1.3. THE SET 8* (G, T) OF SOLVABLE UNIQUENESS SUBGROUPS OF G 511"K" in that result, so that P n C1(Z(f'))^00 is cyclic oforder p dividing q-o. This
contradicts the irreducible action of NT (P) on P /CJ! ( P). Suppose case A.3.15.2
holds. Then L ~ (8)L3(p) and P ~ Epz; here the parabolic N1(P) induces 8L 2 (p)
on P, and in particular the action of T on P is irreducible. This case appears as our
conclusion (2)-using I.1.3 to see that the only cover of L 3 (p) is 8L 3 (p). In cases
(1), (4), (6), and (7) of A.3.15 Pis cyclic, whereas Pis noncyclic, so those cases do
not arise here. Thus it remains to consider the cases in A.3.15.5, with L of Lie type
over F 2 n with n > 1, As P ::::; L, Plies in a Cartan subgroup of L by that result,
so L is of Lie rank at least 2, and hence L is of one of the following Lie types: A 2 ,
B 2 , G 2 ,^3 D4, or^2 F4. As in an earlier case, AutT(P) is isomorphic to a subgroup
of Out(L). In the last three types, OutT(L) consists only of field automorphisms;
so as P is a p-group, NT(P) normalizes each subgroup of P, contradicting the
irreducible action of NT(P) on P/CJ!(P). If Lis 8p4(2n) then Out(L) is cyclic as
n > l; cf. 16.l.4 and its underlying reference. So as NT(P) is irreducible, n is
even and hence conclusion (3) holds. Finally if Lis (8)L 3 (2n) then OutT(L) is the
product of groups generated by a field automorphism and a graph automorphism
of order 2. However the field automorphism acts on each subgroup of P as above,
and any automorphism of P of order 2 is not irreducible on P, so NT(P) is not
irreducible on P. This eliminates (8)L 3 (2n), completing the proof for p > 3.
We have reduced to the case p = 3. Here a priori L/Z(L) can be any group
appearing in the conclusion of Theorem C. To eliminate the various possible cases,
ordinarily we first apply the restriction m3(L) = 2 (as P is noncyclic), and then
the restriction PT= TP; a final sieve is provided by the irreducibility of NT(P)
on P/<I!(P).
Thus from the cases in conclusion (2) of Theorem C: We do not have L ~ L 2 (qe)
for q > 3, as m 3 (L) = 2, and L is not L 2 (3^2 ) ~ A 6 as PT = TP. The latter
argument eliminates U 3 (3); while L ~ L3(3) appears in conclusion (2). The groups
Lg(q) for q > 3 are eliminated when q = -o mod 3 since m3(L) = 2; and when
q = o mod 3, since PT= TP and NT(P) is irreducible on P/<I!(P).
We next turn to conclusion (1) of Theorem C: A 5 is eliminated as m 3 (L) = 2,
and A 6 is impossible since PT = T P as just noted. In A1 there is indeed a subgroup
PT= TP ~ Z 2 /(A4 x A3); but even in Aut(A1) = 81, we see that 84 x 83 fails
the requirement NT(P) irreducible on P /CJ!(P). Finally A 8 ~ L4(2) appears in
conclusion (4) of our proposition, as do the groups L4(2) and Ls(2) arising in
conclusion (4) of Theorem C.
In conclusion (3) of Theorem C, L/Z(L) is of Lie type in characteristic 2. Then
as P ::::; L and PT = TP, P is contained in a proper parabolic of L, and unless
possibly Lis defined over F 2 , we have Pin the Borel subgroup N1(f'nL). The case
where P is contained in a Borel subgroup was treated above among the embeddings
in A.3.15. In the case where Lis defined over F 2 , proper parabolics have 3-rank at
most 1, contradicting P noncyclic.
·This leaves only conclusion (5) of Theorem C, where L/Z(L) is sporadic. Notice
the case L ~ M 11 appears in conclusion ( 4) of our lemma, while L is not J1 as
m 3 (L) = 2. In the other cases, we use [Asc86b] to see that PT= TP rules out
all but M 23 , M 2 4, J 2 , ]4-which contain 2-groups extended by GL2(4), 83 x L3(2),
Z 2 /(8 3 x Z 3 ), 8s x L3(2), respectively. In these cases (even in Aut(J2)) NT(P) is
not irreducible on P/CJ!(P).
This completes the proof of 1.3.4. D