i4.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIANThis contradiction establishes the claim that D = 1. Now by (e) and (g):
(h) 02(I) = 02(K) and Zo = Zz :=:;; Zi.
io27Thus I~ (S5/E16) wr Z2. It follows also that Iz = C1(z) ~ (S4/E 16 ) wr Z 2 , and:
(i) Kz = 02 (Iz) ~ Z3/Q~XZ3/Q~, CRJ02(Kz)) = Zo, and CRZ(02(Kz)/Zo) =
02(Kz).
Set Gt := Gi/Zi and Ci := Ca(Zi). As Ci :::l Gi E He, Ci E He by 1.1.3.1,
so that Qi = 02(Ci) = F*(Ci). Then Qt = F*(Ct) by A.1.8. Let X be thepreimage in Gi of F*(Gt); as Y centralizes Zi by (f), 02 (X) :::;; Y :::;; Ci; so as
Qt= F(Ci), 02 (X) = 1 and hence F(Gt) =Qt. Thus using B.2.14:
(j) E+ ;= Di(Z(T+)) nRtRt:::;; Di(Z(Qt)) =: F+.
Next 02(K) = Ui x U2, where Ui := 02(Ki), and Ei := Ui n Ri is a hyperplane
of Ui. Let Eo := EiE2. Then as I~ S5/E16 wr Z2, Z(To/Zo):::;; Eo/Zo and Iz is
irreducible on Eo/Z 0 ; so as Zo :::;; Zi by (h), we conclude from (j) that
(k) Eo:::;; E:::;; F:::;; Qi:::;; 02(KzT1),
where E and Fare the preimages of E+ and p+ in Gi.
Recall r E T - T1, T1 = To, and case (iii) of C.5.6. 7 holds. Hence by C.5.6. 7,
A := 02 (K) and Ar are the two To-invariant members of A(T 0 ), and An Ar =
[A, Ar] is ofrank 4. Thus as Eo is of rank 6, E 0 i An Ar, so as E'[j :::;; Ar, E'[j i A.
Now Fis normal in Gi, so E'[j:::;; F:::;; 02(KzT1) by (k). Then as E 0 i A and KzTis irreducible on 02(KzT1)/A:
(1) 02(Kz) = Eo[Ei), Kz]:::;; F:::;; Qi:::;; Rz.
It follows from (1) that Zi = Di(Z(Q1)):::;; CRz(02(Kz)), so we conclude from (h)
and (i) that:
(m) Zi = Zo.
Then asp+= Di(Z(Qt)), (1) says Qi:::;; CRz(02(Kz)+), while as Zi = Zo by (m),CRz(02(Kz)+) = 02(Kz) by (i). So we conclude from (1) that:
(n) Qi = 02(Kz)·
From (i), o;(:K:) ~ Q~, so by 14.5.15.1 and (n), V:::;; Z(Qi) = Z(O~)) = Z 0 •
Thus V = Z 0 = Zi :::l Gi, contrary to Gi i M = Na(V). This contradictionfinally completes the treatment of the case F*(J) = 02(!).
Thus it remains to treat the case F*(J) #-02(J). As O(J) = 1 by 14.6.7.3,
there is a component K of I. As Iz is the unique maximal overgroup of T1 in the
minimal parabolic I, I and Iz are described in E.2.2, and in particular I= KoT1,
where Ko := (KT^1 ). On the other hand by 14.6.7.1, K is described in 1.1.5.3; in
particular K = [K, z] with z 2-central in I.
·We consider the possibilities from the intersection of the lists of E.2.2 and
1.1.5.3: First suppose K/0 2 (K) is a Bender group. Then by E.2.2, Iz is the normal-izer of a Borel subgroup B of K 0 , arid centralizes no element of (02(B)/02(Ko))#,
. whereas Iz centralizes the projection of z on 02(B)/02(Ko). Similarly if K/02(K) ~
Sp 4 (2n)' or L 3 (2n), then Nr 1 (K) is nontrivial on the Dynkin diagram of K/02(K)
by E.2.2, so again Iz is the normalizer of a Borel subgroup B of Ko, and hence
n = 1 since Iz centralizes the projection of z on 02(B)/02(K 0 ). Thus K/0 2 (K)
is L 2 (p) with p > 7 a Fermat or Mersenne prime, or K/02(K) is £3(2) or A6 with
Nr 1 (K) nontrivial on the Dynkin diagram of K/02(K).