i4.2. STARTING THE L 2 (2) CASE OF Cf EMPTY 985Let Gi := KiS, Gz := BiYS, and Gi,2 := Gin Gz = SBi. Consider the
amalgam a:= (Gi, Gi,2, Gz), and let Go := (Gi, G 2 ). To establish hypothesis (e)
of F.1.1, we need to show:LEMMA 14.2.15. 02(Go) = 1.PROOF. Assume 02(Go) # 1, and let Mi E M(Go). Then T i. Mi, since
otherwise by 14.2.2.3, M = !M(YT) = Mi, contrary to (Ki, T) = H i. M. Thus
S < T, and hence one of the cases of 14.2.9 holds.
Let Zs := fh(Z(S)). As T normalizes S, Z :::; Zs. By 14.2.9, Ki :::; Kc EC(Mc)· As 02((Kc,T)) :::; NT(Ki) = S, Zs :::; Oi(Z(02((Kc,T)))). Thus as
Ct(G,T) = 0 by 14.2.1.1,
K:::; (K'{):::; Ca(Zs), (!)
so that Zs ::::! (Kc,T). Hence Na(Zs):::; Mc= !M((Ki,T)) by 14.2.10. As \T:
S\ = 2, Sis normal in a Sylow 2-subgroup Ti of Mi, and hence Ti:::; NM 1 (Zs) :::;
Mc. If S < Ti, then Ti E Sylz(Mc), so Mc = !M( (Ki, Ti)) = Mi by 14.2.10, a
contradiction as Mc # M = !M (YT).
So S E Syl2(Mi). Therefore we can embed Ki in some L E C(Go) by 1.2.4.Now Y = 02 (Y) normalizes L by 1.2.1.3, and S :::; Na(Ki) :::; Na(L), so Go =
(KiS, Y) = LYS.
Suppose that L:::; Ca(Z). Then L centralizes V = (ZY), so (L, T) :::; Na(V) =
M, a contradiction as M # Mc = !M ((Ki, T)).
Therefore [L, Z] # 1. In particular, Ki< L, so as Go= LYS, L = [L, Y]. Let
R := 02 (YS). Then R ::::! YT, so C(G, R) :::; M = !M(YT). Moreover if Yi. L
then YB n L =Sn Lis Y-invariant, so Sn L:::; Rand hence RE Syl 2 (LR).
Next CT(02(Mi)) :::; Mi as Mi E M, and as S E Sylz(Mi) and S :::; Mc,
02(Mi):::; 02(Go):::; 02(Mc n Go):::; S by A.1.6, so that
Co 2 (Mc)(02(Mc n Go)):::; CT(02(Go)):::; CT(02(Mi)):::; S:::; Go, (*)
and hence G 0 , Mc, S satisfy the hypotheses of 1.1.5 in the roles of "H, M, TH".
By (*), hypothesis (b) of 1.2.11 is satisfied, and since a generator z for Z is in
V = [V, Y], hypothesis (a) of 1.2.11 is also satisfied. Thus by 1.2.11, either Go E He,
or L is quasisimple.
Assume first that L is quasisimple. Then L is described in 1.1.5.3, and L =
[L, z]. As Ki/0 2 (K 1 ) is aBender group over F2n with n > 1, and Ki E C(CL(z), S),
comparing the list of 1.1.5.3 with that of A.3.12, we conclude that either L/Z(L) is
Sp4(2n), G2(2n),^2 F4(2n), or^3 D4(2n/^3 ), or else Ki/02(Ki) ~ Lz(4) and L ~ ]z,
J 4 , HS, or Ru. Then by A.3.18, 031 (Go)= L, so that Go= LYS =LS.
Suppose first that Lis of Lie type. As YB= SY, either L ~^3 D 4 (2), or n is
even and Y is contained in the Borel subgroup of Lover S. But in the latter case,
Y lies in the parabolic P of L with Ki= P^00 , so Go= (KiS, Y) :::; P < L, contrary
to Go= LS.
Therefore L ~^3 D 4 (2) with Ki/0 2 (Ki) ~ L2(8), or J2, J 4 , HS, or Ru with
Ki/0 2 (Ki) ~ L 2 (4); note that case (2) of 14.2.8 holds. Now 1 # 02(Go) :::;
Cs(L), so 1 # Cz 8 (L) =: ZL is in the center of Go= LS. Thus we may assume
Mi E M(Ca(ZL)). Then by(!), K:::; Ca(ZL) :::; Mi. Further K = 031 (K) since
Ki/0 2 (Ki) ~ L2(2m) for some m. By 1.2.4 and 1.2.8.4, LE C(Mi), and by A.3.18,
L = 0
31
(Mi). Thus K:::; L. However, when Lis^3 D4(2), ]z, J4, HS, or Ru, CL(z)