i4.3. FIRST STEPS; REDUCING (VG1) NONABELIAN TO EXTRASPECIAL 99iLEMMA 14.3.5. Assume L/02(L) ~ L 2 (2)' and HE H(T) with IH: Tl= 3 or
5. Then H :::;: M. ,
PROOF. Assume H i M. By 14.3.3.3, Hi Nc(T) so that Hj0 2 (H) ~ 33 ,
D10, or 3z(2). But the groups G appearing as conclusions in Theorem 14.2.20 areexcluded by Hypothesis 14.3.1.2, so we conclude that the lemma holds. D
LEMMA 14.3.6. Assume L/02(L) ~ L2(2)' and HE H(T,M) such that K :=
02 (H) = (K'[) for some Ki E .C(G, T). Then
(1) If K/02(K) is of Lie type over F2n of Lie rank 1 or 2, then either
(i} n = 1, K/02(K) ~ L3(2) or A5, and T is nontrivial on the Dynkindiagram of K/02(K), or
(ii} M does not contain the Borel subgroup of K over T n K.
(2) If K/02(K) is of Lie type over F2 of Lie rank 2, then K/0 2 (K) ~ L 3 (2)
or A5, and T is nontrivial on the Dynkin diagram of K/0 2 (K).
{3} If K/02(K) is of Lie type over F4, then KT/0 2 (KT) ~ Aut(3p 4 (4)) or
85 wr Z2.
(4) If K/02(K) ~ L4(2) or L5(2), then Tis nontrivial on the Dynkin diagramof K/02(K).
(5) K/02(K) is not A1.
(6) K/02(K) is not Mi2, M22, or M22-
PROOF. Assume that K either satisfies the hypotheses of one of (1)-(4) or
is a counterexample to (5) or (6). Then K/0 2 (K) is either quasisimple, or else
semisimple of Lie type in characteristic 2, and of Lie rank 1 or 2 using TheoremC (A.2.3). Thus as K = (K'[) with K E .C( G, T), using 1.2.1.3 we conclude that
either K/0 2 (K) is quasisimple, or K is the product of two T-conjugates of Ki< K
with Ki/02(Ki) ~ L 2 (2n) or 3z(2n) and n > 1.
Assume the hypotheses of (1). We may assume that (ii) fails, so that Mn K
contains the Borel subgroup B of Kover T n K. Let Ho be the set of subgroups
(P, T), such that Pis a rank one parabolic of Kover B. Then H =(Ho). So as Hi
M, there exists H 0 E Ho with Ho i M. Then Ho= H2B where H2 E H*(T, M).
Since Hypothesis 14.3.1 excludes the groups in Theorem 14.2.7, we conclude that.
n(H 2 ) = 1. Hence K/02(K) is defined over F2. Then from the first paragraph,
K/0 2 (K) is quasisimple. If Tis trivial on the Dynkin diagram of K, then Ho is a
rank one parabolic, so as K/0 2 (K) is quasisimple and defined over F 2 , IH2 : Tl = 3
or 5 from the list of such groups K/0 2 (K) in Theorem C, contrary to 14.3.5. Thus
Tis nontrivial on the diagram, so again from that list, conclusion (i) of (1) holds.This completes the proof of (1).
If (2) fails, then conclusion (ii) of (1) must hold, so B i M. In particular, a
Cartan subgroup of Bis nontrivial, so as K/0 2 (K) is defined over F 2 , we conclude
from the list of Theorem C that K/02(K) ~^3 D4(2) and IB : T n Kl = 7. Now
B :::;: Nc(T) since Out(K/0 2 (K)) is of odd order, so B :::;: M by 14.3.3.3, contrary
to the first sentence of this paragraph. Thus (2) is established.
Assume the hypotheses of (3); then by the first paragraph, K/0 2 (K) is either
quasisimple of Lie rank at most 2, or L 2 (4) x L 2 (4). Let B be the T-invariant Borel
subgroup of K. By (1), B i M, so there exists H 2 E H*(T, M) with H 2 :::;: BT.
Inspecting the groups in Theorem C defined over F 4, either B /02(B) ~ Z3 or E 9 ; or
K/02(K) ~ U3(4) with B/02(B) ~ Z15; or K/02(K) ~^3 D4(4) with B/02(B) ~