1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

11.2. WEAK-CLOSURE PARAMETER VALUES, AND (vNaCVil) 767 H. Let B be a Hall 2'-subgroup of H n M. If A:= CB(V) =f 1, then by 4.4.1 ...

768 il. ELIMINATION OF Ls(2n), Sp4(2n), AND G2(2n) FOR n > 1 the hypotheses of E.6.14, and that lemma supplies a contradictio ...

11.2. WEAK-CLOSURE PARAMETER VALUES, AND (vNa(V1)) 769 Assume Cv(L) = 1. Then X is regular on vt, so by A.1.12, X is normal in a ...

V{ Thus Vi is an £ 2 -conjugate of Vi, and hence we may assume g E Na(Vi). In particular Vf ::::; Vi. Now by 11.1.3, g E Na(L2), ...

11.3. ELIMINATING THE SHADOW OF L4(q) 771 PROOF. Assume that Wo(T, V) :::; 02(H). We will show that 11.3.2 holds, contrary to ou ...

By 11.0.4, Dp ~ L; so as DpT = TDp, Dp is contained in a Cartan subgroup of L. As [Dp, t] =I-1, t induces a field automorphism o ...

11.3. ELIMINATING THE SHADOW OF L 4 (q) 773 As m([B, VH]) = 4n, VH is the 6n-dimensional maximal central extension of the natura ...

By 11.3.7.4, V =Ai. 02 (H). Therefore we have Hypothesis E.2.8 with H, T, HnM in the roles of "H, T, M". Leth EH-Mand set I:= (V ...

11.4. ELIMINATING THE REMAINING SHADOWS 775 LEMMA 11.3.8. There exist F q-structures on P and V, preserved by Y and L, respectiv ...

PROOF. Let HE H*(T, M). If Cv(L) -/= 1, then Lis not SL3(q) by 11.0.3.3, and [Z, H] -/= 1 as Hi_ M = !M(LT), contrary to 11.3.2. ...

11.4. ELIMINATING THE REMAINING SHADOWS 777 Suppose first that K/02(K) is not quasisimple. Then 000 (K) centralizes R2(KT) by 3. ...

LEMMA 11.4.3. (1) Gz = KCaz(K/02(K))Mz. Therefore if K = L1, then CaJK/02(K)) i M. (2) G1 = K(Ca 1 (Vi) n Ca 1 (K/02(K)))M1. PRO ...

il.5. THE FINAL CONTRADICTION 779 (1) or (2) of 11.1.2. In case (1) of 11.1.2, 02,21(K) i. M, so we may choose Ri :::; Hi :S Ri0 ...

so by (2), V2 ::::; Ca(B) ::::; No(V9). Hence D := Cvg (V2) is of corank at most n + 1 in V9, so from the action of Mfr on V9 ei ...

11.5. THE FINAL CONTRADICTION 781 acts on RiX, and hence by a Frattini Argument can be taken to normalize X. Then [X, Bo] ::::; ...

KY/0 2 (KY), so as G 1 = KYM 1 by 11.4.3.2 and M 1 acts on S, we conclude that S::::; 02 (G1), completing the proof of the claim ...

il.5. THE FINAL CONTRADICTION 783 Suppose U is nonabelian. Then as U = (V 3 °^1 ), U does not centralize Vi, so tJ f:. 1and1 f:. ...

also V9 = [V9,L 2 ]. But then V9 = [R 2 ,L 2 ], contradicting V^9 ::; Ri. This final contradiction completes the proof of 11.5. ...

Part 5 Groups over F2 ...

Results in the previous parts have reduced the choices for L, V in the Fun- damental Setup (3.2.1) to the case where L/0 2 ,z(L) ...