998 14. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Lf(G, T) IS EMPTY
02 (Gd) ::::; 8d, and Ai = 02 (Kd8d), so we conclude that (d) = 02(Gd)·. Let
Gd:= Gd/(d). Next Kd E .C(Gd,8d) and JTd: SdJ::::; 2, so Kd::::; Ld E C(Gd) by
1.2.5. As 02 (Gd) = (d), Gd rf_ 1-le, so Ld is quasisimple by 1.2.11 applied with V,
Gd in the roles of "U, H". As the hypotheses of 1.1.6 are satisfied with Gd in the
role of "H", Ld is described in 1.1.5.3. As C 0 d(z) has a subgroup of index at most
2 isomorphic to (8 3 x 83 )/Q§, we have a contradiction to the 2-local structure of
the groups on that list. D
LEMMA 14.3.21. (1) If JT: RUJ = 2, then there exist involutions in T - RU.
(2) No involution in T-RQ is in z^0.
{3) All involutions in RU are in z^0.
PROOF. First RU E 8yl 2 (K) by 14.3.19.8, and K is transitive on A#, while
all involutions in K - A are fused into At, so (3) holds.
. Assume JT : RUJ = 2. As I 2 =LU by G.2.3.2 and Iz/ R ~ 83, LT/ R ~ 83 x Z2.
Further for X of order 3 in fz, CR(X) = 1. Thus Co 2 (LT)(X) = (tx) with tx an
involution in T - RU, proving (1).
It remains to prove (2). So suppose some t E T - RQ is of the form t = zY
for some y E G. Let It := Cr 2 (t), Rt := R(t), and Rt := Rt/V. By 14.3.18,
An At = V, and A, At are the maximal elementary abelian subgroups of R, so that
V = fh([R, t]);::::: D 1 (CR(t)) and R is transitive on [A+, t+]t+; hence
(*) Each coset of V in [R, t] (t) not contained in [R, t] contains a conjugate oft.
We claim that z E QY. First consider the case where [V, t] = 1. Here Rt =
Cr 2 R.(V) :::l I2Rt. Further by (*), each element of [R, t]t is an involution, so
that t inverts [R, t]; hence CR(t) = V and R is transitive on [R, t]t. Thus R is
transitive on the involutions in Rt, so that It/CR(t) ~ 83. As CR(t) = V, we
conclude It ~ 84. Therefore V = [V, 02 (It)] ::::; UY. In particular, z E QY, as
claimed. Now consider the case where [V, t] =I-1. Then by Exercise 2.8 in [Asc94],
R is transitive on involutions in Rt and JCR(t)J = 8, so since D 1 (CR(t)) ::::; V, weconclude CR(t) ~ Q 8. Then as H/Q has no Q 8 -subgroup, z E QY, completing the
proof of the claim.By the claim, z E QY. Thus t E <J? (Guy ( z)). This is a contradiction as t tf:-
RQ02 ( H) which is of index 2 in H. D
LEMMA 14.3.22. D = Z, so U = Q.
PROOF. Notice by 14.3.19.1 that U = Q if D = Z. So we assume Z < D, and
will derive a contradiction.
As A = Ca(A) by 14.3.20, JDJ ::::; 4 by 14.3.19.6. So JDJ = 4, and we taked E D - Z. By 14.3.19.10, CAut(K)(U) ::::; (a,/3) where (a)A = CAut(K)(A) and
(/])At= CAut(K)(At). Thus d =I-a or f3 as A is self-centralizing in G, sod induces
a/3 on K, and hence D = (d) ~ Z4.
As D ::::! H = Ca(d^2 ), Dis a TI-set in G. Then the standard result I.7.5 from
the theory of TI-sets says that X := (DG n T) is abelian. Now Lis transitive on
V# and D::::; CT(V) = 02(LT), so V::::; D1((DL n T)) ::::; X. Then X::::; CT(V)
so Xis weakly closed in 02 (LT), and hence X :::l LT. Then as M = !M(LT),
Na(X) = NM(X) = LNHnM(X) using 14.3.7. As Xis abelian and weakly closed,
we may apply Burnside's Fusion Lemma A.1.35 to conclude n° n T = nNa(X) =
DL is of order 3.