920 i3. MID-SIZE GROUPS OVER F2{6) H = Gi and Gi is the unique member of Hz.
(1) M =LT and V = 02(M).
{8) If n = 5 then QH =UH, H*::::; nt(2), and either
{i) M = L with H* ~ 83 x Z3, or
{ii) M/V ~ 85 with H* = nt(2) ~ 83 x 83.
{9) If n = 6 then H* = nt(2) ~ 83 x 83, and either
{i) M = L with Q H = UH, or
{ii) M/V ~ 85 with QH = UHCT(Li) and ICT(Li)I = 4.PROOF. We saw UH E R2(fI), so as Li ::::1 H, 02(Li) = 1. Hence Li ~ Z3.
Also V3 = [V3, Li], so as UH = (Vl), UH = [UH, Li]. Then (1) follows from
13.5.14.1, and (2) from 13.5.14.2.Choose l and K as in 13.5.14, and let Q := 02 (LT) and L_ := 02 (KQ). By
13.5.14.4, the hypotheses of G.2.4 are satisfied and K = L_. By (1), the hypotheses
of G.2.4.8 are satisfied. Therefore by G.2.4.8, K is an A5-block with V = 02(K)
and UH = 02 (Li) ~ Q~. Now if n = 5, then L = L_ by construction. If n = 6,
then as Q acts on L and L = K, [K, Q] ::::; 02 (K) = V, so L is an A 6 -block.
Thus (3), (4), and (5) are established.We saw V is of order 2 and 02 (H) = 1, so by the Baer-Suzuki Theorem, there
is g EH with I* not a 2-group, where I := (V, VB). Now by 13.5.13.2, we may
apply F.9.5.6 to conclude that 02(I) =Ur:= Vi V:f ~ Q~ and I/02(I) ~I*~ 83.Therefore since Ur ::::; UH and UH ~ Q~, we conclude 02(I) = UH, and hence
I*= IQH/QH ~I/UH~ 83.
Next as CH(UH) = QH, H* = H/QH S Out(UH) ~ Ot(2). As Li ~ Z3
is normal in H, and Li centralizes V, Li centralizes I*. But the centralizer
in Out(UH) ~ Ot(2) of Li is isomorphic to 83, so we conclude I* = CH·(Li).
Therefore either H ~ 83 x 83, or H =I* x Li ~ 83 x Z3 with T = 02(LiT),
and the latter case can only occur when n = 5 and Mv = L ~ A5. In either case
H = QHLiIT = LiIT. Further if we establish (7), then QH n Q = QH n V =Vi,
so IQHI = 8IQHI, and then (8) and (9) follow using (5). So it remains to prove (6)
and (7).As QH acts on V and VB, QH acts on I. Then as I ::::1 H, H acts on
02 (IQH) = 02 (I), and then LiT acts on 02 (I)V =I. Thus I ::::1 LiIT = H.
As K is an A 5 -block by (3), Q = V x CT(K) by C.1.13.c. Further Li ::::; K
by 13.5.14.3, and CT(Li) = ViCT(K). Thus as UH = [UH, Li] ::::; K, CT(Li) ::::;
CT(UH), so [I,CT(Li)] ::::; Gr(UH) =Vi, and therefore [0^2 (I),CT(Li)] = 1 by
Coprime Action. In particular 02 (I) centralizes CT(L). As 02 (I) = [0^2 (I), V]
and V::::; 02 (M), 02 (I) 1: M. Therefore as M = !M(LT), we conclude CT(L) = 1.
We will show next that D := CT(K) = 1. If n = 5, then K = L, so D =
CT(L) = 1; thus we may assume n = 6. Then as Lis an A5-block, we conclude
from C.1.13.b and I.1.6.5 that either D = CT(L) = 1 or IDI = 2 =IQ: v1. Assumethat the latter case holds and set GD := Ca(D). We saw that 02 (I) centralizes
CT(Li) 2 D, so (0^2 (I), K) ::::; GD. Let G"j) := GD/D and To := CT(D) E
_8yb(CM(D)). Let·To STD E 8yl2(GD); then ITD : Toi S IT: Toi = 2, so as
KE .C(GD,To), there exists a unique KD E C(GD) containing K by 1.2.5, and
02 (I) normalizes KD by 1.2.1.3. As 02 (I) = [0^2 (!), V] and K is irreducible on
V, V n 02(KDD) = 1. Let Ti := To n KDD; thus Ti E 8yb(M n KDD). Then
as Na(Q) SM= !M(LT), Ti E 8yl2(NKvD(Q)). Therefore as Q = 02(KTi) =