940 13. MID-SIZE GROUPS OVER F2
So to complete the treatment of the case L 1 :::; K, it remains only to eliminate
case (1) with n = 3. So suppose K s=! L 3 (2). Since L1 :S K which has 3-
rank 1, L/0 2 (L) s=! A 6 by 13.7.5.5, and LiT is a maximal parabolic of KT.
Set F := (LT,K 0 L 2 T,KT) and Go := (F). Here KoT/02(KoT) s=! 83, and by
13.5.4.1, [Ko,L2] :S 02(L2). As M = !M(LT), 02(Go) = 1. Thus if [Q[ > 2^5 , then
(G 0 ,F) is a 0 3 -system as defined in section I.5, so by Theorem I.5.1, Gos=! Sp5(2).
Therefore Z(KT) = 1, whereas z E Z(KT) as H E Hz. Thus [Q[ :S 25 , so L
is an A 6 -block. But then L 1 has just two noncentral 2-chief factors, whereas L1
has noncentral chief factors on each of 02 (Lt), UH, and QH/Hc by 13.7.4.2. This
contradiction completes the proof of (1) and of the lemma in the case L1 :::; K.
It remains to treat the case L 1 1:. K. When m 3 (K) = 2, K = 031 (H)
by A.3.18 and A.3.19, so K/0 2 (K) s=! L3(2n)), n odd, or A5. Let X := L1
if L/0 2 (L) s=! A 6 , and X := L 1 ,+ if L/02(L) s=! A5. Suppose [K,X] = 1.
Then as EndK(UK/CuK(K)) = F2n with n odd, X centralizes UK = [K, UH]·
But then by the Three-Subgroup Lemma, [UH,X,K] = 1; so as V3 = [i/3,X],
K = 02 (K) :::; Cc(Vi) :::; Mv by 13.5.4.4, and hence (KT) :::; M, contrary to
13.3.9. Therefore K = [K, X]. So as X = [X, T], we conclude that X induces inner
automorphisms on K/0 2 (K). Then again as X = [X, T], K/0 2 (K) is not L3(2m)
form> 1, and either:
(a) X :::; K, and hence X < L 1 as L1 1:. K, so that L/0 2 (L) s=! A5, or
(b) K s=! A5 and X is diagonally embedded in KCK·x·(K), with Va pro-
jecting nontrivially on UK.
But now K* s=! A 5 is ruled out, since in both cases (a) and (b), V 3 = [i/ 3 , X], while
either X or its projection on K is a Borel subgroup of K, which has no such
T-invariant submodule of rank 2 on the A 5 -module UK. Therefore case (a) holds
and K s=! L3(2). Now 13.7.5.7 supplies a contradiction, completing the proof. D
LEMMA 13.7.7. [VH,H] 1:_ UH.PROOF. If [VH, HJ '.SUH, then VH = (VH) = VUH. Then by 13.7.3.6,
UH= [VH,QH] = [V,QH][UH,QH] = ViV1 =Vi,.
contrary to 13.5.9. D
13. 7.2. The elimination of the case (VG^1 ) nonabelian. We come to the
main result of this section, which reduces the treatment of Hypothesis 13. 7.1 to the
case where (V^01 ) is abelian. Then in the following section 13.8, that remaining
case is also shown to lead to a contradiction.
THEOREM 13.7.8. Assume Hypothesis 13. 7.1. Then (V^01 ) is abelian.Until the proof of Theorem 13.7.8 is complete, assume G is a counterexample.
Then the set Hi of those HE Hz with VH nonabelian is nonempty, since G 1 E H 1.
For the remainder of the section, let H denote a member of H 1.
Then VH is nonabelian, though UH is abelian by 13.7.2.
LEMMA 13.7.9. (1) cI>(VH) =Vi, VH = ((5,6)), and QH = R1.
(2) L/02(L) s=! A5 rather than A5. In particular [L 1 [ 3 = 3.
{3) H* is faithful on UH/CH(QH)·