i5.3. THE. ELIMINATION OF Mr/CMr(V(Mr)) = Ss wr Z 2 1133of order 2 for i = 1, 2. Further Zi V2 = Zl/2 = (zY+)::::; (ZL) = U 1 , so U 1 = ZiUL.
Then as Zi ::::; Z(S), U1 = ULCu 1 (LS) by B.2.14, and Cu 1 (LS) = Cz 8 (L) =
Cz 8 (Y+) = Zi, so (2) holds.
Next Ji := Ca(Zi) E H+, so S E Sylz(Ji) by 15.3.11.1. Thus L ::::; Li E C(Ji)
by 1.2.4, and A.3.12 says that either L = Li or Li/02(Li) ~ £5(2), M24, or J4.
As S is nontrivial on the Dynkin diagram of L *, it follows that L = Li, and then
(3) folllows from A.3.18.Let K be the S-invariant maximal parabolic of L, and set X := 02 (K) and
P := 02 (XS). Thus XS/ P ~ S 3 wr Z 2 is determined by the end nodes of the
Dynkin diagram of L*. By (1), XS= C1(Zs), so (3) implies (4). Then as Zs ::SJ T,
Tacts on 02 (C1(Zs)) = X and P. Let t E T - S. Then Tacts on UL n Ul,
so if UL n u1-/=- 1, then z::::; UL n Ul, whereas Zs n UL= Zs n V2 = Z2. Thus
UL n Ul = 1, so as UL ::::! XS and Tacts on XS, (5) holds.
Adopt the notation of ( 6) and let P1 := 02 (J). As XS is irreducible on P*,
either Pj ::::; P1, or P =,P1PJ and (6) follows from (5). But in the fo~mer case
Pj = P1, so that (T, £) acts on P1; then as Y+ 1:. CM(V), L ::::; M by 15.3.2.4,
contrary to the choice of L. Thus (6) h.olds. D
We can now complete the proof of Theorem 15.3.25. Let X, W, and P be
defined as in 15.3.32.
Let B be the set of A E A(S) such that A -/=-1, and A is minimal subject
to this property. Choose some A E B. By B.2.5, A E P(I*, UL)· Now B.3.2.6
describes the possible FF-offenders, and the only strong FF-offender is generated
by four transvections; so from B.2.9 we conclude that one of the following holds:
(i) A*~ E 8 is regular on D := {1, ... , 8}.
(ii) A* is generated by a transposition.
(iii) A*= D := ((1, 2), (3, 4), (5, 6), (7, 8)).
(iv) A* =DnL*.In particular in each case, A 1:. P. Further m(A*) = m(UL/CuL(A)) except in
case (iii), where m(A*) = 4 and m(UL/CuL (A))= 3.
Let C := B n Bt fort ET-S. As A 1:. P, AutA(Ul)-/=-1by15.3.32.6, so there
is A+::::; A such that AutA+(Ul) E P*(Autp(Ul), Ul) by B.1.4.4. Then A+ 1:. P
by the previous paragraph applied to u1 in place of u L' so A+ -/=-1 again using
15.3.32.6. Hence by minimality of A, A+ = A. Thus A E C.
Let Xr := XT/0 2 (XT), so that S ~ Ds, and set So:= SnL02(LS). Observe:
(I) In (i), IAI = 2 and So= Ax Z(S).(II) In (ii), IAI = 2 and .A 1:. So.
(III) In (iii), A is the 4-subgroup of S distinct from So.
(IV) In (iv), A= Z(S).
Let B := CA(W), m 0 := m(A), mi := m(A n P), m2 := m(AutAnP(Ul)),
m3 := m(UL/CuL (A)), and m4 := m(U1/Cu1 (A)). Then m(A)::::; mo +mi +m2 +
m(B). Also m(BW) = m(B) + m 3 + m 4. Therefore as m(A) 2: m(BW) since
A E A(S),
mo +mi + m2 2: m3 + m4. (!)
Suppose first that S < T. Then Tis Sylow in GL 2 (3), so T ~ SD16, and hence