3.3. NORMALIZERS OF UNIQUENESS GROUPS CONTAIN Na(T) 59iLEMMA 3.3.16. L is not SL3(2n), Sp4(2n), or G 2 (2n) with n > 1.
PROOF. Assume otherwise. Let TL:= T n Land Mi, i = 1, 2, be the maximal
parabolics of L containing TL. Set Yi := MZ'^0 ; then Yi E £(L, T) with Yi/0 2 (Yi) ~
L2(2n) and LT= (Yi, Y2, T). By 3.3.10.1, Hi :=(Yi, TD) E H(T). Thus by 3.3.15,we may assume that D does not normalize Yi =.: Y, n = 2, and Y < K E C(Hi)
with K ::::l Hi, K/02(K) ~Ji, Kn Di M, and T induces inner automorphisms
on Y/02(Y). Set H~ := Hi/CH 1 (K/02(K)). Then Y ~ L2(4), so 02 (Y) = 1.
By 3.3.6.d, (LT, T) is an MS-pair, and so we· may apply the Meierfrankenfeld-
Stellmacher result Theorem C.1.32. Since n = 2, L/0 2 (L) is SL 3 (4) or Sp 4 (4) or
G2(4). By Theorem C.1.32, L/02(L) is not G2(4), and if L/0 2 (L) ~ Sp 4 (4), then
Lis an Sp4(4)-block.
As T induces inner automorphisms on Y/0 2 (Y), T induces inner automor-phisms on L/02(L) from the structure of Aut(L/02(L)). From the structure of
L/02(L), X := CnnL(Y/02(Y)) is oforder 3, and as X normalizes T, Q := [X, T] is
a 2-group. Now X:::; D:::; Hi, and we saw that K ::::l Hi. As [X, Y] :::; 02 (Y*) = 1
and CAut(K*)(Y*) is of order 2 since K* ~ Ji, we conclude X* = 1. Therefore
Q* = [X*, T*] = 1, so Q = [X, 02(KT)] = 02(0^2 (X0 2 (KT))) ::::l KT. But ifL/02(L) ~ SL3(4), then 02(L)X = 02,z(L) ::::l LT, so that Q = [0 2 (L),X] is also
normal in LT, and hence K:::; No(Q):::; M = !M(LT), contradicting KnD i M.
Therefore Lis an Sp4(4)-block. Now 02(L) is of order at moflt 210 using the
value for 1-cohomology of the natural module in I.1.6; thus Q is of order at most
I02(Y): 02(L)ll02(L)I:::; 26. 2i^0 = 2i^6.
Therefore as 19 divides the order of Ji but not of L 16 (2), K centralizes Q. Thisis impossible as Y :::; K and Y is nontrivial Q02(L)/02(L). This contradiction
completes the proof. DLEMMA 3.3.17. If L ~ A1, then m([V, L]) = 6, eliminating case (5) of 3.3.8.
PROOF. Assume the lemma fails. Then by 3.3.8, m([V,L]) = 4. We work with
two of the three proper subgroups in £(L, T). First, let Mi:= CL(Z)^00 • By 3.3.9,
CL(Z) = CL(Zn [V,L]), so M1 = Cr,(Z) ~ L3(2). Then 1 i= Z:::; 02((TD,Mi)).Second, there is M2 E .C(L,T) with M2f' ~ S5, so by 3.3.10.1, 02((M2,TD)) /= 1.
As LT= (Mi,M2,T) and MiT/02(MiT) is not isomorphic to L2(4) or A5, 3.3.15
supplies a contradiction. D
LEMMA 3.3.18. If L ~ Ln (2) with n = 4 or 5, then [V, L] is not the direct sum
of isomorphic natural modules.
PROOF. Assume otherwise; then [V, L] = Vi EB·· ·EB Vr, where the Vi are isomor-
phic natural modules for L. Therefore T induces inner automorphisms on L/0 2 (L),
and in particular normalizes each parabolic of L containing T n L.
Let Y 1 := CL(Z)^00 , and recall CL(Z) =CL( Zn [V, L]) by 3.3.9. As the natural
submodules Vi are isomorphic, CL(Z n [V, L]) is the parabolic stabilizing a vector
in each Vi, so that Y1 ~ Ln-i(2)/E2n-1, and hence Yi E £(L,T).
Let Wi be the T-invariant 3-subspace of Vi, and set Y2 := NL(Wi)^00 • Then
Y2 ~ L3(2)/ Es or L3(2)/ E54 for n = 4 or 5, respectively, so Y2 E .C(L, T). If some