i2.5. ELIMINATING L5(2) ON THE iO-DIMENSIONAL MODULE 82iLEMMA 12.5.9. LetTi := Cr(V{), and choose notation with Ti E Syh(CM(V{)).
Then(1) IT: Til = 4, and Ti E Syl2(Ca(V{)).
(2) V{ ¢'.Vi°.
(3}Vi^0 nV=Vf.
(4) If g E G - Na(V) with Vi:::; Vg, then Vg E v^01 and [V, Vg] = 1.PROOF. The first part of (1) follows from 12.5.5.3. By 12.5.7, W 0 := W 0 (T, V) =
Wo(Ti, V) and Na(Wo) :::; M, so Ti E Syh(Ca(V{)). Thus (1) holds, and (1) im-
plies (2). Then (2) and 12.5.5.1 imply (3). Finally under the hypothesis of (4), (3)
and A.1.7.1 imply Vg E v^01 , and then 12.5.8.2 implies [V, Vg] = 1. DLEMMA 12.5.10. (1) Either Wi := Wi(T, V) centralizes V, or Wi = R 6.
(2) Ca(Ci(T, V)):::; M.
PROOF. Suppose A:= Vg n M:::; T with [A, V] =/= l, and m(Vg /A) :::; 1. By
12.5.7, m(Vg /A) = 1 and V > I:= Nv(Vg). We now argue much as in 12.5.7:
This time r(G, V) > 3 = s(G, V) by 12.5.6, so A E A2(f', V) by E.3.10. Thereforeeither A centralizes Vi, or AutA(Vi) E A2(Autr(Vi), Vi), so that m(A/CA(Vi)) =
m2(L3/02(L3)) = 2, and hence Vi :::; I as r(G, V) > 3. But in the latter case,
Vi :::; [Vi, A] :::; Vg, contrary to 12.5.9.4. We conclude A centralizes Vi, and similarly
that A centralizes the space V£ of 12.5.7; so again A:::; C := CM 6 (Vi + V£) with
m(C/R6) = 1, and as A E A2(f', V), A:::; CLr(V5) = R5. Thus as L5 is irreducibleon R5, Wi = R5. Hence (1) holds.
By (1), V 6 :::; Ci(T, V), so (2) follows from 12.5.4. D
For the remainder of the section, let HE 1i*(T, M). Recall from 3.3.2.4 that
H is described in B.6.8 and E.2.2.LEMMA 12.5.11. (1) n(H) > 1.
(2) Ki= Li.
PROOF. By 12.5.7and12.5.10, Na(Wo) :SM 2: Ca(Ci(T, V)); so as s(G, V) =
3, (1) follows from E.3.19 with i,j = 0, 1. Suppose Li <Ki, so that in particular
Ki f:. M. Then using the description of the embedding of Mi in Ki in 12.5.3 and
its proof, there is HE H(T) with H:::; KiT, Hf:. M, and either H/02(H) ~ 83,
or Ki/0 2 (Ki) ~ SL2(7)/E 49 and H := 'B1(Ki)T. Thus H E 1i*(T,M) withn(H) = 1, contrary to (1). Thus (2) is also established. D
LEMMA 12.5.12. If H:::; Gi, then n(H) = 2, and a Hall 2' -subgroup of H n M
is a nontrivial 3-group.PROOF. By 12.5.11.1, n(H) > 1. Then applying 12.2.11 to Vi in the role of
"U", the lemma holds. DWe are now in a position to complete the proof of Theorem 12.5.1.
By Theorem 12.4.2.2, Gi f:. M, so we may choose HE 1i*(T, M) n Gi. Hence
by 12.5.12, n(H) = 2 and a Hall 2'-subgroup of H n Mis a nontrivial 3-group. Set
K := 02 (H), so that K f:. M, and X := Ca 1 (Li/02(Li)). Then as Li= Ki :::1 Gi
by 12.5.11.2, and T:::; L, Gi = LiX. In particular, X f:. Mas Gi f:. M, so we may
choose H:::; XT. As m 3 (Li) = 1 and m3(Gi):::; 2, m3(X):::; 1. Therefore m3(X) =