16.5. IDENTIFYING Ji, AND OBTAINING THE FINAL CONTRADICTION 1199(6) Assume that Z(L) = 1, v is the projection of u on L, and there is l E L
with vv^1 an involution in X. Then vv^1 is not 2-central in L'.
PROOF. Part (2) follows from 16.5.8 and inspection of the groups in (E2).
Suppose that L* is of Lie type in characteristic 2. By Proposition 16.5.1 and
our assumption that G is not J 1 , Lis not of Lie rank 1, and by hypothesis L* is
not of Lie rank 2 over F 2 n for some n > 1. Thus from Theorem C, either L* is
of Lie rank 2 over F2, or L* ~ L4(2) or L5(2). By 16.5.8 Lis not L 3 (2) ~ L 2 (7),
by 16.5.9 Lis not U3(3) ~ G2(2)', and by 16.5.3 Lis not A 6 ~ Sp 4 (2)'. Thus (1)
holds since Lis simple by 16.1.2.1.
Next by inspection of Aut(L*) for L* listed in (1) and (2), l0ut(L*)l 2 :::; 2, so
(3) holds. Similarly (4) follows from inspection of Aut(L*). Then (5) follows from(3), (4), and 16.5.7.1. Finally assume the hypotheses of (6). Then the hypotheses
of 16.5.6.5 are satisfied using (5), so (6) follows from that result. D
LEMMA 16.5.11. L* is of Lie type in characteristic 2.
PROOF. Assume otherwise; then L* is in the list of 16.5.10.2.Suppose first that u is 2-central in L. Then by 16.5.10.5, Z(L) -/=- 1, so
applying 16.1.2.1 to the list in 16.5.10.2, L is M 12 , M 22 , h, HS, or Ru. Fur-
thermore using 16.1.5, we find that either CH (X) is of order 2, or L is HS and
CH (X) ~ Z4. Thus by 16.5.6.2, either IRI = 2, or L is HS and R ~ Z4. In any
case (u) = fli(R) and (z) = 01(Tc). Further if L is M22, then as Tc~ R ~ Z 2 ,
Tc = (z) and Z(L) ~ Z2. Hence we conclude from 16.1.2.2 that in each case
(z) = 01 (Tc) = Z(L) ~ Z 2. Then from the structure of the covering group L of
L * in parts (5)-(7) of I.2.2, either:(a) There is a unique v E uZ(L) such that there exists x E 0 2 (X) with x^2 = v.
(b) L* is Ru, and setting Y1 := Ca 2 (x)((02(X))), Y := [Y1, Y1] is of of order 2,
and Y = (u).
In case (a) set Y := (u). Thus in any case Y = (u). Further T normalizes X, and
hence centralizes Y, so T centralizes Z(L)Y = (z, u). Therefore 1 i= u ER n Z(T).
Choose g as in 16.4.2.4; then g E Nc(T) by 16.5.2.1, and g E Na(X) by 16.5.6.3.
Next ICT (T£)1 = 2 in each case, so Z(TL) = (u) and hence Z := 01(Z(T)) =
(z,u) ~ E 4. By our choice of g and 16.4.11.3, R = Tfj and hence u = zY. Then
as g acts on T, g induces an element of order 3 on Z, and in particular (g) acts
irreducibly on Z. This is impossible since g acts on X and hence on Y < Z.
Therefore u is not 2-central in L. Thus as M 2 2 has one class of involutions,
L is not M22·
Inspecting the list of centralizers of non-2-central involutions in 16.1.5 for the
remaining groups in 16.5.10.2, either CH·(X) is of order 2, or L* is M12, M24,
J 2 , HS, or Ru and CH (X) ~ E 4. Arguing as in the previous paragraph, either
IRI = 2 = ITcl, or one of the exceptional cases holds with R ~ E4 ~Tc. ,In any
case, (R) = 1 = (Tc), so R = U.
Assume L ~ J 4 or M 2 4. Then Out(L) = 1, so T =TL x Tc with (Tc) = 1,
and hence z rf. (T) for TE Syl 2 (Cc(z)), and Tc is in the center of 021 (Cc(z)).
Thus if ITcl = 2, then z^0 n TL i= (i) by Thompson Transfer, whereas for each
involution a E TL, a E (CTL(a)) by 16.1.5.9. Thus ITcl > 2, so L ~ M24, and
then U is not centralized by 021 (CL(u)); so as U = R = Tfj, this is contrary to Tc
in the center of 021 (Cc(z)).