13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 897LEMMA 13.4.2. (1) M = Na(L) = Na(V) = Oa(Zv).
(2) Oa(Z)::::; Oa(ZL)::::; Oa(Zv) = M, and M = !M(Oa(ZL)).
(3) For each v EV#, Oa(v) is transitive on conjugates of V containing v. In
particular, V is the unique member of va containing Zv.(4) OM(V) = OM(L/02(L)) = OM(V/Zv ).
(5) L = B(M), and if L/0 2 (L) ~ A 6 , then L = 031 (M).
PROOF. Theorem 12.2.2.3 shows that M = Na(L), and (since Zv i-1) that
V ::::! M. Hence M ::::; Oa(Zv ), and (1) follows as M E M. As Zv ::::; ZL ::::; Z and
M = !M(LT), (2) holds. Since LT controls fusion in V by 13.2.2.5, (3) follows fromA.1.7.1. Observe (5) follows from 12.2.8. Finally OM(L/0 2 (L)) = OM(V/Zv) =
OM(V) by A.1.41, estabilishing (4). DAs M = Na(V) by 13.4.2.1, M := M/OM(V) from Notation 12.2.5.2. Recall
Vi = V n Z, and by 13.3.5.3 there is z E V1 - Zv with L1T ::::; Gz := Oa(z) i M.Fix a choice of z and observe z has weight 2 or 4 in V. Eventually we will see that
there is a unique z E V1 with Gz i M. As usual define
'liz :={HE 1i(L1T): H::::; Gz and Hf:_ M}.In particular Gz E 'liz so 'liz i-0. Recall that R1 is defined in Notation 13.2.1.
LEMMA 13.4.3. (1) L = [L, J(T)].
(2) VZ = VZL and IZ: ZLI = 2.
(3) Z = V1ZL, so L 1 centralizes Z.
PROOF. As 1 =I-Zv ::::; ZL, (1) follows from 3.1.8.3. Then by (1) and Theorem
B.5.1, [VZ,L] = v, so vz = VZL by B.2.14. Then as 1znv: Zvl = 2by13.3.4.1,
(2) and (3) hold. DLEMMA 13.4.4. If H E 'liz and VH E R2(H) with ZL n VH(z) #-1, then
(1) OH(VH)::::; M.
(2) Set L+ := 11 or L 1 ,+, for L/0 2 (L) ~ A 6 or A5, respectively. Then either:
(i) 02 (0H(VH)) = 1 and OH(VH) = 02(H)::::; 02(L1T)::::; R1, or
(ii) L+ = 02 (0H(VH)) ::::! H, and H1 := OH(L+/02(L+)) is of index 2
in H with R1 E Syb(H1).
(3) L2 i H, and if L/02(L) ~ A5, L2,+ i H.
PROOF. As neither L 2 nor L 2 ,+ centralizes z, (3) holds. Next OH(VH) =OH(VH(z)) ::::; Oa(ZL n VH(z)) ::::; M = !M(LT) by 13.4.2.2, so (1) holds.
Set Y := 02 (0H(VH)). By (1), Y ::::; M, and as H E 'liz, Y centralizes z E
Vi-Zv. Thus the hypotheses of 13.3.9 are satisfied, so we can appeal to that lemma.
IfY = 1, then OH(VH) is a 2-group; so as VH E R2(H), OH(VH) = 02(H). Further
LiT ::::; H, so 02(H) ::::; 02 (L1T) by A.1.6, and hence conclusion (i) of (2) holds.
Thus we may assume conclusion (2) of 13.3.9 holds. In particular L+ = Y ::::! H,
so conclusion (ii) of (2) holds. D
LEMMA 13.4.5. Assume H E 1i(T, M) is nonsolvable. Then
(1) There exists KE C(H), and for each such K, KE £j(G, T), K ::::! H, and
K/02(K) ~As, La(2), A5, or A5.