15.1. INITIAL REDUCTIONS WHEN .Cf(G, T) IS EMPTY 1101In the remainder of the section, let Y := as' ( G 1 nM). As G is a counterexample
to Theorem 15.1.15, 15.1.23 says we are in case (4), case (6) with p = 3, or case (7)
of 15.1.7, so that Mo is a {2, 3}-group. In particular Y 1:. Mz by 15.1.18.2.LEMMA 15.1.24. (1) Either case (4) of 15.1. 7 holds, or case (6) of 15.1. 7 holds
with p = 3. In particular, Z1:::; V1.
{2) Lis not an Ls(2)-block.(3) For each Yo = 02 (Yo) :::; Y with Yo 1:. Mz, Vz [Vz, Yo] and IYo
Cy 0 (V2)I = 3. In particular Vz =[Vi, Y].
(4) V1 :S: Cs(Y).
PROOF. By construction of Vz in 15.1.16, and since p = 3 when case (6) of
15.1.7 occurs, Y is of order 3 and Vz = [Vz, Y]. Thus (3) follows from 15.1.18.2.
Further from 15.1.16, in cases (4) and (6) of 15.1.7, Vi~ Z1 and Y centralizes V 1 ,
so (4) will follow once we prove (1).To establish (1), we may assume that case (7) of 15.1.7 holds, and we must
produce a contradiction. Let X 0 be the preimage in M of Z(0^2 (M)), R := 02 (Mn
Mc), and X 1 := 02 ((Rx^0 )). We apply 14.1.1.7 to Mc, X 0 in the roles of "M 1 , Yo"
to conclude X1 = Xo = [Xo, R] and [R, Cx 1 (V)] :S: 02(M). As X1 = Xo, Z1 =
[Z 1 ,Xo]. Then as X1 = [X1,R] and [R,Cx 1 (V)] :S: 02(M), there is a subgroup
X 2 of X 1 of order 3 with Z1 = [Z1,X2]. So as ms(Na(Z1)):::; 2, A.3.18 eliminatesthe possibilities in 15.1.22 of 3-rank 2, leaving the case where L is an Ls(2)-block.
Then by 1.2.1.3, X 2 = 02 (X 2 ) normalizes Land as' (G 1 ) =Los' (Ca 1 (L/0 2 (L))).
Therefore as X2 1:. G1 and ms(Na 1 (Z1)) :S: 2, L =as' (G1).Thus to establish both (1) and (2), it suffices to assume L is an Ls(2)-block.
As usual let U(L) := [0 2 (L), L]. By 15.1.22, ML is the parabolic of L centralizing
Zs := 01 (Z(SL)), and Mn Lis the remaining maximal parabolic of Lover SL.
Let Yo := 02 (M n L), so that as L :5: G1, Yo :5: Y but Yo f:. ML; hence Vi =
[Vz, Yo] and Cy 0 (V2) = 02(Yo) by (3). Thus Vz :S: Z(02(Yo)) :S: U(L), so Vz :S:
[Cu(L)(0 2 (Y 0 )), Yo]=: U2. As Lis an Ls(2)-block, U2 is ofrank 2, so Vz = U2 is of
rank 2. This eliminates cases (6) and (7) of 15.1.7, and in particular completes the
proof of conclusions (1) and (4) as mentioned earlier, though not yet of (2). Thus
case (4) of 15.1.7 holds. Further V1:::; Cs(Yo) by (4), and Cs(Yo) = Cs(L) as Lis
an Ls(2)-block, so L centralizes Vi. Since Z1 :S: V1 by (1), Ca(V1) :S: Ca(Z1) = G1,
and hence LE C(Ca(Vi)). Lett ET - Sand X E Syh(YJ); then Xis of order
3, and by our construction of V 1 and Vz in 15.1.16, V1 = [Vi, X] and [Vz, X] = 1.
As L E C(Ca(V 1 )), YJ = 02 (YJ) acts on L by 1.2.1.3, and as X centralizes Vz =[U(L), Y 0 ], X centralizes L, since L is an L3(2)-block. Then as ms(Na(V1)) :S: 2
and X f:. Ca(V 1 ), arguing as above, we conclude that L = as' (Ca(V 1 )). Indeed
as X centralizes L, so does (XYJ) = YJ. Then U(L) :::; Cs(YJ) = Cs(Lt), and
by symmetry, U(L)t centralizes L. Thus (L, T) acts on W := U(L)U(L)t. Then
setting N := Na(W), L :=:; N^00 :=:;Mc by 14.1.6.3, contrary to the choice of L. This
contradiction completes the proof of (2), and hence of the lemma. D
LEMMA 15.1.25. (1) L =as' (G1).
{2) L is not Mn.
{3) V 1 does not centralize L.(4) IS: Cs(Vi)I = 2.
(5) L is not an A1-block.
{6) IT: SI= 2.