3.3. NORMALIZERS OF UNIQUENESS GROUPS CONTAIN Na(T) 593not contain the unipotent radical of the maximal parabolic determined by the end
node stabilizing a 4-space in the natural module. Thus by B.4.2.11 (again for more
detail see K.3.2.3), J(02(YT)) ::; GT(V), so again J(02(YT)) = J(0 2 (LT)), for
the same contradiction. The proof is complete. DThe next technical result has the same flavor as 3.3.12, and will be used in asimilar way. In particular it will help to eliminate the shadows discussed earlier.
LEMMA 3.3.20. Assume X = 02 (X) is T-invariant with XT/0 2 (XT) ~ 83 ,
and D does not normalize R := 02 (XT). Let Y := (XD), and let 'Y denote the
number of noncentral 2-chief factors for X. Then
{1) (XT,D) E 1i(T) and Y :::! (XT,D) = YTD.
{2) YT/02(YT) ~ L2(P) for a prime p = ±11 mod 24.
{3) 02(X)::; 02(Y), XT/02(YT) ~ Di2, and ID: ND(X)I = 3.
(4) 'Y ?. 3..
{5) If"(::; 4, then:{a) Y has a unique noncentral 2-chief factor W, m(W) ?. 10, and ITI ?.
(b) (02(X)) ::; Z(Y).
{c) If Z(YT) =f. 1, then ITI > 212.
PROOF. Let B := (XT,D). As D does not act on R, R =f. l. Thus XT E 1i(T)
and Tis maximal in XT as XT/R ~ 83. Therefore B E 1i(T) by 3.3.10.l. Also
xTD = xD soy :::! B, establishing (1).Notice using A.1.6 that 02(B) ::; 02(XT) = R. Let Bo be maximal subject to
Bo :::! B and XT n Bo ::; R. Then XT n Bo = R n Bo = T n Bo =: T 0 contains
02(B) and is invariant under XT and D, so To :::! B. Thus To = 02(B). As Ddoes not act on R by hypothesis, T 0 < R.
Set B := B/B 0. As T 0 = XT n Bo< R, R =f. 1 =f. X*. Then as XT/R ~ 83 ,
ITI = 2IRI > 2.
Let Bi be a minimal normal subgroup of B*. By maximality of Bo, XT n Bi 1:.
R = 02 (XT), so X* nBi is not a 2-group. So as JX: 02 (X)I = 3 and X = 02 (X),
X* ::; B'{. Then by minimality of Bi, Bi = (X*D) = Y*. In particular Y* is the
unique minimal normal subgroup of B*, so Y* = F* ( B*); hence T* is faithful on
Y*.
Suppose Y* is solvable. Then Y* ~ E 3 n as Y* is a minimal normal subgroup
of B*. As Y* = (X*D), and D acts on T with X* a simple T-submodule of
Y*, Y* is a semisimple T-module. Therefore as T* is faithful on Y*, <I>(T*) = 1,and as ITI > 2, m(T) > l. Then by (1) and (2) of A.1.31, m(T*) = 2 and
m( CY* ( t*)) ::; 1 for each t* E T*#, so that n = 2 or 3. Further if n = 3, then as
B = YTD by (1), T* D* ~ A4 is irreducible on Y*, contrary to A.1.31.3. Thusn = 2, so T D ::; GL 2 (3). Then as (T) = 1 and D is a subgroup of GL 2 (3)
of odd order normalizing T, D = 1. Hence Y = (XD) = X, contradicting
n=2.
So Y* is not solvable, and hence Y* = F*(B*) =Yi* x · · · x ~* is the directproduct of isomorphic simple groups Yi* permuted transitively by TD. Then (1.a)
of Theorem A (A.2.1) holds, so mq(Y*) ::; mq(B)::::; 2 for each odd prime q, so that
s ::; 2 and Y is an SQTK-group. Thus as D = 02 (D), D normalizes each Yi, so
T is transitive on the Yi. Therefore if T acts on Yi, then s = 1 and Y is simple.
As Y = (XD) and Y is not solvable, D =J 1.