1074 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cf(G, T) IS EMPTY
Suppose first that m(U 0 ) = 2. Again (2) holds. Also T acts on V, Vi, and F,
and T centralizes Za by our choice when m(U 0 ) :::; 2; in particular, T* centralizes
FnVofrank 1. As H* ~ 85 by Theorem 14.7.63; m(C.v-(T)) = 1 = m(Cp(T)), so
as Za:::; F, we conclude that V 2 = Cv-(T) = V n F = Cp(T) = Za, since all these
subspaces are of rank 1, and each successive pair is related by inclusion. Thus (1)
holds. Then we saw that (4) also holds, so that Ua E uL, and hence as v:::; u, also
V:::; Ua, so that FV:::; Un Ua. Then (3) follows from(*), completing the proof of
the lemma in this case.
Therefore we may assume m(U 0 ) = 3, and it remains to derive a contradiction.
This time as F n V = 1 and Za :::; F, we have Za i V. Let E := V1Za, YE :=
(QH, Qa), and Y := 02 (YE)• As H is irreducible on U /Uo and Za i Uo, itfollows from 14.5.15.1 that [Za, QH] = Vi. By the symmetry between ')'1 and a,
[V 1 , Qa] = Za· Then by A.1.14, YE induces GL(E) on E, Na(E) = YECa(E), and
YE :SI Na(E). As Za i Uo and H = Ca(z), Ca(E) = CH(Za) is a 2-group.
As Ri centralizes Uo V by 14.7.65, Rl acts on E. We claim T :::; Na(E),
so suppose otherwise. Then for t E T - R 1 , F 0 := ViZaZ~ is of rank 3, so as
T acts on F with E = V1Za :::; F :::; Un Ua, Fo is contained in Un Ua n U~.
Therefore Yo :=(YE, Y~) induces GL(Fo) on Fo, since AutyE(Fo) is the stabilizer
of E in GL(F 0 ). But then there is an element of order 3 in Cy 0 (z), impossible as
NH(Fo):::; T.
Thus T:::; Na(E) as claimed, so Tacts on 02 (YE) = Y, and further ZaCu 0 (T) =
Cu-(T) = °V2Cu- 0 (T). Therefore·(Z~^1 ) = VZa is of rank 4, as we saw Za i V.
Let I := (L1T, Y), Vi := (Vl), Q1 := 02(I), and J+ := I/Q1. Then
(I, LiT, YT) is a Goldschmidt triple in the sense of Definition F.6.1, so a :=
(LtT+, r+, y+r+) is a Goldschmidt amalgam by F.6.5.1, and hence is described
in F.6.5.2. Next L 1 has at least five noncentral 2-chief factors, one on 02 (Li) and
two each on U and QH/CH(U) using 14.5.21.1. Thus we conclude from F.6.5.2
that Q 1 -f 1. In particular I is an SQTK-group and IE H(T) <;;;;He by 1.1.4.6, sothat VJ E R 2 (J) by B.2.14. As E:::; VJ and Ca(E) is a 2-group, Q 1 = C1(VJ).
We finish much as at the end of the proof of 14.7.32: If y+ acts on Lt, then
as T acts on Y, J+ = L 1 + r+y+, so VJ = (V 1 L+r+y+^1 ) = (Vt + ) = E, impossible as
Li does not act on E. Therefore y+ does not act on Lt, so in particular Lt is not
normal in J+, and so Lt i-1.Suppose Y :::; M. As Y does not act on L 1 but T acts on Y, the projection
of Yon L/02(L) in M/02(L) = L/02(L) x CM(L/0 2 (L))/02(L) is the maximal
parabolic L202(L)/02(L). Then Y = [Y, T n L] :::; L, so E =(Vt) :::; V, whereas
we saw earlier that Za i V. Thus Yi M.Assume next that J(T) :::; Q1. Then J(T) = J(Q 1 ) by B.2.3.3, so that I :::;
Na(J(T)). Then since M = !M(LT) and Yi M, we conclude again using B.2.3.3
that J(T) i 02(LT). Thus Ll = [L1, J(T)] by 14.3.9.2, contradicting J(T) :::; Q1
and Lt -f 1.
Therefore J(J)+ -f 1, so by Theorem B.5.6, either J(J)+ is solvable and the
direct product of copies of 83 , or there is K1 E C(J(I)) with Kt -f 1. In the latter
case, K1 E Lt(G, T), so by 14.3.4.1, Kt is L3(2) or A5.
Let I' := I/031(!). By F.6.11.2, either I! is described in Theorem F.6.18, orI^1 ~ 83. But in the latter case, and in case (1) of F.6.18, T+ is of order 2, so that
T+ = J(T)+, and J+ = (T+l+) = J(J+) ~ 83, contrary to Lt not normal in J+.