12.6. ELIMINATING As ON THE PERMUTATION MODULE 831We next prove (7). Most parts of Hypothesis F.9.1 are easy to check: For
example, (4) implies hypothesis (c) of F.9.1, (1) implies (b), and (d) holds asM = !M(LT) but Hf:. M. Assume the hypothesis of (e); as the conclusion of (e)
holds trivially if [V, VB] = 1, we may assume 1 =f. [V, VB]. Then 12.6.20.4 says VB is
the unipotent radical R of the stabilizer of some 3-subspace of V, while (6) and the
hypothesis of (e) that VB centralizes Vs forces VB ~ R 1. This contradicts R 1:. R 1 ,
and completes the proof of (e). As HE H(T), HE He by 1.1.4.6. If Xis a normal
subgroup of H centralizing i/5, then by (6) and Coprime Action, 02 (X) centralizes
V, so [L, 02 (X)] ~ 02 (L). Hence LT normalizes 02 (X0 2 (L)) = 02 (X). Then
if X is not a 2-group, H ~ Na(0^2 (X)) ~ M = !M(LT), contrary to H f:. M.Therefore X ~ 02(H) = QH, so that hypothesis (a) of F.9.1 holds. Thus we have
established (7).
Next (7) and parts (1) and (3) of F.9.2 imply (2) and (8), respectively. By (2),
Vs ~ 02(G1) n Cc(V2), so as Cc(V2) ~ G1, Vs ~ 02(Cc(V2)) ~ 02(G2). Then
v = (Vl^2 ) ~ 02(G2). Hence (3) holds. If g E G - Mv with V1 < v n VB, then
by 12.6.20.1, V n VB is totally singular; so without loss V2 ~ VB, so that VB ~ G 2.But then (V, VB) is a 2-group by (3). Hence (5) holds. D
LEMMA 12.6.25. The following are equivalent:
(1) UH is abelian.(2) V ~ QH..
(3) V ~ Cc(UH)·
(4) VH is abelian.PROOF. Parts (2) and (3) are equivalent by F.9.3, which applies by 12.6.24.7.
Similarly the hypotheses of F.9.4.3 are satisfied by 12.6.24.6, so (1), (2), and (4)
are equivalent by F.9.4.3. DObserve since H ~ Gz = G1, that if VH is nonabelian, then Vc 1 is also non-
abelian.In the non-quasithin shadows mentioned earlier (such as nt(2)), VH is non-
abelian. Hence these shadows are eliminated in the next result:LEMMA 12.6.26. VH is abelian.PROOF. Assume that VH is nonabelian. Then by 12.6.25, UH also is nonabelian
and V f:. 1. Notice the hypotheses of F.9.5 are now satisfied, so in particular V
is of order 2 and [UH, V*] = Vs by parts (1) and (2) of F.9.5. By 12.6.24.5, the
hypothesis of part (5) of F.9.5 is satisfied. Also m(V) = 6, and by 12.6.24.6,
CH(Vs) = CH(V), so the hypotheses of LL.5.6F.9.5.6ii are satisfied, and hence we
can appeal to that result also. For example if g E H such that I := \V, VB)
is not a 2-group, then by F.9.5.5, I is faithful on U1 := V5 Vf; and by F.9.5.6ii,
U 1 ~ Q§ and J ~ D 6 , D 10 , or D 12. Therefore elements of odd order in H inverted
by V* are of order 3 or 5.
We show first that V ~ 02(CH(Li/02(L1))): Assume otherwise. Then by
the Baer-Suzuki Theorem, for some g E CH(Li/02(L1)), I:= (V, V*B) is not a
2-group. By the previous paragraph, I is faithful on U1 ~ Q§ and J is dihedral
of order 2m, m = 3, 5, or 6. Also as g E CH(Li/02(L1)), as V centralizes Li,
and as 02 (Li) = CLi (i/5), we conclude that I centralizes Li and Li acts on Vtf
with 02(Li) = CLi(V/). Let IL;.:= Outn 1 (U1). Then L1 ~ Eg is faithful on Vs