542 2. CLASSIFYING THE GROUPS WITH IM(T)I = iBy C.5.5, we may choose x E Mo with U"' 1:. Q, and and so as Qi = Q2 with
Q = Qin Q 2 , U"' 1:. Qi. By C.5.6.4, <P(J)"' = <P(J). Then as x also acts on S, x
acts on S1 and hence also on <P. Let G1 := Na(<P), and set So:= (S1,x). Then
So::; Nr(<P). As J(S)::; R, 02 (Lo8)J(8)::; Ns(L) = 81, while Ns(L) is of index
2 in 8, so S1 E (3 by 2.3.8.5b. As NH(<P) E He by 1.1.3.2, and NH(<P) containsL 1:. M, from the definitions in Notations 2.3.4 and 2.3.5, (S1,NH(<P)) E U(G1),
and hence G 1 Er. By 1.1.6, the 2-local G1 satisfies the hypotheses of 1.1.5 in the
role of "H".
As U"' 1:. Qi= 02 (LR), U"' 1:. 02 (LS1 ). Therefore as U"' ::; R:::; SJ, while L"' is
irreducible on U"', U"' n 02 ( G 1) = 1. Notice LS1 E He, so that (S1, L81) E U(LS1 ),
and hence LS 1 Er. Define Hi to consist of the subgroups Hi satisfying:
Hi E He(LS1) n G1, andHi = (L, Si) for some Si E Syb(Hi) containing S1.
Then Hi is nonempty, since LS1 E Hi.
We next claim
(*) LE C(Hi) for any Hi E Hi.It is clear that () holds if Si = S1, so assume instead that Si > S1. Then
as [S : 81[ = 2, [Si[ ::::-: [Sf. Since Hi E He, (S1, Hi) E U(Hi) and Hi E r.
Then by 2.4.3.2, [Si[ = [Sf and Hi E I' 0. As [Si : S1f = 2 and L E .C(Hi, S1),
L::; K E C(Hi) by 1.2.5. Then as Hi E r 0 , K is a xo-block of Hi by 2.3.8.4. Since
no xo-block has a proper A 5 -block, K = L, completing the verification of ().
Now L :::; Gj, and by 1.2.1.1, Gj is a product of C-components Ki, ... , Kn
with L inducing inner automorphisms on Ki/0 2 (Ki) for each i. However using
1.2.1.1, Car(Gj/02(Gj)) = 02(Gj), so as Un02(G1) = 1, Ln02(Gj) = 1.
Hence Autu(Kif0 2 (Ki)) =-/= 1 for some Ki E C(G1). Choose notation so that the
projection LK; of Lon Ki/0 2 (Ki) is nontrivial iff 1 ::; i:::; t. Since Ln0 2 (Gj) = 1,
it follows that L::; LK 1 • • ·LKt· Observe for i:::; t that LK; has a quotient A5, so
that m3(LK;) ::::-: 1.
We claim that t = 1: For t :::; m 3 ( G 1) ::; 2 since G 1 is an SQTK-group, so that
t = 2 if t > 1, and then the proof of 1.2.2 (which does not depend on conjugacy of
the C-components in the lemma) shows that 031 ( G 1) = Ki K 2. Since 031 ( M n Lo)
centralizes <P, 03 ' (MnL 0 )::; 031 (G1):::; KiK 2. Therefore for i = 1or2, there exists
y of order 3 in Lo n Ki with L = [L, y]. Then L = [L, y] ::; Ki, so that L ::; LK;i
and hence L = LK; with LKj = 1forj=-/=1, contrary to our assumption that t = 2.
This contradiction establishes the claim that t = 1. Hence L = LK 1 :::; Ki =: K.
Since Un 02(G1) = 1, m2(K/0 2 (K)) ::::-: m(U) = 4, ruling out cases (c) and (d) of
1.2.1.4, and hence showing that K/0 2 (K) is quasisimple.
Suppose first that F(K) = 02 (K). Now S 1 = Ns(L) normalizes Land hence
normalizes K. Then KS1 E Hi so LE C(KS1) by(). Thus LE C(K) and hence
L = K, contrary to Un 02(G1) = 1.
Thus F*(K) > 02(K), so as K/02(K) is quasisimple, we conclude that K is
quasisimple, and hence K is a component of G 1. Thus K is on the list of 1.1.5.3.
Indeed as K contains the A 5 -block L, we conclude from that list that K is either
of Lie type and characteristic 2 of Lie rank at least 2, but not L 3 (2), or one of
M22, M23, M24, J4, HS, He, or Ru. Let S::; T1 E Syl2(G1); then T1 normalizes
K by 1.2.1.3. Let X E Hen KT1, S1 :S Si E Syb(X), and Y := (L, Si). Then
Si E Syb(Y), and Y E He by 1.1.4.4, so Y E Hi. Then by (*), L is subnormal
in Y, so L E .C(X, Si). Thus we have shown that for each X E Hen KT 1 and