4.4. CONTROLLING SUITABLE ODD LOCALS 62iLEMMA 4.4.9. If K is a component of GB, then !KGB I S 2, and in case of
equality, K ~ L2(2n), Sz(2n), L2(Pe), for some prime p > 3 and e S 2, Ji, or
SU3(8).
PROOF. Since we saw that GB is a QTK-group, this follows from (1) and (2)
of A.3.8; notice we use 4.4.7.1 to guarantee 02 (K) = 1, and I.1.3 to see that the
Schur multiplier of SU3(8) is trivial, and in the remaining cases the multiplier of
K/Z(K) is a 2-group, so that K is simple. D
By 4.4.8, VB centralizes O(GB), and by 4.4.7.1, 02(GB) = 1, so VB is faithfulon E(GB)· Thus there is a component K of GB with [K, VB] =f. 1. Set Ko := (KMB)
and MK:= Mn K. Recall that GB is a quasithin JC-group, and hence so is K by
(a) or (b) of (1) in Theorem A (A.2.1), so that K/Z(K) is described in Theorem
B (A.2.2).LEMMA 4.4.10. (1) K </:.MB.(2) VB S Ka.
(3) CGB(Ko) = O(GB).
PROOF. As [K, VB] =f. 1 and VB S 02(MB), (1) holds. As LB = 02 (LB),
LB acts on K by 4.4.9, so 1 =f. VB = [VB, LB] acts on K. Indeed as Out(K) is
2-nilpotent for each Kin Theorem B, VB induces inner automorphisms on K 0 , so
that VB s KoH where H := CGB (Ka). Then the projection of VB on H is an
MB-invariant 2-group Q. If Q =f. 1, then by 4.4.7.2, MB= NGB (Q); but then Ks
CGB(Q) S MB contrary to (1). Thus Q = 1, giving (2). Now HS CGB(VB) S MB
by 4.4.6.3. Sets:= TB n H. As tb IS Sylow in MB, and H :'.SI MB, sis Sylow in
H, S :'.SI TB, and
[S, LB] s cLB (VB) n H s 02(LB) n H s 02(H) s 02(GB) = 1,
in view of 4.4.7.1. Thus LBTB s NG(S), so if S =f.1 then NG(S) s M by 4.4.6.l;as S centralizes K, this contradicts (1). Thus the Sylow 2-group S of H is trivial,
so (3) holds. DLEMMA 4.4.11. (1) K =Ko :'.Si GB.
(2) LBS MK.
PROOF. Observe Out( Ko) is solvable, since !KGB I S 2 by 4.4.9 and the Schreier
property is satisfied for the groups in Theorem B. Also CGB (Ko) is solvable by
4.4.10.3. Hence LB = L'EJ s K 0. Thus (2) will follow from (1).
Assume K is not normal in GB. By 4.4.9, Ko = KiK 2 where Ki :=KandK2 := K^8 for s E GB - NGB(K), and K is a simple Bender group, L2(Pe), Ji, or
SU 3 (8). But then K has no nonsolvable 2-local MK with 02 (MK) not in the center
of MK, contradicting LB s Mn K 0 • This establishes (1). D
LEMMA 4.4.12. K/Z(K) is not of Lie type and characteristic 2.
PROOF. Assume otherwise. By 4.4.11.1and4.4.10.3, O(GB) = CG(K), so TB
is faithful on K. By 4.4.10.2, VB SK, so QB := 02(MB)nK </:. Z(K). Therefore as
K/Z(K) is of Lie type and characteristic 2 by hypothesis, MB acts on some proper
parabolic of K (e.g. using the Borel-Tits Theorem 3.1.3 in [GLS98]). Hence by
4.4.7.2, MK is a maximal Mwinvariant parabolic of K. Furthermore from Theorem
B, K/Z(K) either has Lie rank at most 2, or is L4(2) or Ls(2) or Sp 6 (2), so as