1120 15. THE CASE .Cr(G, T) =^0respectively. For a:= "(1g and /3 := 'YoY, set Ua := UfI, Za := z~, and V,a :=VY.
Let b := b(r, V). Also we choose a geodesic
'Yo, "/1, .. ·,'Yb =: 'Y
with Vi G~^1 ). As Vis not an FF-module for M by 15.1.8, bis odd by F.7.11.7.
From 15.2.8, QH = G~~). Thus as V S QH by 15.1.11.2, b > 1. As b is odd, G 1 is
a conjugate of H, so G~
1) = 02(G,) =: Q 1.
While Hypothesis F.8.1 does not hold, we can still make use of arguments insection F.8. As in section F.8, define D 1 := U 1 n QH and DH := UH n Q,. By
choice of 'Y, V i Q 1 , so DH < UH. Indeed V does not centralize U 1 , so there is.
g E G with "/lg = 'Y and [V, VB] # 1. But if D 1 = U 1 , then V^9 S Wo(QH, V) S
CH(UH) S CH(V) by 15.2.26.3, a contradiction. Therefore D 1 < U 1 , so we have
symmetry between "/land'"'( (cf. Remark F.9.17).
Next
m(U 1 /D 1 ) = m(U;) s m2(H*) = 2,so by symmetry, m(UH/DH) s 2. Now [UH,D,] s Zs by 15.2.28, so by symmetry
[U 1 , DH] S Z 1. Then [DH, D 1 ] s Zs n Z 1 , while as [V, V 1 ] # 1, we conclude from
15.2.27.3 that ZsnZ 1 =1. Thus [DH,D,] = 1. Next m(D 1 /CD)V)) S m2(M) =1by15.2.25.1, and by symmetry m(DH/CDH(V^9 )) S 1, so
m(UH/CuH(V^9 )) S m(UH/DH) + 1S3.
Therefore VB* induces a transvection on each of the 2-chief factors of H on UH
appearing in 15.1.12, so Vg* S Ht for i = 1 or 2. Hence m(V9 /(VB n QH)) = 1.By symmetry, m(V/(V n Q 1 )) = 1, and
[V n Q,, V^9 n QH] s [D,, DH] = 1.
Therefore as s(G, V) > 1 by 15.2.26.1, E.3.6 says V n Q 1 s Ca(V^9 ), and then by
another application of those lemmas, VB s Ca(V n Q 1 ) s Ca(V), contrary to the
choice of VB.
This contradiction completes the proof of Theorem 15.1.3.15.3. The elimination of Mr/CMr(V(Mr)) = Ss wr· Z2
In this section, we complete our treatment of the groups satisfying Hypothesis
14.1.5, by proving:THEOREM 15.3.1. Assume Hypothesis 14.1.5. Then G is isomorphic to J2, J3,(^3) D4(2), the Tits group (^2) F4(2)', G 2 (2)', or M12.
Observe that the groups in Theorem 15.3.1 have already appeared in Theorem
15.1.3, so that we will be working toward a contradiction. On the other hand, the
shadows of the groups Aut(Ln(2)), n = 4, 5, 89 , A 10 , Aut(He), and L wr Z2 forL ~ 85 or L of rank 2 over F 2 arise, and cause difficulties: Each of these groups
possesses ME M(T) such that V(M) is of rank 4 and AutM(V(M)) = o+(V(M)).
For X S G, we let B(X) denote the subgroup generated by all elements of X
of order 3.