13.4. THE TREATMENT OF THE 5-DIMENSIONAL MODULE FOR A 6 913FF-modules by Theorem B.5.1, so by 13.4.14, we are in case (i) of 13.4.7.l. Ar-guing as in the proof of the previous lemma, m(R2) = 1 and then m(VJ7*) =
1 = m(Vo/Cv 0 (Vt)) and [Vo, Vt]= Zv. By symmetry between Vo and Vt, zt =
[Vo, VJ]= Zv, so g E Na(Zv) = M = Na(V) by 13.4.2.l. Then V = V9::::; Rg, so
by 13.4.15, [V, Vt]= l. Thus as V = V9, also [V, Vo]= l.Let Uo := [Z, K2], so that Uo is a 4-group as case (i) of 13.4.7.1 holds. As
Ho is solvable, z is of weight 4 in V and Zv ::::; U 0 ::::; VH by 13.4.14.l. Therefore
Uo = (Z{f^2 ) and CK 2 r(Uo) = 02(K2T). By 13.4.12.1, Ca(v) :'.S:: M = Na(V)
for each v E V# not of weight 4, so by 13.4.2.3, V is the unique member of v^0
containing v. But up to conjugation under L, (e 1 , 2 , 3 ,4, e 1 , 2 , 5 , 6 ) is the unique maximal
subspace U of Vall of whose nontrivial vectors are of weight 4, so r(G, V) 2; 3.
Let Wo := Wo(T, V). We claim [V, W 0 ] = 1, so that Na(W 0 )::::; M by E.3.34.2:
For suppose A := VY n M with A =/=-l. Assume for the moment that also V ::::; MY.
Then 1 =/=-[V, A] ::::; V n VY, so by the previous paragraph, [V, A]# contains only
vectors of weight 4. We conclude that all involutions of A are of cycle type 23 , and
hence IAI = 2 and IV : Cv(VY)I 2; IV : Cv(A)I = 4. Therefore A < VY when
V ::::; MY-since if A = VY, then we have symmetry between V and VY, so that
2 = IAI =IV: Cv(VY)I = 4.Now assume A = VY; then V f:. MY by the previous paragraph. Therefore
m(A) 2; r(G, V) 2; 3, and hence m(A) = 3 as m2(M) = 3. Further
U = (Cv(B): 1 =/=-B::::; A)= (Cv(D): m(A/D) < 3)::::; MY.
Thus U < V, which is not the case if A is conjugate to R 2. Therefore A is conjugate
to R 1 , and then m(V/U) = 1 so that U = V n MY and m(U/Cu(A)) = 2. But
applying the previous paragraph with U, A in the roles of "A, V", we conclude
that m(U/Cu(A)) = l. This contradiction establishes the claim that Wo::::; Cr(V)
and Na(Wo) :::; M.
Thus Wo is not normal in K2T, as K2 1:. M, and hence Wo 1:. 02(K2T) =
CK 2 r(U 0 ) by E.3.15. Therefore there is D := V"' :::; T for some x E G with
[Uo, D] =/=-l. But ID: CD(Uo)I = 2 as IT: 02(K2T)I = 2, so Uo:::; Ca(CD(Uo)) :::;
Na(V"') as r(G, V) 2; 3. Then as D does not centralize Uo, Zv = [Uo,D] :::; D.
By 13.4.2.3, V is the unique member of va containing Zv, so D = V. But
now [D, U 0 ] =/=-1, whereas U 0 :::; V 0 and [V, V 0 ] = 1 by the first paragraph. This
contradiction completes the proof of 13.4.16. D
Our final lemma shows that the 2-locals Mand Ho resemble the parabolics A
and P 3 of Sp 6 (2), except that z is of weight 2 in V and Zv :::; Soc(Vo). Still withthis information we will be able to obtain a contradiction to our assumption that
Gz is not solvable, completing the proof of Theorem 13.4.l.
LEMMA 13.4.17. (1) z is of weight 2 in V.
. (2) There exists g E H such that [V, V9] = (z) :::; V9.
(3) H 0 /02(Ho) ~ L 3 (2), V 0 is the core of the permutation module, Zv <
Soc(Vo), Zv :::; VH, and VH is not a 5-dimensional module if H/02(H) ~ A5.
(4) H is transitive on Vjf", and each subgroup of VH of order 2 is in Z{;.
(5) r(G, V) 2: 3.
(6) For each g E G - M, V# n V9 consists of elements of weight 2.PROOF. Let G1 :=LT= Mand G2 :=Ho, and form the coset graph r with