1054 14. L 3 (2) IN THE FSU, AND L2(2) WHEN .Cf(G, T) IS EMPTYW 0 (R 1 , D) of Din R 1 is normal in both LT and H, so that H::::; Na(X) ::::; M =
.!M(LT), contrary to Hi M.
It suffices to show that X centralizes U: For then as V::::; U, X::::; Cr(V) = Q
and X::::; Cr(U)::::; QH, so X = W 0 (Q,D) = W 0 (QH,D) is normal in LT and H
using E.3.15. Thus we may assume there is g E G such that A:= D^9 ::::; Ri, but A
does not centralize U. By 14.7.26, A is a TI-subgroup of G, so:
(!) [Cu(a),A]::::; An U for each a EA#.
Suppose first that An U = l. Then by (!),
(*) Cu( a)= Cu(A) for each a EA#.In particular if 1-!=-a E CA(U), then A centralizes U, contrary to our assumption,
so A is faithful on U. Thus A is not cyclic of order 4 by (*), so A ~ E4. Now
as m 2 (Ri) = 1 by 14.7.15.2, An QH -!=-l. Then as A is faithful on U, for each
b EA n Q'.t, Cu(b) is a hyperplane of U in view of 14.7.4.l. However no element of
H - QH centralizes a hyperplane of U, and elements of QH - bCQH (U) centralizehyperplanes of U distinct from Cu(b) by the duality in 14.5.21.1, so again using
(*),we conclude A#~ bCQH(U), a contradiction as A is faithful on U.
Therefore An U -!=-l. Then as [A[ = 4, [An U[ = 2, and hence A induces a
group of transvections on U with center An U by (!). As no element of H - QH
centralizes a hyperplane of U, A::::; QH; hence [A, U] =Vi by 14.7.4.1, so AnU = V1.
Therefore as D is a TI-subgroup of G by 14.7.26, A= D ::::; Z(K) ::::; Ca(U) sinceU = [U, K] ::::; K, contrary to our assumption that A does not centralize U.
Thus the proof of Theorem 14.7.14 is complete.
In the remainder of the subsection, H again denotes an arbitrary member of
'Hz. We deduce various consequences of Theorem 14.7.14 for members of 'Hz.LEMMA 14.7.27. For each HE 'Hz, either Os(H*) = 1 or Os(H*) =Li.
PROOF. Suppose His a minimal counterexample, and let P* := 03 (H*) with
Pa Sylow 3-group of the preimage of P*. Let Po be a supercritical subgroup of P,so that Po ~ Z 3 , E 9 , or 31+^2 by A.1.25.l. Further by definition, Po contains each
subgroup of order 3 in Cp(P 0 ), so if [Po[ = 3, then Pis cyclic.
Suppose first that P 0 ::::; M. Applying 14.7.9 with 02 (PoQH) in the role of "Y"
we conclude that Li = P 0 is of order 3, so Pis cyclic. But then P::::; M by 14.7.8,
so as M = LCM(L/02(L)), P = Cp(L/02(L)) x (P n Li); then as P is cyclic,
P* = Li, contrary to the choice of H as a counterexample.
Thus Poi M, so by minimality of H, H = PoLiT. Let B be of order 3 in Li;
we may assume B acts on P.
Assume first that B i P. Then Li i 03 (H*), so since Li is T-invariant
in H = PoLiT, we conclude that 1 -!=-02(Li). Then by A.1.21.3, Li is faithful
on P 0 /cI>(P 0 ), so H* is the split extension of P 0 , isomorphic to E 9 or 31+^2 , by
LiT* ~ GL 2 (3). However if P 0 is 31+^2 , then this split extension is of 3-rank 3,
contradicting G quasithin. Therefore P 0 ~ E 9. Now q(H*,UH)::::; 2 by 14.5.18.3,and the normal subgroup J := (Q(H, UH)) is either H ~ GL2(3) or Os,z(H).
But the first does not appear in D.2.17, and the second does not satisfy conclusion
(3) of D.2.17, since irreducibles for H* faithful on P 0 have dimension 8 rather than
4.
Therefore B ::::; P. If Pon M-!=-1, we may apply 14.7.9 to 02 (PoQH) in the
role of "Y" to conclude that P 0 ~ 31+^2. But now Theorem 14.7.14 supplies a