11i4 i5. THE CASE .Cf(G, T) = 0LEMMA 15.2.21. (1) If S is not irreducible on K/02(K) then KS = HiH2,
K1 = K2 fort ET-S, and Kc centralizes Zs, so that Kc= 031 (M1).
(2) If K =Kc then CM(V) is a 3^1 -group.
(3) Assume Kc/0 2 (Kc) ~ L 4 (2), conclusion (ii) of 15.2.18.3 holds, and there is
an S-invariant subgroup Yi= 02 (Yi) of N1 with YiS/02(YiS) ~ S3 and Yi i. M1.Then S is irreducible on K / 02 ( K).
PROOF. Assume S is not irreducible on K/0 2 (K). From Notation 15.2.9,
K*T* ~ ot(2), so T* is irreducible on K* and hence S* < T*. But IT: SI= 2by 15.2.17.3, so 02 (KT) ::::; S. As J(T) ::::; S by 15.2.18.2, and IT : J(T)I = 2 by
15.1.12.3, J(T)* = S*. Further J(T)* E Syh(HiH2) by 15.1.12.3, so KS= HiH2,
and hence Kf = K 2 fort ET-S.
Recall Kc = (KMc) = 031 (Mc) is described in 15.2.12.2. If K = Kc, then
(1) holds by 15.2.11.2, so we may assume K < Kc. Then Kc = (LT) for someL E C(Mc) described in 15.2.12.2. Now using A.1.6, 02(KcT) ::::; 02(KT) ::::; S =
Cr(Zs), so as L '/. Ct(G, T) by Hypothesis 14.1.5.1, Kc::::; Ca( Zs) ::::; M1 by 1.2.10
and 15.2.16, completing the proof of (1). ·
Assume in addition the hypotheses of (3); we will obtain a contradiction to
our assumption that Sis not irreducible on K/0 2 (K), and hence establish (3). As
!Yi : 02(Yi)I = 3 and Yi ::::; N1 but Yi i M1, Yi ~ Z3. By hypothesis, case (ii) of15.2.18.3 holds, so N1 = CM(Zi) is of index 6 in M, for some complement Zi to
Zin Zs; in particular, f'i = OCflh) is not T-invariant. Define Y as in 15.2.3, and
set M := M/0 2 (M). If case (iii) of 15.2.3.1 holds, then Y ~ Z 3 is T-invariant, so
YYi/02(YYi) ~ YYi ~ Eg, and hence CM(V) is a 3'-group as m3(M)::::; 2.
Next KS is the maximal parabolic subgroup of KcS determined by the end
nodes of the Dynkin diagram for Kc/0 2 (Kc)· Let Yo := 02 (P), where P is the
minimal parabolic determined by the middle node. If YoT i M, then Y 0 T EH*(T, M), contrary to 15.2.8; hence Y 0 T::::; M, so Yo ::::; CM(V) by 15.2.14.5. Thus
CM(V) is not a 3'-group, so case (ii) of 15.2.3.1 holds by the previous paragraph.
Therefore Y ~ 3i+^2 and Yo= Z(Y). Now S inverts f'i which is not T-invariant, so
as IT: SI = 2, Sis the subgroup of order 2 of T inverting Y, and so S centralizes
Yo, This is impossible, as Kc ::::; M1 by (1), so S E Syh(KcS) by 15.2.18.2, and
then SY0/02(SY0) ~ 83. So (3) is established.
Finally assume K =Kc. Then as MnK = 02 (K) by 15.2.8, and K = 031 (Mc)
by 15.2.12.4, we conclude 031 (Mn Mc) = 1, so (2) holds. D
LEMMA 15.2.22. Assume conclusion (ii) of 15.2.18.3 holds, F*(I) = 02 (I),
and CM(V) is a 31 -group. Assume Sis not irreducible on K/0 2 (K), and there is
Yi = 02 (Yi) ::::; N1 which is S-invariant with YiS/02(YiS) ~ S3. Then [Yi, K2] i.
Yi nK2.
PROOF. Assume [Yi,K2] :S Yin K2. Then fort ET-S, [Y/,Ki] :SY/ n Ki
as K~ =Ki by 15.2.21.1. Next as conclusion (ii) of 15.2.18.3 holds, fh = C.M(Zi)
is of index 6 in M ~ nt (V) for some complement Zi to Z in Zs. Then as Y 1 ::::; N 1
and CM(V) is a 3'-group, Y1 = 03Cih), Y1f't = O(M) ~ Eg, and M has Sylow
3-subgroups isomorphic to E 9 • Define Y as in 15.2.3; as Y ::::! M but Y 1 is not
T-invariant, YYi contains a Sylow 3-subgroup of M, so by a Frattini Argument,
we may take t to act on YYi, and thus YYi = Yi Y/ with [Yi, Y/] ::::; Yi n Yit.
Let X := (Y1,K 1 ); then [X,Xt]::::; XnXt, in view of the commutator relations