888 13. MID-SIZE GROUPS OVER F2We check that Hypothesis C.6.2 is satisfied, with L, B, T, LT, Na(B) in the roles
of "L, R, TH, H, A": For example C.6.2.3 holds, since M = !M(LT). The only
part of Hypothesis C.6.2 which is not evident is that Q := 02(LB) :S B, and
this was established during the proof of C.1.37 using Baumann's Argument B.2.18.
Thus we may apply C.6.3.1 to conclude that there exists x E Na(B) with vx i Q.
Now reversing the roles of V and vx if necessary, we may assume that m(ifx) ;:::
m(V/Cv(Vx)). Further since T contains no strong FF*-offenders on V by B.4.6.13,
B.2.4.3 says V :::; B, so that vx :S B :::; R 1. Also by B.1.4.6, ifx E P(R1, V); then by
parts (13) and (3) of B.4.6, we have the hypotheses of B.2.20, so ifx = J(R1) = B
is the unique member f3 of P(R 1 , V), and m(V/Cv(Vx)) = m(B).
Next since Cv(L) = 1by13.3.2.3,
m(V/Cv(Vx)) = 3 = m(B) = m(B/Cs(V)) = m(B/Cs(Vx)) = m(V)/2.
Therefore as L :=:; (f3,f3^1 ) for suitable l EL, L :=:; (Vx, vxz, V), and
m(Q/(CQ(Vx) n CQ(Vx^1 )):::; 2m(B/Cs(Vx)) = m(V).Hence Q = V x CB ( (Vx, vxz)) = V x CB ( L), and in particular vx centralizes
Cs(L). Also E 8 ~ [V, vx] :S V n vx, so as m(Vx) = 3, Cvx(V) = V n vx. Then
[VxCs(L)[ = [VCs(L)[ = [Cs(V)[ and hence Cs(Vx) = vx x Cs(L). Thereforeas x E Na(B),
<P(Cs(L)) = <P(Cs(Vx)) = <P(Cs(V))x = <P(Cs(L)r.Thus if <P(Cs(L) =I- 1, then x E Na(<P(Cs(L)) :::; M = !M(LT); but then as
M = Na(L), vx :::; 02 (£) :::; 02 (LB) = Q, contrary to the choice of x. Therefore
<J?(Cs(L)) = 1, so that Cs(V) = Q is elementary abelian; and then A(B) =
{Q, Qx} is of order 2 by B.2.21 using B.4.6.6. Hence 02 (Na(B)) :S Na(Q) :S
M, and then Na(B) = 02 (Na(B))T :::; M, contrary to our assumption. This
contradiction completes the proof of (2), and hence of 13.3.10. D
LEMMA 13.3.11. Assume L ~ A5. Then
(1) For each v EV#, {U E v^0 : v EU}= v^0 v.
(2) Vl = Vz° n V and Vl = vp n V.
(3) Vis the unique member of v^0 containing Vi.
(4) v^02 = {U E v^0 : V2:::; U}.(5) If g E G with [Vi, Vi] = 1, then [V, VB] = 1.
PROOF. Part (1) follows from 13.2.6.2 and A.1.7.1. As VkL, k = 2,3, are the
unique classes of subgroups of V of rank k containing a unique singular point,13.2.6.2 also implies (2). Then (2) and A.1.7.1 imply (4), as well as the analogous
statement for Vi and G3. Thus as G3 :S Mv by 13.2.3.2, (3) holds. If [Vi, Vil = 1,
then by (3), Vi acts on V. Therefore as CM"v(Vi) = 1, V :S Ca(Vi) :S Na(VB)again using (3), so that V :S. Ca(VB) again using CMv(V3) = 1. Thus (5) holds. D
LEMMA 13.3.12. Assume L ~ U 3 (3). Then
(1) s(G, V) > 1.
(2) If U :S V with Ca(U) i M, then U is totally isotropic. Hence r(G, V);::: 3.
(3) If r(G, V) = 3, then Ca(Vi) i M.
(4) If g E G with 1 =I-[V, VB]:::; VnVB, then VnVB = [V, VB]= Cv(VB) E vp,
and we may take g E Ca(V n VB), so that Ca(Vi) i M.