2.4. THE CASE WHERE q IS NONEMPTY S41
{R1, Rz} transitively, so that INT(R1)I 2:: 2181 = 21821 > 182 1. This contradiction
eliminates the case Mo < NT(R).
Therefore Mo= NT(R). Then as NT(Mo) ::::; NT(J(Mo)) = NT(R) =Mo, we
conclude Mo= T, and hence ITI = 2181. Recall 81 = \x)Ns(R1) = NM 0 (R1); thus
IB1I = ISi = ITl/2. Then by 2.3.7.1, HE r*, and as we saw KS1 Er, similarly
KS1 Er* with 81 E Sylz(KS1).As Lz E C(KS1, Ns(R1)) and IB1 : Ns(R1)I = 2, Lz ::::; Kz E C(KS1) by
1.2.5. By construction 81 normalizes R 1 , and K centralizes R 1 D 1 ; indeed much
as in the proof of 2.4.23.5, R 1 is the largest subgroup of 81 invariant under L 2
and x, so that Rl = 02(KS1) 2:: 02(K2). As K centralizes R 1 , we conclude that
02(K2) ::::; Z(Kz). Then as mz(K2) 2:: mz(L2) > 1, we conclude from 1.2.1.5 that
Kz is quasisimple, and hence is a component of KS 1. Thus K 2 is described in
2.3.9.7; so as Kz n M contains the Lz(q)-block Lz and Ou 2 (L2) = 1, we conclude
that K = Kz and K/02(K) ~ L3(q). But now mp(KD1) > 2, for pa prime divisor
of q - 1, contradicting K D 1 an SQTK-group.
This contradiction shows that the case s = 2 cannot occur in Theorem 2.4.7.
Hence the proof of Theorem 2.4.7 eliminating L 2 (2n) blocks for n > 1 is at last
complete.2.4.2. The small examples and shadows of extensions of L4(3). In this
subsection, we complete the proof of Theorem 2.4.1. Thus we continue the hy-
potheses and notation from the beginning of this section. By Theorem 2.4.7, the~lock L is of type A 3 or As, and in the latter case Lo n M is a Borel subgroup of
Lo as L 0 S is a minimal parabolic. Therefore Z(L) = 1 by C.1.13.c.
Recall from the beginning of this section 2.4 that Q := 02(H). However in this
new subsection, J(S) is no longer denoted by R, but instead
R := Baum(S).
Recall also from 2.3.8.4 that Li= [Li, J(S)] for each i, so that R normalizes Li by
C.1.16.
LEMMA 2.4.25. If L is an As-block, thens= 1.
PROOF. Assume otherwise, so that s = 2. Recall we defined R = Baum(S)
just above, and set Qi := 02(LiR), I:= CR(L), 81 := Ns(L), and To := NT(S).
By 2.4.5.1, Hypotheses C.5.1 and C.5.2 are satisfied with Sin the role of both "TH"
and "R", for each subgroup Mo of To with S a proper normal subgroup of Mo. As
R denotes Baum(S), U 0 , R play the roles played by "U, S" in section C.5, while I
plays the role of "D1".
By C.5.4.3, Q 2 = U2 x D2 where Dz := CR(L2) and U2 := 02(L2), and
R/Q 2 ~ E 4 is generated by two transpositions in LzR/Q2 ~ Ss. Also from the
proof of C.5.4.3, RQ2 = J(S)Q2 and for A E A(S) with A 1:. Qz, IU2 : Cu 2 (A)I =
IA : (An Q 2 )1. It follows that [A, U1] = 1 so [A, L] ::::; CL(U1) = U1, and hence
A= U 1 x (An I). Thus I/Q 1 is generated by two transpositions in Lzl/Q1 ~ 85,
where Q1 := 02(L2I) = U2 x Do, and Do:= CR(Lo). Thus [U2,I]::::; <P(J)::::; Q1,
and as U2 is the As-module for Lz, it follows that <:!? := Ci!>(l) (81) = <1?2 x Dq,
centralizes O~' (Mn Lz), where <1?2 := Cu 2 (81) ~ Z2 and Dq, := Oi!>(I)nDo (81 ).
Therefore 03 ' (Mn Lo)= 031 (MnL)0^3 ' (MnL 2 ) centralizes<:!?. Observe also that
81 = Ns(<P(I)) = Ns(<P).