16.5. IDENTIFYING Ji, AND OBTAINING THE FINAL CONTRADICTION 1201L', Tc and L, R, CM(u) moves z, so zM is of order 7 or 13. Further A= (zM),
so M+ acts faithfully on zM. Since [£ 5 (2)[ is not divisible by 13, [zM[ = 7, so
M+ ::; 87. As CM+(z) ~ Ot(2), [M+[ = 23 · 32 · 7. But 87 has no subgroup of
index 10.
Therefore A i. L' K'. Hence LT ~ 88 x (z), so from the structure of 88 ,
A(T) = {A,A1,Ai,B} for suitable Bandt E TL, where J(02(CLT(u))) = {A,A1}.
As A i. L'K', Ai = 02(CL'(z))R = J(CTJ,Tf!,(u)), so Ai E AG. Observe thateach member of A(T) is normal in J(T), so by Burnside's Fusion Lemma, I :=
NG(J(T)) is transitive on AG n J(T), and hence Ai E A^1. As [T : J(T)[ = 2, I
is not transitive on A(T), so A^1 = {A,Ai,AD and I induces 83 on A(T). ButJ(T) ~ Z 2 x D 8 x D 8 , so by the Krull-Schmidt Theorem A.l.15, I permutes the
two involutions generating the Frattini subgroups of the D 8 -subgroups, so that
02 (!) centralizes 'P(J(T)). Then as Z(J(T)) ~ E 8 , by Coprime Action, 02 (1) ::;
02 (CG(Z(J(T))) ::; 02 (Gz) = 02 (LT) = L; then I = NL(J(T))T ::; NG(A),
contrary to [A^1 [ = 3. This completes the proof of 16.5.12. DLEMMA 16.5.13. (1) u is not in the center of T.
{2) L* is not L3(2n) or 8p4(2n).PROOF. By 16.5.12, L* ~ Y(2n)', where Y is one of the Lie types A 2 , C 2 ,G 2 ,^2 F 4 , or^3 D 4. Further if n = 1, then L is the Tits group^2 F4(2)' or^3 D4(2) by
16.5.10.1, and so (1) holds by 16.5.10.5. Thus we may assume that n > l. Further
L is not £ 3 (4) by 16.5.4, so by 16.1.2.1, either Z(L) = 1 or L is G2(4).
We first treat the case where u* is a long-root involution. (When L ~· 8p 4 (2n),
either class of root involutions can be regarded as "long" , as the classes are in-terchanged in Aut( L).) Thus u is 2-central in L . Let Z denote the root group
of the projection v of u on £-unless L* is G2(4) with Z(L) =/= 1, where we let
Z := [NL(Z 1 ), Z1] where Z 1 is the preimage in L of the root group of u*. Set
P := NL(Z) and recall the definition of X :=Xu from Notation 16.5.5. As u* is a
long-root involution, either
(a) Pis a maximal parabolic of L, and we check (cf. 16.1.4.1) that X = P^00 ,
or
(b) L ~ L 3 (2n) and X = Cp(Z) if n is odd, while X = 03 (Cp(Z)) if n is even.
In case (b), n > 2 by 16.5.4. Thus in any case X =/= 1 and Z* = CL*T*(X*), so thatV := ZTc = CT(X), and Tc~ R ~ R ::; Z by 16.5.6.2. In particular 'P(R) =
'P(Tc) = l. Choose gas in 16.4.2.4; thus V = ZTc ::; T n H' = NT(K') ::; T9.
Further xg = x by 16.5.6.3, so
v = cT(X)::; cTY(X) = cTY(X^9 ) = cT(X)^9 = V^9 , ·
and hence g E Na(V) n NG(X) =: M. Let To := NT(X) and notice that eitherTo = T, or T is nontrivial on the Dynkin diagram of L ~ 8p4 ( 2n) with To of index
2 in T. Let M := M/CM(V). We can finish much as in the proof of Proposition
16.5.1: For P ~ Z 2 n_ 1 is regular on z# = [V,P]#, Tc= Cv(P) is a TI-set in V
under the action of M by I.7.2.3, NM(Tc)::; NM(P) by 16.4.2.5, and f' E 8yb(M)
acts on P. Thus again we have the hypotheses for a Goldschmidt-O'Nan pair in
Definition 14.1 of [GLS96], so we may apply O'Nan's lemma Proposition 14.2 in[GLS96]. Conclusion (i) of that result is eliminated since g E M - NG(Tc), so
either m(V) = 3 and M ~ Frob 21 is irreducible on V, or Tlf is of order 2n where