544 2. CLASSIFYING THE GROUPS WITH IM(T)I = iit remains to show that A(S) = A(T), or equivalently to establish the assertion
J(S) = J(T) in (1).
As Tc = Cr(Lo) ::::; Q E A(S), <P(Tc) = 1. Further if Lo is an A5-block,
then Q = Tc x U by 0.5.4.3, and this holds when Lo is a product of A3-blocks
as S 4 = Aut(A 4 ). Also 0.5.6.7 says Tc n T(: = 1 for x E Mo - S; hence (3) will
also follow, once we have established the equality IT: SI = 2 in (1). Thus to prove
(1)-(3), it remains to establish (1).Suppose (1) fails. Since we saw that R = J(S), either IT : SI f:. 2 or J(S) f:.
J(T); thus conclusion (a) of 0.6.1.6 does not hold. By 2.4.26, Lo is not an A3-
block, so conclusion (b) of 0.6.1.6 does not hold. Hence conclusion (c) of 0.6.1.6
holds. Define Ai E A(S) as in 0.6.1.4 and set W := AiQ and Sw := Nr(W).By c9nclusion (c) of 0.6.1.6, ISwl 2: ISi, and QY = Ai for some y E Sw, &ince
Nr(Nr(S)) induces D 8 on A(S) = A(Nr(S)). Then A(W) = {Q, QY}, so Gw :=
Na(W) = (Gq n G~)Sw. By 2.4.3.1, Gq E He. As Gq n G~ = Noq (W),
GqnG~ E He by 1.1.3.2; therefore Gw = (GqnG~)Sw E He. From the structure
of Lo, Q ::::; J(S) = R = Ns(W), IS : RI = 2, and NH(W) i M, so Gw i M.
As H EI'{), 2.3.8.5c says Co 2 (M)(R) ::::; R. Then we conclude from 2.3.8.5b that
R E (3. Then as usual (R, Gw) E U( Gw ), so Gw E r. Hence as ISwl 2 ISi,
Gw E I'o by 2.4.3.2, so that Gw E I' 0. Thus Gw satisfies the hypotheses for Hinthis section. In particular as we showed that Q = 02(H) is abelian, by symmetry
between Hand Gw, 02 (Gw) is abelian. This is a contradiction, as W::::; 02(Gw)
and W = AiQ is nonabelian since Q E A(S). This contradiction establishes (1),
and completes the proof of (1)-(3).
By 0.5.6.2, for each x E T - S, R = U 0 Q and [Uo, U 0 ] = Uo n U 0. Thus
Lo= [Lo, U 0 ] and U 0 nQ::::; U 0 , so as UoU 0 and Tc arenormalinR, R = TcxUoU 0.That is, (4) holds. D
REMARK 2.4.28. In the next lemma, we deal with the shadows of extensions of
L4(3) ~ PDt(3) which are not contained in POt(3). In this case, Lis an A5-block.
The subcase where Cr(L) f:. 1 is quickly eliminated using 2.3.9.7: that subcase is
the shadow of Aut(L4(3)), which is not quasithin since an involution in Cr(L) has
centralizer Z 2 x P0 5 (3). The remaining cases we must treat correspond to the two
extensions of L 4 (3) of degree 2 distinct from POt(3), which are in fact quasithin.
. These sub cases are eventually eliminated by using transfer to show G is not simple,
but only after building much of the 2-local structure of such a shadow.
Shadows of extensions· of L 4 (3) will also appear several more times in later
reductions.LEMMA 2.4.29. L is an A3-block. Hence H = LoS where Lo is a product of
two S-conjugates of L.
PROOF. The second statement follows from the first in view of 2.4.26. Weassume Lis not an A 3 -block, and derive a contradiction. Then Lis an A 5 -block, and
s = 1 by 2.4.25. Set Tc:= Cr(L). By 2.4.27, Q::::; J(T) = J(S) = Baum(S) = R,
<P(Tc) = 1, Q = Tc x U, and R = Tc x uux. By 0.5.4.3, R/Q ~ E 4 and
LR/Q ~ S5 = Aut(A 5 ), so that LS= LR. Recall L n Mis a Borel subgroup of L.
Let K := 02 (MnL) and P := 02 (K). Then P ~ Q~, and S =PR= PUUxTccentralizes Tc, so <P(S) = <P(UUx)<P(P)[UUx, P] ::::; P. Therefore Z := Z(P) =
(z) = <P(S) n Z(S). Since Sis of index 2 in T by 2.4.27.l, z E Z(T).