656 5. THE GENERIC CASE: L 2 (2"') IN .Ct AND n(H) >^1
representations of G on r (which is in turn G-isomorphic to the analogous graph
on ra := G/Gx) and G on ra are equivalent. This leads us to write x for Gx
regarded as a point of r a-namely the coset of Gx containing the identity. Thus
Gx is indeed the stabilizer of the point x E I'a.
Let I denote the global stabilizer in M 2 of { 4, 5, 6, 7}. Notice that the repre-
sentation of M 2 on n s;;; e induces a representation of M 2 on !1^2 s;;; I'; this is the
representation implicit in the next lemma.
LEMMA 5.2.19. (1) Gx n M2 = H is the stabilizer in M2 of x = {6, 7} Er, and
the stabilizer in M2 of x = Gx E I'a.
· (2) The representation of M 2 on xM 2 s;;; r a is equivalent to its representation
on !1^2 s;;; r.
(3) (: M 2 --+ M2 restricts to an isomorphism ( : I--+ I.
(4) ((Ix) =Ix and <;(Ix,y) = ln,y·
PROOF. By 5.2.11.l, H is the .stabilizer of x = {6, 7} E I', and hence is a
maximal subgroup of M 2 • Therefore H = GxnM 2 , and thus His also the stabilizer
in M 2 of the coset Gx Era, which we are denoting by x. Therefore (1) holds. Then
(1) implies (2), while 5.2.17 and the definition of I and I imply (3) and (4). D
Using the equivalence of 5.2.19.2, the point fj = {5, 6} E r corresponds to a
pointy E I'a; namely the coset y = Gxt, where t EI has cycle (x,y) on r. Such a
t exists, as I is the global stabilizer of { 4, 5, 6, 7} in M2, and hence induces the full
symmetric group on that subset. The coset y is independent oft by 5.2.19.1.
Recall as in [Asc94] that I( { x, y}) denotes the global stabilizer in I of { x, y }.
LEMMA 5.2.20. (1) Gx,y = L.
(2) Ix is the stabilizer in M 2 of the partition {1, 2, 3}, { 4, 5}, {6, 7} of n, andIx/U ~ 83 x Z2.
(3) lx,y = UB and ((I({x,y})) = I({x,y}).
(4) Gx = (Gx,yJx)·
(5) There is an isomorphism /3 : Gx --+ Go; agreeing with ( on Ix, such that
/3(Gx,y) = Gx,y·
PROOF. By 5.2.11.2, M 1 , 2 is the global stabilizer in M 2 of {5, 6, 7}, so there is
t E (M1,2)4 ::::; In M with cycle (x, y). Then by a remark preceding this lemma, t
has cycle (x, y). As L ::::; G 0 T = Gx, L fixes x, so that L = Lt fixes xt = y, and
then L::::; Gx,y· But LT and G 0 are the only maximal subgroups of Gx containing
L, and by 5.2.19.2, Ti. Gx,y "t. K. So (1) holds.
Parts (2) _and (3) are easy calculations given 5.2.19. As observed earlier, LT
and Go are the only maximal subgroups of Gx containing Land Gx,y = L by (1).
So as Ix i. Go n M2 =Kand Ix i. M1,2, (4) holds.
By 5.2.16 and 5.2.18.1, there is an isomorphism /3 : Gx --+ Go;, which we may
take to map T to T := ((T), B to fJ := ((B), and K and L to the parabolics
K := ((K) and L := Gx,y of 02 (Go;). Now by (2) and (3), Ix= UB(t,r), where
t := (1, 2)(6, 7) and r := (4, 5)(6, 7) on n. In particular Ix= UNKT(B), so
f3(Ix) = /3(U)N(3(K)f3(T)(f3(B)) = UNJ<r(B) =Ix.
Finally let 'Y := i:;-^1 o (3, regarded as an automorphism of Ix, so that 'Y E
Aut(Ix)· Notice INaL(U)(Aut1,,(U)) : Aut1,,(U)I = 2 and U = CAut(I,,)(U), so