934 13. MID-SIZE GROUPS OVER F2
LEMMA 13.6.25. (1) Wi(T, V) ::::] LT for i = 0, 1.
(2) n(H 1 ) > 1 for each H1 E 1i(T, M).(3) Each solvable member of 1i(T) is contained in M.
(4) r(G, V) = 3 and w(G, V) > 1.
PROOF. We first observe that by 13.6.3.1 and 13.6.24.1, r(G, V) = 3. Let
g E G-M with A:= VBnM:::; T and Ba hyperplane of A. Suppose m(V^9 /A):::; 1
but [V,A] =f. 1. Then m(VB/B):::; 2 < r(G, V), so Cv(B):::; Na(V^9 ) and hence
[Cv(B),A]::; VnVB; therefore [Cv(B),A] = 1by13.6.24.2. Thus A E A2(Mv, V),
whereas we compute directly that a(Mv, V) = 1. This contradiction shows that
Wi(T, V) :::; Cr(V) = 02(LT) for i = 0, 1, establishing (1).
By (1), w(G, V) > 1, where w(G, V) appears in Definition E.3.23; this com-
pletes the proof of (4). By (4), min{r(G, V), w(G, V)} > 1, so (2) and (3) follow
from E.3.35.1. DLEMMA 13.6.26. For HE 1iz:
(1) No member of C(H) is contained in M.
(2) 02,p(H) = QH for each prime p > 3.
PROOF. Assume KE C(H). Part (1) follows from 13.3.8.2 with L, M, (KT)
in the roles of "K, MK, Y". Let p > 3 be prime and set X :=OP' (0 2 ,p(H)). By
13.6.25.3, X :::; M, so asp > 3, X = 1 by 13.3.8.2. Hence (2) holds. DLEMMA 13.6.27. There exists KE C(G 1 ) such that one of the following holds:
(1) G1 =KT and K/02(K) ~ h or M23·
(2) K/02(K) ~ L3(4), T is nontrivial on the Dynkin diagram of K/0 2 (K),G 1 = 031 (G 1 )T, and either K = 031 (G 1 ) or 031 (Gi/02(G 1 )) ~ PGL3(4).
PROOF. By 3.3.2 there exists H 1 E 1i*(T, M). By 13.3.5.2, H 1 :::; G1, and by
13.6.25.2, n(H1) > 1. Now we apply Theorem 5.2.3: Hypothesis 13.3.1 rules outconclusions (2) and (3) of that Theorem, so we are left with conclusion (1) of 5.2.3.
In particular K1 := 02 (H 1 ) lies in some K E C(Ca(Z)). As T normalizes K1, it
normalizes K, and as K1 i M, KT E 1i(T, M). Thus n(KT) > 1 by 13.6.25.2, so
in particular K/0 2 (K) ~ A1. So by Theorem 5.2.3.1, either
(a) Ki/02(K1) ~ L2(4) and K/02(K) ~ J2 or M23, or
(b) K1 = K with K/02(K) ~ L 3 (4), and T nontrivial on the Dynkin diagram
of K/0 2 (K) by E.2.2.
Next as Ca(Z):::; G1, KE £(G1, T), so K:::; K+ E C(G~) by 1.2.4. If K/0 2 (K) ~
J2 or M23, we conclude K = K+ from 1.2.8.4. If K/0 2 (K) ~ L 3 (4), then eitherK = K+ or K+/02(K+) ~ M23 by A.3.12, and the latter case is impossible as T
is nontrivial on the Dynkin diagram of K/0 2 (K). Thus K = K+ E C(G 1 ).
By A.3.18, either 03 ' (Gl.) =Kor 0
31
(Gif0 2 (G 1 )) ~ PGL 3 (4). In particularK is the unique member of C(G1) which is not a 3'-group, and 02 ,3(G 1 ) = 1 so
that 02,F(G1) = 02(G1) using 13.6.26.2.
Suppose Ko E C(G1) - {K}. By an earlier observation, Ko is a 3'-group, so
Ko/02(Ko) is Sz(2m). By 13.6.26.1, Ko i M, while by 13.6.25.3, a Borel subgroup
of Ko is contained in M. Therefore (Ko, T) E 1i*(T, M), which is contrary to
Theorem 5.2.3 as we saw above.
Let G\ := Gi/02(G1); we have shown that k = P.*(G 1 ). But Out(K) is a
2-group if k is h or M23, while Out(L3(4)) ~ Z 2 x 83. It follows that either