2.3. PRELIMINARY ANALYSIS OF THE SET ro 523PROOF. Let U E U be of maximal order and Hu E 'He(U, M). Then JVI :S::
IUI ::::: IHl2 for HE ro. D.
The remainder of the proof of Theorem 2.1.1 focuses on the members of r 0. We
need to consider members of r maximal in the two different senses of Notation 2.3.5
because: On one hand, at a number of points in the proof we produce members of
r* (for example in 2.3.7.1), so we need results on the structure of such subgroups.
On the other hand, near the end of the proof, particularly in 2.5.10, we need to work
with those HE r such that U(H) contains a member (U, Hu) with IUI maximal in
U. Thus at that point we choose HE I'.
We often use the following observations to produce members of r 0 : ·
LEMMA 2.3.7. Assume HE r, and let (U, Hu) E U(H) and U :S:: SE Syl 2 (H).
(1) Assume IT : SI = 2. Then H E r. If Hi E r with IHil2 2 ISi, then
IHil2 = ISi, and Hi Er*.
(2) Assume H E I'o and Hi E 7-l(H). Then Hi E I'o, and S is Sylow in Hi
andHinM.
(3) Assume HE r 0 and S:::; Hi Er; when HE I', assume in addition that
IUI is maximal among members of U and that 'He(U, M) n Hi =f 0. Then Hi Ero,
and S is Sylow in Hi and Hin M.
(4) Under the hypotheses of (2) and (3), if HE I', r, then Hi EI', r*,
respectively.
PROOF. Since U :S:: S by hypothesis, S :S:: M by 2.3.2.1.
Assume IT: SI= 2 and Hi Er. As M = !M(T), IHil2::::: ITl/2 = ISi = IHl2,
so HE r, and if IHil2 21s1, then Hi Er, establishing (1).
Now assume the hypotheses of (2); then Hi E 'H(H) ~ r. When H E r,
maximality of ISi forces H1 Er, withs Sylow in Hi, and hence in Hi nM. Thus
(2) and the corresponding part of (4) hold in this case. When HE I'* there is some
(U, Hu) E U(H) ~ U(Hi), with U of maximal order in U, so by the maximality of
ISi subject to this constraint, H1 E I'* and S is Sylow in Hi and in Hi n M. This
completes the proof of (2), along with the corresponding part of (4).
Assume the hypotheses of (3); the proof is very similar to that of (2): Again if
HE r, then as S :S:: Hi, Hi Er by maximality of ISi. Thus we may assume that
H E I'*. Then by hypothesis IUI is maximal in U and there is H2 E 7-le(U, M) n
H 1. Thus (U, H 2 ) E 7i(H1), and hence Hi E r. Then by maximality of IUI and
maximality of ISi subject to that constraint, Hi EI'*. D
The next result 2.3.8 lists various properties of members of r. In particular
part ( 4) of that lemma is the basis for our analysis of the case where r 0 contains a
member of 'He in the next section.
LEMMA 2.3.8. Let HE r, (U,Hu) E U(H), and U :S:: SE Syl2(H). Then
(1) ISi < ITI and SE /3. In particular, S :S:: M, so SE Syb(H n M).(2) 02(Hu) E S2(G).
(3) (U,Hu) E U(Na(02(H))), and Na(02(H)) Er.
(4) If H E r 0 n 'He, then C(H,S) :S:: H n M, so H = (H n M)Li ···Lswith s :::; 2 and Li an L 2 (2n)-block, A 3 -block, or A5-block such that Li :f:. M and
Li= [Li, J(S)].
(5) Assume HE I'o. Then