530 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1by B.4.2.1, and hence R E Syl 2 (LQ). Then (3) will follow once we prove (2).However, we will first establish (5) and the assertion in (4) that Dx normalizes L.
As L = 02 (H), CR(L) ::; Q, so as Q is abelian and R::; LQ, CR(L) ::::; Z(R).
Further by (1), we may apply 2.4.5.2 with Tin the role of "Mo", to conclude that
Q = UCs(L). Then as U::::; Land R::::; LQ, R = (Rn L)CR(L). As R is Sylow
in LQ, Sn L = Rn L; then Mn L = (Rn L)D is a Borel subgroup of L, and
R =(Rn L)CR(L) is D-invariant.Now S normalizes the Borel subgroup (RnL)D over RnL, and hence normalizes
RD. Thus S also normalizes (RD)x = RDx. Also T permutes Q and Qx, and so
acts on Gq n G'Q =: Y. We saw D normalizes R, so as D = 02 (D), D normalizes
the two members Q and Qx of A(R); that is, D ::::; Y, and hence also Dx ::::; Y.
By 2.4.6, Gq normalizes Lo = L; in particular Dx normalizes L, giving the first
assertion in (4). Now MnGq normalizes MnL = (RnL)D as well as Q = UCR(L),
and hence normalizes their product RD. Then as Dx::; MnY, Dx also normalizesRD. As x^2 E S normalizes RD and (RD) x = RDx,
RDDx ::;! \RDDx, S, x) = RDDxT = DDxT.
Hence by a Frattini Argument, we may take x to act on a Hall 21 -subgroup B ofRD Dx containing D.
We can now obtain the conclusions of (5), except possibly for CR(B) = 1: FirstDX normalizes RD n B = D' so D DX is a subgroup of B) and hence D DX = B.
Then T normalizes RD Dx = RB, while R is normalized by D and hence also by
Dx, and further \T,D) =BT. We saw D ::;! DDX = B, so also DX ::;! B, and then
DX[D,DX]::::; Dx. As DX is abelian this shows that no element of DX induces an
outer automorphisms on L/U ~ L 2 (2n), so that B = DDx = D x CB(L/U). Then
since B = D Dx and D is abelian, it follows that B is abelian. By a Frattini
Argument on RB ::;! TB, T = RNr(B). This extension splits once we show
CR(B) = 1.Thus to complete the proof of (5), it remains to show that CR(B) = 1. We
saw that R =(Rn L)CR(L), so [R, D] =[Rn L, DJ= Rn L since Lis an L 2 (2n)-
block; thus also [R, Dx] =Rn U. Further we saw Q n Qx = Z(R) ;:::: CR(L), soR/Z(R) = [R/Z(R),D]. Then also R/Z(R) = [R/Z(R),Dx], so
RnL = [RnL,Dx]::::; [R,Dx] = RnLx;
so as (RnLr = RnLx, RnL = Rn£X. Therefore RnL = [R,Dx], so
[R, B] = [R, DDx] = Rn L. As CR/CR(L) (D) = 1, CR(B) ::; CR(L) ::::; Z(R), so
CR(B) ::;! LRNs(B) = LS = H. Also x normalizes Rand B, and hence also
CR(B), so that CR(B) = 1 by 2.4.4, completing the proof of (5).
Now by Coprime Action, R = [R, B] = R n L, so that R ::::; L. As Q ::; R,
Q = 02(L) = U by 2.4.5.1, so that (2) holds. This also completes the proof of (3)
as mentioned earlier.
So it remains to complete the proof of (4) and establish (6) and (7). As L
is an L2(q)-block, L is indecomposable on U with U/Z(L) the natural module
for L/U. From the cohomology of that module in I.1.6, m(Z(L)) ::::; n. Further
Z(L) = CR(D) with D semiregular on R/Z(L), so Dx is semiregular on R/CR(Dx).
Thus as CR(B) = 1, Dx is semiregular on Z(L)#, so as m(Z(L)) ::::; n, either
Z(L) = 1 or m(Z(L)) = n. In each case (using I.1.6 in the latter) the representation
of L/U on U is determined up to equivalence, and as the Sylow group R = uux of
L splits over U, L also splits over U by Gaschutz's Theorem A.1.39. Therefore Lis