878 13. MID-SIZE GROUPS OVER F2If J(T) ::::;: CT(V) = 02 (LT), then (4) is vacuously true. Thus we may suppose
that there is A E A(T) with A-=/= 1; in particular L = [L, J(T)]. Now the hypotheses
of B.2.10.2 (and hence of B.2.10.1) are satisfied with LT, Tin the roles of "G, R",
so P := PT,LT is a nonempty stable subset of P(LT, V) by B.2.10.1. Similarly using
R1 in the role of "R", J(R 1 ) i Q iff Q := PR 1 ,LT is a nonempty stable subset of
P(LT, V). Moreover by definition in B.2.5, J(T) = Jp(T) and J(R1) = JQ(R1).
If Mv ~ A5, then from B.3.4.1, Jp(T) = R2, so L1 = [L1, Jp(T)], and hence
(4) holds. So assume instead that Mv ~ 86. If JQ(R 1 ) = 1, then (4) follows from
B.3.4.2iv, while if JQ(R 1 ) f. 1, then (4) follows from B.3.4.2v. This completes the
proofof ( 4).
For part (7), we observe that the hypotheses of C.1.37 are satisfied with R1 in
the role of "R" and P := L 1 T, except when Mv ~ 86, when we take P := L1,+T.
Thus conclusion (1) or (2) of C.1.37 holds, giving the alternatives of conclusion (7)
of the present result. (Recall that Lis not a A5-block as 02 ,z(L) = CL(V)).
Next we will prove (8) and (9), so we may assume L/0 2 (L) ~ A 6 • Set R :=
02(L 1 T). We claim first that J(R) = J(Q): If Mv = A5 then R = R 1 by 13.2.1,
and by B.3.4.1, J(R1) :SQ :S R1 so that J(R) = J(R1) = J(Q). If Mv = 86, then
by 13.2.1, IR1 : RI = 2 and R = 02(L1)Q ::::;: LQ. But by B.3.4.2v, if J(R) f. 1
then J(R) i L, so again J(R)::::;: Q and J(R) = J(Q), completing the proof of the
claim. Then by B.2.3.5, Baum(R) = Baum(Q), establishing (8).
Recall Lo = 02 (02,z(L)). Therefore L 0 :SI LT, so that Na(Lo) ::::;: M =
!M(LT). Finally Na(L1) = Ca(Li/02(L1))Na(R) by A.4.2 and a Frattini Argu-
ment, so as Na(R) :S Na(J(R)) ::::;: M by (8), and Ca(L1/02(L1)) normalizes Lo
with Na(Lo) ::::;: M, we conclude Na(L 1 ) ::::;: M, completing the proof of (9).
Finally we prove (10). Let S := Baum(T). If J(T) :S CT(V) = 02 (LT), then
using B.2.3.3, J(T) is a nontrivial characteristic subgroup of S normal in LT, so
(I) holds. Thus we may assume there is A E A(T) with 1 f. A E P. If B is a
hyperplane of A with _BL n T CZ. R 2 , then as B E A 1 (T), (III) holds. Thus we
may assume no such B exists. Therefore by B.3.4.2vi, !Al = 2 for each such A.
In particular, J(T) lies in the subgroup R 2 of T generated by transvections, so
Baum(T) = Baum(R 2 ) by B.2.3.5. Observe that we now have the hypothesis of
C.1.37 with R2 in the role of "R" and P := L 2 T, unless LT/0 2 (LT) ~ 86 , when
we take P := L2,+T· Further conclusion (5) of C.1.37 does not hold, as there are
no FF* -offenders with image of order greater than 2, so only conclusions (1) or (2)
of that lemma can hold. In case (1), (I) holds, and in case (2), L is an A 6 -block
(Again Lis not a A5-block as CL(V) = 02 ,z(L)). Further FF*-offenders of order 2
are not strong by B.3.4.2i, so that A( Q) s;;; A(T) by B.2.4.3, and hence (II) holds.
This completes the proof of (10), and of 13.2.2. D
13.2.2. Results on A5. In this subsection we assume n = 5 and establish