1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

i4.4. FINISHING THE TREATMENT OF (vG1) NONABELIAN ioo7 Conclusion (10) of G.11.2 is impossible, as m3(H*) = m 3 (H):::; 2 since ...

1008 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN L:r(G, T) IS EMPTY 26 is Sylow in both Land Ki. Then (Ki, Cr(L)] ~[Ki, Cr(U)] ~ CK ...

14.4. FINISHING THE TREATMENT OF (VG1) NONABELIAN 1009 axis W, or [W, DJ = Vf = V2 with m([U, d]) S 2 for each d ED. This contra ...

1010 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTY As the preimage W1 of W1 satisfies W1 = Cu(V), Wf = Cwg•(V) = W^9 ...

14.4. FINISHING THE TREATMENT OF (VG1) ·NONABELIAN 1011 hyperplane W of U, so Vx 2 S [U1, W9] SE. Thus X 2 E Y, and hence (4) ho ...

1012 14. Ls(2) IN THE FSU, AND L 2 (2) WHEN .Cf(G, T) IS EMPTY in the remaining maximal 2-local D := (TX2,Xg) ~ S5/Ern/Zg of H o ...

14.5. STARTING THE CASE (vG1) ABELIAN FOR Ls (2) AND L2 (2) 1013 14.5. Starting the case (VG^1 ) abelian for L 3 (2) and L 2 (2) ...

10i4 i4. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTY Next we saw V < (V^01 ) = U :::;l YT, so m 2 (U) ~ 4 since m(V ...

14.5. STARTING THE CASE (vG1) ABELIAN FOR L 3 (2) AND L 2 (2) 1015 (VJ/^1 ) is abelian, so we have symmetry between LT, V and H ...

1016 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTY or p1+^2 by A.1.25. As m 2 (Aut(R)) ::::; 2 and X = [X,J1(T)] by ...

14.5. STARTING THE CASE (VG1) ABELIAN FOR L 3 (2) AND L 2 (2) 1017 PROOF. By 14.5.11.2, H* is a quotient of H+, while by 14.5.12 ...

1018 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY Since V plays the role of "V+" in 14.5.14.2 in the notation of ...

14.5. STARTING THE CASE (vGi) ABELIAN FOR Ls(2) AND L2(2) 1019 Set Yz := 02 (02,z(Y)), and Ye := 02 (CyM(L/02(L)). By (!), Yz ej ...

1020 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY Assume case (2) of Hypothesis 14.3.1 holds. Then (3) follows fr ...

14.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIAN 1021 Assume the hypotheses of (6), and let Zi := Oi(Z(Ti)). As R S Ti by hy- pot ...

1022 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN Lf(G, T) IS EMPTY CQ 1 (u) i CQ 1 (V). Since Qi ::::; QH, CQH(u) i CQH(V), and sin ...

14.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIAN 1023 For IE I, set T1 := T n I and Iz :=In G1. The next two observations are str ...

10Z4 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY Hz/Oz(Hz) ~ 83 by the hypothesis of (6). Then since To = QiTu b ...

i4.6. ELIMINATING L 2 (2) WHEN (VG1) IS ABELIAN 1025 so the lemma is established. D LEMMA 14.6.9. Assume IE I*, IT : QHI > 4, ...

1026 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cr(G, T) IS EMPTY Rz = 02(Na 1 (Rz)) by A.1.6-that is Rz E B2(G1), so setting Qi ...