2.4. THE CASE WHERE I'Q IS NONEMPTY 537At this point, we have obtained strong control over the 2-local structure and
2-fusion of G, which we can use to obtain contradictions via transfer arguments.LEMMA 2.4.22. (1) T/R is not cyclic.
{2) R < 8.
PROOF. If R = 8 then IT: RI = 2 by 2.4.8.1, so T splits over R by 2.4.14, and
hence there is an involution t E T - R. On the other hand if R < 8, there is an
involution tin 8 - R by 2.4.12.1. Thus in any case there is an involution t ET-R.
As T / R is cyclic if 8 = R, it remains to assume T / R is cyclic and derive a
contradiction. By 2.4.21, tG n R = 0. Then by Generalized Thompson TransferA.1.36.2, t tJ. 02 (G), contrary to the simplicity of G. D
By 2.4.22.2, R < 8; so since 8 splits over R by 2.4.12.1, there is an involution
8 - R. It is convenient to use the notation s for this involution; there should
be no confusion with the earlier numerical parameter "s", as in the branch of the
argument for several pages before and after this point, that parameter has the value
1. Let Gs := Ga(s), Ls := GL(s), etc.
We use the standard notation that for x an integer, x 2 denotes the 2-primary
part of x.
LEMMA 2.4.23'. {1) Either Ls is an L 2 (2nl^2 )-block with Us = U(Ls), or q = 4
and Ls~ 84 or 84 x Z2.
(2) Rs is the strong closure of Q in Ts.
{3) Us= 02(Ls) and Nc(Us) :S GQ.
(4) T = RTs, there exists x E Ts - S, and Ts E 8yl2(Gs)·
(5) Assume Z(L) -=f:. 1 and q = 2n > 4. Set Ks := (Ls,L~). Then Ks ~
Sp4(2nl^2 ), GrJKs) = (s), andTs/(s)Rs is cyclic of ordern2 = !Out(Ks)l2·
PROOF. Part (1) follows from 2.4.12.2. By (1), Us= 02 (Ls)· Part (2) follows
from 2.4.21.l.
From (1) and the proof of 2.4.16.3, Us= ((z:)G n Us)· But Nc(Us) permutes
(z:)G n Us and Z is a TI-subgroup of G, so Na(Us) permutes (Z"')G n U and henceNc(Us) ::; Nc(U) = GQ by 2.4.16.3, and as Q = U by 2.4.8.2. This completes the
proof of (3).
By 2.4.12.1, 8/R is cyclic, so (s)R :'.:) T. By 2.4.12.2, R is transitive on the
involutions in sR, so by a Frattini Argument T = RTs, and as SE Syb(GQ) by
2.4.3.1, 8s is Sylow in Nas (Us) by (3). As S < T, there is x E Ts -,.- 8 and by2.4.8.7, U and U"' are the maximal elementary abelian subgroups of R, so A(Rs) =
{Us, U:}. Therefore Na(Rs) = Nc(Us)(x). So using (2), Nc(Ts) :S Nc(Us)(x).
Thus as 8s is Sylow in Nc.(Us) and 8s(x) =Ts, Ts E 8yb(Gs), so that (4) holds.
Assume the hypotheses of (5), and set Ks := (Lx, L~). Let 8 be the set of
subgroups of Ss invariant under Ls. From the action of S and L, Us(x) is the
unique maximal member of 8, and if Y E 8 with Us i. Y, then Y ::; (s)Gu(L).
Therefore as Rs = UsU:' and Gu(L) n Gu(L)"' = 1, (s) is the largest subgroup ofTs invariant under Ls and L~, and hence (s) = 02(KsTs)· As q > 4 by hypothesis,
Ls E C(Gs, 8s), so since !Ts: 8sl = 2 with Ts E 8yb(Gs) by (4), Ls :SK E C(Gs)
by 1.2.5. As x acts on Rs ::; K, x acts on K, so Ks ::; K. Then using (4) and
. A.1.6, 02(K) :S 02(KTs) ::::; 02(KsTs) = (s) ::::; Ga(K). As m2(K) 2:: m2(Ls) > 1,
K is quasisimple by 1.2.1.5. By (1), Ls is an L2(q^112 )-block with Zs= Z(Ls) -=f:. 1,
so as Rs =Usu: is a Sylow 2-subgroup of Ls, we conclude by examination of the