926 i3. MID-SIZE GROUPS OVER F213.6. Finishing the treatment of A
In this section, we complete the treatment of the case in the Fundamental
Setup where L/0 2 (£) ~As, using assumption (4) in Hypothesis 13.3.1 as discussed
earlier. To do so, we treat the only case remaining after the reduction in the previous
section 13.5. We adopt the notational conventions of section B.3 and Notations
12.2.5 and 13.2.1.We will prove a result summarizing the work of both this section and the
previous section:THEOREM 13.6.1. Assume Hypothesis 13.3.1 with L/02(L) ~As. Then G is
isomorphic to U4(2) or L4(3).
The groups in Theorem 13.6.1 have already appeared as conclusions in Theorem13.5.12.1, under the hypothesis that (Vs°^1 ) is nonabelian; we will prove that there
are no examples in the remaining case. (Indeed, as far as we can tell, there are noteven any shadows.) We assume throughout this section that G is a counterexample
to Theorem 13.6.1, and work toward a contradiction. The contribution from the
previous section 13.5 is:
LEMMA 13.6.2. (V^01 ) is abelian, so VH is abelian for each HE Hz.
PROOF. By Theorem 13.5.12.1, (V 3 °^1 ) is abelian; hence (V^01 ) is abelian by
13.5.10. DLEMMA 13.6.3. (1) Oa(z) 1:. M.
(2) Oz(L) = 1.
PROOF. This follows from 13.3.5.2 and the fact that M = !M(LT). 0
Set Q := 02(LT), S := Baum(T), let v E Vi - Vi, and let Gv := Oa(v) andMv := OM(v). In the notation of section B.3, the generator z of Vi is e 1 , 2 , 3 , 4 , and
we may take v = ei,2· Set Qv := 02(Gv), Gv := Gv/(v), Lv := 02 (0L(v)) and
Vv := (zLv). Then Lv/02(Lv) is of order 3, Vv = (v) X [V,Lv], and Vv = [V,Lv] is
a natural module for Lv/0 2 (Lv), of rank 2." By 13.2.6.1,
Tv := Or(v) E Syb(Gv)·
By 13.2.4.2, S::::; Tv, so it follows from B.2.3 that:
LEMMA 13.6.4. S = Baum(Tv) and J(T) = J(Tv)·
Observe also that there is a E z^0 n Ov(Lv) (e.g., a= ei,3,4,s) and a E Z(Tv)·
LEMMA 13.6.5. NaJS) ::S: Mv = OMv(v).PROOF. First Na(S) ::S: M by 13.2.5, and then Mv :::;· Mv by 12.2.6. 0
LEMMA 13.6.6. F*(Gv) = Qv.
PROOF. Assume that Qv < F*(Gv)· Then by 1.1.4.3, z tf. Qv. By 1.1.6 we
can appeal to lemma 1.1.5 with Gv, Tv, Gi, z in the roles of "H, S, M, z". In
particular F*(OaJz)) = 02 (0aJz)) and z inverts O(Gv)· On the other hand,
z E Vv = (v) x [V, Lv], and [V, Lv] centralizes O(Gv) by A.l.26.1, so z centralizes
O(Gv), and hence O(Gv) = 1. Thus there is a component K of Gv. By 1.1.5.3,
K = [K, z] and K is described in 1.1.5.3. Also Lv = 02 (Lv) normalizes K by
1.2.1.3, so as z E (v)Lv, also K = [K, Lv].