5.3. IDENTIFYING RANK 2 LIE-TYPE GROUPS 659Thus it remains to show Na(S) ::::; Mi. But as SE Syb(Li) and Tis in a unique
maximal subgroup of LiT, we conclude from Theorem 3.1.1 that 02 ( (Na(S), Li)) =f=.
- Therefore Na(S) ::::; Mi = !M(LiT) by (1). Thus (3) is established. Then (4)
follows from (3) via a Frattini Argument. D
LEMMA 5.3.2. "( is an M(a)-extension of a (in the sense of Definition F.4.3),
for some extension M(a) of G(a).
PROOF. Let Mo := (M1, M2)· We first verify that"( satisfies Hypothesis A of
the Green Book [DGS85], with Li in the role of "Pt'.
By 5.3.1.1, 02(Mo) = 1. By 5.3.1.2, F*(Mi) = 02(Mi) = 02(Li), so condition
(ii) of Hypothesis A holds. Then as 02 (Mi) = 02 (Li), condition (i) holds by 5.3.1.4.
Condition (iii) follows from 5.3.1.3, and the list of possibilities for Li in 5.2.6. This
completes the verification of Hypothesis A.
As Hypothesis A holds, and q > 2 by 5.2.7.1, case (a) of Theorem A in theGreen Book [DGS85] holds, so that 'Y is an extension of the Lie amalgam a of G(a).
That is, 'Y determines subgroups Mi(a) ~ Mi of Aut(G(a)), with corresponding
completion M(a) := (M 1 (a),M2(a))::::; Aut(G(a)). So the lemma holds. D
orLet Zs := Z(S) and Zi := Z(Li)·
LEMMA 5.3.3. Either(1) The hypotheses of Theorem F.4.31 are satisfied, with G in the role of ''M ",
(2) G(a) ~ Ls(q), and Oa(z) i. Mi,2 for each involution z E Zs.
PROOF. By 5.3.2, 'Y is an extension of the Lie amalgam a, so that M(a) plays
the role of "M" in Theorem F.4.31. Hypothesis ( d) of F.4.31 holds for G in the
role of "M", as T::::; Mi,2 and TE Syl2(G). Hypothesis (e) holds as G is simple.Hypothesis (a) follows from the fact that LiT is a uniqueness subgroup by 5.3.1.1.
Further ED is transitive on zf. Thus if Zi =f=. 1, each involution in Zi is conjugate
under Mi to some z E Z(LiT), and therefore Oa(z) ::::; Mi using 5.3.1.1. Similarly if
G(a) ~ Ls(q), then ED is transitive on zff, so if Oa(zo)::::; Mi for some zo E zff,then Ca (z) ::::; M 1 for all z E zff. Hence the first statement in Hypothesis ( c) holds,
and either Hypothesis (b) holds, or conclusion (2) of 5.3.3 holds. If G(a) is Sp4(q),
then each involution z in Zs is fused into Z(T) under ED, and hence Oa(z) E He
by 1.1.4.6. This completes the verification of Hypothesis ( c). Therefore either thehypotheses of F.4.31 are satisfied, so that conclusion (1) of 5.3.3 holds, or conclusion
(2) of 5.3.3 holds. · D
THEOREM 5.3.4. Either
(1) G ~ G(a), or
(2) G(a) ~ Ls(q), and Oa(z) i. Mi,2 for each involution z E Zs.
PROOF. If 5.3.3.l holds, we may apply Theorem F.4.31 to conclude G ~ M(a).Since G is simple, we must in fact have M(a) ~ G(a). D
By Theorems 5.2.9, 5.2.10, and 5.3.4, Theorem 5.2.3 holds unless possibly G is
of type L 3 (q) and conclusion (2) of 5.3.4 holds. We will finish by showing (in 5.3.7
below) that the latter case leads to a contradiction.
Thus in the remainder of this section, we assume G is of type Ls(q) and con-
clusion (2) of 5.3.4 holds.