660 5. THE GENERIC CASE: L2(2n) IN .C.t AND n(H) >^1
Pick z E z# and set Gz := Ca(z). Set Vi := 02(Li) and observe S = V1 V2 =
J(T) using 5.3.2 and F.4.29.6. Similarly by F.4.29.2, if t ET - S, then t induces a
field automorphism on Li, so [Zs, t] =/= 1; that is, S = Cr(Zs).
LEMMA 5.3.5. Na(Zs) = Ml,2·
PROOF. Set Gz :=Ca( Zs). Then S =Vi V2 E Syb(Gz), as we just observed.
As T::::; Na(Zs), F(Gz) = 02(Gz) by 1.1.4.6, and therefore also F(Gz/Zs) =
02 (Gz/Zs) by A.1.8. Hence as S/Zs is abelian, S/Zs = 02(Gz/Zs), so S =
02 (Gz). Then as A(S) = {Vi, l/2} with Vi '.SJ T, Na(Zs) ::::; Na(Vi) = Mi as
Mi EM by 5.3.1.1. On the other hand by 5.3.1.3, Mi,2 = Na(S)::::; Na(Zs). D
LEMMA 5.3.6. (1) Vi is weakly closed in T with respect to c:
(2) z<§ n Vi= z~i is of order q + 1.PROOF. We saw A(T) = {Vi, l/2} ~nd Vi '.SJ T; in particular, Vi° n T ~ A(T),
so to establish (1) we only need to show V2 <f_ vp. But if V2 E vp then as Vi '.SJ T
and Na(T) controls fusion of normal subgroups of T by Burnside's Fusion Lemma,
V2 is in fact conjugate to V 1 in Na(T), and hence in 02 (Na(T)) as T normalizes V1
and l/2. This is impossible as IA(S)I = 2, establishing (1). By (1) and Burnside's
Fusion Lemma, Mi controls fusion in Vi, so (2) follows. D
LEMMA 5.3.7. Gz::::; Mi,2-PROOF. As Zs '.SJ T and M 1 , 2 is transitive on zff, we may take z E Z(T).
Therefore F*(Gz) = 02 (Gz) =: R by 1.1.4.6. Next unless q = 4 and Mi/Vi ~ 85
for i = 1 and 2, S = V 1 l/2 = 02 (CM;(z)) for each i. Assume for the moment
that the exceptional case does not hold. Then as S E Syl 2 (Ca(Zs)) by 5.3.5,
R::::; S by A.1.6, so Zs = Z(S) ::::; S""h(CaJR)) = D1(Z(R)) =: ZR· If Zs = ZR
then R::::; Na(Zs) ::::; M 1 , 2 by 5.3.5, and the lemma holds; so assume instead that
Zs< ZR.
Let G denote our target group G(a) ~ L 3 (q) and Mi the subgroups Mi(a)
in 5.3.2. Recall we have a corresponding isomorphism of amalgams (3 : i' :=
(M 1 , M 1 , 2 M 2 ) -+ 'Y· Thus S ~Sis isomorphic to a Sylow 2-group of L 3 (q), so Vi
and V2 are the maximal elementary abelian subgroups of S. Therefore Vin ZR> Zs
for i = 1 or 2, so that R::::; Cs(Vi n ZR) =Vi. Then Vi ::::; Caz (R) ::::; R, so Vi = R.
But then by 5.3.6.1, Gz ::::; Na(Vi) = Mi, so that Gz = Gz n Mi ::::; Mi,2 using (3,
and the lemma holds.
It remains to treat the exceptional case where q = 4 and Mi/Vi ~ 85 for
i = 1 and 2. Let Gz := Gz/ (z), so that F*(Gz) = 02(Gz) by A.1.8. Now Mi
is determined up to isomorphism, so in particular T is isomorphic to a Sylow 2-
subgroup of M22· Therefore J(T) = Q ~ E15 with Q ~ Q~ and Cr(Q) ::::; Q. Let
Vz := (Z~·). As Zs::::; Z(T), Vz is elementary abelian by B.2.14, so <P(Vz)::::; (z).
Suppose first that Vz is abelian, and therefore elementary abelian. Then Vz ::::;
Cr( Zs)= S using an earlier observation. As V 1 and V2 are the maximal elementary
abelian subgroups of S, Vz ::::; Vi for i = 1 or 2. If Vz =Zs, then the lemma holds
by 5.3.5, so we may assume Zs < Vz ::::; Vi. But Vi = Ca(A) for each hyperplane
A of Vi through Zs, so Vz =A or Vi, and in any case Vi '.SJ Gz. Hence the lemma
holds, again since Gz n Mi ::::; Mi,2-