13.8. FINISHING THE TREATMENT OF A 6 957Thus L/02(L) ~ A5, so that Lo and Li,+ = X are the two T-invariant sub-
groups of 3-rank 1 in Li. As usual Ki. M by 13.8.12.1, so that K does not act on
·Lo in view of 13.2.2.9. Then CL 1 (K) = X rather than L 0 , so that Lo s K. Now
X = [X, t] fort E T n L ::::; Cr(Lo/0 2 (L 0 )), so [K, t*] = 1. Also T acts on L 0 ,
and hence is trivial on the Dynkin diagram of K, so H ~ L 3 (2) x 83 • As earlier,
UH = [UH, X], so an H-chief factor W in UH is the tensor product of natural mod-
ules for the factors, as usual using B.4.5 and the fact that u; E Q(H*, UH). As Lo
centralizes Vs, L 0 T* is the stabilizer of a point in these natural modules. Then as
v; ::::; 02(LJ:T*) and v; is quadratic on UH, v; is of order 2, contrary to 13.8.18.4.
So (1) is established.By (1), Li is contained in each K E C(H), so there is a unique K E C(H).
Then by 13.8.14, K = F(H), Bo (3) holds as Out(K) is a 2-group for each of
the groups listed in (2); if K ~ L3(2) that H ~ Aut(L 3 (2)) by 13.8.19. Part (4)
follows from (1) and 13.7.5.2. DLet W be a proper H-submodule of UH and set UH := UH /W. As UH = (V JI)
and Li is irreducible on V3 ~ E4, it follows that V3 ~ E4 is LiT-isomorphic to V3.
By 13.8.21, UH = [UH, K] and K = F(H) is simple, so that H is faithful on
UH..
LEMMA 13.8.22. Assume K is nontrivial on W. Then H* is faithful on W,
case {2) of 13.8.8 holds, and either.
{1) Ai s W, u; contains an FF"-offender on the FF-module UH, and either
u; contains a strong FF"-offender on UH, or w::::; DH and [W, u;J =Ai.
{2) Ai i. W, u; contains an FF"-offender on the FF-module W, and either
u; contains a strong FF"-offender on w, or UH= WDH and [UH, U'Y] =Ai, sothat Ai SUH.
PROOF. As K* is nontrivial on Wand K* = F*(H*) is simple, H* is faithful
on W. As H* is also faithful on UH, no member of H* induces a transvection on
UH, so case (2) of 13.8.8 holds.
Suppose Ai s W. Then using F.9.13.6, [DH, U 7 ] S Ai= 1, so DH< UH andm(UH/CoH(U 7 )) S m(UH/DH)::::; m(UH/DH) S m(U;),
so by B.1.4.4, u; contains an FF*-offender on the FF-module UH. Indeed ei-
.ther u; contains a strong FF*-offender, or all inequalities are equalities, so thatm(UH/DH) = m(UH/DH), and hence W S DH. In the latter case, [W,U 7 ] S
[DH, U 7 ] =Ai, so that (1) holds.
So assume Ai i. W. Then [DH n W, U 7 ] s W n Ai= 1, so
m(W/Cw(U 7 )) S m(W/(DH n W)) S m(UH/DH)::::; m(U;)
since case (2) of 13.8.8 holds. So by B.1.4.4, u; contains an FF*-offender on
the FF-module W. Further if u; does not contain a strong FF*-offender, then
all inequalities are equalities, so that m(W/(DH n W)) = m(UH/DH) and hence