15.1. INITIAL REDUCTIONS WHEN .Cr(G, T) IS EMPTY 1085that Mc :SM. Then since Mc i. M, A.5.6 gives
There is no 1 =f. X::::; V with X =((Zn X)MnM^0 ) :SJ M. (*)
Suppose first that (M1, V1) satisfies case (6) of D.2.17; we will derive a con-
tradiction. For then P := 02 (M 1 ) = P 1 x P 2 x P 3 , with Pj ~ Z 3 for each j,
and V1 = U1 EB U2 EB Ua, where Uj := [V, Pj] is of rank 2 for Pj the preimage of
Pj. As m3(M1) > 1, Mi and V1 are normal in M by the second paragraph of the
proof, so M permutes X := {P1, P2, P3}. Then T fixes some member of X, say P 1 ,
so that Tacts on [V,P 1 ] = U1, and on U := U 2 EB U 3. Thus 1 =I Z 1 :=Zn U 1
and 1 =I Zu := Zn U. So P1 ::::; CG(Z1) ::::; Mc = !M(CG(Z)), and similarlyP2P3::::; CM(Zu)::::; Mc. Then V1 = (Zf1,ZC^2 P^3 ) = ((ZnVi)McnM), contrary to
(*). This completes the proof that no (Mi, Yi) satisfies conclusion (6) of D.2.17.
Next suppose for the moment that (M 1 , Vi) satisfies conclusion (3) of D.2.17; in
this case, we show that M 1 T acts irreducibly on V 1. Again by the second paragraph,
M 1 and V 1 are normal in M. As case (3) of D.2.17 holds, M 1 ~ Z 2 /E 9 , with
m(V1) = 4 and O(M1) inverted in Mi. Thus AutM(V1)::::; NGL(V1)(M1) ~ ot(V1).
Assume now that M 1 T acts reducibly on V1. Then Autr(Vi) ~ Z2 or E 4 , and
in either case Zn V 1 ~ E4 and M = (CM(z) : z E z# n V1), so M ::::; Mc as
Mc= !M(CG(Z)). This contradiction completes the proof that if (Mi, Vi) satisfiescase (3) of D.2.17, then MiT acts irreducibly on 1/i.
Next we introduce a basic case division for the proof of the lemma: Let Zi :=
Zn Wi, and suppose that Wi = (Zfi) for some i, and that either Cv(MJ) =I 1, or
a> 1. Then Cz(Ki) =f. 1, so that Ki ::::; CG(Cz(Ki)) ::::; Mc= !M(CG(Z)). Then
wi is generated by zfi ~ zf1nMc' so as wi :SI M since Ki = (MiM), we have acontradiction to (*). Thus we conclude that either
(i) a= 1 and V =Vo, or
(ii) For each i, (Zfi) < Wi.We first assume that case (i) does not hold; then case (ii) holds, and we will
show that conclusion (6) is satisfied in this case. Choose notation so that W1 =
V 1 EB · · · EB Vb; then b ::::; 2 by paragraph two. As (ii) holds, (Zf^1 ) < W1, so KiT
acts reducibly on W 1 , and hence M 1 Nr(V1) acts reducibly on V1. Now in D.2.17,
M 1 acts reducibly only in case (3), and in case (1) when m(V1) = 4. But earlier
we showed MiNr(V 1 ) is irreducible on V1 in case (3), so Mi ~ 83 with m(V1) = 4,and in particular q(M1, V1) = 2 and AutM(Vi)::::; NGL(V1)(M1) = nt(Vi).
Since (Cv 1 (Nr(V1))M^1 ) < V1, m(Cv 1 (Nr(Vi))) = 1, so IAutr(V1)I > 2. Then
as AutM(V1) ::::; Dt(V1), Autr(V1) ~ E4, so AutM(Vi)) is either Dt(V1) or 83 x Z2.
However in the latter case, M = KiCM(Z1), so as W1 > (Zf^1 ), also W1 > (Zf'1),
contrary to V = (ZM).Therefore AutM(V 1 ) = nt(V1). Now as Mi = Aut(M1), Mo := Ntiil(M1) =
Mix Ctiil 0 (M1), with m 3 (C.M 0 (M1))::::; 1 using A.1.31.1. Supposes> 1. Then by
symmetry, M2 ~ 83, so Mo =Mi x M2 x C.Mo (M1M2), and then O(M1)0(M2) =
031 (Mo) by A.1.31.1. This is a contradiction as AutM(V 1 ) = nt(V 1 ) and [V1, M 2 ] =- Therefore s = a = 1, and hence Ki = Mi = MJ and W1 = V1 = Vo. If
V = Vo, then case (i) holds, contrary to our assumption, so we may assume that
Cv(MJ) =f. 0. Now CM(V 0 ) and CM(Cv(MJ)) lie in Mc = !M(CG(Z)), and
IM: CM(Vo)CM(Cv(MJ))TI divides IAutM(Vo) : K1TI = 3; then as Mi. Mc, we