830 12. LARGER GROUPS OVER F2 IN .Cj (G, T)
PROOF. Assume Zv = Cv(L) -=/= 1. Let HE 1-l*(T, M), K := 02 (H), VH :=
(zH), and H* := H/CH(VH)· By 12.6.20.3and12.6.22.3, min{r(G, V),w(G, V)} >1, so each solvable member of 1-l(T) is contained in M by E.3.35.1 and E.l.13. In
particular H is not solvable. By 12.2.9.1, Ca(Z) s M, so by Corollary 12.3.2,
1 -=/= VK := [VH, K] is the sum of at most two As-modules for K* S::! As. As
H =KT, 02 (H) = CH(VH) = CT(VH)· Let H'Af be the Borel subgroup of H over
T; then HM= H n M by 3.3.2.4. Now 02 (HM) = 031 (HM) S 031 (M) = L by
12.6.1.5.
Next if Wo := Wo(T, V) s CT(VH), then H s Na(Wo) s M by 12.6.22.l and
E.3.34.2, contrary to Hi M. Therefore Wo i CT(VH), so there is A:= Vg ST
with A-=/= l. Now by 12.6.22.1, As W 0 (T, V) s CT(V) = Q. Thus as 02 (HM) S
L, HMS LT, so that AS 02(HM)·
Let B :=An 02(H); then m(A/B) = m(A) =:ks 2 = m2(H). Asks 2,
VH S Ca(B) S Na(A) by 12.6.20.3, so [A, VH] S An VH. However 02(H'Af) is
not quadratic on VH, so A < 02 (H'Af ). Therefore k = 1, so VH induces a group
of transvections with axis B on A. Thus IVH : CvH(A)J = 2 from the action of
Lf' S::! Ss on V. This is impossible, since as A* s 02(H'Af ), m(VH/CvH(A)) > 1
from the action of involutions in As on its permutation module. This contradiction
completes the proof of 12.6.23. D
12.6.2. The amalgam setup, and the elimination of VH nonabelian.
With 12.6.23 in place, we can begin to use some of the techniques from section
F.9, which are more representative of the arguments for the F 2 -case in the three
chapters of this part.
By 12.6.23, Zn V = Vi is of order 2. Let z be the generator of V 1. Recall from
Notation 12.2.5 that Gz = Ca(z) and Gz := Gz/V 1. Now z is of weight 4 in V,
so Gz i M by 12.6.21. Recall that L1 = 0^2 (NL(V1)) = 02 (CL(z)) = Lz, Vs is
the 5-subspace V 1 ..L of V, L1 S::! Eg /Em, and L1f'/02 ( L1 f') acts on Vs as DJ ( 2) or
ot(2) on its natural module.
We now make a definition which, will be repeated in many later sections of the
chapters on the F 2 case: let
1-lz :={HE 1-l(L1T): Hi Mand HS Gz}.
As Gz i M, Gz E 1-lz, and hence 1-lz -=/= 0.
For the remainder of the section, H will denote an element of 1-lz.
Set QH := 02(H), UH:= WsH), VH := (VH), and H* := H/QH.
LEMMA 12.6.24. (1) L1T is irreducible on Vs.
(2) UH S 02(H). In particular, V5 S 02(G1).
(3) Vs 02(G2).
(4) Gs s Na(V) = Mv.
(5) If g E G with V 1 < V n Vg, then (V, Vg) is a 2-group.(6) CaCVs) = Ca(V)R1 = CM(V)R1 and Ca(Vs) = Ca(V) = CM(V).
(7) Hypothesis F.9.1 is satisfied for each H E 1-lz, with Vs in the role of ''V+ ".
(8) QH = CH(UH), so H* = AutH(UH)·
PROOF. Part (1) is standard and an easy calculation. Since m(Vs) = 5 > 3,
(4) follows from 12.6.20.1. By (4), Ca(i/5) S M; so as R1 = CMvCVs) and Mv
contains no transvection with axis Vs, (6) holds.