966 i3. MID-SIZE GROUPS OVER F2
CQH (Vi), we conclude that He s CQH (Vi) = Q n QH S Q. Thus He centralizes
V, and hence He also centralizes (VH) = VH. Therefore as Ai :::; VH by 13.8.32.3,
He s Ca(Ai) = G'Y, since H = Gi by 13.8.27.2; thus [He, U'Y] s Hen U'Y. But
m(U'Y n QH) = 4 by 13.8.32.1, and by 13.8.32.2, m(U'Y n VH) ~ 3. So m( (U'Y n
He)VH/VH) s 1. Thus as [He, U'Y] s HenU'Y, K has at most one noncentral chief
factor on He/VH, and by G.6.4, that factor is 4-dimensional if it exists. But this
contradicts 13.8.31.3. This completes the proof of Theorem 13.8.28.
By Theorem 13.8.28, case (a) of 13.8.27.1 holds: Namely L/02(L) ~ A5, H* ~
A 6 or 86 , and UH is the natural module for K~ on which Li has two noncentral
chief factors, or its 5-dimensional cover.
LEMMA 13.8.33. Case (2) of 13.8.8 holds; that is, D'Y < u'Y'
PROOF. Assume instead that case (1) of 13.8.8 holds. By 13.8.27.3, DH= UH,
D'Y = u'Y, v induces a group of transvections with center Vi on u'Y, v'Y i QH,
and we have symmetry between ')'i and')', (cf. the first part of Remark F.9.17),
so v; induces a transvection on UH with center Ai, and Ai s UH. As usual
choose g := 9b E (LT,H) with ')'i9 = '/'· By F.9.13.7, [UH,U'Y] = 1. Therefore
U'Y s Ca(Vi) s Mv by 13.5.4.4. By 13.8.5, H =KT and H* ~ 85. In particular,
we can appeal to 13.8.6 and adopt the notation of that lemma. As [V, U'Y] -=/=-1,
we may pick g so that [Vf, VJ-=/=-1. Thus as [Vi, Vf] s [UH, U'Y] = 1, 13.5.4.4 says
Vi= [V, Vf] and -Vf = ((5, 6)), so that LT~ 85.
Notice that if m(UH) = 4 then Ai s 1f 3 h f~r some h EH. Assume instead for
the moment that m(UH) = 5. Then Ai is of weight 2, while by 13.8.6.1, V3 consists
of vectors of weight 4, so Ai is not contained in an H-conjugate of Vs. Thus as
Vs = V n UH by 13.8.4.5, we conclude from 13.8.4.4 that b > 3 when m(U H) = 5.
We claim that UL is abelian; the proof will require several paragraphs. Assume
UL is nonabelian. Then b = 3 by (1) and (3) of 13.8.4, so by the previous paragraph,
m(UH) = 4 and Ai S V 3 h for some h EH. Thus Vi= Vih is orthogonal to Ai in
Vh, so V1 is orthogonal to A?-
1
in V, and 1 =[UH, U'Y] =[UH, Ujj] =[UH, Ui'i-h-1
].
h-1 1
Now V.f = Aq- = Vl for some y E L, so as H = G1 by 13.8.27.2, we conclude
that Ui'i-h-
1
= U]',-. Finally LiT is transitive on the points of V distinct from Vi
and orthogonal to Vi, and T is transitive on the points of V not orthogonal to Vi;
so since we are assuming UL is nonabelian, [UH, UJr] -=/=-1 for some l E L with V{
not orthogonal to Vi, and hence for all such V{ by transitivity of T on this set.
Therefore Uj; -=/=-1: for otherwise UL S QH, and hence UL S Qk, so that by 13.7.3,
[UH, UJr]:::; Vin V{ = 1, contrary to our choice of l.
Choose l with l^2 E Q. Then as V3 = v n UH, W2 := v n UH n u1 is a
complement to V1 in Vi, and to Vi V{ in V. Further X1 := 02 (CL 1 (V1 Vl)) acts on
UH and UJr, with W2 = [W2,Xi]. By 13.8.27, Li has two nontrivial chief factors
on UH, so [UHV/V,Xi] = UHV/V ~ E4, and hence [UJrV/V,Xi] = UJ,.V/V ~
E4. Then X1 is irreducible on UJrV/V, so as Uj; -=/=-1, Uj; = 02 (Li) ~ E4 and
UJr n QH = Vl = UJr n v.
Next by 13.7.3.7, IHe: (Hen H^1 )1s2, and as UL :SQ s Na(He) by 13.7.3,
[Uk,HenH^1 ] s UknHe s uJ:rnQH = uJ:rnv s VH·