1056 i4. L 3 (2) IN THE FSU, AND L 2 (2) WHEN .Cf(G, T) IS EMPTY
PROOF. Let Ko := (KT). By 14.7.30, K 0 LiT E Hz, so without loss H =
K 0 LiT. We assume T does not act on Kand derive a contradiction. By 1.2.1.3,
Ko= KKt fort E T-NT(K). Then by 14.7.30 we may apply F.9.18.5 to conclude
that K/02(K) is L2(2n), Sz(2n), or L3(2).
Suppose first that K* ~ L 3 (2). Then by 1.2.2, Ko = 0
31(H), and so Li :::; Ko.
Therefore there is an overgroup Hi of LiT in H with Hi/02(Hi) ~ 83 wr Z2,
and hence by Theorem 14.7.29, Hi:::; M. But then 02 (Hi) =[Li, Hi]:::; L, so that
m3(Hi n L) = 2, contrary to m3(L) = 1.
So K* is L 2 (2n) or Sz(2n). Let Bo be the preimage of the Borel subgroup of
K 0 containing T 0 := T n K 0 , and B := 02 (B 0 ). Then Bo is the unique maximal
overgroup of LiTnK 0 in K 0 , so LiT normalizes B. Hence as Bis solvable, B:::; M
by Theorem 14.7.29, so B acts on Li. However if K is Sz(2n), then B acts on
no subgroup Li of Aut(K) with JLi : 02(Li)I = 3, so that [K 0 , Li] = 1. Hence
Li ::::) H*, contrary to 14.7.7.
We now interrupt the proof of 14.7.32 briefly, to observe that we can use the
previous argument to establish three further results:
LEMMA 14.7.33. If HE Hz and KE C(H), then K/02(K) is not Sz(2n).PROOF. By the reduction above, we may assume that T normalizes K, and
take H = KLiT using 14.7.30; then we repeat the argument for that reduction
essentially verbatim. D
Then using 14.7.33 and 1.2.1.4:
LEMMA 14.7.34. If HE Hz and KE C(H), then m 3 (K) = 1 or 2.
By 14. 7.28 and 14. 7.34:
LEMMA 14.7.35. For each HE Hz, 031(H) = QH.
Now we return to the proof of 14.7.32. Recall we had reduced to the case
where K/02(K) ~ L 2 (2n) and Bo= Kon M. Then 3 divides the order of K, so
Li:::; 0
31
(H) =Ko by 1.2.2, and hence Li:::; 02 (MnK 0 ) = B, son is even. As Li
is T-invariant, Li is diagonally embedded in KKt. Also Li/0 2 (Li) is inverted by
a suitable tL ET n L, so either tL induces a field automorphism on both K and
Kt, or t L interchanges K and K*t.
By 1.2.4, K :::; Ki E C(Gi); then as K < K 0 , 1.2.8.2 says that Ki is not
T-invariant and either K = Ki, or n = 2 and Ki/0 2 (Ki) ~ Ji or L 2 (p) for
p^2 == 1 mod 5. In the latter cases we replace H by Hi := (Ki, LiT) and obtain a
contradiction from the reductions above. Therefore KE C(Gi) and K* ~ L 2 (2n)
with n even. Again by 1.2.2, Ko= 0
31
(Gi), so C := Cc 1 (Ko/02(Ko)) = 02(Gi)
by 14.7.35.
Next as M = LCM(L/0 2 (L)), B = LiBe, where Be:= 02 (CB(L/0 2 (L))) is
of index 3 in B. Further [Be, tL]:::; 02 (Be), so that n = 2 and tL does not induce
a field automorphism on both K and Kt; hence tL interchanges K and Kt.
As n = 2, Out(K 0 ) is a 2-group, so as C = 02 (Gi) we conclude
Gi = KoT = H, ( *)
and hence ·u H = (VG^1 ) = U. As before, our convention will be to also abbreviate