il.5. THE FINAL CONTRADICTION 783Suppose U is nonabelian. Then as U = (V 3 °^1 ), U does not centralize Vi, so
tJ f:. 1and1 f:. [Vi, U] :S: Vi by 11.5.5.2. As U j LiT, tJ = 02(Li) by(!). Therefore
[V, U] =Vi, which is impossible since by the previous paragraph, V::;: 02 (Gi), so
[V, U] :S: Vi by 11.5.5.2.Thus U is abelian. If [V, U] f:. 1 then as [Vi, U] = 1, tJ = Z(L 1 ) by (!). In this
case set W := U =UH. On the other hand if [V, U] = 1, set W := WH, where
WH := (V^01 ). As U j G1, [U, W] = 1. As V :S: 02(Gi), W :S: 02(Gi), so as we are
assuming that Wis nonabelian, and as [Vi, W] = 1, again W = Z(Li). Therefore
in either case, W = Z(Li), so [V, W] = [V, Z(L 1 )] =Vi, and hence <I>(W) = V 1.Choose an element y E L so that (Z(Li), Z(Li)Y) ~ L 2 (q), and I := (W, WY)
contains Xi = Cx(Li/0 2 (L 1 )). Observe that I= (W, iiJY) for each 1 f:. iiJ E W,
and V =Vi EB Vt Then as [W, 02(I)] :S: W n 02(I),Q := (W n 02(J))(WY n 02(I)) j I,
.with [0 2 (J), I] 5, Q. Since I= (W, WY) and (W) =Vi, [WnWY, I] :S: V 1 Vl :S: V.
Also (W) =Vi :S: V, and
Q/(W n WY)V =(WV n Q)/(W n WY)V x (WYV n Q)/(W n WY)V,
with (WV n Q)/(W n WY)V = CQ/(WnWY)v(w), as I= (W, iiJY).
We claim that Q::::: 02(Gi) =:Qi: It follows from G.1.6 that Q/(W n WY)V
is a sum of natural modules for I, so that CQ(Xi) :S: (W n WY)V ::;: Q 1. Thus
as Q = [Q, Xi]CQ(Xi), it remains to show [Q, Xi] ::;: Qi. Next (Q) ::;: (W n
WY)V::;: Q 1 and Q::;: 02 (I)::;: 02 (LT), so using parts (4) and (5) of 11.5.6, [Q,Xi]
centralizes KY/0 2 (KY). Next Li is transitive on Vi.° n (V - Vi), so by a Frattini
Argument, Mi= LiNM 1 (V{). Thus as NM 1 (V{) acts on [Q,Xi], Gi = KYMi =
KYNM 1 ([Q,Xi]), so as [Q,Xi] centralizes KY/0 2 (KY), the claim is established.
Next Lis generated by three conjugates Jli, 1 :S: i :S: 3, of I under L 1. As 02 (L)acts on W, it acts on WY^1 i for each i, so [0 2 (L), L] is the product of [02(L), W] ::;:
Q::;: Qi and [0 2 (L), WY^1 i] ::;: Qi, so [0 2 (L), L] ::;: Q 1. Then as the Schur multiplier
of Sp4(q) is trivial by I.1.3, 02(L) = [02(L), L] :S: Q1.
Now as Z(Li) = W ::::; Qi :SJ Li, Qi = R1 or W. However in the latter
case as 02 (L) :S: Qi, Qi E Syb(IQ 1 ), and then by C.1.29, there is a nontrivial
characteristic subgroup Q 0 of Qi normal in IQ 1. But then Qo j (I, L1T) =LT,
so Gi ::;: Ne( Qi) ::;: Nc(Qo) ::;: M = !M(LT), contradicting 11.5.4. Therefore Qi =
Ri, so R 1 n LQi =Qi. Then by C.l.32, either there is a nontrivial characteristic
subgroup Q 0 of Qi normal in LQ 1 , or Lis an Sp 4 (4)-block. The former case leads
to the same contradiction as before.
Therefore Lis an Sp 4 (4)-block. Since Hi(_E, V) = 0 by I.1.6, and we have seenthat 02 (L) = [02(L), L], we conclude from C.1.13.b that 02(L) = V.
We now treat the case that [V, U] = 1. Here we recall that W = WH and
W = Z(L 1 ), so [W,L1] :S: 02(L) = V, and hence [W,Li] = [V,L1] =Vi :S: U.
Therefore K = (Lf) centralizes W/U, so W =UV. But then as [V, U] = 1 and Uand V are abelian, W = W H is abelian, contrary to our assumption.
Therefore [V, U] f:. 1. In this case we recall that W = U. As 1 f:. [V, U],
WH i. Cr;'.T(Vi) = Z(Li), so as WH is LiX-invariant, WH = Ri and there exists
g E Gi with VY ::;: R 1 but VY "t Z(Li). Thus conjugating in Li if necessary,V 1 :S: Cv(VY) :S: Vi :S: [V, VY]. But WH :S: Q1 :S: Nc(V), so VY j WH, and
thus [V, VY] ::;: V n VY ::;: Cv(VY). Hence Vi = V n VY is conjugate to V:f under
LY, so we may choose g E Na(V2). By 11.1.3, g acts on L2, so as V = [V, L2],