1078 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN L:r(G, T) IS EMPTYTheorem 14.7.63, U is the £ 2 (4)-module for H*. By 14.7.74.2, QH = 02(K), so by
14.7.73.3, QH/U is a 6-dimensional indecomposable for H*. Thus hypothesis (Ru2)
is satisfied. Therefore G is of type Ru, completing the proof of the Theorem. D
14.8. The QTKE-groups with Lr(G, T) =/= 0
We now come to a major watershed in this work: We complete the treatmentof the case where L f ( G, T) is nonempty. We begin with the following preliminary
result:THEOREM 14.8.1. Assume Hypothesis 13.3.1. Then one of the following holds:
(1) L/02(L) ~ A6 and G ~ Sp5(2) or U4(3).
{2) L/0 2 (L) ~ A5 and G ~ U4(2) or L4(3).
{3) L/0 2 (L) ~ £3(2) and G ~ Sp5(2), G2(3), HS, or Ru.PROOF. First by 13.3.2.1, L/0 2 (L) ~ A5, £3(2), A5, A5, or G2(2)'. By Theo-
rem 13.3.16, L/0 2 (L) is not G 2 (2)'. If L/0 2 (£) ~ A 5 , then (2) holds by Theorem
13.6.1. If L/02(L) is A6 or A 6 , then G is Sp5(2) or U4(3) by Theorem 13.8.1, so
(1) holds. This leaves the case where L/0 2 (L) ~ £ 3 (2). Then G is not U4(3), as in
that case there is no L E .C(G, T) with L/02(L) ~ £3(2). Further if G ~ Sp5(2),then (3) holds, so we may assume G is not Sp 6 (2). Therefore Hypothesis 14.3.1.1
is satisfied. Let U := (V).^01 ). If U is nonabelian then G is G 2 (3) or HS by Theo-
rem 14.4.14, so that (3) holds. Thus we may assume U is abelian. Then Theorem
14.7.75 shows that G ~Ru, so that (3) holds, completing the proof. D
We can now easily deduce our main result Theorem D (14.8.2) below from Theo-
rem 14.8.1. Theorem 14.8.1 assumes Hypothesis 13.3.1, and some major reductions
are concealed in Hypothesis 13.3.1, so we briefly recapitulate those reductions; they
take place in the proof of 13.3.2. In Hypothesis 13.3.l we assume that Lt( G, T) =/= 0.This rules out the groups in Theorem 2.1.1, so that IM(T)I > 1, and allows us to
appeal to the theory based on Theorem 2.1.1. The groups excluded in Hypothesis
13.1.1 are also excluded in Hypothesis 13.3.1, so we are able to apply Theorem
13.1.7 to conclude that K/0 2 (K) is quasisimple for each KE L1(G, T). By 1.2.9,
Lj(G, T) =/= 0, and we pick LE Lj(G, T). In particular, Hypothesis 12.2.l is satis-. fied. The proof of Theorem 12.2.2 discusses how previous work leads to the groups
in conclusions (1) and (2) of 12.2.2; Hypothesis 12.2.3 excludes these groups, but
they are also excluded in Hypothesis 13.3.1, so Hypothesis 12.2.3 is also satisfied.
This allows us to appeal to the work in chapter 12 which restricts the choice for
the pair L, Vin the Fundamental Setup to those listed in 13.3.2.THEOREM 14.8.2 (Theorem D). Assume that G is a simple QTKE-group, with
TE Syl2(G), and L1(G, T) =/= 0. Then one of the following holds:
(1) G is a group of Lie type over F2n, n > 1, of Lie rank 2, but G ~ U5(2n)
only for n = 2.
{2) G is £4(2), £5(2); Ag, M22, M23, M24, He, or J4.{3) G is Sp5(2), U4(2), L4,(3), G2(3), HS, or Ru.
PROOF. Since the groups excluded in parts (2) and (3) of Hypothesis 13.3.1 ap-pear as conclusions in Theorem D, we may assume that parts (1)-(3) of Hypothesis
13.3.l are satisfied. Now choose LE Lj(G, T) with L/0 2 (L) not A5 if possible.