768 il. ELIMINATION OF Ls(2n), Sp4(2n), AND G2(2n) FOR n > 1the hypotheses of E.6.14, and that lemma supplies a contradiction, completing the
proof of the claim that r(G, V) 2 m.
Assume the hypotheses of (2). By 11.0.3.4, CM(U) :::; Mv. Then CM(U) =
CM (V) since if Y is of odd prime order in Mv, then m(V/ Cv (Y)) 2 2m; notice we
use ll.0.4"and A.1.41 to conclude C:Mv(L) = Z(L), and to exclude diagonal outer
automorphisms. Now (2) follows from E.6.12 and the fact that r(G, V) 2 m > 1.We have shown r(G, V) 2 m. Further in case of equality, we may pick U with
k = m, Ca(U) i. M, and 0
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(C:Mv(U)) =f. 1 by (2). But then U = Cv(i) for asuitable root involution i E L, anq up to conjugacy either:
(i) L ~ Sp4(q) or G2(q) and Vs :::; U, or
(ii) L ~ SL3(q) and U = V2.Case (i) contradicts 11.1.4, so that (3) holds.
Let A:::; VYnTwith m(VY /A)=: j < m-1. To prove (4), we must show that
A:::; TL, so we may assume A =f. 1. Then for B:::; A with m(VY / B) < m = s(G, V),
A E Am-1(Mv, V) ~ A 2 (Mv, V), using E.3.10. Hence (using B.4.6.9 in case Lis
G2(q)) A:::; L, so W1(T, V) :::; TL. Then as Vi = Cv(TL), (4) holds.
Assume next that L ~ G 2 (q) and j := m(VY /A) = 0 or 1 with A =f. 1. By
B.4.6.3, there is Eqs ~Ai :::1 N1(Vi) with Cv(Ai) E V3. As Lis G2(q), m = 2n,
so as n 2 2, A E Am-j(T, V) ~ An+i(T, V) by the previous paragraph. Hence
by B.4.6.9, A:::; Aq for some h EL, and if j = 0, then A= Aq. Further if j = 1
and A:::; Ri, then by B.4.6.12, A:::; Ai. Thus Cv(Wi(Ri, V)) 2 Cv(Ai) =Vs, so
Ca(Ci(Ri, V)):::; Ca(V3):::; M by 11.1.4. That is, (6) holds.
Now take j = 0. Thus A= Aq, so without loss A= Ai. Let D:::; A with fJ a
long root group of L. Then m(VY / D) = m(A/ D) = 2n = m < r( G, V) by (3), so
Cv(D) :::; Na(A). SetE := (Cv(D) : D:::; A, fJ is a long root subgroup of L).
Then by B.4.6.3, m(V/E) = n and [E,A] =Vs. We just saw ED := Cv(D) :::;
Na(A), so E acts on A, and hence Vs = [E,A] :::; A. Thus V 3 :::; V n VY,so as V is a TI-set under M by 11.0.3.4, g tj. M. Furthermore D = CA(ED),
so m(A/CA(ED)) = m(A/D) = 2n. Therefore by B.4.6, the image of ED in
LY /02(£9) is contained in a long root group, and [ED, A] =: AD E V~ =: V 2 (A).
But as A= Ai, also [ED,A] E V2 =: V2(V). So if we define A(V3, V) := V 2 (V)nV3,
we seeA(V3, V) = A(V3, V)Y := A(V3,A).
Define
L(V3, V) := (L(I) : IE A(V3, V)).By(*) and 11.1.3, L(V3, V) = L(V3, V)Y =: L(V3, A). But we check that L(V3, V) =
L, so by Theorem 4.2.13,
M = !M(L(V3, V)) = !M(L(V3,A)) =MY,
contradicting g tj. M. Together with E.3.16.1, this establishes (5).
LEMMA 11.2.3. (1) If Cv(L) = 1, then Na(Vi) = Ca(Vi)NM(Vi)·
(2) If Ca(Vi) :::; M, then Na(Vi) :::; M.D
PROOF. Set Y := Na(Vi) and Y* := Y/Ca(Vi)· Now Cr(Vi) = TL and
T = (!)TL where f induces a field automorphism on L, so T is cyclic. Hence by
Cyclic Sylow 2-Subgroups A.1.38, Y = O(Y)T.