1198 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTIC(b) of 16.5.2.2 holds, and replacing T by the subgroup T^1 defined there, we mayassume T 0 := TaTL = Nr(R) is of index 2 in T, and take g E Na(To) with
K9 = K'. Set T 1 := T or T 0 in the respective cases, and set Zc := Z(T1) n Tc
and ZL := Z(T1) n TL. Thus g E Na(T 1 ), so g acts on Z(Ti) and Ti = Nr(R).
By hypothesis, TLTc = TL x Ta and Cr(TL) = TcZ(TL) with IZ(TL)I = 2, so
ZL = Z(TL) = CrL (T), and as Z(T1) ::; Cr(TL),By 16.4.2.1, Zc n Z!J = 1, so as g acts on Z(Ti) and IZ(Ti) : Zcl = IZLI = 2 by
(*),we conclude Zc is also of order 2. Hence T centralizes ZL x Zc. Then since R
is normal in Ti, 1 -=I-Rn Z(T 1 ) is central in T by (*), so that Rn Z(T) -=I-1; thus
case (a) of 16.5.2.2 holds, and hence Ti= T.As Ti = T, Z(T) = ZL x Zc is of rank 2 and g E Na(T). In particular
R = T n K9 = (T n K)9 = T{!;. Also as Z!J n Zc = 1, g induces an element of
order 3 on Z(T) so either Zb or Zb
2
is not equal to ZL, and replacing g^2 by g of
necessary, we may assume Z!J -=I-ZL. As Tc ::::] T, also R = T{!; ::::] T, so RnL ::::] T.
Hence as Z(T) n R = Z!J is of order 2 and does not lie in L, Rn L = 1. Thus
[TL, R] ::; TL n R = 1, so R* ::; Cr* (TL)= Z1. Therefore as IZ11 = 2 and R ~ R*,R is of order 2, and hence Tc = Zc. Then as IT* : T£1 ::; 2, IT: TLI ::; 4, so that
[T, T] ::; TL. By Proposition 16.5.1 and our assumption that G is not Ji, Lis not
L2(2n), so by inspection of the groups in (E2), TL is nonabelian. Therefore as ZL
is of order 2, ZL = Z(T) n [T, T] ::::] Na(T). This is impossible, as (g) is irreducible.on Z(T) ~ E 4. ' D
LEMMA 16.5.8. (1) L is not L 2 (p), p an odd prime, or L 3 (3).
(2) Lis not Mu, M22, M23.
PROOF. If Lis a counterexample to (1) or (2), then L has one class of involu-tions, Z(L) = 1, Out(L) is of order at most 2, and Cr* (TL) is of order 2; hence the
lemma follows from 16.5. 7. DLEMMA 16.5.9. L is not U3(3).PROOF. Assume otherwise. Then L has one class of involutions, X ~ SL 2 (3),
and CH (X) ~ Z4 or Qg. Thus R* ~ R is of 2-rank 1 by 16.5.6.2, and Tc ~ R
by 16.4.11.1. Then Z := Di(Z(T)) ~ E 4 , with u E Z := (z, v), where z E Tc and
v E TL. Choose g as in 16.4.2.4; then T{!; = R by 16.4.11.3. As Rn Z(T) -=I-1,
g E Na(T) by 16.5.2.1, and as T{!; = R, g is nontrivial on Z. Then as Z is of rank2 we conclude that g induces an element of order 3 on Z. This is impossible, as
g E Na(X) by 16.5.6.3, so g acts on Z(X) = (v). D
LEMMA 16.5.10. Assume L* is not of Lie type of Lie rank 2 over F 2 ,, for some
n > 1. Then
(1) If L* is of Lie type in characteristic 2, then Lis^2 F 4 (2)',^3 D 4 (2), L 4 (2), or
L5(2). ·(2) If L is not of Lie type and characteristic 2, then L ~ Mi 2 , M 22 , M 24 ,
J2, J4, HS, or Ru; and if L* ~ M22, then Z(L)-=/-1.
(3) IT*: T£1::; 2.
(4) I Cr* (TL) I ::; 2.