1134 15. THE CASE .Cf(G, T) =^0(I)-(IV), and also one of the analogous conclusions on Ul. Then inspecting (I)-
(IV), we conclude that either:
(a) A* is in case (ii) and AutA(Ul) is in case (i), or vice versa; or:
(b) A* and AutA(Ul) are in case (iv).
However in case (a), the tuple of parameters (mo, mi, m2, m3, m4) is (1, 0, 2, 1, 3),
contrary to (!). Similarly in case (b), the tuple is (1, 2, 2, 3, 3), again contrary to
(!).
Thus T = S. So as So and Z(S) are normal in S, their preimages are normal
in T. Then inspecting (I)-(IV), we conclude that A* and AutA(Ul) always appear
in the same case of (i)-(iv). In cases (i), (ii), and (iv), we calculate the tuple of
parameters to be (1, 2, 2, 3, 3), (1, 0, 0, 1, 1), and (1, 2, 2, 3, 3), again contrary to (!).We conclude A* is in case (iii). In particular, A i. S 0 ; so since A is an arbitary
member of B, it follows that J(So)::::; Cr(UL)· Thus J(So) = J(Cr(UL)) :::l (T, L),
again contrary to L i. M = !M(YT) by 15.3.2.4 since Y+ i. CM(V).
This completes the proof of Theorem 15.3.25.Theorem 15.3.25 has reduced the treatment of Hypothesis 15.3.10 to the case
F*(I) -=j:. 02 (J). As O(I) = 1 by 15.3.11.8, there is a component L of I, and
Z(L) = 02(L). As F*(Mr) = 02(Mr) by 15.3.11.7, Li. M. Thus to complete our
proof that M = !M(Y+S), it remains to determine the possibilities for L, and then
to eliminate each possibility.
Set Lo := (L^8 ), SL := s n L, and ML :=Mn L. Let z denote a generator of
z.LEMMA 15.3.33. (1) If L 1 is a Y+S-invariant subgroup of Lo with F*(L1) =
02(L1), then Li::::; Mr.
(2) The hypotheses of 1.1.5 are satisfied with I, Mc is the roles of "H, M".
(3) If K E C(J) then K i. M, K = [K, z] is described in 1.1.5.3, O(K) = 1,
and
PROOF .. Choose L 1 as in (1); if L 1 i. M, then L 1 Y+S E 1-i+, contrary to The-
orem 15.3.25, so (1) holds. Next let HE M(I), so in particular H = Na(0 2 (H)).Then H E 1-i+, so S E Syb(H) by 15.3.11.1. Thus S = T n H and 02 (H) ::::;
02(I n Mc) by A.1.6, so
Co 2 (Mc)(02(I n Mc))::::; Cr(02(H)) ~ T n H = S ~I,
and hence (2) follows since Ca(z) = Mc by 15.3.4. Then (2) and 1.1.5 imply (3),
since we saw O(J) = 1. · DObserve that 15.3.33.3 applies to L in the role of "K". We now begin to
eliminate the various possibilities for Lin 1.1.5.3.
LEMMA 15.3.34. L/Z(L) is not Sz(2n). Hence m 3 (L)?: 1.
PROOF. If L/Z(L) ~ Sz(2n), then Y+S acts on a Borel subgroup B of L 0 ,
so B = Mr n Lo by 15.3.33.1, since B is a maximal S-invariant subgroup of L 0.
By 15.3.11.4, Mr = Nr(Y+), so as automorphisms of L 0 /0 2 (L 0 ) of order 3 acting
on B are nontrivial on B/0 2 (L 0 ), we conclude Y+ ~ Cr(B) = Cr(L 0 ). Thus
Lo~ Nr(Y+) =Mr by 15.3.11.4, contrary to our choice of L. D