944 i3. MID-SIZE GROUPS OVER F2on Ui; but this contradicts the structure of FF* -offenders in conclusions (ii)-(iv) of
B.5.1.l. This completes the proof of (1) and (2).Suppose K < K 0. Then by 1.2.l.3 and (1), K* ~ L 3 (2) or A 5 , and by 1.2.2.1,
Ko= 031 (H), so Li:::; Ko. Then as Tacts on Li, we conclude K* ~ A5 and Li is
diagonally embedded in K 0. Since B* :::::; Ri, B* induces inner automorphisms onK, so by B.4.2, UK is the natural module for K rather than the A 5 -module, and
\13+ = [Ulf, B] :::; U"j(. Now as Tacts on 113, Tacts on UK and hence also on K,
contrary to assumption. This establishes (3).
Next 1 =/=-[U"j(,B]:::; v 3 + n U"j( by 13.7.12.2. Then as Li acts on K, and acts
irreducibly on \13+,
V3+ = [113 nut, Li]:::; U"j(,
so Ulf = (V 3 +H) = U"j( since UK is H-invariant by (3), and hence (4) holds. D
LEMMA 13.7.16. {1) T is faithful on K.
(2) [Uo, K] = l.
{3) f~h E Irr+(K*,UH)·
PROOF. By parts (2) and (4) of 13.7.15 and A.l.41, CH(K) is of odd order,
so (1) holds. Also (3) follows from 13.7.15.2 if (2) holds, so it remains to prove (2).
Assume (2) fails. Then as Uo :::; Z(QH), K E .C1(G, T), so by 13.5.2.1, K*
is A5, L3 ( 2), A6, or A.6. Further K acts nontrivially on U 0 and Ulf, so K has
at least two noncentral chief factors on UH. On the other hand by 13.7.12.3, Ufr*
contains an FF-offender D on UH, and by (1), D is faithful on K, so by B.l.5.1,
AutD(UK) is an FF*-offender on the FF-module UK. Then by Theorems B.5.6 and
B.5.1.1, K* ~ L3(2) and UK is the sum of two isomorphic natural modules for
K. Then as Ufr S Q :::; 02(L1T), Li(T n K) is the stabilizer of a line in each
irreducible and Autuj,(UK) = 02(AutL 1 (UK)).
By 13.5.2.3, Ui := [Uo, K] is a natural module, so Ui is isomorphic to Ui, and so
Cu 1 (Li)= 1. But Zi :=Zn Ui is of order 2, and as Autug H (U 1 ) = 02 (AutL 1 (U1)),
Zi S [Uo, Ufr]. Then as Ui S UH S NQ(Ufr) .by 13.7.3.3, Z1 S Ufr, so Z1 is
centralized by \VH, V_k) =I and by T. Thus L :::; (J, T) :::; Ca(Zi), contrary to
Cu 1 (Li)= l. D
LEMMA 13.7.17. {1) Zr n Uo =Vi.
{2) If Uo S [UH,A]l/3, then Uo =Vi.PROOF. First QH and I = (VH, V_k centralize Uo n Ufr =: U1, so Lo :=
(J, QH) S Ca(Ui). However QH = Ri by 13.7.9.1, and LT :::; (Ri,l), so LT=
LoQ. Further Q and Lo act on Ui,· so LT= LoQ:::; Na(Ui).
If Ui =/=- 1 then Na(Ui) SM= !M(LT). But then by 13.7.16.2, K:::; Ca(Ui):::;M, contrary to 13.3.9. Thus Ui = 1.
Next by 13.7.10.5, DH= (DH nDH9)Vi, and by 13.7.10.7, EH= Zr, so
Zr n Uo =EH n Uo =DH n Uo = (DH n DH9 )Vin Uo
=(DH n DH9 n Uo)Vi = (UH9 n Uo)Vi = "f/i Vi= Vi,
establishing (1). By 13.7.3.2, UH S Q n VH, so [A, UH] S Zr by 13.7.10.3. By
13.7.10.9, 113 S Zr. Thus if Uo:::; [UH,A]l/3 then U 0 :::; Zr, so (1) implies (2). D