10.2. WEAK CLOSURE PARAMETERS AND CONTROL OF CENTRALIZERS 747CE(BK) = CE(Di) = E2 are of rank n. Thus m = n by (*), and D 1 = BK as
BK::; Di and IDil = 2n -1=2m -1 = IBKI·
Next recall from E.2.3.2 that S normalizes Land K. Therefore as D 1 =BK,Si := [S, D1] ::; L n K. Since either L or K is a block, and Cz(L 0 ) = Cz(H) = 1,
we conclude Si is special of order 2an; then it follows that both Land Kare blocks,
with 02(L) and 02 (K) of rank 2n, and Si is Sylow in both Land K.
Next Li and L2 commute by C.1.9, so [Si, D2] ::; [Li, D 2 ] = l. So as D 2
centralizes Si E Syb(K), D 2 centralizes K from the structure of Aut(K). Similarly
S2 := [S, D2] E Syl2(L2) and S2 centralizes K. But S2 = Sf for t E T - Ti, so
S 2 is Sylow in the block Kt and Kt -j. K. Hence 02 (H) ~ KKt = K x Kt. since
Cvx (K) = 1. Setting Ki := K and K 2 := Kt, SiDi is Borel in both Li and Ki.
Set Mi := Na(SiDi). As L2 centralizes Li, L2Ti ::; Mi, with Ti = T n Mi.
Similarly K2Ti ::; Mi. Embed Ti ::; T+ E Syb(Mi). Recall that S ::; Ti, so
S = Baum(T+), and hence T+::; Na(S)::; M by 10.2.l. If Ti< T+ then T+ is also
Sylow in M, so M = !M(LoT+) by 1.2.7.3. However LoT+ = (L 2 , T+) ::; Mi, so
K2 ::; Mi ::; M, contradicting K2 i. M.
This contradiction shows that Ti = T+ is Sylow in Mi. Hence L 2 ::; L+ E
C(Mi) by 1.2.4. Now K2 = 02 (K2) normalizes L+ by 1.2.1.3, so as D 2 ::; L 2 ::; L+,
also K2 = [K 2 , D 2 ] ::; L+, and hence L 2 < L+. As L2 and K 2 are distinct members
of £( L+, Ti) and both are blocks of type L 2 ( 2n) with trivial centers, we conclude
from A.3.12 that 02(L+) = 1 and L+ ~ (S)La(2n). Now L+ normalizes SiDi,
and so in fact centralizes Si Di since Si is special of order 2an. Therefore for p a
prime divisor of 2n - 1, mp(DiL+) > 2, contradicting Mi an SQTK-group. This
completes the elimination of case (1) of 10.l.l when n(H) > 1, and hence establishes
10.2.3.. D
LEMMA 10.2.4. Assume that L ~ La(2), but case (5) of 10.1.1 does not hold,
so that LTi ~ La(2). Let P be one of the two maximal subgroups of LoT containing
T. Set X := 02 (P), assume HE 1i(XT), and set K := (XH). Then one of the
following holds:
(1)K=X.
(2) K = KiKJ. with Ki E C(H), Ki/0 2 (K1) ~ L2(2m) for some even m or
L2(P) for some odd prime p, ands ET-Nr(K1). · ·
(3) KE C(H) and KT/0 2 (KT) ~ Aut(Lk(2)), k = 4 or 5.
. PROOF. As X E 3(G, T), K is described in 1.3.4 wit]} p = 3. Further
XT/0 2 (XT) ~ Sa wr Z2, which reduces the list to the cases appearing in the
lemma. D
LEMMA 10.2.5. Na(T1)::; M.
PROOF. If J(T) ::; Cr(V) then the lemma follows from 3.2.10.8, so we may
assume J(T) i. Cr(V). Then one of the first three cases of 10.l.1 holds by 10.1.2.l.
If case (1) or (2) of 10.l.1 holds then S ::; Ti by 10.1.2.6, so S = Baum(Ti) and
then Na(Ti) ::; M by 10.2.l.
Thus we may assume case (3) of 10.l.l.3 holds, so L = LT ~ La(2). Let Hi
and H 2 be the two maximal subgroups of L 0 T containing T. Thus Xi:= 02 (Hi) E