12.2. GROUPS OVER F2, AND THE CASE V A TI-SET IN G 805PROOF. Let R := Cr(U). By a Frattini Argument, R = UNR(P), and then
R = U x Cr(PU). Also [Cr(PU), v] :S .Cv(PU) = 1, so Cr(PU) = Cr(I) and
R = U x Cr(I). Thus Cr(UV) = CR(V) = (Un V) x Cr(I). Next from the
structure of AutaL(U) (I /U), Cr(U n V) = VCr(U), so Cr(U n V) =UV x Cr(I).
Next Q :S Cr(UnV), so by the previous paragraph, Q = Cr(I) xV = VCQ(U).
By 12.2.24, Lis an Ln(2)-block for 3 :S n :S 5, so by C.1.13, m(Q/VCr(L)) :S
m(H^1 (L/02(L), V)). If n of. 3, then H^1 (L, V) = 0, by (6) and (8) of I.1.6 so that
Q = V x Cr(L) If n = 3, the same conclusion holds since Q = VCQ(U) and tJ isof elementary abelian of order 4 in L, ruling out the indecomposable in B.4.8.2.
Now [Cr(L), U] :S Cu(L) = 1, so Cr(L) :S R. Thus CQ(U) = Cr(L) x
Cv(U) = Cr(L) x (Un V), so as Cr(U n V) = UQ, we conclude R = CuQ(U) =
UCQ(U) = U x Cr(L). We have shown:R = Cr(U) = U x Cr(I) = U x Cr(L). (*)
Next LT and I act on To:= Cr(L) n Cr(I), so if T 0 # 1 then I::; Na(T 0 )::; M =
!M(LT), contrary to I 1:. M. Therefore T 0 = 1. But by(*), (Cr(L)) = (R) =
(Cr(I)), so (R) :S To= 1, and hence R is elementary abelian.
From 12.2.20, T = (T n L)Q, and we saw Q = V x Cr(L) with <I>(Cr(L)) = 1,
so T = (T n L) x Cr(L) and Z = Cv(T) x Cr(L). So as \Cv(T)\ = 2, Cr(L)is a hyperplane of Z. Next as Cr(L) :S Z, from (*) we see that [Cr(I), T] ::;
[U, T] n Cr(I) :S Cu(I) = 1; thus Cr(I) :S Z by (*) since R is elementary abelian.
Then Cr(I) is also a hyperplane of Z, since \Cr(L)\ = \Cr(I)\ by (*). Hence as
To =· 1, \Cr(I)\ = \Cr(L)\ :S 2. Thus \R : U\ :S 2 in view of(*). Also we saw
Cr(U n V) =UV x Cr(I), with J(UV) = U, so R = J(Cr(U n V)) ::::1 YT, since
YT acts on Cr(U n V) by an observation just before 12.2.25; in particular,Y :S Na(R).
We suppose for the moment that Cr(L) = 1. Then Q = V, so 02 (M) :S V
by A.1.6. Therefore as L ::::1 M, V = 02 (M) = F*(M). Thus T :S L and
M = LCM(L/V) by 12.2.20. Then as EndL(V) = F2, CM(L/V) = CM(V) = V,
so M = L. Thus (1) will hold once we show that Cr(L) = 1. Also Cr(L) = 1implies U = Cr(U) by (*). But Na(U) E He by 1.1.4.6, so that Ca(U) E He
by 1.1.3.1, so that Ca(U) = U. Thus U = Ca(U) also follows once we establish
Cr(L) = 1.
Assume first that n > 3. Then Y E £1(G, T), so Y :S YR E C(Na(R)) by
1.2.4, with YR E £(G,T). Then YR :S Yo E £j(G,T) by 1.2.9. If Yo/02(Yo)
is quasisimple, then applying Theorem 12.2.2.3 to restrict the list of A.3.12, we
conclude that either Yo/02(Yo) S:! Lm(2) for some n - 1 :S m :S 5, or n = 4 and
Yo/0 2 ,3(Yo) S:! A1. If Y 0 /0 2 (Yo) is not quasisimple, then from A.3.12, n = 4 and
Yo/02(Yo) S:! SL2(7)/E49. As Y :S YR :S Yo, we conclude that either YR/02(YR) S:!
Lk(2) with n - 1 :S k :S m :S 5; or n = 4 and either YR/02,3(YR) S:! A1, orYR/02(YR) S:! SL2(7)/E49. However YR :S Na(R), so that YR/RS:! Ln-1(2) is a
T-invariant subgroup of YRR/ R; so we conclude that either Y = YR ::::J Na(R),
or n = 4 and YRR/ RS:! A1. Assume this last case holds. We showed \R : U\ :S 2,
so m(R) :S 7. Then as Y has two isomorphic 3-dimensional composition factors
on U, we conclude that U = [R, YR] is the 6-dimensional permutation module for
YRR/R. This is impossible, as in the A 7 -module, U/V is dual to Vas a Y-module.
This contradiction shows that Y =YR ::::1 Na(R). Then using (*), Cr(L) =
CR(Y) ::::J Na(R). Now I normalizes Ras R = Crr(U), so if Cr(L) # 1 then