i2.1. A PRELIMINARY CASE: ELIMINATING Ln(2) ON n EEl n* 79iparabolic of K* determined by the end nodes of the Dynkin diagram. Therefore
R* := 02(X*) S Ti, and hence as it is the unipotent radical of P*, R* is weakly
closed in T* with respect to H* by l.2.5. Since Cr(K*) s Cr(X*) :::;; Ti, a Frattini
Argument shows that NH(Ti)* = NH•(Ti). Thus T 0 S NH(Ti)* = NH•(Ti) SNH• (R) =NH (P) =NH (X*), so
K~T* = H* = (T~,X*T*) s NH·(X*) = X*T*,
a contradiction. This completes the proof of 12.1.4.LEMMA 12.1.5. (1) Let v generate Cv 1 (Ti). Then Ti E Syh(Gv)·
(2) M controls fusion of involutions in V.
D
PROOF. If n = 4, then (1) follows from 12.1.4. If n = 5, then by 12.1.3,
Wo s Q s Ti, so Ne(Ti) s Ne(Wo) by E.3.15. As M = !M(Ne(Q)) by 1.4.1,
Ne(Wo) S M by E.3.34.2. Thus Nev (Ti) S Nev (Wo) S Mv, so as Ti is Sylow in
Mv, (1) also holds in this case.
By (1), [Gv[2 = [T[/2 for any v E Oi, while [Gz[2 = [T[ for z E 02. Finally by12.1.1, [Gv[2 < [T[/2 for v E 03. Thus the distinct M-classes of involutions in V
are in different G-classes, so (2) holds. D
LEMMA 12.1.6. (1) For v E Oi, (Vevj is abelian.
(2) If Vi n V^9 I= 1, then [V, V9] = 1.
PROOF. Let v E Oi. By 12.1.5.2, M = Ne(V) is transitive on G-conjugates
of v in V, so by A.1.7.1, Gv is transitive on G-conjugates of V containing v. Thus
(2) follows from (1), so it suffices to establish (1).
We may as well choose v to generate CVi (Ti). By 12.1.5.1, Ti E Syh(Gv)·
By lemma H.9.1, u := [V, Lv] = Vi EB U2, where U2 := v..L n Vi. Let V2 generate
Cv 2 (Ti). Then z := vv2 generates Cv(T), and v2 E U2.
By 1.1.6, the hypotheses of 1.1.5 are satisfied with Gv, Gz in the roles of "H,
M". But as z EU, [O(Gz), z] = 1 by A.1.26, so O(Gv) = 1 as z inverts O(Gz) by
1.1.5.2.
Suppose first that Gv ~He. Then as O(Gv) = 1, there is a component K of
Gv, and by 1.1.5.3, K = [K, z] 1:. M. As Ti E Syh(Gv), Lv S Kv E C(Gv) by
1.2.4. Now z E U = [U, Lv] S Kv, so as K = [K, z], we conclude K = [K, Kv] = Kv
from 1.2.1.2. Thus v E Lv s K. Set K := K/0 2 (K). Then Un Z(K) = (v),
so U ::S! L~, with U* = Vt EB U:J. the sum of the natural module and its dual
for Lv/02(Lv) ~ Ln-i(2). As no group on the list of 1.1.5.3 has such a subgroup
invariant under a Sylow 2-group Ti, we have a contradiction.
This contradiction shows that Gv E He. Let Qv := 02(Gv), and Gv := Gv/(v).
Now Ti E Syl 2 (Gv), and Lv is irreducible on Vi, so Hypothesis G.2.1 holds with
(v), Vi, Ti, Gv in the roles of "Vi, V, T, H". Then by G.2.2.1, Vi S Z(02(Gv)) =
Z(Qv)· Similarly as Lv is irreducible on U2, [Qv, U2] S (v) n U2 = 1, so that
U2 S Z(Qv)· In particular U = Vi.U2 S Qv, so for any g E Gv, U2 centralizes U^9.
Hence by 12.1.2.2, U2 s Ce(U^9 ) = CM9(V^9 ).
Suppose first that V S Qv. Then for all g E Gv, V^9 S Qv S M = Ne(V).
By the previous paragraph, V9 centralizes U2, so V^9 acts on Vi and V 2 ; hence by
symmetry, V acts on V[ and v;r Then as v ~ Vi,
[Vi, Vil s [Vi, Qv] n Vi s (v) n Vi = 1,
so that V1::; CMB(V/) = CM9(V^9 ). Then V^9 s CM(Vi) = Ce(V), so (1) holds.