7.7. MINI-APPENDIX: r > 2 FOR Ls(2).2 ON 3 E!l S 707This completes the proof of Lemma 7.7.6 and Proposition 7.7.2.7.7.4. Proof of Proposition 7.7.3. Now that Proposition 7.7.2 is estab-
lished, we work under Hypothesis (HO), and it remains to obtain a contradiction,
establishing Proposition 7.7.3.
We are in a position to exploit Thompson factorization: First, lemma B.2.14tells us that
U := (!11(Z(To))H) E R2(H),
so setting H := H/CH(U), we.have 02 (H) = 1. Further
V = (Cv(To)L0
) ::::; U,
soCH(U)::::; CH(V)::::; MH.
We saw early in the proof of 7.7.6 that J(H) 1. M, so J(H)* # 1.
Next J(H)* is described in Theorem B.5.6. In particular as J(H) 1. M, either
03 (J(H)) 1. Mf.r or some component K of J(H)* is not contained in Mf.r.
Assume the first case holds. ThenX := 03(J(H)) = X~ x · · · x X,i
with Xi ~ Z3 and [U, X] = U1 EB · · · EB Ud where Ui := [U, Xi] is of rank 2.
Further d ::::; 2 so that L^0 = 02 (L^0 ) acts on each Ui. As J(T) ~ L^0 T 0 and L^0
acts on Ui, L^0 acts on Cui (J(T)) ~ Z2, so [Ui, L^0 ] = 1. Then 1 = [U, X, L^0 ], and
[X, L^0 ] = 1 which says [X, L^0 , U] = 1. So by the Three-Subgroup Lemma we have
[L^0 ,U,X] = 1. But recall v_ = [L^0 ,V]::::; [L^0 ,U]. Thus X centralizes V 0 V_ = V,
contradicting X 1. M.Therefore some component K.'.f. of J(H)* is not contained in Mf.r, so taking
KE C(H) with K.'.f. = K* and setting Ko:= (KT^0 ), H = KoL^0 To by minimality of
H. Similarly by a Frattini Argument, H = CH(U)NH(Cr 0 (U)), so that K/02(K)
is quasisimple by 1.2.1.4 and minimality of H.LEMMA 7.7.7. Hypothesis C.2.3 is satisfied with Q in the role of^11 R".
PROOF. Recall C(G, Q) ::::; M, so C(H, Q) ::::; MH < H. By A.4.2.4, Q E
Syb(Co), where Co := CMH(L^0 /02(L^0 )) ~ MH; then Co 2:: (QMH), so Q is also
Sylow in the latter group. Also Q E B2(MH) by C.1.2.4, so that Q E B2(H) by
C.1.2.3. Thus we have verified Hypothesis C.2.3. DLEMMA 7.7.8. Q::::; NH(K).
PROOF. Assume otherwise. Then by C.2.4, Q n KE Syl2(K), and as K 1. M,
K is a xo-block. Further as K* is quasisimple and K < Ko, we conclude from
the list in A.3.8.3 that K* ~ L 2 (2n) with n 2:: 2. Then by C.2.4, Kon M is the
Borel subgroup B normalizing Q n K 0 • Let Y be a Cartan subgroup of B. By
7.7.4, IY: Ycl ::::; 3 and Na(Yc) ::::; M because Ye# 1 since Ko is the product of
two conjugates of K. On the other hand, YT 0 = ToY and To acts on L, so also
YcTo = T 0 Yc. Then as H 1. M, NH(Yc) 1. M by 4.4.13.1. This contradiction.
completes the proof. D
Now that Q ::::; NH(K) by 7.7.8 and K/02(K) is quasisimple, we may apply