724 8. ELIMINATING SHADOWS AND CHARACTERIZING THE J4 EXAMPLEREMARK 8.3.4. The second case of lemma 8.3.3 in fact arises in the shadows of
G = Aut(Ln(2)), n = 6 and 7. In those shadows, His the parabolic determined by
the node(s) complementary to those determining the maximal T-invariant parabolicM. Further w = 1, and U = Cv(R) is the centralizer of aw-offender. In most
earlier cases in this chapter, we were able to use elementary weak closure· arguments
to show that the configuration correponding to a shadow is the unique solution of
the Fundamental Weak Closure Inequality FWCI, and then obtain a contradictionto the fact that Na (U) is an SQTK-group. But here, as in o1:1r treatment of the cases
corresponding to the Fischer groups, we instead use the fact that G is quasithin to
show that Ca(U) :::;: M for suitable subgroups U of V, and then use weak closure
to obtain a contradiction.LEMMA 8.3.5. Na(W 0 (S, V)):::;: M 2: Ca(C1(S, V)).
PROOF. By 8.3.3, W 1 (S, V) = W1(Q, V). As Wo(S, V):::;: W1(S, V) and M =
!M(Na(Q)), the lemma follows from E.3.16. D
LEMMA 8.3.6. If HE 'He with SE Syb(H) and n(H) = 1, then H:::;: M ..
PROOF. Since s(G, V) = 3 by 8.3.2, this follows from 8.3.5 using E.3.19 withO; 1 in the roles of "i, j". D
As usual we wish to show that Ca(U) :::;: M for various subspaces U of V.
Usually these subspaces will contain a 2-central involution, so it will be useful to
establish some restrictions on the centralizers of such involutions.
Let z be a generator for Cv(T); in the notation of subsection H.4.4, we maytake z to the involution x1,1 generating V1,10Vi, 1. Set Gz := Ca(z), Mz := CM(z),
X := 02 (CL 0 (z)), and Kz := (X^0 z). Note that 02(XT) = S.
LEMMA 8.3.7. Gz = KzMz, where either
(i) Kz = KK^8 for some K E C(Gz) ands E T - Nr(K) with K/0 2 (K) ~
L2 (p), p prime, or(ii} Kz/02(Kz) ~ L4(2) or L5(2).
PROOF. Let P E Syl3(X); then XE B(G, T), P ~ E 9 , and Autr(P) ~ D 8 •
We apply 1.3.4 to Gz E 'H(XT) in the role of "H". If X <I Gz, define Kz := X;
otherwise 1.3.4 gives X :::;: Kz := (KT), where K E C(Gz) is described in one of
the cases of 1.3.4. Notice case (3) of 1.3.4 is ruled out, as there Autr(P) is cyclic.
Similarly case (2) of 1.3.4 and case (4) with Kz/0 2 (Kz) ~ M 11 are eliminated,
as in those cases Autr(P) contains a quaternion subgroup. We may assume the
lemma fails. Thus neither of the remaining possiblities in case ( 4) of 1.3.4 holds, so
case (1) of 1.3.4 holds and we may take Kz = KK^8 with K/0 2 (K) ~ L 2 (2n) and
n 2: 4 even, asp= 3 and L 2 (4) ~ L2(5).
Note in either case that Kz :::) Gz. Set Yz := CaJX/02(X)). As TE Syb(G)
acts on X, TnYz E Syb(Yz), so by A.4.2.4, S = TnYz. If K/0 2 (K) ~ L 2 (2n), Xis
characteristic in NKz(TnKz) and TnKz = 02(Kz)02(X)), so by Sylow's Theorem
xNa(Kz) = xKz. This holds trivially if Kz = x. Hence by a Frattini Argument,
Gz = KzNaz (X) = KzNaz (Yz). Then as S E Syb(Yz), Gz = KzYzNaz (S)
by another Frattini Argument. As J(T) :::;: Q :::; S, Na(S) :::;: M by 3.2.10.8, so
Gz = KzYzMz. Next Yz = XY, where Y := 03 (Yz) is a 3'-group as m3(Gz) = 2.
As SE Syl2(Yz), SE Syl2(YS).