i3.8. FINISHING THE TREATMENT OF A 6 947Recall we may choose 9b with ('Yo, 'Yi)gb = bb-i, "!)· Then U'"Y := Ufl, V'"Y :=
VJ/, Q'"Y := Qfj, and Ai:= vr. Further DH:= CuH(Uy/Ai), EH:= CvH(U'"Y/Ai),
D'"Y := Cu ... /[fH), and E'Y := Cv'Y(UH)· We will appeal extensively to lemmas F.9.13
and F.9.16.Set UL:= (UJ;), and Q := 02(LT).
LEMMA 13.8.4. (1) b ::'.:'. 3 is odd.
(2) UL:::; Q = 02(LT).
(3) If b > 3, then UL is abelian.
(4) If b = 3, then Ai :::; Vh for some h E Ji.
(5) Vi= V n UH <UH, and VH/UH is a quotient of the F2H*-permutation
module on H* /(H n M)* with [VH/UH,H] # 0.
(6) v; is quadratic on VH/UH and [JH·(7) v <UL.
PROOF. We have already observed that (1) holds. Part (2) follows from 13.7.3.3.
Parts (3) and (4) follow from parts (1) and (2) of F.9.14.
By 13.7.7, [VH,H] i UH, so vi UH. Then as Vi:::; vnuH with Vi of index 2
in V, V3 = VnUH. By 13.7.3, HnM acts on VUH/UH ~ V/(VnUH) = V/Vi ~ Z2,
so as VH = (VH) and [VH, H] j;_ UH, (5) holds.
As V, is abelian and VH and V'Y normalize each other by F.9.13.2, (6) follows.As Vi:::; UH, V = (V.f):::; UL, and as V j;_ UH:::; UL, V <UL. Thus (7) holds. D
LEMMA 13.8.5. If some element of H* induces an F2-transvection on UH, then
(1) H =KT with KE C(H).
(2) Either
(a) H* ~ 85, L/0 2 (L) ~ A5, and Li has two noncentral chief factors on(b) H* ~ 81 or L4(2), and L/02(L) ~ A5.
(3) UH is a natural module for H* or the 5-dimensional cover of such a module
for H* ~ 85.PROOF. Let t* E T* induce an F2-transvection on UH. If K* = [K*, t*] # 1
for some KE C(H), then as t* is an F2-transvection, we conclude from G.6.4 thatK* is Ln(2) or An and UK/CuK(K) is a natural module, where UK:= [UH,K].
Hence the lemma follows from parts (1) and (3) of 13.7.6, using I.1.6.1 in the latter
case.
So we may assume instead that K := (tH) is solvable, and we derive a
contradiction. By B.1.8, K* =Ki x · · · K;, Kt~ L2(2), withs:::; 2 since m 3 (H):::;
2, and UK= [UH,K*] = Ui EB··· EB Us, where Ui :=[UH, Ki]~ E4. Then Li acts
on each Ki. Thus ifs= 2, then as m 3 (H) :::; 2, Li:::; K*. This is impossible as T
normalizes a subgroup of order 3 of Li, whereas T is irreducible on K = (tH) by
construction.
Hences= 1 and K =Ki :::l H. This time we conclude from the T-invariance
of Li that either
(a) L/02(L) ~ A5 so that [Lib= 3, and either Li= 02 (K) or [K, Li] = 1,
or
(b) L/0 2 (L) ~ A5 so that Li has 3-rank 2, and hence 02 (K) =Lo or Li,+·