i4.6. ELIMINATING L 2 (2) WHEN (vGi) IS ABELIAN 1033order divisible by each prime dividing 2m - 1, and one of these primes is largerthan 5. On the other hand H n J 0 is a solvable overgroup of T n J 0 in J 0 , and hence
is 2-closed, so H acts on NJ 0 (T n J 0 ) = B, contrary to the previous paragraph
applied to HB in the role of "H+".
Now KE 3(G, T) by 14.6.12, so by 1.3.4, either K = Kz, or Kz = (K'[) forsome Ki E C(Gi) with Ki/02(Ki) quasisimple, and Kz is described in 1.3.4.
Supose K = Kz. If p = 5 then (1) holds, so we may assume that p = 3.
Now k contains all elements of order 3 in Ca 1 (K) since m3(K) = 2 and Gi is an
SQTK-group. Thus if J E C(Gi) then J is a 3'-group, which is impossible by the
claim, so we conclude from 1.2.1.1 that Gi is solvable. Then Gi is a {2, 3}-group
by the first paragraph applied to Gi in the role of "H". Therefore (1) holds when
K=Kz.
Thus we may assume that K < Kz = (K'[) with Ki E C(Gi). As Kif0 2 (Ki)is quasisimple, Kz is described in part ( 4) or (5) of F.9.18. Comparing the lists of
1.3.4 and F.9.18, we conclude that either:
(i) Kz = KiK2 with Kz := Kf for s E T - Nr(Ki), and either Ki ~ L 2 (2m)
with 2m = 1 mod p, or p = 3 and Ki~ L3(2).
(ii) p = 3 and KiT/02(KiT) ~ Aut(Ln(2)), n = 4 or 5.
Notice that the Sp 4 (2n)-case in 1.3.4.3 is excluded, as here Autr(P) is noncyclic
by 14.6.12.
Observe that Kz = QP' (Gi): in case (i), this follows from 1.2.2, and in case
(ii) from A.3.18. Furthermore when p = 5 we have case (i) with Ki ~ L 2 (2m) for
m divisible by 4, so that Kz = 0
31
(Gi) by 1.2.2. Thus Ca 1 (Kz) is a 3'-group, andif p = 5, th~n Ca 1 (Kz) is a {3, 5}'-group. Therefore applying the first paragraph
to H0 2 ,p(Gi) in the role of "H+", we conclude F(Gi) = 1, and by the second
paragraph, Ci has no Suzuki components. Therefore as Ca 1 (Kz) is a 3'-group,
Kz = F*(Gi).
As F*(Gi) = Kz, conclusion (2) of the lemma holds in case (ii), so we may
assume case (i) holds. Similarly conclusion (3) holds if Ki ~ L3(2), since Nr(Ki) is
trivial on the Dynkin diagram of Ki because T acts on K. Thus we may assume that
Ki~ L 2 (2m). Applying the first paragraph to BT in the role of "H+", where Bis a
Borel subgroup of Kz over TnKz, we conclude that m = 2 if p = 3, and that m = 4
if p = 5. If p = 3, then as T induces Ds on K, Ci ~ 85 wr Z2, so conclusion
(3) holds. Finally if p = 5 we showed BT E 1-l(T,M), so for B3 E Syls(B),
B 3 T E 1-l(T, M) by 14.6.1.4. Applying 14.6.12 to B 3 T in the role of "H", we
conclude B3T/02(B3T) ~ ot(2). Therefore Ci ~ Aut(L2(l6)) wr Z2, so that
conclusion (4) holds. This completes the proof of 14.6.14. D
LEMMA 14.6.15. p = 3.
PROOF. Assume otherwise. Then by 14.6.12.1 we may assume p = 5, and
it remains to derive a contradiction. As p = 5, conclusion (2) of 14.6.13 holds,
so we may choose 'Y as in 14.6.13.2; in particular z"Y ::::; u H,i· Also since p = 5,
case (1) or (4) of 14.6.14 holds. Let Uz := [U, Kz]. As UH ::::; U and K ::::; Kz,
UK= [UH,K] :=::; Uz·
In the next several paragraphs we assume K < Kz and establish some pre-
liminary results in that case. First case (4) of 14.6.14 holds, so Gi = KzT and
Kz = KiKf for Ki E C(Gi) with Ki/02(Ki) ~ L2(l6) ands ET-Nr(Ki). We