i3.3. STARTING MID-SIZED GROUPS OVER F 2 , AND ELIMINATING U 3 (3) 89iWe recall from Notation 13.3.3 that since Cv(L) = 1 by Hypothesis 13.3.13,
we have m(Vi) = i. In particular by 13.3.4.1, Vi = Zn V is of order 2, and from
13.3.2,3, V3 = [V3,Li] = (V 2 L^1 ). Also Gi = Na(Vi) = Ca(Vi) and Gi E He as G is
of even characteristic.LEMMA 13.3.14. Assume Hypothesis 13.3.13. Then Qi does not centralize Vi.
PROOF. We assume that [Vi, Qi] = 1 and derive a contradiction. The bulk of
the proof proceeds by a series of reductions labeled (a)-(g).
Set U := (V 3 °^1 ) and Gi := Gi/Qi. Set L+ := L1,+ if L/0 2 (L) ~ A5, and
L+ :=Li otherwise. Then 02(L+T) =Riis of index 2 in T, and as V 3 = [V3,Li] =
(v:}^1 ), V3 = [V3,L+] = (Vf'+).Observe that Hypothesis G.2.1 is satisfied with V3, Gi, L+ in the roles of "V,
H, L". Therefore by G.2.2.1, U :S Qi. As V 3 = (V 2 L+), U = (V 2 °^1 ). Then as
[Vi, Qi]= 1:
(a) U :S Sli(Z(Qi)).Observe that as Cv(L) = 1 by Hypothesis 13.3.13, Gi :t_ M by 13.3.6. Set
Y := 02 (Ca(U)). Then Y ::::; Ca(V3) ::::; Ca(Vi) ::::; Gi. Also Y j Gi, and
Y n M ::::; Mv by 12.2.6.
(b) Y is solvable with mp(Y) ::::; 1 for each odd prime p.
For if Y = 1 then certainly (b) holds, so we may assume Y =/= 1. Consider any T-invariant subgroup Yo = 02 (Yo) of Y n M. As C Mv (Vi) is a 2-group, Yo centralizes
V. Then [L, Yo] ::::; CL(V) = 02,z(L), so L = L^00 centralizes Y 002 (L)/0 2 (L).
Hence as Yo is T-invariant, L acts on 02 (Y 0 02(L)) =Yo. Therefore if Yo=/= 1, then
Na(Yo)::::; M = !M(LT). In particular Y 1. M, as otherwise Gi::::; Na(Y)::::; M.
Now if L ~ L3(2), then V = V3 ::::; U, so Y::::; Ca(U) ::::; Ca(V3) = Ca(V) ::::; M,
contrary to the previous paragraph. Similarly if L ~ As, then Ca(V3) ::::; M by
13.2.3.2, for the same contradiction. Therefore we may assume that L is A 6 or
U3(3).
Let g E L2 - Gi be of order 3. Then m(V3 Vi) = 4, so V3 Vi = V if L ~ A5,
and V3 Vi is not totally isotropic if Lis U3(3). In the latter case Ca(V3 Vi) ::::; M
by 13.3.12.2, while if L is A5, then Ca(V3 Vi) = Ca(V) ::::; M. So in either case,
Ca(V3 Vi) ::::; M.
Set Yi:= 02 (Y n Y^9 ) and YM := 02 (Y n M). Then Yi::::; Ca(V3V:f)::::; M, so
Yi ::::; YM· Further Y centralizes V2, so Y ::::; Gf ::::; Na(Y9), and hence Yi j Y;
then by symmetry, Yi j Y9. Next T::::; Mi ::::; Na(YM), so using YM in the role
of "Yo" above, L acts on YM and Na(YM)::::; M; hence YM = Yl£::::; Yi as g E L2.
We conclude YM =Yi, so if Yi =/= 1, then Y ::::; Na(Yi) ::::; M, contrary to the first
paragraph. Therefore Yi = 1, so that Y n yg is a 2-group.
Set Ch:= G2/02(G2). As Y j Gi while Ca(V2)::::; Gi, Y j Ca(Vi), so that
02 (Y) ::::; 02 (G 2 ), and hence Y+ := (Y, Y^9 ) j Ca(Vi). Then as Y and Y^9 are
normal in Y+, and Y n yg is a 2-group normal in Y+, 1\ = Y x Y9.
Therefore since G 2 is an SQTK-group, mp (Y) = 1 for each odd prime p. Further
ifY is not solvable, then by 1.2.1.1, there is KE C(Y), and as Y j G 2 , KE C(G 2 ).
Then as g is of order 3, g acts on K by 1.2.1.3, contradicting Y n yg a 2-group.
This contradiction completes the proof of (b).
(c) 02(L'f-) =/= 1.