76Z 11. ELIMINATION OF Ls(2°), Sp4(2°), AND G2(2°) FOR n > 1Next Li = L'f S Ca(Vi) = Gi, so K = [K, Li] S Gi. As Xi is faithful on
Vi, the product KXi is semidirect. Thus for each prime p dividing q - 1, K doesnot contain all elements of order p centralizing Li/Oz(Li), so applying A.3.18 we
conclude that in case (i):
(*) q = 4, K* ~ L 3 (4), and K* Xi ~ PGL 3 (4) with Xi inducing outer auto-
morphisms on K*.
We will return to case (*), after treating case (ii). There q = 4 so that JXi I =- If K* ~ Ji then K = (Li,NK(T)) S Musing Theorem 3.3.1, contradicting
K f:. M. Thus K/Oz(K) is Lz(p) or Lz(25) or L3(5), so Out(K*) is a 3'-group,
and hence XJ.' centralizes K*. If K/Oz(K) ~ Lz(25) or L3(5), then some t E TnK
induces an outer automorphism on Li/Oz(Li), sot induces a field automorphism onL/Oz(L), impossible as [t,Xi] S Oz(XiT). Thus K/Oz(K) ~ Lz(p), so conclusion
(2) will hold in this case, once we show L/Oz(L) is not SL3(4). But in that case,
XiOz(L) = Oz,z(L) :::! M, so Y := oz(XiT) ~ LT, and hence Na(Y) SM=
!M(LT). Then as [K, Xi] S Oz(K) ST, K normalizes oz(YOz(K)) = Y so that
KS Na(Y) SM, contrary to Ki. M.Thus to complete the treatment of the case i = 1, we assume (*) holds; as
this is the first requirement of conclusion (3), it remains to establish the remaining
assertions of (3). This argument will require several pages.
Our strategy will be to use Kand L to construct a third group I, and obtain
a triple L = (Li,Lz), K = (Li,Ko), and I:= (Lz,Ko)-where Ko is essentially
the maximal parabolic of K over T n K other than Li (T n K). We will be able to
· exploit some symmetry in this triangle of subgroups.
Let Ko denote the member of £( G, T) n K distinct from Li and K-that is,
Ko/Oz(K) is normal in the maximal parabolic of K/Oz(K) stabilized by XT which
is distinct from NK(Li). In particular, Ko/Oz(Ko) ~ Lz(4), and Ko E He. Set
S := Oz(XT), Hi:= KoSX, Hz:= LzSX, and Hi,z = SX.
Assume that there is no nontrivial normal subgroup of T normal in H :=
(Hi, Hz). Then Hypothesis F.1.1 is satisfied with Ko, Lz, Sin the roles of "Li, Lz,
S", so by F.1.9, a := (Hi, Hi,z, Hz) is a weak BN-pair of rank 2. As S ~ Hi,z,
a appears on the list of F.1.12. Indeed a must be one of the (untwisted) caseswhere the nonabelian chief factor of Hi and Hz is isomorphic to Lz(4). As Lz has
at least two noncentral 2-chief factors, a is not the PGL 3 (4), SL 3 (4), or Sp 4 (4)
amalgam, so a is the Gz(4)-amalgam. By construction Vi is Hi,z-invariant, and
hence plays the role of the long root group of Gz ( 4) normal in a maximal parabolic.The parabolic Hi stabilizing this long root group is irreducible on Oz(Hi)/Vi, and
Hi /Oz(Hi) has two A5-modules on this section; but in our construction Ko has
a natural Lz(4)-chief factor on Oz(Ko).
This contradiction shows that Oz(H) =f=. 1, and hence HT E H(T) ~He using
1.1.4.6. Now by 1.2.4, Lz SI E C(HT), and I:::] HT by 1.2.1.3 as Lz is T-invariant,
so also IE He. Similarly K 0 S Io E C(HT). We conclude from(*) thatXK := XnK=XnLi =XnKo.
But as [Vi, Li] = 1 while Vi = [Vi, Xz] from the action of L on V, X n Li =f=. Xz;
so XK does not centralize Lz/Oz(Lz), and hence [I, XK] f:. Oz(I). Thus [I, Io] f:.
Oz(I), so Ko S 10 =I. Therefore
x = (X n Li)(X n Lz) = XK(X n Lz) s I,