The non E–uniqueness case

U. Meierfrankenfeld G. Stroth

preliminary version of September 30, 2003

In this paper we continue the investigations of [E]. Further we will assume that we are not in the PU-case and not in the amalgam-case. Then the result was that there is some amalgam (P, L) with b = 1. In this paper we will investigate this case. We will prove Theorem. Assume that we are in the non E-uniqueness case. Then one of the following holds

(i) We have P U (ii) There is some amalgam with common point stabilizer (iii) E(G) ≅ M_{22}, M_{23}, M_{24}, L_{r}(q) or A_{n}.

For the remainder of this paper we assume not to be in (i) or (ii). For convenience of the reader we repeat some facts. We have that Y_{P} is an F-module with offender Y_{L}. In O_{p} of the point stabilizer there is never an over offender. Let I_{P} be a quasi simple module for a p-component of N. If I_{P} has a non uniqueness element, then N/O_{p}(N) is quasi simple. If x ∈ I_{P}, [S, x] = 1, then x is an uniqueness element. If T ∈ Syl_{p}(B), B = L ∩ C̃ and [x, T] = 1, then either x is an uniqueness element or T ∈ Syl_{p}(C I_{G}(x)). If D = 1 = E, then T = S. Now we start the proof.

(1.1) Proposition. Suppose that H_{L} has more than one component (including L_{2}(2), Sz(2) and U_{3}(2)), then E(G) ≅ A_{m} for suitable m.

Proof. Let I_{L} be some quasi simple module for some component of H_{L}. Then I_{L} is an F-offender on Y_{P}. If p ≥ 3, then by quadratic action Y_{L} is a direct sum [GBSM]. Hence [I_{L}, Y_{P}] contains no uniqueness elements

If p = 2 and |Y_{P} : Y_{P} ∩ Q_{L}| > 2 we get the same conclusion with [GBSM].

So let p = 2 and |Y_{P} : Y_{P} ∩ Q_{L}| = 2. Then I_{L} induces transvections on Y_{P}. If H_{P} ≅ L_{n}(q) or Sp(2n, q) all elements in I_{P} are uniqueness elements.

Let F be a (r, r)-group inverted by I_{P}. Then

[Y_{L}, F] = ⟨C I_{[Y_{L},F]}(x)|xϵF^{♯}⟩

If C I_{[Y_{L},F]}(x) ≠ 1, then [I_{P} , C I_{[Y_{L},F]}(x)] ≠ 1, and so F ≤ P, a contradiction.

Hence not all elements in I_{P}, are uniqueness elements and so N_{P} ≅ N ≅ A_{n} or Ω^{±}(2n, 2).

Now |I_{L} : I_{L} ∩ Q_{P} | = 2 and then all further components have to centralize I_{L}, and so [I_{L}, I_{P}] contains no uniqueness elements.

So all together we get

(∗) [I_{L}, Y_{P}] contains no uniqueness elements.

From (∗) we get that N/O_{p}(N) is quasi simple.

Inspection of the modules in [GBSM] implies H_{P} ≅ O^{±}(2n, 2) or Σ_{n}.

Let first H_{P} ≅ O^{±}(2n, 2). Then I_{L} corresponds to a transvection. Let H_{L_{1}}, H_{L_{2}} be two components with corresponding modules I_{L_{1}}, I_{L_{2}}. Then there are singular elements in [Y_{P} , I_{L_{1}} ⊗ I_{L_{2}}] and so there are uniqueness elements. This shows that we have exactly two components. Further |[Y_{L}, Y_{P}]| = 4. Let x be the unique singular element in [Y_{L}, Y_{P}]. Set U = C I_{P} (x). Then B ≤ U and so Q_{U} ≤ Q_{B}. We have [Q_{B}, B ∩ H_{L_{1}}] ≤ Q_{B}. As I_{L_{1}}^{Q_{U}} = I_{L_{2}}, we get that Q_{B} interchanges the two components. Hence B ∩ H_{L_{1}} is a 2-group. This shows H_{L_{2}} ≅ Σ_{3} and H_{L_{1}}H_{L_{2}} ≅ Σ_{3} × Σ_{3}. Now O^{2}(C I_{B}(I_{P})) ⊴ ⟨H_{P} , H_{L}, B⟩, and then C I_{B}(I_{P}) = Q_{P}. This shows P/Q_{P} ≅ O^{±}(2n, 2).

We have I_{P} D_{P} = Y_{P} and I_{P} ∩ ϕ(Q_{P}) = 1. But all elements in Z(S) are uniqueness elements and D_{P} ≤ Q_{L}. Further D_{L} × I_{L} = Q_{L}. This now shows Y_{L} = I_{L} is of order 16 and then Q_{P} = Y_{P}.

Now we see S ∈ Syl_{p}(L). Then L/Q_{L} ≅ O^{+}(4, 2) and |S| = 2^{7}. This gives P/Q_{P} ≅ Σ_{5} ≅ O^{−}(4, 2).

Let now i be a non 2-central involution in P like [I_{P} , I_{L_{1}}]. Then C I_{L}(i) ≅ C I_{P} (i) ≅ (V_{4} × Σ_{4}) : 2. Furthermore C I_{P} (i) contains a fours group of non 2-central involutions. As P ≅ C̃, we get O(C I_{G}(i)) = 1. From this we see C I_{G}(i) ≅ (V_{4} × A_{6}) : 2 and N(O_{2}(C I_{G}(i))) ≅ (A_{4} × A_{6}) : 2.

Let now ρ ∈ N(O_{2}(C I_{G}(i))), o(ρ) = 3, [ρ, A_{6}] = 1. Then using the structure of the centralizer of a 2-central involution, we see that

N(⟨ρ⟩) ≅ (Z_{3} × A_{7}) : 2

Now we can also determine the structure of the normalizer of an element of order 5 and so we see a subgroup (A_{5} × A_{5}) : 2. Hence with [MF] we get an A_{10}-amalgam and so E(G) ≅ A_{10}.

Let now H_{P} ≅ A_{n}. Then Y_{L} is a transposition group. Then we have that U = C I(F ix(Y_{L})) ∩ N(Y_{L}Q_{B}) ≅ Z_{2} ≀ Σ_{k}, where Y_{L} corresponds to ⟨(12), ⋯, (2k − 1, 2k)⟩.

We have [U, Q_{B}] ≤ Y_{L}Q_{P} ≤ Q_{B}. Hence U ≤ B. As |[Y_{P} , I_{L_{1}}]| = 2, we have [B ∩ H_{L_{1}}, [Y_{P} , I_{L_{1}}]] = 1. As U acts transitively on the components of H_{L}, we see [U, B ∩ H_{L_{1}}] ≤ C I_{U}([Y_{L}, Y_{P}]) ≤ Q_{B}. Hence B ∩ H_{L_{1}} is a 2-group and then again H_{L_{1}} ≅ Σ_{3} and we have Σ_{3} ≀ Σ_{k}. Now [C I_{P} (Y_{L}), B ∩ H_{L}] ≤ Q_{B} and so O^{2}(C I_{P} (Y_{L})) ⊴ ⟨H_{L}, H_{P} , B⟩. This implies P/Q_{P} ≅ Σ_{n}. Then ϕ(Q_{P}) ≤ D_{P} and ϕ(Q_{L}) ≤ D_{L}. As Q_{P} = (Q_{P} ∩Q_{L})Y_{P}, we get Q_{P} = Y_{P}. As D_{L}×I_{L} = Q_{L} and D_{P} ≤ Q_{L}, we see D_{P} ≤ D_{L} and then D_{P} = 1. But then P = C̃ and S ≤ L. Again any element in Z(S) is a uniqueness element. Hence D_{L} = 1. This gives I_{L} = Q_{L}. In particular n = 2k or 2k + 1 and as in the A_{10}-case we construct an amalgam which with [MF] implies E(G) ≅ A_{4k} or A_{4k+2}.

(1.2) Lemma. Let H_{L} be solvable. Then H_{L} is either an elementary abelian 3-group, SL_{2}(3), or D_{2r} and H_{P} ≅ L_{n}(2).

Proof. As Y_{P} acts quadratically, we see p ≤ 3. Let first p = 3. Then [O_{p′}(H_{L}), Y_{P}] is a 2-group. As H_{L} = [O_{p′}(H_{L}), Y_{P}]Y_{P} we get that H_{L} has to be a 2-group. By quadratic action we see that there are transvections in N on Y_{P}. Hence H_{P} ≅ SL_{n}(3) or Sp(2n, 3). In both cases all elements in I_{P} are uniqueness elements. As in (1.1) we get |I_{P} : I_{P} ∩ Q_{L}| = 3. Quadratic action now implies O_{p′}(H_{L}) ≅ Q_{8} and H_{L} ≅ SL_{2}(3).

Let now p = 2. Assume |Y_{P} : Y_{P} ∩ Q_{L}| > 2. Then the dihedral lemma provides us with transvections or Y_{P}. Hence H_{P} ≅ L_{n}(2), Sp(2n, 2), O^{±}(2n, 2) or Σ_{n}. In the first two cases all elements in I_{P} are uniqueness elements and we get a contradiction.

Let now D_{1} × ⋯ × D_{e} be a direct product of dihedral groups acting on Y_{L} with Y_{P} as a Sylow 2-group. As there is no fours group of transvections in H_{P} we see |[D_{1}, Y_{L}]| = 4. Hence O_{2′}(H_{L}) is a 3-group. Let now x ∈ Y_{P} with |[Y_{L}, x]| = 2. Let U = C I_{O′_{2}(H_{L})}(Y_{L}). Then |[O_{2′}(H_{L}), x]U/U| = 3. This shows Y_{P} O_{2′}(H_{L})U/U ≅ Σ_{3} × ⋯ × Σ_{3}. As B is maximal in L we have U ≤ B and then [U, Y_{P}] = 1. In particular [O_{2′}(H_{L}), Y_{P}] ≅ Z_{3} × ⋯ × Z_{3}, the assertion.

Let finally |Y_{P} : Y_{P} ∩ Q_{L}| = 2. Again I_{L} induces transvections. If H_{P} ≅ Σ_{n}, Sp(2n, 2) or O^{±}(2n, 2), we see as above that r = 3, I_{P} induces transvections and so O_{2′}(H_{L}) is cyclic. As B is maximal this shows |O_{2′}(H_{L})| = 3, the assertion.

So let H_{P} ≅ L_{n}(2). Then we just have uniqueness elements. In particular Y_{P} cannot invert some group of type (r, r) and also cannot act on an extraspecial group r^{1+2}. This again shows O_{2′}(H_{L}) is cyclic and then as before |O_{2′}(H)| = r.

(1.3) Lemma. Assume that all elements in I_{P} are uniqueness elements. Then we have

(i) If [I_{P}^{g} , I_{P}] ≤ I_{P} ∩ I_{P}^{g}, then [I_{P}^{g} , I_{P}] = 1, g ∈ G.

(ii) Let g ∈ L with R = ⟨I_{P} , I_{P}^{g} ⟩ ≰ L ∩ C̃. Set V = (I_{P} ∩ Q_{L})(I_{P}^{g} ∩ Q_{L}). Then [Q_{R} ∩ I_{P} , V] = 1 and C I_{V} (I_{P}) = [I_{P} , V] = I_{P} ∩ Q_{L}

(iii)Let g ∈ L with [I_{P} , I_{P}^{g}] ≤ Q_{L}. Then ⟨I_{P} , I_{P}^{g} ⟩ acts quadratically on Y_{L}

Proof. (i) Let [I_{P}^{g} , I_{P}] ≠ 1. Then there is 1 ≠ x ∈ I_{P} ∩ I_{P}^{g}. By uniqueness we have C̃ = C̃^{g}. Hence g ∈ C̃ and so I_{P}^{g} ≤ Q. In particular I_{P}^{g} ≤ Z(Q), the assertion.

(ii) We have I_{P} ∩ I_{P}^{g} = 1. So V ◁ ⟨I_{P} , I_{P}^{g} ⟩ and [I_{P} , V] = I_{P} ∩ Q_{L} = C I_{V} (I_{P}). Consider C I_{V} (Q_{R}I_{P}). Then 1 ≠ C I_{V} (Q_{R}I_{P}) ≤ I_{P} ∩ V. Hence C I_{I_{P} ∩V} (I_{P}^{g} ∩ Q_{R}) ≠ 1. Then I_{P}^{g} ∩ Q_{R} ≤ L ∩ C̃. This shows I_{P}^{g} ∩ Q_{R} ≤ N(I_{P}). Hence [I_{P}^{g} ∩ Q_{R}, I_{P} ∩ Q_{R}] ≤ I_{P}^{g} ∩ I_{P} = 1.

(iii) Let X = [Y_{L}, I_{P}]. Then I_{P}^{g} ≤ N(X). If [I_{P}^{g} , X] ≠ 1, then we get [Y_{L}, I_{P}]∩[Y_{L}, I_{P}^{g}] ≠ 1. Now I_{P} ≤ N_{G}(I_{P}^{g}) and I_{P}^{g} ≤ N_{G}(I_{P}) and so (i) applies.

From now on we assume that H_{L} has just one component.

(1.4) Lemma. Let H_{L} be nonsolvable. Then H_{L} ≅ L_{n}(q), Sp(2n, q), G_{2}(q), A_{m} or 3A_{6}. In any case I_{L}/C I_{I_{L}}(H_{L}) is the natural module.

Proof. We have that I_{P} is an F -module with offender I_{L}. Assume first that I_{P} is a self dual module. Then

|[I_{P} , I_{L}]| = |I_{P} /C I_{I_{P}} (I_{L})|.

Hence I_{L} is a dual offender. Application of [GBSM] shows that I_{L} is an F -module or H_{L} ≅ 3A_{6} acting on the 6-dimensional module. As I_{P} is in the O_{p} of the point stabilizer we have the assertion.

So it remains:

(1) H_{P} ≅ L_{n}(q), I_{P} is the natural module or the exterior square

(2) H_{P} ≅ O^{+}(10, q), I_{P} is the half spin module

(3) H_{P} ≅ Sp(2n, q), D_{P} ∩ I_{P} ≠ 1, I_{P} /D_{P} ∩ I_{P} is the natural module

(4) H_{P} ≅ 3A_{6} or A_{7}

(5) H_{P} ≅ G_{2}(q), I_{P} is the natural module

(6) H_{P} ≅ Σ_{n}, D_{P} ∩ I_{P} ≠ 1.

In case (4), (2) and (1) with the exterior square there is a unique offender and so I_{P} is an offender on I_{L}, again the assertion.

Suppose H_{P} ≇ Σ_{n}. Then all elements in I_{P} are uniqueness elements. Let first |I_{P} Q_{L}/Q_{L}| = 2. As H_{L} is non solvable, there is some g ∈ H_{L} with I_{P}^{g} ≤ S and I_{P} Q_{L} ≠ I_{P}^{g} Q_{L}. By (1.3)(iii)⟨I_{P} , I_{P}^{g} ⟩ acts quadratically on Y_{L}. If H_{L} is sporadic or a Lie type group in odd characteristic we get with [GBSM] that H_{L} = ⟨C I_{H_{L}}(u)|u ∈ [I_{P} , I_{L}]^{♯}⟩, a contradiction. Hence H_{L} is alternating, the assertion, or H_{L} is of Lie type in characteristic 2 or 3A_{6}, and again we have the assertion.

If I_{P} Q_{L}/Q_{L} is odd and H_{P} is not of Lie type in characteristic p, we get again H_{L} = ⟨C I_{H_{L}}(u)|u ∈ [I_{P} , I_{L}]^{♯}⟩, a contradiction as before.

So H_{P} is of Lie type in characteristic p. Let I_{P} be V (λ) for some weight λ. As L ∩ C̃ is a maximal parabolic we see that I_{P} contains a root subgroup. If |[I_{P} , I_{L}]| > q we get L = ⟨C I_{L}(u)|u ∈ [I_{P} , I_{L}]^{♯}⟩ or H_{L} ≅ L_{n}(q) and I_{L} is the natural module, the assertion.

So we have |[I_{P} , I_{L}]| = q and then I_{P} is an F -module offender on I_{L}, the assertion.

Let finally H_{P} ≅ Σ_{n} and D_{P} ∩ I_{P} ≠ 1. We further have that I_{L} is an over offender and so n = 2m, |I_{L} : I_{L} ∩ Q_{P} | = 2^{m} = |Q_{L} : Q_{L} ∩ Q_{P} |.

Suppose that I_{L} is not an F -module with offender in the O_{p} of the point stabilizer. Then J(Q_{B}) = J(Q_{L}). Hence B = N_{P} (Y_{L}). Then Q_{L}Q_{P} = I_{L}Q_{P} and so [Y_{L}, H_{L}] = I_{L}. We have that [H_{L}, I_{L}]/C I_{[H_{L},I_{L}]}(H_{L}) is irreducible. So C I(H_{L}) is trivial on [H_{L}, I_{L}]. Assume 2n ≥ 10. Then O^{2}(B ∩ H_{P}) ≤ H_{L}. So D_{L} ∩ I_{P} ≤ D_{P}. This shows D_{L} = 1. In particular Q_{L} = Y_{L}. The point stabilizer of H_{L} contains 2^{n−1}Σ_{n}. As I_{P} is in the O_{2} of the point stabilizer we see that H_{L} is not a symmetric group.

We have |I_{P} /I_{P} ∩Q_{L}| ≥ 16. Hence H_{L} ≅ 3M_{22}, 3U_{4}(3) or a group of Lie type in characteristic two. As I_{P} is in the O_{2} of the point stabilizer we get that H_{L} is a group of Lie type in characteristic two. We have that B is a maximal parabolic subgroup in H_{L}. The structure of H_{L} and the fact that O^{2}(H_{P} ∩B) is subnormal in B, implies n = 5 and H_{L} ≅ O^{−}(6, 2), n = 8, H_{L} ≅ O^{+}(8, 2) or n = 8 and H_{L} ≅ L_{6}(2). As there is no 2^{7}Σ_{8} in L_{6}(2) : 2 or O^{+}(8, 2), we see H_{L} ≅ O^{−}(6, 2), H_{P} ≅ Σ_{10}, |I_{P} : I_{P} ∩ Q_{L}| = 2^{5} and |I_{P} | = 2^{9}. But I_{P} acts quadratically on I_{L},which is impossible in the O_{2} of the point stabilizer.

So we have that I_{L} is an F -module and then [GBSM] implies the assertion or n = 4. In the case n = 4, we have |I_{P} | = 2^{7} and H_{P} induces Σ_{8}. Then we have a quadratic group of order 8 in the O_{2} of the point stabilizer and so H_{L} is a group of Lie type, 3M_{22} or 3U_{4}(3). Further there is some involution i ∈ I_{P} with |[I_{L}, i]| = 16. This shows H_{L} is a group of Lie type. Now we see that I_{L} acts strongly quadratic. Then [GBSM] gives a contradiction.

(1.5) Lemma. We have N = H_{P}.

Proof. Let N ≠ H_{P}. Then all elements in I_{P} are uniqueness elements. Further [I_{L}, H_{P}^{g}] ≠ 1 for all g ∈ L ∩ C̃. If H_{L} ≅ L_{n}(q) or Sp(2n, q) and I_{L} is the natural module, then I_{L} induces GF (q)-transvections on Y_{P}, so H_{P} ◁ N. If H_{L} ≅ G_{2}(q) or 3A_{6}, there is some x ∈ I_{L} with [x, Y_{P}] = [I_{P} , I_{L}] and again H_{P} ◁ N. Let H_{L} ≅ A_{m} and I_{L} be the natural module. As all elements in I_{P} are uniqueness elements, we get |I_{P} : I_{P} ∩ Q_{L}| = 2. Hence I_{L} induces transvections. This also applies for the solvable case H_{L} ≅ D_{2r}. The assertion now follows with (1.2) and (1.4).

(1.6) Lemma. If H_{L} ≅ 3A_{6}, then G ≅ M_{24}

Proof. We have |I_{P} Q_{L}/Q_{L}| = 4 and |I_{L}Q_{P} /Q_{P} | = 16. Further I_{L} is generated by transvections over GF (2). As I_{L} is a dual offender and an over offender on I_{P}, we see with [GBSM] that H_{P} ≅ L_{n}(2). Further we have that Q_{P} Q_{L} = I_{P} Q_{L} and then [I_{L}, Q_{P}] = [I_{P} , I_{L}].

We have that N(O^{p}(C I_{B}(I_{P} I_{L}))) ≥ ⟨H_{P} , H_{L}⟩ and so C I_{B}(I_{P} I_{L}) is a p-group. Now C I_{B}(I_{P}) ≤ C I_{B}(I_{P} Q_{L}/Q_{L}) ∩ C I_{B}([I_{P} , I_{L}]). This gives O^{p}(C I_{B}(I_{P})) = 1 and then C I_{B}(I_{P}) = Q_{P} and P/Q_{p} ≅ L_{n}(2).

Let g, h ∈ H_{L} will ⟨I_{P} , I_{P}^{h}, I_{P}^{g} ⟩/O_{2}(⟨I_{P} , I_{P}^{h}, I_{P}^{g} ⟩) ≅ 3A_{6}. Set

V = ⟨I_{P} ∩ Q_{L}, I_{P}^{g} ∩ Q_{L}, I_{P}^{h} ∩ Q_{L}⟩.

By (1.3) V ◁ L, V is abelian and [V, I_{P}] = I_{P} ∩ Q_{L}. Let g̃ ∈ H_{P} with

R̃ = ⟨I_{P} , Ig̃ ; P ⟩Q_{L}/Q_{L} ≅ Σ_{4}. Then by (1.5) [Ig̃ ; P ∩ Q_{L}, I_{P} ∩ Q_{R}] = 1 and

[I_{P} , Ig̃ ; P ∩ Q_{L}] = I_{P} ∩ Q_{L}.

So Ig̃ ; P ∩ Q_{L} induces a full transvection group in C I([I_{P} , I_{L}]). As Ig̃ ; P ∩ Q_{L} ≤ V, we get

V Q_{P} = C I_{P} (I_{P} ∩ Q_{L}, I_{P} /I_{P} ∩ Q_{L}).

So B = N_{P} (V Q_{P}) is a parabolic subgroup in P.

Now I_{P} ∩ Q_{L} is an irreducible B-module. As [I_{P} , I_{L}] ≤ I_{P} ∩ Q_{L}, we get [I_{P} , I_{L}] = I_{P} ∩ Q_{L}. This shows |I_{P} | = 16 and V = I_{L}. We have Y_{L} = (Y_{L} ∩ D_{L}) × I_{L}. As there are uniqueness elements, we get Y_{L} ∩ D_{L} = 1. So Y_{L} = I_{L}. As I_{L} ≰ ϕ(Q_{L}), we have Q_{L} = I_{L} and so I_{P} = Q_{P}. Now with [MF] we get G ≅ M_{24}. (Recall that there is U ≤ S,U elementary abelian of order 2^{6}, U ≠ I_{L}. It is easy to see that N_{P} (U)/O_{2}(N_{P} (U)) ≅ L_{3}(2) and so ⟨N_{P} (U), N_{L}(U)⟩/U ≅ Σ_{3} × L_{3}(2)).

(1.7) Lemma. Let H_{L} ≅ D_{2r}, r prime. Then r = 3. Further H_{P} ≅ Σ_{m}, Sp(2n, 2), O^{±}(2m, 2) or L_{n}(2). If H_{P} ≅ L_{n}(2), then E(G) ≅ L_{m}(2) or M_{24}.

Proof. I_{L} induces transvections on I_{P}. Let N_{P} ≅ Σ_{m}, O^{±}(2m, 2) or Sp(2n, 2). Then |I_{L}| = 4 and so H_{L} ≅ Σ_{3}.

Let now H_{P} ≅ L_{n}(2). Let H_{L} = ⟨I_{P} , I_{P}^{g} ⟩. Set V = ⟨I_{P} ∩ Q_{L}, I_{P}^{g} ∩ Q_{L}⟩. Then we may apply (1.3). Let X ⊴NQ_{L} minimal with respect to [X, O^{2}(N)] ≰ I_{P}. As Q_{P} Q_{L}/Q_{L} is cyclic, we get that also XQ_{L}/Q_{L} is cyclic. Let X/X_{0} ≅ X̄, X_{0} a maximal Q_{L}N-module. Then we receive |X̄/C IX̄(V)| ≤ 2. Hence we can calculate

|[X̄, V]| ≤ |[X, V]I_{P} /I_{P} | < |I_{L}/I_{P} ∩ V | = |I_{P} ∩ V | = |[I_{P} , V]| ≤ |V Q_{P} /Q_{P} |.

This shows [X̄, V] = 1. But then [X̄, N] = 1, a contradiction. So we have [Q_{P} , O^{p}(N)] ≤ I_{P}. This shows ϕ(QP) ≤ D_{P} ≤ Q_{L}, as

[D_{P} , V] ≤ V ∩ Q_{P} = I_{P} ∩ Q_{L}.

But I_{P} ∩ D_{P} = 1 and so [D_{P} , V] = 1 and so D_{P} ≤ Q_{L}. As Q_{P} = D_{P} I_{P} we see Q_{P} ≤ I_{P} Q_{L}. Also Q_{P} = I_{P} (Q_{P} ∩ Q_{L}) and so Q_{P} = D_{P} × I_{P}. Now ϕ(Q_{P}) = ϕ(Q_{P} ∩ Q_{L}) = ϕ((Q_{P} ∩ Q_{L})V) ⊴ ⟨P, H_{P} ⟩. So we have ϕ(Q_{P}) = 1 and then Q_{P} = Y_{P}. As before {V D_{P}, Y_{P} } are the unique elementary abelian subgroups of order |Y_{P} | is V Q_{P}. So P ∩ L = N_{P} (V Q_{P}) is the hyperplane stabilizer in P and Q_{B} = V Q_{P}. We have O^{2}(B) ⊴ L. This shows D_{P} = C I_{Q_{B}}(O^{2}(B)) = C I_{Q_{L}}(O^{2}(B)) ⊴ L. Hence D_{P} = 1. Now C̃ = P and |V | = 2^{2(n−1)}, V = Q_{L}.

Let first n > 3. Then we have exactly two irreducible L_{n−1}(2)-modules in V. This gives

L/V ≅ Σ_{3} × L_{n−1}(2).

If n > 4 we find a second elementary abelian subgroup U which is normalized by Σ_{3} × L_{n−2}(2) in L and by L_{n−2}(2) in P. This shows that we have an (L_{n}(2), L_{n}(2))-amalgan and so G ≅ L_{n+1}(2).

Let now n = 4. Again we have U, |U| = 16 and N_{P} (U)/O_{2}(N_{P} (U)) ≅ L_{3}(2). If N_{L}(U) ≰ P, we may argue as above. Let N_{L}(U) = N_{L∩P} (U). Consider a second subgroup W in L, W elementary abelian of order 2^{6} with N_{L}(W)/O_{2}(N_{L}(W)) ≅ Σ_{3} × Σ_{3}. In P we see N_{P} (W) ≰ L. Now we get N_{G}(W)/W ≅ L_{3}(2) × Σ_{3} or 3Σ_{6}. In the first case we see N_{G}(U) ≰ P. So we have N_{G}(W)/W ≅ 3Σ_{6}. By (1.4) we get G ≅ M_{24}.

Let finally n = 3. Again C̃ = P = 2^{3}L_{3}(2), |Q_{L}| = 16. Let r ≠ 3. Then L/V ≅ (Z_{3} ×D_{10}) : 2. This gives B/V ≅ Z_{3} ·Z_{4}. Then there is no involution in S \ Q_{P} Q_{L}. But this contradicts the structure of P.

So we are left with n = 3 and L/V ≅ Σ_{3} × Σ_{3}. Then any elementary abelian subgroup of order 8 in Q_{B} is normalized by some Σ_{3}. Further there is some different from Q_{P} which is centralized by Σ_{3} in L/V. So we get the L_{4}(2)- amalgam, the assertion.

(1.8) Lemma. H_{L} ≇ G_{2}(q).

Proof. Consider [I_{P} , I_{L}]. In this group we find some B-invariant subgroup W of order q^{2}, with ⟨C I_{H_{L}}(x)|x ∈ W^{♯}⟩ = H_{L}. Thus W does not contain uniqueness elements outside of C I_{I_{L}}(T). Now [GBSM] yields H_{P} ≅ Σ_{n} and so q = 2. All elements in I_{P} now have a 2-dimensional or 3-dimensional commutator. So I_{L} corresponds in H_{P} to the group ⟨(12)(34), (34)(56), (56)(78)⟩. In I_{L} we see exactly 3 elements with a 2-dimensional commutator and 4 with a 3-dimensional one. But in the group above there are exactly 6 elements with a 2-dimensional commutator on the permutation module, a contradiction.

(1.9) Lemma. If H_{L} ≅ L_{n}(q), n > 2, then E(G) ≅ L_{m}(q) for some m.

Proof. We have that Y_{P} is normal in the point stabilizer. Hence

|I_{P} : I_{P} ∩ Q_{L}| = q^{n−1} and Q_{B} = Q_{P} Q_{L} = Y_{P} Q_{L} = I_{P} Q_{L}.

So [Q_{B}, Y_{L}] = [Q_{P} , Y_{L}] = [Y_{P} , I_{L}] = [Y_{P} , I_{L}].

This shows I_{P} = [Q_{P} , H_{P}]. We have that I_{L} is a strong dual offender on I_{P}. So by [GBSM] we get H_{P} ≅ L_{m}(q_{0}), Sp(2m, q_{0}), 3A_{6} or A_{7} with |I_{L} : I_{L} ∩ Q_{P} | ≥ 4.

Assume first H_{P} ≅ 3A_{6}. Then we have q = 2 and n = 3. Now |[I_{P} , I_{L}]| = 16. But this contradicts the action of L_{3}(2) on I_{L}.

Now we have that all elements in I_{P} are uniqueness elements or H_{P} ≅ L_{3}(2) and |I_{P} | = 16.

We first consider the last case. Then |I_{L} : I_{L} ∩ Q_{P} | = 4. Hence again n = 3 and q = 2. Now [Q_{L}, I_{P}] = [I_{L}, I_{P}]. Because of |[I_{L}, I_{P}]| = 4 we have C I_{I_{L}}(H_{L}) ≠ 1. As [C I_{I_{P}} (H_{P}), T] ≠ 1, we get [Z(T), H_{L}] = 1. As Z(S) ∩ Z(T) ≠ 1, this is a contradiction.

So all elements in I_{P} are uniqueness elements. As usual we define V and so [I_{P} , V] = I_{P} ∩ Q_{L} ◁ B.

Assume that I_{P} is self dual. Then

[I_{P} , V]^{⊥} = C I_{I_{P}} (V) = I_{P} ∩ Q_{R}

by (1.3)(ii). Hence B normalizes I_{P} ∩Q_{R} and so (I_{P} ∩Q_{R})Q_{L}/Q_{L} ≨ Q_{B}/Q_{L}. But then I_{P} ∩ Q_{R} = I_{P} ∩ Q_{L}, a contradiction to n > 2.

Hence I_{P} is not self dual. If H_{P} ≅ Sp(2m, q_{o}), then D_{P} ∩ I_{P} ≠ 1.

If H_{P} ≅ A_{7}, then q = 4 and so n = 2, a contradiction.

It remains H_{P} ≅ SL_{m}(q_{0}) or Sp(2m, q_{0}). In both cases Y_{P} Q_{L} = Q_{P} Q_{L} and I_{P} = [Q_{P} , H_{P}]. Let now g_{1}, ⋯, g_{m} ∈ H_{P}, ⟨I_{P}^{gi}|i = 1, ⋯, m⟩ = H_{P} and ⟨I_{P}^{gi}, I_{P}^{gj}⟩ ≅ SL_{2}(q) for g_{i} ≠ g_{j}. Set W = ⟨I_{P}^{gi} ∩ Q_{L}|i = 1, ⋯, m⟩. Then W ◁ L. As D_{P} ≤ Q_{L} and Q_{P} = D_{P} Y_{P} we see Q_{L} = D_{L}W = D_{P} W. So (Q_{P} ∩Q_{L})W ◁L. Hence ϕ(Q_{P} ∩Q_{L})⊴⟨P, L⟩, which shows ϕ(Q_{P} ∩Q_{L}) = 1. In particular Q_{L} is elementary abelian and then also Q_{P} is elementary abelian and Q_{P} = Y_{P}.

In Q_{P} W we have that Q_{P} and D_{P} W are the unique maximal elementary abelian subgroups of order |Q_{P} |. Hence N_{P} (Q_{P} W) ≤ L. So L ∩ P is a maximal parabolic subgroup in P. This shows S ∈ Syl_{p}(L). It follows D_{L} = 1. Hence W = Q_{L}. As O^{p}(H_{L} ∩ B) ≤ H_{P} we see q_{0} = q.

Then I_{L} induces transvections. If H_{P} ≅ Sp(2m, q) we get |I_{L} : I_{L} ∩ Q_{P} | = q. Hence q^{n−1} = q and so n = 2, a contradiction.

So we have H_{P} ≅ SL_{m}(q). It follows O^{p}(H_{P} ∩ B/Q_{B}) ≅ SL_{n−1}(q) × SL_{r}(q) and F^{∗}(L/Q_{L}) ≅ SL_{n}(q) × SL_{r}(q). By order we see |Q_{P} | = q^{m}. Now P = C̃ and there is some further elementary abelian subgroup U with N_{P} (U)/O_{p}(N_{P} (U)) contains SL_{n−1}(q). The structure of L tells us that U ◁ H_{L}. Hence N_{G}(U) > N_{P} (U). Then N_{G}(U)/U ≅ SL_{m}(q) and we have some A_{m}-Dynkin diagram for G. This shows G ≅ L_{m+1}(q).

(1.10) Lemma. If H_{L} ≅ Σ_{m} or Sp(2m, q), then all elements in I_{P} are uniqueness elements.

Proof. Suppose false. We have that Y_{P} /Y_{P} ∩ Q_{L} is contained in a transvection group.

Let first H_{L} ≅ Sp(2n, q), q > 2. Then there is some F -offender A, |A| = q, with C I_{Y_{P}} (A) = C I_{Y_{P}} (a) for all a ∈ A^{♯}. Let I_{P} be a self dual module. Then |[A, I_{P}]| = |A| > 2. As [A, I_{P}] in totally isotrop, we have a contradiction.

Hence we have H_{P} ≅ 3A_{6} and |I_{P} | = 2^{6}. Then |[I_{P} , I_{L}]| = 16 and so C I_{I_{L}}(H_{L}) ≠ 1. Hence B does not act irreducibly on [I_{L}, I_{P}]. But we know that B ∩ H_{P} acts irreducibly, a contradiction.

So we are left with H_{L} ≅ Σ_{n} or Sp(2n, 2). If H_{L} ≅ Sp(2n, 2), then we have |Y_{P} : Y_{P} ∩ Q_{L}| = 2. Hence H_{P} ≅ O^{±}(2m, 2) or Σ_{m} on the natural module.

If H_{L} ≅ Σ_{n}. Then I_{L} in generated by elements i with |[I_{P} , i]| ≤ 4. Again we get H_{P} ≅ O^{±}(2m, 2) or Σ_{m}, or H_{P} ≅ L_{3}(2) on 2^{4} or 3A_{6} on 2^{6}. As I_{P} is generated by transvections on I_{L} the last two cases are impossible.

Now assume first H_{P} = O^{±}(2m, 2) and I_{P} is the natural module. Let H_{L} ≅ Σ_{n} and |I_{P} : I_{P} ∩ Q_{L}| > 2. Then there is some x ∈ I_{L} with |[I_{P} , x]| = 4 and [I_{P} , x] contains a singular element y. So C I_{H_{L}}(y) ≤ P. We have C I_{H_{L}}(y) ≅ Z_{2} × Σ_{n−2} or Σ_{4} × Σ_{n−4}. As |Y_{P} /Y_{P} ∩ Q_{L}| ≥ 4, we get C I_{H_{L}}(y) ≅ Z_{2} ×Σ_{4} or Σ_{4} ×Σ_{4}. But Y_{P} Q_{L}/Q_{L} ≤ O_{2}(C I_{H_{L}}(y)) is a transvection group, a contradiction. So we have |Y_{P} /Y_{P} ∩ Q_{L}| = 2. Choose g ∈ H_{L} with R = ⟨I_{P} , I_{P}^{g} ⟩, R/Q_{R} ≅ Σ_{3}. Then R ≰ C̃. In particular I_{P} ∩ I_{P}^{g} contains no singular elements. This shows |I_{P} ∩ I_{P}^{g} | ≤ 4. As there is no fours group consisting of transvections to a hyperplane in H_{L}, we get I_{P} ∩ I_{P}^{g} ≠ 1.

Let |I_{P} ∩I_{P}^{g} | = 4. Then I_{P} = (I_{P} ∩I_{P}^{g})⊥(I_{P} ∩I_{P}^{g})^{⊥}. So [I_{P} ∩Q_{L}, I_{P}^{g} ∩Q_{L}] = 1 and I_{P}^{g} ∩ Q_{L} induces transvections on (I_{P} ∩ I_{P}^{g})^{⊥}. Now |I_{P}^{g} ∩ Q_{L}/I_{P} ∩ I_{P}^{g} | ≤ 2 and then |I_{P} | ≤ 16.

Let |I_{P} ∩ I_{P}^{g} | = 2. Then again |I_{p}^{g} ∩ Q_{L} : C I_{I_{P}^{g} ∩Q_{L}}(I_{P} ∩ Q_{L})| = 2. As [I_{P}^{g} ∩ Q_{L}, I_{P}] ≅ I_{P} ∩ Q_{L}/(I_{P} ∩ I_{P}^{g}) we get |I_{P}^{g} ∩ Q_{L} : I_{P} ∩ I_{P}^{g} | = 2, a contradiction. So we have |I_{P} | = 16 and H_{P} ≅ Ω^{−}(4, 2) ≅ A_{5}.

Let now H_{P} ≅ A_{m}. Then I_{P} is a central extension of the natural module. Let x ∈ Y_{L} ∩ I_{P} with [T, x] = 1. Let further [I_{P} , C I_{H_{P}} (x)] ≰ Q_{L}. If C I_{G}(x) ≰ C̃, then T ∈ Syl_{2}(C I_{G}(x)). Then [I_{P} , C I_{H_{P}} (x)] ≤ O_{p}(C I_{G}(x)), and so I_{P} ≤ O_{p}(C I_{H_{L}}(x)). This now implies |I_{P} : I_{P} ∩ Q_{L}| = 2 and then x ∈ [I_{P} , I_{L}](D_{L} ∩ Y_{L}) \ (D_{L} ∩ Y_{L}). The same also applies for C I_{G}(x) ≤ C̃ as I_{P} ≤ O_{2}(C̃ ∩ L). If |[I_{P} , I_{L}]| ≠ 2 there is always such an x. So we have |[I_{P} , I_{L}]| = 2. If x ∉ [I_{P} , I_{L}] we can argue as above. Hence C I_{I_{P}} (T) ≤ [I_{P} , I_{L}](D_{P} ∩ I_{P}). This shows |D_{P} ∩ I_{P} | = 2 and |[I_{P} , I_{L}](D_{P} ∩ I_{P})| = 4, or D_{P} ∩ I_{P} = 1 and |[I_{P} , I_{L}]| = 2.

Let |D_{P} ∩ I_{P} | = 2. As there is no fours group of uniqueness elements in Z(T), we get m = 2^{r}. Then there is x ∈ D_{P} [I_{P} , I_{L}] \ (D_{P} ∪ [I_{P} , I_{L}]) with x ∈ D_{L} ∩ Y_{L}, a contradiction as before.

So we have D_{P} ∩ I_{P} = 1 and I_{P} acts irreducibly. As C I_{I_{P}} (T) = [I_{P} , I_{L}], we get m = 2^{r} + 2. We have ϕ(Q_{P}) ≤ Q_{L}, so [ϕ(Q_{P}), I_{L}] = 1. This gives as D_{P} ∩ Y_{P} = 1 that ϕ(Q_{P}) = 1. Hence Y_{P} = Q_{P}. Further |Q_{P} : I_{P} | ≤ 2. Now Z(T) = Z(S) is of order two. This gives |Y_{L} : I_{L}| ≤ 2. As I_{L} ∩ ϕ(Q_{L}) = 1, we get ϕ(Q_{L}) = 1 and so Q_{L} = Y_{L}.

Now P = C̃, E = 1, and then T = S. Then I_{P} ∩ I_{L} is a hyperplane in both I_{P} and I_{L}. Any element conjugated to [I_{P} , I_{L}] is a uniqueness element. But then I_{P} ∩ I_{L} contains too many of them, so that L ≤ C̃, a contradiction.

(1.11) Lemma. Let H_{L} ≅ Σ_{m}, so H_{L} ≅ Σ_{3}.

Proof. By (1.10) all elements in I_{P} are uniqueness elements. We have B ∩ H_{L}/Q_{L} ≅ Z_{2} × Σ_{m−2}, as I_{P} is normal in the point stabilizer. Further we now see m ≡ 2( mod 4) for m > 3.

We have [Q_{P} , I_{L}] = [I_{P} , I_{L}], and so Q_{P} = D_{P} I_{P}. Consider

V = ⟨I_{P}^{g} ∩ Q_{L}|g ∈ H_{L}, such that ⟨I_{P}^{g} , Ig̃ ; P ⟩ ≅ Σ_{3} pairwise ⟩.

Then V ⊴ L and

[I_{P} , V] = I_{P} ∩ V = I_{P} ∩ Q_{L}.

Hence V is the full transvection group to the hyperplane I_{P} ∩Q_{L}. This shows H_{P} ≅ L_{m}(2).

Now we have Q_{L} = D_{L}V, Q_{P} = D_{P} I_{P}. Further D_{P} ≤ Q_{L} and then Q_{P} = (Q_{P} ∩ Q_{L})I_{P}. This shows ϕ(Q_{L} ∩ Q_{P}) ≤ D_{L} ∩ D_{P} and so ϕ(Q_{L} ∩ Q_{P}) = 1. Hence Q_{P} is elementary abelian. Then also Q_{L} is elementary abelian and N_{P} (Q_{L}) = L ∩ P. Thus H_{P} ∩ L/O_{2}(H_{P} ∩ L) ≅ L_{n−1}(2). If O^{2}(H_{P} ∩ L) ◁ L, then H_{L} acts on [Q_{L}, O^{2}(H ∩ L)], which implies H_{L} ≅ Σ_{3}, a contradiction. So we have O^{2}(H_{P} ∩ L) ◁̸L. Then O^{2}(H_{P} ∩ L) ≅ Z_{3}. In particular n = 3. As O^{2}(H_{P} ∩ L) ◁ ◁B, we see H_{L} ≅ Σ_{6}. But O^{2}(H_{L} ∩ P) has two nontrivial modules in V, a contradiction.

(1.12) Lemma. If H_{L} ≅ Σ_{3}, then E(G) ≅ L_{n}(2) or M_{24}

Proof. By (1.11) all elements in I_{P} are uniqueness elements. By (1.7) we have H_{P} ≅ Σ_{m}, Sp(2m, 2), O^{±}(2m, 2) or L_{m}(2). Let H_{P} ≇ L_{m}(2). Then |I_{L}| = 4. Further Q_{P} = D_{P} Y_{P} with D_{P} ≤ H_{L}. Let R = ⟨I_{P} , I_{P}^{g} ⟩ with g ∈ H_{L}. By (1.3) we now get V = (I_{P} ∩ Q_{L}) × (I_{P}^{g} ∩ Q_{L}) and Q_{L} = D_{L} × V.

Hence I_{P}^{g} ∩ Q_{L} induces transvections on I_{P}. This shows |I_{P}^{g} ∩ Q_{L}| = 2. But then |I_{P} | = 4, a contradiction.

So we have H_{P} ≅ L_{m}(2). Now the assertion follows with (1.7).

(1.13) Lemma. If H_{L} ≅ Sp(2n, q) ≇ Σ_{3}, then E(G) = M_{22}, M_{23} or L_{r}(q).

Proof. By (1.11) all elements in I_{P} are uniqueness elements. Further I_{P} Q_{L}/Q_{L} is contained in a transvection group. Set again R = ⟨I_{P} , I_{P}^{g} ⟩ with R ⋬ C̃. As I_{P} is in the O_{p} of the point stabilizer and acts quadratically, we get |I_{P} Q_{L}/C_{L}| = q (recall that L ∩ C̃ acts on I_{P} Q_{L}/Q_{L}). So we have R/Q_{R} ≅ SL_{2}(q) and I_{P} ∩ Q_{R} = I_{P} ∩ Q_{L}. By (1.3) we now get V = (I_{P} ∩ Q_{L}) × (I_{P}^{g} ∩ Q_{L}). Further V is a direct sum of natural modules for R/Q_{R}. On I_{P} we have that V induces an offender and a dual offender. This shows with [GBSM] that H_{P} ≅ Sp(2n, q_{0}) with q_{0}^{x} = q and |I_{P} | = q^{2}, or H_{P} ≅ A_{7} and |I_{P} | = 16, or H_{P} ≅ SL_{n}(q_{0}).

Let first H_{P} ≇ SL_{n}(q_{0}). We first show m = 1. Let m > 1. Then [Y_{L} ∩ Q_{P} , Q_{P}] = [Y_{L}, Y_{P}] ⋬ N, as q > 2. So (Y_{L} ∩ Q_{P})Y_{P} ⋬ N. Set

W = ⟨(Y_{L} ∩ Q_{P})^{N}⟩.

Let W ≤ Y_{P} Q_{L}. Then [W, Y_{L}] ≤ [Y_{P} , Y_{L}] ≤ Y_{P}. Hence [W, O^{p}(N)] ≤ Y_{P}, a contradiction. So W ≰ Y_{P} Q_{L}. Choose g ∈ N with (Y_{L} ∩ Q_{P})^{g} ≰ Y_{P} ∩ Q_{L}. Then [(Y_{L} ∩ Q_{P})^{g}, Y_{L} ∩ Q_{P}] = [Y_{P} , Y_{L}] = [Y_{P} , Y_{L}^{g}] ⊴ ⟨P ∩ L, g⟩ = P, a contradiction.

So we have m = 1, i.e. H_{L} ≅ SL_{2}(q). Let D_{H} = C I_{L}(H_{P}). We have ϕ(Q_{L}) ≤ D_{H}, as [Q_{L}, I_{P}] ≤ V. As O_{P} Q_{L} = Y_{P} Q_{L} there is just one nontrivial module in Q_{P}. Hence ϕ(Q_{P}) ≤ D_{P}. As D_{P} ∩ D_{H} ◁ ⟨P, H⟩, we have D_{P} ∩D_{H} = 1. Thus Q_{P} ∩Q_{H} is elementary abelian and so Q_{P} = Y_{P} (Q_{P} ∩Q_{H}) is elementary abelian too, which implies Q_{P} = Y_{P}.

We have |Y_{P} : C I_{Y_{P}} (Y_{L})| = q = [Y_{L}, Y_{P}]. This Q_{P} = D_{P} × [Y_{P} , N]. This shows that [Q_{L}, H_{L}] = [Y_{L}, H_{L}] is the natural module. As C I_{Q_{L}} ≤ D_{L}Y_{L} we get D_{P} ≤ C I_{Q_{L}}(Y_{P}) ≤ [Y_{P} , Y_{L}]D_{L} and then [D_{P} , H_{L} ∩ P] ≤ [I_{P} , Y_{L}] ∩ D_{P} = 1. Thus D_{P} ≤ D_{L} and so D_{P} = 1. This gives Q_{P} = I_{P} and then P = C̃. Now D = 1 = E, so L and P have a common Sylow p-subgroup. In particular Q_{L} = Y_{L}. Further |Q_{L}| ≤ q^{3} and S/Q_{P} is an extension of an elementary abelian group by a cyclic group. This shows H_{P} ≅ A_{7} or A_{6} and q = 4. Further H_{L} ≅ A_{5}. Now Q_{B} = Q_{L}Q_{P} and so B = N_{P} (Q_{B}) involves Σ_{3}. This shows L/Q_{L} ≅ Σ_{5} or (A_{5} × Z_{3}) : 2 in case of H_{P} ≅ A_{7}.

Let first H_{P} ≅ A_{6}. If P ≅ 2^{4}A_{6}, then L ≅ 2^{4}Σ_{5} and so E(G) ≅ M_{22} by [MF]. If P ≅ 2^{4}Σ_{6}, then L ≅ 2^{4+1}Σ_{5}. Then we also find a (2^{4}A_{6}, 2^{4}Σ_{5})-amalgam and again E(G) ≅ M_{22}.

Let |Q_{P} | = 2^{4+1}, then also |Q_{L}| = 2^{4+1}. But S/Q_{L} ≅ D_{8}, which implies |Y_{P} : Y_{P} ∩ Q_{L}| = 4, a contradiction.

Let now H_{P} ≅ A_{7}. Then L ≅ 2^{4}(A_{5} × Z_{3}) : 2. By [MF] then E(G) ≅ M_{23}.

So assume now H_{P} ≅ SL_{n}(q_{0}). If D_{L} ∩I_{L} = 1, we get H_{L} ≅ SL_{2}(q) as above. The same also applies for q > 2. Hence we first assume H_{L} ≅ Sp(2m, 2) and D_{L} ∩ I_{L} ≠ 1. Then V is the full transvection group and so H_{P} ≅ SL_{n}(2), |V | = 2^{2(n−1)}. Further D_{L}∩I_{L} ≤ [I_{L}∩Q_{P} , Q_{P}]. This shows D_{L}∩I_{L} ≤ ϕ(Q_{P}). But ϕ(Q_{P}) ≤ D_{P}, and so D_{L} ∩ I_{L} = 1.

So we also have m = 1. Again I_{P} Q_{L} = Q_{P} Q_{L} and I_{P} is the unique nontrivial H_{P} -module in Q_{P}. As L = (P ∩ P^{g} ∩ L)H_{P} with suitable g ∈ H_{L}, we see (Q_{P} ∩ Q_{L})V ⊴ L. Thus ϕ(Q_{P} ∩ Q_{L}) ◁ ⟨P, L⟩, which shows ϕ(Q_{P} ∩ Q_{L}) = 1. Hence Q_{P} = Y_{P}.

We have [I_{P} , I_{L}] ∩ C I_{I_{L}}(L) = 1, thus I_{L} is the natural module. As B acts non trivially on I_{P} /I_{P} ∩ Q_{L} and [I_{P} , I_{L}] as well, we see Q_{P} = D_{P} × I_{P} and SL_{n}(q_{0}) = SL_{n}(q). Now Q_{P} ∩ Q_{L} = D_{P} (I_{P} ∩ Q_{L}) = D_{P} (I_{P} ∩ V). This Q_{L} = D_{P} V which shows Q_{L} = Y_{L}. This implies Q_{L} = D_{L} × V.

Let first n = 2. Then |Q_{L}| ≤ q^{3} ≥ |Q_{P} |, as D_{L} ∩ D_{P} = 1. But now all elements in Q_{P} are uniqueness elements, a contradiction. So we have n > 2. Then D_{P} V and Q_{L} are the unique maximal elementary abelian subgroups in Q_{P} V. This shows N_{P} (Q_{P} V) ≤ L. Hence L ∩ P is a maximal parabolic subgroup. This shows D_{P} = C I_{Q_{B}}(O^{p}(H_{P} ∩ B)). Then

[D_{L}, O^{p}(H_{P} ∩ B)] ≤ D_{L} ∩ I_{P} ≤ D_{L} ∩ V = 1.

So D_{L} ≤ D_{P}, D_{L} = 1. Now P is an extension of the natural module by an automorphismgroup of SL(n, q). In particuler P = C̃ and then D = E = 1 and L, P have a common Sylow p-subgroup. We now have F^{∗}(L/Q_{L}) ≅ SL_{2}(q) × SL_{n−2}(q).

Let first n > 4. Then we see in L a second elementary abelian subgroup W with |W | = q^{n} and N_{L}(W) involves SL_{2}(q) × SL_{n−3}(q). In N we see that N(W) involves SL_{n−1}(q). Thus N_{G}(W)/W ≅ SL_{n}(q) and W is the natural module. Now we have an (SL_{n}(q), SL_{n}(q))-amalgam which intersects in the point stabilizer and has an A_{n} Dynkindiagram, so E(G) ≅ SL_{n+1}(q).

Let next n = 4, 3. As q > 2, we can use the Cartan subgroup to find W with |W | = q^{4} as before.

It n = 2 we have H_{P} ≅ SL_{2}(q) and we have an (SL_{2}(q), SL_{2}(q))-amalgam and the assertion follows with [Cher]

(1.14) Proof of the theorem. Let G be a counterexample. By (1.1) H_{L} possesses exactly one component. If H_{L} is solvable, so either by (1.2) we have the situation of (1.1) or H_{L} ≅ SL_{2}(3), Σ_{3} or D_{2r}, r > 3.

By (1.7) we get H_{L} ≅ SL_{2}(3) or SL_{2}(2). So H_{L} ≅ SL_{2}(q). Together with (1.4) we see that in any case (nonsolvable or not)we have H_{L} ≅ SL_{n}(q), Sp(2n, q), G_{2}(q), A_{m} or 3A_{6}. By (1.6) we have H_{L} ≇ 3A_{6}. By (1.8) H_{L} ≇ G_{2}(q). By (1.9) H_{L} ≇ SL_{n}(q) for n > 2. As we may consider SL_{2}(q) as Sp(2, q) we see with (1.13), (1.11) and (1.12) that H_{L} ≇ Sp(2n, q), or Σ_{m}. Hence there is no counterexample.

References

[Cher] A. Chermak, preprint

[E] U. Meierfrankenfeld, B. Stellmacher, G. Stroth, Non E–uniqueness, the trychonomie theorem, preprint

[GBSM] B. Baumeister, U. Meierfrankenfeld, G. Stroth, The great book of small modules, in preparation

[MF] U. Meierfrankenfeld, preprint