922 i3. MID-SIZE GROUPS OVER F2Next as y E L 2 - H, z -=f-zY E Vi::::; QH::::; Ca(j) so that j E Ca(zY) =HY,
and similarly jY EH. Therefore [j,jY]::::; Z(QH) n Z(Q'k) = 1. Thus (jL^2 ) =:Bis
abelian with ifi(B) =Vi; so as Bis the E 8 -subgroup generated by the transvections
inf', B ~ Z~ x Z 2. Let A:= fh(B). Then A is of order 2 and normal in L2f',
so A= ((1,2)(3,4)(5,6)). Further for a EA·_ Vi, [a, VJ= Vi, so Vis transitiveon A - Vi, and hence CM(a) = C.M(a) ~ Z2 x 84 for each such a. Therefore
B = 02 (CM(a)) and X := 02 (0^2 (CM(a))) ~ Z~ with Vi = ifi(X). As before by
Thompson Transfer, there is r E G with ar = z. Then 02 (CM(a)Y ::::; 02 (H), so
as UH is Sylow in 02 (H) by 13.5.15, xr::::; UH. Then Vi= ifi(XY::::; ifi(UH) =Vi
of rank 1. This contradiction completes the proof. DPROPOSITION 13.5.17. If n = 5 and Li :::] H, then G ~ U4(2) or L4(3).
PROOF. By 13.5.15.6, H := Gi is the unique member of 1-lz. By 13.5.15.7,
V = 02 (M). By 13.5.15.8, QH = UH =: U ~ Q§, and either M = L with
H ~ 83 x Z 3 , or M/V ~ 85 and H ~ 83 x 83. Let TL := T n LE 8yl2(L), so
that IT: TLI = 1 or 2. Define I as in the proof of 13.5.15. Observe that Hypothesis
F.1.1 is satisfied with I, L, T in the roles of "Li, L 2 , 8": In particular, recall
that during the proof of 13.5.15 we showed that I :::] Hand H = LiIT, so that
02((I,L,T)) = 1asH1:. Mand M = !M(LT).
Therefore "Y := (H, LiT, M) is a weak BN-pair by F.1.9. As T n I is self-
normalizing in I, the hypotheses of F.1.12 are satisfied; so as I/0 2 (!) ~ 83, while
L/0 2 (L) ~ A 5 does not centralize Z, we conclude from F.1.12 that "Y is of type
U4(2) when M = L, and "Y is of type 06(2) when M/V ~ 85.Next we verify the hypotheses of Theorem F.4.31: Let G 0 := (M,H). Then
the inclusion "Y--+ Go is a faithful completion of"'/- As ME M, M = Na(V). We
saw H = Gi = Ca(z). Thus hypotheses (a) and (b) ofF.4.31 hold. Hypotheses (c)
and (d) are vacuously satisfied, and hypothesis (e) holds as G is simple.We now appeal to Theorem F.4.31, and conclude as G is simple that either
M =Land G ~ U4(2) or M/V ~ 85 and G ~ L4(3). D
We mention that the shadows of extensions of U 4 (2) and ,L. 4 (3) were essentially
eliminated during the proof of Theorem F .4.31.13.5.2.2. Obtaining a contradiction in the remaining cases. During the remain-der of the section we assume that G is a counterexample to Theorem 13.5.12. Thus
appealing to 13.5.16 and 13.5.17, it follows that:
LEMMA 13.5.18. Assume L/0 2 (L) is not A 6 • Then Li is not normal in any
HE 1-lz such that UH is nonabelian.
LEMMA 13.5.19. Let Y :=Lo if L/02(L) ~ A5, and Y :=Li otherwise. Let
HE 1-lz and Hi:= NH(Y). Then
(1) Hi= NH•(Y*), and
(2) V::::; 02(Hi).
PROOF. Recall that QH = CH(UH ); hence as Y = 02 (YQH ), (1) holds. Notice
(2) holds when L/0 2 (L) ~ A5, since Na(Lo) ::::; M by 13.2.2.9. Therefore we may
assume L/02(L) is A5 or A5, and V 1:. 02(Hi). Hence Hi 1:. M, so Hi E 1-lz.
As V 1:. 02 (Hi), by the Baer-Suzuki Theorem there is h E Hi such that I* is
not a 2-group, where I:= (V, Vh). By 13.5.13.2, we may apply F.9.5.6 to conclude
that (V/) is nonabelian. Thus UH 1 is nonabelian, contrary to 13.5.18. D