1194 i6. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTICChoose g as in 16.4.2.4; that is, so that Nr(K') :::; TB. Then R :::; T/j :::; K' and
RE Syb(K'), so R = Tf::, establishing (3). D16.5. Identifying Ji, and obtaining the final contradiction
In this final section, we first see that G ~Ji when L/Z(L) is a Bender group.
Then we eliminate all other possiblities for L appearing in (E2), to establish our
main result Theorem 16.5.14.Recall R is faithful on L by 16.4.2.1 and H* = H/K. Set U := Di(R); as
K' E b..o,U:::; KL.
Let u denote an element of U#, and set Uc:= Di(Tc). By 16.4.11.3, if we choose
gas in 16.4.2.4, then R = T/j and hence U = U/j; in particular, Uc E U^0.PROPOSITION 16.5.1. If L* is a Bender group, then G ~Ji.
. PROOF. By hypothesis, L* ~ L2(2n), U3(2n), or Sz(2n). Set UL := Di(TL)·
Then there is X :::; NL(TL) with X ~ Z 2 n_ 1 and X regular on Ul#· Now either
Z(L) = 1, or from 16.1.2, L* ~ Sz(8), with U[, = D 1 (Tl) from I.2.2.4. Thus as
U:::; KL,
U:::; UcUL =: V = Uc x [V, X],
with X regular on [V, X]# and Uc = Cv(X).
We claim that there is g E M := Na(V) with KB = K'. Suppose first that
L* ~ L 2 (2n) or Sz(2n). Recall by 16.4.6 that if L ~ L 2 (4) ~ L 2 (5), then no
involution in U induces an outer automorphism on L, so that V = (U^0 nT). In the
remaining cases, no outer automorphism of L is an FF* -offender on UL, so that
V = J(T). Thus in each case, V is weakly closed in T with respect to G, so by
Burnside's Fusion Lemma A.1.35, M controls fusion in V, and hence there is g EM
with U/j = U and hence KB = K', as claimed. So we may assume L ~ U 3 (2n).
Choose gas in 16.4.2.4, so that Nr(K') :::; TB, and as observed earlier, R = T/j
and U = U/j. Fix u EU#. Set Yu:= 03 (Lu) if n f:. 3 and Yu:= Lu if n = 3; then
Yu :::; L' by 16.4.8. But in either case TL :::; Yu, so TL :::; L' and hence TL = Tl as
Nr(K'):::; TB. Thus UL:::::: Uf, so as U = U/j,
VB= (ULUc)B = ULU/j = ULU = V,
completing the proof of the claim.
Pick gas in the claim, and set Af+ := M/Ca(V). As Uc = V n Kand K
is tightly embedded in G, Uc is a TI-set in V under the action of M by I.7.2.3.
Further x+ = NL(V)+ :SI (MnH)+ = NM+(Uc) since Na(Tc):::; H by 16.4.2.5,
and x+ is regular on [V,X]#, while T+ E Syl 2 (M+) acts on x+. So we have a
Goldschmidt-O'Nan pair in the sense of Definition 14.1 of [GLS96]; hence we may
apply O'Nan's lemma Proposition 14.2 in [GLS96] with Af+, x+, V in the roles
of "X, Y, V": Observe that conclusion (i) of that result does not hold, since there
M normalizes Uc-whereas here g EM - Na(Uc). Similarly conclusions (ii) and