8.2. DETERMINING LOCAL SUBGROUPS, AND IDENTIFYING J 4 719If j is L2(2), then conclusion (a) of (3) holds for k = 1, and Pis the sum of
copies of the natural module J with Endp 2 r(J) = F 2 , so (5) follows from 27.14 in
[Asc86a] in this case.
Suppose Lis M22· Then P /[P, Lu] is of rank 2, so as Pis the sum of copies
of J, it follows that j is L 2 (2), so that (3a) and (5) hold in this case by the
previous paragraph. In the remaining cases for L, B is the sum of k copies of the
natural irreducible module A for Lu, so P is the sum of 2k copies of A. :Further~ := EndF 2 tu(A) is F4, F4, F2, respectively; and by 27.14 in [Asc86a], P has the
structure PA of a ~-module for Lu"'Ei, where "'E. := CGL(F) (Lu) = GL(8) for some
2k-dimensional ~-module e, and PA= A0 e as a Lu"'E.-module. Then j::; "'E., and
among the possibilities for j listed in 8.2.2.5, the only ones which are subgroups
of GL2k(~) are j f::! L2(2k), or Dia in the case k = 1 and~= F 4. :Further J is
F2j-isomorphic to 8 by parts (3) and (10) of E.2.14. This completes the proof of
(3) and (5).
Suppose V is the code module for M24. Then by (3), Luj f::! A 6 x D 2 m form := 3 or 5. Therefore as m 3 (N) ::; 2 since N is an SQTK-group, m = 5.^1
Next f'Lu/02(f'Lu) f::! S5/E54 with A= 02(Lu), where each involution in Tis
fused into A under M, and there is an involution inf' - Lu. Therefore there is an
involution t ET - Lu02(LuT). Assume Tacts on 02 (1). Then as I= (V, Vh) =
02 (I)(T n I) since T n IE Syh(I), while V :::'.] T, I= 02 (I)V, so that Tacts on
I. Extend the earlier "dot notation" to Yr := YT by defining Yr := Yr/ 02 (Yr),
and let v E V - B. Then s := i or iv centralizes j, Thus j acts on C p ( s), whereas
by (5), Cp(s) is of 2-rank 6, while all irreducibles for j on Pare of rank 4. This
contradiction completes the proof of (6).
It remains to establish (1). As Ku :::'.] N, we must show that Ku= Lu. FirstAutLu(U)::; AutKu(U), and by 8.2.7 and 8.2.6.3, either CKu(U) = 02(Ku), or V
is the code module for M24 and CKu(U) = 02(Ku)0^2 (02,3(Lu)). If Lis M24 on
the cocode module then AutLu(U) = GL(U), so Ku= LuCKu(U) = Lu02(Ku),and hence Lu = Ku in this case. Thus we may assume one of the first three cases
holds, so m(U) = 4 by 8.2.5.
Suppose case (a) of (3) holds. Then by (6), one of the first two cases holds. Now
m3(J) = 1 = m3(Lu ), Lu ::; Ku with [Ku, I] ::; 02(Ku ), and N is an SQTK-group,
so m3(Ku) = 1. Also AutLu(U) ~ A5, and Autr(U) acts on AutLu(U). The proper
overgroups of AutrLu(U) in GL(U) have 3-rank at least 2, so as m3(Ku) = 1, we
conclude again that AutKu(U) = AutLu(U) and Ku= Lu.
Finally assume case (b) of (3) holds. As 02 (Ku) acts on I, X := 02 (1) =
02 (I0 2 (Ku )). Thus as [Ku,I] ::; 02(Ku ), Ku acts on X, and hence also on
02 (X)U = P. Now F15 = Endp 2 x(J), and Pis the sum of e := m(P)/4 = 2
or 3 copies of J, so by 27.14 in [Asc86a], Ku/CKu(P) ::; GL(fl), where fl is an
e-dimensional space over F15. Arguing as in the previous paragraph, m 5 (I) = 1 =
m5(Lu) so that m 5 (Ku) = 1. Then inspecting the overgroups of AutrLu(fl), we
conclude as before that Ku = Lu. This completes the proof of the lemma. D
LEMMA 8.2.9. (1) Tacts on 02 (!), and H =IT.
(2) T normalizes VA.
(3) V is not the code module for L ~ M24.(^1) We just eliminated the shadow of 001, where m = 3 in the 2-local N.