14.5. STARTING THE CASE (vG1) ABELIAN FOR Ls (2) AND L2 (2) 101314.5. Starting the case (VG^1 ) abelian for L 3 (2) and L 2 (2)
In this section, and indeed in the remainder of the chapter, we assume:
HYPOTHESIS 14.5.l. Hypothesis 14.3.1 holds and U := (VG^1 ) is abelian.
As U is abelian and <J?(V) = 1, U is elementary abelian. Recall from the
discussion after 14.3.6 that Hypothesis 12.8.1 holds. In particular G 1 "f:. M by
12.8.3.4, so that V < U. Recall also the definitions of Gi, Li, and Vi, for i :::; dim(V),
from Notation 12.8.2.LEMMA 14.5.2. If g E G with 1 f= v n V9' then [V, VB] = l.
PROOF. As (V^01 ) is abelian by Hypothesis 14.5.1, the results follows from the
equivalence of (2) and (3) in 12.8.6. D14.5.1. A result on XE H(T) with X/0 2 (X) = L 2 (2). Recall that under
case (2) of Hypothesis 14.3.1 where L/0 2 (L) ~ £ 2 (2)', 14.3.5 says there existsno X E H(T, M) such that X/02(X) ~ L2(2). In this subsection, we establish
a result providing some restrictions on such subgroups in case (1) of Hypothesis
14.3.1, where L/0 2 (L) ~ £ 3 (2). Namely we prove:THEOREM 14.5.3. Suppose Y = 02 (Y):::; G 1 is T-invariant with YT/0 2 (YT)
~ L2(2). Then
(1) Either Y:::; M, or case (1) of Hypothesis 14.3.1 holds and [V2, Y] = l.
(2) IfYL1 = L 1 Y, then Y:::; M.
(3) (V{) is not isomorphic to E 8 •Until the proof of Theorem 14.5.3 is complete, assume Y is a counterexample.
LEMMA 14.5.4. (1) Y "f:. M.
(2) Case (1) of Hypothesis 14.3.1 holds, namely J;/02(L) ~ L3(2).
PROOF. Assume (1) fails, so that Y :::; M. Then conclusions (1) and (2) of
Theorem 14.5.3 are satisfied. Further Y acts on V by 14.3.3.6. Thus as Vi :::; V,
m( (V{)) :::; m(V) :::; 2, so that conclusion (3) of 14.5.3 holds. This contradicts our
assumption that we are working in a counterexample.
Thus (1) is established. Then (1) and 14.3.5 imply (2). DSet X := L2, and H := (X, Y, T). Notice that H 1:. G1 since X 1:. G1. Set
VH := (V 1 H), QH := 02(H), iI := H/QH, and H* := H/CH(VH)· Observe that
(H, XT, YT) is a Goldschmidt triple (in the language of Definition F.6.1), so by
F.6.5.1, a := ext, t, YT) is a Goldschmidt amalgam, and so is described in F.6.5.2.
LEMMA 14.5.5. QH f= l.
PROOF. Assume QH = l. By 1.1.4.6, XT and YT are in He, and so satisfy
Hypothesis F.1.1 in the roles of "L 1 , £ 2 ", with Tin the role of "S". Then a is a
weak BN-pair of rank 2 by F.1.9, and the hypothesis of F.l.12 is satisfied, so that
a is described in case (vi) of F.6.5.2. Then as X has at least two noncentral 2-chief
factors (from V and the image of 02(L2) in L/02(L) ~ L3(2)), by inspection of
that list, a is isomorphic to the amalgam of G2(2)', G2(2), M12, or Aut(M12),
and X has exactly two such factors. In each case, Z = fh(Z(T)) is of order 2, so
V1 =Z.