i4.6. ELIMINATING L 2 (2) WHEN (VG1) IS ABELIAN 1025so the lemma is established. DLEMMA 14.6.9. Assume IE I*, IT : QHI > 4, and m(UH/CuH(QH)) ~ 4.
Then ITI > 211 , and
LT= 02 (L n I)T102(LT).PROOF. Observe first that by the duality in 14.5.21.1,
m(QH/CqH(UH)) = m(UH/CuH(QH)) =: m,
with Z :::; Cu H ( Q H), so that IQ HI ~ 22 m+i ~ 29 since m ~ 4 by hypothesis. As we
·also assume IT: QHI > 4, ITI > 211 , establishing the first conclusion of 14.6.9. By
14.6.6.1, T1 =Tu or To, so IT: T1I:::; 4 by 14.6.3.1, and hence IT1I > 29.
Thus we may assume that LT> 02 (L n I)T 102 (LT), and it remains to derive
a contradiction. Then by 14.6.6.4, C(I, T 1 ) :::; Iz. As we are working toward
a contradiction, we may also assume that I is minimal under· inclusion; that is,
I E I*. Then by 14.6.6.5, Iz is the unique maximal subgroup of I containing T1.
Since T1 is not normal in I by 14.6.6.7, I is a minimal parabolic in the sense of
Definition B.6.1.We first treat the lengthier case where F*(I) = 02 (J). Here since T 1 E Syb(I)
by 14.6.6.2, and I is a minimal parabolic, we may apply C.1.26: Since C(I, T 1 ) :::;
Iz < I, we conclude that I= T1 Ki··· Ks, where Ki is a xo-block of I not contained
in Iz, and T1 is transitive on the Ki. Furthers= 1 or 2 as I is an SQTK-group, andthe action of J(T1) on 02 (K) is described in E.2.3. Also Ki is not an L 2 (2n)-block
for n > 1, as Iz = C1(z) is the unique maximal overgroup of T1 in I, whereas when
Ki is an L 2 (2n)-block, the center of that overgroup is Z(I). Thus Ki is a block of
type A3 or A5.
Observe using 14.6.5 and 14.6.6.2 that:
(a) If 1 f- S :::; T1 with S :::;I I, then Na(S) E I* and NT(S) = T1 E
Syl2(Na(S)).
Since T1 < T by 14.6.6.1, we may chooser E NT(T1) -T1 with r^2 E T1. Then by
(a),.(b) r acts on no nontrivial subgroup S of T1 normal in I.
Set K :=Ki·· ·Ks, so that I= KT1. Assume first that K is not the product
of two A5-blocks. As F*(I) = 02(J), this assumption establishes part (i) of the
hypothesis of Theorem C.6.1, with I, T1(r), T1 in the roles of "H, A, TH", while (a)
gives part (iv) of that hypothesis, and (ii) and (iii)· are immediate. If K is an A 3 -
block then IT1I :::; 16 since case (a) or (b) of C.6.1.6 must hold, contrary to IT1I > 29in the first paragraph of the proof. Therefore K is an A 5 -block or a product of two
A3-blocks. In either case by C.1.13.c, 02 (!) = D x 02 (K), where D == CTr(K),
and by C.6.1.4, Dis elementary abelian, so that D:::; D1 := fh(Z(J(T 1 )). Then we
conclude from the action of J(T1) on 02 (K) described in E.2.3, that JD1: DJ = 4.
As D n Dr is normalized by KT 1 =I and r, D n Dr= 1 by (b), so that JDJ :::; 4.
But now JT1I:::; 4JAut(K)J 2 :::; 29 , again contrary to the first paragraph.
Therefore K = Ki x K2 is the product of two A 5 -blocks. Set Kz := 02 (Iz)
and Rz := 02(Iz). Then KzTJ/Rz ~ 83 wr Z2, and J(Rz) = J(02(I)) us-
ing E.2.3.3 and B.2.3.3. So applying (a) to J(Rz) in the role of "S", we ob-
tain T1 E Syb(Na(J(Rz))); hence T1 E Syb(Na 1 (Rz)). Thus as Rz = 02(Iz),