900 13. MID-SIZE GROUPS OVER F2Let Ht := H 0 /0 3 1(Ho). As CH 0 (Vo) is a 3'-group, H+ is a quotient of H*.
Observe that H+ is described inF.6.11. By F.6.6, 02 (Ho) = (K1, Kz), so 02 (Ho) =
()(Ho).
Suppose Ho :s; M. Then 02 (H 0 ) :s; ()(M) = L by 13.4.2.5, so as each Kiis T-invariant, Ki :s; Lk(i) for some k(i) := 1 or 2 by 13.3.4.3. By 13.4.3.3, L1
centralizes Z; and if L / 02 ( L) ~ A 6 then Lo :s; L 1 , so Lo centralizes Z. Therefore
as [Z,Ki] =f. 1 for each i by (b), it follows that K1 = Kz = Lz if L/02(L) ~ A5,
while K 1 = K 2 = L 2 ,+ if L/0 2 (L) ~ A 6. But K1 =f. Kz, as otherwise H1 = TK1 =
TK2 =Hz, contrary to hypothesis. Hence Ho i M.
As Ho i M, Hk i M for some k := 1 or 2. Therefore CzL (Hk) = 1, as
otherwise lh :s; Cc(CzL(Hk) :s; M = !M(LT). But Cz(Hk) is a hyperplane of
Z by (b), while 1 =f. Zv :s; ZL and ZL is also a hyperplane of Z by 13.4.3.2. We
conclude that m(Z) = 2 and ZL = Zv and Cz(Hk) are of rank 1. As Kj =
0
31
(CH 0 (Cz(Kj))) by (c) and K1 =f. Kz, Cz(H1) = Cz(K1) =f. Cz(K2) = Cz(Hz).
Then we conclude from (b) that Z = Cz(H 1 ) x Cz(H2) with m(Cz(Hi)) = 1 for
each i. As Vi = Zn V is of rank 2, Z = Vi :s; V. Thus we have established the
initial conclusions of (1), so it remains to show that one of conclusions (i)-(iv) of
(1) holds.
Suppose first that [Ki,K2] = 1. Then 02 (H 0 ) = KiK2, so KiCH 0 (Vo) is
normal in Ho for i = 1,2. Furthermore CH 0 (Vo) normalizes Ki by (c), and we
saw CH 0 (Vo) is a 3'-group, so Ki= 031 (KiCH 0 (Vo)) :::1 Ha. It follows that Ho=
H1H2 and [K1, Kz] :s; 02(K1) n 02(K2) :s; 02(Ho); hence Ho/02(Ho) ~ 83 x 83
or Z2/ Eg. Set Vi := [Vo, Ki]· As Z = Cz(H1) x Cz(H2) with /Cz(Hi)/ = 2,
Vi= (Cz(H 3 -i)H;) ~ E4 is centralized by H3-i, so we conclude that case (i) of (1)
holds. Thus we may assume from now on that [Ki, K2] =f. 1; under this assumption,
we will show that one of (ii)-(iv) holds.
We first consider the case where H 0 is not solvable, which will lead to (ii) or (iv).
By 1.2.1.1, there is Ko E C(Ho) with K 0 =f. 1. Then by 13.4.5.1, Ko :::1 H 0 , Ko E
Cj(G, T), and K 0 /0 2 (K 0 ) is listed in 13.4.5.1. In particular Ko is not a 3'-group, so
that Kt =f. 1; hence Ht is described in F.6.18 by F.6.11.2. Also Kt is a quotient
of Ko/02(K 0 ), so comparing the possibilities for Ko/0 2 (Ko) in 13.4.5 with the
possible quotients Kt in cases (3)-(13) of F.6.18, we conclude Kt must be L 3 (2),
A6, or A5, with Ht= KtT+ appearing in case (6) or (8) ofF.6.18. Furthermore if
Ko/02(Ko) ~ A5, then as CH 0 (Vo) is a 3'-group, K 0 ~ Ko/02(Ko) ~ A5, contrary
to 13.4.5.4. Thus in any case CK 0 (Vo) = 02(Ko) = 031(Ko), so Ko/02(Ko) ~
K 0 ~Kt, and Ka/02(K 0 ) is not A 6 •
As Ht= KtT+, Ho= KaT031(Ho), so as Ko :::1 Ho, Ko= 031 (Ho). ThusKi :s; Ko for i = 1, 2, so Ko = 02 (H) by F.6.6, and hence Ho = KaT. Since
K1 =f. Kz, Hi and H2 are the minimal parabolics of H over T.
By 13.4.5.3, we may choose a T-invariant I E Irr +(Ko, Vo) in the FSU. By
13.4.5.4, CK 0 (I) = CK 0 (Vo), so that K 0 = Ko/CK 0 (I). By 13.4.5.3, I is either a
natural module for K 0 or a 5-dimensional module for K 0 ~ A 6 • As Hi. and H2 are
the minimal parabolics of H*, some Ki (say K 1 ) centralizes Zn I, so as [ Z, K 1 ] =f. 1
by hypothesis (b), IZ > ICz(K 0 ). Thus K 0 is nontrivial on V 0 /I. Hence if
J(T) i 02 (H 0 ), then by Theorem B.5.1, [Vo, Ko] is the sum of two isomorphic
natural modules for K 0 ~ L 3 (2); since m(Z) = 2, this contradicts [Z, K 1 ] =f. 1.