14.6. ELIMINATING L 2 (2) WHEN (vG1) IS ABELIAN 1023For IE I, set T1 := T n I and Iz :=In G1.
The next two observations are straightforward from the definitions:LEMMA 14.6~4. If I EI* and T1 :::; J E I, then J EI* and TJ = T 1.
LEMMA 14.6.5. If I E I and I :::; J E ?i, then J E I. If further I E I*, thenJ E I* and TJ = T1.
Recall from Definition F.6.1 the discussion of Goldschmidt triples.LEMMA 14.6.6. Assume IE I*, and let L 1 := 02 (L n J). Then
(1) T1 is either Tu or To.
(2) T1 E 8yb(I).
(3) If In Mi G1 then L = L102(L) and LT= L1T10 2 (LT).
(4) Either C(I, T1):::; Iz, or L = L102(L) and LT= L1T102(LT).
(5) If IE I* then either Iz is the unique maximal subgroup of I containing T 1 ,
or L = L102(L) and LT= L1T10 2 (LT).
(6) Assume ITI > 29 and L = L102(L). Assume further that there existsH2 with To :::; H2 :::; CH(u), H2/02(H2) ~ 83, H2 i M, and H2 has at least
two noncentral 2-chief factors. Then setting I2 := 02 (H2)T1, I1 := L1T 1 , and
Io :=(Ii, I2), we have Io EI* and (Io, Ii, h) is a Goldschmidt triple.
('l) T1 is not normal in I.PROOF. We first establish (1) and (2). Let T1 :::; 8 E 8yl 2 (I) and set Qi :=
02(Gi). Now Na(Tu) =To by 14.6.3.4, so as Tu is of index 2 in T 0 by 14.6.3.1,
Ns(Tu) = Tu or To. In the first case, 8 = Tu = T1, so that (1) and (2) hold.
In the second case I is not contained in M or Gi, so that T1 < T by 14.2.2.5,
and hence T1 =To since IT: Toi = 2 by (Ul). Then as Na(To) = T by 14.6.3.2,
Ns(T1) :::; Nrn1(To) = T1, so that 8 = T1 =To, and so (1) and (2) hold in this
case also.
Next we prove (3), so assume X :=In M i Gi. As Lis transitive on V#,M = L(MnGi) and IM: MnGil = 3 is prime, so M = X(MnGi). Next Tu:::; T1,
so by 14.6.3.3, L = [L, a] for some a E Qin T1 :::; X. As LQi :::l L(M n Gi) = M,
(ax):::; LQi n X. If ax~ Qi, then as M = X(Gi n M) and Qi :::l G 1 , aM ~ Q 1so that (aM) is a 2-group and hence a E 02 (M), contradicting L = [L, a]. Thus
ax g; Qi, so as Qi is of index 3 in LQi and L = 02 (LQi), L :::; (aX)0 2 (LQi).
Then as L = 0^2 (LQ1), 02 ( (ax)) :::; L n X, so that L = (L n X)02(L) = L10 2 (L),
and as L = [L, a], LT= L1T102(LT). Hence (3) holds.
Next suppose there is 1-::/= C char T1 with N1(C) i Iz. As T1 < T by (1), T 1
is proper in Nr(T1):::; Na(C). Then as IE I*, Na(C) rf_ I by 14.6.4, and hence
Na(C):::; M since N1(C) i Iz. Therefore In Mi G1, so (4) follows from (3).
Next assume I E I* and let Y be a maximal subgroup of I containing T 1.
Then by minimality of I, Y is contained in Gi or M, so that Y is Iz or In M by
maximality of Y. Thus (5) also follows from (3).
Assume the hypotheses of (6), and set Ii := L1T1. By (2), T1 E 8yb(I), so that
T1 E 8yl2(I1). As L = L102(L), we conclude from 14.6.3.3 that Ii/02(Ii) ~ 83.
Next since T1 :::; To ::; H2 using (1) and the hypothesis for (6), I 2 := 02 (H 2 )T 1
is a subgroup of H 2 with 02 (I 2 ) = 02 (H 2 ). Also 02 (H 2 ) centralizes ii, and hence
also v,, so as Tu E 8yl2(CH(u)) by 14.6.3.4, Tu E 8yb(0^2 (H2)Tu)· Thus as Tu ::;