9SS 14. L 3 (2) IN THE FSU, AND L2(2) WHEN L.:r(G, T) IS EMPTYWe prove:
THEOREM 14.2.20. Let HE 1-i*(T,M). Then
(1) If H/0 2 (H) ~ D 10 or 8z(2), then H/0 2 (H) ~ 8z(2) and G ~^2 F4(2)'.(2) If H/02(H) ~ L2(2) then G ~ Mi2 or G2(2)'.
Again we assume that H satisfies one of the hypotheses. of Theorem 14.2.20,
and we begin a series of reductions.
Let G 1 := H, G 2 :=YT, and Go := (G1, G2). Then G1 n G2 = T. We checkeasily that G 1 , G 2 , T satisfy Hypothesis F.1.1 in the roles of "L1, L2, 8": For
example since H 1:. M, 02(Go) = 1 by 14.2.2.3. By F.1.9, a := (G1, T, G2) is a
weak EN-pair of rank 2. Set K := 02 (H).
LEMMA 14.2.21. H = Mc, M = YT, and one of the following holds:
(1) a is the amalgam of^2 F4(2) or of the Tits group^2 F4(2)'.
(2) a is the amalgam of Mi2 or of Aut(M12).
(3) a is the amalgam of G2(2)' or of G2(2).
PROOF. Since T = Nai (T), the hypothesis ofF.1.12 holds. Since G2/Ca 2 (V) ~
L 2 (2), while Gi/0 2 (G 1 ) is D10, 8z(2), or L2(2) with G1 centralizing Z, we con-
clude from the list of F.1.12 that either a appears in conclusions (1)-(3) of 14.2.21,
or a is the amalgam of 8p 4 (2). However in the latter case, IZI = 4, contrary to
14.2.2.6.
Thus it remains to show Mc= Hand M =YT. If Mc= H, then Mn Mc= T,
so CM(V) = CT(V), and then M =YT by 14.2.2.1. So it suffices to show Mc= H. ·
Let Kc:= 02 (Mc)· If K =Kc, then H =KT= KcT =Mc, so we may assumeK <Kc, and it remains to derive a contradiction.
Let Q := 02(Mc)· Then Q::; QH by A.1.6, and F*(Mc) = Q by A.1.8, so
Z(QH)::; CMJQ)::; Q::; QH. (*)
Suppose first that a is the amalgam of G2(2)', G2(2), or M 12. Then QH is
abelian, so QH = Q by(*). Hence if a is the G 2 (2)'-amalgam, then Q ~ Qs * Z 4
is the central product of Qs and Z 4. Therefore 02 (Aut(Q)) ~ A 4 ~ AutK(Q),
so K = Kc, contrary to our assumption. Hence a is the amalgam of G 2 (2) or
Mi2, so Q = QH ~ Q§, and hence Out(Q) ~ Ot(2). Then as K < Kc, Kc ~
8L2(3) *8L2(3). Next YT~ D12/Z~, so V::; E::; Q, where Es~ E :SI YT. Hence
Na(E) ::; M by 14.2.2.3. But Eis a maximal totally singular subspace of Q, sofrom the structure of Kc, 84 ~ AutMc (E) is the stabilizer in GL(E) of z. Then since
Y does not centralize Z, Auta(E) ~ L 3 (2), contradicting Na(E)::; M = Na(Y).
Assume next that a is the Aut(M 12 )-amalgam. Then QK := [QH, K] ~ Q§
and Q H ~ E4 wr Z2. Therefore we conclude from (*) that either Q is Q H or Q K,
or else Q ~Es is the maximal abelian subgroup of QH distinct from QK.
Assume this last case holds. Then Q ~ Em and 84 ~ H/Q ::; Mc/Q, with
Mc/Q contained in the stabilizer L 3 (2)/Es in GL(Q) of the point Z of Q. Further
T/Q ~ Ds is Sylow in Mc/Q, so as K < Kc, we conclude Mc/Q ~ L3(2) acts
indecomposably on Q. But then Mc E Lt(G, T), contrary to 14.2.1.1.
So Q = QH or QK, and therefore QK = J(Q) :SI Mc. In either case, Q§ ~
QK :SI Mc. Then as above, K <Ke, implies Kc~ 8L 2 (3) * 8L2(3). Now from the
structure of Aut(M12), J(T) ~Em is normal in YT, so Na(J(T))::; M = !M(YT).