i4.7. FINISHING L 3 (2) WITH (VG1) ABELIAN 106iTHEOREM 14.7.40. Assume H E Hz such that H = KLiT for some K E
C(H) with K/02,z(K) of Lie type over Fzn for some n > 1. Then H* ~ 85 and
UH/ Cu H ( K) is the Lz ( 4 )-module.
Until the proof of Theorem 14. 7.40 is complete, assume the hypotheses of the
Theorem. By 14.7.30, we may apply F.9.18.4 to conclude that
(*) K* is a Bender group, (8)L3(2n), 8p4(2n), or G 2 (2n).
Let B 0 be the Borel subgroup of K* containing T 0 := T* n K* and let B :=
02 (B 0 ). As K is defined over F 2 n with n > 1, and LiT = TLi, Li acts on B,
so by Theorem 14.7.29, B ~ H n M ~ NH(Li). Then as M = LCM(L/0 2 (L)),
BLi = BeLi, where Be := 02 (CBL 1 (L/02(L))). Also Li/0 2 (Li) is inverted by
some t ET n L, and [t, Be] ~ 02(L) n Be ~ 02(Be), so Be02(LiB)/0 2 (LiB)
is the unique t-invariant complement to Li02(LiB)/02(LiB) in LiB/0 2 (LiB).Choose Xi E 8yl3(Li) with Xi inverted by t.
LEMMA 14. 7.41. Either
(1) Li i K, m3(K) = 1, Be= B, Li ::::1 H, and Li is inverted in CH·(K*),
or(2) Li ~ K* ~ Lz(4), U3(8), or (8)L3(4).
PROOF. Suppose first that Li i K. Then B* /02(B*) is at-invariant comple-
ment to Xi in XiB* /0 2 (B*), so as B(:;02(B*)/0 2 (B*) is the unique such com-
plement, Be = B. Thus Xi (t) centralizes B* /0 2 (B*), so from the structure ofAut(K) for K on the list in (), either Xi induces inner automorphisms on K,
or K Xi~ PGL3(4). As Li i K, K is not GL3(4) by 14.7.28. As q(H*, UH)~ 2
by Notation 14.7.1, Theorems B.4.2 and B.4.5 eliminate the case K* Xi~ PGL 3 (4).
Thus Li~ K*CH·(K*/02(K*)) =: Y*, and as Li i K, B(Y) i K, where Y is
the preimage of Y* in H. Therefore m3(K) < 2 by A.3.18, so that m3(K) = 1 by
14.7.34. Then as t centralizes B* /0 2 (B*), t also induces an inner automorphismon K, from the structure of Aut(K) for K in () of 3-rank 1. Indeed the pro-
jection oft on K* then lies in 02(B*) ~ Ri, so we conclude Li= [Li, te] for some
te E CT(K*), and hence Li centralizes K*. Therefore Li ::::1 H* as H = KLiT,
so H normalizes 02 (LiQH) =Li, and hence (1) holds.
So assume instead that Li ~ K. As Tacts on Li, Li ~Band B/02(B) =
Li0 2 (B)/0 2 (B) x Be0 2 (B)/02(B). Then as t inverts Li/02(Li) but [t, Be] ~
02 (Be) with Be of index 3 in B, we conclude (2) holds from the structure ofAut(K) for K on the list of(*). D
LEMMA 14.7.42. If Li i K then H* ~ 85 x 83, UH= [UH,K] E9CuH(K), and
[UH, K] is the tensor product of the 83-module and the 85-module.
PROOF. Assume Li i K. Then by 14.7.41, Li ::::1 H, Li is inverted in CH* (K*),
m 3 (K) = 1, and B =Be. Also B::::; HnM = NH(V) by 14.3.3.6, so B centralizesV as EndL/0 2 (L) (V) ~ Fz.
As Li is inverted in CH ( K), each H-chief factor W on UH is the sum W =
Wi E9 W2 of a pair of isomorphic K-modules wi. Indeed since u~ E Q(H, UH)
by Notation 14.7.1, arguing as in the proof of F.9.18.6, U~ is an FF*-offender on
Wi and W2, so that K* is L2(2n), 8L3(2n), 8p4(2n), or G2(2n) by Theorem B.4.2.
As m 3 (K) = 1, the last two cases are eliminated, and n is odd if K* ~ 8L 3 (2n).