noo 15. THE CASE .Cr(G, T) = 0In case (2c), ML is some proper parabolic. In any case, let
Po := { (P^8 ) : P is a minimal parabolic and P i ML}.
Now either L+s+ ~ Aut(L5(2)) and F*(L) = 02(L), or Po ~ P. In the latter
case, conclusion (3) or (7) of 15.1.22 holds by (*) and (!), so we may assume the
former case holds. Here L = (Pc, Pe), where Pe is the parabolic generated by the
two end-node minimal parabolics, Pe E P, and PcS/02(PcS) ~ Aut(L3(2)). As
Pe E P, Pe is contained in Me E {ML,M n L} by(*). Then as PeS is a maximal
subgroup of LS, Me= Pe.
If Pc centralizes Z, then Pc :::; Mc as Ca(Z) :::; Mc, so Pc= ML by maximality
of PcS in LS. Then Pe= MnL by the previous paragraph, so that conclusion (3) of
15.1.22 holds. Thus we may assume that [Z, Pc] -=f. 1. Now W := (zPc) E R2(Pc) by
B.2.14, so as PcS/0 2 (PcS) ~ Aut(L3(2)), and the latter group has no FF-module
by Theorem B.5.1, we conclude that J(S) :::; Cpcs(W) = 02(PcS). Therefore
J(S) = J(0 2 (PcS)) by B.2.3.3, and hence Pc :::; Na(J(S)) :::; M by 15.1.17. As
Mn L < L and Pc is a maximal S-invariant subgroup of L, we conclude that
Pc = Mn L, and then Pe = ML by the previous paragraph, so again conclusion (3)
of 15.1.22 holds.
Finally we assume that case (d) of 15.1.20.3 holds, so that Lis a component of
G 1 with L/Z(L) sporadic, and Z(L) = 02 (L).
Suppose first that L/Z(L) is HS or Ru. Then there is K E C(LS, S) n L
with F*(K) = 02 (K) and K/0 2 (K) ~ L 3 (2). Further 02 (K) ~ Z~ or 23 +8,
respectively, so KS is not among the conclusions of C.1.34. Hence by C.1.34, there
is a nontrivial characteristic subgroup C of S normal in K. Then as S :::;] T by
15.1.18.3, (K, T) :::; N := Na(C). Then K :::; N^00 :::; Mc by 14.1.6.3; but this is
impossible, as Ki CL(Z(SL)), whereas CL(Z(SL)) =ML by 15.1.20.3.
Therefore L/Z(L) is a Mathieu group, J 2 , or He, and ML = CL(Z(SL)) by
15.1.20.3. Assume first that L/Z(L) is not Mn, and set K := (ML, P). Then from
the structure of Aut(L), either K = L, or L/Z(L) ~ M 22 and K/Z(L) ~ A 6 /E15.
Moreover in the latter case, K > ML as we saw in our treatment of M 22 during the
proof of 15.1.20. Thus in any case there is PEP with Pi ML, and as IZtl = 2
in these groups, (!) completes the proof that conclusion (5) holds.
It remains to treat the case L/Z(L) ~ M 11 , where L ~ Mn by I.1.3, and
L :::;] Gi by 1.2.1.3. Then Out(L) = 1, so that Gi = L x Ca 1 (L); in particular
J(S) = J(Cs(L)) x J(SL), and hence Na 1 (J(S)) :::; Na 1 (J(SL)). Further J(SL) ~
Ds, so that NL(J(SL')) = SL, and hence 02 (Na 1 (J(S)) centralizes L, so that
conclusion (6) holds.
This completes the proof of 15.1.22. DLEMMA 15.1.23. If case (6) of 15.1. 7 holds, then p = 3, so M ~ S 3 wr Z2.
PROOF. Assume case (6) of 15.1.7 holds. If p = 3, then M ~ S 3 wr Z 2 by15.1.10. So we may assume that p = 5, and it remains to derive a contradiction.
Then Mo~ D 10 x Dw. Hence there is a 5-group Y:::; CM(Z1) with SY= YS and
Vi= [Vi, Y]. Set Go:= Na 1 (L) and G 0 := Go/Ca 0 (L). By 15.1.19.7, YVi acts onL, and V2 acts nontrivially on L. Thus as Y is faithful and irreducible on Vi, YVi is
faithful on L. Thus comparing the list in 15.1.22 to the possibilities for L/0 2 (L) in
A.3.15, we conclude L ~^2 F4(2)', and Auty(L) :::; Autp(L), where P := CL(Z(SL))
with P/02(P) ~ Sz(2). This is impossible, as Autys(L) does not act irreducibly
on an E15-subgroup Autv 2 (L) of P. D