12.g. THE FINAL TREATMENT OF Ln(2), n = 4, 5, ON THE NATURAL MODULE 857In particular U+ := {O, e5,6, e5,7, e6,7} ::::; 1f 2 ..L· is T-invariant but does not contain V 2 ,
so by 12.8.11.4, U+ n E = 1. Then by 12.8.11.1, [U+, X] ::::; U+ n E = 1.
Next 02 (C.H(V2)) ~ A4 x Z 3 , so by an earlier observation, Zij -=/= 1. Thus(*) implies m(E) ::::: 3, with equality only if m(X) = 3. By 12.8.11.2, E is totally
isotropic so that m(E) ::::; 3 = m(X). However we showed in the previous paragraph
that X::::; Ct(U+), which is a contradiction as Ct(U+) ~ D 8.
This completes the proof of 12.8.13. D
12.9. The final treatment of Ln(2), n = 4, 5, on the natural module
In this section _we prove:
THEOREM 12.9.1. Assume Hypothesis 12.2.1 with L/0 2 (L) ~ Ln(2), n = 4 or
5. Then n = 4, and G is isomorphic to L 5 (2) or M24.
We recall that the QTKE-groups G ~ L5(2) and M24 appear as conclusions in
Theorem 12.2.13. In proving Theorem 12.9.1 we verify that Hypothesis 12.8.l holds
and apply Theorem 12.2.13 to establish 12.8.3.4. Two groups appear as shadows:
The sporadic group Co3 has a 2-local LE .Cj(G, T) with L ~ L4(2)/E 2 4; but Co3
is neither quasithin nor of even characteristic, in view of the involution centralizerSp6(2)/Z 2 , and is essentially eliminated in 12.9.3 below. Similarly the sporadic
Thompson group F3 contains L E .Cj(G, T) with L ~ L5(2)/ E 2 s, but F3 is not
quasithin in view of the involution centralizer Ag/2^1 +8, and is eliminated in 12.9.4.
Furthermore in many groups of large rank there is LE .C1(G, T) which is not
maximal, but satisfies the rest of the hypothesis of Theorem 12.9.1: namely in many
groups of Lie type over F2, as well as in the sporadic groups F 3 , the Baby Monster,and the Monster. In addition the Conway group Co 2 has a 2-local L not containing
a Sylow group with structure L4(2)/(E 24 x 21 +6). These groups are of course not
quasithin, and the configurations are also eliminated in 12.9.3 and 12.9.4.The proof of Theorem 12.9.1 involves a series ofreductions. Assume G, L afforda counterexample to Theorem 12.9.1, and choose the counterexample so that n = 5
if that choice is possible. Neither Ag nor the groups appearing in conclusions (1)
arid (2) of Theorem 12.2.2 contain LE .Cj(G, T) with L/02(L) ~ Ln(2) for n = 4
or 5. Thus Hypothesis 12.2.3 is satisfied, we can pick Vas in Theorem 12.2.2.3, and
Na(L) =:ME M(T). Then Theorems 12.5.l and 12.6.34 eliminate cases (c) and
(d) of Theorem 12.2.2.3, so we conclude that Vis the natural module for L/0 2 (L).
As G, L affords a counterexample to Theorem 12.9.1, G is neither L 5 (2) nor M 24.Also G is not L 6 (2) as G is quasithin, and G is not Ag as we observed earlier. Thus
Hypothesis 12.8.1 is satisfied, so we can appeal to the results of section 12.8, and
adopt the conventions of Notation 12.8.2 of that section. Recall that G 1 1:. M by
12.8.3.4, so that G1 E 1-iz.
LEMMA 12.9.2. If n = 4, then there is no K E .Cj(G, T) with K/0 2 (K) ~
L5(2), M24, or J4.
PROOF. Assume such a K exists. By Remark 12.2.4, Hypothesis 12.2.1 is
satisfied with Kin the role of "L" and conclusion (3) of Theorem 12.2.2 holds, so
K/0 2 (K) ~ L 5 (2). This is a contradiction as n = 4 by hypothesis, contrary to our
choice of n = 5 if such a choice is possible. D