5.2. USING WEAK EN-PAIRS AND THE GREEN BOOK 657as (3(T) = T = ((T), "!(t) E 02 (Ix)t, so"/ E Inn(Ix). Thus adjusting (3 by the
inner automorphism of Gx which acts on Ix as "/-1, we may choose (3 = ( on Ix,
proving (5). DLEMMA 5.2.21. G = (M,M2) = (Gx,I).
PROOF. Let Y := (M,M2). If Y < G, then by induction on the order of
G, Y ~ M23. In particular, Y has one class of involutions; while by (1) and (2)
of 5.2.12, Ne(T) :::;; Ce(z) :::;; Y. Thus Y is a strongly embedded subgroup of G
(see I.8.1), so by 7.6 in [Asc94], Y has a subgroup of odd order transitive on theinvolutions in Y. Now Y has
i := 3. 5. 11. 23
involutions, but no subgroup of odd order divisible by i. This contradiction shows
G = (M, M2). But M = LM1,2 and M2 = (K, I), so
G = (M,M2) = (LT,K,I) = (Gx,I),
completing the proof. DLEMMA 5.2.22. I= (I({x,y}),Ix)·
PROOF.. Notice Ix contains the kernel U B of the action of I on A := { 4, 5, 6, 7}.
Further Ix contains elements inducing (4, 5) and (6, 7) on A, while I( {x, y}) contains
an element inducing (5, 6). So as the symmetric group on A is generated by these
three transpositions, the lemma holds. D
We are now in a position to complete the proof of Theorem 5.2.10, by appealing
to the theory of uniqueness systems in section 37 of [Asc94]. Namely write re for
the graph on re= G/Gx with edge set (x,y)G = (Gx,Gxt)G, and let rr be the
subgraph with vertex set xI and edge set (x, y)I. By 5.2.19.2, rr is isomorphic to
the subgraph r1 := { 4, 5, 6, 7}^2 of r.
Observe that u := ( G' I' re' r I) is a uniqueness system in the sense of [Asc94].
Namely by 5.2.21, G = (Gx,I); by 5.2.20.4, Gx = (Gx,y,Ix); and by 5.2.22, I =
(I( { x, y} ),Ix)· This verifies the defining conditions for uniqueness systems (see (U)
on page 198 of [Asc94]). Similarly U := (G,l, r, r1) is a uniqueness system.
Now by 5.2.19 and 5.2.20, (3 : Gx , Gx and ( : I , I define a similarity of
uniqueness systems, as defined on page 199 of [Asc94]. Next we will apply Theorem
37.10 in [Asc94], to prove this similarity is an equivalence.
In applying Theorem 37.10, we take L in the role of the group "K" in the
Theorem, and take t, h E I to be elements acting on [2 as
t := (1, 2)(5, 7), h := (1, 2)(5, 6).
Then t, h E M1, 2 :::;; Ne(L), and by construction t has cycle (x, y), th= (1, 2)(6, 7) E
K:::;; Gx, and ((h) E M1,2 :::;; N 0 (L), so that hypothesis (2) of Theorem 37.10 holds.
Next L = Gx,y by 5.2.20.1, so trivially Gx,y = (Ly, Ix,y), which is hypothesis (3) of
37.10. Finally L n I= BU, and from the structure of the L 2 (4)-block L, we check
that CAut(L)(BU) = l; this verifies hypothesis (1) of 37.10.
Therefore U is equivalent to U. It remains to check that r I is a base for U
in the sense of p.200 of [Asc94]: for then as G = M 23 is simple, Exercise 13.l in