658 5. THE GENERIC CASE: L2(2n) IN .Ct AND n(H) > 1
Recall from page 200 of [Asc94] that rr is a base for U if each cycle in the
graph r is in the closure of the conjugates of cycles of rr. But each triangle in r
is conjugate to one of:
{6,7},{5,6},{5,7}.or {6,7},{5,7},{4,7},which are triangles of rr. So it remains to show r is triangulable in the sense of
section 34 of [Asc94]; that is, that each cycle of r is in the closure of the triangles,
or equivalently the graph r is simply connected. This is the crucial advantage of
working with r as opposed to re; one can calculate in r to check it is triangulable.
As r is of diameter 2, by Lemma 34.5 in [Asc94], it suffices to show each r-gon
is in the closure of the triangles, for r ::; 5. For r = 2, 3 this holds trivially, and we
now check the cases with r = 4 and 5, using the 4-transitivity of M 23 on 8.
It follows from 34.6 in [Asc94] that 4-gons are in the closure of triangles:
Namely a pair of points at distance 2 are conjugate to {1, 2} and {3, 4}, whose
common neighbors are {1, 3}, {1, 4}, {2, 4}, {2, 3}-forming a square in r, which
is in particular connected. Finally it follows from Lemma 34.8 in [Asc94] that
5-gons are in the closure of the triangles: for if Xo, X1, X2, X3 is a path in I' with
d(xo, X2) = d(xo, X3) = d(x1, X3) = 2, then Up to conjugation under Q, Xo = {1, 2},
x1 = {2,3}, x2 = {3,4}, and X3 = {4,a} for some a E 8 - {1,2,3,4}. Then as
xo, x2, and X3 are all connected to {2, 4}, it follows that x~ n x~ n xt =f. 0, in the
language of [Asc94].
Thus the proof of Theorem 5.2.10 is complete.
5.3. Identifying rank 2 Lie-type groups
In this section, we complete the proof of Theorem 5.2.3. Recall the definition
of groups of type Xr(q) and type M 23 appearing before the statement of Theorem
5.2.9. If G is of type M23, then conclusion (2) of Theorem 5.2.3 holds by Theorem
5.2.10. Therefore by Theorem 5.2.9, we may assume that one of conclusions (1)-(3)
of Theorem 5.2.9 holds. Thus G is of type Xr(q) for some even q > 2 and some
Xr of Lie rank 2. Recall from 5.2.7 that a= (G 1 , G1,2, G 2 ) is an Xr(q)-amalgam,
where Gi = LiBD, G1,2 = SBD, and S = T n £1 = T n £ 2 = 0 2 (G 1 ,2). We
write G(a) for the corresponding group Xr(q) of Lie type defining the amalgam.
To establish Theorem 5.2.3, we must show G ~ G(a).
Set Mi:= Na(Li) and Ml,2.:= Mi nM2, and let 'Y := (M1,M1,2,M2) be the
corresponding amalgam.
LEMMA 5.3.l. (1) Li E C(G,T) and Mi= !M(LiT) with M 1 =f. M 2.
(2) F*(Mi) = 02(Mi) = 02(Li)·
(3) Na(S) = Ml,2 = NM;(S).
(4) Mi = LiM1,2·
PROOF. By the hypothesis of Theorem 5.2.3, £ 1 = L E £,*(G,T), and by5.2.7.2, £2 = 02 (H) so that £2 1:. Ml. By 5.2.8 and our assumption that G is
not of type M23, L2 = K E C(G, T). Thus (1) holds by 1.2.7. Hence F(Mi) =
02(Mi) by 1.1.4.6. By 5.2.7.3, 02(GiT) = 02(Li), so 02(Mi) = 02(Li) using
A.1.6, completing the proof of (2).
To prove (3), it will suffice to show Na(S) ::; Mi for i = 1 and 2. For then
Na(S)::; Ml,2· On the other hand NM;(S) is maximal in Mi, and M 1 and M 2 are