Introduction to finite group theory

20. The Higman–Sims group 41
Our immediate aim is to prove that ˛ is an automorphism of the graph .
20.5 Lemma. For any pair of ovals O i , O j and any line l in P 2 .4/ the following
assertions hold:
1) ˛.O i / is an oval;
2) ˛2.O i / D O i ;
3) O i \ l 1 D ¿ ” ˛.l 1 / 2 ˛.O i /;
4) O i \ O j D ¿ ” ˛.O i / \ ˛.O j / D ¿.
Proof. It is easy to compute that
˛.O/ D ˚ ax 1 C x 2 C x 3 ; a 1 x 1 C x 2 C x 3 ; x 1 C ax 2 C x 3 ;
where
x 1 C a 1 x 2 C x 3 ; x 1 C x 2 C ax 3 ; x 1 C x 2 C a 1 x 3
D xA O;
0 1
a 1 a
A D @ 1 a a A :
1 1 a 1
1) Write O i in the form O i D xB O, where xB 2 PSL 3 .4/. Then
˛.O i / D P 2 .4/ n S ˛. xB Nv/
Nv2O
D P 2 .4/ n S .B > / 1 ˛.Nv/
Nv2O
D .B > / 1 ˛.O/ D .B > / 1 A O;
that is, ˛.O i / is an oval.
2) We have ˛2.O i / D B.A > / 1 A O D xB O D O i since .A > / 1 A OD O,
as can easily be proved.
3) O i \ l 1 D ¿ ” Nv … l for all Nv 2 O i ” .l/ … .Nv/ for all Nv 2 O i
” ˛.l 1 / 2 ˛.O i /.
4) Suppose that O i \ O j D ¿, butNv 2 ˛.O i / \ ˛.O j /. By3)wehave
˛.Nv/ \ O i D ¿ and ˛.Nv/ \ O j D ¿. Since the ovals O i , O j and the line
l D ˛.Nv/ n f1g are pairwise disjoint, their complement in P 2 .4/ contains 4 points,
say y 1 , y 2 , y 3 , y 4 . By Exercise 20.6, there exists a point x 2 l such that the number
of lines connecting x to the points y k is at least 3.
Denote three of these lines by l 1 , l 2 , l 3 . Since l 2 L i , at most one of them
can lie in L i (by Lemma 20.3). Analogously at most one of them can lie in L j .
Therefore one of them, say l 1 D l.x;y 1 /, does not lie in L i [ L j and hence it
intersects O i and O j . Then l 1 contains 6 points: x, y 1 , two points from l 1 \ O i ,
and two from l 1 \ O j (see Lemma 20.2), a contradiction. The converse implication
follows from 2).