11.1. THE SUBGROUPS Na(V;) FORT-INVARIANT SUBSPACES V; OF V 763so m3(I) = 2. Also .C(G, T) n I contains two members £ 2 , Ko with L 2 X/0 2 (L 2 ) ~
KoX/02(Ko) ~ GL2(4); inspecting the list of A.3.14, we conclude I/0 2 (!) ~
SL3(4), Sp4(4), or G2(4) and I= (Ko, £ 2 ). Furthermore 02 (H) = (K 0 , £ 2 , X) ~I,
so I= 02 (HT) and HT= IT.Suppose first that L ~ SL3(4); we must eliminate this case as part of our proof
that (3) holds. This subcase will require approximately a page of argument.
First
X1 =X2,
and X1 is Sylow in 02,z(L). Thus I/0 2 (I) is not SL 3 (4) or else X 1 = X 2 =Cx(L2/02(L2)) = Cx(Ko/02(Ko)), which is not the case in K X ~ PGL3(4).
In the remaining cases the subgroup X 1 = X 2 of the Cartan group X of I is
inverted by a 2-element projecting on the center of the Weyl group (D 8 or D 12 ) of
I/02(I), so this element is not in L2X. Thus N1(X2) i. L2X =In M. Therefore
as X1 = X2, Gx 1 := Na(X1) i. M.
Let Lx 1 := NL(Xi)°'^0 , so that L = 02(L1)Lx 1 and X1 ~ Lx 1. We now show
that it suffices to prove Qx 1 := [02(Lx 1 ), Lx 1 ] i=-1: For then 02(LxJ i=-1, so that
Theorem 4.2.13 says M = !M(LxJ. Then as Gx 1 "i. M by the previous paragraph,
02(GxJ = 1. Next as Qx 1 i=-1, Lx 1 has a 2-chief section of rank at least 6, so
m2(GxJ ::::'.'. m(QxJ > 3. Therefore mp(Op(GxJ) ~ 2 for each odd p by A.1.28,
so Lx 1 ~ Cx 1 := Ca 1 (O(F(GxJ)). As 02(Gx 1 ) = 1, F*(Cx 1 ) = EZ(Cx 1 ),
where E := E(GxJ). Further using (1) of Theorem A (A.2.1), IJGx1 I ~ 3 for
each component J of G Xu so G')( 1 normalizes J. Hence E = C'X 1 as J satisfies the
Schreier Conjecture. Then Lx 1 ~ E, so that Qx 1 projects nontrivially on some
component Kx 1 of Gx 1. As G is quasithin, m2,3(E) = 2, so Kx 1 is the unique
component not centralized by Lx 1 , and hence Lx 1 ~ Kx 1 , so X1 ~ Z(KxJ.
However Kx 1 /Z(KxJ appears in Theorem B (A.2.2), and inspecting such groups
for a 2-local containing a subgroup L with L/0 2 (L) ~ £ 3 (4) and [0 2 (L), L] i=-1,
we conclude Kx 1 /Z(Kx 1 ) ~ J4. This is contradiction as X1 ~ Z(KxJ but the
multiplier of J4 is trivial. This completes the proof that to eliminate L/0 2 (£) ~
SL3(q), it is sufficient to show Qx 1 i=-1.
So we assume Qx 1 = 1, and it remains to derive a contradiction. We set up the
apparatus to apply lemma G.2.5. Set U := (V^01 ) and G 1 := Gi/V1. It is straight-
forward to check that Hypothesis G.2.1 is satisfied, with Na(Vi), 02 (NL(V1)),
G 1 X 1 T, U, Vin the roles of "G1, Li, H, U, V". Therefore U ~ Z(02(G1)) by
G.2.2.
Let P be a Sylow 3-subgroup of KX containing X with XK = Z(P), so that
P ~ 31+^2. As Z(P) = XK = X n £ 1 is nontrivial on V, Pis faithful on U; so as
P ~ 31+^2 , 1 # [Cu(X1), X] = [Cu(X1), XK]· If Y := (Cu(X1), X] ~ 02(LT), then
Lx 1 = (X~x^1 ) is nontrivial on 02(Lx 1 ), which we saw above suffices; so we may
assume Yi. 02 (LT). In particular U "i. 02(LT) so the hypotheses of G.2.5 are also
satisfied. Thus by G.2.5.1, Ri = U02(LT). Furthermore by G.2.5.2, L ~ J ~LT,
where J plays the role of "]" in G.2.5-with the structure of 02(J) described
in detail in the remaining parts of G.2.5. Let L+ := N Li (X1)^00 • As 1 = Qxu