15.3. THE BLIMINATION OF Mf/CMr(V(Mr)) = Ss wr Z 2 1151Let Bi:= Gp(Di)V. Then Bi~ E54 and I has four orbits on Bf: zI and vr,
and orbits of length 12 and 36 on Bi - V. Further Bi :::; TK, and Ei := Bin U is of
rank at most m(U) = 4, while m(B*) :::; m(H*) = 2, so Bi is a 4-group in Tk and
Ei = Go(Bi). Then Bi= GTK(Ei) is invariant under NH(Ei) = NH(Bi) ~ 84 , soas NH(Bi) does not act on V*, Na(Bi) does not act on V. Thus as v ~ z^0 , the
two orbits of I on V# are fused to its two orbits on Bi~ V, so all involutions in Bi
are fused to z or v, completing the proof of (2). DWe now eliminate the shadows of.Aut(L 5 (2)) and Aut(He), and establish
Theorem 15.3.50.
First the involution t of 15.3.60.2 is in T-TK, since TK = UGT(V).by 15.3.59.2.
By 15.3.59.3, IT : TKI = 2, so as G is simple, t^0 n TK f=. 0 by Thompson Transfer.
Thus t E z^0 U v^0 by 15.3.61.2. However this contradicts 15.3.60.2 and 15.3.61.l.
This contradiction completes the proof of Theorem 15.3.50.15.3.4. The case (VMc) abelian. By Theorem 15.3.50, VH is abelian for
each HE H(T, M). This will allow us to use weak closure in 15.3.63, and to verify
Hypothesis F.9.8. Then Hypothesis F.9.8 eventually leads to a contradiction.
LBMMA 15.3.62. (1) Mc is transitive on {VB : g E G and Z :S VB}.
(2) If V n VB f=. 1, then [V, VB]= l.PROOF. Part (1) folllows from 15.3.46.1 using A.l.7.1, since Mc = Ga(Z) by
15.3.4. If g E G-M and V n VB f=. 1, then as V n VB is totally singular by 15.3.46.2
and M is transitive on singular vectors, we may take Z :::; V n VB. Therefore
VB :S VM 0 :S Ga(V) by (1) since VM 0 is abelian by Theorem 15.3.50. D
LEMMA 15.3.63. Assume r(G, V) ~ 3. Then
(1) W1(T, V):::; GT(V), so w(G, V) > 1.
(2) n(H) > 1 for each HE H(T, M).
PROOF. Assume W 1 (T, V) does not centralize V, and let A be a hyperplane of
VB with A:::; T and A f=. l. In particular Vi. MB by 15.3.46.4, so as r(G, V) ~ 3
by hypothesis, m(VB /Gvg (V)) = m(A) + 1 > 2, and hence m(A) = 2 = m2(M).
As M is solvable, a(M, V) = 1 by E.4.1, so there is a hyperplane B of A with
GA(V) :::; B such that 1 f=. [Gv(B), A]=: VB. As r(G, V) ~ 3 and m(VB /B) = 2,
Gv(B) :::; MB, so 1 f=. VB :::; V n VB, contrary to 15.3.62.2. Thus [V, W 1 (T, V)] = 1,
establishing (1).
By A.5.7.2, M = !M(NM(GT(V))), while r(G, V) > 1 < w(G, V) by our
hypotheses and (1). Thus (2) follows from E.3.35.l. D
Recall that Hypothesis F.9.1 holds by 15.3.48.1and15.3.49.3 .. Further 15.3.62.2
gives part (f) of Hypothesis F.9.8, while case (ii) of part (g) of Hypothesis F.9.8
holds by 15.3.47. Thus Hypothesis F.9.8 holds, so we conclude from F.9.16.3 that:
LEMMA 15.3.64. q(H*, f~fH) :S 2.
LEMMA 15.3.65. (1) If HE H*(T, M), then n(H) = 1.
(2) r(G, V) = 2.PROOF. By 15.3.46.3, r(G, V) ~ 2, so if (2) fails then r(G, V) ~ 3, and hence
n(H) > 1 for H E H(T, M) by 15.3.63.2. Thus (1) implies (2), so it remains to
establish (1).