1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

(jair2018) #1
942 13. MID-SIZE GROUPS OVER F2

and DHg ::; Z1 by (5), so that EHg ::; Z1::; VH, and hence EHg ::; Q n He by (2)
and 13.7.3.2. But Z1 ::; EHg, so EHg = Z1, and then by symmetry EHg = Z1 =

EH::; VH. Then [EHg, VH]::; Vi by 13.7.3.5, completing the proof of (7).

Next there exists l E L with Li = 02 (1)0 2 (.Ei) ~ A4. We saw Q acts on I,
so Li has k + 1 noncentral 2-chief factors, where k is the number of noncentral

2-chief factors of I. One of those k factors is Vi V.f!, and by (2) and (3) there are

k -1 = m(0 2 (I)/Z1)/2 = m(A/Z1) factors on 02(I)/Z1. Now
m(A/Z1) = m(A/EHg) = m(A/(A n QH)) = m(A*)
by (7), so that (8) holds. As V3 n V:f is a complement to Vi in V3, and V1 1:. Ufr by
(1), (9) holds.

Since EHg ::; Q by (7), (10) is immediate from the definitions of A and EH9· D

LEMMA 13.7.11. (1) DH< UH.


(2) 1-/-Ufr* and Ufr ::; A.

PROOF. Recall Ufr::; A, UHg = (UH)^9 , and DHg = (DH)^9. Therefore DH=
UH iff DIJr = UHg iff UHg::; QH = CH(UH); and hence (1) and (2) are equivalent.

Thus we may assume that UH = DH, and it remains to derive a contradiction.

By 13.7.10.6, UH = DH centralizes A, so A ::; QH. Thus by 13.7.10, A = An
QH = EHg, while by 13.7.10.7, EHg =EH::; VH; so using symmetry we conclude
v;;-n Q = VH n Q. Let A be the graph on the points of v obtained by joining
non-orthogonal points. Then A is connected, so VH n Q = Vjj n Q for all x E L.
Therefore L acts on VH n Q. Now (VH) = V 1 by 13.7.9.1, so as L does not act
on Vi, (VH n Q) = 1. Also by 13.7.9.1, VH = Z(VH)Vo with Vo extraspecial and
IVH: VH n QI= 2; so we conclude (Z(VH)) = 1 and Vo~ Ds. But now, VH has


just two maximal elementary abelian subgroups, one of which is Z(VH)V; so both

are normal in 02 (H)T = H, and hence (VH) = VH = Z(VH)V is abelian, contrary
to our choice of Gas a counterexample to Theorem 13.7.8. D


Recall that Uo = CuH(QH), and from 13.7.4 and 13.7.9.2 that U"Ji = UH/Uo is


H-dual to QH/He and H* is faithful on U"Ji.

LEMMA 13.7.12. (1) A* centralizes (QH n Q)He/He of corank 2 in QH/He.
(2) [U"Ji,A*]::; v 3 +.
(3) For FE {A*,Ufr*}, rFu ,H ::; 1 2 rpu+, 'H so F contains FF*-offenders on
each of these FF-modules.

PROOF. As g E Na(Q), Q acts on A, so by 13.7.10, [A, QH n Q] ::; An QH =
EHg, and EHg ::; He by 13.7.10.7. Further by 13.7.9.1, QH = R 1 ~ E 8 and VH is of
order 2, so VH =He by parts (2) and (7) of 13.7.3. Thus IQH: (QH n Q)Hel = 4,
completing the proof of (1). Then since V3 centralizes Q and Cv 3 ( QH) = Vi is of
index 4 in V3, v;+ corresponds to ( Q H n Q) He/ He under the duality between U"Ji
and QH/He, so part (2) is the dual of (1). By 13.7.10.6, [DH,A] = 1, and


m(Ufr*) = m(Ufr/DHg) = m(Ufr/DIJr) = m(UH/DH),


so rF,UH::; 1 for FE {A, Ufr}, keeping in mind that 1-/-UJ/::; A* by 13.7.11.2.


Then rpu+ >H ::; 1 also holds using 13.7.4.1. Thus both modules are FF-modules for

H, and B.1.4.4 shows that F contains FF-offenders on the modules. D

LEMMA 13.7.13. Ifm(UH) = 4 then [UH,L1] <UH.