i4.7. FINISHING Ls(2) WITH (VG1) ABELIAN 10778yb(K). That is, (1) holds. Further as 8 :::; 02 (Li) and [f' : 02 (Li)[ = 2, (2)
holds.Let B E 8yh(Li). By 14.7.72.2 and H.6.3.3, [Cs(B)[ = 8, so as P/Cp(U) =
[P/Cp(U), B] and Cu(B) =Vi using 14.7.72.3, (8n8k)/U = C 8 (B)U/U is of order
4. As 8 = [8,Li], Li is indecomposable on P/U, so we conclude (3) holds. D
LEMMA 14.7.74. (1) M =Land 8 = 02 (M).
(2) H =KT and 02(H) = 02(K).
PROOF. By 14.7.72.2, 8/V is the Steinberg module which is a projective L-
module, so Q/V = Qc/V EB 8/V, where Qc/V = CQ;v(L). Now [Qo,Li] :::; Vand Li contains the Sylow 2-subgroup 02(Li) of K*, so by Gaschiitz's Theorem
A.1.39, [Qc, K] :::; U. Then as QcU centralizes V, QcU centralizes (VK) = U, so
Qc centralizes (UL) = 8.
Let QL := CQ(L), so that QL :s; Qc. Then [QL, K] :s; [Qo, K] :s; U. Further
in the unique nonsplit L-module extension W of V in I.1.6 whose quotient is atrivial L-module, 02(Li) does not centralize a vector in W - V (cf. B.4.8.3),
so QL Vi = CQ 0 u(Li). Therefore since Li contains a Sylow 2-subgroup of K,
{JLU = QL x U with K centralizing QL again using Gaschiitz's Theorem A.1.39.
Then K centralizes QL by Coprime Action. So since K f:. M = !M(LT), we
conclude Q L = 1..Let B E 8yl3(Li), and set QB := CQ 0 (B). Then Qc = VQB, so <T?(QB) =
<T?(Qc) ::::) LT. But Lis irreducible on V, and QB n V = Vi, so <T?(Qc) n V =
- Then [<T?(Qo),L] :::; <T?(Qo) n V .= 1, so that <T?(Qo) :s; CQ(L) = 1 by the
previous paragraph. Since also CQ 0 (L) = 1, m(Qc/V):::; dimHi(L, V) = 1, with
[Qc, 02 (Li)] =Vin case of equality (again cf. B.4.8.3).
Suppose V < Qc. By 14.7.73.1, 02(Li) = 802 (K), so as we saw that 8
centralizes Qc, [Qc, 02(K)] = [Qc, 02(Li)] = V. Then as Vi = [U, 02(K)],[QcU, 02(K)] = [Qc, 02(K)]Vi = V. However, K normalizes QcU, and hence
also normalizes [QcU, 02 (K)] = V, so H = KT :::; Nc(V) :::; M, contrary to
HE 1iz.This contradiction shows that Q c = V. Hence Q = 8 = 02 (L) :::; 02 ( M) by
14.7.72.1, so 02 (L) = 02 (M) by A.1.6. By 14.7.72.4, V char 8, so that V ::::) M.
Thus M = LT by 14.7.67.5, so as LT= L0 2 (LT), 02 (LT) = 02(M) = 02(L),
and hence (1) holds.
Finally using (1) and 14.7.72.2, 4[QH[ = [Ri[ = 4[8[ = 2i^3 , so [QH[ = 2^11 =[0 2 (K)[ by 14.7.73.3, and hence (2) holds. D
Under the hypotheses of this section, we can· now identify Gas Ru.
THEOREM 14.7.75. Assume Hypothesis 14.3.1 holds with L/02(L) ~ L 3 (2) and
(VG^1 ) abelian. Then G ~ Ru.
PROOF. We verify that G is of type Ru as defined in section J.l. Then the
Theorem follows from Theorem J .1.1.
By 14.7.74.1, M =Land 8 = 02 (L). Thus as L acts on V and M E M,
L = Nc(V) with L/8 ~ L 3 (2). By 14.7.72.4, 8 is special with center V. Of course
Vis the natural module for L/8, and by 14.7.72.2, 8/V is the Steinberg module.
Thus hypothesis (Rul) is satisfied.