes in the natural module a subspace of codimension two. This gives |Ṽ_{1} : Ṽ_{1}∩Q_{α}| ≤ r^{4}. Then |Ṽ_{1} : C IṼ_{1}(x)| ≤ 2r^{4} for x ∈ V_{α−1}. Hence V_{α−1} induces transvections on T, i.e. |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r. But then |V_{α−1} : C I_{V_{α−1}}(Ṽ_{1})| ≤ r, contradicting the fact that V_{α−1} involves more than one natural module.

Let T ≅ V (λ_{3}) be a submodule of Ṽ_{1}/C IṼ_{1}(K). Now r^{2} ≤ |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r^{5}. Hence there are at most two such modules involved. Then m_{p}(K) ≥ 2, in particular r ≠ 2. If there are two such modules involved, then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}. But then for some t ∈ T there is |[t, V_{α−1}]C IṼ_{1}(K)/C IṼ_{1}(K)| = r^{4}, contradicting (1.50). Now let T_{α−1} be the corresponding module of K_{α−1}. We have [T ∩ Q_{α−1}, T_{α−1}] = 1, otherwise Z_{α−1} ≤ T. As m_{p}(K) ≥ 2 we have p | r^{2} − 1 and so every t ∈ T is centralized by some ω ∈ K with o(ω) = p. Now we have [T, T_{α−1}] ≤ T ∩ T_{α−1} and a contradiction with (1.50).

It remains the case that V (λ_{2}) is a submodule. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}. If Ṽ_{1}/C IṼ_{1}(K) ≅ V (λ_{2}), then C IṼ_{1}(S) is centralized by L_{2}(r)×L_{2}(r) and so p | r̸^{2} − 1, i.e. m_{p}(K) = 1. But now some 1 ≠ U ≤ [Ṽ_{1}, V_{α−1}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

Hence we have |V_{α−1} : C I_{V_{α−1}}(K)| = r^{5} and so some natural module is involved on which Ṽ_{1} induces transvections. So |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ r and then |Ṽ_{1} : C IṼ_{1}(z)| ≤ 2r, for z ∈ V_{α−1}, a contradiction.

Let now n = 2. There are just natural or dual modules involved. Suppose there are at least two such modules involved. As |V_{α−1} : C I_{V_{α−1}}(K)| ≤ r^{3}, we get that Ṽ_{1} has to induce transvections on one of these modules. Hence |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ r and so |Ṽ_{1} : C IṼ_{1}(z)| ≤ 2r for z ∈ V_{α−1}. This shows r = 2. But now |V_{α−1} : C I_{V_{α−1}}(K)| = 2, a contradiction.

So we may assume that Ṽ_{1}/C IṼ_{1}(K) is the natural module. If m_{p}(K) = 1, we get a contradiction as before. So assume m_{p}(K) = 2, i.e. p | r^{2} − 1. Now every x ∈ Ṽ_{1} is centralized by some ω ∈ K, o(ω) = p. We have C IṼ_{1}(K)∩[Ṽ_{1}, V_{α−1}] = 1. If V_{α−1} contains a group R with |[Ṽ_{1}/C IṼ_{1}(K), R]| = r, then F ≅ E_{p^{3}} acts on [R, Ṽ_{1}] and so some 1 ≠ x ∈ [Ṽ_{1}, V_{α−1}] is centralized by E ≤ F, |E| = p^{2}, a contradiction. Hence we get that ⟨V_{α−1}, V_{α−1}^{g}⟩ = O^{2}(P), for some g ∈ K and some parabolic P of K. Now we see |[O^{2}(P), Ṽ_{1}]Z_{1}/Z_{1}| = r^{4}. Hence we have that Ṽ_{1}/Z_{1} is the natural module. But then every x ∈ Ṽ_{1} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

K ≅ Ω̸_{8}^{−}(r) (4)

By (1.22) Ṽ_{1} just involves V (λ), λ some fundamental weight. Furthermore just V (λ_{1}) or V (λ_{3}) are involved by (1.15). Let V (λ_{1}) be a submodule. As every element in V (λ_{1}) is centralized by Ω_{6}^{−}(r) or Sp_{6}(r), we get p | r̸^{2} − 1. Hence m_{p}(K) = 1. If there are two natural modules involved, then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}. So for some module Ṽ_{1} centralizes a subgroup of index r^{2}. By (1.45) we get |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ r. Now |Ṽ_{1} : C IṼ_{1}(z)| ≤ 2r for z ∈ V_{α−1}, a contradiction. Hence we have that there is exactly one natural module involved. If V (λ_{3}) is involved then we get |V_{α−1} : C I_{V_{α−1}}(K)| = r^{6}, but a group of that large order cannot act quadratically on the natural module.

So we have Ṽ_{1}/C IṼ_{1}(K) is the natural module. But now some ω ∈ C I(K), o(ω) = p, centralizes Ṽ_{1}. Hence there is some ω_{1}, o(ω_{1}) = p, which induces some field automorphism on K. But this contradicts (4.12).

Let now V (λ_{3}) be a submodule. Then |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}. In particular we have Ṽ_{1}/C IṼ_{1}(K) ≅ V (λ_{3}). By (1.20) Ṽ_{1}/Z_{1} ≅ V (λ_{3}). Furthermore as V (λ_{3}) is not an F - module by (1.18), we see Z_{α−1} ≤ Ṽ_{1} and so m_{p}(K) ≥ 2. Hence p | r^{2} − 1. This shows that V_{α−1} does not contain elements of type a_{2} or c_{2}, as those are centralized by L_{2}(r^{2}) × L_{2}(r), Sp_{4}(r), respectively, and so they are centralized by F ≅ E_{p^{2}}. But now as |V_{α−1} : C I_{V_{α−1}}(K)| ≥ r^{4}, we see that there is some t ∈ V_{α−1} with |Ṽ_{1} : C IṼ_{1}(t)| ≥ r^{8}. This would imply |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≥ r^{7}, a contradiction.

(5) The lemma holds

Suppose false. By (1.40), (1.27) and (1) - (4) we just have to treat the case K ≅ G_{2}(r). Now Ṽ_{1}/C IṼ_{1}(K) is the 6 - dimensional natural module. Then every x ∈ Ṽ_{1} is centralized by some ω ∈ M with o(ω) = p. This implies [Ṽ_{1} ∩Q_{α−1}, V_{α−1}] = 1. Hence V_{α−1} offends as an F - module. This yields with (1.37) that |V_{α−1} : C I_{V_{α−1}}(K)| = r^{3}. As [Ṽ_{1}, V_{α−1}] ∩ C IṼ_{1}(K) = 1, we see that for ⟨V_{α−1}, V_{α−1}^{g}⟩ = O^{2}(P), P some parabolic, we get |[Ṽ_{1}, O^{2}(P)]Z_{1}/Z_{1}| = r^{6}. Hence Ṽ_{1}/Z_{1} is the natural module. But now every x ∈ Ṽ_{1} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

(7.15) Lemma. Assume that L_{2}/Q_{2} is as in (7.1)(3)(b) and Ṽ_{1} ≰ Q_{α−1}. If K is some normal p-subgroup of L_{1}/Q_{1}, then [K, V_{α−1}] = 1.

Proof. Suppose false. We have Ṽ_{1} = ⟨t | |V_{α−1} : C I_{V_{α−1}}(t)| ≤ 2⟩. As p > 3, we see that [Ṽ_{1}, K_{α−1}] = 1. But now by (7.7) m_{p}(K) = m_{p}(K_{1}) = 1. Furthermore with (7.10) we get a contradiction.

Let now K be any component of L_{1}/Q_{1} with [V_{α−1}, K] ≠ 1. We choose t ∈ V_{α−1} with K^{t} ≠ K. As 3 ∉ σ(M) by (7.6) we have m_{3}(K) ≤ 1, i.e. K ≅ L_{2}(r), L_{3}(r), U_{3}(r), Sz(r) or J_{1}. If m_{p}(K) ≥ 2, there is some 1 ≠ u ∈ [V_{1}, t] centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. So we may assume K ≅ L_{2}(r), Sz(r) or J_{1}.

Let K_{α−1} × K̃_{α−1} be the components corresponding to K × K^{t} in L_{α−1}/Q_{α−1}. Suppose there is z ∈ Ṽ_{1} \ Q_{α−1} with [z, K_{α−1} × K̃_{α−1}] = 1. Then as [K_{α−1} × K̃_{α−1}, [V_{α−1}, z]] = 1, we get C I_{G}([V_{α−1}, z]) ≤ M_{α−1}. But this is impossible as there is some p - element in L_{1} centralizing some 1 ≠ u ∈ [V_{α−1}, z]. So we have that Ṽ_{1} acts faithfully on K_{α−1} × K̃_{α−1}. This implies |V_{α−1} : C I_{V_{α−1}}(K)| > 2 and so by (1.11) K ≅ L_{2}(r), r even. Now |V_{α−1} : C I_{V_{α−1}}(K)| = r^{2}, but this is impossible as |C I_{K×K^{t}}(t)|_{2} = r > 2. So we have

(7.16) Lemma. If L_{2}/Q_{2} is as in (7.1)(3)(b), then Ṽ_{1} ≤ Q_{α−1}.

(7.17) Lemma. [V_{α−1}, K] = 1 for every component K of L_{1}/Q_{1}, if L_{2}/Q_{2} is as in (7.1)(3)(b)

Proof. By (7.16) we just have to handle the case Ṽ_{1} ≤ Q_{α−1}. Now V_{α−1} induces transvections to Z_{α−1} on Ṽ_{1}. Hence K ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2) or A_{n}. Furthermore Ṽ_{1}/C IṼ_{1}(K) is irreducible. If m_{p}(K) = 1, we get some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, with [Ṽ_{1}, E] = 1, a contradiction to Z_{α−1} ≤ Ṽ_{1}.

As 3 ∉ σ(M), we see that K ≅ L_{6}(2), L_{7}(2), p = 7 or A_{10}, A_{11}, p = 5. But in both cases (p = 7, p = 5), there is ω ∈ M, o(ω) = p with [ω, K] = 1 and [ω, Ṽ_{1}] = 1. Now every x ∈ Ṽ_{1} is centralized by ω_{1} ∈ K with o(ω_{1}) = p. This gives some 1 ≠ x ∈ [Ṽ_{1}, V_{α−1}] with ⟨ω, ω_{1}⟩ ≤ C I_{G}(x), and so Z_{α} ≤ C I_{G}(x) ≤ M, a contradiction.

(7.18) Proposition. L_{2}/Q_{2} is as in (7.1)(1) or (7.1)(3)(a).

Proof. Let K be one of the exceptional components of L_{1}/Q_{1} in (4.9), [K, V_{α−1}] ≠ 1. Then [K, V_{α−1}] ≤ K. This will also handle the case of (4.9)(ii). Now |V_{α−1} : C I_{V_{α−1}}(K)| ≤ 4. Let Ṽ_{1} ≤ Q_{α−1}. Then V_{α−1} induces transvections on Ṽ_{1} and so [O_{r}(K), Ṽ_{1}] = 1 and we have the exceptional case (4.9)(ii). But then (4.10) shows Z_{α} ≤ C I_{G}([Ṽ_{1}, V_{α−1}]) ≤ M, a contradiction. Thus we have Ṽ_{1} ≰ Q_{α−1}. If t ∈ Ṽ_{1} with [t, K_{α−1}] = 1, then we see [K_{α−1}, [t, V_{α−1}]] = 1. Now m_{p}(K) = 1. But then m_{p}(N_{L_{1}/Q_{1}}(K)/C I_{L_{1}/Q_{1}}(K)) = 1, which contradicts (7.10).

Hence we have that Ṽ_{1} acts faithfully on K_{α−1}. Now |Ṽ_{1} : C IṼ_{1}(t)| ≤ 8, t ∈ V_{α−1}. Application of (1.41) shows that we are in (4.9)(ii) andK/O_{r}(K) ≅ L_{2}(7) or A_{5}. As |Ṽ_{1} : C IṼ_{1}(t)| ≤ 8, we see with (1.51) that [O_{r}(K), Ṽ_{1}] = 1, the same contradiction as above.

From now on we have that O^{2}(L_{2}/C I_{L_{2}}(Z_{2})) ≅ L_{2}(q). Furthermore there is C ≅ ZZ_{q−1}, C ≤ L_{1} ∩ L_{2}. We let x be a Zsigmondy prime, x | q − 1 or x = 9 if q = 64, and ω ∈ C with o(ω) = x.

We have |Z_{1}| = q and m_{p}(C I_{L_{1}}(Z_{1})) ≥ 3. Furthermore if ω ≠ 1, then ω ∉ C I_{L_{1}}(Z_{1}).

If b > 2, then we know by (7.4) that V_{1} ≤ L_{α−1} and V_{α−1} ≤ L_{2} but V_{α−1} ≰ M. Hence the main contradiction in that what follows will come from V_{α−1} ≰ M.

We now fix some notation. By (7.4) V_{1} ≰ Q_{α−1}. Hence there is some component or normal p-subgroup K_{α−1} of L_{α−1}/Q_{α−1} with [V_{1}, K_{α−1}] ≠ 1. Let K be the corresponding group in L_{1}/Q_{1}. Set Ṽ_{α−1} = [V_{α−1}, K_{α−1}]. Until further notice assume [V_{1}, K_{α−1}] ≤ K_{α−1}.

(7.19) Lemma. Let b > 2. Then K^{ω} = K.

Proof. Suppose false. Let T = K_{1} × … × K_{y} be an orbit of ⟨ω⟩, y = x, or y = 3 and x = 9. Then ω centralizes in T a subgroup isomorphic to K, or x = 9 and m_{3}(K) = 1. Now (1.63) implies m_{p}(K) = 1, and so K ≅ L_{2}(r), Sz(r), J_{1}, or x = 9 and y = 3.

We have Ṽ_{α−1} = ⟨t | |V_{1} : C I_{V_{1}}(t)| ≤ qu⟩, where u = r, or u = 4 for K ≅ L_{2}(7), u = 2 in the remaining cases. We show

Ṽ_{α−1} ≤ L_{1} (1)

On V_{1}/Z_{1} there is some E ≅ E_{p^{y}}. If t ∈ Ṽ_{α−1} \ L_{1}, then [t, Z_{1}] = Z_{3} and so |V_{1} : C I_{V_{1}}(t)Z_{1}| ≤ u. Now we have that no F ≅ E_{p^{2}}, F ≤ E, centralizes a group of order greater then u. This implies u ≠ 2.

In E there is some F such that |C I_{V_{1}/Z_{1}}(F)| ≥ 2^{(y−2)d(p)}, d(p) the minimal dimension of a cyclic ZZ_{p} - module. Let u = 4, then we get y = 3 and d(p) = 2. But this implies p = 3, a contradiction to o(ω) = 3, or 9.

Let now u = r and y ≥ 4. Then we may choose p with d(p) ≥ m, r = 2^{m}. Now |C I_{V_{1}/Z_{1}}(F)| ≥ 2^{2m} > u, a contradiction.

Hence y = 3, and so d(p) ≤ m. Let z ∈ Ṽ_{1}. Then we may assume |Ṽ_{α−1} : C IṼ_{α−1}(z)| ≤ qr. Let [z, T_{α−1}] ≠ 1. We have q = 4 or q = 64.

If [K_{1}, V_{1}] involves at most two nontrivial irreducible modules, we see that [K_{3}, [K_{1}, V_{1}]] = 1 and so |C I_{V_{1}/Z_{1}}(F)| ≥ r^{2} = 2^{2m}, a contradiction.

So we have |[Ṽ_{α−1}, z]| ≥ r^{4} and then r^{3} ≤ q. This implies q = 64 and r = 4, K_{1} ≅ L_{2}(4), p = 5. But now |C I_{V_{1}/Z_{1}}(F)| ≥ 16 > 4 = u. So we may assume [Ṽ_{1}, T_{α−1}] = 1. Then [Ṽ_{α−1}, Ṽ_{1}] ≤ Z_{α−1} and so obviously Ṽ_{α−1} ≤ L_{1}.

By (1) we now have Ṽ_{α−1} ≤ L_{1}. As [V_{1}, Ṽ_{α−1}] ≠ 1, we get Ṽ_{α−1} ≰ Q_{1}. Let 1 ≠ t ∈ Ṽ_{α−1}. If [T, t] = 1, then [t, V_{1}] contains elements z centralized by F ≅ E_{p^{2}}, F ≤ T and so V_{α−1} ≤ C I_{G}(z) ≤ M, a contradiction. Hence Ṽ_{α−1}/Ṽ_{α−1} ∩ Q_{1} acts faithfully on T. We have that

Ṽ_{α−1} = ⟨t | |V_{1}/Z_{1} : C I_{V_{1}/Z_{1}}(t)| ≤ uq⟩,

u as above. Assume that [[Ṽ_{1}, K_{1}], K_{y−1} × K_{y}] = 1. Then we get again z ∈ [[Ṽ_{1}, K_{1}], t] with V_{α−1} ≤ C I_{G}(z) ≤ M, a contradiction.

Hence as y ≥ 3, we see that there are at least 2^{y−2} nontrivial irreducible K_{1} - modules involved. Now if K_{1} ≅ L̸_{2}(7), then uq ≥ u^{2^{y−2}}. Suppose q ≠ 64. Then q ≤ 2^{y−1} by Fermat’s lemma. So

2^{y−1} ≥ u^{2^{y−2}−1} ≥ 2^{2^{y−1}−2}.

Hence y − 1 ≥ 2^{y−1} − 2 and then y = 3, u = 4 = q. So in any case y = 3. But now

[K_{3}, [Ṽ_{1}, K_{1}]] = 1

As an element of order 3 acts transitively on {K_{1}, K_{2}, K_{3}}, we get

[[Ṽ_{1}, K_{2}], K_{1}] = 1

too, a contradiction.

So assume q = 64. Then y = 3 and we have at least 4 nontrivial irreducible modules involved. Now

64u ≥ u^{4}

and so u = 4, as u > 2. Hence K_{1} × K_{2} × K_{3} ≅ L_{2}(4) × L_{2}(4) × L_{2}(4). Choose ν ∈ L_{1} ∩ L_{2} with o(ν) = 7. Then [ν, K_{1} × K_{2} × K_{3}] = 1. By (1.63) N_{G}(⟨ν⟩) ≤ M, a contradiction.

So it remains K_{1} ≅ L_{2}(7). Then there are at least 3^{y−2} nontrivial irreducible modules involved. Hence

4q ≥ 2^{3^{y−2}}

This implies y + 1 ≥ 3^{y−2} or q = 64. But in both cases y = 3. Now by the same argument as above, we have that there are at least 9 nontrivial irreducible modules involved. So

2^{9} ≤ 4q ≤ 2^{8},

a contradiction.

(7.20) Lemma. Let b > 2. If K_{α−1} is sporadic or a group of Lie type G(r), r odd, then m_{p}(K_{α−1}) = 1, Ṽ_{α−1} ≤ M, [Ṽ_{α−1} ∩ L_{1}, K] = 1 and [Ṽ_{1}, K_{α−1}] ≤ Q_{α−1}, or K/Z(K) ≅ L_{2}(5), L_{2}(7), L_{2}(9), U_{3}(3), P Sp_{4}(3), or 2G_{2}(3)′.

Proof. Let |V_{1} : C I_{V_{1}}(K_{α−1})| = 2^{4}. Then

Ṽ_{α−1} = ⟨t | |V_{1} : C I_{V_{1}}(t)| ≤ 2^{u}q⟩

Furthermore u ≤ 4 by (1.12) or K/Z(K) ≅ U_{4}(3). Suppose K is none of the groups listed in the lemma.

Ṽ_{α−1} ≤ M (1)

If K ≅ 3·U_{4}(3), then m_{3}(K) = 5 and some E ≅ E_{9} centralizes a subgroup of order 2^{6} in Ṽ_{1}/C IṼ_{1}(K). So we have (1).

So assume (1) to be false. We have u ≥ 2. Let u = 2. Then we get p = 3 and m_{3}(L_{1}) = 3. Application of (1.12) shows K ≅ M_{12}, M_{22}, 3 · M_{22}, M_{24}, or J_{2}. In particular m_{3}(K) = 2. Now (7.19) shows that ω induces an automorphism on K normalizing a Sylow 2 - subgroup. Hence we get [ω, K] = 1, as o(ω) ≠ 3, by (1.9). But now (1.63) implies N_{G}(⟨ω⟩) ≤ M, if o(ω) ≠ 1. Hence ω = 1 and so q = 2. Let now z ∈ Ṽ_{1}, [z, K_{α−1}] ≠ 1, which exists as V_{α−1} ≰ M. Then |V_{α−1} : C I_{V_{α−1}}(z)| ≤ 8 or K ≅ 3 · M_{22} and |V_{α−1} : C I_{V_{α−1}}(z)| ≤ 32. Application of (1.25) shows K ≅ M_{22}, or 3 · M_{22} and |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = 2 in the former. But then |Ṽ_{1} : C IṼ_{1}(t)| = 2, which implies t ∈ M, a contradiction.

So we are left with K_{α−1} ≅ 3·M_{22} and Ṽ_{α−1} involves just 12 - dimensional modules by (1.13) anyway. Furthermore |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 16 and by (1.25) |Ṽ_{α−1} : C IṼ_{α−1}(K)| ≥ 8. As Z_{1} ≰ V_{α−1}, we have V_{α−1} ∩ Q_{1} = 1. The 12 - dimensional 3 · M_{22} - module W is not an F_{1} - module. Otherwise an offending group A is contained in 3 · M_{22}, as |W : C I_{W} (x)| = 2^{6} for x ∈ Aut(3 · M_{22}) \ 3 · M_{22}. Hence as W is not an F - module by (1.18), we see that |A| = 8 and |W : C I_{W} (A)| = 16 = |W : C I_{W} (a)|, a ∈ A^{♯}. Now U = ⟨C I_{3·M_{22}}(a) | a ∈ A^{♯}⟩ ≅ 2^{4} · A_{6}. But then we get that |W : C I_{W} (O_{2}(U))| = 16, a contradiction to (1.18) again.

Hence Ṽ_{α−1} is not an F_{1} - module for 3 · M_{22} and we get that |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 8. So we see that |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = 8, |Ṽ_{α−1} : C IṼ_{α−1}(K)| = 16. But now [Ṽ_{1}, Z(K_{α−1})] = 1 and then |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| = 2^{2f}, as Z(K_{α−1}) acts fixed point freely. But this contradicts |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| = 32. This proves (1).

Now assume that for some t ∈ V_{α−1} ∩L_{1} we have [K, t] ≠ 1. If K/Z(K) ≅ U_{4}(3), and Z(K) ≠ 1, we get with (1.55) that t centralizes F ≅ E_{27} in K. Now some 1 ≠ u ∈ [Ṽ_{1}, t] is centralized by E ≅ E_{9}, implying Z_{α} ≤ M, a contradiction.

Suppose m_{p}(K) ≥ 2. If K ≅ G(r), then |Ṽ_{1} : C IṼ_{1}(K)| = 2. As |Ṽ_{1}/Z_{1} : C I_{Ṽ_{1}/Z_{1}}(Z_{α})| > 2, we see that Ṽ_{1} ∩ C I_{L_{α−1}/Q_{α−1}}(K_{α−1}) ≠ 1. Hence there is some component K_{α−1}^{(1)} with [K_{α−1}^{(1)}, K_{α−1}] = 1 and [K_{α−1}^{(1)}, Ṽ_{α−1}] = 1. The same holds for Ṽ_{1}. Furthermore as M ≠ M_{α−1}, we get with (4.2) that Z_{α−1}∩Ṽ_{1} = 1 and so |Ṽ_{1} : C IṼ_{1}(t)| ≤ 2, a contradiction.

So let now K be sporadic. Suppose |Ṽ_{1} : C IṼ_{1}(t)| ≤ 2^{u}. Then we get with (1.25) and (1.11) that u ≥ 3 and K ≅ 3 · M_{22}, in which case u = 4. Hence |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 16. Now Ṽ_{1} involves just one nontrivial irreducible module, the 12 - dimesional module. If m_{p}(K) = 1, we get that [Ṽ_{1}, E] = 1 for some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, and so Z_{α} ≤ C I([Ṽ_{1}, V_{α−1}]) ≤ M, a contradiction. This proves p = 3 and so q = 2. But now as before we get a contradiction.

So we have |Ṽ_{1} : C IṼ_{1}(t)| > 2^{u}, in particular Z̃_{α−1} ∩ Ṽ_{1} ≠ 1. In this case by (4.2) |Ṽ_{1} : Ṽ_{1}∩Q_{α−1}| ≤ 2^{u}, and so |Ṽ_{1}/Z_{1} : C I_{Ṽ_{1}/Z_{1}}(Z_{α−1})| ≤ 2^{u}. As Z_{α−1} ≰ M, we see u ≥ 2, p = 3 in case of u = 2. If p = 3, then as before we see q = 2. Furthermore we get K ≅ 3 · M_{22} in any case and |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 16 and Ṽ_{1} just involves 12 - dimensional modules. We have [ω, K] = 1 and |Ṽ_{1}Z_{2}/Z_{2} : C I_{Ṽ_{1}Z_{2}/Z_{2}}(ω)| ≤ 16. Hence |Ṽ_{1} : Ṽ_{1} ∩ Z_{2}| ≤ 16 for ω ≠ 1. But K ≴ L_{2}(q), so we see ω = 1. Now we may argue as before.

Hence we have [Ṽ_{α−1} ∩ L_{1}, K] = 1. If m_{p}(K) ≥ 2, then [Ṽ_{α−1} ∩ L_{1}, V_{1}] is centralized by K, so V_{1} induces transvections on Ṽ_{α−1}, a contradiction.

As Z_{α} ≰ M, we have |Z_{α} : Z_{α} ∩ Q_{2}| = 2 for q = 2 and so Ṽ_{α−1} ≤ Q_{2} in any case. Now we have [Ṽ_{1}, Ṽ_{α−1}] = 1 and so Ṽ_{1} ≤ C I(K_{α−1}).

(7.21) Lemma. Let b > 2. If K ≅ A_{n}, n = 7 or n ≥ 9, then m_{p}(K_{α−1}) = 1, Ṽ_{α−1} ≤ M and [Ṽ_{α−1} ∩ L_{1}, K] = 1. Furthermore [Ṽ_{1}, K_{α−1}] ≤ Q_{α−1}.

Proof.

q = 2 (1)

Suppose q > 2. We have [ω, K] = 1. Hence m_{p}(K) = 1 by (1.63) and so p ≠ 3. Now [Ṽ_{1}, ω] = Ṽ_{1} and as n ≤ 11, we get that |Ṽ_{1}/Z_{1} : C I_{Ṽ_{1}/Z_{1}}(ω)| ≤ 2^{5}, and so |Ṽ_{1} : Ṽ_{1} ∩Z_{2}| ≤ 2^{5}. As K ≴ L_{2}(q), we get q = 4. But also K ≴ SL_{3}(4), a contradiction.

Suppose next p = 3

|Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2 (2)

Suppose false. Then |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ 4. Suppose furthermore that Ṽ_{1} has a natural submodule T. In such a module every element is centralized by an elementary abelian group of order 9 in K, or K ≅ A_{7}.

In the former [Ṽ_{α−1} ∩ L_{1}, K] = 1. As [Ṽ_{α−1} ∩ Q_{1}, V_{1}] = 1, we get that Ṽ_{1} induces transvections on Ṽ_{α−1}, a contradiction to |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ 4.

Thus we have K ≅ A_{7}. Again for some t ∈ Ṽ_{α−1} we have [K, t] ≠ 1. Now |[Ṽ_{1}, t]| ≤ 8. This implies that Ṽ_{1}/C IṼ_{1}(K) is irreducible. But m_{3}(K) = 2. So there is some ν, o(ν) = 3, with [Ṽ_{1}, ν] = 1. Now also every x ∈ Ṽ_{1} is centralized by E ≅ E_{9} in L_{1}, a contradiction.

So we have that Ṽ_{1}/C IṼ_{1}(K) contains no natural submodule. By (1.12) this implies |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 8. Let now z ∈ Ṽ_{1}. Then we see that |Ṽ_{α−1} : C IṼ_{α−1}(z)| ≤ 16. This shows that Ṽ_{α−1}/C IṼ_{α−1}(K_{α−1}) is a spin module and n = 9. Now we see that Z_{α−1} ≤ [t, Ṽ_{1}] for every 1 ≠ t ∈ Ṽ_{α−1} ∩ L_{1}. As every element in the spin module is centralized by some 3 - element in A_{9}, we get with (4.2) M = M_{α−1}. But this contradicts Z_{α} ≰ M.

p > 3 (3)

If p = 3, then by (2) |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2. Now Ṽ_{α−1} ≤ M. As p = 3 , we have L_{2}/Q_{2} ≅ Σ_{3} and so Ṽ_{α−1} ≤ L_{1}. As Z_{α} ≰ M, we get [Ṽ_{α−1}, K] ≠ 1. But Ṽ_{α−1} has to induce transvections. This implies that [Ṽ_{α−1}, Ṽ_{1}] is centralized by E ≅ E_{9} in L_{1}, a contradiction.

Now by (3) we have n ≤ 11.

|Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 4 (4)

Suppose false. Then we get |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ 8. Hence by (1.12) Ṽ_{1} involves just natural modules or |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = 8 and spin modules are involved.

As [Ṽ_{1}, K_{α−1}] ≠ 1, we see that [Ṽ_{α−1} ∩ L_{1}, K] ≠ 1. This shows that there are at most two natural modules and at most one spin module involved. In both cases we see [C I_{L_{1}/Q_{1}}(K), Ṽ_{1}] = 1. This now shows m_{p}(K) = 2, i.e. p = 5, and some ν with o(ν) = 5 centralizes Ṽ_{1}. Furthermore n = 10, or n = 11. But in the spin module for A_{n} we get |[Ṽ_{1}, t]| > 2^{4} for t ∈ Ṽ_{α−1} ∩ L_{1}, [t, K] ≠ 1, a contradiction. Hence just natural modules are involved. In the natural module for A_{10} or A_{11} every element is centralized by some 5 - element in K. This implies that every x ∈ Ṽ_{1} is centralized by E ≅ E_{25} with Γ_{E,1}(G) ≤ M, which contradicts Z_{α} ≰ M. This proves (4).

Now as p > 3 we get from (4) that Ṽ_{α−1} ≤ M. Suppose [Ṽ_{α−1}∩L_{1}, K] ≠ 1. This is obviously true if m_{p}(K) = 2. We have that |Ṽ_{1}/Z_{1} : C I_{Ṽ_{1}/Z_{1}}(Z_{α−1})| > 4 by (3). So by (4) there is some 1 ≠ t ∈ Ṽ_{1}, [t, K_{α−1}] = 1. Hence p | |C I(Ṽ_{1})|.

This implies that Z_{α−1} ∩ Ṽ_{1} = 1 by (4.2). Now by (4) |[Ṽ_{1}, t]| ≤ 4 for t ∈ Ṽ_{α−1} ∩ L_{1}, [K, t] ≠ 1. So Ṽ_{1} just involves spin modules or natural modules, by (1.30). Furthermore there are at most two natural modules involved. But if we have a natural submodule, then as above every x ∈ Ṽ_{1} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, which contradicts Z_{α} ≰ M.

So we have the spin module as a submodule and now Ṽ_{1}/C IṼ_{1}(K) is the spin module. This implies m_{p}(K) = 2, otherwise we may argue as before. Hence p = 5, K ≅ A_{10} or A_{11}. As [ω, K] = 1 and [ω, Ṽ_{1}] = Ṽ_{1} for q > 2, we see q = 2, and so for z ∈ Ṽ_{1} we have |Ṽ_{α−1} : C IṼ_{α−1}(z)| ≤ 8. But then as |[Ṽ_{α−1}/C IṼ_{α−1}(K_{α−1}), z]|^{2} = |Ṽ_{α−1}/C IṼ_{α−1}(K_{α−1})|, we get |Ṽ_{α−1}/C IṼ_{α−1}(K_{α−1})| = 2^{6}, a contradiction.

So we have m_{p}(K) = 1 and [L_{1} ∩Ṽ_{α−1}, K] = 1. Finally [Ṽ_{1}, K_{α−1}] ≤ Q_{α−1} follows as in (7.20).

(7.22) Lemma. Let b > 2, K ≅ G(r), r even. Then m_{p}(K) = 1, and [Ṽ_{α−1} ∩ L_{1}, K] = 1.

Proof. If ν ∈ K, o(ν) = p, then C I_{Z_{2}}(ν) = Z_{1} by (4.2). We will assume 1 ≠ [Ṽ_{α−1} ∩ K] ≤ K.

Set ω_{1} = ω, or ω^{3} in case of q = 64. Let [ω_{1}, K] = 1, o(ω_{1}) > 1. Then by (1.63) m_{p}(K) = 1. Hence we have K ≅ L_{2}(r), L_{3}(r), U_{3}(r), or Sz(r). Furthermore [ω_{1}, Ṽ_{1}] = Ṽ_{1}. We get for t ∈ Ṽ_{α−1} that |[Ṽ_{1}, t]| ≤ r or r^{2} in case of K ≅ L_{3}(r). This now implies with (1.26) and (1.27) that K ≅ L_{2}(r) or L_{3}(r) and by (1.26), (1.27) just natural or natural and dual modules are involved. There is E ≅ E_{p^{2}} with E ≤ N(K) and E ∩ K = 1. Now E acts on C I_{Ṽ_{1}/C IṼ_{1}(K)}(S ∩ K) and so some 1 ≠ x ∈ [Ṽ_{1}, t] is centralized by F ≅ E_{p^{2}} with Γ_{F,1}(G) ≤ M, and so Z_{α} ≤ M, a contradiction (Recall that ⟨K, E⟩ ≤ C I_{L_{1}/Q_{1}}(Z_{1}) by (7.2)). So we have

If q > 2, then [ω_{1}, K] ≠ 1 (1)

As [ω, Z_{1}] ≠ 1, we have ω ∉ K. Hence (1.65)(i), (iii), or (v) holds. We have that (1.65)(vi) is not possible as in that case q = 4 but 3 ∈ σ(M). Now application of (1.64) shows q ≤ r, or q = r^{2}, K ≅ (S)U_{n}(r), Ω_{2n}^{−}(r), ^{2}E_{6}(r), or ^{3}D_{4}(r).

If t ∈ Ṽ_{α−1} ∩ L_{1}, [t, K] ≠ 1, then |[Ṽ_{1}/Z_{1}, t]| ≤ |Ṽ_{1} : C IṼ_{1}(K_{α−1})|u, where u = r, or u = r^{2} for K ≅ (S)U_{n}(r), Ω_{2n}^{−}(r), ^{2}E_{6}(r), or u = r^{3} for K ≅^{3}D_{4}(r). Suppose p | r^{2} − 1.

(2) Ṽ_{1} involves some nontrivial irreducible module which is not a fun- damental module.

Suppose false. We have that V (λ) for every fundamental weight λ has to be involved. Application of (1.23) shows that K ≅ (S)L_{n}(r), n ≤ 4, (S)U_{n}(r), n ≤ 4, Sp_{4}(r), Sp_{6}(r), or Sz(r).

If K ≅ Sp_{6}(r), then |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r^{6} and then Ṽ_{1} cannot act quadratically on the natural module.

Let K ≅ Sp_{4}(r). As no group of order r^{3} act quadratically on both 4 - dimensional modules, we see |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r^{2}. But now Ṽ_{1} contains some element x whith |Ṽ_{α−1} : C IṼ_{α−1}(x)| ≥ r^{4}, which implies |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| = r^{3}, a contradiction.

Let K ≅ (S)U_{4}(r). Then because of the natural Ω_{6}^{−}(r) - module we get |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r^{3}. Now Ṽ_{1} contains x which does not induce a transvection on the natural SU_{4}(r) - module. Hence |Ṽ_{α−1} : C IṼ_{α−1}(x)| ≥ r^{6}. But then |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| = r^{4}, a contradiction.

Let K ≅ (S)L_{4}(r). Because of the natural Ω_{6}^{+}(r) - module, we have |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r^{3} ≥ |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)|. Hence Ṽ_{1} consists of transvections. As the natural and the dual module are involved, we see |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| ≥ r^{7}, a contradiction.

So we have K ≅ L_{2}(r), SL_{3}(r), SU_{3}(r), or Sz(r). Furthermore just natural or dual modules are involved.

Suppose first there is E ≅ E_{p^{2}}, E ≤ L_{1}, [K, E] = 1. If [Ṽ_{1}, ν] = 1 for some ν ∈ E^{♯}, we get that Z_{α−1} ∩ Ṽ_{1} = 1 and so |[Ṽ_{1}, t]| ≤ |Ṽ_{1} : C IṼ_{1}(K_{α−1})|. This implies K ≅ L_{2}(r) and Ṽ_{1}/C IṼ_{1}(K) is the natural module. As [E, Ṽ_{1}] ≠ 1, we get p | r − 1. But now t acts on F ≅ E_{p^{3}} and some 1 ≠ U ≤ [Ṽ_{1}, t] is normalized by Ẽ ≤ F, |Ẽ| = p^{2}, a contradiction.

So we have that E acts faithfully on Ṽ_{1}/C IṼ_{1}(K). In particular Ṽ_{1}/C IṼ_{1}(K) is not irreducible and so K ≅ L_{2}(r) or SL_{3}(r). Furthermore r = q and p | r − 1. If K ≅ SL_{3}(r), we get that ω centralized F ≅ E_{p^{2}} with Γ_{F,1}(G) ≤ M, and so N_{G}(⟨ω⟩) ≤ M, a contradiction. So we have K ≅ L_{2}(r) and C I_{L_{1}/Q_{1}}(K) ≲ GL_{2}(r). Hence t ∈ Ṽ_{α−1} ∩ L_{1} normalizes F ≅ E_{p^{3}}, a contradiction as before.

Suppose now m_{p}(K) = 1. Then m_{p}(C I_{L_{1}/Q_{1}}(K)) = 1 and there is some ν ∈ L_{1} with o(ν) = p which induces a field automorphism on K. By (4.12) we have K ≅ L_{2}(r) and C I_{L_{1}/Q_{1}}(K) involves L_{2}(r) too. But then t acts on E ≅ E_{p^{3}}. This implies that there is some 1 ≠ U ≤ [Ṽ_{1}, t] normalized by F ≅ E_{p^{2}} with Γ_{F,1}(G) ≤ M, a contradiction. We have that C I_{L_{1}/Q_{1}}(K) acts faithfully on Ṽ_{1}/C IṼ_{1}(K).

We are left with m_{p}(K) = 2. Hence K ≅ SL_{3}(r), p | r−1, or K ≅ SU_{3}(r),

p | r + 1. In the first case ω centralizes E ≅ E_{p^{2}} with E ≤ K, and so N_{G}(⟨ω⟩) ≤ M, a contradiction. Hence K ≅ SU_{3}(r) and Z_{α−1} ∩ Ṽ_{1} ≠ 1. But t ∈ K acts on some E_{p^{2}} and so t acts on some E_{p^{3}}, a contradiction.

p | r̸^{2} − 1 (3)

Suppose false. By (2) we may assume that some T not isomorphic to V (λ)

is involved in Ṽ_{1}. By (1.14) Ṽ_{1} is not strongly quadratic. By (1.15) Ṽ_{1} is an SC - module or K ≅ L_{2}(r), U_{3}(r), or Sz(r). Now application of (1.22) shows that for K ≅ G_{l}, V (λ), λ ≠ λ_{1}, λ_{l}, has to be involved in Ṽ_{1}. Application of (1.38) and (1.26) show K ≅ L_{2}(r), or SL_{3}(r). But then we may argue as above to get that [Ṽ_{1}, Ṽ_{α−1}] contains some 1 ≠ x with C I_{G}(x) ≤ M. Now Z_{α} ≤ M, a contradiction.

If 1 ≠ [Ṽ_{α−1} ∩ L_{1}, K] ≤ K, then m_{p}(K) = 1 (4)

Otherwise by (3) p | r̸^{2} − 1. Let K ≅ (S)L_{n}(r). Then we get for r > 2 that n ≤ 4 and m_{p}(K) = 1. If r = 2, then p ≠ 3 and so n ≤ 7 and m_{p}(K) = 1 or p = 7 and n = 6 or 7.

Suppose m_{p}(K) = 2. Then Ṽ_{1} is either strongly quadratic or an SC - module by (1.15). Now Z_{2} is not centralized by ν ∈ K with o(ν) = 7. If n = 7 this implies that V (λ_{3}) is involved. By (1.23) V (λ_{4}) is not involved and so also V (λ_{5}) is not involved, a contradiction.

Let n = 6. Then assume that V (λ_{3}) is involved. Now neither V (λ_{2}) nor V (λ_{4}) are involved. Furthermore by (1.23) |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ 2^{7} and so by (1.15) and (1.14) just V (λ), λ a fundamental weight are involved in Ṽ_{1}. But we have that neither V (λ_{2}) nor V (λ_{4}) are involved. Hence both V (λ_{1}) and V (λ_{5}) are involved. Now Ṽ_{α−1} ∩ L_{1}/C I_{Ṽ_{α−1}∩L_{1}}(K) is a transvection group in K. This implies |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≤ 2^{5}. As [ω, K] = 1, we get q = 2. Hence for z ∈ Ṽ_{1}, we have [Ṽ_{α−1}, z]| ≤ 2^{6}, contradicting (1.23).

So we have that V (λ_{3}) is not involved. Then both V (λ_{2}) and V (λ_{4}) are involved. Now each t ∈ Ṽ_{α−1} with [K, t] ≠ 1 induces an element on V (λ_{2}) which centralizes a subspace of codimension 4. But these are exactly the transvections in L_{6}(2) and we get the same contradiction as before.

Let K ≅ (S)U_{n}(r), we get K ≅ (S)U_{4}(r) or U_{5}(4). Now m_{p}(K) = 1. If K ≅ Sp_{2n}(r), then n ≤ 3 and again m_{p}(K) = 1. If K ≅ Ω_{8}^{−}(r) or G_{2}(r), then as p | r̸^{2} − 1, we get m_{p}(K) = 1 If 1 ≠ [Ṽ_{α−1} ∩ L_{1}, K], then [Ṽ_{α−1} ∩ L_{1}, K] ≰ K (5)

Suppose false. By (4) we have m_{p}(K) = 1. Furthermore p | r̸^{2} − 1. Let E ≅ E_{p^{2}} with E ≤ C I_{L_{1}}(Z_{1}), [K, E] ≤ K and E ∩ K = 1. Let T_{1} be a submodule of K in Ṽ_{1}/C IṼ_{1}(K). Then [E, T_{1}] ≠ 1. Suppose [E, T_{1}] ≤ T_{1}, then as p | r̸^{2} − 1 by (3) we have ν ∈ E, o(ν) = p with [ν, T_{1}] = 1 and ν_{1} ∈ E, o(ν_{1}) = p, which induces a field automorphism on K. But now we may assume that ν_{1} acts on [T_{1}, t] and so ⟨ν, ν_{1}⟩ centralizes some element in [Ṽ_{1}, Ṽ_{α−1}]^{♯}, a contradiction.

So we have [E, T_{1}] ≰ T_{1}. As p | r̸^{2} − 1 by (3), we see that there are at least three modules isomorphic to T_{1} involved in Ṽ_{1} or K ≅ (S)U_{n}(r), n = 4, or 5, p | r^{2} + 1, and there are exactly two such modules.

Let first K ≅ Ω_{8}^{−}(r). By (1.22) and (1.23) T_{1} is the natural module. As |Ṽ_{1} : C IṼ_{1}(K_{α−1})| > r, we get with (1.45) that |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| ≥ r^{9} and so |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≥ r^{7}, a contradiction.

Let K ≅ Sp_{6}(r). We have |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ r^{2}. As a transvection group is of order r, we see that |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≥ r^{5}. Hence also |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ r^{5} and Ṽ_{1} centralizes in T_{1} a subgroup of index r^{2}. But this contradicts (1.46).

Let now K ≅ Sp_{4}(r). Then p | r^{2} + 1 and so there are four nontrivial irreducible modules involved. These have to be natural and |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r^{3}. But Ṽ_{α−1} ∩ L_{1} has to induce transvections on T_{1} and so |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≤ r. Hence for z ∈ Ṽ_{1} we have |Ṽ_{α−1} : C IṼ_{α−1}(z)| ≤ r^{2}, a contradiction.

Let K ≅ (S)L_{n}(r). If n = 6 or 7, then p ≥ 5 and so there are at least four nontrivial irreducible modules involved. By (1.47) these are natural modules. Let n ≤ 5, then also by (1.47) these are natural.

Suppose T_{1} to be the natural module. Assume N_{E}(T_{1}) = 1. Then we have at least six nontrivial irreducible modules involved and so n ≥ 5. Let n = 5. Then |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≤ 2^{6} and q = 2 by (1). This implies that Ṽ_{1} induces transvections on Ṽ_{α−1}. Hence |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2^{4}. But now |Ṽ_{1} : C IṼ_{1}(t)| ≤ 2^{5}, a contradiction.

If n > 5, then p ≠ 7, and so there are at least 8 modules involved. Again Ṽ_{1} induces transvections. So |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2^{6}, a contradiction again.

Now we have |N_{E}(T_{1})| = p. As all elements in T_{1}^{♯} are conjugated under K, we get that p | r̸^{m} − 1, 0 < m < n. This implies that there are n modules involved. Furthermore Ṽ_{α−1} ∩ L_{1} induces transvections on T_{1}. This implies |Ṽ_{α−1}∩L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| ≤ r^{n−1}. This now yields |Ṽ_{α−1}∩L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K)| = r^{n−1} = |Ṽ_{1} : C IṼ_{1}(K_{α−1})| and q = r.

We have n > 2 as p | r̸^{2} − 1. Then by (1.16) t fixes any T_{1}^{ν}. Hence t acts as some s ∈ K on T_{1} ⊕ … ⊕ T_{p}. But now s inverts some μ ∈ K which is centralized by F ≅ E_{p^{2}}. Hence F normalizes some 1 ≠ U ≤ [[μ, Ṽ_{1}], t], a contradiction.

Let finally K ≅ (S)U_{n}(r), n = 4 or 5. Now |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ r^{4} and so there are at most three natural modules involved. Now |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≥ r^{4}. But then Ṽ_{1} contains some x with |T_{1} : C I_{T_{1}}(x)| = r^{4} and so |Ṽ_{α−1} : C IṼ_{α−1}(x)| ≥ r^{8}, a contradiction as q ≤ r^{2}.

[Ṽ_{α−1} ∩ L_{1}, K] = 1 (6)

Suppose [Ṽ_{α−1} ∩ L_{1}, K] ≠ 1. Then by (5) there is some t ∈ Ṽ_{α−1} ∩ L_{1} with K^{t} ≠ K. Now [Ṽ_{α−1} ∩ L_{1}, K × K^{t}] ≤ K × K^{t} by (1.11). Suppose first |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K × K^{t})| > 2. Then again by (1.11) K ≅ L_{2}(r), r even, and Ṽ_{1} involves the natural Ω_{4}^{+}(r) - module. Furthermore |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K × K^{t})| ≤ 2r.

Let |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K × K^{t})| = 2. If m_{p}(K) ≥ 2, then t acts on an elementary abelian group of order p^{4} in K × K^{t}, which implies Z_{α} ≤ M, a contradiction. Hence K ≅ L_{2}(r), Sz(r), or L_{3}(2).

In both cases we have [ω_{1}, K × K^{t}] ≠ 1, as N_{G}(⟨ω_{1}⟩) ≰ M. Hence in the first case q ≤ r. This then implies q = r and |Ṽ_{1} : C IṼ_{1}(K_{α−1})| = r. We see that Z_{α−1} ≤ Ṽ_{1} and so Z_{2} ≤ Ṽ_{1}. Hence p | |̸C I_{L_{1}/Q_{1}}(K × K^{t})| as Ṽ_{1}/C IṼ_{1}(K × K^{t}) is an irreducible ⟨K, t⟩ - module.

In the second case, as [C I_{Ṽ_{α−1}∩L_{1}}(K × K^{t}), Ṽ_{1}] = 1, we get |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| ≤ 2q. This implies q = r , K ≅ L_{2}(r), or q = 2 = r, K ≅ L_{3}(2). Now also in this case as |[t, Ṽ_{1}]| ≥ r^{2}, we see that Z_{α−1} ≤ Ṽ_{1} and we get that Ṽ_{1}/C IṼ_{1}(K × K^{t}) is a direct sum of two natural modules and so as p | r̸ − 1,

we also get p | |̸C I_{L_{1}/Q_{1}}(K × K^{t})|.

We see that t inverts ν ∈ K×K^{t} with o(ν) = p. If E ≤ C I(ν) with E ≅ E_{p^{2}} and [t, E] ≤ Q_{1}, then E acts on [[ν, Ṽ_{1}], t], and so Z_{α} ≤ M, a contradiction. Hence there is ν̃, o(ν̃) = p, with [ν, ν̃] = 1 and ν̃^{t} = ν̃^{−1}. Furthermore there is ν_{1}, o(ν_{1}) = p, with [ν_{1}, ⟨ν, ν̃⟩] = 1 and [ν_{1}, t] ∈ Q_{1}. Hence we may assume that [ν̃, [[ν, Ṽ_{1}], t]] = 1 and so ⟨ν̃, ν_{1}⟩ acts on 1 ≠ U ≤ [Ṽ_{1}, t], a contradiction.

m_{p}(K) = 1 (7)

Suppose m_{p}(K) ≥ 2. Then [Ṽ_{α−1} ∩ L_{1}, Ṽ_{1}] = 1. By (1.65) and (1.64) we have that Ṽ_{1} induces transvections on Ṽ_{α−1} and K ≅ L_{n}(r), Sp_{2n}(r), Ω_{2n}^{±}(r), or SU_{n}(r), or |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| = r^{2} and K ≅ Ω_{2n}^{−}(r) and also Ṽ_{1} is the natural module.

Suppose next p | r^{2} − 1. Then we see that p | |̸SL_{n−1}(r)|, |Sp_{2n−2}(r)|, |Ω_{2n−2}^{±}(r)|, or |SU_{n−2}(r)|, respectively. As m_{p}(K) ≥ 2 this implies K ≅ SU_{3}(r), p | r + 1. Furthermore q = r^{2}. Now we have ω_{2} ∈ L_{2} ∩ L_{1}, o(ω_{2}) = p. But then as ω_{2} ∉ C I_{L_{1}}(Z_{1}), we see that ω_{2} centralizes E ≅ E_{p^{2}} in C I_{L_{1}}(Z_{1}) with Γ_{E,1}(G) ≤ M. Hence N_{G}(⟨ω_{2}⟩) ≰ M, a contradiction.

So we have p | r̸^{2} − 1. Hence K ≅ L_{6}(2), or L_{7}(2), and p = 7. But then

7 | |C I_{K}(C I_{Ṽ_{1}/C IṼ_{1}(K)}(S ∩ K))| , a contradiction.

(7.23) Lemma. Let b > 2, K be an exceptional component, i.e. O_{r}(K) ≠ 1. Then Ṽ_{α−1} ≤ M and [Ṽ_{α−1} ∩ L_{1}, K] = 1. Furthermore [Ṽ_{1}, K_{α−1}] ≤ Q_{α−1} and L_{1} is not as in (4.9)(ii).

Proof. We have |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 4. Suppose we do not have (4.9)(ii). Then |Ṽ_{1} : C IṼ_{1}(K_{α−1})| ≤ 2 by (1.12) and so Ṽ_{α−1} = ⟨t | |[t, Ṽ_{1}] : [t, Ṽ_{1}] ∩ Z_{α−1}| ≤ 2⟩. This immediately shows Ṽ_{α−1} ≤ M. Now let t ∈ Ṽ_{α−1} ∩ L_{1}, [K, t] ≠ 1. By (4.10) we may assume that Ṽ_{α−1} ∩L_{1}/Ṽ_{α−1} ∩Q_{1} acts faithfully on V_{1}. Hence V_{1} centralizes a subgroup of index four in Ṽ_{α−1}, a contradiction. Thus [Ṽ_{α−1} ∩ L_{1}, K] = 1.

Suppose now that we have (4.9)(ii). Set Ĝ_{2} = ⟨Q_{1}, Z_{α}⟩. Then Ĝ_{2}/Ĝ_{2} ∩Q_{1} contains L_{2}(q). Furthermore there is V̂_{1} ≤ V_{1}, |V_{1} : V̂_{1}| ≤ 4 such that V̂_{1}≤ G_{1}, [Ĝ_{2}, V̂_{1}] ≤ Z_{2}. Now [Q_{1}∩Q_{3}, V̂_{1}] = 1 and ⟨Q_{1}∩Q_{2}, Q_{2}∩Q_{3}⟩/(Q_{1}∩Q_{3}) involves nontrivial L_{2}(q) - modules. This shows q ≤ 4. Now let x ∈ Q_{3} ∩ Q_{2} \ Q_{1}. Then |[V_{1}, x]| ≤ 16. We have [O_{r}(L_{1}), V_{1}] ≠ 1 by (1.51). So r = 5, as |[O_{r}(L_{1}), x]| = r^{2}. Furthermore |[O_{5}(L_{1}), V_{1}]Z_{1}/Z_{1}| ≤ 2^{12} as there is ν ∈ O_{5}(L_{1}), o(ν) = 5, |[ν, V_{1}]| = 16. But in GL_{12}(2) there is no subgroup of type 5^{3}L_{2}(5).

This shows Q_{1} ∩ Q_{2} = Q_{1} ∩ Q_{3}. Now we have a subgroup G̃_{2} ≤ G_{2} with Q_{1} ∈ Syl_{2}(G̃_{2}) and G̃_{2}/O_{2}(G̃_{2}) ≅ L_{2}(q). Further no nontrivial characteristic subgroup of Q_{1} is normal in G̃_{2}. Hence by [Ste, Theorem 1] O_{2}(G_{2}) involves exactly one nontrivial irreducible G̃_{2} - module. Hence this is Z_{2}. Then V_{1} ≤ O_{2}(G̃_{2}) is normal in ⟨L_{1}, G̃_{2}⟩ = G, a contradiction.

(7.24) Lemma. Let b > 2, K_{α−1} be a component of L_{α−1}/Q_{α−1} with 1 ≠ [K_{α−1}, V_{1}]. Then [K_{α−1}, V_{1}] ≤ K_{α−1}.

Proof. Let z ∈ V_{1} with K_{α−1}^{z} ≠ K_{α−1}. Set Ṽ_{α−1} = [V_{α−1}, K_{α−1}^{z} × K_{α−1}]. Suppose first |V_{1} : C I_{V_{1}}(K_{α−1}^{z} × K_{α−1})| = 2. Then Ṽ_{α−1} ≤ M. Let K^{(1)} be the component of L_{1}/Q_{1} corresponding to K_{α−1}^{z}. As Z_{α} ≰ M there is some component K^{(2)} with [K × K^{(1)}, K^{(2)}] = 1 and [Ṽ_{1}, K^{(2)}] = 1. Hence Z_{α−1} ∩ Ṽ_{1} = 1. This implies that for t ∈ Ṽ_{α−1}∩L_{1}, we have |Ṽ_{1} : C IṼ_{1}(t)| ≤ 2. Suppose [t, K × K^{(1)}] ≠ 1, then we may asume [t, K] ≠ 1 and [t, Ṽ_{1}] is centralized by K^{(1)} ×K^{(2)}, contradicting Z_{α} ≤ M. Hence we have [Ṽ_{α−1} ∩L_{1}, K ×K^{(1)}] = 1. This implies [V_{1}, Ṽ_{α−1} ∩ L_{1}] = 1. Now V_{1} induces transvections on Ṽ_{α−1}, a contradiction.

Now we have K_{α−1} ≅ L_{2}(r), r even, and |V_{1} : C I_{V_{1}}(K_{α−1}^{z} × K_{α−1})| ≤ 2r by (1.11). Furthermore Ṽ_{α−1} involves natural Ω_{4}^{+}(r) - modules. Let ω ∈ L_{1} ∩L_{2}, as before. Let q > 2. If [ω_{1}, K × K^{(1)}] = 1, ω_{1} = ω, or q = 64 and ω_{1} = ω^{3}, then by (1.63) K ≅ L_{2}(4) and p = 5. Suppose C I_{V_{1}}(K_{α−1}^{z} × K_{α−1}) ≤ Q_{α−1}. Then |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(Z_{α})| ≤ 8, contradicting p = 5 and Z_{α} ≰ M. Hence there is some component K_{2}, [K × K^{(1)}, K_{2}] = 1 and [Ṽ_{1}, K_{2}] = 1. This implies Z_{α−1} ∩ Ṽ_{1} = 1 and so |Ṽ_{1} : C IṼ_{1}(t)| ≤ 8 for t ∈ Ṽ_{α−1} ∩ L_{1}, [K, t] ≠ 1. Then |Ṽ_{α−1} ∩ L_{1} : C I_{Ṽ_{α−1}∩L_{1}}(K × K^{(1)})| ≤ 2 and so |Ṽ_{α−1} : C IṼ_{α−1}(u)| ≤ 2 for u ∈ Ṽ_{1}, as Ṽ_{α−1} ≤ L_{1} by q > 2. This now shows [Ṽ_{1}, K_{α−1} ×K_{α−1}^{z}] = 1 and so [Ṽ_{1}, Ṽ_{α−1}] = 1. As no element in Z_{α−1} is centralized by some element of order 5 in K ×K^{(1)}, we get that |[t, V_{1}]| ≤ 8 for t ∈ Ṽ_{α−1}. Hence [K ×K^{(1)}, [t, V_{1}]] = 1 and so as Z_{α} ≰ M , we have [Ṽ_{α−1}, V_{1}] = 1, a contradiction.

This now implies q = 2 or [ω_{1}, K × K^{(1)}] ≠ 1. Hence in all cases q ≤ r. As no p - element in K × K^{(1)} centralizes some element in Z_{2} \ Z_{1}, we get that p | r +1. Let now t ∈ Ṽ_{α−1} ∩L_{1}, [t, K ×K^{(1)}] ≠ 1. Suppose there is some ν ∈ C I(K ×K^{(1)}), o(ν) = p. If [ν, Ṽ_{1}] = 1, then Z_{α−1} ∩Ṽ_{1} = 1 and so |[t, Ṽ_{1}]| ≤ 2r. This implies K^{t} = K^{(1)}. But then there is some 1 ≠ U ≤ [Ṽ_{1}, t], U is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction to Z_{α} ≰ M. So we have [ν, Ṽ_{1}] ≠ 1. In particular there are at least two nontrivial irreducible K × K^{(1)} - modules involved. Again K^{t} = K^{(1)}. Now for x ∈ Ṽ_{1} we have |Ṽ_{α−1} : C IṼ_{α−1}(z)| ≤ 2r which implies [Ṽ_{1}, K_{α−1} × K_{α−1}^{z}] = 1, as Ṽ_{α−1} involves two nontrivial irreducible modules. But then C I_{G}([Ṽ_{1}, Ṽ_{α−1}]) ≤ M_{α−1}. Now C I_{K×K^{(1)}}(t) centralizes [t, Ṽ_{1}] and so C I_{K×K^{(1)}}(t) ≤ M_{α−1}. But then with (4.2) we get M = M_{α−1}, a contradiction.

We get that p | |̸C I_{L_{1}/Q_{1}}(K ×K^{(1)})|. Hence there is some ν ∈ L_{1}, o(ν) = p, ν induces a field automorphism of order p on K × K^{(1)}. Now if t ∈ K × K^{(1)}, we get that t acts on E ≅ E_{p^{3}} and so there is some F ≤ E, |F | = p^{2} and 1 ≠ U ≤ [Ṽ_{1}, t] with F ≤ N(U). This implies Z_{α} ≤ M, a contradiction. So we have ν^{t} = ν^{−1}. As all involutions in (K × K^{(1)})⟨ν⟩t are conjugate, we again have that t normalizes some E ≅ E_{p^{3}}, the same contradiction as before.

So we have shown that [Ṽ_{α−1}∩L_{1}, K ×K^{(1)}] = 1. We get that [K, [t, V_{1}]] = 1 for t ∈ Ṽ_{α−1} ∩L_{1}. Hence Z_{α−1} ∩[t, V_{1}] = 1. This implies |[t, V_{1}]| ≤ 2r and so [K × K^{(1)}, [t, V_{1}]] = 1. As Z_{α} ≰ M, we see [Ṽ_{α−1} ∩ L_{1}, V_{1}] = 1. Now we have the contradiction |V_{1} : C I_{V_{1}}(K_{α−1} × K_{α−1}^{z})| ≤ 2, as |Ṽ_{α−1} : C IṼ_{α−1}(V_{1})| ≤ r.

(7.25) Lemma. Let b > 2 and K be a normal p - subgroup in L_{1}/Q_{1}. Then [Ṽ_{α−1} ∩ L_{1}, K] = 1 and m_{p}(K) = 1, Ṽ_{α−1} ≤ M and [Ṽ_{1}, K_{α−1}] = 1.

Proof. Let t ∈ Ṽ_{α−1} ∩ L_{1} with [t, K] ≠ 1. If m_{p}(K) ≥ 3, then by [GoLyS, (11.18)] t acts on E ≅ E_{p^{3}} and so some 1 ≠ U ≤ [Ṽ_{1}, t] is normalized by F ≤ E, |F | = p^{2}, contradicting Z_{α} ≰ M. Hence m_{p}(K) ≤ 2. Let Ṽ_{1} ∩ C I_{L_{α−1}/Q_{α−1}}(K_{α−1}) ≠ 1. Let u ∈ Ṽ_{1} and K_{α−1}^{(1)} be a component or normal p-subgroup of L_{α−1}/Q_{α−1} with [u, K_{α−1}^{(1)}] = 1 and [K_{α−1}^{(1)}, K_{α−1}] = 1. Then by (7.19) - (7.24) [K^{(1)}, Ṽ_{α−1} ∩ L_{1}] = 1 where K^{(1)} is the group corresponding to K_{α−1}^{(1)} in L_{1}/Q_{1}. Hence if m_{p}(K) = 2 we again get E ≅ E_{p^{3}} normalized by t, a contradiction.

So we may assume m_{p}(K) = 1. Now we have [K_{α−1}^{(1)}, Ṽ_{α−1}] = 1 and then [K^{(1)}, Ṽ_{1}] = 1. Hence Z_{α−1} ∩ Ṽ_{1} = 1, and so t induces a transvection on Ṽ_{1}, i.e. p = 3. So we have |Ṽ_{1}/C IṼ_{1}(K)| = 4. But then again some E ≅ E_{9} centralizes Ṽ_{1}, a contradicition.

So we have |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ 4. We get that Ṽ_{α−1} = ⟨t | |[Ṽ_{1}, t] : [Ṽ_{1}, t] ∩ Z_{α−1}| ≤ 2⟩ by [GoLyS, (24.1)] and [GoLyS, (25.12)]. Hence Ṽ_{α−1} ≤ M. As Z_{α} ≰ M this implies p = 3, |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = 4 and m_{p}(L_{1}) = 3. As p = 3, we get L_{2}/Q_{2} ≅ Σ_{3} if q = 2. Hence in any case Ṽ_{α−1} ≤ L_{1} and so |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| = 2, contradicting |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = 4 and [GoLyS, (26.11)].

So we have [Ṽ_{α−1} ∩ L_{1}, K] = 1. If m_{p}(K) ≥ 2, then [Ṽ_{α−1} ∩ L_{1}, V_{1}] = 1. Now V_{1} induces transvections to the hyperplane Ṽ_{α−1} ∩ L_{1}. Hence |V_{1} : C I_{V_{1}}(K)| = 2 by [GoLyS, (26.11)] and p = 3. But now Ṽ_{α−1} ≤ L_{1} and so [Ṽ_{α−1}, V_{1}] = 1, a contradiction.

That [Ṽ_{1}, K_{α−1}] = 1 follows as in (7.20).

(7.26) Proposition. b = 2

Proof. Suppose b > 2. Let K_{1} × … × K_{l} be the components or normal p-subgroups of L_{α−1}/Q_{α−1} with [V_{1}, K_{i}] ≠ 1, i = 1, … l. Hence [V_{1}, K_{j}] = 1 for every component K_{j}, j ∉ {1, …, l}.

Suppose there is x ∈ V_{1} \ Z_{2}, [x, Z_{α}] ≤ Z_{α} and C I_{G}(x) ≤ M. We have that ⟨Q_{1}, Z_{α}⟩ normalizes ⟨x, Z_{2}⟩. As [Q_{1}, x] ≤ Z_{1}, we get x ∈ Z(⟨Q_{1}, Z_{α}⟩) or [Q_{1}, x] = Z_{1} and so all elements in xZ_{2} are conjugate under ⟨Q_{1}, Z_{α}⟩. Hence in any case we get ⟨Q_{1}, Z_{α}⟩ = O_{2}(⟨Q_{1}, Z_{α}⟩)C I_{⟨Q_{1},Z_{α}⟩}(x). But [Z_{1}, O_{2}(⟨Q_{1}, Z_{α}⟩)] = 1. So we get G_{2} ≤ M, a contradiction. We have

(1) There is no x ∈ V_{1} \ Z_{2} with C I_{G}(x) ≤ M and [Z_{α}, x] ≤ Z_{α}

l > 1 (2)

Suppose l = 1. Then by (7.19) - (7.25) we get K ≅ G(r), r even, m_{p}(K) = 1. Let ω_{1} = ω, or q = 64 and ω_{1} = ω^{3}.

Assume K ≅ L_{n}(r). Then we get n ≤ 4, or 5 ≤ n ≤ 7 and r = 2. Let q > r. Then by (1.65) and (1.64) [ω_{1}, K] = 1, as [ω_{1}, Z_{1}] ≠ 1 and so (1.65)(ii) is not possible. Now [ω_{1}, Ṽ_{1}] = Ṽ_{1}. We have |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(Z_{α})| ≤ 2^{m_{2}(K)}. Let first [ω, K] = 1. As C I_{Ṽ_{1}/Z_{2}}(ω) = 1, we get that |Ṽ_{1}/Z_{1}| = q^{u} for some u. If n ≤ 4, then |Ṽ_{1}/Z_{1}| ≤ r^{4}q, and so |Ṽ_{1}/Z_{1}| ≤ q^{4}. As K is not isomorphic to a subgroup of SL_{3}(q), we see |Ṽ_{1}/Z_{1}| = q^{4}. Hence K is isomorphic to a subgroup of SL_{4}(q) and then K ≅ L_{4}(r), q ≥ r^{2}, a contradiction.

So let 5 ≤ n ≤ 7. Then |Ṽ_{1}/Z_{2}| ≤ 2^{6}, 2^{9}, 2^{12}, respectively. Now K ≲ L_{4}(q), or L_{5}(q). But both is not possible. Hence [ω, K] ≠ 1 and so n ≤ 4, [ω_{1}, K] = 1, q = 64. Choose ν ∈ L_{1} ∩ L_{2}, o(ν) = 7, with [ω, ν] = 1. If [ν, K] = 1, we get o(ω_{1}ν) = 21, [ω_{1}ν, K] = 1, and we may argue as before. So assume [ν, K] ≠ 1. Then as r < 64, we get 7 | r − 1 and so r = 8. Hence ω induces a field automorphism of order 3 on K. But [ν, ω] = 1. This is impossible as ν has to induce a diagonal automorphism of order 7.

So we have q ≤ r. Let t ∈ Ṽ_{α−1} ∩ L_{1} with [Ṽ_{1}, t] ≠ 1. Then |[t, Ṽ_{1}]| ≥ r^{n} and so |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≥ r^{n−1}. We have that there is a natural submodule T_{1} ≤ Ṽ_{1}. Let t_{1} ∈ T_{α−1}. Then [t_{1}, Ṽ_{1}] ≠ 1, and |[t_{1}, Ṽ_{1}]| < r^{n}. Hence t_{1} ∉ L_{1} This shows that Ṽ_{1} induces transvections on T_{α−1}, i.e. r = q and |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = r^{n−1}. Now Z_{α−1} ≤ T_{1}. This implies T_{1} = [Ṽ_{1}, Ṽ_{α−1} ∩ L_{1}]Z_{1}/Z_{1}. This now yields that there are at most two natural modules involved. If p | r̸^{2} − 1, then there is some ν, o(ν) = p, with [ν, Ṽ_{1}] = 1, a contradiction.

So we have p | r^{2} − 1 and then n ≤ 3. As no p - element in L_{1} centralizes

some s ∈ Z_{α−1}^{♯}, we see n = 2 and p | r + 1. But now L_{2}(r)⟨ν⟩, ν a field automorphism of order p, acts on the natural module T_{1}. Hence every element is centralized by some p - element, contradicting Z_{α−1} ≤ T_{1}.

So we have shown [Ṽ_{α−1} ∩ L_{1}, Ṽ_{1}] = 1. Now Ṽ_{1} induces transvections on Ṽ_{α−1}, i.e. |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| ≤ r^{n−1}, and Ṽ_{α−1}/C IṼ_{α−1}(K_{α−1}) is the natural module. Assume p | r̸ − 1. Then for every ν ∈ C I_{L_{1}/Q_{1}}(K), o(ν) = p, we have [ν, Ṽ_{1}] = 1. As |Ṽ_{1}Z_{2} ∩ Q_{α−1}| ≥ r^{3}, there is some x ∈ Ṽ_{1} \ Z_{2} with [x, Z_{α}] ≤ Z_{α}, x is centralized by some E ≅ E_{p^{2}} in L_{1} with Γ_{E,1}(G) ≤ M. Hence C I_{G}(x) ≤ M, a contradiction to (1).

So p | r − 1 and then K ≅ L_{2}(r), r = q. Now p | |C I_{L_{1}/Q_{1}}(K)|, furthermore

we have some μ ∈ L_{1} ∩ L_{2} with o(μ) = p. As [μ, Z_{1}] = Z_{1}, we get p | |C I_{K}(μ)|

and p | |C I_{C I_{L_{1}/Q_{1}}(K)}(μ)|. Hence μ ∈ E ≅ E_{p^{3}} with E ≤ L_{1}. This implies N_{G}(⟨μ⟩) ≤ M, a contradiction.

Suppose next K ≅ (S)U_{n}(r), n ≤ 5, or Ω_{8}^{−}(r). Let q > r^{2}. Assume first [ω, K] ≠ 1. Then by (1.65) and (1.64) [ω_{1}, K] = 1, q = 64. Hence r = 2 or 4 and m_{3}(K) ≤ 2. This implies K ≅ (S)U_{n}(4), n ≤ 5. Let ν ∈ L_{1} ∩ L_{2} with o(ν) = 7 and [ν, ω] = 1. Then we see [ν, K] = 1. Hence [⟨ν, ω_{1}⟩, K] = 1 and [⟨νω_{1}⟩, Ṽ_{1}] = Ṽ_{1}. So set ω_{2} = ω or νω_{1}. Then we get [ω_{2}, K] = 1 and [ω_{2}, Ṽ_{1}] = Ṽ_{1}. Now |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r, r^{4}, r^{4}, r^{6}, respectively. Hence in any case we get K ≲ SL_{3}(q), a contradiction.

So we have q ≤ r^{2}. Suppose next [t, Ṽ_{1}] ≠ 1 for some t ∈ Ṽ_{α−1} ∩ L_{1}. As [t, Ṽ_{1}] is a K - module, we see K ≅ Ω_{6}^{−}(r) or Ω_{8}^{−}(r) and |[t, Ṽ_{1}]| = r^{6}, or r^{8}. Now q = r^{2} and |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = r^{4}, or r^{6}. In the last case ω centralizes a subgroup of order p^{2} in K, a contradiction (recall ⟨ω, K⟩ = ZZ_{x} × K and so m_{x}(⟨ω, K⟩) = 4). So we have K ≅ Ω_{6}^{−}(r). But now some elementary abelian subgroup of order r^{4} in K acts quadratically on the natural module, a contradiction.

So we have [Ṽ_{1}, Ṽ_{α−1} ∩ L_{1}] = 1. Now we see that |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = r by (1.45) and Ṽ_{1}/C IṼ_{1}(K) is the natural SU_{n}(r) - module or Ω_{2n}^{−}(r) - module by (1.22), (1.23) and (1.26). If p | r̸^{2} − 1, we get some ν ∈ C I(K) with o(ν) = p and [Ṽ_{1}, ν] = 1. Now we may argue as before to get a contradiction with (1).

We are left with K ≅ SU_{3}(r), p | r − 1, q = r^{2}. Hence there is some μ ∈ L_{1} ∩ L_{2} with o(μ) = p. Now we get a contradiction as before.

Let K ≅ Sp_{2n}(r), n ≤ 3. Let q > r. Then [ω_{2}, K] = 1, ω_{2} = ω, or q = 64 and K ≅ Sp_{4}(r), r < 64. In the latter let ν ∈ L_{1} ∩ L_{2} with o(ν) = 7 and [ν, ω] = 1. If [ν, K] ≠ 1, then 7 | r − 1 and so r = 8. But then ω induces a field automorphism, contradicting [ω, ν] = 1. So [νω_{1}, K] = 1, and then [ω_{2}, K] = 1, where ω_{2} = νω_{1}. In any case we have |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r^{3}, or r^{6}. In the former we get the contradiction Sp_{4}(r) ≲ L_{2}(q). In the latter assume that q ≥ r^{2}. Then we get Sp_{6}(r) ≲ L_{3}(q), a contradiction. So we have K ≅ Sp_{4}(r), q ≤ r, or K ≅ Sp_{6}(r), q < r^{2}.

Suppose [t, Ṽ_{1}] ≠ 1 for some t ∈ Ṽ_{α−1} ∩ L_{1}. Then |[t, Ṽ_{1}]| = r^{4} in the first case while r^{6} ≤ |[t, Ṽ_{1}]| < r^{8} in the second case. In particular [t, Ṽ_{1}] involves the natural module just once. Let T_{α−1} be the natural module. Then in the first case as |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| = r^{3} there is some t_{1} ∈ T_{α−1} ∩ L_{1} with 1 ≠ |[Ṽ_{1}, t_{1}]| ≤ r^{3}, a contradiction. In the second case |Ṽ_{1} : Ṽ_{1} ∩ Q_{α−1}| > r^{4}. Now by (1.46) |T_{α−1} : C I_{T_{α−1}}(Ṽ_{1})| ≥ r^{3} and so there is some t_{1} ∈ T_{α−1}∩L_{1} with 1 ≠ |[Ṽ_{1}, t_{1}]| ≤ r^{3}, a contradiction again. So in any case [Ṽ_{1}, Ṽ_{α−1} ∩ L_{1}] = 1 and then |Ṽ_{1} : Ṽ_{1} ∩Q_{α−1}| = r and Ṽ_{1}/C IṼ_{1}(K) is the natural K - module. Now p | r̸ − 1 as m_{p}(K) = 1 and so some ν ∈ C I_{L_{1}/Q_{1}}(K), o(ν) = p, centralizes Ṽ_{1}. But then as in the L_{n}(r) - case we get a contradiction to (1).

Let finally K ≅ Sz(r), ^{3}D_{4}(r), G_{2}(r), or ^{2}F_{4}(r). Assume q > r or q > r^{5} in case of ^{3}D_{4}(r). Then we get with (1.65) and (1.64) that [ω_{1}, K] = 1. Let q = 64 and [ω, K] ≠ 1. Choose ν ∈ L_{1} ∩ L_{2} with o(ν) = 7 and [ν, ω] = 1. Let [ν, K] ≠ 1. Then we get K ≅ Sz(8), G_{2}(8), ^{2}F_{4}(8) or ^{3}D_{4}(2). In any case ω induces an outer automorphism and so [ω, ν] ≠ 1, a contradiction. Set ω_{2} = νω_{1}, ω_{2} = ω for [ω, K] = 1, then [ω_{2}, K] = 1. But now |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r, r^{5}, r^{3}, r^{5} and so Sz(r) ≲ L_{2}(q), ^{3}D_{4}(r) ≲ L_{2}(q), G_{2}(r) ≲ L_{3}(q), ^{2}F_{4}(r) ≲ L_{4}(q), a contradiction. For t ∈ Ṽ_{α−1} ∩ L_{1}, we have |[Ṽ_{1}, t]| ≤ r^{2}, r^{10}, r^{4}, r^{6}, respectively. We get from (1.27), (1.40) and (1.22) that K ≅ G_{2}(r). But G_{2}(r) ≴ SL_{4}(r), a contradiction. Hence [Ṽ_{1}, Ṽ_{α−1} ∩ L_{1}] = 1. Then |Ṽ_{α−1} : C IṼ_{α−1}(Ṽ_{1})| ≤ r (r^{5} in case of ^{3}D_{4}(r)). This again contradicts (1.18), (1.40), (1.22) and (1.27). Hence we have (2).

Let now l > 1. [V_{1}, K_{1}] ≠ 1 ≠ [V_{1}, K_{2}]. We have Ṽ_{α−1} = [V_{α−1}, K_{1}], Ṽ_{α−1}^{(1)} = [Ṽ_{α−1}, K_{2}].

Ṽ_{α−1}^{(1)} = 1 (3)

Suppose Ṽ_{α−1}^{(1)} ≠ 1. Let K × K^{(1)} be the corresponding components of L_{1}/Q_{1}. Then by (7.19) - (7.25) we have [Ṽ_{α−1} ∩ L_{1}, K × K^{(1)}] = 1. Furthermore m_{p}(K) = m_{p}(K^{(1)}) = 1. We may assume m_{3}(K_{1}) ≤ 1, hence K_{1} ≅ Sz(r), L_{2}(r), U_{3}(r), L_{3}(r), J_{1}, or solvable, or exceptional. Let |V_{1} : C I_{V_{1}}(K_{1} × K_{2})| = 2. Then we get Ṽ_{α−1} ≤ L_{1}. Let t ∈ Ṽ_{α−1}, [t, V_{1}] ≠ 1. Then some E ≅ E_{p^{2}}, E ≤ K × K^{(1)} acts on some 1 ≠ U ≤ [V_{1}, t]. Hence Z_{α} ≤ M, a contradiction. From (1.12) we get that K_{1} ≅ Sz(r), L_{2}(r), U_{3}(r), or L_{3}(r), r even. Suppose now q > r or q > r^{2} in case of K_{1} ≅ U_{3}(r). Then by (1.64) and (1.65) [ω_{2}, K] = 1, ω_{2} = ω if [ω, K] = 1, ω_{2} = ω_{1}ν, o(ν) = 7, for q = 64. As before Ṽ_{α−1}^{(1)} ≰ M. Hence some t ∈ Ṽ_{α−1} \ M centralizes in V_{1}/Z_{2} a subgroup of index r or r^{2} in case of K ≅ L_{3}(r). Then |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r or r^{2}. This implies Sz(r), L_{2}(r), U_{3}(r) ≲ L_{1}(q) or L_{3}(r) ≲ L_{2}(q), respectively. But all these cases are impossible.

So q ≤ r, q ≤ r^{2} in case of K ≅ U_{3}(r). Now |Ṽ_{α−1} : C IṼ_{α−1}(V_{1})| ≤ r, or r^{2}. This implies K_{1} ≅ L_{2}(r), U_{3}(r), or L_{3}(r) and Ṽ_{α−1}/C IṼ_{α−1}(V_{1}) is the natural K_{1} - module. But this contradicts [Ṽ_{α−1}, K_{2}] = Ṽ_{α−1}.

By (3) we know [[V_{α−1}, K_{1}], K_{2}] = 1. Set Ṽ_{α−1}^{(i)} = [V_{α−1}, K_{i}], i = 1, 2. By (7.19) - (7.25) we know [Ṽ_{α−1}^{(1)} ∩ L_{1}, K] = 1. Let |V_{1} : C I_{V_{1}}(K_{1})| = 2. Then there is a fours group W ≤ V_{1} such that W is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, and |W ∩ Q_{α−1}| ≥ 2, W ∩ Z_{1} = 1. But now [W ∩ Q_{α−1}, Z_{α}] ≤ Z_{α}, which contradicts (1). So we have that K_{1}, K_{2} are sporadic, alternating or groups of Lie type in even characteristic. As m_{3}(K) ≥ 2 for K sporadic or alternating we may assume K_{2} ≅ Sz(r), L_{2}(r), L_{3}(r), or U_{3}(r), r even.

Suppose for some t ∈ Ṽ_{α−1}^{(1)} ∩ L_{1} we have [K^{(1)}, t] ≠ 1. Let [Ṽ_{α−1}^{(1)} ∩ L_{1}, C I_{L_{1}/Q_{1}}(K^{(1)})] = 1, then tQ_{1} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M. This shows that some 1 ≠ U ≤ [V_{1}, t] is normalized by E, contradicting Z_{α} ≰ M. So we have [Ṽ_{α−1}^{(1)} ∩ L_{1}, C I_{L_{1}/Q_{1}}(K^{(1)})] ≠ 1. We know that [Ṽ_{α−1}^{(2)} ∩ L_{1}, K^{(1)}] = 1. Suppose there is t_{1} ∈ Ṽ_{α−1}^{(2)} ∩ L_{1} with [K, t_{1}] ≠ 1. Then ⟨K^{(1)}, [C I_{L_{1}/Q_{1}}(K^{(1)}), Ṽ_{α−1}^{(1)} ∩ L_{1}]⟩ centralizes [Ṽ_{1}^{(1)}, t_{1}], a contradiction to Z_{α} ≰ M.

Hence we get [Ṽ_{α−1}^{(2)}∩L_{1}, K ×K^{(1)}] = 1 and so [Ṽ_{α−1}^{(2)}∩L_{1}, V_{1}] = 1. We have Ṽ_{α−1}^{(2)} ≰ M. Now as before we see that Ṽ_{α−1}^{(2)}/C IṼ_{α−1}^{(2)}(K_{2}) is the natural module.

But then t in its action on Ṽ_{α−1}^{(2)}/C IṼ_{α−1}^{(2)}(K^{(1)}) is centralized by E ≅ E_{p^{2}}, which shows that E normalizes some 1 ≠ U ≤ [V_{1}, t], a contradiction to Z_{α} ≰ M.

So we have [Ṽ_{α−1}^{(1)} ∩ L_{1}, V_{1}] = 1. Now Ṽ_{α−1} ≰ M. Let q > 2. If K_{1} is sporadic or alternating, then [ω, K_{1}] = 1 (recall K_{1} ≅ J̸_{1} as m_{3}(K_{1}) ≥ 2). Hence as |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(Ṽ_{α−1}^{(1)})| ≤ 2^{5} and Z_{2} ∩ Ṽ_{1} = Z_{1}, we get K_{1} ≲ GL_{2}(4), a contradiction. Hence K ≅ G(r_{1}), r_{1} even.

Suppose q = 2. Then V_{1} induces transvections on Ṽ_{α−1}^{(1)}. Hence K ≅ G(2) or A_{n}. But in case of K ≅ A_{n} we get |V_{1} : C I_{V_{1}}(K_{1})| = 2, a contradiction as above.

Hence in any case K_{1} ≅ G(r_{1}), r_{1} even. Let first K_{1} be untwisted and q > r_{1}. Then [ω_{2}, K_{1}] = 1. Let K_{1} ≅ L_{n}(r_{1}). Suppose |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r_{1}^{4}. Then |Ṽ_{1}/Z_{1}| ≤ q^{3}, i.e. K_{1} ≲ L_{3}(q). But this implies n ≤ 3 and so |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r_{1}^{2} and then K_{1} ≲ Γ_{1}(q), a contradiction. Hence we have r_{1} = 2 and 5 ≤ n ≤ 7. If n = 5, then |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ 2^{6} and as q ≥ 4, we get L_{5}(2) ≲ L_{3}(4) or L_{2}(8), a contradiction. So let 6 ≤ n ≤ 7. Now m_{3}(K) = 3 and so q ≥ 8. This now implies L_{6}(2) ≲ L_{3}(q) or L_{7}(2) ≲ L_{3}(q), or L_{7}(2) ≲ L_{4}(8), a contradiction again.

Let next K_{1} ≅ Sp_{2n}(r_{1}), n ≤ 3. If n = 2, we get Sp_{4}(r_{1}) ≲ L_{2}(q), a contradiction. If n = 3, we get Sp_{6}(r_{1}) ≲ L_{5}(q). For s | r_{1}^{2} − 1, s prime, we

have an elementary abelian subgroup of order s^{3} in Sp_{6}(r_{1}). Hence s | q − 1. Now choose s as a Zsigmondy prime. Then we get r_{1}^{2} ≤ q. This now implies Sp_{6}(r_{1}) ≲ L_{3}(q), a contradiction. Hence we have r_{1} = 8. But now even Sp_{6}(r_{1}) ≲ L_{4}(q). As Sp_{6}(r_{1}) ≴ L_{3}(q), we get q = 16 and so Sp_{6}(8) ≲ L_{4}(16).

But the order of Sp_{6}(8) does not divide the order of L_{4}(16), a contradiction.

Let K_{1} ≅ G_{2}(r_{1}). Then we get K_{1} ≲ L_{2}(q), a contradiction. For K_{1} ≅ 2F_{4}(r_{1}) we would get K_{1} ≲ L_{3}(q) and for K_{1} ≅ Sz(r_{1}) we would get K_{1} ≲ ΓL_{1}(q), a contradiction.

Hence in all these cases we get that q ≤ r_{1}. Now |Ṽ_{α−1} : C IṼ_{α−1}(V_{1})| ≤ q ≤ r_{1}. This implies that V_{1} induces transvections and so K_{1} ≅ SL_{n}(r_{1}) or Sp_{2n}(r_{1}), q = r_{1}. Furthermore Ṽ_{α−1}^{(1)}/C IṼ_{α−1}^{(1)}(K_{1}) is irreducible.

So let now K_{1} ≅ SU_{n}(r_{1}), or Ω_{8}^{−}(r_{1}), q > r_{1}^{2}. By (1.65) and (1.64) [ω_{2}, K_{1}] = 1, ω_{2} = ω, or the element of order 21 in L_{1} ∩ L_{2} for q = 64 and [ω, K] ≠ 1. Now |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r_{1}, r_{1}^{4}, or r_{1}^{6}. This implies K_{1} ≲ ΓL_{1}(q), L_{2}(q), or L_{3}(q), a contradiction in all cases. Hence q ≤ r_{1}^{2} and so we have that Ṽ_{α−1}^{(1)}/C IṼ_{α−1}^{(1)}(K_{1}) is the natural SU_{n}(r_{1}) - module or Ω_{8}^{−}(r_{1}) - module. Furthermore in any case q = r_{1}^{2}. As ⟨ω, K⟩ ≅ K × ZZ_{x}, we see that (r_{1} + 1)^{3} | |̸K_{1}|. Hence K_{1} ≅ SU_{3}(r_{1}) and Ṽ_{α−1}^{(1)}/C IṼ_{α−1}^{(1)}(K_{1}) is the natural module.

Let finally K_{1} ≅ ^{3}D_{4}(r_{1}), q > r_{1}^{5}. Then by (1.65) and (1.64) [ω_{2}, K_{1}] = 1 Now |V_{1}/Z_{2} : C I_{V_{1}/Z_{2}}(ω_{2})| ≤ r_{1}^{5}. But then K_{1} ≲ ΓL_{1}(q), a contradiction.

So we have q ≤ r_{1}^{5}, and |Ṽ_{α−1}^{(1)} : C IṼ_{α−1}^{(1)}(V_{1})| ≤ r_{1}^{5}, a contradiction to (1.40). This shows

(4) Ṽ_{1}/C IṼ_{1}(K) is irreducible. If [s, K] ≠ 1, s an involution, then there is 1 ≠ U ≤ [Ṽ_{1}, s] with E ≤ N_{G}(U), E ≅ E_{p^{2}} and Γ_{E,1}(G) ≤ M.

Let now t_{1} ∈ Ṽ_{α−1}^{(2)} ∩ L_{1}. Then by (4) [t_{1}, K] = 1, as Z_{α} ≰ M. Now by (7.22) [Ṽ_{α−1}^{(2)} ∩ L_{1}, K × K^{(1)}] = 1 and then [Ṽ_{α−1}^{(2)} ∩ L_{1}, V_{1}] = 1. In particular Ṽ_{α−1}^{(2)} ≰ M. But then Ṽ_{α−1}^{(2)}Q_{2} ≤ Ṽ_{α−1}^{(1)}Q_{2}. Hence there is t_{2} = t_{1}t ∈ L_{1} with t ∈ Ṽ_{α−1}^{(1)} \ L_{1} and t_{1} ∈ Ṽ_{α−1}^{(2)}. By (4) we get [t_{2}, Ṽ_{1}] = 1. But now [t, Ṽ_{1}] ≤ Z_{α−1} ≤ Z_{2}. Hence Ṽ_{1}Z_{2} ≤ L_{2}, a contradiction to (1).

Now by (7.26) we have ⟨V_{1}, V_{3}⟩/O_{2}(⟨V_{1}, V_{3}⟩) ≅ L_{2}(q) or L_{2}/Q_{2} if L_{2} is solvable.

Furthermore V′_{1} = Z_{1}. Set Y = (V_{1} ∩ O_{2}(⟨V_{1}, V_{3}⟩))(V_{3} ∩ O_{2}(⟨V_{1}, V_{3}⟩)) = (V_{1}∩Q_{2})(V_{3}∩Q_{2}). Hence Y ≤ L_{2}. Set U = V_{1}∩V_{3}. If U = Y then V_{1}∩Q_{2}≤ L_{2} and so [V_{1} ∩Q_{2}, L_{2}] = Z_{2} and [Q_{2}, O^{2}(L_{2})] = Z_{2}. This now implies with (7.3) and (1.20) that |Q_{2} : Z_{2}| ≤ q. Then V_{1} = Z_{2}Z_{0}. Now O^{2}(L_{1}) normalizes Z_{2} and so Z_{2} ≤ ⟨L_{1}, L_{2}⟩, contradicting (4.14).

So we have U ≠ Y and by (1.27)(iii) Y/U is a direct sum of natural modules. Set L̂_{2} = ⟨V_{1}, V_{3}⟩.

(7.27) Lemma. If x ∈ U with C I_{G}(x) ≤ M, then x ∈ Z_{1}.

Proof. By way of contradiction we may assume x ∉ Z_{2}. Now [x, L̂_{2}] ≤ Z_{2}. We have C IL̂_{2}(x) ≤ M ∩ L̂_{2}. This implies that Z(L̂_{2}) ∩ ⟨x, Z_{2}⟩ ≠ 1. But L̂_{2} ≤ L_{2} and so L_{2} = L̂_{2}S. Now Z(L_{2}) ≠ 1, contradicting (7.3).

From now on we fix the following notation : X ≤ V_{3}, X ∩Q_{1} = 1, |X| = q, XQ_{1}/Q_{1} ≤ S/Q_{1}, ω ∈ L_{1} ∩ L_{2}, o(ω) = x, [X, ω] = X. Furthermore let K be some component of L_{1}/Q_{1} with [X, K] ≠ 1.

(7.28) Lemma. C I_{V_{1}/Z_{1}}(Y) = Z_{2}/Z_{1}.

Proof. Suppose false. Let s ∈ C I_{U}(Y). If s ∉ Z_{2}, we may assume that s ∈ Z(V_{1}). This now implies s ∈ Z(L̂_{2}) and so Z(L_{2}) ≠ 1, a contradiction. Thus C I_{U/Z_{1}}(Y) = Z_{2}/Z_{1}. Let U_{1} ≤ V_{1}, [U_{1}, Y] ≤ Z_{1}, U_{1}∩U = 1, [ω, U_{1}] = U_{1}, |U_{1}| = q. Then we may assume that for some t ∈ X we have [t, V_{1}] ≤ Z_{2}U_{1}. In particular [X, [t, V_{1}]] ≤ Z_{1}. So we have that X acts quadratically on V_{1}/Z_{1} and for t ∈ X, |[V_{1}/Z_{1}, t]| ≤ q^{2}.

Z(Y) = Z_{2} (1)

We have Z(Y) ∩ V_{1} = (Z(Y) ∩ Z(V_{1}))Z_{2}, Z(Y) ∩ V_{3} = (Z(Y) ∩ Z(V_{3}))Z_{2}. As Z(V_{1}) ∩ Z(V_{3}) ≤ Z(L̂_{2}) = 1, we see that

Z(Y) = (Z(Y) ∩ Z(V_{1}))(Z(Y) ∩ Z(V_{3}))

Now Z(Y) is a direct sum of natural modules and so [Z(Y), V_{1}] = Z(Y) ∩ Z(V_{1}). In particular Z(Y) ≤ Q_{1}. This shows [Z(Y), V_{1}] = Z_{1}. Hence Z(Y) ∩ Z(V_{1}) = Z_{1}, and so Z(Y) = Z_{2}.

This now implies

[t, Ṽ_{1}/Z_{1}] ∩ Z_{2}/Z_{1} ≠ 1, and q ≤ |[t, V_{1}/Z_{1}]| ≤ q^{2}. (2)

q > 2 (3)

Suppose q = 2. We first show [K, Y] ≤ K. So assume there is some y ∈ Y with K^{y} ≠ K. Let first ⟨y⟩ = X. Then [C I_{K×K^{y}}(y), [V_{1}, y]] = 1, as |[V_{1}, y]| = 4, a contradiction to Z_{2}/Z_{1} ≤ [V_{1}, y] by (2). So we have [K, t] ≤ K for ⟨t⟩ = X. Now C I_{K×K^{y}}(y) = K_{1} acts on V̂_{1} = [V_{1}, y] ≤ Q_{2}. Furthermore [t, V̂_{1}] ≤ Z_{2}/Z_{1}. So we may assume that t induces a transvection on some nontrivial irreducible K_{1} - module W in V_{1}. Now (1.30) implies K_{1} ≅ L_{n}(2), Sp_{2n}(2), Ω_{2n}^{±}(2), or A_{n}, and W is the natural module. In any case a 3 - element in K_{1} centralizes Z_{2}/Z_{1}. This now implies that 3 ∉ σ(M). But as K_{1} ≅ K, we see m_{3}(K) = 1, whence K ≅ K_{1} ≅ L_{3}(2) or A_{5}, and p = 7, p = 5, respectively. As [t, K] ≠ 1, [t, K^{y}] ≠ 1, we get [[K, V_{1}], K^{y}] = 1. Now there is ν ∈ L_{1}, o(ν) = p with [ν, K × K^{y}] = 1. As [K, V_{1}]/Z_{1} is irreducible, we get [ν, [K × K^{y}, V_{1}]] = 1. But Z_{2}/Z_{1} ≤ [K × K^{y}, V_{1}], a contradiction. So we have shown that [K, Y] ≤ K.

Now by (1.30) K/Z(K) ≅ A_{n}, L_{n}(s), Sp_{2n}(s), Ω_{2n}^{±}(s), s ≤ 4, U_{n}(s), G_{2}(s), U_{4}(3), or K/O_{r}(K) ≅ L_{2}(4), L_{2}(7), [O_{r}(K), V_{1}] = 1, or K is solvable. The cases K/O_{r}(K) ≅ L_{2}(4), or L_{2}(7) are not possible by (1.51).

Suppose first that 3 ∈ σ(M). Then 3 | |̸C I_{K}(C I_{V_{1}/Z_{1}}(S ∩ K))|. Hence we get K ≅ L_{2}(4) or O_{3}(K) ≠ 1. If K ≅ L_{2}(4), then |Ṽ_{1}| = 16. If O_{3}(K) ≠ 1, then |[[t, O_{3}(K)], Ṽ_{1}]| ≤ 16. In both cases Z_{2} ≤ Ṽ_{1} and some 3 - element in L_{1} centralizes Ṽ_{1}, a contradiction.

Hence we may assume p > 3. Now K/Z(K) ≅ A_{n}, n ≤ 11, L_{n}(4), n ≤ 4, L_{n}(2), n ≤ 7, Sp_{2n}(s), n ≤ 3, s ≤ 4, Ω_{8}^{−}(s), s ≤ 4, U_{4}(2), G_{2}(2), or O_{p}(K) ≠ 1.

Suppose first m_{p}(K) = 1. Then there is E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and E ≤ C I_{L_{1}}(K). But as Ṽ_{1} induces at most two nontrivial irreducible modules and p | 4̸ − 1, we get [E, Ṽ_{1}] = 1, a contradiction to Z_{2}/Z_{1} ≤ Ṽ_{1}.

So we have m_{p}(K) > 1. This implies K ≅ A_{10}, A_{11}, L_{4}(4), Sp_{4}(4), Sp_{6}(4), p = 5 in this cases, or K ≅ L_{6}(2), L_{7}(2), p = 7, or O_{p}(K) ≠ 1.

If m_{p}(K) = 2, then there is some ν ∈ C I_{L_{1}/Q_{1}}(K), o(ν) = p. Again [ν, Ṽ_{1}] = 1, a contradiction. So we are left with m_{p}(K) ≥ 3. As p > 3, we have O_{p}(K) ≠ 1 and t centralizes E ≅ E_{p^{2}} in K. Now [[t, Ṽ_{1}], E] = 1, but Z_{2}/Z_{1} ≤ [t, Ṽ_{1}], a contradiction. This proves (v).

As q > 2 by (3), we have a quadratic fours group on V_{1}. This implies that K ≅ G(r), r even, 3·U_{4}(3), sporadic, alternating or solvable, or [O_{r}(K), V_{1}] = 1 by (1.12), (1.51) and (1.11). The last case is impossible by (4.10).

We first again show [ω, K] ≤ K. Suppose false. Let K_{1} ×…×K_{y} = K^{⟨ω⟩}. Suppose y > 3. Then we have K_{1} ≅ L_{2}(r), r even, or L_{3}(2). Furthermore let ⟨ν⟩ ≤ L_{1} ∩ L_{2}, o(ν) = q − 1. Then gcd(o(ν), |K|) = 1, as N_{G}(⟨ν⟩) ≰ M. Let μ ∈ ⟨ν⟩, with [K_{1}, μ] ≤ K_{1}, then C I_{K_{1}}(μ) is a 2 - group as for every odd prime u, u | |K_{1}|, we get u ∈ σ(M). But this is not possible for outer automorphisms of K. Hence we have K^{⟨ν⟩} = K_{1} × … × K_{q−1}. Now t ∈ X acts on E ≅ E_{p^{q−1}}. But then some F ≅ E_{p^{2}} centralizes some x ∈ Z_{2} \ Z_{1}, a contradiction.

So we have y = 3, and then q = 4 or q = 64. In both cases 3 ∉ σ(M). Let first q = 64. As [ω^{3}, Z_{1}] ≠ 1, we get ω^{3} ∉ K_{1} × K_{2} × K_{3}. But as [ω^{3}, K_{1}] ≤ K_{1}, we get that K_{1} is a 3′ - group. This shows K_{1} ≅ Sz(r). As [t, K_{1}] ≠ 1, we get with (1.64) r ≤ q. Hence r = 8 as K_{1} possesses an automorphism of order 3. Let now μ ∈ L_{1} ∩ L_{2} with o(μ) = 7. Then [μ, K_{1}] ≤ K_{1} and ⟨μ, K_{1}⟩ ≅ ZZ_{7} × K_{1}, as [μ, Z_{1}] ≠ 1 and [K_{1}, Z_{1}] = 1. But then m_{7}(⟨μ, K_{1}, K_{2}, K_{3}⟩) = 4, which shows 7 ∈ σ(M), and N_{G}(⟨μ⟩) ≤ M, a contradiction.

So we are left with q = 4. Now m_{3}(K) ≤ 1 and so K ≅ L_{2}(r), L_{3}(r), U_{3}(r), or Sz(r), r even. We have |[V_{1}, t]| ≤ 16. Now (1.26) and (1.27) imply r ≤ 16 for K ≅ L_{2}(r) or L_{3}(r), r ≤ 4 for K ≅ U_{3}(r) or Sz(r). This shows K ≅ L_{2}(r), r ≤ 16, L_{3}(2), L_{3}(8), or U_{3}(4). But as C I_{K_{1}}(ω) ≅ K, we get with (1.63) that K ≅ L_{2}(r), r ≤ 16, or L_{3}(2). As [ω, S]Q_{1}/Q_{1} = Y is elementary abelian, K ≅ L_{3}(2) is not possible.

If [t, K_{3}] = 1, then K_{3} acts on [V_{1}, t]. As |[V_{1}, t]| ≤ 16 and Z_{2}/Z_{1}∩[V_{1}, t] ≠ 1, we get [K_{3}, [V_{1}, t]] ≠ 1. This shows K ≅ L_{2}(4). If [K_{1}, t] ≠ 1 ≠ [K_{2}, t], then as t ∈ Z(S) we have t ∈ K_{1} and so by (1.27) again K ≅ L_{2}(4).

In both cases there are at most two nontrivial irreducible K - modules involved. Hence [K_{3}, [V_{1}, K_{1}]] = 1. But then as ω acts on {K_{1}, K_{2}, K_{3}}, we also get [K_{2}, [V_{1}, K_{1}]] = 1. Now as |[V_{1}, t]| ≤ 16, we may assume [t, K_{3}] = 1 and so K_{3} centralizes [V_{1}, t], a contradiction.

So we have [K, ω] ≤ K and then also [Y, K] ≤ K, as [Y, ω] = Y. Now as [ω, K] ≠ 1, we see that K is solvable or K ≅ G(r) by (1.9). Furthermore by (1.64) and (1.65) q ≤ r or K is twisted and q = r^{2} or K ≅^{3}D_{4}(r). Recall that (1.65)(ii) and (iv) are not possible as [ω, Z_{1}] ≠ 1 and so ω ∉ K. Furthermore (1.65)(vi) is not possible as 3 ∈ σ(M) in this case. Now application of (1.48) shows K ≅ SL_{n}(r), Sp_{2n}(r), SU(n, r), Ω_{2n}^{±}(r), G_{2}(r), Sz(r), r^{2}, r = q or q^{2}, or ^{3}D_{4}(r). If K ≅ ^{3}D_{4}(r) then by (1.40) q^{2} > r^{5}. Furthermore we get from (1.65) and (1.64) q = 64 and r ≤ 4, or q = 8 and r = 2. As ω acts on K we see r = 4 in case of q = 64. Finally by (1.22) and (1.23) Ṽ_{1} involves just natural modules.

(4) p | r̸^{2} − 1, if K is nonsolvable

Suppose p | r^{2} − 1. Assume that V_{1} is an SC - module for K. Let V (λ_{2}) not be involved in V_{1}, then some p - element centralizes Z_{2} or K ≅ L_{2}(r) or L_{3}(r). Let V (λ_{2}) be involved. Then as |[Ṽ_{1}, t]| ≤ r^{2}, we see K ≅ L_{3}(r), L_{4}(r), Sp_{4}(r) or U_{4}(r). Hence in any case K ≅ L_{2}(r), L_{3}(r), L_{4}(r), Sp_{4}(r), or U_{4}(r). In case of U_{4}(r) both the natural U_{4}(r) - module and the natural Ω_{6}^{−}(r) - module are involved. Hence |[V_{1}, t]| ≥ r^{4}. This implies q = r^{2} and so there is some ν ∈ L_{1} ∩ L_{2} with o(ν) = p. But then m_{p}(C I_{⟨K,ν⟩}(ν)) ≥ 3, contradicting N_{G}(⟨ν⟩) ≰ M. If K ≅ Sp_{4}(r), we get that both natural modules are involved. Now |[V_{1}, t]| ≥ r^{3}. But r ≥ q, a contradiction. Let K ≅ L_{4}(r), then also [V_{1}, t]| ≥ r^{3}, as V (λ_{2}) is involved and V_{1} has to involve some further nontrivial irreducible module. But again r ≥ q gives a contradiction.

Let K ≅ L_{3}(r). Then |[V_{1}, t]| = r^{2} and so q = r. Furthermore there are exactly two (nonisomorphic) irreducible modules involved. As p | r̸ − 1, we

get p | r + 1 and m_{p}(K) = 1. Now there is some μ ∈ C I(K), o(μ) = p with [Ṽ_{1}, μ] = 1. But then [μ, Z_{2}] = 1, a contradiction.

Let finally K ≅ L_{2}(r). Then Ṽ_{1} involves exactly one nontrivial irreducible module, as Ṽ_{1} is an SC - module. Again some μ with o(μ) = p centralizes Ṽ_{1}. But [X, V_{1}] ≤ [X, K] = Ṽ_{1}, and so Z_{2} ∩ Ṽ_{1} > Z_{1}, a contradiction.

So assume now that Ṽ_{1} is not an SC - module. Then we have that K does not contain elementary abelian subgroups of order q^{2}. This shows K ≅ L_{2}(r), U_{3}(r) or Sz(r). Let 1 ≠ y ∈ C I_{Y} (K). Suppose [y, Ṽ_{1}] = 1. Then some p - element μ ∈ C I(K) centralizes Ṽ_{1} and so Z_{2}, a contradiction. So we have [y, Ṽ_{1}] ≠ 1. In particular there are at least two nontrivial irreducible modules involved. Hence K ≅ L_{2}(q) or U_{3}(\sqrt{q}) and both modules are natural modules. But as ω acts on K, we get that |X| = q and so K ≅ L_{2}(q), X ∈ Syl_{2}(K). As ω acts on ⋂_{x∈X^{♯}}[V_{1}/Z_{1}, x] = T_{1}, we have T_{1} = Z_{2}/Z_{1} or

T_{1} = [V_{1}/Z_{1}, t]. As K admits some field automorphism ν of order p which normalizes X and so also T_{1}, we get the latter. Now [V_{1}/Z_{1}, t] = [V_{1}/Z_{1}, X]. Hence Ṽ_{1}/Z_{1} is a direct sum of two natural modules. But now again ν centralizes some s ∈ Z_{2} \ Z_{1}, a contradiction to (4.2).

So we have C I_{Y} (K) = 1. Hence |Y : Y ∩ Q_{1}| ≤ r. As [Y ∩ Q_{1}, V_{1}] ≤ Z_{1}, we get V_{3} ∩ Q_{1} = V_{1} ∩ V_{3}. This implies |Y : V_{1} ∩ V_{3}| ≤ r^{2}. As r < q^{2} and |Y : Y ∩ Q_{1}| = q^{u} for some u, we get |Y : Y ∩ Q_{1}| = r = q. In particular either Ṽ_{1}/C IṼ_{1}(K) is irreducible or a direct sum of two natural L_{2}(q) - modules. Furthermore |Y/V_{1} ∩ V_{3}| = q^{2} and so by (1) |V_{1}/Z_{1}| ≤ q^{4}. Hence K ≅ Sz(q) or L_{2}(q) and |V_{1}/Z_{1}| = q^{4}. But now again some ν ∈ L_{1}, ν inducing some field automorphism, o(ν) = p, centralizes some s ∈ Z_{2} \ Z_{1}, a contradiction.

(5) K is a p–group

Suppose K to be nonsolvable. By (4) p | r̸^{2} − 1. This now implies K ≅ L_{n}(r), n ≤ 4, L_{n}(2), 5 ≤ n ≤ 7, Sp_{2n}(r), n ≤ 3, SU_{n}(r), n ≤ 5, Ω_{8}^{−}(r), G_{2}(r) or ^{3}D_{4}(r).

Suppose m_{p}(K) = 1, K ≅ ^{3̸}D_{4}(r) or SU_{n}(r). Then there is μ ∈ C I_{L_{1}/Q_{1}}(K), o(μ) = p. As p | r̸^{2} − 1 and there are at most two nontrivial irreducible modules involved, we see that [μ, Ṽ_{1}/Z_{1}] = 1, contradicting Z_{2} ∩ Ṽ_{1} > Z_{1}, by (2).

So we have m_{p}(K) > 1 or K ≅ SU_{n}(r) or ^{3}D_{4}(r). As q > 2, we see that L_{n}(2) is not possible. Hence in any case K ≅ SU_{n}(r) or ^{3}D_{4}(r). Let K ≅ SU_{n}(r). Then also m_{p}(K) = 1. Furthermore p | r^{2} + 1, Ṽ_{1}/C IṼ_{1}(K) is a direct sum of two natural modules and so r^{2} = q. Now m_{x}(⟨ω, K⟩) ≥ 4 or K ≅ U_{3}(\sqrt{q}). The former is out as N_{G}(⟨ω⟩) ≰ M, the latter is out as X ≤ K and |X| = q.

So we are left with^{3}D_{4}(4) or^{3}D_{4}(2). As X ≤ S/Q_{1} we see that Z(S∩K)∩ X ≠ 1. But |Z(S ∩ K)| ≤ 4 and ω acts on Z(S ∩ K), so [ω^{3}, Z(S ∩ K)] = 1 and [X, ω^{3}] = X, a contradiction. This proves (4).

Now we have that K is a normal p–group. As q > 2, we get m_{p}(K) ≥ 3. Let U = [K/Φ(K), X]. If |C I_{U}(t)| ≥ p, there is t_{1} ∈ X with [C I_{U}(t), t_{1}] ≠ 1. Now [t_{1}, [t, V_{1}]] = 1 by quadratic action. Hence [C I_{U}(t), t_{1}] centralizes some z ∈ Z_{2} \ Z_{1}, a contradiction. This shows [U, t] = U, which contradicts |X| ≥ 4.

(7.29) Lemma. a) Z_{1} = Z(V_{1}).

b) Let q > 2. Then [K, ω] ≤ K.

Proof. a) follows from (7.28) as Z_{2} ≰ Z(V_{1}).

b) Suppose K_{1}×…×K_{y} = K^{⟨ω⟩}, y ≥ 3. Let first y > 3. Then m_{p}(K_{1}) = 1 for every odd prime p. Now let ⟨ν⟩ ≤ L_{1} ∩ L_{2}, ω ∈ ⟨ν⟩, o(ν) = q − 1. Let μ ≠ 1, μ ∈ ⟨ν⟩, [K_{1}, μ] ≤ K_{1}. Then [K_{i}, μ] ≤ K_{i}, i = 1, …, y. As N_{G}(⟨μ⟩) ≰ M, we get C I_{K_{1}}(μ) is a 2 - group. But K_{1} ≅ L_{2}(r), Sz(r) or J_{1} and so μ induces a field automorphism. Then C I_{K_{1}}(μ) is not a 2 - group. So we have K_{1}^{⟨ν⟩} = K_{1} × … × K_{q−1}. Suppose y = 3. Let q = 64 and o(ω) = 9. As [Z_{1}, ω^{3}] ≠ 1, we have ω^{3} ∉ K_{1} ×…×K_{y}. Now 3 ∉ σ(M). Hence 3 | |̸K_{1}|. This implies K_{1} ≅ Sz(r). Now ω^{3} induces a field automorphism and so r = r_{1}^{3}. Hence 7 | |K_{1}|. Let now μ ∈ ⟨ν⟩, o(μ) = 7. Suppose [μ, K_{1}] ≤ K_{1}.

Then 7 | |C I_{K_{1}}(μ)| and so there is E ≅ E_{7^{4}}, μ ∈ E, E ≤ L_{1}, a contradiction

to N_{G}(⟨μ⟩) ≰ M. Hence K_{1}^{⟨ν⟩} = K_{1} × … × K_{21}. But as C I_{K_{1}}(ω^{3}) is not a 2 - group this contradicts N_{G}(⟨ω^{3}⟩) ≰ M. Hence in any case

(1) {K^{⟨ν⟩} = K_{1} × … × K_{q−1}, ⟨ν⟩ ≤ L_{1} ∩ L_{2}, o(ν) = q − 1, K_{1} ≅ \atop L_{2}(r), Sz(r) or J_{1}}

q = 4 (2)

Suppose q > 4. As [ν, S ∩ K^{⟨ν⟩}] = Y ∩ K^{⟨ν⟩} we see that there is t ∈ Y ∩ K^{⟨ν⟩} such that C I(t) ≥ F ≅ E_{p^{q−3}}. Now F acts on [t, V_{1}]. We have |[t, V_{1}] : [t, V_{1}] ∩ V_{3}| = q. Furthermore by (7.28) [t, V_{1}] ∩ V_{3} ≠ 1.

There is E ≤ F, E ≅ E_{p^{2}} and |C I_{[t,V_{1}]}(E)| ≥ 4^{q−5}. As q ≥ 8, we have 4^{q−5} > q and so C I_{[t,V_{1}]}(E) ∩ V_{3} ≠ 1, which contradicts (7.3).

C I_{Y} (K_{1} × K_{2} × K_{3}) = 1 (3)

Let y ∈ C I_{Y} (K^{⟨ν⟩}), y ≠ 1. Then K^{⟨ν⟩} acts on [y, V_{1}] and |[y, V_{1}] : C I_{[y,V_{1}]}(t)| ≤ 16, for t ∈ Y ∩K^{⟨ν⟩}. Now (1), (1.27) and (1.25) imply K_{1} ≅ L_{2}(r), r ≤ 16. But we have [t, K_{1}] ≠ 1 ≠ [t, K_{2}] and so t inverts an elementary abelian group of order 17^{2}, 7^{2} for r = 16, r = 8, respectively. As |[y, V_{1}] : C I_{[y,V_{1}]}(t)| ≤ 16, this is impossible. Hence K ≅ L_{2}(4), p = 5. Now let ρ ∈ K_{1}, o(ρ) = 5. Then we see |[[t, V_{1}], ρ]| ≤ 2^{8} and so [V_{1}, t] involves at most two nontrivial irreducible L_{2}(4) - modules. Hence we may assume [[V_{1}, t], K_{1}, K_{3}] = 1. As ω acts on {K_{1}, K_{2}, K_{3}} the 3 - subgroup lemma implies [[[V_{1}, t], K_{1}], K_{2} × K_{3}] = 1. As [t, [K_{1}, [V_{1}, y]]] ≤ V_{1} ∩ V_{3} this contradicts (7.3).

Now by (1), (2) and (3) |Y : Y ∩Q_{1}| ≤ r^{2}, (64 for K ≅ J_{1}). As [x, V_{1}] = Z_{1} for every x ∈ Z_{2} \ Z_{1}, (7.28) implies that V_{1}/H is extraspecial for every hyperplane H of Z_{1}. Now V_{3} ∩ Q_{1} = V_{3} ∩ V_{1} is elementary abelian and so |V_{1}/Z_{1}| ≤ q^{2}|Y : Y ∩ Q_{1}|^{2} ≤ 16r^{4} (2^{16} for K ≅ J_{1}). Let r = 2^{m}. Then (L_{1}/Q_{1})^{(∞)} ≲ O^{±}(4m + 4, 2), (O^{±}(16, 2) for K ≅ J_{1}). But if L_{2}(r) × L_{2}(r) × L_{2}(r) or Sz(r) × Sz(r) × Sz(r) acts on V_{1}/Z_{1}, then |V_{1}/Z_{1}| ≥ r^{6}, and so 4m+4 ≥ 6m. This implies m = 2 and we have L_{2}(4) ≅ K_{1}. As 7^{3} | |̸O_{16}^{±}(2)|, we get K_{1} ≅ J̸_{1}.

So we have |V_{1}/Z_{1}| = 2^{12} and K_{1} ≅ L_{2}(4). Now V_{1}/Z_{1} = [V_{1}/Z_{1}, K_{1}] ⊕ [V_{1}/Z_{1}, K_{2}] ⊕ [V_{1}/Z_{1}, K_{3}]. Hence for some t ∈ Y \ Q_{1}, we have [t, V_{1}/Z_{1}] is centralized by ρ ∈ L_{1}, o(ρ) = 5 = p. This implies [t, V_{1}/Z_{1}] ∩ Z_{2}/Z_{1} = 1. But as [t, V_{1}/Z_{1}] ∩ C I_{V_{1}/Z_{1}}(Y) ≠ 1, we get a contradiction to (7.28).

(7.30) Lemma. C I_{Y} (K) = 1.

Proof.

[K, Y] ≤ K (1)

This is clear by (7.29) if q > 2. So let q = 2, y ∈ Y with K^{y} ≠ K. Then [C I_{Y} (K), K^{y}] = 1. Let ỹ ∈ C I_{Y} (K)^{♯}, W = [V_{1}, ỹ]. Then K × K^{y} acts nontrivially on W. As W ≤ Q_{2}, we see |[W, y]| ≤ 4. Now [C I_{K×K^{y}}(y), [W, Y]] = 1, contradicting (7.28).

We fix some notation. Let again ỹ ∈ C I_{Y} (K)^{♯}, W = [V_{1}, ỹ] and W̃ = [K, W]. Then we have |[t, W̃]| ≤ q^{2} and |W : W ∩ V_{1} ∩ V_{3}| ≤ q.

Let K be sporadic or alternating, K ≅ A̸_{5}, A_{6} or A_{8}. Then q = 2, otherwise [ω, K] ≠ 1 implies [ω, Z_{1}] = 1.

Now |[W̃ , t]| ≤ 4. This implies with (1.25) K/Z(K) ≅ A_{n}. Suppose [C I_{Y} (K), W̃] = 1. Then we get that no p - element centralizes C IW̃ (Y). This implies p > 3 and so n ≤ 11. As |W̃ : W̃ ∩V_{1}∩V_{3}| ≤ 2, we get |W̃ : C IW̃ (Y)| ≤ 2|Y : C I_{Y} (K)|. As |Y : C I_{Y} (K)| ≤ 2^{5} this implies |W̃ : C IW̃ (Y)| ≤ 2^{6}. As C IW̃ (Y) = Z_{2}/Z_{1}, we see |W̃ | ≤ 2^{7}. Hence we get K ≅ A_{7}, |W̃ | = 16 and |Y : C I_{Y} (K)| = 4. As C I_{V_{1}/Z_{1}}(K) = 1, we get W = W̃ and so |[V_{1}/Z_{1}, ỹ]| = 16. Furthermore we have Z_{2} ≤ W̃ and so if ỹ_{1} ∈ C I_{Y} (K), 1 ≠ ỹ_{1} ≠ ỹ, then [W, ỹ_{1}] = [W, ỹ]. But this is impossible as for w ∈ V_{1} \ Q_{2} we have [Y, w](V_{1} ∩ V_{3}) = V_{1} ∩ Q_{2}. So we get |Y : C I_{Y} (K)| = 2 and then |Y : Y ∩ Q_{1}| ≤ 2^{3}. This implies |V_{1}/Z_{1}| ≤ 2^{8} and L_{1}/Q_{1} ≲ O_{8}^{±}(2). As m_{p}(L_{1}/Q_{1}) ≥ 3, p ≠ 3, this impossible.

So we have [C I_{Y} (K), W̃] ≠ 1. Let ỹ_{1} ∈ C I_{Y} (K) with W̃^{y_{1}} ≠ W̃. Then as |W̃ : C IW̃ (ỹ_{1})| = 4, we get |W̃ | = 4. But then [K, W̃] = 1, a contradiction.

Let K be a group of Lie type in odd charcteristic which is not a group of Lie type in even characteristic too. If q = 2, we get a contradiction with (1.30). So we have q > 2. Suppose first that ω induces a field automorphism on K. Then E(C I_{K}(ω)) ≅ G(p^{h}), where K ≅ G(p^{f}). If p ∈ σ(M), we get m_{p}(C I_{K}(ω)) = 1. Hence E(C I_{K}(ω)) ≅ L_{2}(p) or ^{2}G_{2}(3)′ . Now K ≅ L_{2}(p^{f}) or ^{2}G_{2}(3^{f}). But [X, ω] = X and so o(ω) = 3 or o(ω) = 7 in case of K ≅ 2G_{2}(3^{f}). Now |[W̃ , t]| ≤ 16, or |[W̃ , t]| ≤ 2^{6}. But this contradicts (1.29). So p ∉ σ(M). In particular m_{p}(K) ≤ 3. Hence f ≤ 3 and so o(ω) = 3 or 9. Furthermore K ≅ L_{2}(p^{3}). But now again |[W̃ , t]| ≤ 16, a contradiction.

So we have that ω does not induce a field automorphism. Now application of (1.10) shows that o(ω) = 3, 9 or K ≅ ^{2}G_{2}(r). In the latter we get |[W̃ , t]| ≤ 64, a contradiction to (1.29). Hence K ≅ ^{2̸}G_{2}(r) and now q = 4 or q = 64.

If q = 64, then there is ν ∈ L_{1} ∩ L_{2} with o(ν) = 7. As X ≤ K, we see [ν, K] ≤ K. Hence ν induces a field automorphism on K by (1.10). Now as before K ≅ L_{2}(p^{7}), a contradiction.

So we are left with q = 4, o(ω) = 3. Now |[W̃ , t]| ≤ 16. By (1.29) K ≅ L_{3}(3), L_{2}(25), L_{2}(p) or 3 · U_{4}(3). As 3 ∉ σ(M), we have K ≅ 3̸ · U_{4}(3). As L_{3}(3) and L_{2}(25) do not possess an automorphism of order 3 normalizing a Sylow 2 - subgroup, we get K ≅ L_{2}(p), p ≡ 3, 5 (mod 8). We have |Y : Y ∩ C I(K)| = 4. Suppose [W̃ , C I_{Y} (K)] = 1. Let X = ⟨t, t_{1}⟩. We have C IW̃ (Y) = C IW̃ (X). Now C IW̃ (t) ≰ V_{1} ∩ V_{3} and so |C IW̃ (t) : C I_{C IW̃ (t)}(t_{1})| ≤ 4. Hence |W̃ : C IW̃ (Y)| ≤ 2^{6}. As C IW̃ (Y) ≰ Z_{2}/Z_{1} by (7.28), we get |W̃ | ≤ 2^{8}.

But L_{2}(p) ≴ GL_{8}(2), as p > 7. So we have [W̃ , ỹ_{1}] ≠ 1, for some ỹ_{1} ∈ C I_{Y} (K). This implies |W̃ | = 4, a contradiction. So we have shown

(2) K ≅ G(r), r even, or a normal p-subgroup, or O_{r}(K) ≠ 1 for some r, r odd, and K/O_{r}(K) ≅ L_{3}(r), or L_{2}(r).

If O_{r}(K) ≠ 1, then we have q ≤ 4. So |[W̃ , t]| ≤ 16. By (1.51) [O_{r}(K), W̃] = 1. Hence [[ỹ, V_{1}], O_{r}(K)] = 1. But then [Z_{2}, O_{r}(K)] = 1, so either we have (4.9)(ii) or we may consider L_{1} as a component L_{3}(r), r odd. The latter contradicts (2). The former is a contradiction to (4.10). So

(3) K ≅ G(r), r even, or a normal p–subgroup

Let first K ≅ G(r). We have |[W̃ , t]| ≤ q^{2}. Let K be untwisted. Then application of (1.65) and (1.64) shows q ≤ r. Now we get with (1.48) that K ≅ L_{n}(r), Sp_{2n}(r), Ω_{2n}^{+}(r), or G_{2}(r). Suppose next [W̃ , C I_{Y} (K)] = 1. Then C IW̃ (S ∩ K) ≤ Z_{2}/Z_{1}. Hence p | |̸C I_{K}(C IW̃ (S ∩ K))|.

Let V (λ_{2}) be involved in W̃. Then by (1.23) K ≅ L_{3}(r) or L_{4}(r). If K ≅ L_{4}(r), then as |[W̃ , t]| ≤ r^{2}, we get that W̃ ≅ V (λ_{2}). Now C IW̃ (S ∩ K) is centralized by L_{2}(r) in all cases but K ≅ L_{2}(r) or L_{3}(r). This shows p | r̸^{2} − 1.

Assume now p | r̸^{2} − 1. Then K ≅ L_{n}(r), n ≤ 4, L_{n}(2), n ≤ 7, Sp_{4}(r), Sp_{6}(r), G_{2}(r).

If for some ỹ ∈ C I_{Y} (K), [W̃ , y] ≠ W̃, then |[W̃ , y]| ≤ q^{2} ≤ r^{2} and so K ≅ L_{2}(q). So we have

(4) If K ≅ G(r), untwisted, then K ≅ L_{n}(r), n ≤ 4, L_{n}(2), n ≤ 7, Sp_{4}(r), Sp_{6}(r) or G_{2}(r). If [W̃ , C I_{Y} (K)] ≠ 1, then K ≅ L_{2}(q).

Suppose next that K is twisted, K ≅ ^{3̸}D_{4}(r). Let q ≤ r. Then by (1.48) K ≅ SU_{n}(r), Ω_{2n}^{−}(r) or Sz(r). Furthermore q = r.

Let K ≅ S̸z(r) or SU_{3}(r). If p | r^{2} − 1, then as W̃ involves the natural

module, we see that p | |C I_{K}(Z_{2})|, a contradiction. So p | r̸^{2} − 1. This shows K ≅ SU_{4}(r), SU_{5}(4), or Ω_{8}^{−}(r).

Let now q > r. Then by (1.65) and (1.64) q = r^{2} and K ≅ SU_{n}(r) or Ω_{2n}^{−}(r). Now let K ≅ S̸U_{3}(r), SU_{4}(r), and SU_{5}(r). Then by (1.48) W̃ involves just natural modules and so C IW̃ (S ∩ K) is centralized by L_{2}(r). As p | r^{2} − 1 or K ≅ Ω_{8}^{−}(r), we get |[W̃ , C I_{Y} (K)]| ≠ 1. But for ỹ ∈ C I_{Y} (K), we have |[W̃ , ỹ]| ≤ r^{4}, a contradiction. We have (5) If K ≅ G(r), then K ≅ L_{n}(r), n ≤ 4, L_{n}(2), n ≤ 7, Sp_{4}(r), Sp_{6}(r), G_{2}(r), q ≤ r, SU_{n}(q), SU_{n}(\sqrt{q}), n ≤ 5, Sz(q), Ω_{8}^{−}(q), Ω_{8}^{−}(\sqrt{q}), or ^{3}D_{4}(r). If [W̃ , C I_{Y} (W̃] ≠ 1, then K ≅ L_{2}(q).

As V_{1}/H is extraspecial for every hyperplane H of Z_{1} by (7.28), we get that |W | ≤ q^{2}|Y : C I_{Y} (K)|, if [C I_{Y} (K), W] = 1.

(6) K ≅ L_{2}(r), or K is a normal p–subgroup

Suppose false. Then [C I_{Y} (K), W] = 1. Let first K ≅ SU_{n}(q), n ≤ 5, Ω_{8}^{−}(q) or Sz(q). Then |W | ≤ q^{2}q, q^{2}q^{4}, q^{2}q^{4}, q^{2}q^{6}, q^{2}q, respectively. But then by (1.48) we have K ≅ SU_{4}(q) ≅ Ω_{6}^{−}(q), |W | = q^{6}, or K ≅ Ω_{8}^{−}(q) and |W | = q^{8}. In both cases W is the natural module for K. Now choose t ∈ X^{♯}. As t ∈ Z(S ∩ K) we get that |[W, t]| = q^{2}. Hence W = [V_{1}, ỹ] for every ỹ ∈ C I_{Y} (K)^{♯}. So |C I_{Y} (K)| ≤ q. Now |V_{1}/Z_{1}| ≤ q^{12} or q^{16}. As |[V_{1}, y]| = q^{6}, q^{8}, we get |V_{1}/Z_{1}| = q^{12} or q^{16} and V_{1}/W̃ is the natural module too. Hence |C I_{Y} (K)| = q. Now C I_{L_{1}/Q_{1}}(K) acts on C I_{V_{1}/Z_{1}}(S) and so C I_{L_{1}/Q_{1}}(K) ≅ L_{2}(q), as it contains p - elements and a Sylow 2 - subgroup of L_{2}(q). Now we get that p | q^{2} −1. But as W̃ is the natural module Z_{2}/Z_{1} is centralized by L_{2}(q), contradicting (4.2).

As ω acts fixed point freely on X ≤ Z(S ∩K), we see that K ≅ SU_{n}(\sqrt{q}), Ω_{8}^{−}(\sqrt{q}) are not possible, since |Z(S ∩ K)| = \sqrt{q}.

Let now K ≅ L_{n}(r), n ≤ 4, Sp_{4}(r), Sp_{6}(r) or G_{2}(r). Then |W | ≤ q^{2}r^{2}, q^{2}r^{4}, q^{2}r^{3}, q^{2}r^{6}, q^{2}r^{3}. As q ≤ r we have K ≅ G̸_{2}(r). Furthermore W involves exactly one nontrivial irreducible module which is the natural module or K ≅ L_{4}(r), r = q and W̃ is the orthogonal module, or K ≅ Sp_{6}(r), r = q, and W̃ is the spin module. As Z_{2} ≤ W̃ we get W̃ ∩ [V_{1}, ỹ_{1}] ≠ 1 for all ỹ_{1} ∈ C I_{Y} (K). Hence W̃ = [[V_{1}, ỹ_{1}], K]. Set

W_{1} = ⟨[V_{1}, ỹ_{1}] | ỹ_{1} ∈ C I_{Y} (K)⟩

Now [W_{1}, K] = W̃. As C I_{V_{1}/Z_{1}}(K) = 1 we see by (1.20) that |W_{1} : W̃ | ≤ r. But |W_{1}(V_{1} ∩ V_{3}) : V_{1} ∩ V_{3}| = |C I_{Y} (K)|. As |W̃ : W̃ ∩ V_{1} ∩ V_{3}| ≤ q, we have |C I_{Y} (K)| ≤ qr. This now implies |Y : Y ∩Q_{1}| ≤ qr^{3}, qr^{5}, qr^{4}, qr^{7}, respectively.

As p | r̸^{2} − 1, we get that | ⨁_{i=0}^{p−1} W̃^{μ^{i}}| ≥ |W̃ |^{3} for μ ∈ C I_{L_{1}/Q_{1}}(K), o(μ) = p.

So we have |V_{1}/Z_{1}| ≥ r^{9}, r^{12}, r^{12}, r^{18}, respectively. As |V_{1}/Z_{1}| ≤ q^{2}|Y : Y ∩ Q_{1}|^{2}, we get |V_{1}/Z_{1}| ≤ q^{4}r^{6}, q^{4}r^{10}, q^{4}r^{8}, q^{4}r^{14}.

Let K ≅ (S)L_{3}(r). If |W | = r^{3}, then |Y : Y ∩ Q_{1}| ≤ qr^{2} and so |V_{1}/Z_{1}| ≤ q^{4}r^{4} ≤ r^{8}, a contradiction. So we have W > W̃. By (1.20) K ≅ L_{3}(2). Now |V_{1}/Z_{1}| ≤ 2^{10}. But m_{7}(O_{10}^{±}(2)) = 1, contradicting p = 7.

Let K ≅ (S)L_{4}(r). We have |W̃ | = r^{4}. By (1.20) W̃ = W. Now |C I_{Y} (K)| ≤ q and then |V_{1}/Z_{1}| ≤ q^{4}r^{8}. This shows q = r and |V_{1}/Z_{1}| = r^{12}. If q > 2 then ⟨ω, K⟩ ≅ ZZ_{x} × K and so m_{x}(⟨ω, K⟩) = 4, whence x ∈ σ(M), a contradiction to N_{G}(⟨ω⟩) ≰ M. This shows q = 2 and p = 7. But m_{7}(O_{12}^{±}(2)) ≤ 2, a contradiction.

Let K ≅ Sp_{4}(r). Then |V_{1}/Z_{1}| = r^{12} = q^{12} and |C I_{Y} (K)| = q^{2}, |W | = q^{5}. But in this case W_{1} = W. Hence [V_{1}, ỹ_{1}] = W for all ỹ_{1} ∈ C I_{Y} (K)^{♯}. This implies |C I_{Y} (K)| ≤ q, a contradiction.

Let finally K ≅ Sp_{6}(r). Then r = q and |W : W̃ | = q. But then again |C I_{Y} (K)| ≤ q, which implies |Y : Y ∩ Q_{1}| ≤ q^{7}, a contradiction as |V_{1}/Z_{1}| ≤ q^{16}.

Let now K ≅ L_{n}(2), 5 ≤ n ≤ 7, p ≠ 3 . We have q = 2. Furthermore |W | ≤ 2^{8}, 2^{11}, 2^{14}, respectively. Hence in any case just natural modules are involved. By (1.20) we get |W | = 2^{n} or n = 7 and |W | = 2^{14} = |W̃ |. Suppose the former. Then Z_{2} ≤ W and so W = [V_{1}, ỹ_{1}] for every ỹ_{1} ∈ C I_{Y} (K)^{♯}. Now |C I_{Y} (K)| ≤ 2 and so |Y : Y ∩ Q_{1}| ≤ 2^{7}, 2^{10}, 2^{13}, respectively. Furthermore |V_{1}/Z_{1}| ≤ 2^{16}, 2^{22}, 2^{28}. As p | |̸C I_{K}(Z_{1})|, we get n ≠ 6, p = 31 for n = 5 and p = 127 for n = 7, But |O_{16}^{±}(2)|_{31} = 31, |O_{28}^{±}(2)|_{127} = 127^{2}, a contradiction.

Hence W = W̃, |W | = 2^{14}. But |[W, u]| = 4 for every u ∈ Y with [K, u] ≠ 1. This implies that u induces a transvection on the natural module. Now |Y : C I_{Y} (K)| ≤ 2^{6}. Hence |W | ≤ 2^{8}, a contradiction.

Let finally K ≅ ^{3}D_{4}(r). By (1.40) q > r^{2}. By (1.65) and (1.64) q = r^{3} or q = 64 and r = 2. As [ω, X] = X, and X ≤ Z(S ∩K), we get a contradiction since |Z(S ∩ K)| = r.

(7) If K is a normal p–subgroup, then q = 2, and K ≅ ZZ_{5} or p = 3 and m_{3}(K) ≤ 2.

Let K be a normal p-subgroup, Y_{1} ≤ Y, C I_{Y_{1}}(K) = 1. Let |Y_{1}| ≥ 16. Then there is K_{1} ≤ K, K_{1} ≅ D_{2p} × … × D_{2p}, |K_{1}| = p^{n}|Y_{1}|, where |Y_{1}| = 2^{n} and Y_{1} ∈ Syl_{2}(K_{1}) by [GoLyS, (24.1)]. For t ∈ Y_{1}, we have [W, t] ≤ V_{1} ∩ V_{3}. Let t ∈ Y_{1} with p^{3} | |C I_{K_{1}}(t)|. Then by (7.3) we get [W, t] = 1, a contradiction. So we have |Y_{1}| ≤ 8, i.e. q ≤ 8. If q = 8, we can choose t such that K_{2} ≅ D_{2p} × D_{2p} acts on [W, t]. Now for t_{1} ∈ Y ∩ K_{2}, we get 1 ≠ [[W, t], t_{1}] ≤ Z_{2}. We may choose t_{1} with C I_{K_{2}}(t_{1}) ≥ D_{2p}. As [Y ∩ C I_{K_{2}}(t_{1}), Z_{2}] = 1, we get some x ∈ Z_{2} \ Z_{1}, with p | |C I_{K_{1}}(x)|, a contradiction to (4.2).

So we have q ≤ 4. Let q = 4. Then p ≥ 5. We have ⟨X, ω⟩ ≅ A_{4}. As |W : W ∩ V_{3}| = 4, we see that m_{p}(K) = 2. Now for a critical subgroup C of K we get Ω_{1}(C) ≅ ZZ_{p} × ZZ_{p} or E_{p^{1+2}}. But no such group admits A_{4}, a contradiction.

Now q = 2, |W : W ∩ V_{3}| = 2. Again m_{p}(K) ≤ 2. As |[W, t]| ≤ 4, we see p ∈ {3, 5}. If p = 5, we get C I_{K}(t) = 1 and so K ≅ ZZ_{5}. This proves (7).

Now by (6) and (7) we have K_{1}×…×K_{l} ≤ L_{1}/Q_{1}, K_{i} ≅ L_{2}(r_{i}), q ≤ r_{i}, or K_{i} is a normal p–subgroup, i = 1, …, l. Furthermore C I_{Y} (K_{1} × … × K_{l}) = 1. Finally at most one of the K_{i} is a p-subgroup.

If q = 2, then K ≅ L̸_{2}(r) (8)

Suppose false. Then as |[W, t]| ≤ 4, we see K ≅ L_{2}(4). But as q ≠ r, we get that |W | ≤ rq^{2} = 16. So W = W̃ and then [W̃ , C I_{Y} (K)] = 1. In particular W̃ = [V_{1}, ỹ_{1}] for every ỹ_{1} ∈ C I_{Y} (K)^{♯}. This now implies |C I_{Y} (K)| = 2, and then |Y : Y ∩ Q_{1}| ≤ 8. Now |V_{1}/Z_{1}| ≤ 2^{8}. Hence L_{1}/Q_{1} ≲ O_{8}^{±}(2). As p^{3} | |L_{1}/Q_{1}|, we see p = 3. Now V_{1}/W ≅ W and so an element of order 5 acts fixed point freely on V_{1}/Z_{1}. As 3 | |̸C I_{L_{1}}(Z_{2})|, we see that |Z_{2}^{L_{1}}| = 135 and so all involutions in V_{1} are conjugate. But there is some ρ ∈ L_{1}, o(ρ) = 3, with |C I_{V_{1}}(ρ)| = 2^{5}. Now Ω_{1}(C I_{V_{1}}(ρ)) ≥ Z_{1}, and so there are involutions i centralized by ρ. But this contradicts i ∼ j, j ∈ Z_{2} \ Z_{1}, and 3 | |̸C I_{L_{1}}(j)|.

Now by (7) and (8) we get that K_{i} ≅ L_{2}(r_{i}), r_{i} ≥ q > 2, for all i.

l = 2 (9)

Assume l > 2. Let t_{i} ∈ K_{i}. Then |[V_{1}, t_{1}] : [V_{1}, t_{1}] ∩ V_{3}| ≤ q. Hence [[V_{1}, t_{1}], t_{2}] ≤ V_{1} ∩ V_{3} and |[[V_{1}, t_{1}], t_{2}]| ≤ q^{2}. This implies K_{3} ≅ L_{2}(q). Now interchange K_{1}, K_{2}, K_{3}, which implies K_{i} ≅ L_{2}(q) for all i. But then there is some μ ∈ L_{1} ∩ L_{2} with o(μ) = p. This gives N_{G}(⟨μ⟩) ≤ M, as [μ, K_{i}] ≤ K_{i}. But this contradicts L_{2} ≰ M.

By (9) we get Y ≤ K_{1} × K_{2}, K_{i} ≅ L_{2}(r_{i}), i = 1, 2. Hence there is some ρ ∈ L_{1}, ρ ∉ K_{1} × K_{2} with o(ρ) = p. Let t_{1} ∈ K_{1}, t_{2} ∈ K_{2}. We may assume [ρ, t_{i}] = 1. Set W = [V_{1}, t_{1}]. Assume [Y ∩ K_{1}, W] = 1. Then we have |W | ≤ q^{2}r_{2} ≤ r_{2}^{3}. As K_{2} acts on W we get with (1.32) W̃ = [W, K] is of order r_{2}^{2} or r_{2} = u^{3} and |W̃ | = u^{8}. We have that ρ acts on C IW̃ (S∩K) ≤ Z_{2}/Z_{1}. Now there is ρ_{1} ∈ K_{2} with ⟨ρ_{1}⟩ acts transitively on C IW̃ (S ∩ K)^{♯}. This shows that some p - element in ⟨K_{1} × K_{2}, ρ⟩ centralizes some x ∈ Z_{2} \ Z_{1}, contradicting (4.2).

So we have [W, Y ∩ K_{1}] ≠ 1. By (4) K_{2} ≅ L_{2}(q), |W | = q^{4}. Now we may interchange K_{1} and K_{2} and so K_{1} ≅ K_{2} ≅ L_{2}(q). As V_{1}/C I_{V_{1}}(t_{1}) ≅ W, we see |V_{1}/Z_{1}| ≥ q^{8}. But |Y : Y ∩ Q_{1}| ≤ q^{2} and so |V_{1}/Z_{1}| ≤ q^{2}|Y : Y ∩ Q_{1}|^{2} ≤ q^{6}, a contradiction.

(7.31) Lemma. We have q = 2 and [Y, K] ≰ K.

Proof. Let q > 2. As C I_{Y} (K) = 1 by (7.30) and [Y, ω] = Y, [Q_{2}, ω] ≤ Y, we see that Y ≤ K. Now C I_{L_{1}/Q_{1}}(K) acts on C I_{V_{1}/Z_{1}}(Y) = Z_{2}/Z_{1}. If there is ν ∈ C I_{L_{1}/Q_{1}}(K) with o(ν) = p and [Z_{1}, ν] = 1, then p | q − 1. But now there is some μ ∈ L_{1} ∩ L_{2} with o(μ) = p. Then [μ, Z_{1}] ≠ 1. Hence p | |C I_{K}(μ)|, p | |C I_{C I_{L_{1}/Q_{1}}(K)}(μ)|, and so m_{p}(C I_{L_{1}}(μ)) ≥ 3, which contradicts N_{G}(⟨μ⟩) ≰ M. Hence we have shown that C I_{L_{1}}(Z_{1})/Q_{1} ≲ Aut(K).

We are going to prove the same result for q = 2. Let K_{1} ≤ C I_{L_{1}/Q_{1}}(K). By (7.30) [Y, K_{1}] ≠ 1. Hence we may assume that K is nonsolvable. Let V_{1}^{(1)} ≤ V_{1}/Z_{1} be an irreducible K - submodule, such that N_{Y} (V_{1}^{(1)}) is maximal. Suppose Y ≠ N_{Y} (V_{1}^{(1)}). Then there is y_{1} ∈ Y \ N_{Y} (V_{1}^{(1)}) with V_{1}^{(1)}^{y_{1}} = V_{1}^{(2)} ≠ V_{1}^{(1)}. Set [V_{1}^{(1)} ⊕ V_{1}^{(2)}, y_{1}] = Ṽ. Then |Ṽ : Ṽ ∩ V_{3}| ≤ 2. Let y_{2} ∈ Y \ ⟨N_{Y} (V_{1}^{(1)}), y_{1}⟩. We have |Ṽ : C IṼ (y_{2})| ≤ 4. As |V_{1}^{(1)}| > 4 (K is nonsolvable), we get C I_{V_{1}^{(1)}⊕V_{1}^{(2)}}(y_{2}) ≠ 1. Hence W_{1} = (V_{1}^{(1)} ⊕ V_{1}^{(2)})^{y_{2}} ∩ (V_{1}^{(1)} ⊕ V_{1}^{(2)}) ≠ 1. Now W_{1} is a K - module. If W_{1} ≠ V_{1}^{(1)} ⊕ V_{1}^{(2)}, then W_{1} ≅ V_{1}^{(1)} and y_{2} ∈ N_{Y} (W_{1}). But N_{Y} (V_{1}^{(1)}) ≤ N_{Y} (V_{1}^{(1)} ⊕ V_{1}^{(2)}) and so N_{Y} (V_{1}^{(1)}) ≤ N_{Y} (W_{1}). By the maximality we now get y_{2} ∈ N(V_{1}^{(1)} ⊕V_{1}^{(2)}). So we have shown |Y : N_{Y} (V_{1}^{(1)})| ≤ 2. By (7.28) N_{L_{1}/Q_{1}}(K) acts on V_{1}^{(1)} ⊕ V_{1}^{(2)}, as Z_{2}/Z_{1} ≤ V_{1}^{(1)} ⊕ V_{1}^{(2)}. In particular C I_{K_{1}}(V_{1}^{(1)} ⊕ V_{1}^{(2)}) = 1. Let U = S ∩ K_{1}. Then by (7.30) [U, Y] = 1. As |C I_{V_{1}^{(i)}}(Y)| = 2, we get |U : C I_{U}(V_{1}^{(1)} ⊕ V_{1}^{(2)})| = |Y : N_{Y} (V_{1}^{(1)})| ≤ 2. Hence K_{1} is solvable. Let E ≤ K_{1} with E ≅ E_{p^{3}}. Then there is F ≤ E, |F | = p with [F, V_{1}^{(1)} ⊕ V_{1}^{(2)}] = 1. But this contradicts Z_{2}/Z_{1} ≤ V_{1}^{(1)}⊕V_{1}^{(2)} and (4.2). We have m_{p}(K_{1}) ≤ 2 and so |Y : C I_{Y} (K_{1})| ≤ 4. By (7.30) we get |Y : Y ∩ Q_{1}| ≤ 4 and so |V_{1}/Z_{1}| ≤ 2^{8}, if |Y : N_{Y} (V_{1}^{(1)})| = 2. If Y = N_{Y} (V_{1}^{(1)}), then m_{p}(K_{1}) = 1, and so |Y : Y ∩ Q_{1}| = 2, which shows |V_{1}/Z_{1}| ≤ 2^{6}.

Suppose first |V_{1}/Z_{1}| ≤ 2^{6}. As m_{p}(L_{1}/Q_{1}) ≥ 3, we get L_{1}/Q_{1} ≲ O_{6}^{−}(2). Now as K is nonsolvable 5 | |K|. But in O_{6}^{−}(2) no 5 - element is centralized by a 3 - element, a contradcition. So we have L_{1}/Q_{1} ≲ O_{8}^{±}(2) and m_{3}(K_{1}) = 2. But in O_{8}^{±}(2) no 5 - or 7 - element τ with |[V_{1}/Z_{1}, τ]| ≥ 2^{6} is centralized by E ≅ E_{9}, a contradiction again. So we have

C I_{L_{1}}(Z_{1})/Q_{1} ≲ Aut(K) (1)

In particular m_{p}(K) ≥ 2.

If q > 2, then K ≅ G(r) (2)

Suppose first that K is sporadic or alternating. Then as [ω, K] ≠ 1 and K has no outer automorphism of odd order, we see that K ≅ A_{5} or J_{1} by (1.9), contradicting m_{p}(K) ≥ 2. Suppose that K is a normal p-subgroup. Now m_{p}(K) ≥ 3. Let t ∈ X^{♯}. Then |[V_{1}, t] : [V_{1}, t] ∩ V_{3}| = q. As C I_{V_{1}/Z_{1}}(Y) = Z_{2}/Z_{1}, we see Z_{2}∩[V_{1}, t] ≠ 1. Let |Y : Y ∩Q_{1}| ≥ 2^{5}. Then as [t, V_{1}, s] ≤ V_{1}∩V_{3} for s ∈ Y, we see that there is some 1 ≠ u ∈ V_{1} ∩ V_{3} \ Z_{1}, u is centralized by E ≅ E_{p^{2}} by [GoLyS, (25.12)]. But this contradicts (7.3). So we have |Y | ≤ 2^{4}. Now |V_{1}/Z(V_{1})| ≤ 2^{16}. If q = 4 or 16, we have p ≠ 3. Let |Y | = 16. Then we see that m_{p}(O_{16}^{±}(2)) ≥ 5, a contradiction. Hence q ≤ 8. Let q = 4, then |Y | = 4 and so |V_{1}/Z(V_{1})| ≤ 2^{8}. But now m_{p}(O_{8}^{±}(2)) ≤ 2, a contradiction. So we have q = 8, |Y | = 8 and |V_{1}/Z(V_{1})| ≤ 2^{12}. Now we get p = 3. By (7.28) Z(V_{1}) = Z_{1} and so |V_{1}/Z_{1}| ≤ 2^{12}. As m_{3}(O_{3}(K)) = 6, we get some ρ ∈ O_{3}(K), |C I_{V_{1}}(ρ)/Z_{1}| = 2^{10}. Hence C I_{V_{1}}(ρ) ∩ Z_{2} ≱ Z_{1}, a contradiction.

(3) If q = 2, then K ≅ G(r), r even

Choose t ∈ X. Then |[t, V_{1}] : [t, V_{1}] ∩ V_{3}| = 2. Suppose there is E ≅ E_{p^{3}}, E ≤ C I(t), then there is u ∈ V_{3}∩V_{1}, u ∉ Z_{1}, u centralized by F ≤ E, |F | = p^{2}. But this contradicts (7.3). Furthermore |V_{1} : C I_{V_{1}}(t)| ≤ 2|Y |.

Let first K be sporadic. Then by (1.25) K ≅ M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, HiS, Co_{2} or Co_{1}. Now in any case m_{p}(K) ≥ 3. This shows with (1.4) p = 3 and K ≅ Co_{1} or Co_{2}. But then C I(t) involves Sp_{6}(2) or U_{4}(3). In both cases m_{3}(C I(t)) ≥ 3, a contradiction.

Let K ≅ A_{n}. Then also m_{p}(K) ≥ 3, and then p = 3 by (1.4). As m_{3}(K) ≥ 3, we get n ≥ 9. Furthermore we get n ≤ 18. We have that |Y : Y ∩ Q_{1}| ≤ 2^{⌊ \frac{n}{2} ⌋}. Hence |V_{1}/Z_{1}| ≤ 2^{n+2}. As 3 | |̸C I_{K}(Z_{2})|, we get with (1.9) that |K|_{2′} ≤ 2^{n+2} − 1, a contradiction.

Let now K be a group of Lie type in odd characteristic which is not a group of Lie type in even characteristic too. By (1.41) we get K ≅ 3 · U_{4}(3), L_{3}(3), U_{4}(3) or L_{4}(3). As m_{3}(K) ≥ 3, we get K ≅ L̸_{3}(3). By (1.10) C I_{K}(Z_{2}) ∈ Syl_{2}(K), as 3 | |̸C I_{K}(Z_{2})|. But |Y : Y ∩ Q_{1}| ≤ 2^{6} and so |V_{1}/Z_{1}| ≤ 2^{14} and |K|_{2′} > 2^{14} − 1, a contradiction.

Let now K be a normal p-subgroup. Let |Y : Y ∩ Q_{1}| ≥ 16. Then by [GoLyS, (24.1)] there is y ∈ Y \ Q_{1}, m_{p}(C I_{K}(y)) ≥ 3. We have |[y, V_{1}] : [y, V_{1}] ∩ V_{3}| = 2. By (7.3) we see [y, V_{1}] ∩ V_{3} = 1. But then [Y, [y, V_{1}]] = 1 and so Z_{2} = [y, V_{1}] by (7.28), a contradiction. So we have |Y : Y ∩ Q_{1}| ≤ 8 and then |V_{1}/Z_{1}| ≤ 2^{8}. This now implies p = 3.

Let first |V_{1}| = 2^{7}. Then L_{1}/Q_{1} ≲ O_{6}^{−}(2). As V_{1} ∩ V_{3} is elementary abelian we get |V_{1} ∩ V_{3}| ≤ 8 and so |Y : Y ∩ Q_{1}| = 8. This now implies Σ_{3} × Σ_{3} × Σ_{3} ≲ L_{1}/Q_{1}. For y ∈ Y \ Q_{1}, we have |[V_{1}, y]Z_{1}/Z_{1}| ≤ 8. But then some 3 - element in L_{1} centralizes Z_{2}, a contradiction to (4.2).

Hence we have |V_{1}| = 2^{9} and L_{1}/Q_{1} ≲ O^{+}(8, 2). Let Q̂_{1} = C I_{Q_{1}}(V_{1}). Then Q̂_{1} ∩ V_{1} = Z_{1}, Q̂_{1} ≤ Q_{2} and C I_{G}(x) ≤ M for every x ∈ Q̂_{1}^{♯}. We have [V_{3} ∩ L_{1}, Q̂_{1}] ≤ Q̂_{1} ∩ V_{3} ≤ Z_{1}. Let u ∈ Q̂_{1}, u ∉ Q_{3}. Then u acts on an elementary abelian group of order 3^{3}. Suppose [V_{3} ∩ L_{1}, u] = Z_{1}. We have |[V_{3}, u]| ≤ 4. But then there is some 3 - element ν in L_{3} with [ν, Z_{1}] = 1, a contradiction to (4.2). So we have [V_{3} ∩ L_{1}, Q̂_{1}] = 1. As [Q̂_{1}, V_{1} ∩ L_{3}] = 1, we see that V_{1} ∩ L_{3} acts on [V_{3}, Q̂_{1}]. By (7.28) we get [V_{3}, Q̂_{1}] ≤ Z_{3}. Now as [V_{3} ∩ L_{1}, Q̂_{1}] = 1, we get |Q̂_{1} : Q̂_{1} ∩ Q̂_{3}| ≤ 2. As M ≠ M_{3}, we have |Q̂_{1}| = 2 and so Q̂_{1} = Z_{1}. This implies Q_{1} = V_{1}. Now choose x ∈ V_{3} ∩ L_{1}, such that x centralizes E ≅ E_{9} in O_{3}(L_{1}/Q_{1}). We have |[x, V_{1}]Z_{1}/Z_{1}| = 16 and E acts faithfully on [x, V_{1}]. Hence [x, V_{1}]Z_{1} is elementary abelian of order 32 and if we choose x with x^{2} = 1, then ⟨x, [x, V_{1}], Z_{1}⟩ = F is elementary abelian of order 64. Furthermore F is the only elementary abelian group of order 64 in ⟨V_{1}, x⟩. This shows that E ≤ N_{G}(F) and so N_{G}(F) ≤ M.

Now let ρ ∈ L_{2} with o(ρ) = 3. Set F_{1} = ⟨V_{1} ∩ V_{3}, x, x^{ρ}⟩. Then |F_{1}| = 2^{7}. Hence F ≤ F_{1}. As |V_{1} ∩ V_{3} : C I_{V_{1}∩V_{3}}(x)| = 2, we see that |Z(F_{1})| = 2^{4}. Hence there is exactly one elementary abelian group of order 64 in F_{1}. This implies ρ ∈ N_{G}(F) ≤ M, a contradiction.

K ≅ G(r), r even (4)

By (2) and (3) it is enough to treat the case that K ≅ G(r), r odd, and K is not a group of Lie type in characteristic two. Furthermore q > 2. As [ω, Z_{1}] = Z_{1}, we get that ω induces an automorphism on K which normalizes a Sylow 2 - subgroup of K. In case of q = 64 let ν ∈ L_{1} ∩ L_{2} with o(ν) = 7. Then also ν induces an automorphism on K which normalizes a Sylow 2 - subgroup.

Suppose that ν or ω induce field automorphisms. Let K_{1} = C I_{K}(ω). Then m_{p}(K) ≤ 1. Let r = p^{f}, p ∈ σ(M). Then we get K_{1} ≅ L_{2}(p) and K ≅ L_{2}(p^{f}). Now m_{2}(L_{1}/Q_{1}) = 2, and so |Y : Y ∩ Q_{1}| = 4. Hence |V_{1}/Z_{1}| ≤ 2^{8} and so K ≲ O_{8}^{±}(2). This implies p = 3. But then also f = 3 and so o(ω) = 3 , contradicting 3 ∈ σ(M).

So we have p ∉ σ(M). Now m_{p}(K) ≤ 3. But as f ≥ 3, we get K ≅ L_{2}(r) = L_{2}(p^{3}) again. This contradicts m_{u}(K) ≥ 3 for u ∈ σ(M).

So we have shown that neither ω nor ν induces a field automorphism. Application of (1.10) shows that o(ω) = 3 or K ≅ ^{2}G_{2}(r).

Suppose K ≅ ^{2}G_{2}(r). Then |Y : Y ∩ Q_{1}| ≤ 8 and so |V_{1}/Z_{1}| ≤ 2^{12}, K ≲ O_{12}^{±}(2). But then by order consideration we get r = 3, a contradiction as L_{2}(8) · 3 ≅ ^{2}G_{2}(3).

So we have q = 4, o(ω) = 3. This now implies that 3 ∉ σ(M). Let m_{2}(L_{1}/Q_{1}) ≤ 6. Then we get K ≲ O_{16}^{±}(2) and so either K ≅ L_{2}(r) or K contains a nonabelian subgroup of order u^{3} for some odd prime u. The latter now implies r = 3^{f}. As m_{3}(K) ≤ 3, we get with (1.5) K ≅ L_{3}(3), U_{3}(3), P Sp_{4}(3), or G_{2}(3). But as 3 ∉ σ(M) and m_{p}(K) = 1 for p ≠ 3, p odd, this is a contradiction.

Let K ≅ L_{2}(r). Then |Y | = 4 and so |V_{1}/Z_{1}| ≤ 2^{8}. As r = p^{f}, f ≥ 3, we get a contradiction.

So we have m_{2}(L_{1}/Q_{1}) > 6. Application of (1.3) shows that m_{3}(K) = 3, K ≅ Sp_{6}(r), Ω_{8}^{−}(r), L_{6}(r), U_{6}(r), L_{7}(r), or U_{7}(r). Now 3 | r̸ and 3 | r̸ − 1,

3 | r̸ + 1 in case of K ≅ U_{6}(r) or U_{7}(r). Hence never K possesses a diagonal automorphism of order 3. As ω does not induce a field automorphism, we see ⟨ω, K⟩ ≅ ZZ_{3} × K, a contradiction to m_{3}(ZZ_{3} × K) = 4 and 3 ∉ σ(M).

p | r̸^{2} − 1 (5)

Let K ≅ L̸_{2}(r), Sz(r), or U_{3}(r). Then by (1.53) there is some minimal parabolic P of K such that Y ≤ O_{2}(P) and P/O_{2}(P) ≅ L_{2}(r) in case of 2F_{4}(r). Hence P/O_{2}(P) ≅ L_{2}(r) or U_{3}(r) and so by (7.28) C I_{V_{1}/Z_{1}}(O_{2}(P)) ≤ Z_{2}/Z_{1}. If [ω, P] ≤ P, then there is some ν ∈ P, o(ν) = p with [Z_{2}, ν] = 1, a contradiction to (4.2) and M ≠ M_{3}. Hence we may assume [ω, P] ≰ P. Then we have K ≅ Ω_{8}^{+}(r) and ω induces a graph automorphism of order 3. In particular 3 ∉ σ(M). But this contradicts m_{3}(Ω_{8}^{+}(r)) ≥ 4.

So assume K ≅ L_{2}(r), Sz(r) or U_{3}(r). As m_{p}(K) ≥ 2, we see K ≅ U_{3}(r), p | r +1. Now |Y : Y ∩Q_{1}| ≤ r and so |V_{1}/Z_{1}| ≤ r^{4}. But then |[V_{1}, t]Z_{1}/Z_{1}| ≤ r^{2}, and so by (1.26) |V_{1}/Z_{1}| ≥ r^{6}, a contradiction.

By (5) K ≅ L_{n}(r), n ≤ 4, L_{n}(2), 5 ≤ n ≤ 7, U_{n}(r), n ≤ 5, Sp_{2n}(r), n ≤ 3, G_{2}(r), ^{3}D_{4}(r), ^{2}F_{4}(r), Ω_{8}^{−}(r). But we have m_{p}(K) ≥ 2. This implies K ≅ L_{6}(2) or L_{7}(2), p = 7, ^{2}F_{4}(r), p | r^{2} + 1, or ^{3}D_{4}(r), p | r^{6} − 1.

If K ≅ ^{2}F_{4}(r), or ^{3}D_{4}(r), then by (1.53) there is a parabolic P with Sz(r) ≲ P/O_{2}(P), or L_{2}(r^{3}) ≲ P/O_{2}(P) and Y ≤ O_{2}(P). But then we get the same contradiction as in (5).

So we are left with K ≅ L_{6}(2) or L_{7}(2) and p = 7. Then q = 2. But as m_{7}(L_{1}) ≥ 3, this contradicts (1).

(7.32) Proposition. If b = b_{2}, then [Z_{2}, Z_{α}] = 1.

Proof. Suppose [Z_{2}, Z_{α}] ≠ 1. Then by (7.31) we have q = 2 and [K, Y] ≰ K. Let y ∈ Y, K^{y} ≠ K. As Y ≤ S, we see that K possesses abelian Sylow 2 - subgroups. Hence K ≅ L_{2}(r), or J_{1}. Now C I_{K^{y}×K}(y) ≅ K acts on [y, V_{1}]. If 1 ≠ t ∈ C I_{K^{y}×K}(y)∩Y, then |[[y, V_{1}], t]| ≤ 4. By (7.30) Z_{2}/Z_{1} ≤ [y, V_{1}] and so C I_{K^{y}×K}(y) acts faithfully on [y, V_{1}]. Application of (1.27) shows K ≅ L_{2}(4). By (7.30) we see |Y : Y ∩ Q_{1}| ≤ 8 and so |V_{1}/Z_{1}| ≤ 2^{8}. Now L_{1}/Q_{1} ≲ O_{8}^{±}(2). Furthermore p = 3. But then 3 | |C I_{L_{1}/Q_{1}}(K^{y} × K)|. As 5^{2} | |L_{1}/Q_{1}|, we have L_{1}/Q_{1} ≲ O_{8}^{+}(2). Now we have |Z_{2}^{L_{1}}| = 135 as no 3 - element centralizes Z_{2}. But as E ≅ E_{27} is contained in L_{1}, we get some 3 - element ρ with |C I_{V_{1}}(ρ)| ≥ 2^{5}. Now ρ centralizes some involution i ∉ Z_{1}. As all involutions in V_{1} \ Z_{1} are conjugate under L_{1} (there are exactly 270 such involutions), we get i ∼ j ∈ Z_{2} \ Z_{1}, a contradiction.

§ 8 The case [Z_{2}, Z_{α}] = 1

In this final chapter we consider the case b = b_{2}. By (7.32) we know [Z_{2}, Z_{α}] = 1, α ∼ 1. We are going to work under

(8.1) Hypothesis. L_{1} is not as in (4.9)(ii).

The construction of L_{1} in (4.4) now implies L_{1} = C I_{L_{1}}(Z_{1})(L_{1} ∩ L_{2}), m_{p}(C I_{M}(Z_{1})) ≥ 3. In particular Z_{1} ≤ Z_{2}, or we are in the situation of (4.13).

Suppose first b > 1. Then b ≥ 3 and we may apply (5.6). Remember that in the case of (4.13) we have the situation of (5.6) anyway.

Set

W_{1} = ⟨Z_{2}^{L_{1}}⟩ and V_{1} = ⟨(C I_{Z_{2}/Z_{1}}(Q_{1}) mod Z_{1})^{L_{1}}⟩.

We have

[V_{1}, Q_{1}] ≤ Z_{1}.

Now W_{1} = V_{1} in case of E(L_{2}/Q_{2}) ≅ L_{2}(q) and Z_{2} is a direct sum of at most two natural modules.

If E(L_{2}/Q_{2}) ≅ A_{9}, E(L_{2}/Q_{2}) ≅ L_{2}(q) and Z_{2} is the orthogonal module, or E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q) and Z_{2} is the orthogonal module, then W_{1} involves at least three nontrivial irreducible L_{1} - modules as has been seen in § 5.

If E(L_{2}/Q_{2}) ≅ L_{2}(4) × L_{2}(4) and Z_{2} is a sum of two orthogonal L_{2}(4)

- modules, then J(S) ≰ Q_{1}. Then m_{p}(L_{1}) ≥ 3. Now by (5.4) for every x ∈ C I_{Z_{2}}(Q_{1})^{♯} we have C I_{G}(x) ≤ M, a contradiction, as ⟨C I_{L_{2}}(x)|1 ≠ x ∈ C I_{Z_{2}}(Q_{1})⟩ = L_{2}, or we have (5.4)(i). In this case m_{p}(⟨J(S)^{L_{1}}⟩) ≤ 1. Set X = ⟨J(S)^{L_{1}}⟩Q_{1}/Q_{1}. As m_{p}(N(J(S))) ≥ 2 we get with (1.63) that J(S) ≰ Q_{2}. Let U ∈Syl_{3}(⟨J(S)^{L_{1}}⟩) with |U ∩ G_{2}| = 9. Then m_{p}(N_{L_{1}}(U)) ≥ 2. As 3 ∉ σ(M), we have m_{3}(U) ≤ 3. If m_{p}(C I(U)) ≥ 2, then we get G_{2} = N_{G_{2}}(U ∩ G_{2})(G_{2} ∩ M) ≤ M by (1.63). So we have m_{p}(C I(U)) ≤ 1 and then p = 5 or 13 and m_{p}(C I(U)) = 1. Further m_{p}(X) = 1 and m_{p}(L_{1}) = 3. Let C be a critical subgroup of U. Then C ≅ E_{27} or E_{3^{1+4}}. Let K_{1} be a component of L_{1}/Q_{1}. Then we have C ∩ K_{1} = 1 or C ≤ K_{1}. If K_{1} ≤ X, we get C ≤ K_{1} and so K_{1} possesses an outer automorphism of order p. Now with (1.67) and m_{3}(K_{1}) = 3, we see that K_{1} is a group of Lie type in characteristic two. Now as m_{p}(K_{1}) = 1, this automorphism has to be a field automorphism. But this contradicts (4.12). So we have F^{∗}(L_{1}/Q_{1}) ∩ X ≤ F (L_{1}/Q_{1}) This shows C I_{X}(F (X)) ≤ F (X) and m_{r}(F (X)) ≤ 3 for any odd prime r. So C cannot act nontrivially on any Sylow r-subgroup of F (X) for r ≠ 3, which shows C ≤ F (X). But now some elementary abelian group F of order p^{3} normalizes C and so some F_{1} ≤ F, |F_{1}| = p^{2} centralizes U ∩ L_{2}, a contradiction.

In all cases [W_{3}, W_{α}] ≤ W_{3} ∩ W_{α} and we have quadratic action.

(8.2) Hypothesis. There is a component K_{α} of L_{α}/Q_{α} with 1 ≠ [Z_{2}, K_{α}] ≤ K_{α}, m_{p}(K_{α}) ≥ 2.

Set Ṽ_{α} = [V_{α}, K_{α}]. Let K_{3} be the corresponding component to K_{α} in L_{3}/Q_{3}.

(8.3) Lemma. Assume (8.1) and (8.2). Then if K_{α}/Z(K_{α}) is a group of Lie type in odd characteristic it is a group of Lie type in even characteristic too.

Proof. Suppose false.

K_{α}/Z(K_{α}) ≅ U_{4}(3) (1)

If |W_{3} : C I_{W_{3}}(K_{α})| ≥ 4, then (1) follows from (1.12). So assume |W_{3} : C I_{W_{3}}(K_{α})| ≤ 2. The same applies for |W_{α} : C I_{W_{α}}(K_{3})|. Let x ∈ Ṽ_{α}. Then |[W_{3}, x] : [W_{3}, x] ∩ Z_{α}| ≤ 2. Suppose [x, K_{3}] ≤ Q_{3}, x ∉ Q_{3}. As m_{p}(K_{3}) ≥ 2 and p | |̸C I_{K_{3}}(u)|, for u ∈ Z_{α}^{♯} by (4.2) and (5.7), we get |[W_{3}, x]| = 2. As

p||C I_{L_{3}}(Ṽ_{3})| we have that p||C I_{L_{α}}(Ṽ_{α})|. But [W_{3}, x] is centralized by E ≅ E_{p^{2}}, Γ_{E,1}(G) ≤ M_{3}, and so (4.2) and (5.7) yield a contradiction. We have x ∈ Q_{3}.

So we get |Ṽ_{α} : Ṽ_{α} ∩ L_{2}| ≤ 2. As Z_{2} cannot induce transvections on Ṽ_{α}, we get [Ṽ_{α} ∩ L_{2}, Z_{2}] ≠ 1. Let x ∈ Ṽ_{α} ∩ L_{2}. Then again |[Z_{2}, x] : [Z_{2}, x] ∩ Z_{α}| ≤ 2. By (5.7) we have Z_{α} ∩ Z_{β} = 1 for all β ∈ Δ(2). This now implies E(L_{2}/Q_{2}) ≅ L_{2}(4) or L_{2}(4) × L_{2}(4), or L_{2} is solvable. Furthermore |Z_{α}| ≤ 4. Hence |[W_{3}, x]| ≤ 8 for x ∈ Ṽ_{α}. Application of (1.41) shows [Ṽ_{α}, K_{3}] ≤ Q_{3} and so [Ṽ_{α}, Z_{2}] ≤ Z_{2} ∩ Ṽ_{α}. Now we see that there is y ∈ Z_{2}, [y, K_{α}] ≠ 1, |Ṽ_{α} : C IṼ_{α}(y)| ≤ 4, and so again K_{α}/Z(K_{α}) ≅ U_{4}(3) by (1.41).

L_{2}/Q_{2} ≅ F_{10}, F_{20}, orΣ_{3}. (2)

By (1) 3 ∈ σ(M) and so m_{3}(L_{2}) ≤ 1. Let ω ∈ L_{3} ∩ L_{2}, o(ω) > 3. We have [ω, K_{3}] ≤ K_{3} by (1.63) and then [ω, K_{3}] = 1. But this contradicts N_{G}(⟨ω⟩) ≰ M.

Now by (2) |Z_{α}| ≤ 4. For x ∈ Ṽ_{α} we have |[W_{3}, x]| ≤ 2^{6}. Suppose |W_{3} : C I_{W_{3}}(K_{α})| ≥ 4. Then by (1.12) just 12 - dimensional modules are involved. Now let x ∈ Ṽ_{α}, 1 ≠ [x, K_{3}] ≤ K_{3}. Then W_{3} involves exactly one irreducible nontrivial module, and by (5.5) L_{2}/Q_{2} ≅ Σ_{3}. (Remember that (5.5) may be applied. If m_{p}(L_{1}) = 2, then by the construction in (4.13) L_{1} < L̂_{1}, L̂_{1} satisfies (3.2). Furthermore K_{1} is also a component of L̂_{1}.) But then 3||C I_{K_{3}}(Z_{2})|, a contradiction to (4.2) and (5.7).

Let K_{3}^{x} ≠ K_{3}. Then x acts fixed point freely on a group of order 3^{4}. Hence there is ρ ∈ K_{3} × K_{3}^{x}, o(ρ) = 3, |[W_{3}, ρ]| = 4. But then again there is exactly one nontrivial irreducible K_{3} - module involved, a contradiction to (4.2) and (5.7).

Suppose |W_{3} : C I_{W_{3}}(K_{α})| = 2. Then |[W_{3}, x]| ≤ 2^{3}, and so K_{3}^{x} ≠ K_{3}, but x inverts E ≅ E_{3^{4}}, a contradiction.

So we have [Ṽ_{α}, K_{3}] ≤ Q_{3}. Let Ṽ_{α} ≰ Q_{3}. Then [K_{3}, [V_{3}, Ṽ_{α}]] ≤ Z_{3} and so by the 3 - subgroup lemma we get [Ṽ_{3}, Ṽ_{α}] ≤ Z_{3}. Hence a p-element centralizes Ṽ_{3}. If [Ṽ_{3}, Ṽ_{α}] ≠ 1, then we have a contrdiction to (4.2) and (5.7). So we have [Ṽ_{3}, Ṽ_{α}] = 1. Now as above we arrive at a contradiction. So we see Ṽ_{α} ≤ Q_{3} and [Ṽ_{α}, Z_{2}] ≤ Z_{2} ∩ Ṽ_{α}. By (2) Z_{2} induces transvections on Ṽ_{α}, a contradiction to (1.18).

(8.4) Lemma. Assume (8.1) and (8.2). Then K_{α}/Z(K_{α}) is not sporadic

Proof. Suppose K_{α}/Z(K_{α}) to be sporadic. Let ω ∈ L_{2}∩L_{3}, o(ω) = r > 1, N_{G}(⟨ω⟩) ≰ M. Then as m_{p}(K_{3}) ≥ 2, we get [ω, K_{3}] ≤ K_{3} with (1.63) and [K_{3}, ω] ≠ 1. Now (1.9) implies o(ω) = 3. This implies |Z_{α}| ≤ 4.

Let |W_{3} : C I_{W_{3}}(K_{α})| = 2. Then for x ∈ Ṽ_{α} we have |[W_{3}, x]| ≤ 8. Suppose 1 ≠ [K_{3}, x] ≤ K_{3}. Then by (1.25) W_{3} in volves exactly one nontrivial irreducible module, K_{3}/Z(K_{3}) ≅ M_{22} and L_{2}/Q_{2} ≅ Σ_{3} by (5.5). Furthermore by (1.30) |Ṽ_{3}/C IṼ_{3}(K_{3})| = 2^{10}. But now 3||C I_{K_{3}}(Z_{2})|, a contradiction as 3 ∈ σ(M).

If K_{3}^{x} ≠ K_{3}, there is some 5- or 7-element t in K_{3}^{x}×K_{3} with |[W_{3}, t]| = 16 or 2^{6}, and again there is just one nontrivial irreducible K_{3} - module in W_{3}. Hence we have [Ṽ_{α}, K_{3}] = 1.

Let |W_{3} : C I_{W_{3}}(K_{α})| ≥ 4. Then by (1.12) K_{α}/Z(K_{α}) ≅ M_{12}, M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2} or Sz. Now K_{α} ≅ 3 · M_{22} or |W_{3} : C I_{W_{3}}(x)| ≤ 16 for x ∈ Ṽ_{α}. In the latter as before [Ṽ_{α}, K_{3}] = 1. Let K_{α} ≅ 3·M_{22} and |W_{3} : C I_{W_{3}}(x)| ≤ 64. Then as 11||K_{α}|, we see [K_{3}, x] ≤ K_{3}. But then there is also just one nontrivial module involved by (1.25). This is the 12 - dimensional module by (1.13). Furthermore by (5.5) L_{2}/Q_{2} ≅ Σ_{3}. But then 3 ∈ σ(M) and 3||C I_{K_{3}}(Z_{2})|, a contradiction. So in any case [Ṽ_{α}, K_{3}] = 1.

Let x ∈ Ṽ_{α} \ Q_{3}. Then [K_{3}, [V_{3}, x]] = 1. This shows with (4.2) and (5.7) that [V_{3}, Ṽ_{α}] = 1. Hence Ṽ_{α} ≤ Q_{3}. We now have that Z_{2} induces transvections on Ṽ_{α}, contradicting (1.18).

(8.5) Lemma. Assume (8.1) and (8.2). Then O_{r}(K_{α}/Z(K_{α})) ≠ 1 for some r, or K_{α}/Z(K_{α}) is a group of Lie type in even characteristic.

Proof. Suppose false. Then K_{α}/Z(K_{α}) is an alternating group A_{n}, n ≥ 9, or n = 7.

|Z_{α}| ≤ 4 (1)

This follows from (1.63) as in (8.4).

(2) There is no submodule Ṽ_{α}^{(1)} of Ṽ_{α}, such that [K_{α}, Ṽ_{α}^{(1)}] ≠ 1 and every v ∈ Ṽ_{α}^{(1)} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α}.

Suppose false. As M_{3} ≠ M_{α} by (5.7), we have [Ṽ_{α}^{(1)}, Ṽ_{3}^{(1)}] = 1. Let x ∈ Ṽ_{α}^{(1)}\Q_{3}. Then [V_{3}, x] is normalized by K_{3} and so there is some t ∈ [V_{3}, x], with p||C I_{K_{3}}(t)|. But t ∈ Ṽ_{α}, and so by (4.2), M_{3} = M_{α}, a contradiction.

We have Ṽ_{α}^{(1)} ≤ L_{2}. Furthermore Z_{3} ∩ [Ṽ_{α}^{(1)}, Z_{2}] = 1. This shows that L_{2}/Q_{2} is nonsolvable, in particular 3 ∉ σ(M). Hence E(L_{2}/Q_{2}) ≅ L_{2}(4) and Z_{2} is the orthogonal module, or E(L_{2}/Q_{2}) ≅ L_{2}(4) × L_{2}(4) and Z_{2} is the orthogonal O^{−}(4, 4) - module. But Ṽ_{α}^{(1)} ≤ Q_{3} and so Ṽ_{α}^{(1)}Q_{2}/Q_{2} ≤ E(L_{2}/Q_{2}). This then yields [Ṽ_{α}^{(1)}, Z_{2}] ∩ Z_{3} ≠ 1, a contradiction.

(3) K_{α} ≅ A_{7}, A_{10} or A_{11}, where p = 5 in the last two cases.

Otherwise p = 3. Let Ṽ_{α}^{(1)} be an irreducible submodule of V_{α}/C I_{V_{α}}(K_{α}). By (2) Ṽ_{α}^{(1)} is not the permutation module. Hence by (1.12)(iii) |W_{3} : C I_{W_{3}}(K_{α})| ≤ 8 and so |W_{3} : C I_{W_{3}}(x)| ≤ 32 for every x ∈ V_{α}. Let [x, K_{3}] ≠ 1. Then by (1.30) there is exactly one nontrivial submodule involved in Ṽ_{α}. Now by (5.5) Z_{2} is an F - module and so |Z_{α}| = 2. Hence |W_{3} : C I_{W_{3}}(x)| ≤ 16. By (2) and (1.30) we have |W_{3} : C I_{W_{3}}(x)| = 16 and so by (1.12)(iii) K_{α} ≅ A_{9} and the spinmodule is involved. Now Z_{α} ≤ Ṽ_{3}. But then Z_{α} is centralized by a 3 - element in K_{3}. Hence we have M_{3} = M_{α} by (4.2), contradicting (5.7). So we have [Ṽ_{α}, K_{3}] = 1. Then Ṽ_{α} ≤ L_{2} and there is y ∈ Z_{2} with [y, K_{α}] ≠ 1 and |Ṽ_{α} : C IṼ_{α}(y)| ≤ 4. But then again (2) and (1.30) yield a contradiction.

K_{α} ≅ A_{7} (4)

Let K_{α} ≅ A_{10} or A_{11}. For x ∈ Ṽ_{α} we have |W_{3} : C I_{W_{3}}(x)| ≤ 2^{7}. Suppose |W_{3} : C I_{W_{3}}(K_{α})| ≥ 16. Then by (1.12) just permutation modules are involved. If x ∈ Ṽ_{α} induces some s ∈ K_{3} on Ṽ_{3}, we see that at most three permutation module are involved. But 5||C I_{L_{1}}(K_{1})| and so every v ∈ Ṽ_{α}^{(1)} is centralized by E ≅ E_{25}. This contradicts (2).

So we have |Ṽ_{α} : C IṼ_{α}(K_{3})| ≤ 2 for |W_{3} : C I_{W_{3}}(K_{α})| ≥ 16. Now for z ∈ Ṽ_{3}, we have |Ṽ_{α} : C IṼ_{α}(z)| ≤ 8 and if [Ṽ_{3}, K_{α}] ≠ 1, we see that there are at most three permutation modules involved, the same contradiction as before. Hence [Ṽ_{3}, K_{α}] = 1 and so [Ṽ_{3}, Ṽ_{α}] = 1.

If |W_{3} : C I_{W_{3}}(K_{α})| ≤ 8, we have |W_{3} : C I_{W_{3}}(x)| ≤ 32 for x ∈ Ṽ_{α}. As before |Ṽ_{α} : C IṼ_{α}(K_{3})| ≤ 2, the same contradiction as above. Hence we have [Ṽ_{α}, Ṽ_{3}] = 1. Now [[Ṽ_{α}, V_{3}], K_{3}] = 1. But then [Ṽ_{α}, V_{3}] = 1, i.e. Ṽ_{α} ≤ L_{2}. This shows there is z ∈ Z_{2} with |Ṽ_{α} : Ṽ_{α} ∩ C I(z)| ≤ 4. Now Ṽ_{α} involves at most two permutation modules, a contradiction.

Finally we have to treat the case K_{α} ≅ A_{7}. Let x ∈ Ṽ_{α} with [x, K_{3}] = 1. Suppose [[x, V_{3}], K_{3}] ≠ 1. Then |[V_{3}, x]| ≥ 16 and so [V_{3}, x] ∩ Z_{α} ≠ 1. But every s ∈ [x, V_{3}] is centralized by a 3 - element in K_{3}. This contradicts (4.2). Hence we have [K_{3}, [x, V_{3}]] = 1. This shows [x, V_{3}] = 1, i.e. x ∈ Q_{3}.

Let Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) be the permutation module. If x ∈ Ṽ_{α}^{(1)} with [K_{3}, x] ≠ 1, then x corresponds to a transvection, otherwise some 1 ≠ v ∈ [x, Ṽ_{3}] is centralized by E ≅ E_{9} in K_{3} and by an element of order three in K_{α}, contradicting (4.2). Furthermore Z_{3} ∩ Ṽ_{α}^{(1)} = 1, i.e. |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(Ṽ_{3})| = 2. By (2) there is some 3 - element ρ ∈ C I_{L_{3}/Q_{3}}(K_{3}) with (Ṽ_{3}^{(1)})^{ρ} ≠ Ṽ_{3}^{(1)}. This implies that Z_{α} ∩ [Ṽ_{α}^{(1)}, (Ṽ_{3}^{(1)})^{⟨ρ⟩}] ≠ 1. But [C I_{K_{3}}(Ṽ_{α}^{(1)}), [(Ṽ_{3}^{(1)})^{⟨ρ⟩}, Ṽ_{α}^{(1)}]] = 1, contradicting (4.2) and (5.7). Hence we have shown Ṽ_{α}^{(1)} ≤ L_{2}. But then [Ṽ_{α}^{(1)}, Z_{2}] ∩ Z_{3} ≠ 1, a contradiction.

Let now Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) be the 4 - dimensional module. Then again Ṽ_{α}^{(1)} ≰ L_{2} and so there is x ∈ Ṽ_{α}^{(1)}, [x, K_{3}] ≠ 1. We have that two such modules are involved by (2). Hence |[x, Ṽ_{3}]| ≥ 16. Now let ρ be as before.

Then every element in [Ṽ_{3}^{(1)}, ρ] is centralized by a 3 - element in L_{3}. As Z_{α} ∩ [[Ṽ_{3}^{(1)}, ρ], x] ≠ 1, this contradicts (4.2) and (5.7).

Now we treat the general case. Let x ∈ Ṽ_{α}. We may assume that neither the 6 - dimensional nor the 4 - dimensional module is a submodule of Ṽ_{α}/C IṼ_{α}(K_{α}). Hence by (1.12) |W_{3} : C I_{W_{3}}(K_{α})| ≤ 2. This implies |Ṽ_{3} : C IṼ_{3}(x)| ≤ 8. By (1.30) [x, K_{3}] = 1. Now Ṽ_{α} ≤ L_{2} and so Z_{2} induces transvections on Ṽ_{α}, contradicting (1.18) again.

(8.6) Lemma. Assume (8.1) and (8.2). Then O_{r}(K_{α}/Z(K_{α})) ≠ 1 for some r.

Proof. By (8.5) we may assume K_{α}/Z(K_{α}) ≅ G(r), a group of Lie type over a field GF (r) of characteristic two.

Let x ∈ Ṽ_{α} with [x, K_{3}] ≠ 1. Then |W_{3} : C I_{W_{3}}(x)| ≤ |Z_{α}||W_{3} : C I_{W_{3}}(K_{α})|.

(1) If W_{3} is an SC - module, then it involves some irreducible module which is not V (λ) for some basic weight λ

Suppose false. Let first p|r^{2} − 1. If ω is a p - element, with N_{G}(⟨ω⟩) ≤ M in some minimal parabolic of K_{α}, such that ω centralizes C I_{U_{α}}(S) for any composition factor U_{α} of W_{α}, then we have that ω centralizes Z_{2}, a contradiction to (4.2).

Hence V (λ) for any basic weight λ is involved. Application of (1.23) shows K_{α}/Z(K_{α}) ≅ (S)L_{n}(r), n ≤ 4, (S)U_{n}(r), n ≤ 4, Sp_{4}(r), or Sp_{6}(r).

If K_{1} ≅ Sp_{6}(r), then by (1.64) L_{2} is solvable or q ≤ r. We see that |W_{3} : C I_{W_{3}}(K_{α})| ≥ r^{5}, but such a group does not act quadratically on the natural module.

Let K_{1} ≅ Sp_{4}(r). As no group of order greater than r^{2} acts quadratically on both natural modules, we have |W_{3} : C I_{W_{3}}(K_{α})| ≤ r^{2}. Let Ṽ_{α}^{(1)} be a natural submodule of Ṽ_{α}/C IṼ_{α}(K_{α}). Then every v ∈ Ṽ_{α}^{(1)} is centralized by some p - element in L_{α}. Hence [Ṽ_{α}^{(1)}, Ṽ_{3}^{(1)}] does not contain any element centralized by E ≅ E_{p^{2}} in L_{1}, and vice versa.

Let [Ṽ_{α}^{(1)}, K_{3}] = 1. We have |[v, V_{3}]| ≤ r^{4} for v ∈ Ṽ_{α}^{(1)}. Now if [K_{3}, [v, V_{3}]] ≠ 1, then |[v, V_{3}]| = r^{4} and Z_{α} ∩ [v, V_{3}] ≠ 1. But every x ∈ [v, V_{3}] is centralized by some p - element in K_{3}, contradicting (4.2) and (5.7). So we have [V_{3}, Ṽ_{α}^{(1)}] = 1, and so Ṽ_{α}^{(1)} ≤ L_{2}. Now we see [Ṽ_{α}^{(1)}, Z_{2}] ∩ Z_{3} ≠ 1, a contradiction. Hence we have [Ṽ_{α}^{(1)}, K_{3}] ≠ 1 and we get that [Ṽ_{α}^{(1)}, K_{α}]Z_{α}/Z_{α} is the natural module. Furthermore there is some p - element ω ∈ L_{α}, with (Ṽ_{α}^{(1)})^{ω} ≠ Ṽ_{α}^{(1)}. But [Ṽ_{α}^{(1)}, Ṽ_{3}] ≤ Ṽ_{α}^{(1)} ∩ Ṽ_{3} and so [Ṽ_{α}^{(1)} ∩ Q_{3}, Ṽ_{3}] = 1. This implies |Ṽ_{3} : C IṼ_{3}(K_{α})| = r and so |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{3})| = r^{2}. But by symmetry |Ṽ_{α} : C IṼ_{α}(K_{3})| = r, a contradiction.

Suppose next K_{α} ≅ (S)U_{4}(r). Then the natural SU_{4}(r) - module and the natural SO_{6}^{−}(r) - module are involved. Now for x ∈ Ṽ_{α}, we have |W_{3} : C I_{W_{3}}(x)| ≥ r^{4} (if x is a unitary transvection) or |W_{3} : C I_{W_{3}}(x)| ≥ r^{6}. As a group of order r^{4} does not act quadratically on the SO_{6}^{−}(r) - module and the U_{4}(r) - module by (1.39), we get |W_{3} : C I_{W_{3}}(x)| < r^{4}|Z_{α}|. Suppose |Ṽ_{α} : C IṼ_{α}(K_{3})| > r. Then we get r^{2} < |Z_{α}| and then by (1.65) and (1.64) r^{2} = q, |Z_{α}| = q^{2}, and so we get some ω ∈ L_{2} ∩L_{3}, o(ω) = q −1, [Z_{3}, ω] = Z_{3}, [ω, K_{3}] ≤ K_{3}. Now as p|q − 1, and ⟨ω, K_{3}⟩ possesses abelian Sylow p - subgroups of order p^{3}, we get N_{G}(⟨ω⟩) ≤ M, a contradiction.

Hence we have |Ṽ_{α} : C IṼ_{α}(K_{3})| ≤ r. Let Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) be an irreducible K_{α} - module (i.e. the natural SU_{4}(r) - module or the natural SO_{6}^{−}(r) - module). In both cases every v ∈ Ṽ_{α}^{(1)} is centralized by some p - element in K_{α}. Hence [C I_{Ṽ_{α}_{(1)}}(K_{3}), Ṽ_{3}^{(1)}] = 1. As |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{3})| ≤ r, we see that Ṽ_{3}^{(1)} induces a group of transvections on Ṽ_{α}^{(1)}. This shows that r = 2 and Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) is the natural SO_{6}^{−}(2) - module. But then [Ṽ_{α}^{(1)}, Ṽ_{3}^{(1)}] is centralized by A_{6} in K_{α} and K_{3} as well. As 3 ∈ σ(M), a contradiction.

So we have K_{α} ≅ (S)L_{3}(r) or (S)L_{4}(r). If K_{α} ≅ SL_{4}(r), then the natural module, the dual module and the natural SO_{6}^{+}(r) - module are involved. Now by (1.39) |W_{3} : C I_{W_{3}}(K_{α})| ≤ r^{3}. This shows that for x ∈ V_{α}, |W_{3} : C I_{W_{3}}(x)| ≤ r^{3}|Z_{α}|. As p|r^{2} − 1, we get p = 3 for r = 2. Now application of (1.65) shows that (1.65)(i) holds. Then by (1.64) q ≤ r and we get |W_{3} : C I_{W_{3}}(x)| ≤ r^{5}. Hence x induces a transvection on the natural module. Furthermore |W_{3} : C I_{W_{3}}(K_{α})| ≥ r^{2}. By (1.16) W_{3} normalizes Ṽ_{α}^{(1)} where Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) is the natural module for SL_{4}(r) or SO_{6}^{+}(r). In the latter every v ∈ Ṽ_{α}^{(1)} is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α} in K_{α}, a contradiction. So we have the natural SL_{4}(r) - module. Now |[W_{3}, Ṽ_{α}^{(1)}]| ≤ r^{5}. This implies |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{3})| ≤ r. As before [Ṽ_{α}^{(1)}, K_{3}] = 1. As above there is some ρ ∈ L_{α}, o(ρ) = p with (Ṽ_{α}^{(1)})^{ρ} ≠ Ṽ_{α}^{(1)}. As|[W_{3}, Ṽ_{α}^{(1)}]| ≤ r^{5} we see U_{α} = (Ṽ_{α}^{(1)} ⊕ (Ṽ_{α}^{(1)})^{ρ})Z_{α}/Z_{α} ≤ L_{α}/Z_{α}. Furthermore every x ∈ U_{α} is centralized by some p - element in L_{α}. Hence [U_{α} ∩ C I(K_{3}), U_{3}] = 1. Now |[U_{α}, U_{3}]Z_{3}/Z_{3}| = r^{4} and some group E ≅ E_{p^{3}} with E ≤ C I(Z_{3}) acts on [U_{α}, U_{3}]Z_{3}/Z_{3}. Hence there is some v ∈ [U_{α}, U_{3}]^{♯}, v centralized by F ≤ E, |F | = p^{2}, a contradiction to (4.2) and (5.7).

Let finally K_{α} ≅ (S)L_{3}(r). Then the natural module and the dual module are involved. Furthermore p|r − 1. Hence in both modules every element is centralized by a p - element in K_{α}. Now let Ṽ_{α}^{(1)} be as above. Then there is ρ ∈ L_{α} with o(ρ) = p and (Ṽ_{α}^{(1)})^{ρ} ≠ Ṽ_{α}^{(1)}. If U_{α} = Ṽ_{α}^{(1)} ⊕ (Ṽ_{α}^{(1)})^{ρ} is normal in L_{α} we argue as before, as again every v ∈ U_{α} is centralized by some p - element. Hence we have three natural modules involved and at least one dual module. This shows that for x ∈ Ṽ_{α} we have |W_{3} : C I_{W_{3}}(x)| ≥ r^{4}. Hence as |W_{3} : C I_{W_{3}}(K_{α})| ≤ r^{2}, we get |Z_{α}| = r^{2}. As q ≤ r by (1.65) and (1.64) we see that q = r and |Z_{α}| = q^{2}. But then for ω ∈ L_{3} ∩ L_{2}, o(ω) = p, we have m_{p}(⟨ω, K_{3}⟩) = 3, and so N_{G}(⟨ω⟩) ≤ M_{3}, a contradiction.

So we have seen that p |r̸^{2} − 1. This implies that K_{α} ≅ L_{6}(2), or L_{7}(2), p = 7. Let K_{α} ≅ L_{7}(2). Then we may assume that V (λ_{3}) is involved. By (1.23) all other modules involved are natural modules or dual modules and so V (λ_{4}) and V (λ_{5}) are not involved. Then 7||C I_{K_{α}}(Z_{α−1})| , a contradiction.

Let K_{α} ≅ L_{6}(2). If V (λ_{3}) is involved we have that V (λ_{1}) and V (λ_{5}) are involved too. As |Z_{α}| ≤ 4, we get |W_{3} : C I_{W_{3}}(K_{α})| ≥ 2^{6}. Let x ∈ Ṽ_{α}\C IṼ_{α}(K_{3}). Suppose x does not induce a transvection on the natural module . Then we get that |W_{3} : C I_{W_{3}}(x)| ≥ 2^{10} and so |W_{3} : C I_{W_{3}}(K_{α})| ≥ 2^{8}. Hence we may asume that W_{3}/C I_{W_{3}}(K_{α}) contains transvections.

If V (λ_{3}) is not involved, then V (λ_{2}) and V (λ_{4}) are involved and so again |W_{3} : C I_{W_{3}}(K_{α})| ≥ 2^{6}. Let x ∈ K_{α} be an involution which does not induce a transvection on the natural module. Then x inverts some element ν of order 5 in K_{α}. Let P be a parabolic with P/O_{2}(P) ≅ L_{5}(2) such that C I_{V (λ_{2})}(O_{2}(P)) = W is the module V (λ_{2}) for L_{5}(2). Now let P_{1} be a parabolic in L_{5}(2) such that P_{1}/O_{2}(P_{1}) ≅ A_{8} and C I_{W} (O_{2}(P_{1})) is the orthogonal A_{8} - module while W/C I_{W} (O_{2}(P_{1})) is the natural L_{4}(2) - module. This shows |[V (λ_{2}), ν]| = 2^{12} and so |[V (λ_{2}), x]| ≥ 2^{6}. Hence also transvections are involved in W_{3}/C I_{W_{3}}(K_{α}).

Let w be a transvection in W_{3}/C I_{W_{3}}(K_{α}). Then ⟨W_{3}^{C I_{Kα}(w)}⟩ = C I_{K_{α}}(w). But then [[w, W_{α}], C I_{K_{α}}(w)] = 1 and so just natural modules are involved, a contradiction. This proves (1).

(2) W_{3} is not an SC - module

Suppose false. Then by (1) and (1.38) we have that K_{α} ≅ (S)L_{3}(r), and by (1.22) V (λ_{1}) ⨂ V (λ_{1})^{σ} is involved, σ an automorphism of order 2 of GF (r). Let x ∈ Ṽ_{α} with [x, K_{3}] ≠ 1. Then we see that |W_{3} : C I_{W_{3}}(x)| ≤ r^{2}|Z_{α}|. As q ≤ r, we get |W_{3} : C I_{W_{3}}(x)| ≤ r^{4}. By (1.14) we have |W_{3} : C I_{W_{3}}(K_{α})| ≤ r and so |W_{3} : C I_{W_{3}}(x)| ≤ r^{3}. This shows |W_{3} : C I_{W_{3}}(x)| = r^{3} and |Z_{α}| = r^{2}. Hence as r ≥ 4, we have q = r and so for ω ∈ L_{2} ∩ L_{3}, o(ω) = p , we see that m_{p}(⟨ω, K_{3}⟩) = 3. But this contradicts N_{G}(⟨ω⟩) ≰ M_{3}.

(3) W_{3} is a strong quadratic module

Suppose false. Let first q ≤ r. Then |Z_{α}| ≤ r^{2} and so |W_{3} : C I_{W_{3}}(x)| ≤ |W_{3} : C I_{W_{3}}(K_{α})| · r^{2}. Let u = 2^{m_{2}(K_{3})}. Then we get |W_{3} : C I_{W_{3}}(K_{α})| > u/r^{2} by (2).

Application of (1.15) shows K_{α}/Z(K_{α}) ≅ SL_{3}(r), SU_{3}(r), or Sp_{4}(2)′. If K_{α}/Z(K_{α}) ≅ S̸p_{4}(2)′, then |W_{3} : C I_{W_{3}}(K_{α})| ≤ r and if K_{α}/Z(K_{α}) ≅ Sp_{4}(2)′, then |W_{3} : C I_{W_{3}}(K_{α})| = 4.

Suppose first K_{α}/Z(K_{α}) ≅ S̸p_{4}(2)′. Now |W_{3} : C I_{W_{3}}(x)| ≤ r. Let x ∈ Ṽ_{α} with [x, K_{3}] ≠ 1, then |W_{3} : C I_{W_{3}}(x)| ≤ r^{3}. Then we see with (1.26) that either W_{3} involves exactly one nontrivial irreducible K_{3} - module, or every irreducible K_{3} - module involved is an SC - module. In the latter we argue as above.

Hence by (5.5) we have that Z_{2} is an F - module and then |Z_{α}| ≤ r. Remember (5.5) is applicable as L_{1} is contained in L̃_{1} satisfying (5.2) and K_{3} ≅ L̸_{3}(2) or A_{2^{m}}. Now by (2) K_{α} ≅ SU_{3}(r) and by (1.26) just the natural module is involved. By (1.20) we have Ṽ_{α}Z_{α}/Z_{α} is the natural module. But now every v ∈ Ṽ_{3} is centralized by a p - element in L_{3}. As Z_{α} ∩ [x, Ṽ_{3}] ≠ 1, this is a contradiction to (4.2) and (5.7).

We are left with K_{α}/Z(K_{α}) ≅ A_{6}. As |W_{3} : C I_{W_{3}}(K_{α})| = 4, we see |W_{3} : C I_{W_{3}}(x)| ≤ 16 for x ∈ Ṽ_{α}. Let Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) be an irreducible submodule of Ṽ_{α}. Then |Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α})| = 16, or 64. Every v ∈ Ṽ_{α}^{(1)} is centralized by a 3 - element in K_{α}. Hence we have [Ṽ_{α}^{(1)}, K_{3}] ≠ 1 and |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{3})| = 4. The same applies for Ṽ_{3}^{(1)} and so there are exactly two such modules involved and for ρ ∈ C I_{L_{α}/Q_{α}}(K_{α}), o(ρ) = 3, we have U_{α} = (Ṽ_{α}^{(1)})^{ρ} ⊕ Ṽ_{α}^{(1)} ≠ Ṽ_{α}^{(1)}. If Ṽ_{α}^{(1)} is the natural module, every v ∈ U_{α} is centralized by a 3 - element, a contradiction as Z_{3} ∩ U_{α} ≠ 1. Hence Ṽ_{α}^{(1)} is the 3 · A_{6} - module. But now [U_{α}, U_{3}] ≥ ⟨Z_{3}, Z_{α}⟩ and [U_{3}, U_{α}] is normalized by E_{9} in L_{α} and L_{3} as well, a contradiction to M_{3} ≠ M_{α} by (5.7).

We have q > r. Suppose r = 2 and (1.65)(ii) holds. Now 3 ∉ σ(M). This shows K ≅ L_{6}(2) or L_{7}(2), p = 7 and q = 4. Further ω ∈ K. Hence [ω, Z_{1}] = 1. This shows with (5.6) that |Z_{2}| = 16 and |Z_{1}| = 2. By (2) and (1.15) we now have a contradiction. So (1.65) (ii) does not hold. Then by (1.64) and (1.65) K_{α}/Z(K_{α}) ≅ (S)U_{n}(r), ^{2}E_{6}(r), Ω_{2n}^{−}(r), q = r^{2}, ^{3}D_{4}(r), or 2F_{4}(r). Let u be as before. Then |W_{3} : C I_{W_{3}}(K_{α})| > u/r^{4}. Now application of (1.15) shows K_{α}/Z(K_{α}) ≅ Ω_{8}^{−}(r), SU_{4}(r), SU_{5}(r), or ^{3}D_{4}(r). Hence |W_{3} : C I_{W_{3}}(K_{α})| ≤ r^{3}, r^{2}, r^{2}, r, respectively.

Suppose first that any module involved in W_{3} is an SC - module. Then by (1.23) V (λ_{1}), V (λ_{2}) and V (λ_{3}) are involved in case of K_{α} ≅ Ω_{8}^{−}(r). But now |W_{3} : C I_{W_{3}}(x)| ≥ r^{16}, a contradiction. In case of K_{α} ≅ SU_{4}(r), we have V (λ_{1}) and V (λ_{2}) are involved. But now we may argue as in (1).

Hence we have that W_{3} involves exactly one nontrivial irreducible module. By (5.5) Z_{2} is an F - module and as q > 2, we get |Z_{α}| ≤ q. Hence |W_{3} : C I_{W_{3}}(K_{α})| > u/q. Now the assertion follows with (1.15).

p |r̸^{2} − 1 (4)

Suppose false. By (3) and (1.14) just V (λ), λ some fundamental weight is involved in W_{3}. Hence V (λ) for every weight λ is involved. This shows K_{α} ≅ (S)L_{n}(r), (S)U_{n}(r), Sp_{2n}(r) or F_{4}(r).

Let q ≤ r. Then if every V (λ) which is involved in W_{3} is an SC - module, we see K_{α} ≅ (S)L_{n}(r), n ≤ 7, (S)U_{n}(r), n ≤ 6, Sp_{2n}(r), n ≤ 3. We may assume that K_{α} ≅ (S)L_{n}(r), n ≥ 5 or (S)U_{n}(r), n ≥ 5. All other cases can be handled as in (1).

Now in SL_{5}(r), SL_{6}(r), SL_{7}(r), we have |W_{3} : C I_{W_{3}}(x)| ≥ r^{8}, r^{16}, r^{32}, respectively, by (1.23). But we have in any case |W_{3} : C I_{W_{3}}(x)| ≤ r^{8}, r^{11}, r^{14}. Hence we have K_{α} ≅ SL_{5}(r) and |W_{3} : C I_{W_{3}}(K_{α})| = r^{6}. But then just natural modules can be involved, as W_{3} acts quadratically. If K_{α} ≅ SU_{5}(r) or SU_{6}(r) we get |W_{3} : C I_{W_{3}}(x)| ≥ r^{8}, or r^{16}. But |W_{3} : C I_{W_{3}}(x)| ≤ r^{6}, or r^{11}.

Hence we have that there is some V (λ) which is not an SC - module. This shows that W_{3} involves exactly one nontrivial irreducible module. Hence there is just one weight. This implies K_{α} ≅ SU_{3}(r), and we argue as above.

Let now q > r. As p|r^{2} − 1 we have r ≠ 2 by (1.65). Hence (1.64) and (1.65) show q = r^{2} and K_{α} ≅ (S)U_{n}(r). By (1.23) n ≤ 6, and we just have to handle n = 5 or 6. Now |W_{3} : C I_{W_{3}}(x)| ≤ r^{8}, r^{13}, respectively. This gives K_{α} ≅ (S)U_{5}(r) and |W_{3} : C I_{W_{3}}(K_{α})| = r^{4}. Now let t ∈ W_{3} be an unitary transvection. Then ⟨W_{3}^{C I_{Kα}(t)}⟩ = O^{2′}(C I_{K_{α}}(t)). Hence [[t, W_{3}], C I_{K_{α}}(t)] = 1 and so just natural modules are involved in W_{3}, a contradiction.

[Ṽ_{α}, K_{3}] = 1 (5)

Otherise by (4) we have K_{α} ≅ L_{6}(2) or L_{7}(2), p = 7. Now |W_{3} : C I_{W_{3}}(x)| ≤ 2^{11} or 2^{14}, respectively. But then we get a contradiction as before.

Let now x ∈ Ṽ_{α}, [[x, V_{3}], K_{3}] ≠ 1. We have that [x, V_{3}] is an SC - module. Let first q ≤ r. Then |[x, V_{3}]| ≤ ur, u as before. This now implies with (1.23) K_{α} ≅ (S)L_{n}(r), (S)U_{n}(r), Sp_{2n}(r), Ω_{2n}^{±}(r) and just natural modules are involved. Hence there is Ṽ_{α}^{(1)} such that Ṽ_{α}^{(1)}/C I_{Ṽ_{α}_{(1)}}(K_{α}) is the natural module. Choose x ∈ Ṽ_{α}^{(1)}. Suppose some w ∈ [x, V_{3}] is centralized by E ≅ E_{p^{2}} in K_{α}. Then we get a contradiction to (4.2) and (5.7). This implies that K_{α} ≅ (S)L_{n}(r), n ≤ 4, Ω_{6}^{−}(r), Sp_{4}(r), or L_{6}(2). In any case [x, V_{3}] ∩ Z_{α} = 1. Hence we have K_{α} ≅ SL_{4}(r) or L_{6}(2). Now |V_{3} : C I_{V_{3}}(K_{α})| = r^{4} or 2^{6}. But as x ∈ Ṽ_{α}^{(1)} we get [V_{3}, x] ∩ C I_{Ṽ_{α}_{(1)}}(K_{α}) ≠ 1, a contradiction.

So we have [Ṽ_{α}^{(1)}, V_{3}] = 1 and so Ṽ_{α}^{(1)} ≤ L_{2}. But now in any case Ṽ_{α}^{(1)} ∩ Z_{3} ≠ 1, a contradiction as in the natural module every element is centralized by some element of order p.

So we have q > r. Suppose first that r = 2 and we have (1.65)(ii), (iii) or (vi). Then 3 ∉ σ(M). This shows K_{α} ≅ L_{6}(2) or L_{7}(2), p = 7. Furthermore |Z_{1}| = 2. Then we get the same contradiction as before. Now (1.64) and (1.65) yield K_{α} ≅ SU_{n}(r), Ω_{2n}^{−}(r), ^{2}E_{6}(r), ^{2}F_{4}(r), or ^{3}D_{4}(r). Again [x, V_{3}] is an SC - module and so ^{3}D_{4}(r) and ^{2}F_{4}(r) are not possible by (1.22) and (1.23). If K_{α} ≅ ^{2}E_{6}(r), then by (1.22) and (1.23) V (λ_{1}) is involved. But dimV (λ_{1}) = 54, while |W_{3} : C I_{W_{3}}(K_{α})| ≤ q^{20}, a contradiction.

Hence we have that K_{α} ≅ SU_{n}(r), or Ω_{2n}^{−}(r) and just natural modules are involved. But now we argue as before, or K_{α} ≅ SU_{4}(r), [x, V_{3}] is the natural module. Then [x, V_{3}] ≥ Z_{α}. But every v ∈ [x, V_{3}] is centralized by a p - element in K_{3}, a contradiction to (4.2) and (5.7). Hence also in this case we get Ṽ_{α}^{(1)} ≤ L_{2} and then we have the same contradiction as before.

(8.7) Lemma. Assume (8.1). Let K_{α} be a component of L_{α}/Q_{α} or K_{α} normal in L_{α}/Q_{α} with O_{r}(K_{α}) ≠ 1 for some odd r. If 1 ≠ [Z_{2}, K_{α}] and m_{p}(K_{α}) ≥ 2, then [Z_{2}, K_{α}] ≰ K_{α}.

Proof. By (8.6) we just have to treat the case O_{r}(K_{α}) ≠ 1. Then Ṽ_{α} = ⟨t||[W_{3}, t] : [W_{3}, t] ∩ Z_{α}| ≤ 2⟩, by [GoLyS, (24.1)], [GoLyS, (25.12)] and (1.12).

Suppose first |Ṽ_{α} : C IṼ_{α}(K_{3})| ≥ 4. Then r = p ∈ σ(M). By [GoLyS, (24.1)] there is D_{2p} × D_{2p} ≅ U_{3} ≤ K_{3}Ṽ_{α}, Ṽ_{α} ∩ U_{3} ∈ Syl_{2}(U_{3}). If |[V_{3}, t]| > 2, then there is ρ ∈ U_{3}, o(ρ) = p, C I_{Z_{α}}(ρ) ≠ 1. But this contradicts (4.2) and (5.7). Hence |[W_{3}, t]| = 2 and so p = 3 and |[ρ, V_{3}]| = 4. Now [ρ, V_{3}] is centralized by E ≅ E_{9} in L_{3}. Choose x ∈ [ρ, V_{3}] \ C I(t). Then |[x, V_{α}]| = 2 and so 3||C I_{L_{α}}([x, V_{α}])|, contradicting (4.2) and (5.7).

So we have |Ṽ_{α} : C IṼ_{α}(K_{3})| ≤ 2. The same applies for Ṽ_{3}. Hence for x ∈ Ṽ_{α} with [K_{3}, x] = 1, we see that [V_{3}, x] = 1. This shows |Ṽ_{α} : Ṽ_{α} ∩ L_{2}| ≤ 2. Let [Ṽ_{α} ∩ L_{2}, Z_{2}] ≠ 1. Then for x ∈ Ṽ_{α} ∩ L_{2}, [x, Z_{2}] ∩ Z_{3} ≠ 1. As |[Z_{2}, x] : [Z_{2}, x] ∩ Z_{α}| ≤ 2 we see that |[Z_{2}, x] ∩ Z_{3}| ≤ 2. This shows with (1.31) and (5.6) that |Z_{3}| = 2 and E(L_{2}/Q_{2}) ≅ L_{2}(4) or L_{2} is solvable. Let t ∈ Ṽ_{α} with |[W_{3}, t] : [W_{3}, t] ∩ Z_{α}| ≤ 2 and t ∉ L_{2}. Now |[Ṽ_{3}, t]| ≤ 4. So we get 3 ∈ σ(M) and there is ρ ∈ K_{3}, o(ρ) = 3 with |[V_{3}, ρ]| ≤ 16. Hence [V_{3}, ρ] is centralized by some 3 - element in L_{3}, a contradiction as above. We have Ṽ_{α} ≤ L_{2} or [Ṽ_{α} ∩ L_{2}, Z_{2}] = 1 and so Z_{2} is generated by elements inducing transvections on Ṽ_{α}. Hence 3 ∈ σ(M) and then L_{2}/Q_{2} ≅ Σ_{3}. Now let ρ_{1} ∈ K_{α}, z ∈ Z_{2} with ρ_{1}^{z} = ρ_{1}^{−1}. Then |[Ṽ_{α}, ρ_{1}]| = 4. Further there is E ≤ L_{α}, E ≅ E_{9} with [E, [Ṽ_{α}, ρ_{1}]] = 1. But [Ṽ_{α}, Z_{2}] = Z_{3}, and so [Z_{3}, E] = 1. This shows E ≤ M_{3}, contradicting M_{α} ≠ M_{3} by (5.7).

(8.8) Lemma. Assume (8.1). Let K_{α} be a component of L_{α}/Q_{α} with m_{p}(K_{α}) ≥ 2. Then [K_{α}, Z_{2}] = 1

Proof. Suppose [K_{α}, Z_{2}] ≠ 1. As m_{p}(K_{α}) ≥ 2, we get from (1.11) and (8.7) that |W_{3} : C I_{W_{3}}(K_{α})| = 2. As we may interchange the roles of 3 and α, we get |Ṽ_{α} : C IṼ_{α}(K_{3})| ≤ 2 too. As for v ∈ Ṽ_{α}, we have |[Ṽ_{3}, v] : [Ṽ_{3}, v]∩Z_{α}| ≤ 2 and v acts on E ≅ E_{p^{4}}, for [v, K_{3}] ≠ 1, we see [Ṽ_{α}, K_{3}] = 1, and then [Ṽ_{α}, V_{3}] = 1. This implies Ṽ_{α} ≤ L_{2}. Now we get a contradiction as in (8.7).

(8.9) Lemma. Assume (8.1). Let K^{(α)} = K_{1}^{(α)} × K_{2}^{(α)} × K_{3}^{(α)} ≤ L_{α}/Q_{α} where K_{i}^{(α)} is either a component or a normal p-subgroup and m_{p}(K_{i}^{(α)}) = 1. Then we may assume that [K_{3}^{(α)}, Z_{2}] = 1.

Proof. Let [K_{i}^{(α)}, Z_{2}] ≠ 1 for i = 1, 2, 3. Set Ṽ_{α} = [V_{α}, K^{(α)}]. Assume that Ṽ_{α}/C IṼ_{α}(K^{(α)}) = Ṽ_{α}^{(1)} ⊕ Ṽ_{α}^{(2)} with [K_{1}^{(α)} × K_{2}^{(α)}, Ṽ_{α}^{(2)}] = 1 and

[Ṽ_{α}^{(1)}, K_{3}^{(α)}] = 1. Let x ∈ Ṽ_{α}^{(2)}. Suppose [K^{(3)}, x] ≠ 1. Suppose that [Ṽ_{3}, x] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α}. But [Ṽ_{3}, x] contains an element which is centralized by ρ ∈ K^{(3)} with o(ρ) = p. This contradicts (4.2) and (5.7). So we may assume that [Ṽ_{3}, x] is not centralized by such an elementary abelian group. This shows that [V_{3}, K_{3}^{(α)}] ≰ K_{3}^{(α)}. Suppose first that |V_{3} : C I_{V_{3}}(K_{3}^{(α)})| > 2. Then by (1.11) K_{3}^{(α)} ≅ L_{2}(r) and [⟨K_{3}^{(α)}^{V_{3}}⟩, V_{α}] = [K_{3}^{(α)}, V_{α}] is centralized by K_{1}^{(α)} × K_{2}^{(α)}. Then we see that [Ṽ_{3}, x] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α} , a contradiction. Suppose that |V_{3} : C I_{V_{3}}(K_{3}^{(α)})| = 2. If |V_{3} : C I_{V_{3}}(K^{(α)})| > 2, then we may assume that [[K_{1}^{(α)}, V_{α}], K_{3}^{(α)}^{V_{3}}] = 1 and some y ∈ [K_{1}^{(α)}, V_{α}] acts nontrivially on Ṽ_{3} and then also on K^{(3)}. So we just can interchange roles.

This means that we may assume that |Ṽ_{3} : C IṼ_{3}(K^{(α)})| = 2. But then also by symmetry |Ṽ_{α} : C IṼ_{α}(K^{(3)})| = 2. This shows |[Ṽ_{3}, Ṽ_{α}] : [Ṽ_{3}, Ṽ_{α}] ∩ Z_{α}| ≤ 2 ≥ |[Ṽ_{3}, Ṽ_{α}] : [Ṽ_{3}, Ṽ_{α}]∩Z_{3}|. Hence we see that |[Ṽ_{3}, Ṽ_{α}]| ≤ 4. But this contradicts K_{3}^{(α)} is not normalized by Ṽ_{3}.

Hence [Ṽ_{α}^{(2)}, K^{(3)}] = 1. This implies by the same reason as before [Ṽ_{α}^{(2)}, V_{3}] = 1, and so Ṽ_{α}^{(2)} ≤ L_{2}. But then Z_{3} ∩ [Ṽ_{α}^{(2)}, Z_{2}] = 1 and so [Ṽ_{α}^{(2)}, Z_{2}] = 1, which implies [K_{3}^{(α)}, Z_{2}] = 1, a contradiction.

Hence we have that [[K_{i}^{(α)}, Ṽ_{α}], K_{j}^{(α)}] ≠ 1 for every pair {i, j}. As m_{p}(K_{i}^{(α)}) = 1, we get that m_{3}(K_{3}^{(α)}) ≤ 1. Hence K_{3}^{(α)} ≅ L_{2}(r), U_{3}(r), L_{3}(r), Sz(r), J_{1}, or a p-group. The same applies for K_{2}^{(α)}. We have |W_{3} : C I_{W_{3}}(K^{(α)})| = 2, if r is odd. In this case we have that |Ṽ_{α} : C IṼ_{α}(K^{(3)})| ≤ 2. Furthermore let t ∈ Ṽ_{α} with [t, K^{(3)}] = 1 and [t, V_{3}] ≠ 1. Then |[t, V_{3}]| = 2. Hence there is U ≤ C I(K^{(3)}) with p||U| and [U, Ṽ_{3}] = 1. As [t, V_{3}] ≤ Ṽ_{α}, we get that [t, V_{3}] is centralized by a p - element in L_{α} and E ≅ E_{p^{3}} in L_{3}, a contradiction to (4.2) and (5.7). This implies |Ṽ_{α} : Ṽ_{α} ∩ L_{2}| = 2. Let t ∈ Ṽ_{3} ∩ L_{2} with [Z_{2}, t] ≠ 1. Then |[Z_{2}, t] : [Z_{2}, t] ∩ Z_{α}| ≤ 2. As [Z_{2}, t] ∩ Z_{3} ≠ 1, we now see |Z_{3}| = 2, and so |[Z_{2}, t]| ≤ 4. This shows that Z_{2} contains elements z with |[Ṽ_{α}, z]| ≤ 8, a contradiction as Ṽ_{α} involves at least four nontrivial irreducible K_{3}^{(α)} - modules. So we have |W_{3} : C I_{W_{3}}(K^{(α)})| > 2, and then r is even.

Now we choose notation such that K_{3}^{(α)} ≅ G(r_{3}), K_{2}^{(α)} ≅ G(r_{2}) with r_{3} ≥ r_{2}. We see that for v ∈ Ṽ_{α}, [v, K^{(3)}] ≠ 1, we get |[v, Ṽ_{3}] ≥ r_{3}^{4}. Hence we get |Z_{α}| ≥ r_{3}^{3}, r_{3}^{2} (in case of K_{3}^{(3)}/Z(K_{3}^{(3)}) ≅ L_{3}(r_{3})). Suppose there is some ω ∈ L_{α−1} ∩ L_{α}, o(ω) = q − 1, i.e. L_{2} is nonsolvable, or SU_{3}(2) ≤ L_{2}/Q_{2}. Now |Z_{α}| ≤ q^{2}, and so r_{3}^{3} ≤ q^{2}, or r_{3}^{2} ≤ q^{2}. Let [ω, K_{3}^{(α)}] ≤ K_{3}^{(α)} and [ω, K_{3}^{(α)}] ≠ 1. Then application of (1.64) and (1.65) shows that K_{3}^{(α)} ≅ SL_{3}(r_{3}) = SL_{3}(q), SU_{3}(r_{3}) = SU_{3}(\sqrt{q}), or L_{3}(2). If K_{3}^{(α)} ≅ L_{3}(2), then q = 4 and ω ∈ K_{3}^{(α)}. Hence [ω, Z_{α}] = 1. Now we see |Z_{α}| = 2 < 4 = r_{3}^{2}, a contradiction.

In case of K_{3}^{(α)} ≅ SU_{3}(\sqrt{q}), we get four natural modules involved and so |[v, Ṽ_{3}]| ≥ q^{4}, i.e. |Z_{α}| > q^{2}, a contradiction.

Hence we have K_{3}^{(α)} ≅ SL_{3}(q) and there are exactly four natural modules involved. Now K_{1}^{(α)} × K_{2}^{(α)} ≲ L_{2}(q) × L_{2}(q). But |W_{3} : C I_{W_{3}}(K^{(α)})| = q^{2}, a contradiction.

Hence let now [ω, K_{3}^{(α)}] ≰ K_{3}^{(α)}. Then K_{3}^{(α)} ≅ Sz(r), L_{2}(r), or L_{3}(2), as m_{p}(K_{3}^{(α)}) = 1. Let ⟨(K_{3}^{(α)})^{⟨ω⟩}⟩ = X_{1} × ⋯ × X_{t}, X_{1} ≅ K_{3}^{(α)}. As no X_{i}, X_{j}, i ≠ j can centralize [X_{1}, V_{α}], otherwise we argue as before, we see that |Z_{α}| ≥ r^{2^{t−1}−2}. But we may assume that 2^{t−1}−1 ≥ q−1 or 2^{2t−2} ≥ r^{2^{t−1}−2} ≥ 2^{2^{t−1}−2}. So we have t = 3. Now |Z_{α}| ≥ 4^{2}, q = 4 for K_{3}^{(α)} ≅ L_{3}(2) or q = 64 and K_{3}^{(α)} ≅ L_{2}(4), L_{2}(8) or L_{2}(16). If q = 64 there is ν ∈ L_{3} ∩ L_{2}, o(ν) = 7. This shows [K_{3}^{(α)}, ν] ≰ K_{3}^{(α)}, as [ω, ν] = 1. But then we have 63 conjugates of K_{3}^{(α)}, a contradiction. So we are left with K_{3}^{(α)} ≅ L_{3}(2), q = 4. By the choice of r_{3}, we may assume K^{(α)} = X_{1} × X_{2} × X_{3}. But now there are 9 natural modules involved and so we get |Z_{α}| ≥ 2^{7} > q^{2}, a contradiction.

We are left with [K_{3}^{(α)}, ω] = 1. But as r_{2} ≤ r_{3}, we also get [K_{2}^{(α)}, ω] = 1, or [K_{2}^{(α)}, ω] ≰ K_{2}^{(α)} and r_{2} < r_{3}. The latter implies q = 4 and so K_{2}^{(α)} × K_{3}^{(α)} ≲ C I_{K^{(α)}}(ω), a contradiction to (1.63). Hence [ω, K_{2}^{(α)}] = 1. By (1.63) K_{2}^{(α)} ≅ K_{3}^{(α)} ≅ A_{5}, p = 5. But now K_{1}^{(α)} is of the same shape as K_{2}^{(α)} and so again [K_{1}^{(α)}, ω] = 1, a contradiction to (1.63).

We are left with L_{2} solvable. Now |Z_{α}| ≤ 4. Hence K_{3}^{(α)} ≅ L_{3}(2) ≅ K_{2}^{(α)}. This shows that also K_{1}^{(α)} has cyclic Sylow 3 - subgroups and so we also get K_{1}^{(α)} ≅ L_{3}(2). Now we have 6 modules involved. This shows |[x, Ṽ_{3}]| ≥ 2^{6} > 4|Z_{α}| = 16, a contradiction.

(8.10) Lemma. Assume (8.1) and let K_{1}^{(α)} × K_{2}^{(α)} ≤ L_{α}/Q_{α}, K_{i}^{(α)} components or normal p-subgroups with m_{p}(K_{i}^{(α)}) = 1, i = 1, 2. Then we may assume [K_{2}^{(α)}, Z_{2}] = 1.

Proof. Assume K_{1}^{(α)}^{W_{3}} ≰ K_{1}^{(α)} ×K_{2}^{(α)}. Then we have that K_{1}^{(α)}^{W_{3}} ×K_{2}^{(α)} fulfills (8.9). So either [K_{1}^{(α)}, Z_{2}] = 1 or [K_{2}^{(α)}, Z_{2}] = 1. Hence we may assume that [K_{1}^{(α)}×K_{2}^{(α)}, W_{3}] ≤ K_{1}^{(α)}×K_{2}^{(α)}. Suppose first [[K_{1}^{(α)}, V_{α}], K_{2}^{(α)}] = 1. Let v ∈ [K_{1}^{(α)}, V_{α}] = Ṽ_{α}^{(1)} with K_{1}^{(3)} ≥ [v, K_{1}^{(3)}] ≠ 1. First of all [v, Ṽ_{3}^{(1)}]∩Z_{α} = 1. Furthermore, suppose that there is ν ∈ K_{1}^{(3)}, o(ν) odd, ν^{v} = ν^{−1}, and ρ ∈ L_{3}, [ρ, ⟨ν, v⟩] = 1, o(ρ) = p, ρ ∉ K_{2}^{(3)}. Then ρ acts on [[ν, Ṽ_{3}^{(1)}], v] and so some U ⊆ [v, Ṽ_{3}^{(1)}] is normalized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}, and centralized by some p - element in L_{α}. This shows with (4.2) M_{3} = M_{α}, contradicting (5.7).

As m_{p}(K_{1}^{(3)}) = 1, there is some p - element ρ ∉ K_{2}^{(3)}, ρ induces an outer automorphism on K_{1}^{(3)}. This implies with (1.67) that K_{1}^{(3)} ≅ G(r), a group of Lie type. As m_{p}(K_{1}^{(3)}) = 1, we see that ρ has to induce a field automorphism. Now by (4.12) K_{1}^{(3)} ≅ L_{2}(r) or Sz(r), r even. Hence we may assume v ∈ C I_{K_{1}^{(3)}}(ρ), a contradiction.

So we have shown that [v, K_{1}^{(3)}] = 1. Assume now [v, K_{2}^{(3)}] ≠ 1. We may choose notation such that m_{3}(K_{2}^{(3)}) ≤ 1. Hence by (1.5) K_{2}^{(3)} ≅ L_{2}(r), U_{3}(r), Sz(r), L_{3}(r), J_{1} or a p-group. Now as before there has to be some outer automorphism ρ of K_{2}^{(3)} with o(ρ) = p. This again has to be a field automorphism and so it centralizes v, a contradiction as before. We have

(1) If [[K_{1}^{(α)}, V_{α}], K_{2}^{(α)}] = 1, then for Ṽ_{α}^{(1)} = [V_{α}, K_{1}^{(α)}], we have [Ṽ_{α}^{(1)}, K_{1}^{(3)}× K_{2}^{(3)}] = 1, or (K_{1}^{(3)})^{v} = K_{2}^{(3)} for some v ∈ Ṽ_{α}^{(1)}, |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{1}^{(3)})| = 2.

We now show

[[K_{1}^{(α)}, V_{α}], K_{2}^{(α)}] ≠ 1 (2)

By (1) we may assume |Ṽ_{α}^{(1)} : C I_{Ṽ_{α}_{(1)}}(K_{1}^{(3)} × K_{2}^{(3)})| ≤ 2. Choose t ∈ Ṽ_{α}^{(1)} with [t, K_{1}^{(3)} × K_{2}^{(3)}] = 1. Then K_{1}^{(3)} × K_{2}^{(3)} acts on [t, V_{α}]. Suppose t ∉ L_{2}. We may also use (1) with 3 and α interchanged. This together with (8.3) - (8.9) shows that |Ṽ_{3}^{(1)} : Ṽ_{3}^{(1)} ∩ Q_{α}| ≤ 2. Hence |[Ṽ_{3}^{(1)}, t] : [Ṽ_{3}^{(1)}, t] ∩ Z_{α}| ≤ 2. By (4.2) and (5.7) this shows |[Ṽ_{3}^{(1)}, t]| ≤ 2. Hence [t, Ṽ_{3}^{(1)}] = 1 and so [K_{1}^{(3)}, [V_{3}, t]] = 1. If [V_{3}, t] ≠ 1 we have C I_{L_{α}}(Ṽ_{α}^{(1)}) contains E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α}. So we get [V_{3}, t] = 1. This shows Ṽ_{α}^{(1)} ≤ L_{2}.

By (4.2) and (5.7) we have [Ṽ_{α}^{(1)}, Z_{2}] ∩ Z_{3} = 1. But this would imply [Ṽ_{α}^{(1)}, Z_{2}] = 1, a contradiction.

[Ṽ_{α}^{(1)}, K_{1}^{(3)} × K_{2}^{(3)}] = 1 (3)

Suppose false. We choose notation such that m_{3}(K_{2}^{(3)}) ≤ 1. In particular K_{2}^{(3)} ≅ L_{2}(r), L_{3}(r), U_{3}(r), Sz(r), J_{1}, or a p-group. Suppose first that |W_{3} : C I_{W_{3}}(K_{2}^{(α)})| = 2. Choose x ∈ Ṽ_{α}, [x, K_{1}^{(3)} × K_{2}^{(3)}] = 1. Now |[x, V_{3}] : [x, V_{3}] ∩ Z_{α}| ≤ 2. As K_{1}^{(3)} × K_{2}^{(3)} acts on [x, V_{3}], we see |[x, V_{3}]| ≤ 2 and [K_{1}^{(3)} × K_{2}^{(3)}, [x, V_{3}]] = 1. Then there is some p - element ρ ∈ L_{α} with [Ṽ_{α}, ρ] = 1. Let x ∉ L_{2}. So [x, V_{3}] is centralized by ρ ∈ L_{α} and by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}. Application of (4.2) and (5.7) provides us with a contradiction.

Now for x ∈ Ṽ_{α} ∩ L_{2} we have |[Z_{2}, x] : [Z_{2}, x] ∩ Z_{α}| ≤ 2. Furthermore if [Z_{2}, x] ≠ 1, then [Z_{2}, x] ∩ Z_{3} ≠ 1. This implies |Z_{3}| = 2 and so |[Z_{2}, x]| ≤ 4. Application of (1.43) shows that in any case for z ∈ Z_{2} with [z, K_{1}^{(α)} × K_{2}^{(α)}] ≠ 1 we get |[Ṽ_{α}, z]| ≤ 4|Ṽ_{α} : Ṽ_{α} ∩ L_{2}|. By (1.12) |Ṽ_{α} : Ṽ_{α} ∩ L_{2}| ≤ 2 or K_{2}^{(α)} ≅ L_{2}(r), U_{3}(r), Sz(r) or L_{3}(r), r even, and |Ṽ_{α} : Ṽ_{α} ∩ L_{2}| ≤ r or r^{2} in case of K_{2}^{(α)} ≅ L_{3}(r). Set r = 2 in the first case. As there are at least two nontrivial K_{2}^{(α)} - modules involved, we see that z has to induce a transvection on one of these modules. Now we get K_{2}^{(α)} ≅ L_{2}(4), L_{3}(r) , r even, or a 3group. In any case, as p |r̸ − 1 in case of K_{2}^{(α)} ≅ L_{3}(r), some ρ ∈ L_{α} with o(ρ) = p centralizes Ṽ_{α}. Hence [Z_{2}, x] ∩ Z_{3} = 1 and then [Z_{2}, x] = 1. This implies that |Ṽ_{α} : C IṼ_{α}(Z_{2})| = r^{2} and K_{2}^{(α)} ≅ L_{3}(r) and exactly two natural modules are involved. Then K_{1}^{(α)} is a subgroup of L_{2}(r). But now by (2) |Ṽ_{α} : C IṼ_{α}(K_{1}^{(α)})| ≤ r, a contradiction. So we have

(i) |W_{3} : C I_{W_{3}}(K_{2}^{(α)})| ≥ 4

Application of (1.12) now again implies K_{2}^{(α)} ≅ L_{2}(r), U_{3}(r), Sz(r) or L_{3}(r), r even. Next we show

(ii) K_{1}^{(α)} × K_{2}^{(α)} ≅ L_{2}(r) × L_{2}(r) and Ṽ_{α}/C IṼ_{α}(K_{1}^{(α)} × K_{2}^{(α)}) is the O^{+}(4, r) - module, L_{2} is nonsolvable and E(L_{2}/Q_{2}) ≅ G(q) with q = r.

Let α + 1 ∈ Δ(α) with [Z_{α+1}, K_{1}^{(3)} × K_{2}^{(3)}] ≠ 1. Suppose first [Ṽ_{3} ∩ L_{α+1}, Z_{α+1}] = 1. Then |Ṽ_{3} : C IṼ_{3}(Z_{α+1})| ≤ r or r^{2} in case of L_{3}(r). As [Z_{α+1}, Ṽ_{3}] ≠ 1, this implies |Ṽ_{3} : C IṼ_{3}(Z_{α+1})| = r^{2} and there are exactly two L_{3}(r) - modules involved. Now K_{1}^{(α)} ≲ L_{2}(r), which contradicts |Ṽ_{3} : C IṼ_{3}(K_{1}^{(α)})| = r^{2}.

So we have [Ṽ_{3} ∩ L_{α+1}, Z_{α+1}] ≠ 1. Then even [Ṽ_{3} ∩ L_{α+1}, Z_{α+1}] ∩ Z_{α} ≠ 1. Furthermore for x ∈ Z_{α+1} with [x, Ṽ_{3}] ≠ 1, we see that |Ṽ_{3} ∩ L_{α+1} : C I_{Ṽ_{3}∩L_{α+1}}(x)| ≥ r. If L_{2} is solvable this implies r = 2 and then K_{1}^{(α)} ≅ K_{2}^{(α)} ≅ L_{3}(2). But now there is some p - element ρ ∈ L_{3} with [ρ, Ṽ_{3}] = 1. As [x, Ṽ_{3}] ∩ Z_{α} ≠ 1, this contradicts (4.2) and (5.7).

We have L_{2} is nonsolvable and so q ≥ r. Suppose q > r (q > r^{2} for K_{2}^{(α)} ≅ U_{3}(r)). Let ω ∈ L_{2} ∩ L_{3}, o(ω) = x, x a Zsigmondy prime or x = 9 for q = 64. Suppose [ω, K_{2}^{(3)}] ≰ K_{2}^{(3)}. If [K_{2}^{(3)}, V_{3}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}, then we see M_{3} = M_{α}, a contradiction. Let K_{2}^{(3)}×…×K_{2i}^{(3)} =

K_{2}^{(3)}^{⟨ω⟩} with o(ω) = y, y a Zsigmondy prime in q − 1. Then i = y. Now for x ∈ V_{α}^{(1)} with [K_{2}^{(3)}, x] ≠ 1, we see that |[V_{3}^{(1)}, x]| ≥ r^{2^{y−2}}. But as y|2^{y−1} − 1, we see that |[V_{3}^{(1)}, x]| ≤ r^{2}2^{y−1}. So 2^{y−1} ≥ r^{2^{y−2}−2}. This shows y = 3, q = 4 and so r = 2. Then K_{2}^{(3)} ≅ L_{3}(2). But now we see that there are at least 6 nontrivial irreducible modules involved, a contradiction. So we have q = 64. We see that there are at least 8 nontrivial modules involved and so r^{8} ≤ qr^{2}. Now again r = 2 and K_{3}^{(2)} ≅ L_{3}(2). But now we have at least 9 nontrivial irreducible modules involved, a contradiction. So we have [ω, K_{2}^{(3)}] ≤ K_{2}^{(3)} and then [ω, K_{2}^{(3)}] = 1. Hence by (1.63) either K_{1}^{(3)} × K_{2}^{(3)} ≅ A_{5} × A_{5} or [ω, K_{1}^{(3)}] ≠ 1. Assume the latter. By (i) and (1.12) we have that K_{1}^{(3)} is either a group of Lie type over a field of characteristic two or a sporadic or alternating group. Suppose K_{1}^{(3)} is not a group of Lie type over a field of characteristic two. Then by (1.9) q = 4. This shows r = 2 and so K_{2}^{(3)} ≅ L_{3}(2). Hence m_{3}(K_{1}^{(3)}) = 2. Now 3 ∉ σ(M) and so ω ∈ K_{1}^{(3)}. But then ω is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, if m_{p}(L_{1}) ≥ 3. If m_{p}(L_{1}) = 2, then by the construction in (4.13) we have L_{1} ≤ L̃_{1} with m_{p}(L̃_{1}) = 3 and K_{1}^{(3)} is a component in L̃_{1} too. Hence ω is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M and E ≤ L̃_{1}. As ω ∈ K_{1}^{(3)} we have [ω, Z_{1}] = 1. Hence O_{2}(C I_{M}(ω)) ≠ 1. As p ≠ 3, we may apply (1.63). This shows N_{G}(⟨ω⟩) ≤ M , a contradiction to (1.66).

Now K_{1}^{(3)} ≅ G(r_{1}), r_{1} even. Assume q ≤ r_{1}. Let t ∈ Z_{α+1} be as before. Suppose K_{1}^{(3)} ≅ L̸_{3}(r). Then there are at least two nontrivial irreducible modules involved and r_{1}^{2} ≤ |[Ṽ_{3}, x]| ≤ qr < q^{2} ≤ r_{1}^{2}. Suppose K_{1}^{(3)} ≅ L_{3}(r), then there are at least three nontrivial irreducible modules involved and so r_{1}^{3} ≤ |[Ṽ_{3}, x]| ≤ qr^{2} < q^{3} ≤ r_{1}^{3}. In both cases we have a contradiction.

Assume q > r_{1}. By (1.66) we see that ω ∉ K_{1}^{(3)}. Otherwise as above we get N_{G}(⟨ω⟩) ≤ M. Now (1.64) and (1.65) imply q = r_{1}^{2} and K_{1}^{(3)} ≅ U_{n}(r_{1}), Ω_{2n}^{−}(r_{1}), ^{2}F_{4}(r_{1}) or ^{3}D_{4}(r_{1}). The last two cases are not possible as neither 3D_{4}(r_{1}) nor ^{2}F_{4}(r_{1}) do possess SC - modules by (1.22). In the other two cases we get r_{1}^{2} = q = r, a contradiction.

So we are left with r = 4 and K_{1}^{(3)} × K_{2}^{(3)} ≅ A_{5} × A_{5} and 5 ∈ σ(M). Now some t ∈ Ṽ_{3} acts on E ≅ E_{25} and C I_{E}(t) = 1, and centralizes some ρ ∈ L_{α} with o(ρ) = 5 and [E, ρ] = 1. This shows that there is U ≤ [t, Ṽ_{α}], |U| = 16, U is centralized by a 5 - element in L_{α}. But |[t, Ṽ_{α}] : [t, Ṽ_{α}] ∩ Z_{3}| ≤ 4, and so we get with (4.2) M_{3} = M_{α}, a contradiction to (5.7).

We have shown q = r or q ≤ r^{2} for K_{2}^{(3)} ≅ U_{3}(r). In the latter |Ṽ_{3} :

C IṼ_{3}(t)| ≤ r^{2} for every t ∈ Z_{α+1}, a contradiction as at least two nontrivial irreducible K_{2}^{(3)} - modules are involved in Ṽ_{3} and (1.26). The same applies for K_{2}^{(3)} ≅ Sz(q). Hence we have K_{2}^{(3)} ≅ L_{2}(q) or L_{3}(q). Suppose we have K_{2}^{(3)} ≅ L_{3}(q). Then there are at most three nontrivial irreducible modules involved. This shows K_{1}^{(3)} ≲ L_{3}(q). As m_{p}(K_{1}^{(3)}) = 1, we get K_{1}^{(3)} ≅ L_{2}(q). By interchanging the roles of K_{1}^{(3)} and K_{2}^{(3)} we may assume K_{2}^{(3)} ≅ L_{2}(q). Let t ∈ Z_{α+1}. Then |[Ṽ_{3}, t]| ≤ q^{2}. If [t, K_{2}^{(3)}] ≠ 1 we get the assertion.

So assume [t, K_{2}^{(3)}] = 1. Then |[t, Ṽ_{3}]| = q^{2}. As |Z_{α}| ≤ q^{2} we get |[u, Ṽ_{α}]| ≤ q^{3} for u ∈ Ṽ_{3} with [u, K_{2}^{(α)}] ≠ 1,. In particular there are at most three nontrivial irreducible L_{2}(q) - modules involved. This implies K_{1}^{(3)} ≅ SL_{3}(q) or L_{2}(q). As p|q^{2} − 1, we see in the former that [t, Ṽ_{3}] is centralized by some p - element in L_{3}. But |Z_{α} ∩ [t, Ṽ_{3}]| ≠ 1, a contradiction.

So we have K_{1}^{(3)} ≅ L_{2}(q) ≅ K_{2}^{(3)}. Now we get Ṽ_{3}/C IṼ_{3}(K_{1}^{(3)} × K_{2}^{(3)}) ≅ V_{1}^{(3)} ⊕ V_{2}^{(3)}, V_{1}^{(3)} is the O_{4}^{+}(q) - module and [V_{2}^{(3)}, K_{2}^{(3)}] = 1, V_{2}^{(3)} is the natural K_{1}^{(3)} - module. Let u ∈ Ṽ_{3}. Then we get |[u, Ṽ_{α}]| = q^{3} and so Z_{3} ≤ Ṽ_{α}. But in V_{1}^{(α)} ⊕ V_{2}^{(α)} any subgroup of order q^{2} contains elements centralized by a p - element, as this is true for any subgroup of order q in C I_{V_{1}^{(α)}}(W_{3}). This is a contradiction to (4.2) and (5.7). Hence (ii) holds.

We now can prove (3). Suppose we are in the situation of (4.13). Then there is a p-element in M_{3} centralizing Ṽ_{3}. But as Z_{α} ∩ [Ṽ_{3}, t] ≠ 1 we have a contradiction to (4.2) and (5.7). So we have m_{p}(L_{1}) ≥ 3. Now as p ||̸L_{2}∩L_{3}|, we have p|q +1. Suppose there is some ν with o(ν) = p such that ν induces a field automorphism on K_{3}^{(1)}. Then by (4.12) we have a subgroup isomorphic to D_{2p} in Out(K_{3}^{(1)} × K_{3}^{(2)}). So we have some ρ with o(ρ) = p such that [ρ, Ṽ_{3}] ≤ C IṼ_{3}(K_{3}^{(1)} × K_{3}^{(2)}). Hence [[Ṽ_{3}, K_{3}^{(1)} × K_{3}^{(2)}], ρ] = 1 as [ρ, Z_{3}] = 1. So we see [ρ, Ṽ_{3}] = 1, the same contradiction as before. This contradiction proves (3).

Let now t ∈ Ṽ_{α}. By (3) [t, K_{1}^{(3)} × K_{2}^{(3)}] = 1. We argue as before. Let [[t, Ṽ_{3}], K_{1}^{(3)}×K_{2}^{(3)}] ≠ 1. Then we may assume that t ∈ Z_{α+1} and so |[t, Ṽ_{3}]| ≤ rq or r^{2}q. In particular q > r, q > r^{2} for K_{2}^{(3)} ≅ U_{3}(r). Hence L_{2} is nonsolvable and for ω ∈ L_{2}∩L_{3}, o(ω) = x, x as above, we see that [ω, K_{2}^{(3)}] = 1 as before. Now K_{1}^{(3)} ≅ G(r_{1}) and also [K_{1}^{(3)}, [V_{3}, t]] ≠ 1, so q > r_{1}, q > r_{1}^{2} for K_{1}^{(3)} ≅ U_{n}(r_{1}), Ω_{2n}^{−}(r_{1}), or ^{2}E_{6}(r_{1}), and q > r_{1}^{3} for K_{1}^{(3)} ≅ ^{3}D_{4}(r_{1}). Hence we see [ω, K_{1}^{(3)}] = 1, too. Application of (1.56) shows K_{1}^{(3)} ≅ K_{2}^{(3)} ≅ A_{5} and p = 5. Now |[V_{3}, t] : [V_{3}, t] ∩ Z_{α}| ≤ 4. But there is ρ ∈ K_{1}^{(3)} × K_{2}^{(3)}, o(ρ) = 5, with |C I_{[V_{3},t]}(ρ)| ≥ 16, a contradiction to (4.2) and (5.7).

Hence we have Ṽ_{α} ≤ L_{2}. Now again q ≥ r^{2} (q ≥ r^{4} for K_{2}^{(α)} ≅ U_{3}(r)).

As there is t ∈ Z_{2}, with |[Ṽ_{α}, t]| ≤ q, we get as before K_{1}^{(3)} ≅ K_{2}^{(3)} ≅ A_{5}, p = 5. Now for u ∈ Ṽ_{α}, we have |[Z_{2}, u] : [Z_{2}, u] ∩ Z_{α}| ≤ 4. Hence |Z_{α}| ≤ 4, as Z_{3} ≤ [Z_{2}, u]. This shows q ≤ 4. But this contradicts q ≥ r^{2} = 16.

(8.11) Proposition. Assume (8.1). Then b = 1.

Proof. Let K_{α} be a component or normal p-subgroup in L_{α}/Q_{α} with [K_{α}, Z_{2}] ≠ 1. By (8.8) we have m_{p}(K_{α}) = 1. Assume first [Ṽ_{α}, K_{3}] ≠ 1 too. By (8.10) we have [K_{α}, Z_{2}] ≤ K_{α} and [Ṽ_{α}, K_{3}] ≤ K_{3}.

|Ṽ_{α} : C IṼ_{α}(K_{α})| ≥ 4 (1)

Otherwise for v ∈ Ṽ_{α}, we have |[Ṽ_{3}, v] : [Ṽ_{3}, v] ∩ Z_{α}| ≤ 2. This implies m_{p}(C I_{L_{3}}(K_{3})) = 1. Hence K_{3} admits an outer automorphism ρ of order p. This yields that K_{3} is a group of Lie type. As m_{p}(K_{3}) = 1, we get that ρ induces a field automorphism. Hence we get K_{3} ≅ L_{2}(r), or Sz(r) by (4.12). In any case we may assume that v is centralized by E ≅ E_{p^{2}} in L_{3}. As p ||̸C I_{L_{3}}(u)| for u ∈ Z_{α}^{♯}, we get |[Ṽ_{3}, v]| = 2. This contradicts (1.27).

(2) K_{3} ≅ G(r), a group of Lie type in even characteristic

By (1) we see that K_{3} admits a quadratic fours group. Now application of (1.12) shows K_{3}/Z(K_{3}) ≅ U_{4}(3), M_{12}, M_{22}, M_{24}, J_{2}, Co_{1} , Co_{2}, Sz, or A_{n}, n ≤ 11. In any case there is E ≅ E_{p^{2}}, E ≤ C I_{L_{3}/Q_{3}}(K_{3}). As m_{3}(K_{3}) ≥ 2, we have p > 3. Suppose |W_{3} : C I_{W_{3}}(K_{α})| = 4, then we see that |[v, Ṽ_{3}]| = 4 as before. Now (1.30) yields K_{3}/Z(K_{3}) ≅ A_{n}, as K_{3}/Z(K_{3}) ≅ U̸_{4}(3) by m_{p}(K_{3}) = 1. Suppose |W_{3} : C I_{W_{3}}(K_{α})| > 4. Then K_{3}/Z(K_{3}) ≅ A_{n} or M_{22}. In any case, K_{3}/Z(K_{3}) ≅ A_{n} or M_{22}, there is E ≅ E_{p^{2}}, E ≤ C I_{L_{3}/Q_{3}}(K_{3}). Now [E, Ṽ_{3}] ≠ 1. Hence there are at least three nontrivial irreducible modules T_{1}, T_{2}, T_{3} involved, which are centralized by a p - element. This shows that [v, ⟨T_{1}, T_{2}, T_{3}⟩] ∩ Z_{α} = 1. Now |[v, T_{1}]| = 2 and so we have that |Ṽ_{3} : Ṽ_{3} ∩ Q_{α}| = 2. But then |[v, ⟨T_{1}, T_{2}, T_{3}⟩]| = 2, a contradiction.

Let [Ṽ_{α}, K_{3}] ≠ 1. Then m_{p}(C I_{L_{3}/Q_{3}}(K_{3})) ≥ m_{p}(L_{3}) − 1. (3)

Suppose false. Then K_{3} admits a field automorphism. By (4.12) K_{3} ≅ L_{2}(r), or Sz(r) and there is K̃_{3} ≤ C I_{L_{3}/Q_{3}}(K_{3}), K̃_{3} ≅ K_{3}. If [K̃_{3}, Ṽ_{3}] = 1, we have Ṽ_{3} ∩ Z_{α} = 1. Hence K_{3} ≅ L_{2}(r) and there is just one natural module involved (1.27). As we have some ρ ∈ L_{3}, ρ induces a field automorphism on K_{3}, we see that there is some v ∈ [Ṽ_{3}, Ṽ_{α}], which is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M, a contradiction.

So we have [K̃_{3}, Ṽ_{3}] ≠ 1. As r ≥ 8, we get |Z_{α}| ≥ 8. Hence L_{2} is nonsolvable. Let ω ∈ L_{1} ∩ L_{2}, o(ω) odd. Then [ω, K_{3}] ≠ 1 or [ω, K̃_{3}] ≠ 1 by (1.63). In particular q ≤ r. Now for z ∈ Z_{2} \ Q_{α}, we have |Ṽ_{α} : C IṼ_{α}(z)| ≤ qr. This shows that q = r and |[Ṽ_{α}, z]| = r^{2}. Now K̃_{α} acts on [z, Ṽ_{α}] and induces the natural module. Furthermore some ρ_{α} of order p with [ρ_{α}, Z_{α}] = 1 induces a field automorphism on K_{α} and acts on [Ṽ_{α}, z]Z_{α}/Z_{α}. Now every v ∈ [Ṽ_{α}, z] is centralized by some p - element ρ̃_{α} ∈ L_{α} with N_{G}(⟨ρ̃_{α}⟩) ≤ M_{α}. But [z, Ṽ_{α}]∩Z_{3} ≠ 1, as can be seen from the action of Ṽ_{α} ∩L_{2} on Z_{2} by (5.6). This shows M_{3} = M_{α} with (4.2), contradicting (5.7).

[Ṽ_{α}, K_{3}] = 1 (4)

Let first p|r^{2} − 1. Then K_{3} ≅ Sz(r), L_{2}(r), L_{3}(r), or U_{3}(r). By (3) m_{p}(C I_{L_{3}/Q_{3}}(K_{3})) ≥ m_{p}(L_{3}) − 1. Suppose there is ρ ∈ M_{3} with [ρ, Ṽ_{3}] = 1 and N_{G}(⟨ρ⟩) ≤ M_{3}. Then Ṽ_{3} ∩ Z_{α} = 1. Hence K_{3} ≅ L_{2}(r) and Ṽ_{3}/C IṼ_{3}(K_{3}) is the natural module, or K_{3} ≅ L_{3}(r) and there are at most two nontrivial modules involved, which have to be the natural module and its dual. In both cases we get p|r − 1. Hence as m_{p}(K_{3}) = 1, we see K_{3} ≅ L_{2}(r), and there is v ∈ [Ṽ_{3}, Ṽ_{α}] centralized by E ≅ E_{p^{2}} in M_{3}, a contradiction.

So we have m_{p}(L_{3}) ≥ 3 and [ρ, Ṽ_{3}] ≠ 1 for all ρ ∈ L_{3} with o(ρ) = p. Choose ρ with [ρ, K_{3}] = 1. Now [C IṼ_{3}(ρ), K_{3}] ≠ 1. This shows Z_{α} ∩ [C IṼ_{3}(ρ), K_{3}] = 1 and so again K_{3} ≅ L_{2}(r) or L_{3}(r). Let Ṽ_{3}^{(1)} be a nontrivial module in [C IṼ_{3}(ρ), K_{3}]. Then Ṽ_{3}^{(1)} is the natural module. We have |[Ṽ_{3}^{(1)}, Ṽ_{α}^{(1)}]| = r in case of K_{3} ≅ L_{2}(r) and so we see that any element in Ṽ_{3}^{(1)} is centralized by E ≅ E_{p^{2}}. As the same holds for Ṽ_{α}^{(1)}, we get a contradiction to (4.2) and (5.7).

So we have p |r̸^{2} − 1. Choose ρ with o(ρ) = p and [K_{3}, ρ] = 1 or in case of m_{p}(L_{3}) = 2 with [Ṽ_{3}, ρ] = 1. Now C IṼ_{3}(ρ) involves at least three nontrivial irreducible K_{3} - modules T_{i}, i = 1, 2, 3. As T_{i} ∩ Z_{α} = 1, we see |[T_{i}, t]| ≤ r^{2}, i = 1, 2, 3, for t ∈ Ṽ_{α}, and so K_{3} ≅ L_{3}(r), L_{4}(r), L_{5}(2), L_{6}(2), L_{7}(2), Sp_{4}(r), Sp_{6}(r) or Ω_{8}^{−}(r). In the last three cases by (1.46) and (1.45) |Ṽ_{3} : C IṼ_{3}(K_{α})| ≤ r and then |[Ṽ_{α}, T_{1} ⊕ T_{2} ⊕ T_{3}]| ≤ r, a contradiction.

In case of K_{3} ≅ L_{3}(r), we have |Ṽ_{3} : C IṼ_{3}(K_{α})| ≤ r^{2}, a contradiction too. Hence we have K_{3} ≅ L_{4}(r), L_{5}(2), L_{6}(2), or L_{7}(2).

Suppose K_{3} ≅ L_{4}(r). Then |Ṽ_{3} : C IṼ_{3}(K_{α})| = r^{3}. Furthermore p||L_{3}(r)|. Now every element in T_{1} is centralized by E ≅ E_{p^{2}}. But then [Ṽ_{3}, Ṽ_{α}] contains some v ≠ 1, centralized by E ≅ E_{p^{2}} in L_{3} with Γ_{E,1}(G) ≤ M_{3} and some p - element in L_{α}, contradicting (4.2) and (5.7).

Let K_{3} ≅ L_{5}(2), L_{6}(2), or L_{7}(2). Now |[T, t]| ≤ 2^{2}, 2^{3}, 2^{4}, respectively. By (1.47) we see that T is the natural module, or K_{3} ≅ L_{7}(2) and |[T, v]| = 2^{4}. Now there are exactly three nontrivial irreducible modules involved and so p = 7. But m_{7}(L_{7}(2)) = 2. Hence in any case T_{i} are natural modules. Furthermore p |̸ |L_{n−1}(2)|, n = 5, 6, 7. This shows p = 31 for n = 5, or p = 127 for n = 7. In particular there are n modules T_{1}, …, T_{n}. Now Ṽ_{3} induces transvections and so |Ṽ_{3} : Ṽ_{3} ∩ Q_{α}| ≤ 2^{n−1}, a contradiction.

We will assume Ṽ_{α} ≰ L_{2}. By symmetry also Ṽ_{3} ≰ L_{α+1}. Hence there is a component K_{3}^{(1)} with [K_{3}^{(1)}, Ṽ_{α}] ≠ 1 ≠ [K_{α}, Ṽ_{3}]. Let K_{α}^{(1)} be the corresponding component of L_{α}/Q_{α}. Hence [K_{3}, Ṽ_{α}] = 1 = [K_{α}^{(1)}, Ṽ_{3}].

(5) One of K_{3} or K_{3}^{(1)} is a group of Lie type over a field of characteristic two. If K_{3} or K_{3}^{(1)} is not a group of Lie type then K_{3}/Z(K_{3}) or K_{3}^{(1)}/Z(K_{3}^{(1)}) is isomorphic to A_{n}, 7 ≤ n ≤ 11, n ≠ 8, M_{12}, M_{22}, M_{24}, or J_{2}. In particular m_{3}(K_{3}/Z(K_{3})) ≥ 2 or m_{3}(K_{3}^{(1)}/Z(K_{3}^{(1)})) ≥ 2.

Assume false. Let |W_{3} : W_{3} ∩ Q_{α}| = 2. Then for v ∈ Ṽ_{α}, we have |[v, V_{3}] : [v, V_{3}] ∩ Z_{α}| ≤ 2. As m_{p}(K_{3}^{(1)}) = 1 we have |[v, V_{3}]| ≥ 4. We may assume v ∈ Z_{α+1}. Now by (8.3) - (8.10) we have F^{⋆}(L_{3}/Q_{3}) = K_{3}^{(1)}C I_{F^{⋆}(L_{3}/Q_{3})}(v). Hence F^{⋆}(L_{3}/Q_{3}) = K_{3}K_{3}^{(1)} and m_{p}(L_{3}) ≥ 3. Furthermore there is a fieldautomorphism induced. Now by (4.12) we get that v centralizes some E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3} which acts on [v, V_{3}]. So we have a contradiction. Hence |W_{3} : W_{3} ∩Q_{α}| > 2. By the same argument we have that |W_{α} : W_{α} ∩Q_{3}| > 2. Now if K_{3} is not a group of Lie type we have K_{3}/Z(K_{3}) ≅ A_{n}, n ≤ 11, M_{12}, M_{22}, M_{24}, or J_{2}. If the same applies for K_{3}^{(1)} we have m_{3}(K_{3}^{(1)} × K_{3}) ≥ 4, a contradiction. So one of both is a group of Lie type.

(6) L_{2} is nonsolvable

Choose notation such that m_{3}(K_{3}) ≤ 1. By (5) K_{3} is a group of Lie type and so K_{3} ≅ L_{2}(r), L_{3}(r), U_{3}(r) or Sz(r). Now for v ∈ Ṽ_{α} we have |[v, V_{3}]| ≤ 2r, or 2r^{2} in case K_{3} ≅ SL_{3}(r). But [K_{3}, [v, V_{3}]] ≠ 1. This implies K_{3} ≅ L_{3}(2), p = 7 and |[v, V_{3}]| = 8. Now p | |̸C I_{L_{3}/Q_{3}}(K_{3}×K_{3}^{(1)})|. Application of (4.12) gives a contradiction.

We are now going to prove that Ṽ_{α} ≤ L_{2}. So assume false. Choose notation such that G(r_{1}) ≅ K_{3} is a group of Lie type. Furthermore if K_{3}^{(1)} is a group of Lie type too we assume K_{3}^{(1)} ≅ G(r_{2}), r_{1} ≥ r_{2}. By (6) q > 2.

Suppose q ≤ r_{1}. Then we get |[v, V_{3}]| ≤ |W_{3} : C I_{W_{3}}(K_{α})|r_{1}. This shows K_{3} ≅ L_{2}(r_{1}), |[v, V_{3}]| = r_{1}^{2}; K_{3} ≅ L_{3}(r_{1}), |[v, V_{3}]| = r_{1}^{3}; K_{3} ≅ L_{4}(r_{1}) or K_{3} ≅ Sp_{4}(r_{1}), |[v, V_{3}]| = r_{1}^{4}; K_{3} ≅ Sp_{6}(r_{1}). Suppose [v, V_{3}] ∩ Z_{α} ≠ 1. There is some ρ ∈ C I_{M_{3}}(K_{3} × K_{3}^{(1)}) with o(ρ) = p. As [ρ, [V_{3}, v]] ≠ 1, we get with (4.13) that we may assume that ρ ∈ L_{3}. Now ρ acts on [v, V_{3}] and we see o(ρ)|r_{1} − 1. As m_{p}(K_{3}) = 1, this implies K_{3} ≅ L_{2}(r_{1}). But again every x ∈ [v, V_{3}]^{♯} is centralized by some p - element in K_{3} × ⟨ρ⟩, contradicting (4.2) and (5.7).

We are left with K_{3} ≅ L_{4}(r_{1}), |[v, V_{3}]| = r_{1}^{4}, or K_{3} ≅ Sp_{6}(r_{1}), |[v, V_{3}]| = r_{1}^{6}. Now [K_{α}, Ṽ_{α}] involves the natural module. We have m_{3}(K_{3}^{(1)}) ≤ 1. This implies with (5) K_{3}^{(1)} ≅ L_{2}(r_{2}), L_{3}(r_{2}), U_{3}(r_{2}), or Sz(r_{2}). Let ρ ∈ C I_{L_{3}/Q_{3}}(K_{3}) × K_{3}^{(1)}) with o(ρ) = p. Then [ρ, [V_{3}, v]] = 1, as o(ρ) | r̸_{1} − 1.

This shows that [[K_{3}^{(1)}, V_{3}], ρ] = 1. Hence [K_{3}^{(1)}, V_{3}] ∩ Z_{α} = 1 in any case. Now let u ∈ V_{3}. Then |Ṽ_{α} : C IṼ_{α}(u)| ≤ r_{2}q. As K_{α} acts on [K_{α}^{(1)}, V_{α}], we get that |[K_{α}^{(1)}, V_{α}] : C I_{[K_{α}^{(1)},V_{α}]}(u)| ≤ r_{2}. But K_{α}^{(1)} acts on [[K_{α}^{(1)}, V_{α}], u], a contradiction.

We have q > r_{1}. Suppose there is ω ∈ L_{2} ∩ L_{3}, o(ω) = x, x a Zsigmondy prime, or x = 9 for q = 64 with [ω, K_{3}] ≠ 1. If [K_{3}, ω] ≰ K_{3}, we may assume [V_{α}, K_{3}^{⟨ω⟩}] = 1. Now V_{3} is generated by elements x with m_{p}(C I_{M_{3}}(x)) ≥ 2 and so C I_{G}(x) ≤ M_{3}. Hence [x, Ṽ_{α}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}. But there is x ∈ Ṽ_{3} with m_{p}(C I_{L_{α}}(x)) ≥ 2 and so some 1 ≠ y ∈ [x, Ṽ_{α}] is centralized by some p-element in L_{α}, contradicting (4.2) and (5.7). So we have [K_{3}, ω] ≤ K_{3}. As N_{G}(⟨ω⟩) ≰ M_{3}, we see that (1.65)(ii) is not possible. In these cases ω would be centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}. As [Z_{1}, ω] = 1 and p ≠ 3, we get a contradiction with (1.63).

Now with (1.64) and (1.65) we get q = r_{1}^{2}, K_{3} ≅ U_{n}(r_{1}), n ≤ 5, Ω_{8}^{−}(r_{1}), 2F_{4}(r_{1}), or ^{3}D_{4}(r_{1}). As [v, V_{3}] is an SC - module for K_{3}, the cases K_{3} ≅ 2F_{4}(r_{1}) and K_{3} ≅ ^{3}D_{4}(r_{1}) are not possible by (1.22). Hence we have K_{3} ≅ Ω_{6}^{−}(r_{1}), |[V_{3}, v]| = r_{1}^{6}; K_{3} ≅ Ω^{−}(8, r_{1}), |[V_{3}, v]| = r_{1}^{8}. In both cases [V_{3}, v] ∩ Z_{α} ≠ 1. Let ρ ∈ C I_{L_{3}/Q_{3}}(K_{3} × K_{3}^{(1)}) with o(ρ) = p. Then [ρ, [V_{3}, v]] ≠ 1. This implies p|r_{1} − 1, contradicting m_{p}(K_{3}) = 1. Hence we have [ω, K_{3}] = 1.

Let now [ω, K_{3}^{(1)}] ≠ 1. Again [ω, K_{3}^{(1)}] ≤ K_{3}^{(1)}. Suppose K_{3}^{(1)} is not a group of Lie type. Then by (5) and (1.9) o(ω) = 3. Now K_{3} ≅ L_{3}(2) and so ω ∈ K_{3}^{(1)}. But then C I_{M}(ω) ≥ E, E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{3}. As C I_{M}(ω) ≥ K_{3}, we get N_{G}(⟨ω⟩) ≤ M_{3} by (1.63), a contradiction. Now with (1.64) and (1.65) we see r_{2}^{2} = q and K_{3}^{(1)} ≅ U_{n}(r_{2}), Ω_{8}^{−}(r_{2}), ^{2}F_{4}(r_{2}) or ^{3}D_{4}(r_{2}). Let v ∈ [K_{3}^{(1)}, V_{3}]. Then we see that |[v, V_{α}]| ≤ r_{2}^{6}, r_{2}^{6}, r_{2}^{10}, r_{2}^{6}, respectively. As [v, V_{α}] is an SC - module for K_{3}^{(1)}, we see with (1.22) that K_{3}^{(1)} ≅ Ω_{6}^{−}(r_{2}), |[v, V_{α}]| = r_{2}^{6}, or K_{3}^{(1)} ≅ Ω_{8}^{−}(r_{2}), |[v, V_{α}]| = r_{2}^{8}. But now we get the same contradiction as before. This implies [ω, K_{3} × K_{3}^{(1)}] = 1. By (1.63) we get K_{3} × K_{3}^{(1)} ≅ L_{2}(4) × L_{2}(4) and p = 5. Now |[v, V_{3}] : [v, V_{3}] ∩ Z_{α}| ≤ 4. As E_{25} acts on [v, V_{3}], we get some ρ ∈ L_{3} with o(ρ) = 5 and N_{G}(⟨ρ⟩) ≤ M_{3}, and C I_{Z_{α}}(ρ) ≠ 1, contradicting (4.2) and (5.7). So we have shown

Ṽ_{α} ≤ Q_{3} ≤ L_{2} (7)

We now investigate [Ṽ_{α}, Z_{2}]. Suppose L_{2} to be solvable. Then we have transvections on Ṽ_{α}. This implies K_{α} ≅ A_{n}, L_{n}(2), Sp_{2n}(2) or Ω_{2n}^{±}(2). Furthermore as K_{α} does not possess an outer automorphism of order p, we get [z, Ṽ_{α}] is centralized by E ≅ E_{p^{2}} with Γ_{E,1}(G) ≤ M_{α}, where [z, Ṽ_{α}] ∩ Z_{3} ≠ 1. But this contradicts (4.2) and (5.7). So we have L_{2} is nonsolvable. Now there is z ∈ Z_{2} with 1 ≠ [z, Ṽ_{α}] ≤ Z_{3}. As E ≅ E_{p^{2}} acts on [v, V_{3}] we get some 1 ≠ u ∈ Z_{3} centralized by some p - element in L_{α}, contradicting (4.2) and (5.7). This final contradiction proves the proposition.

(8.12) Proposition. b = 1

Proof. By (8.11) we just have to show that (8.1) holds. Hence assume L_{1}/O_{2}(L_{1}) is an extension of an r - group R by L_{2}(r) times L_{1} ∩ L_{2}. Now we have |Z_{2} : Z_{2} ∩ Q_{α}| ≤ 4 ≥ |Z_{α+1} : Z_{α+1} ∩ Q_{3}|. Choose α + 1 ∈ Δ(α) with [Z_{2}, Z_{α+1}] ≠ 1.

(1) We may assume [Z_{α+1} ∩ Q_{3}, Z_{2}] ≠ 1

Otherwise we have [Z_{α+1} ∩ Q_{3}, W_{3}] = 1. Hence W_{3} ∩ Q_{α} centralizes a subgroup of index four in Z_{α+1}. If [W_{3} ∩ Q_{α}, Z_{α+1}] = 1, then Z_{α+1} centralizes a subgroup of index four in W_{3}. Now by (1.51) either Z_{α+1} ≤ Q_{3} or O_{r}(L_{3}/Q_{3}) centralizes W_{3}. By (4.10) the latter is not possible as this would imply L_{2} ≤ M_{3}. So we have [Z_{α+1}, W_{3}] = 1, a contradiction. This implies [W_{3} ∩ Q_{α}, Z_{α+1}] ≠ 1. In particular W_{3} ∩ Q_{α} centralizes a subgroup of index four in Z_{α+1}. By (1.43) we get E((L_{2}/Q_{2})/F (L_{2}/Q_{2})) ≅ L_{2}(4), L_{3}(2), L_{2}(4)×L_{2}(4), L_{3}(2)×L_{3}(2), or L_{2}/Q_{2} is solvable. In any case Z_{α+1} is generated by elements z such that |W_{3} : C I_{W_{3}}(z)| ≤ 16. Application of (1.51) gives [O_{r}(L_{3}/Q_{3}), W_{3}] = 1, a contradiction to (4.10).

[Z_{α+1} ∩ Q_{3}, Z_{2} ∩ Q_{α}] ≠ 1 (2)

Otherwise Z_{α+1} ∩Q_{3} centralizes a subgroup of index four in Z_{2} ∩Q_{α}. Now any z ∈ Z_{2} centralizes a subgroup of index 16 in W_{α}. Application of (1.51) and (4.10) yields a contradiction as Z_{2} ≰ Q_{α}.

(3) L_{2} is nonsolvable and |W_{α} ∩ Q_{3} : W_{α} ∩ Q_{2}| ≥ 16

Otherwise we have that Z_{2} = ⟨z||W_{α} : C I_{W_{α}}(z)| ≤ 32⟩. By (1.51) [O_{r}(L_{α}/Q_{α}), W_{α}] = 1, contradicting (4.10). Now application of (1.7), (2.1) together with (1.12) shows that E(L_{2}/Q_{2}) is as in (1.7)(i) or (iii).

|Z_{α+1} ∩ Q_{3} : Z_{α+1} ∩ Q_{2}| ≥ 4 (4)

Otherwise z ∈ Z_{2} ∩Q_{α} centralizes a subgroup of index eight in Z_{α+1}. Now we apply (1.26) and (1.27). This shows that E(L_{2}/Q_{2}) ≅ L_{2}(8), L_{2}(8) × L_{2}(8), L_{2}(4) × L_{2}(4), A_{6}, 3 · A_{6}, 3 · A_{6} ∗ 3 · A_{6}. As |W_{α} ∩ Q_{3} : W_{α} ∩ Q_{2}| ≥ 16, we get E(L_{2}/Q_{2}) ≅ L_{2}(8)×L_{2}(8), L_{2}(4)×L_{2}(4), or 3·A_{6} ∗3·A_{6}. Now in any case by quadratic action we have that E(L_{2}/Q_{2}) = X_{1}X_{2}, Z_{2}/C I_{Z_{2}}(X_{1}X_{2}) = ([X_{1}, Z_{2}] ⊕ [X_{2}, Z_{2}])C I_{Z_{2}}(X_{1}X_{2})/C I_{Z_{2}}(X_{1}X_{2}). But now again Z_{2} = ⟨z||W_{α} : C I_{W_{α}}(z)| ≤ 32⟩, contradicting (1.51) and (4.10).

Now we know :

|Z_{2} : C I_{Z_{2}}(Z_{α+1} ∩ Q_{3})| ≤ 4|Z_{α+1} ∩ Q_{3} : Z_{α+1} ∩ Q_{2}| ≤ |Z_{α+1} ∩ Q_{3} : Z_{α+1} ∩ Q_{2}|^{2}.

Hence we may apply (1.35). This now yields the following cases :

(i) E(L_{2}/Q_{2}) ≅ L_{2}(q), q ≥ 16, Z_{2} is the natural module

(ii) E(L_{2}/Q_{2}) ≅ L_{2}(16), Z_{2} is the orthogonal module

(iii) E(L_{2}/Q_{2}) ≅ L_{2}(q) × L_{2}(q), q ≥ 4, Z_{2} is a direct sum of two natural modules, or q = 4 and Z_{2} is a direct sum of two orthogonal modules.

(iv) E(L_{2}/Q_{2}) ≅ L_{2}(16) × L_{2}(16), Z_{2} is a direct sum of two orthogonal modules.

(v) E(L_{2}/Q_{2}) ≅ L_{2}(4) × L_{2}(4), Z_{2} = V_{1} ⊕ V_{2}, V_{i}, i = 1, 2 is a direct sum of two natural modules.

(vi) E(L_{2}/Q_{2}) ≅ L_{3}(2) × L_{3}(2), Z_{2} = V_{1} ⊕ V_{2}, V_{i}, i = 1, 2 is a direct sum of the natural module and its dual.

(vii) E(L_{2}/Q_{2}) ≅ 3 · A_{6} ∗ 3 · A_{6}, Z_{2} = V_{1} ⊕ V_{2}, V_{i} is the direct sum of the natural A_{6} - module and its dual.

(viii) E(L_{2}/Q_{2}) ≅ SL_{3}(4) and Z_{2} is a sum of the natural module and its dual.

(ix) E(L_{2}/Q_{2}) ≅ A_{9}.

If E(L_{2}/Q_{2}) possesses two components, then the projection of W_{α} ∩ Q_{3}/W_{α} ∩ Q_{2} onto each of these components is of order at least 16 by (1.51) and (4.10). Hence we just have (i), (iii), (viii) or (ix), as for L_{2}(16) on the orthogonal module no Sylow 2 - subgroup acts quadratically by (1.27)(iv). As W_{α} ∩ Q_{3} ≤ L_{2} ∩ L_{3} and W_{α} ∩ Q_{3} ≰ Q_{2} we see that E(L_{2}/Q_{2}) ≅ A̸_{9}.

So we have Z_{2} is either the natural module or a direct sum of two natural modules. In the latter Z_{1} is centralized by O_{r}(L_{1}/Q_{1}) and C I_{L_{2}}(x) ≰ M, for some x ∈ Z_{1}^{♯}. This contradicts (4.10). So we have

(5) E(L_{2}/Q_{2}) ≅ L_{2}(q), q ≥ 16, Z_{2} is the natural module.

But then 1 ≠ [Z_{α+1} ∩Q_{3}, Z_{2} ∩Q_{α}] ≤ Z_{α} ∩Z_{3}. Hence M_{α} = M_{3}. By (1.51) and (4.12) C I_{G}(x) ≤ M_{3} for every x ∈ Z_{1}^{♯}. Hence application of (5.7) gives a contradiction. This proves the proposition.

(8.13) Proposition. b > 1

Proof. By (5.7) we just have to show that Z_{2} is one of the modules in (5.6) or the exceptional case of (5.7) holds. Assume b = 1. Then Z_{2} ≰ Q_{1}. Application of (1.62) shows

(⋆) There is some g ∈ L_{1} such that for X = ⟨Z_{2}, Z_{2}^{g}⟩ we have X/O_{2}(X) ≅ D_{2r}, r odd, L_{2}(r), Sp_{4}(r), Sz(r), r even, and for Y = (Z_{2}∩O_{2}(X))(Z_{2}^{g}∩ O_{2}(X)) we have |Z_{2} : C I_{Z_{2}}(Y)| ≤ |Y : Y ∩ Q_{2}|^{2}, or Z_{2} is an F - module.

For what follows we will assume that Z_{2} is neither an F - module nor one of the modules from (5.6). This in particular implies [Z_{2}, Y, Y] ≠ 1. Suppose first that L_{2} is nonsolvable. As [Z_{2}, Y, Y] ≠ 1, we get (1.35)(5), (6), (8), (9), (11), (12), (13), or (14). Now (1.35)(5) is (5.6)(ii), (1.35)(9) is (5.6)(iii). If we have (1.35)(14), then Q_{1} ≤ Q_{2} and so [Z_{2}, Q_{1}] = 1, i.e. Z_{2} ≤ Q_{1}. Hence b > 1.

It remains (1.35)(6), (8), (11), (12), or (13). As m_{p}(C I_{L_{1}}(Z_{1})) ≥ 3, we get q = 4 in (8) and q = 2 in (11).

Suppose we have (1.35)(11), (12), or (13). As |Z_{2} : C I_{Z_{2}}(Y)| ≤ |Y : Y ∩ Q_{2}|^{2}, we see that |Y : Y ∩ Q_{2}| = 8 or we have (1.35)(13). Let first |Y : Y ∩ Q_{2}| = 8. As X cannot act transitively on (Y/Z_{2} ∩ Z_{2}^{g})^{♯}, otherwise Y would be abelian, we now see that X/O_{2}(X) is dihedral. Hence for every t ∈ Y we have |Z_{2} : C I_{Z_{2}}(t)| ≤ 8. But for t ∈ Y ∩ L′_{1}, we have |Z_{2} : C I_{Z_{2}}(t)| ≥ 16, a contradiction.

Hence we have (1.35)(13) and |Y : Y ∩ Q_{2}| > 8. We have furthermore |Y : Y ∩ Q_{2}| ≤ 64. In any case there is t ∈ Y with |Z_{2} : C I_{Z_{2}}(t)| ≥ 2^{7}. In particular X/O_{2}(X) is not dihedral. Suppose X/O_{2}(X) ≅ L_{2}(r). As X cannot act transitively on (Y/Z_{2} ∩ Z_{2}^{g})^{♯}, we get r ≤ 8. Now again |Z_{2} : C I_{Z_{2}}(t)| ≤ 2^{6}, a contradiction. If X/O_{2}(X) ≅ Sz(8), then for t ∈ Y \ Q_{2}, we have [S ∩ X, t](Z_{2} ∩ Z_{2}^{g}) = Z_{2} ∩ O_{2}(X). Hence [t, Z_{2} ∩ O_{2}(X)] = 1, and so |[t, Z_{2}]| ≤ 8, a contradiction. Let finally X/O_{2}(X) ≅ Sp(4, r). As not all elements in (Y/Z_{2} ∩ Z_{2}^{g})^{♯} are conjugate, we see that X/O_{2}(X) ≅ Sp_{4}(2). Now |[Z_{2}, t]| ≤ 2^{7} for every t ∈ Y, as |Z_{2} : Z_{2} ∩ O_{2}(X)| = 4. This implies |Y : Y ∩Q_{2}| = 16, otherwise Y contains t ∈ E(L_{2}/Q_{2}) which does not belong to either of the components and then |[t, Z_{2}]| ≥ 2^{8}. But |Y : Y ∩ Q_{2}| > 8 yields |[t, Z_{2}]| ≤ 2^{5} for every t ∈ Y, a contradiction.

Suppose now (1.35)(6) or (8), q = 4. If |Y : Y ∩ Q_{2}| ≤ 4, then we get X/O_{2}(X) is dihedral and so for t ∈ Y, |Z_{2} : C I_{Z_{2}}(t)| ≤ 4. This proves that we have (8). Now |Y : Y ∩ Q_{2}| ≤ 16. We have that E(L_{2}/Q_{2}) = X_{1} × X_{2}, X_{i} ≅ L_{2}(4) and Z_{2} = Z_{2}^{(1)} ⊕ Z_{2}^{(2)}, [Z_{2}^{(1)}, X_{2}] = [Z_{2}^{(2)}, X_{1}] = 1, and Z_{2}^{(i)} is a direct sum of two orthogonal X_{i} - modules, i = 1, 2. This implies for t ∈ Y ∩ E(L_{2}/Q_{2}), that |[t, Z_{2}]| ≥ 16, and |[t, Z_{2}]| ≥ 2^{8}, for t ∉ X_{1} ∪ X_{2}. As |Y/Y ∩ Q_{2}| ≤ 16, we have |Y/Z_{2} ∩ Z_{2}^{g}| ≤ 2^{8}. This shows that X/O_{2}(X) ≅ Sp_{4}(2), L_{4}(2), or dihedral. Hence for every t ∈ Y, we have |[Z_{2}, t]| ≤ 2^{5}. But then we get |Y : Y ∩ Q_{2}| ≤ 8. Now we have that X/O_{2}(X) is dihedral and then |[Z_{2}, t]| ≤ 8 for t ∈ Y. Hence |Y ∩E(L_{2}/Q_{2})| = 1. Now |Y : Y ∩Q_{2}| = 4, and |[Z_{2}, t]| ≤ 4 for t ∈ Y. But Y contains some t with [X_{1}, t] ≠ 1 ≠ [X_{2}, t] and so |[Z_{2}, t]| ≥ 16, a contradiction.

We are left with (1.35)(15) and (16). We have |Y : Y ∩ Q_{2}| ≤ 8. Hence either Y acts quadratically on Z_{2} and |Z_{2} : C I_{Z_{2}}(Y)| ≤ 4, i.e. Z_{2} is an F - module, or X/O_{2}(X) is dihedral and |Z_{2} : C I_{Z_{2}}(t)| ≤ 8 for every t ∈ Y. Furthermore |Y : Y ∩ Q_{2}| ≥ 4. Suppose the latter. Then L_{2} is a {2, 3} - group. Furthermore we may assume that (1.35)(16)(iii), (v), or (vi) holds. In any case Z_{2} is not an F - module. Hence |[Z_{2}, t]| ≥ 4 for t ∈ Y \ Q_{2}. If we have (1.35)(16)(iii), we have |Y : Y ∩ Q_{2}| = 4 and |Z_{2} : C I_{Z_{2}}(Y)| = 8. Now |Z_{2} : C I_{Z_{2}}(O_{3}(L_{2}/Q_{2}))| ≤ 2^{6}, contradicting |Z_{2}| = 2^{8}. Let L_{2} be as in (1.35)(16)(v). Then S/Q_{2} ≲ Sp_{4}(3) · 2. But Y does not contain some t with [t, O_{3}(L_{2}/Q_{2})] = O_{3}(L_{2}/Q_{2}). Hence again |Y : Y ∩ Q_{2}| = 4. Let Y = ⟨y_{1}, y_{2}⟩(Y ∩Q_{2}). Then W = [O_{3}(L_{2}/Q_{2}), y_{1}] ≅ 3^{1+2}. Now |[W, Z_{2}]| = 16. As [Z_{2}, Y, Y] ≠ 1, we see [Y, [Z_{2}, W]] ≠ 1. Hence [C I_{Z_{2}}(W), Y] = 1. But then [Y, O_{3}(L_{2}/Q_{2})] = W, a contradiction.

So we are left with (1.35)(16)(vi). By (1.8)(iv) |Y : Y ∩ Q_{2}| = 4. But now |Z_{2} : C I_{Z_{2}}(Y)| < |Y : Y ∩ Q_{2}|^{2}. This contradicts (1.36). This final contradiction proves the lemma.

Now the main theorem follows from (8.12), (8.13), (7.32) and (6.39) .

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