2.4. THE CASE WHERE r~ IS NONEMPTY 533(1) Z(L) = 1, Z(R) = CR(t), R is transitive on t[R, t], and [R, t] is transitive
on tZ(R); in this case we say CR(t) is of type L2(q).
(2) Z(L) = 1, n is even, and CR(t) is a Suzuki 2-group of order q^312 with
[fh(CR(t))[ = q^112 ; in this case we say CR(t) is of type U 3 (q^112 ).
(3) Z(L) ~ Eq, and CR(t) is a Suzuki 2-group of orderq^2 with [0 1 (CR(t))[ = q;
in this case we say CR(t) is of type Sz(q).
PROOF. Since t E T - S, t serves in the role of the element "x" in Notation
2.4.10; in particular, we may apply 2.4.8. As RB '.';! TB by 2.4.8.5, by a Frattini
Argument we may choose t to normalize B. Also by 2.4.8.5, R '.';!RB and CR(B) =
1, so as t^2 E R, [B, t^2 ] ::; B n R = 1 and hence tis an involution. In particular R(t)
splits over R.
We recall from Notation 2.4.10 that Qt = Qx is independent of the choice of x E
T-S, so by 2.4.8.3, m(R/Z(R)) = 2m(CR/Z(R)(t)) and CR/Z(R)(t) = [R/Z(R), t].
Let Rt denote the preimage of CR/Z(R)(t), so that Rt contains CR(t). By 2.4.8.7,
Q and Qt are the maximal elementary abelian subgroups of R, so Z(R) = 01(Rt),
and hence Cz(R)(t) = n1(CR(t)).
Assume that Z(L) -=/= 1. Then Z(L) ~ Eq is a TI-subgroup of G by 2.4.11.3,
while [Z(R)[ = 22 n = [Z(L)[^2 by 2.4.8.4, so Z(R) = Z(L) x Z(L)t. Thus by Exer-
cise 2.8 in [Asc94], R is transitive on the involutions in tR, and Rt= Z(R)CR(t).
As B = D x Dt by 2.4.8.6, CB(t) is a full diagonal subgroup of B, and so
CB(t) is regular on Cz(R)(t)# = Z(CR(t))#. Further CR(t) is nonabelian, so
that [CR(t),CR(t)] = Z(CR(t)); thus CR(t) is a Suzuki 2-group oforder q^2 , so.that
conclusion (3) holds.
Now assume instead that Z(L) = 1. Set (TB)* := TB/CrB(Z(R)). As t
normalizes B and B = D is regular on Z(R)# by 2.4.8.6, either t* = 1 or
m(Z(R)) = 2m(Cz(R)(t)), and in either case CB(t) = CB(t)* is regular on
Cz(R)(t)#. Assume first that t* = 1. Then as Rt/Z(R) = [R/Z(R), t], with
01 (Rt) = Z(R), t inverts an element r of order 4 in each coset of Z(R) in Rt.
So as r is of order 4, CR(t) = CRt(t) = Z(R), ani::l conclusion (1) holds. Fur-
ther [R : CR(t)[ = [Rt[, so R is transitive on tRt and hence on the involutions
in tR. Now assume instead that t* -=/= 1, so that m(Z(R)) = 2m(Cz(R)(t)). By
Exercise 2.8 in [Asc94], [CR(t)[ = q^312 and R is transitive on the involutions in
tR. As CB(t) is transitive on Cz(R)(t)# = Z(CR(t))#, and CR(t) is nonabelian
so that [CR(t),CR(t)] = Z(CR(t)), CR(t) is a Suzuki 2-group of order q^312. Thus
conclusion (2) holds. D
NOTATION 2.4.15. In the remainder of our treatment of the case s = 1, we
define Z as follows: If Z(L) = 1, set Z := Z(R), while if Z(L)-=/= 1 set Z := Z(L).
LEMMA 2.4.16. (1) Z ~ E2n and Z '.';! 8.
(2) For x ET - S, either
(a) Z(L) = 1 and zx = Z = Z(R), or
(b) Z(L)-=/= 1 and Z(R) = Z x zx.
(3) u = ((zx)L) = (Z^0 n U).PROOF. Part (1) follows from 2.4.8.4. Next x normalizes J(S) = R, so conclu-
sion (a) of (2) holds when Z(L) = 1 as Z = Z(R) in that case. If Z(L) -=/= 1 then
Z = Z(L) is a TI-subgroup of G by 2.4.11.3, and Z-=/= zx by 2.4.4, so conclusion
(b) of (2) holds as [Z(R)I = [Z[^2.