606 4. PUSHING UP IN QTKE-GROUPSOM(M+/0 2 (M+)), but just cover M modulo OM(M+/02(M+)) for some M E
:E(M+)· For example in the FSU, take M+ := L 0 , and M_ := LoOM(Lo/02(Lo))T,
or more generally M E :E(M) with M ~ LoOM(Lo/02(Lo))T and LT= LM-.
In the remainder of the section pick M E :E(M+) and define 'TJ = 'TJ(M+, M)
to be the set of all subgroups I of M such that IOM(M+/02(M+)) = M and
M+ ~ I0 2 (M+) with 02 (I) =f. 1. We wish to show that each IE 'T/ is a uniqueness
subgroup; thus we consider the set of counterexamples to this conclusion, and define
μ = μ(M+, M_) to consist of those IE 'T/ such that H(I, M) =f. 0, where
H(I, M) :={HE H(I): Hi. M}.
Finally define a relation ;Son 'I] by Ii ;S I2 if 02(Ii) ~ 02(I2) and IinM+ ~ I2nM+.Let μ = μ(M+, M_) consist of those I E μ such that 02(I) is not properly
contained in 02 (Ii) for any I 1 E μ such that I ;S Ii.
We begin to study this set μ* of "maximal" members of μ.
LEMMA 4.1.3. Let IE 'TJ, I~ Io ~ M_, and Ii ~Io with 1 =f. 02(fi). AssumeIo= Ii01 0 (M+/02(M+)) and M+ n Io~ Ii02(M+)· Then
(1) Ji E 'T/·
(2) If IEμ*, I ;S Ii, and 02(I) < 02(Ii), then M = !M(fi).
PROOF. By hypothesis IE 'I] and I~ Io ~ M_, so from the definition of 'I],M = IOM(M+/02(M+)) ~ IoOM(M+/02(M+)) ~ M, (*)
and hence all inequalities in (*) are equalities. Again from the definition of 'T/,M+ ~ I02(M+) ~ Io02(M+)·
Next as Io= Ii01 0 (M+/02(M+)) by hypothesis, and (*) is an equality,
M = IoOM(M+/02(M+)) = fiOM(M+/02(M+)) ~ M,
and again this inequality is an equality. As M+ ~ I 0 02(M+) and M+ n Io ~
Ii02(M+), M+ =(Io nM+)02(M+) ~ Ii02(M+)· Then as 02(Ii) =f. 1 by hypoth-
esis, Ii E 'T/, and hence (1) holds.
Assume the hypothesis of (2). If M =f. !M(Ii), then 'H(Ii, M) =f. 0, so that
Ii E μ. As I ;S Ii and 02(I) < 02(I1), this contradicts I E μ*, establishing
(~. D
The next two results are used to establish Hypothesis C.2.8 in various situa-
tions; see 4.2.4 for one such application. Hypothesis C.2.8 allows us to apply the
pushing up results in chapter C of Volume I.
LEMMA 4.1.4. Suppose I E μ*, and let R := 02 (I) and H E H(I, M). Set
H+ := 02 (M+ n H). Then
(1) R ~ OM(M+/02(M+)).
(2) O(G, R) ~ M.
(3) M+ = H+0 2 (M+) and H+ ::::! H n M.
(4) RE Syl2(GH(H+/02(H+))) n Syh(OHnM(H+/02(H+))).
(5) R = 02(NH(R)) so that RE B2(H), and 02(H) ~ 02(H n M) ~ R.(6) F*(H n M) = 02(H n M).
PROOF. Let I+:= 02 (M+ n I). As M+ ~ I0 2 (M+) by definition of 'I], while
M+ = 02 (M+) by Hypothesis 4.1.1, M+ = I+0 2 (M+). Therefore (1) follows from