12.4. SOME FURTHER REDUCTIONS 813and V/Zv is a natural module for L/0 2 (£). Then as the 1-cohomology of the
dual of V/Zv in I.1.6 is 1-dimensional, Zv = (z) is of order 2. As M = !M(LT),
Gz:SM.
On the other hand, when L/0 2 (£) ~ £ 5 (2), we have m(V) = 10 by hypothesis;and as we are in a counterexample to the theorem, Ca ( Z n V) ::::; M. As dim(V) =
10, Zn V is of order 2, and in this case we take z to be a generator for Zn V. Thus
· Gz ::::; M in this case also.
We begin a series of reductions.LEMMA 12.4.3. (1) Ca(Z) :SM.
(2) L = [£, J(T)].
PROOF. As Gz ::::; M and z E Z, (1) holds; then (2) follows from 12.2.9.2. D
We are already in a position to GOmplete the proof of part (2) of Theorem
12.4.2:
LEMMA 12.4.4. L/02(£) is not £ 5 (2).
PROOF. Assume L/02(L) is £5(2). Then L has two orbits on V#, represented
by z and some further involution t.
We claim that t ¢ z^0. First L = 031 (M) by 12.2.8. Next Lz/0 2 (Lz) ~
£3(2) x Z3, so as Gz ::::; M, m3(G';°) = l. However Lt/02(Lt) ~ A6 is of 3-rank 2,so t ~ zG, establishing the claim.
It follows from the claim that Lis transitive on z^0 n V, so as Gz ::::; M, while
Vis a TI-set under M by 12.2.6, Vis the unique member of v^0 containing z by
A.l.7.3.
Next m(Mv) = 3, sos( G, V) ;::::: 3 by Theorem E.6.3. By 12.4.3.1, Ca(Z) ::::; M,
so by 12.4.1, there exists g E G with 1 -=/= [V, VB] ::::; V n VB. Conjugating in Mv ifnecessary, we may assume VB ::::; T. Let A := VB. Interchanging the roles of A and
V if necessary, we may assume m(A) 2:: m(V/Cv(A)): Then by B.l.4.4, A contains
a member of P(Mv, V). Therefore by B.4.2.11, Cv(A) = [V, A] is a 6-dimensional
subspace of A, and A of rank 4 is the unipotent radical of the maximal parabolic
of Lover f' stabilizing [V, A]. In particular, [V, A] is T-invariant, so the generator
z of Z n V is in [V, A] ::::; V n VB. This contradicts our earlier observation that z isin a unique conjugate of V, completing the proof. D
By 12.4.4, L/0 2 (£) ~ £ 3 (2) or G 2 (2)', so as we are in a counterexample to
the Theorem, Zv-=/= l. Hence V ::::) M by Theorem 12.2.2.3. Then M normalizes
Cv(L) = Zv = (z), so since ME M,
M = Ca(z) = Na(V).
When L ~ £ 3 (2), let Ebe the T-invariant 4-subgroup of V, choose v E E - Zv,and let £1 := 02 (CL(E)) and Rl := CT(E).
LEMMA 12.4.5. If L ~ £3(2), then
(1) [Z,L] = L
(2) ZQ := fh(Z(Q)) = ZV.
(3) R1 :=A for each A E A(R1) with Ai. Q.
PROOF. Observe that (3) holds by B.4.8.2. By 1.4.1.4, ZQ = R2(LT), so
V = [ZQ, L] by B.5.1.l. Then ZQ = Cz(L)V by B.4.8.4, so (2) holds. Again By
B.4.8.2, Zn V = Zv, so (2) implies (1). D