2.4. THE CASE WHERE rQ IS NONEMPTY 531determined up to isomorphism in each case. The parabolics P in cases (a) and (b)of (4) exhibit such extensions, so this completes the proof of (4). Part (4) implies
(7).
When Z(L) =/= 1 the fact that Dx is semiregular on Z(L) = CR(D) shows that
B = D x Dx. When Z(L) = 1, Dis regular on Z(R)#, so D ~ Autv(Z(R)) is
self-centralizing in GL(Z(R)); thus as B is abelian, AutB(Z(R)) = Autv(Z(R)).
This completes the proof of (6), and hence also of 2.4.8. D
REMARK 2.4.9. The cases (a) and (b) of 2.4.8.4 were treated separately in
sections 3 and 4 of [Asc78a]. However many of the arguments for the two cases
are parallel, so we give a common treatment here where possible.NOTATION 2.4.10. During the remainder of the treatment of the cases= 1, x
. denotes an element of T - S. By 2.4.8.1, IT : SI = 2, so as S acts on L, H, U, the
conjugates Lx, Hx, ux are independent of the choice of x.
LEMMA 2.4.11. (1) GQ = Na(L).
(2) GQ = !M(L).
(3) If Z(L) =/= 1 then Z(L) is a TI-subgroup of G with GQ = Na(Z(L)).
PROOF. Recall GQ :::; Na(L) by 2.4.6; so as Q = U = 02(L) by 2.4.8.2,
Na(L) :::; GQ, so (1) holds.
As Q = 02 (LS) while S E Sylz(GQ) by 2.4.3.1, Vl 0 Q (L, 2) = {Q}. Then as
Lis irreducible on Q/Z(L) and indecomposable on Q, if 1 =/=VE VlaQ(L,2) theneither V = Q or V:::; Z(L).
Let XE 7-l(L); to prove (2), we must show that L '.':1 X, so assume otherwise.
Let P := 02(X). Then 1 =/=Po:= Np(Q):::; GQ, so Po E VlaQ(L,2). Thus by
the previous paragraph, either Po = Q or Po :::; Z ( L). In either case Po :::; Q, so
that NpQ(Q) = P 0 Q = Q, and then PQ = Q so that P =Po. If P = Q, then
X:::; Na(Q) = GQ, contrary to assumption; hence P:::; Z(L). This shows that (2)
holds when Z(L) = 1. Thus for the rest of the proof, we may assume Z(L) =!= 1,
since this is also the hypothesis of (3). In particular, case (b) of 2.4.8.4 holds.
Next we claim that Ca(v) :::; GQ for each v E Z(L)#. Assume otherwise;
then we may choose Ca(v) in the role of "X" in the previous paragraph. As B
is transitive on Z(L)# by 2.4.8.4, we may assume S:::;; X. Thus H = LS:::;; X,
so by 2.3.7.2, X E I'o and S E Syl2(X). Then by 1.2.4., L :::;; K E C(X). As
the Sylow 2-subgroup S of X normalizes L, but we are assuming L is not normal
in X, L < K. Now 02 (K) :::;; 02 (X) = P :::; Z(L) by the previous paragraph,
so K = [K, L] centralizes 02 (K). Also m2(K) 2 m2(L) > 1, so we concludefrom 1.2.1.5 that K is quasisimple, and hence K is a component of X. Thus K is
described in 2.3.9.7. Since 1 =!= v EL n Z(X) :::;; Z(K), Z(K) is of even order, so
K/02(K) is £3(4) or A5 and Z(K) = 02(K). If K/Z(K) is A5, then K ~ SL2(9)
by l.2.2.1, a contradiction as L :::; K with m2(L) 2 4. Thus K/02(K) ~ £3(4),
L/Q ~ L 2 (4), and Z(K) = Z(L) =Pas Lis irreducible on Q/Z(K) and we sawZ(K) :::;; P :::; Z(L). In particular, P '.':1 H. Further Z(L) ~ E4 since n = 2 and
case (b) of 2.4.8.4 holds. Observe since K/Z(K) ~ £3(4) that L = NK(Q), so
as R is Sylow in L by 2.4.8.3, R is Sylow in K. Now consider x E T - S as in
Notation 2.4.10. By parts (2) and (3) of 2.4.8, A(R) = {Q, Qx}, so NK(Qx) is the
maximal parabolic of K over R distinct from L. Therefore NK(Qx)/Qx ~ £ 2 (4)
and P = Z(K) = Z(NK(Qx)). Hence Lx = (RNa(Q'")) 2 NK(Qx), so we conclude