792 12. LARGER GROUPS OVER F2 IN .Cj(G, T)
So assume instead that Vi Qv. Then by the Baer-Suzuki Theorem, there is
g E Gv such that I := (V, V9) is not a 2-group. We showed U2 :::'.: Ca(V^9 ), and by
symmetry Uf centralizes V, so U2Uf (v) :::; Z(I) and Vf :::; Cr 1 (U2). But Cr 1 (U2)
is the group of transvections on Vi with axis U2, so as V1 is dual to V2, Cr 1 (U2) is
the group of transvections on Vi with center (v). Hence [V, V/] :::'.: U2(v) = U2(z),
and then [V1U2V/Uf,J]:::; U2Uf(z). Therefore 02 (1) :::'.: Ca(V1U2) = Ca(V) using
12.1.2.2, contradicting I not a 2-group. This completes the proof of 12.1.6. D
LEMMA 12.1.7. Wo := W 0 (T, V) centralizes V, so that w := w(G, V) > 0.
PROOF. Assume that W 0 i Cr(V). Then n = 4 by 12.1.3, and there is
A := V^9 ::::: T with A -I-1.
Suppose first that V :::; Na(A). Then interchanging the roles of A and V
if necessary, we may assume m(A/CA(V)) 2: m(V/Cv(A)). Then by B.1.4.4, A
contains a member of P(T, V), which is J(T) by B.4.9.2iii. Thus A is the unipotent
radical of the stabilizer in L of a 2-subspace of Vi, so that [Vi, A] is of rank 2. As
A normalizes Vi, and Vi :::; V:::; Na(A), 1 "I-[Vi, A] :::; Vin A, contrary to 12.1.6.2.
Therefore we may assume that V i Na(A). As r(G, V) 2: 4 by 12.1.2.2,
m(A) 2: 4 by E.3.4, so that m(A) = 4 = m 2 (L'f). Then by lemma H.9.3.3, we may
take A to be one of the groups denoted there by Ai for 0 :::; i :::; 2. As r( G, V) 2: 4,
we conclude from E.3.32 that
i\A:(V) = I\A(V)::::: U := Nv(A).
As we are assuming U < V, i "I- 0 by lemma H.9.3.4. Let B := NA(Vi). Then
B =An L has rank 3, as A is A1 or A 2. Set Ui :=Vin U. By 12.1.6.2,
1 =Vin A 2: [Ui,B].But for any b E f3#, Cv;(b):::; Ui by(*), so Cv;(b) = Ui = Cv(B). However this is
not the case as B contains at least one transvection on V 1 , but not all elements of
f3# induce transvections on V1. This contradiction completes the proof. D
LEMMA 12.1.8. W1 := W 1 (T, V) centralizes V, so that w > 1.
PROOF. Assume that W1 i Cr(V). As w > 0 by 12.1.7, w = 1. Thus there
is a w-offender A := Nvg (V) :::; T with A a hyperplane of V9 and A "I-1. Now
Vi Na(V^9 ) by E.3.25. As r(G, V) 2: n by 12.1.2.2, m(A) 2: n-1 by E.3.28.3. As
r(G, V) 2: n, E.3.32 says that
f'n-2,A(V) = f'n-2,A:(V) :::'.: U := Nv(V^9 ). (*)
Let Ui :=Vi nu and B := NA(Vi.). Then m(A/B) :::'.: 1, so m(V9 /B) :::'.: 2. Also
[Ui, B] ::::: Vin V^9 = 1 by 12.1.6.2, so that ui ::::: Cv; (B).
Suppose m(A) = n -1. Then by(*), Cv;(b) :::'.: Ui for each b E f3#, so that
ui = Cv; (b) for each such b. However, if b is a transvection in L, then ui is
a hyperplane of Vi, so that B must be the full group of transvections with axis
Ui for i = 1 and 2. This is not the case as V 1 is dual to V 2 • Thus B contains
no transvections, and hence dim(Ui) = n - 2 and B lies in the unipotent radical
Ri centralizing ui. However m(R1 n R 2 ) = 4 and R 1 n R 2 contains a 4-group
R with each member of R# a transvection, so as B contains no transvections,
m(B) :::; m((R1nR2)/ R) = 2. Thus as m(B) 2: n-2, we conclude n = 4 and A> B,
so Cu 1 (A) = 1 and hence U1 is faithful on V9. But U 1 centralizes the subspace
B of codimension 2 in V^9 ; this forces U 1 to induces a group of transvections on