1164 i5. THE CASE .C.r(G, T) = r/JSet K := (Xr^0 ) and recall KE 3(G, T). As Mi= Nc(X), Xis not normal in Io
since Io i. Mi by 15.4.19.2.Observe that either:
(i) KE C(Io) with KTr/0 2 (KTr) ~ Aut(Ln(2)), n = 4 or 5, or
(ii) K = KiKi for some Ki E C(Io) and r E Tr-Nrr(Ki), with Ki/02(Ki) ~L 2 (2n), n even, or L 2 (p) for some odd prime p.
This follows from 1.3.4, since K/0 2 (K) is not L3(3), Mn, or Sp4(2n) because Tr
induces D 8 on X/02(X) since XS/02(X8) ~ 83 x 83 by 15.4.12.6, while Tr doesnot act on Li. Also K = 031 (Io) by A.3.18 or 1.2.2, so I::::; KTr, and hence without
loss Io = KTr.
Suppose first that F*(K)-/= 02 (K), so that K is a product of components of I 0.
By an earlier remark, To is also Sylow in Nc(02(Io)), so we may apply 15.4.17 with
I 0 , Nc(0 2 (Io)), 02(Io) in the roles of "I, H, B" to conclude that for z E z# andMz E M(Cc(z)), the hypotheses of 1.1.5 are satisfied with I 0 , Mz, z in the roles of
"H, M, z". Thus K is described in 1.1.5.3, and Z is-faithful on K. Suppose first
that case (ii) holds. As Z is noncyclic and in the center of Tr, while Tr induces Ds
on X/0 2 (X), Ki is not L2(P) for p odd. Thus Ki ~ L2(2n), so as L2(4) ~ L2(5)
and n is even, n ;::::: 4. Further as C ( G, R) ::::; Mi, a Borel subgroup B of K is
contained in Mi, and hence B = 02 (B) acts on the 4-subgroup Vi of 15.4-.12.6; this
is impossible, as when n;::::: 4, B does not act on a 4-subgroup of 02 (B). Thus (i)
holds, in which case we again have a contradiction to Z 2-central, noncyclic, andfaithful on K.
Therefore F*(K) = 02(K). Let Ix :=Ion Mi. Recall C(G, R) ::::; Mi, so that
Hypothesis C.2.3 is satisfied with Ix, Io in the roles of "MH, H". If case (ii) holds,
then as R centralizes X/02(X), R normalizes Ki, so it follows from C.2.7.3 that
either Ki is a block of type L2(2n) or A 5 , or Ki/0 2 (Ki) ~ L 3 (2).
Let Vr := (ZK) so that Vr E R2(Io) by B.2.14, and set I 0 := I 0 /0 2 (I 0 ).
As K/02(K) is semisimple, 02(Io) = Cr 0 (Vr). As Li = [Li, J(T)], Vr is an FF-
module for I 0. Then as Cr 0 (Z) is a 2-group by Theorem 15.4.8, it follows from thedescription of the modules in C.2.7.3 and C.1.34 for the groups in cases (i) and (ii),
that Ki is an L2(2n)-block. But once again a Borel subgroup B of K is contained
in Mi, and hence B = 02 (B) acts on the 4-subgroup Vi of case (ii) of 15.4.12.6,
so we conclude that Ki is an L 2 (4)-block. Finally by 15.4.14, Cz(X) -/= l, so
Cz(X)::::; Z(Io) from the structure of K. Thus Io::::; Cc(Cz(X))::::; Mi= !M(XT),
contrary to 15.4.19.2. DRecall I E H by 15.4.19.1, so To E Sylz(I) by 15.4.20. Then (I, Gi, G 2 ) is a
Goldschmidt triple. Set Qi:= 02 (Gi) and I+:= I/03'(I).
LEMMA 15.4.21. If F*(I) = 02(I), then I= LiL2To with Li :::! I.
PROOF. Assume F*(I) = 02(I) and let Vr := (Z^1 ). As Cr(Vr) ::::; Cr(Z) =To
by Theorem 15.4.8, and Vr E R2(I) by B.2.14, we conclude that C 1 (Vr) = 02 (I).
Let I := I/02(I), so that I+ is a quotient of I. As Li = [Li, J(T)], Vr is an
FF-module for I and Li centralizes 03 (F(I)) by B.1.9.
We claim a:= (Gt, Tft, Gi) is a Goldschmidt amalgam. For if not, byF.6.11.2,
Qi = Q2 and I+ ~ 83 with 03'(I)* -/= 1. Then Qi = 02 (I) and I is solvable by
F.6.11.1, so as Li centralizes 03 (F(I*)), I is a {2, 3}-group by F.6.9, contradicting