\baselineskip=20pt
\magnification=\magstephalf
\def\a{{\alpha}}
\def\be{{\beta}}
\def\o{\overline}
\def\w{\widetilde}
\def\b{{\alpha^\prime}}
\def\G{{\cal G}}
\def\O{\Omega}
\def\L{{\cal L}}
\def\M{{\cal M}}
\def\O{{\Omega}}
\def\owP{{\o{\w P}}}
\def\P{{\cal P}}
\def\U{{\cal U}}
\def\K{{\cal K}}
\def\d{{\delta}}
\font\tenrm=cmr10
\long\def\fussnote#1#2{{\baselineskip=10pt
\setbox\strutbox=\hbox{\vrule height 7pt depth 2pt width 0pt}%
\tenrm
\footnote{#1}{#2}}}
\centerline{{\bf The P!-Theorem}}
\bigskip\bigskip
\centerline{Ch.W. Parker, University of Birmingham}
\centerline{G. Parmeggiani, University of Padova}
\centerline{B. Stellmacher, University of Kiel}
\bigskip\bigskip
Let $H$ be a finite group and $p$ be a prime dividing the order of $H$. Then
$H$ is of {\bf characteristic p} if $C_H(O_p(H)) \leq O_p(H)$; and $H$ is of
{\bf local characteristic p} if every $p$-local subgroup of $H$ is of
characteristic $p$. Moreover, $H$ is a {\bf ${\cal K}_p$-group} if the
simple sections of the
$p$-local subgroups are "known" simple groups\fussnote{$^1$}{Which
means, they are groups of prime order, groups of Lie type, alternating
groups or one of the 26 sporadic groups.}.
Every group with a self-centralizing cyclic Sylow $p$-subgroup, as
for example the alternating group $A_p$, is of local characteristic $p$, and these groups
are particular examples of groups with a strongly $p$-embedded subgroup.
Apart from such groups, all
groups of Lie type in characteristic $p$ of rank at least $2$ and
some sporadic groups (for suitably chosen $p$) have local
characteristic $p$.
Therefore it would be a major
contribution to a revision of the classification of the finite simple
groups to give a classification of all finite groups of local
characteristic $p$ that do not have a strongly $p$-embedded
subgroup. This is the goal of a project initiated by
U. Meierfrankenfeld. For an overview of this project see [MSS1].
The part of the project our paper deals with uses the following
hypothesis:
\bigskip
{\bf Q!-Hypothesis.} $H$ is a finite ${\cal K}_p$-group of local
characteristic $p$, $S \in
Syl_p(H)$ and $Z:= \O_1(Z(S))$. There exists a maximal $p$-local
subgroup $\w C$ of $H$ with $N_H(Z) \leq \w C$ such that for $Q :=
O_p(\w C)$
$$C_H(x) \leq \w C\hbox{ for every }1\ne x \in Z(Q).\quad \hbox{\bf (Q-Uniqueness)}$$
\bigskip
In the subdivision given in [MSS1] this hypothesis refers to the $E!$-case, see
[MSS1, Lemma 2.4.2], and we will prove the $P!$-Theorem, as it was announced in
section 2.4.2 of [MSS1]. To state this result we need some further
notation.
\bigskip
Throughout this paper $S \in Syl_p(H)$, and $Z$, $\w C$ and $Q$ are as
in the above hypothesis. Moreover
$$C := C_H(Z),\; B(T) := \O_1(Z(J(T))) \hbox{ ($T$ a $p$-subgroup)},\; X^0
:= \langle Q^X\rangle\hbox{ ($X$ a subgroup)}.$$
A subgroup $P \leq H$ is called {\bf minimal parabolic} (with respect to $p$),
if $P$ is not $p$-closed and every Sylow $p$-subgroup of $P$ is contained in a unique maximal subgroup of $P$.
\noindent Let $X$ and $M$ be subgroups of $H$, and let $T$ be a $p$-subgroup
of $H$:
$$\eqalign{&Loc_M(X):= \{ U \leq M \mid X \leq U\hbox{
and }
C_M(O_p(U)) \leq O_p(U)\},\cr
&{\cal M}_M(X) \hbox{ is the set of maximal elements of
}Loc_M(X).\cr
&{\cal L}_M(T) := \{U \in Loc_M(T) \mid T \in
Syl_p(U)\},\cr
&{\cal P}_M(T) := \{ P \in {\cal L}_M(T) \mid \hbox{ $P$ is
minimal parabolic}\},\cr}$$
According to (1.2) below every element $U \in Loc_M(X)$ contains a unique
maximal elementary abelian normal subgroup $Y_U$ satisfying $O_p(U/C_U(Y_U)) = 1$.
Let $P \in \P_H(S)$ and $B(P) := \langle B(S)^P\rangle$. Then $P$ is said to
be of {\bf type} $L_3$, if $p$ is odd, $O_p(P) =
Y_P \leq B(S)$, $B(P)/Y_P \cong SL_2(p^m)$, and $Y_P$ is a natural
$SL_2(p^m)$-module for $B(P)/Y_P$.
\bigskip
{\bf Hypothesis I.} The Q!-Hypothesis holds, and there exists $P \in \P_H(S)$ such that $P \not \leq \w C$ and $Y_M \leq Q$ for every $M\in
\M_H(P)$.
\bigskip
In this paper we prove:
\bigskip
{\bf P!-Theorem.} Assume Hypothesis I. Let $P^* := P^0O_p(P)$ and $Z_0 := \Omega_1(Z(S \cap
P^*))$. Then the following hold:
(a) $P^*/O_p(P) \cong SL_2(p^m)$ and $Y_P$ is a natural
$SL_2(p^m)$-module for
$P^*/O_p(P)$.
(b) $Z_0$ is normal in $\w C$; in particular $P \cap \w C$ is the
unique maximal subgroup of $P$ containing $S$.
(c) Then either $P$ is the unique
element of ${\cal P}_H(S)$ not in $\w C$, or every element of
$\P_H(S)\setminus \P_{\w C}(S)$ is of type $L_3$.
\bigskip
The proof of the P!-Theorem uses the Structure Theorem, which was proved in
[MSS2]. To state this result we need some further notation.
Let
$$\o{\cal L}_H(S) := \{ U \in {\cal L}_H(S) \mid C_H(Y_U) \leq U\}.$$
For
$U,\w U \in \o{\cal L}_H(S)$ define
$$U << \widetilde U \iff U = (U \cap \w U)C_U(Y_U).$$
Then (1.5) below shows that $<<$ is a partial order on $\o{\cal L}_H(S)$.
Let
$${\cal L}_H^*(S) = \{L\in \o{\cal L}_H(S)\mid \hbox{ $L$ is
maximal with respect to $<<$}\}.$$
Note that ${\cal M}_H(S) \subseteq \o{\cal L}_H(S)$ and ${\cal L}^*_H(S)
\subseteq {\cal M}_H(S)$, if $H$ has local characteristic $p$.
\bigskip
{\bf Structure-Theorem.} Assume the Q!-Hypothesis. Suppose that
there exists $M \in {\cal
L}_H^*(S)\setminus \{\w C\}$ such that $Y_M \leq Q$. Then for $M_0 :=
M^0C_S(Y_{M})$ and $\o M := M/C_M(Y_M)$ one of the following holds:
(a) $F^*(\o M) = \o M_0^\prime$, $\o M_0 \cong SL_n(p^m)$, $n \geq 2$,
$Sp_{2n}(p^m)$, $n\geq 2$, or $Sp_4(2)^\prime$ (and $p = 2$), and
$[Y_{M},M_0]$ is the corresponding natural module for $\o M_0$. Moreover,
either $C_{M_0}(Y_{M}) = O_p(M_0)$ or $p = 2$ and $M_0/O_p(M_0) \cong 3Sp_4(2)^\prime$.
(b) $P_1 := M_0S \in {\cal P}_H(S)$, $Y_{M} = Y_{P_1}$, and there exists a
a normal subgroup $P_1^*\leq P_1$ containing $C_{P_1}(Y_{P_1})$ but not $Q$ such
that
(i) $\o P_1^* = K_1 \times \cdots \times K_r$, $K_i \cong
SL_2(p^m)$, $Y_M = V_1 \times \cdots \times V_r$, where
$V_i := [Y_M,K_i]$ is a natural $K_i$-module,
(ii) $Q$ permutes the components $K_i$ of (i) transitively,
(iii) $O^p(P_1^*) = O^p(M_0)$, and $P_1^*C_M(Y_M)$ is normal in $M$,
(iv) $C_{P_1}(Y_{P_1}) = O_p(P_1)$, or $r > 1$, $K_i \cong SL_2(2)$ (and $p =
2$) and $C_{P_1}(Y_{P_1})/O_2(P_1)$ is a $3$-group.
\bigskip
We will refer to property (b) (ii) of the Structure Theorem as {\bf
Q-transitivity}. As a corollary of the Structure- and the P!-Theorem we get:
\bigskip
{\bf Corollary.} Assume Hypothesis I. Then for every $L \in Loc_H(P)$ the
following hold, where $\o L := L/C_L(Y_L)$ and $L_0 = L^0C_S(Y_L)$:
(a) $F^*(\o L) = \o L_0^\prime$, $\o L_0 \cong SL_n(p^m)$, $Sp_{2n}(p^m)$ or $Sp_4(2)^\prime$ (and $p = 2$),
and $[Y_L,L_0]$ is the corresponding natural module.
(b) Either $C_{L_0}(Y_{L}) = O_p(L_0)$, or $p = 2$, $L_0/O_p(L_0) \cong
3Sp_4(2)^\prime$ and $LC_H(Y_L) \in \L_H^*(S)$.
\bigskip
{\bf Acknowledgement.} We would like to thank A. Chermak for pointing out a
mistake in a previous version of Lemma (3.7) and also the referee for
his helpful suggestions.
\vfill\eject
{\bf 1. Elementary Properties.}
\bigskip
{\bf (1.1)} Let $X = S_k$ and $V$ be the non-central irreducible
constituent of the $GF(2)$-permutation module for $X$.
(a) Let $k = 2m+1$ and set $t_i := (2i-1,2i)$ and $d_i =
(2i-1,2i,k)$, $i = 1,...,m$. Then $X = \langle t_i,d_i \mid i =
1,...,m\rangle$.
(b) Let $t$ be a transposition of $X$ and $x \in X$ such that $[V,t,x]
= 0$. Then $k = 4$ or $t^x = t$.
(c) Let $k\ne 4$, $t_1,...,t_m$ be a maximal set of commuting
transpositions and $V_0 = C_V(t_1,...,t_m)$. Then $C_X(V_0) = \langle t_1,...,t_m\rangle$.
\bigskip
Proof. (a): It is well known that $\O := \{(k,k+1) \mid k = 1,...,2m\}$ is a
generating set for $X$. Thus the claim follows from the fact that
$$t_m^{d_m} = (2m,2m+1)\hbox{ and } t_i^{d_id_{i+1}} = (2i,2i+1),\ i = 1,...,m-1.$$
(b): Let $W = \langle v_1,...,v_k\rangle$ be the $GF(2)$-permutation
module for $X$ with basis $\{v_1,...,v_k\}$, where $v_ix := v_{ix}$
for $x \in X$. Set
$$W_0 := \langle \sum_{i=1}^kv_i\rangle,\;W_1 := \langle v_i+v_j\mid i,j\in
\{1,...,k\}\rangle \hbox{ and }\o W_1 := (W_1 + W_0)/W_0.$$
Then $V = \o W_1$. Let $t = (i,j)$ and $t^x = (r,s)$, so
$$\langle \o v_i + \o v_j \rangle = [\o W_1,t] = [\o W_1,t^x] =
\langle \o v_r +\o v_s\rangle.$$
It follows that $v_i+v_j + v_r + v_s \in W_0$, and either $\{i,j\} =
\{r,s\}$ and $t = t^x$, or $k = 4$.
(c): This is a direct consequence of (b).
\bigskip
{\bf (1.2)} Let $U$ be a finite group of characteristic p, $T \in Syl_p(U)$
and $T \leq \w U \leq U$. Then the following hold:
(a) There exists a unique maximal elementary
abelian normal $p$-subgroup $Y_U$ of $U$ such that $O_p(U/C_U(Y_U)) = 1$.
(b) $Y_{\w U} \leq Y_U$.
(c) $\O_1(Z(T)) \leq Y_U$.
(d) If $U = \w UC_U(Y_U)$ then $Y_U = Y_{\w U}$.
(e) If $O_p(U) = C_T(Y_U)$ then $Y_U = \O_1(Z(O_p(U)))$.
\bigskip
Proof. (a): Let $\Omega$ be the set of all elementary abelian normal
$p$-subgroups $X$ of $U$ satisfying $O_p(U/C_U(X)) = 1$. For the existence of
a unique maximal element in $\Omega$ it suffices to show that the product of
two elements of $\Omega$ is again in $\Omega$.
Let $A_1,A_2 \in \Omega$ and $A = A_1A_2$. Then $A \leq C_U(A_1)
\cap C_U(A_2)$ and thus $A$ is elementary abelian.
Let $C_U(A) \leq D \leq U$ such that $D/C_U(A) = O_p(U/C_U(A))$.
Then $DC_U(A_i)/C_U(A_i)$ is a $p$-group since $C_U(A)\leq C_U(A_i)$.
Hence $D \leq C_U(A_1) \cap C_U(A_2) = C_U(A)$.
(b): Set $V =
\langle (Y_{\w U})^{U}\rangle$. By the definition of $Y_{\w U}$,
$O_p(U) \leq C_U(Y_{\w U})$ and so $Y_{\w U} \leq \Omega_1(Z(O_p(U)))$ as
$U$ is of characteristic $p$. Hence also $V$ is in $\Omega_1(Z(O_p(U)))$;
i.e. $V$ is elementary abelian.
Let $C_{U}(V)\leq D \leq U$ such that $D/C_{U}(V) = O_p(U/C_{U}(V))$. Then
$$D = (D \cap T)C_{U}(V) \leq (D \cap T)C_{U}(Y_{\w U}).$$
Hence $O_p(\w U/C_{\w U}(Y_{\w U})) = 1$ gives $T \cap D \leq C_{U}(Y_{\w U})$
and thus $D =
C_{U}(V)$. Since $V$ is elementary abelian we conclude that $V \in \Omega$ and
thus $Y_{\w U} \leq V \leq Y_{U}$.
(c): This follows from (b) with $\w U := T$.
(d): According to (b) it suffices to show that $Y_U \leq Y_{\w U}$. But this
is clear since $U/C_U(Y_U) \cong \w U/C_{\w U}(Y_U)$ and thus $O_p(\w U/C_{\w
U}(Y_U)) = 1$.
(e): Let $Y := \O_1(Z(O_p(U)))$. Then $Y_U \leq Y$ by the definition of
$Y_U$. Let $C_U(Y) \leq D \leq U$ such that
$D/C_U(Y) = O_p(U/C_U(Y))$. Since $C_U(Y) \leq
C_U(Y_U)$ we get $DC_U(Y_U)/C_U(Y_U) \leq O_p(U/C_U(Y_U)) = 1$, and so $D \leq
C_U(Y_U)$. It follows that $D/O_p(U)$ is a $p^\prime$-group and
\break $O_p(U/C_U(Y)) = 1$, so $Y \leq Y_U$.
\bigskip
{\bf (1.3)} Let $U$ be a finite group of characteristic $p$, $T \in Syl_p(U)$
and $P \in \P_U(T)$. Then the following hold:
(a) $U = \langle {\cal P}_{U}(T)\rangle N_U(T)$.
(b) For every normal subgroup $N$ of $P$ either $O^p(P) \leq N$ or $T \cap N \leq O_p(P)$.
(c) For every normal subgroup $T_0$ of $T$ either $T_0 \leq O_p(P)$ or $O^p(P) =
[O^p(P),T_0]$.
(d) $Y_P = \Omega_1(Z(O_p(P)))$ or $[\O_1(Z(O_p(P))),O^p(P)] = 1$.
\bigskip
Proof. (a): We proceed by induction on $|U|$. Set $U_0 = \langle {\cal P}_{U}(T)\rangle N_U(T)$, and note that
$N_U(T)$ normalizes $\langle {\cal P}_{U}(T)\rangle$, so $U_0$ is a subgroup
of $U$. By induction all proper subgroups of $U$ containing $T$ are in $U_0$. If $U
\ne U_0$, then $U_0$ is the unique maximal subgroup of $U$ containing $T$. But
then $U \in {\cal P}_U(T)$ and thus $U = U_0$, a contradiction.
(b): By the Frattini argument $P = N_P(N\cap T)N$. As $T$ is in a unique
maximal subgroup of $P$ at least one of $NT$ and $N_P(N\cap T)$ is not a
proper subgroup of $P$. This gives (b).
(c): Let $P_0 = [O^p(P),T_0]$ and $P_1 = [O^p(P),T_0]T_0$. Then $P_1$ is normal in $P$. Hence, by (b)
either $O^p(P) \leq P_1$ and thus $P_0 \leq O^p(P_1) = O^p(P) \leq P_0$, or $T_0 \leq
O_p(P)$.
(d): If $C_T(Y_P) = O_p(P)$, then $Y_P = \Omega_1(Z(O_p(P))$ follows from
(1.2)(e). In the other case (c) gives $[\O_1(Z(O_p(P))),O^p(P)] = 1$.
\bigskip
{\bf Hypothesis and Notation.} For the rest of this section the Q!-Hypothesis
holds. We use the notation given in the introduction.
For $L_1,L_2 \in \L_H(S)$ we define
$$L_1 << L_2 \iff L_1 = (L_1 \cap L_2)C_{L_1}(Y_{L_1}).$$
\bigskip
{\bf (1.4)} Let $L,\w L \in \L_H(S)$ such that $L << \w L$. Then $L^0 \leq \w
L^0$.
\bigskip
Proof. Note that $C_L(Y_L) \leq \w C$. Hence $C_L(Y_L)$ normalizes $Q$ and
$Q^L = Q^{L\cap \w L}$.
\bigskip
{\bf (1.5)} $<<$ is a partial ordering on $\o \L_H(S)$.
\bigskip
Proof. By (1.2) $L_1 = (L_1 \cap L_2)C_{L_1}(Y_{L_1})$ implies that
$Y_{L_1} = Y_{L_1\cap L_2} \leq Y_{L_2}$. This gives the reflexivity and
anti-symmetry. Assume now that $L_1 << L_2$ and $L_2 << L_3$. Then
$$L_1 \cap L_2 \leq (L_2 \cap L_3)C_{L_2}(Y_{L_2})\hbox{ and } Y_{L_1} \leq
Y_{L_2}.$$
It follows that $C_{L_2}(Y_{L_2}) \leq C_H(Y_{L_1}) = C_{L_1}(Y_{L_1})$ and
thus $L_2 = (L_2 \cap L_3)C_{L_2}(Y_{L_1})$. Hence
$$L_1 \cap L_2 = (L_1 \cap L_2 \cap L_3)C_{L_2}(Y_{L_1}).$$
This shows $L_1 = (L_1 \cap L_3)C_{L_1}(Y_{L_1})$ and the transitivity of
$<<$.
\bigskip
{\bf (1.6)} Every $p$-subgroup of $H$ contains at most one conjugate of
$Q$; in particular $Q$ is the only conjugate in $\w C$.
\bigskip
Proof. Let $g \in H$ and $Q^g \leq S$. It suffices to show that $Q^g =
Q$. As $Z \leq C_{\w C^g}(Q^g) = Z(Q^g)$, Q-Uniqueness shows
that $S \leq \w C^g$, so $S \leq \w C \cap \w C^g$. Now Sylow's
Theorem shows that $\w C$ and $\w C^g$ are conjugate by an element of
$N_H(S)$. As by the definition of $\w C$, $N_H(S) \leq N_H(Z) \leq
\w C$ we conclude that $\w C = \w C^g$ and thus also $Q = Q^g$.
\bigskip
{\bf (1.7)} Let $P$ be a subgroup of $H$ with $Q \leq O_p(P)$.
Then $P \leq \w C$.
\bigskip
Proof. This is a direct consequence of (1.6).
\vfill\eject
{\bf 2. Pushing Up}
\bigskip
{\bf Hypothesis and Notation.} In this section the Q!-Hypothesis holds. In
addition, $P\leq H$ is a minimal parabolic subgroup of characteristic p and $T \in Syl_p(P)$.
We set $\o P := P/C_P(Y_P)$ and
$$\eqalign{B(T)&:= C_T(\O_1(Z(J(T))))\hbox{ and } Z_0 := \O_1(Z(J(T))),\cr
{\cal U}(P) &:= \{A \mid A \leq P, \o A\hbox{ an elem. abelian $p$-group, and
} |A/C_A(Y_P)| \geq |Y_P/C_{Y_P}(A)|\},\cr
U(P) &:= \langle A \mid A \in {\cal U}(P)\rangle \hbox{ and } B(P) := \langle
B(T)^P\rangle.\cr}$$
Moreover $\K(P)$ denotes the set of all $B(T)$-invariant subgroups $K \leq P$ satisfying:
(i) $\o K$ is normal in $\o{U(P)}$,
(ii) $L :=KB(T)$ is minimal parabolic of characteristic p and $O_p(P) \leq T \cap L \in Syl_p(L)$,
(iii) $\o K \cong SL_2(p^m)$ and $[Y_P,K]/C_{[Y_P,K]}(K)$ is a natural
$SL_2(p^m)$-module for $\o K$, or $p = 2$, $\o K \cong S_{2^n+1}$ and
$[Y_P,K]$ is a natural $S_{2^n+1}$-module for $\o K$.
Note that trivially $C_P(Y_P) \in {\cal U}(P)$ and so $C_P(Y_P) \leq
U(P)$. Then recall from (1.3) that either $U(P) = C_P(Y_P)$ or $P = U(P)T$ and
similarly $B(P) = B(T) \leq O_p(P)$ or $P = B(P)T$.
Let $K = SL_2(p^m)$ and $V$ be an irreducible $GF(p)K$-module. Set
$F:=End_K(V)$. By Schur's Lemma, $F$ is a finite field, so $V$ is an
$FK$-module. We say that $V$ is a {\bf natural $SL_2(p^m)$-module} for $K$
if $dim_F(V) = 2$.
\bigskip
{\bf (2.1)} Suppose that $\o{U(P)} \ne 1$ and $A \in {\cal U}(P)$. Then there exist subgroups
$U_1,...,U_r$ of $U(P)$ such that the following hold:
(a) $\o{U(P)} = \o U_1 \times \cdots \times \o U_r$, $\o U_i
\cong SL_2(p^m)$ or $S_{2^n+1}$ (and $p = 2$).
(b) Either $[Y_P,U_i]/C_{[Y_P,U_i]}(U(P))$ is a natural $SL_2(p^m)$-module for $\o
U_i$, or $[Y_P,U_i]$ is a natural $S_{2^n+1}$-module for $\o U_i$, $i = 1,...,r$.
(c) $Y_P = C_{Y_P}(U(P))\prod_{i=1}^r[Y_P,U_i]$ and $[Y_P,U_i,U_j] = 1$
for $i \ne j$.
(d) $\o T$ acts transitively on $\{\o U_1,...,\o U_r\}$.
(e) $[Y_P,A,A] = 1$ and $|\o A| = |Y_P/C_{Y_P}(A)|$. In particular
$|E| \leq |Y_P/C_{Y_P}(E)|$ for every elementary abelian $p$-group $E \leq \o P$.
(f) $\o A = \o{A \cap U_1} \times \cdots \times \o{A\cap U_r}$ and $A \cap
U_iC_P(Y_P) \in \U(P)$, $i = 1,...,r$.
(g) $\o{A \cap U_i} \in Syl_p(\o U_i)$ if $\o U_i \cong SL_2(p^m)$ and
$\o{A\cap U_i} \ne 1$.
(h) $\o{A \cap U_i}$ is generated by a set of commuting transpositions if
$\o U_i \cong S_{2^n+1}$.
\bigskip
Proof. See [Cher].
\bigskip
{\bf (2.2)} ${\cal A}(T) \subseteq \U(P)$ and $\o{J(T)} = \o{B(T)} \leq \o{U(P)}$.
\bigskip
Proof. Assume that $J(T) \leq C_P(Y_P)$. Then clearly $ {\cal A}(T) \subseteq \U(P)$
and $Y_P \leq Z_0$; in particular $B(T) \leq C_P(Y_P)$ and $1 = \o{J(T)} = \o{B(T)} \leq \o{U(P)}$.
Assume now that $J(T) \not\leq C_P(Y_P)$. Let $A \in {\cal A}(T)$ such that $\o A \ne 1$. The
maximality of $A$ gives $C_{Y_P}(A) = A \cap Y_P$. Hence
$$|C_A(Y_P)||Y_P||C_{Y_P}(A)|^{-1} = |C_A(Y_P)||Y_P||A\cap Y_P|^{-1} =
|C_A(Y_P)Y_P| \leq |A|$$
and $A \in \U(P)$; in particular $\o{J(T)} \leq \o{U(P)} \ne 1$.
We now use the notation given in (2.1). In addition we set $Y_i :=
[Y_P,U_i]$ and $\w Y_P := Y_P/C_{Y_P}(U(P))$. Then (2.1)(c) implies
$$(*)\quad \w Y_P = \w Y_1\times \cdots \times \w Y_r\hbox{ and }[Y_i,U_j] =
1\hbox{ for }i \ne j.$$
Assume first that $\o U_i \cong SL_2(p^m)$. Then (2.1)(f) and (g) show
that $\o{J(T)}\in Syl_p(\o{U(P)})$, and (2.1)(b), (e) and (f) that $[Y_i,J(T)]
\leq Y_i \cap Z_0$ and $|Y_i/Y_i\cap Z_0| = p^m$; in particular
$Y_i\cap Z_0 \not\leq C_{Y_i}(U_i)$. As $B(T)$ centralizes $Y_i \cap
Z_0$, we get from ($*$) that $\o{B(T)} \leq N_{\o P}(\o U_i)$.
Let $F := End_{\o U_i}(\w Y_i)$. Then the elements of $N_{\o P}(\o
U_i)$ induce field automorphisms on $F$ and semi-linear transformations on
$\w Y_i$. As $\w{Y_i \cap Z_0}$ is a $1$-dimensional $F$-subspace
centralized by $B(T)$, we conclude that the elements of $B(T)$ act
$F$-linear on $\w Y_i$, so $\o{B(T)} \leq (\o{J(T)}\cap \o U_i)C_{\o
P}(\o U_i)$ by (2.1)(g). It follows that $\o{B(T)} \leq \o{J(T)}$ since $C_{\o
P}(\o{U(P)}) \leq \o{U(P)}$, whence $\o{B(T)} = \o{J(T)}$.
Assume now that $\o U_i \cong S_{2^n+1}$. Recall that any two
transpositions of $S_m$ commute if they generate a $2$-group. Hence,
by (2.1)(h) $\o{J(T)}\cap \o U_i$ is generated by a maximal set of
commuting transpositions, and as above, by (2.1)(e) and (f) $[Y_i,J(T)]
\leq Y_i \cap Z_0$ and $\o{B(T)}\leq N_{\o P}(\o U_i)$. Now (1.1)(c) shows that $\o{B(T)} \leq (\o{J(T)}\cap \o U_i)C_{\o
P}(\o U_i)$ and, again as above, $\o{B(T)} = \o{J(T)}$.
\bigskip
{\bf (2.3)} Suppose that $\o{U(P)} \ne 1$. Then $\K(P) \ne \emptyset$, and for
every $K \in \K(P)$ and $L :=KB(T)$:
(a) $U(L)/C_L(Y_L) \ne 1$; i.e. $L$ satisfies the hypothesis of (2.1).
(b) $Y_L \leq Y_P$ and $[Y_L,K] = [Y_P,K]$.
(c) $B(T) \leq O_p(P)$ or $L = [K,B(T)](T\cap L)$.
(d) There exists $U_i$ as in (2.1) such that $\o K = \o U_i$.
\bigskip
Proof. We first show that $\K(P) \ne
\emptyset$. Let $U_1,...,U_r$ be as in (2.1) and fix $U \in
\{U_1,...,U_r\}$. By (2.1) and (2.2) $\o{J(T)} = \o{B(T)} \leq N_{\o P}(\o U)$ and $B(T) \leq
N_P(UC_P(Y_P))$. Among all subgroups $K_0\leq UC_P(Y_P)$, which are $B(T)$-invariant and satisfy
($*$) $\o K_0 = \o U$ and $O_p(P) \leq T\cap K_0B(T) \in Syl_p(K_0B(T))$,
\noindent we choose $K$ minimal and set $L = KB(T)$. According to (2.1)(a) there exists
$C_L(Y_P)(T \cap L) \leq L_0 \leq L$ such that $\o L_0$ is the unique maximal
subgroup of $\o L$ containing $\o{T\cap L}$. Hence, the minimality of $K$
implies that $L_0$ is the unique
maximal subgroup of $L$ containing $T \cap L$, so $L$ is minimal parabolic.
Moreover, $L$ is of characteristic $p$ since $O_p(P) \leq O_p(L)$. This shows
that $K \in \K(P)$.
Now let $K \in \K(P)$. Then (d) follows from (2.1)(a). Let $L = KB(T)$. From (1.3)(d) we get $\O_1(Z(O_p(L))) = Y_L \leq \O_1(Z(O_p(P))) = Y_P$, so
$Y_L = C_{Y_P}(O_p(L))$. Since $[\o K,\o{O_p(L)}] = 1$ the $P\times Q$-Lemma
gives $[Y_L,K] \ne 1$ and thus by (2.1)(b) $[Y_L,K] = [Y_P,K]$. This is (b).
From (1.3)(c) we get either $L = [K,B(T)](T\cap L)$ or $B(T) \leq O_p(L)$. In
the latter case $[\o K,\o{B(T)}] = 1$, and (2.1)(d) implies $B(T) \leq
C_T(Y_P)$. This shows (c) since $C_T(Y_P) = O_p(P)$ by (1.3)(c).
According to (2.1)(d) and (f) there exists $A\in \U(P)$
such that $\o A \ne 1$ and $\o A \leq \o{T\cap K}$. Since $C_T(Y_P) = O_p(P) \leq L$ and $\o A$ is a $p$-group we may assume
that $A \leq T\cap L$. Set $A_0 = C_A(Y_L)$. By (2.1)(e)
$$|\o A_0| \leq |Y_P/C_{Y_P}(A_0)| \leq |Y_P/Y_LC_{Y_P}(A)| =
|Y_P/C_{Y_P}(A)||Y_L/C_{Y_L}(A)|^{-1} = |\o A||Y_L/C_{Y_L}(A)|^{-1}$$
and $|Y_L/C_{Y_L}(A)| \leq |A/A_0|$. It follows that $U(L) \ne C_L(Y_L)$, and
(a) holds.
\bigskip
{\bf (2.4)} Suppose that $\o{U(P)} \ne 1$. Let $A \in \U(P)$ and $A_1 \leq P$
such that $[Y_P,A,A_1] = 1$. Then
$$[Y_P,A_1] \leq [Y_P,A][C_{Y_P}(A),A_1].$$
\bigskip
Proof. We apply (2.1) and choose the subgroups $U_1,...,U_r$ as in (2.1). Let
$V_i := [Y_P,U_i]$.
By (2.1)(c)
$$[Y_P,A_1] = [C_{Y_P}(A),A_1]\prod_{i=1}^r[V_i,A_1].$$
Hence, it suffices to show that
($*$) $[V_i,A_1] \leq [V_i,A][C_{Y_P}(A),A_1]$.
If $\o{A\cap U_i} = 1$, then by
(2.1)(c) and (f) $V_i \leq C_{Y_P}(A)$, and ($*$) is obvious. Hence, we may
assume that $\o{A\cap U_i} \ne 1$. Then $[V_i,A,A_1] = 1$ shows that
$A_1$ normalizes $U_i$ and $V_i$.
Assume first that $\o U_i \cong
SL_2(p^m)$. By (2.1)(g) $\o{A\cap U_i} \in Syl_p(\o U_i)$, so $[V_i,A,A_1] =
1$ implies $A_1 \leq
AC_P(V_i)$, and ($*$) follows.
Assume now that $\o U_i \cong S_{2^n+1}$. By (2.1)(h) $\o{A\cap U_i} = \langle
t_1,...,t_s\rangle$, $t_1,...,t_s$ commuting transpo\-si\-tions of $S_{2^n+1}$; in particular
$$C_{\o U_i}(\o A) = C_{\o
U_i}(\o{A\cap U_i}) = \langle t_1,...,t_s\rangle \times X,\ X\cong
S_{2^n+1-2s}\hbox{ and }[V_i,X] = [C_{V_i}(A),X].$$
Since $[V_i,t_j,A_1] = 1$ for $j = 1,...,s$ we get $\o A_1 \leq C_{\o U_i}(\o
A)C_{\o P}(V_i)$. Hence,
$$[V_i,A_1] \leq [V_i,A][C_{V_i}(A),A_1] \leq
[V_i,A][C_{Y_P}(A),A_1],$$
and again ($*$) follows.
\bigskip
{\bf (2.5)} Suppose that $T = S$, $\o{U(P)} \ne 1$ and $P \not \leq \w
C$. Let $K \in \K(P)$. Then the following hold:
(a) $Z(P) = Z(U(P)) = 1$.
(b) $Y_P = \times_{\{\o K \mid K\in \K(P)\}} [Y_P,\o K]$, and $[Y_P,\o K]$ is an natural $\o
K$-module.
(c) $Q$ acts transitively on $\{\o K\mid K \in \K(P)\}$.
(d) $\o K \cong SL_2(p^m)$ or $p = 2$ and $\o K =\o{U(P)} \cong S_5$.
(e) If $\o K \cong SL_2(p^m)$ and $A \leq P$ with $[Y_P,A,A] = 1$, then $[Y_P,K,A] =
[Y_P,K,a]$ for all $a \in A\setminus C_P([Y_P,K])$. Moreover, either
$|A/C_A([Y_P,K])| = 2\,(=p)$ or $\o A \leq \o KC_{\o A}(\o K)$.
\bigskip
Proof. (a): It suffices to show that $C_{Y_P}(U(P)) = 1$ since
$\O_1(Z(P)) \leq Y_P$. If $C_{Y_P}(U(P)) \ne 1$, then there exists $1\ne x
\in C_{Y_P}(U(P))\cap Z(Q)$, and by $Q$-Uniqueness
$U(P) \leq C_H(x) \leq \w C$. Since
also $S \leq \w C$ we get that $P = U(P)S \leq \w C$, a contradiction.
(b): This follows from (a) and (2.1)(c).
(c): By (b) and (2.1)(c),(d) together with (2.3)(d)
$$Y_P = [Y_P,K_1]\times \cdots \times [Y_P,K_r],$$
where $K_i \in \K(P)$ and $\O := \{\o K\mid K\in \K(P)\} = \{\o K_1,...,\o K_r\}$.
Assume that $Q$ is not transitive on $\O$. Then
there exist $1\ne x \in Z(Q) \cap Y_P$ and $K_i \in \{K_1,...,K_r\}$ such
that $[K_i,x] = 1$. Again by $Q$-Uniqueness $K_i \leq \w C$ and thus $P = \langle
K_i,S\rangle \leq \w C$, a contradiction.
(d): We use (2.1) and (2.3)(d). Assume that $\o K \cong S_{2^n+1}$, $n \geq 2$ (and $p = 2$). The
action of $U(P)$ on $Y_P$ shows that there exists $1\ne x \in Z(Q) \cap Y_P$
such that $\o{C_K(x)} \cong S_{2^n}$. On the other hand by $Q$-Uniqueness $C_H(x) \leq \w C$
and thus $[C_K(x),Q] \leq Q$. Since $S_{2^n}$ is not a 2-group we
get $\o K^Q = \o K$, and $\o P \cong S_{2^n+1}$ follows with (c). Moreover
$\o Q$ is a normal $2$-subgroup of $\o{C_K(x)}$.
If $n = 2$, then (d) follows. In the other cases $\o Q = 1$ and thus $Q \leq
C_S(Y_P) = O_2(P)$. But this contradicts (1.7).
(e): By (b) $V := [Y_P,K]$ is a natural $SL_2(p^m)$-module for $\o
K$. Assume first that $V^A = V$. Then again (b) implies that
$\o K^{\o A} = \o K$. Since $V$ is a faithful irreducible $\o
K$-module we conclude that $C_{\o A}(\o K) = C_{\o A}(V)$.
Let $V_0 := [V,A]$ and $F:= End_{\o K}(V)$. Recall that the elements
of $\o A$ induce semi-linear transformations on the $F$-vector space
$V$. Thus, if $V_0$ contains a $1$-dimensional $F$-subspace, then $\o A \leq \o KC_{\o P}(\o
K)$. In the other case no element of $\o A^\sharp$ induces an
$F$-linear transformation on $V$. As $\Gamma L(V)/GL(V)$ has cyclic
Sylow $p$-subgroups, we get in this case that $|A/C_A(V)| =
p$. Moreover, the quadratic action of $A$ on $V$ shows that the
elements of $A^\sharp$ induce field automorphisms of order $2$ in $F$, so $p = 2$.
Assume now that $V^A \ne V$. Then the quadratic action of $A$ gives
$$\langle V^A\rangle = V \times V^a\hbox{ for }a\in A\setminus N_A(V);$$
in particular $|A/N_A(K)| = p\ (= 2)$. Since
$$[V,N_A(K)] \leq C_V(A) \leq C_V(a) = 1$$
we get $N_A(K) \leq C_A(V)$ and $|A/C_A(V)| = p$. Now again (e) is obvious.
\bigskip
{\bf (2.6)} Suppose that neither $\O_1(Z(T))$ nor $B(T)$ is normal in
$P$. Then $\o{B(P)} = \o{U(P)} \ne 1$ and $\o{B(T)} = \o{J(T)}\ne 1$.
\bigskip
Proof. According to (1.3) $C_T(Y_P) = O_p(P)$ since $\O_1(Z(T))$ is not normal
in $P$. Hence $B(T) \not\leq C_P(Y_P)$ since also $B(T)$ is not normal in
$P$. It follows with (2.2) that $\o{B(T)} = \o{J(T)} \leq \o{U(P)} \ne 1$, and
(2.1) gives $\o{B(P)} = \o{U(P)}$.
\bigskip
{\bf (2.7)} Suppose that neither $\O_1(Z(T))$ nor $B(T)$ is normal in
$P$. Then $Z_0 \leq \O_1(Z(J(O_p(P))))$ and
$$[\O_1(Z(J(O_p(P)))),J(T)] \leq Z_0 \cap Y_P;$$
in particular $[\O_1(Z(J(O_p(P)))),O^2(P)] \leq Y_P$.
Moreover, if in addition $\o K \cong SL_2(p^m)$ for $K \in \K(P)$,
then $B(T) \in Syl_p(O^p(K)B(T))$.
\bigskip
Proof. By (2.6) $\o{U(P)} \ne 1$ and $\o{J(T)} = \o{B(T)} \ne 1$.
Let $A \in {\cal A}(T)$ such that $\o A \ne 1$ and $Z_1 :=
\O_1(Z(J(O_p(P))))$. Then by (2.1) $[Y_P,A] \leq C_{Y_P}(J(T)) \leq Z_0$, and
(2.1)(e) gives $Y_PC_A(Y_P) \in {\cal A}(T)$. This shows that
$$Y_PC_A(Y_P) \in {\cal A}(O_p(P))\subseteq {\cal A}(T).$$
Hence $Z_1 \leq Y_PC_A(Y_P)$ and $Z_0 \leq Z_1$. It follows that $[Z_1,A] \leq
Y_P\cap Z_0$ and thus $[Z_1,J(T)] \leq Y_P\cap Z_0$. Since $O^p(P) \leq \langle J(T)^P
\rangle$ by (1.3) we get $[Z_1,O^p(P)] \leq Y_P$.
Assume now that $\o K \cong SL_2(p^m)$, where $K \in \K(P)$. By (2.2) and
(2.1)(d), (g) we can choose $A$ such that $\o A \cap \o K \in Syl_p(\o K)$; in
particular
$$\langle \o A\cap \o K,(\o A\cap \o K)^{\o g}\rangle = \o K\hbox{
for some }g\in K.$$
Set $L = KB(T)$, $W = [Y_L,K]$,
$Z_0^* := Z_0 \cap Z_0^g$ and $L_0 = C_L(Z_0^*)$. Then $B(T) \leq L_0$ and $L
= L_0C_L(Y_P)$. Since $L$ is minimal parabolic and by (1.3) $C_T(Y_P) = O_p(P)$ we get
(1) $L = L_0O_p(P)$, and $L_0$ is normal in $L$.
By (2.3) $L$ satisfies the hypothesis of (2.1), and $W =
[Y_P,K]$. As $[Z_0,K] = [Z_0,K,K] \leq W$, $Z_0W$ is normal in $L$, and (2.1)(b),(g), applied to $L$,
gives $Z_0W = Z_0Z_0^g$, $C_W(T \cap L) = W\cap
Z_0$ and $|WZ_0/Z_0| = p^m$; in particular $Z_0^* \cap W = C_W(L)$. It follows that
$$|Z_0^*W/Z_0^*| = |W/W\cap Z_0^*| = p^{2m} \hbox{ and } |Z_0W/Z_0^*| = |Z_0Z_0^g/Z_0^*| \leq p^{2m}.$$
This shows that $Z_0^*W = Z_0W$ and $Z_0 = Z_0^*C_W(T\cap L)$; in particular
(2) $B(T) = C_{T\cap L}(Z_0) =C_{T\cap L}(Z_0^*)$.
By (1) and (2) $B(T) \in Syl_p(L_0)$ and $O^p(K) \leq O^p(L) \leq L_0$, so $B(T) \in Syl_p(O^p(K)B(T))$.
\bigskip
{\bf (2.8)} Suppose that neither $B(T)$ nor $\O_1(Z(T))$ is normal in $P$ and
$Z(P) = 1$. Then $O_p(P) \leq B(T)$.
\bigskip
Proof. By (2.7) $Z_0Y_P$ is normal in $P$. Hence, $R :=
[Z_0Y_P,O_p(P)]$ is a normal subgroup of $P$ in $Z_0$. But then by (2.6) and
(1.3) $O^p(P)$
centralizes $R$, and $Z(P) = 1$ implies $R = 1$. This gives $O_p(P) \leq B(T)$.
\bigskip
{\bf (2.9)} Suppose that neither $B(T)$ nor $\O_1(Z(T))$ is normal in $P$.
Then there exist subgroups $L_1,...,L_k \leq P$ such that for $i = 1,...,k$ and $\hat L_i = L_i/C_{L_i}(Y_{L_i})$:
(a) $L_i$ is minimal parabolic of characteristic $p$ and $O_p(P)B(T) \in
Syl_p(L_i)$.
(b) $\hat L_i \cong SL_2(p^m)$, and $Y_{L_i}/C_{Y_{L_i}}(L_i)$ is a
natural $SL_2(p^m)$-module for $\hat L_i$.
(c) $[Y_{L_i},O^p(L_i)] = [Y_P,O^p(L_i)]$.
(d) $L_1,...,L_k$ are conjugate under $T$, $\langle L_1,...,L_k\rangle T
= P$, and $\cap_{i=1}^kO_p(L_i) = O_p(P)$.
(e) $[Y_P,B(P)] \cap Z_0 = \prod_{i=1}^k[Y_{L_i},B(T)]$ and $[Y_{L_i},B(T),L_j] =
1$ for $i\ne j$.
\bigskip
Proof. By (2.6) $\o{U(P)} \ne 1$, and we are allowed to apply (2.1) and (2.3)
to $P$. Let $K \in \K(P)$, and set $L = KB(T)$ and $\hat L =
L/C_L(Y_L)$. Then (2.3) shows that $L$ satisfies (2.1) and $[Y_L,O^p(L)] =
[Y_P,O^p(L)]$.
Assume first that $\o K \cong SL_2(p^m)$. Then (2.1)(f),(g) gives
$$\o L = \o K \times C_{\o{B(T)}}(\o K)\hbox{ and } \o{B(T)} \cap \o K \in
Syl_p(\o K);$$
in particular $O_p(P)B(T) \in Syl_p(L)$ and $[O_p(L),O^p(L)] \leq O_p(P)$. Now (a) -- (d) follow for $k = 1$, and (e) is a consequence of
(2.1)(b).
Assume now that $\o K \cong S_{2^n+1}$ (and $p = 2$). Then $\o{K \cap
B(T)}$ is generated by a maximal set $\{\o t_1,...,\o t_{2^{n-1}}\}$ of
transpositions, where $t_1,...,t_{2^{n-1}} \in K$. For every $t_i$ there
exists $d_i\in K$ such that $\o d_i$ has order 3 and
$$\langle \o d_i,\o{K\cap B(T)}\rangle = \langle \o d_i,\o t_i\rangle \times
\langle \o t_j \mid i\ne j\rangle\hbox{ and } \langle \o d_i,\o t_i\rangle \cong SL_2(2).$$
Note that the subgroups $\langle \o d_i,\o t_i\rangle$, $i = 1,...,2^{n-1}$,
are conjugate under $\o{T\cap K}$ and that by (1.1)
$$\langle \o d_i,\o t_i\mid i=1,...,2^{n-1}\rangle = \o K.$$
Note further that by (2.1)(b)
$$(*)\quad [Y_P,K] \cap Z_0 = [Y_P,\langle
t_1,...,t_{2^{n-1}}\rangle]\hbox{ and }[Y_P,t_i,d_j] = 1\hbox{ for }i\ne j.$$
We now choose $L_1 \leq \langle d_1,B(T)\rangle$
minimal with respect to
$$O_2(P)B(T) \leq T_1:= T\cap L_1 \in Syl_2(L_1)\hbox{ and }\o L_1 = \langle \o
d_1,\o{B(T)}\rangle.$$
Then $L_1$ is a minimal parabolic subgroup of
characteristic $2$. Moreover $O_2(P)B(T) = T_1$ and $[O_2(L_1),O^2(L_2)] \leq O_2(P)$, and (a) follows for
$L_1$. Since $Y_{L_1} \leq \O_1(Z(O_2(L_1))) \leq \O_1(Z(O_2(P)))$ we get from (1.3)(d) that $Y_{L_1}
\leq Y_P$. It follows that $|[Y_{L_1},L_1]| = 4$, and (b) and (c) hold for
$L_1$ since $O^2(\o L_1) \cong C_3$.
Finally, for every $i \in \{1,...,2^{n-1}\}$ there exists a $T$-conjugate $L_i$
of $L_1$ with $d_i \in L_i$, and $\langle \o L_1,...,\o L_{2^{n-1}}\rangle
\o{B(T)}= \o L$. Since $L$ is minimal parabolic we get $\langle
L_1,...,L_{2^{n-1}}\rangle B(T)= L$. Similarly, since $P$ is minimal parabolic
(2.1)(d) and (2.3)(d) imply (d); and (e) follows from (d) and ($*$).
\bigskip
{\bf Notation.} Let
$$\P_0 := \P_H(S)\setminus (\P_{N_H(B(S))}(S) \cup \P_{\w C}(S))\hbox{ and
}\P_0^* := \{P^g \mid P \in \P_0,\,\, g\in N_H(B(S))\},$$
and let $\P$ be the set of all subgroups $X \leq H$ satisfying:
(i) $X$ is minimal parabolic of characteristic p and $B(S) \in Syl_p(X)$,
(ii) $\langle X,S\rangle = P$ for some $P \in \P_0$,
(iii) $X/C_X(Y_X) \cong SL_2(p^m)$ and $Y_X/C_{Y_X}(X)$ is a natural
$SL_2(p^m)$-module for $X/C_X(Y_X)$.
\smallskip
\noindent Let $\P^* := \{X^g \mid X \in \P,\,\,g\in N_H(B(S))\}$,
$G := \langle X \mid X\in \P^*\rangle$ and $L :=
GN_H(B(S))$.
\bigskip
{\bf Theorem 1.} One of the following holds:
(a) $L \in \L_H(S)$ and $\P_H(S) = \P_L(S) \cup \P_{\w C}(S)$.
(b) $\P_H(S) = \P_{N_H(B(S))}(S) \cup \P_{\w C}(S)$.
(c) $O_p(P) = Y_P$ and $Z(P) = 1$ for every $P \in \P^*$.
\bigskip
Proof. We may assume that neither (a) nor (b) holds. Then $\P_0 \ne \emptyset
\ne \P_0^*$. Let $P^* \in \P_0^*$ and set $Z_0 := \O_1(Z(B(S)))$ .
(1) $P^*$ satisfies the hypotheses of (2.1), (2.8) and (2.9), and, after a
suitable conjugation, also that of (2.5).
By the definition of $\P_0^*$ there is $P_0 \in \P_0$ and $g \in N_H(B(S))$
such that $P_0^g = P^*$. Hence, it suffices to show the claim for $P_0$.
From the choice of $P_0$ and the definition of $\w C$ follows that neither $B(S)$ nor $Z$ is normal in $P_0$. Hence,
$P_0$ satisfies the hypotheses of (2.6) and (2.9), and by (2.6) also those of
(2.1) and (2.5). Finally, by (2.5) $P_0$ satisfies the hypothesis of (2.8).
(2) $Z(P^*) = 1$ and $O_p(P^*) \leq B(S)$.
This follows from (1), (2.5) and (2.8).
Let $P_0 \in \P_0$. According to (2.9) and (2) there exists a subset
$$\O(P_0) := \{L_1,...,L_k\}\subseteq \P$$
such that the subgroups $L_1,...,L_k$ satisfy (2.9)(a) -- (e) (with respect to
$P_0$ and $S$). We fix this notation. From (2), (2.1)(c) and (2.9)(e) we get
(3) $Z_0 = \prod_{i= 1}^k[Y_{L_i},B(S)]$.
Next we prove:
(4) $L = \langle N_H(B(S)), P_0\mid P_0 \in \P_0\rangle$.
Let $\w L := \langle N_H(B(S)), P_0\mid P_0 \in \P_0\rangle$. By the definition
of $\P^*$ we have $L \leq \w L$. On the other hand, for $P_0 \in \P_0$ by
(2.9)(d) $ P_0 \leq GS$ and so also $\w L\leq L$.
(5) $O_p(G) = 1 = O_p(L)$.
From (4) we get
$$\P_H(S) = \P_L(S) \cup \P_{\w C}(S).$$
Hence, $O_p(L) = 1$ since (a) does not hold. As $G$ is normal in $L$ we also
have $O_p(G) = 1$.
\smallskip
In the following let
$$\Delta^*:= \cup_{P_0\in \P_0}\O(P_0).$$
We now apply the amalgam method to $G$ with respect to the subgroups in $\P^*$
and use the standard notation, see for example [DS] or [KS]. For the
convenience of the reader we repeat some of the notation:
$\Gamma = \{ Px\mid x\in G,\,\, P\in
\P^*\}$ is the set of vertices, and two vertices are adjacent, if they are
different and have non-empty intersection. $\P^*$ is a (maximal) set
of pairwise adjacent vertices (where the elements of $\P^*$ are understood as
cosets), and every pair of adjacent vertices is conjugate (under $G$) to a pair
of vertices from $\P^*$. For a vertex $\delta \in \Gamma$ the
stabilizer of $\delta$ in $G$ is denoted by $G_\delta$. Moreover
$$Q_\delta = O_p(G_\delta) \hbox{ and } Z_\delta = \langle \O_1(Z(X))\mid X\in
Syl_p(G_\delta)\rangle.$$
A critical pair $(\delta,\delta^\prime)$ of vertices satisfies
$Z_\delta \not\leq Q_\delta$ with the distance
$d(\delta,\delta^\prime)$ being minimal. This distance is denoted by $b$.
\bigskip
Note that by (2.9)(b) $Z_\delta =
Y_{G_\delta}$ for every $\delta \in \Gamma$.
Since by (1.3)(b) $C_{B(S)}(Y_P) = O_p(P)$ for every $P \in \P^*$ we get from (2.1)(g):
(6) $Z_\a Q_\b \in Syl_p(G_\b\cap G_{\b-1})$ and $Z_\b Q_\a \in Syl_p(G_\a
\cap G_{\a+1})$ for every critical pair $(\a,\b)$.
Let $(\a,\b)$ be a critical pair with $G_\a \in \P^*$. Then
there exists $T_1 \in Syl_p(G_\a)$ such that $G_\a = \langle
T_1,Z_\b\rangle$. Thus, possibly after conjugation in $G_\a$, we may assume
\centerline{($*$) $(\a,\b)$ is a critical pair such that $G_\a \in \P^*$ and $G_\a = \langle
B(S),Z_\b\rangle$.}
\noindent In the steps (7), (8) and (9) below $(\a,\b)$ is a critical pair satisfying
($*$). Further we set $R_\rho :=[Z_\rho,Q_\a]$ for every $\rho \in
\P^*$. Note that by (2.1)(e) and (g) $R_\rho \leq Z(B(S))$. We first show:
(7) Let $\rho \in \P^*$ and $b > 1$ or $Z_\rho \leq Q_{\b-1}$. Then
$R_\rho \leq Z(G_\a)$.
Assume first that $Z_\rho\leq Q_{\b-1}$. Then by (6) $Z_\rho \leq Z_\a Q_\b$ and
$$[Z_\rho,Z_\b] \leq [Z_\a,Z_\b] \leq Z_\a.$$
Hence, $\ Z_\rho Z_\a$ is normal in $\langle B(S),Z_\b\rangle = G_\a$; so also
$[Z_\rho,Q_\a] = R_\rho$ is normal in $G_\a$. Since $R_\rho \leq Z(B(S))$ we
get $R_\rho \leq Z(G_\a)$.
Assume now that $Z_\rho \not \leq Q_{\b-1}$. Then $(\rho,\b-1)$ is a
critical pair, and (6) gives $[Z_\rho,Z_{\b-1}] = [Z_\rho,Q_\a] = R_\rho$. If $b > 1$, then $R_\rho$ is centralized by $\langle B(S),Z_\b\rangle =
G_\a$.
Next we show:
(8) Let $\rho \in \P^*$. Suppose that $b > 1$ or $Z_\rho \leq
Q_{\b-1}$. Then either $Q_\a = Q_\rho$ or $Q_\a Q_\rho = B(S)$.
Let $T := Q_\a Q_\rho$. Assume that $Q_\a \leq Q_\rho$ but $Q_\a \ne
Q_\rho$. Then the action of $G_\a$ on $Z_\a$ shows that
$$Z_\rho \leq C_{Z_\a}(T) = Z_0,$$
so $B(S) \leq Q_\rho$, a contradiction. Hence, we may assume now that $Q_\rho
< T < B(S)$.
There exists $x \in G_\a$ such that $(\a+1)^x \in \P^*$ and $(\a,\b^x)$ is
a critical pair; so by (6) $B(S) = Z_\b^xQ_\a$. If $(\rho,\b^x)$ is not a critical pair, we get $Z_\b^x \leq
Q_\rho$ and thus $T = B(S)$, a contradiction. Hence, also $(\rho,\b^x)$ is a
critical pair, and by (6) $B(S) = Z_\b^xQ_\rho$ and $T = Q_\rho(Z_\b^x\cap T)$.
Let $t \in Z_\b^x$ such that $t \in T\setminus Q_\rho$. Then there exists $y
\in Z_\rho$ such that $[t,y] \ne 1$, and by (7) $[t,y] \in Z(G_\a)$. On the
other hand, according to (6) (applied to $(\rho,\b^x)$ and $(\a,\b^x)$) there
exists $y^\prime \in Z_\a$ such that $[t,y] = [t,y^\prime]$. The action of
$Z_\b^x$ on $Z_\a$ gives $[t,y^\prime] \not\in Z(G_\a)$, a contradiction.
\smallskip
We now let $N_H(B(S))$ act on $\Gamma$ in the following way:
Let $g \in N_H(B(S))$ and $\delta \in \Gamma$, so $\delta = Py$ for some $P\in \P^*$ and
$y \in G$. Then
$$g:\,\delta \mapsto \delta^g:= P^gy^g.$$
(9) For every $P\in \Delta^*$ there exists a critical pair $(\delta,\delta^\prime)$
satisfying ($*$) such that $G_\delta = P$.
There exists $P_0 \in \P_0$ such that $P \in \O(P_0) \subseteq
\Delta^*$. Hence, there exist $\d_1,...,\d_k\in \P^*$ such that
$$\O(P_0) = \{G_{\d_1},...,G_{\d_k}\}.$$
Note that by (2.9)(d) the subgroups in $\O(P_0)$ are conjugate under
$S$. We will show that there exists a critical pair
$(\d_i,\d_i^\prime)$ for some $i \in\{1,...,k\}$. The ($*$)-property then can be achieved by a
suitable conjugation in $G_{\d_i}$ and the claim for the other $\d_j$ by the
action of $S$.
Hence, we may assume that $Z_{\d_i}\leq
Q_{\b-1}$ for all $i=1,...,k$. If there exists $j \in \{1,...,k\}$ such that
$Q_{\d_j} = Q_\a$, then $(\d_j,\b^x)$ is a critical pair, where $x \in G_\a$
such that $B(S)^{x^{-1}} \leq G_{\a+1}$. Thus, we may also assume that $Q_\a
\ne Q_{\d_i}$ for all $i = 1,...,k$. Now (7) and (8) give
$$R_{\d_i} = [Z_{\d_i},B(S)] \leq Z(G_\a),\,\, i=1,...,k,$$
and by (3)
$$Z_0 = \prod_{i=1}^k[Z_{\d_i},B(S)] = \prod_{i=1}^k R_{\d_i} \leq Z(G_\a),$$
a contradiction.
\smallskip
(10) There exists $\rho \in \P^*$ and $P \in \Delta^*$ such that $Q_\rho^q\ne O_p(P)$ for all $q \in Q$.
Assume that (10) does not hold. Let $P_0 \in \P_0$ and $\O(P_0) = \{L_1,...,L_k\}$. By (2.9)(d)
$$\cap_{i=1}^k O_p(L_i)= O_p(P_0).$$
Now let $\rho \in \P^*$ and $L_i \in \O(P_0)$. Then there exists $q \in Q$ such that
$Q_\rho^q = O_p(L_i)$; in particular $O_p(P_0) \leq Q_\rho^q$. Since
$O_p(P_0)$ is $Q$-invariant we get
$$O_p(P_0)\leq Q_\rho\hbox{ for all }\rho \in\P^*\hbox{ and all }P_0\in
\P_0.$$
Note that $\P^*$ is invariant under $N_H(B(S))$. Hence
also
$$O_p(P^*) \leq Q_\rho\hbox{ for all }\rho \in \P^*\hbox{ and all
}P^*\in \P_0^*.$$
It follows that
$$O_p(P^*) \leq \cap_{L_i \in \O(P_0)} O_p(L_i)= O_p(P_0) \hbox{ for all }P_0 \in
\P_0\hbox{ and all }P^* \in \P_0^*.$$
This shows that $O_p(P^*) = O_p(P_0)$ for all $P^* \in \P_0^*$ and all $P_0\in
\P_0$, and by (4)
$O_p(P_0)$ is normal in $L$, a contradiction to (5).
\smallskip
By (10) there exists $\rho \in \P^*$ and $P \in \Delta^*$ such that
$Q_{\rho}^q \ne O_p(P)$ for all $q \in Q$, and by (9) there exists a critical
pair $(\a,\b)$ satisfying ($*$) such that $G_\a = P$. We fix this notation
with the additional property
that $P_0:= \langle P,S\rangle \in \P_0$ and $P\in \O(P_0)$.
(11) There exists $q \in Q$ such that $(\rho^q,\a)$ is a critical pair; in
particular $b = 1$.
Suppose that $b > 1$ or $Z_{\rho^q} \leq Q_{\b-1}$ for all $q \in Q$. Then (8) shows that $B(S) = Q_\a
Q_\rho^q$ for all $q \in Q$. Hence $[Z_\rho^q,Q_\a] = [Z_\rho^q,B(S)]$ and by
(7)
$$R := \prod_{q\in Q}[Z_\rho^q,B(S)] \leq Z(G_\a);$$
in particular $R$ is a $Q$-invariant and non-trivial subgroup of $Z(G_\a)$.
Hence, $Q$-Uniqueness gives $G_\a=P \leq \w C$. But then also $P_0 \leq \w C$, which
contradicts $P_0 \in \P_0$. This shows that $b = 1$
and there exists $q \in Q$ such that $(\rho^q,\a)$ is a critical pair.
\smallskip
(12) Let $\gamma \in \P^*$ such that $G_{\gamma}\leq P_0$. Then $Y_{G_{\gamma}}\leq Y_{P_0}$; in particular $Z_\a
\leq Y_{P_0}$ and no $Q$-conjugate of $G_{\rho}$ is contained in $P_0$.
Since by (2) $O_p(P_0)\leq B(S)$, we have $\O_1(Z(Q_{\gamma}))\leq
\O_1(Z(O_p(P_0)))$. Hence (1.3)(d) and (2) yield $Y_{G_{\gamma}}\leq Y_{P_0}$.
This gives, together with (11), that there exists $q\in Q$ such that
$G_{\rho}^q$ is not contained in $P_0$, and, since $Q\leq S\leq P_0$,
no $Q$-conjugate of $G_{\rho}$ is contained in $P_0$.
\smallskip
Let $\mu := \rho^q$ be as in (11). Then (6) and $b = 1$ give
$$B(S) = Z_\mu Z_\a (Q_\a \cap Q_\mu);$$
in particular
$$\Phi(Q_\a) = \Phi(Q_\a\cap Q_\mu) =\Phi(Q_\mu).$$
This gives $[Q_\a,Z_\mu ]=[Z_\a,Z_\mu ]\leq Z_\a$. Hence (2),
(1.3)(b) and (12) yield
$$[O_p(P_0),O^p(G_\a)]\leq [Q_\a,O^p(G_\a)]\leq [Q_\a,\langle Z_\mu
^{G_\a}\rangle]\leq Z_\a \leq Y_{P_0}.$$
From $G_\a \in \O(P_0)$ and (2.9)(d) we get $[O_p(P_0),O^p(P_0)] \leq
Y_{P_0}$. Now $Z(P_0) = 1$ yields $Y_{P_0} =O_p(P_0)$, and
(2.1) and (2.9) applied to $P_0$ give $B(S)=Y_{P_0}\langle Z_\mu ^S\rangle$.
From (2.1) and (3) it follows that $\Phi(B(S)) = Z_0$; in particular
$$\Phi(Q_\a)= \Phi(Q_\mu) \leq Z(G_\a)\cap Z(G_\mu).$$
Assume that $\O(P_0) = \{P\}$. Then $Z(G_\a) = 1$ and $Z_\a =Q_\a$ is a natural
$G_\a/Q_\a$-module. In particular
$$B(S) = Z_\a Z_\mu\hbox{ and } Z_\a\cap Z_\mu = Z_0.$$
Thus, also $Q_\mu = Z_\mu$, and the action of $Z_\a$ on $Z_\mu$ also shows that
$Z(G_\mu)= 1$.
Let $\lambda \in \P^*$. If $Q_\lambda^q \ne Q_\a$ for
all $q \in Q$, then, as for $\rho$ and $\mu$, $Q_\lambda = Z_\lambda$ and $Z(G_\lambda) = 1$. If
$Q_\lambda^q = Q_\a$ for some $q \in Q$, then $Z_\a = Z_\lambda^q = Q_\lambda^q$, and the
action of $Z_\mu$ shows that $Z(G_\lambda^q) = Z(G_\lambda) = 1$. Hence, (c)
holds in the case $\O(P_0) = \{P\}$.
Assume now that $\O(P_0) \ne \{P\}$ and choose $L_i \in \O(P_0) \setminus
\{P\}$; i.e. $L_i = G_\nu$ for some $\a\ne \nu\in \P^*$.
Since $[Z_\mu,Q_\a] = [Y_P,B(S)]$ and by (2.9)(e) $[Y_{L_i},B(S)] \ne
[Y_P,B(S)]$ we get from $b = 1$ and (6) that $Z_\nu \leq Q_\mu \cap Q_\a$. Hence,
$$R_0 := [Z_\nu,B(S)] = [Z_\nu,Q_\a\cap Q_\mu]\leq Z(G_\a)\cap
Z(G_\mu).$$
Let $U = N_H(R_0)$. Then $U$ is of characteristic $p$ and $\langle
G_\a,G_\mu\rangle \leq C_H(R_0)$. Thus
$$O_p(U) \cap Q_\mu = O_p(U) \cap B(S) = O_p(U) \cap Q_\a,$$
so $O_p(U) \cap B(S)$ is normal in $G_\a$ and $[O_p(U)\cap B(S),Z_\mu] =
1$. Note that $[O_p(U),Z_\mu] \leq O_p(U) \cap B(S)$. Since $O^p(G_\a) \leq \langle
Z_\mu^{G_\a}\rangle$ we get that $[O_p(U),O^p(G_\a),O^p(G_\a)] = 1$. This
contradicts the fact that $U$ is of characteristic $p$.
\bigskip
{\bf Corollary 1.} Suppose that the cases (a) and (b) of Theorem 1 do not hold. Let $P \in
\P_H(S)\setminus \P_{\w C}(S)$
such that $\O_1(Z(B(S)))$ is not normal in $P$. Then $\o{B(P)} \cong
SL_2(p^m)$, and $O_p(P)$ is a natural $SL_2(p^m)$-module for
$\o{B(P)}$. Moreover, either $N_H(B(S)) \leq N_H(O_p(P))$, or $P$ is of type
$L_3$.
\bigskip
Proof. By the choice of $P$ and the definition of $\w C$, $P$ satisfies the hypothesis of (2.6). Hence
$\o{U(P)}\ne 1$ and by (2.5)(a) $Z(P)=1$. Thus (2.8) gives
$O_p(P)\leq B(S)$. Applying (2.9) and Theorem 1 (c) we get that $\o{B(P)} \cong
SL_2(p^m)$ and that $O_p(P)=Y_P$ is a natural $SL_2(p^m)$-module
for $\o{B(P)}$. Hence either $P$ is of type $L_3$ or $p=2$.
Assume that $p = 2$. Suppose that $N_H(B(S))$ is not contained in $N_H(Y_P)$
and pick $x\in N_H(B(S)) \setminus N_H(Y_P)$. Then $B(S)=Y_PY_P^x$
and ${\cal A}(S)=\{Y_P, Y_P^x\}$. Since $N_H(B(S))$ acts on ${\cal A}(S)$
we get $O^2(N_H(B(S))) \leq N_H(Y_P)$ and thus also $N_H(B(S)) \leq
N_H(Y_P)$, a contradiction.
\vfill\eject
{\bf 3. P-Uniqueness}
\bigskip
Throughout this section we assume Hypothesis I. In particular, the Structure
Theorem applies to all $M \in {\cal L}_H^*(S)$ with $P \leq M$. In
addition, among all $P$ satisfying
Hypothesis I we choose $P$ maximal (with respect to inclusion).
\bigskip
{\bf Local P!-Theorem.} Let $P^* = U(P)$ and $P \leq M \in \L_H^*(S)$. Then
one of the following holds:
(a) Case (a) of the Structure Theorem holds for $M$, $P^* = P\cap M_0$ and
(i) $P^*/O_p(P) \cong SL_2(p^m)$ and $Y_P$ is a natural $SL_2(p^m)$-module,
(ii) ${\cal P}_{M}(S) = \{P\}\cup {\cal P}_{M \cap \w C}(S)$,
(iii) $M \cap \w C = N_{M}(\Omega_1(Z(S \cap P^*)))$.
(b) Case (b) of the Structure Theorem holds for $M$, and
(i) ${\cal P}_{M}(S) = {\cal P}_P(S) \cup {\cal P}_{M \cap \w C}(S)$, in
particular $P = O^p(M_0)S$,
(ii) $M\cap \w C \leq N_{M}(\Omega_1(Z(S \cap P^*)))$,
(iii) ${\cal M}_H(P) =\{M\}$.
\bigskip
Proof. We discuss the two cases of the Structure Theorem separately.
Assume
first that case (a) of the Structure Theorem holds for $M$.
Let $\o M := M/C_M(Y_M)$, $S_0 := S \cap M_0$ and $Z_0 := \O_1(Z(S_0))$.
The $p$-local structure of $M_0/O_p(M_0)$ shows:
(+) There exists a unique $U \in {\cal} {\cal P}_{M_0}(S_0)$ such that
$[Z_0,U] \ne 1$; in particular ${\cal P}_{M_0}(S_0) = \{U\} \cup
{\cal P}_{M_0\cap \w C}(S_0)$.
(++) $U/O_p(U) \cong SL_2(p^m)$, and $Y : = C_{Y_{M}}(O_p(U))$ is a
natural $SL_2(p^m)$-module for $U/O_p(U)$.
Since $Q\leq S_0$ from (1.7) it follows $N_H(S_0) \leq \w C$, hence
(+) gives $N_H(S_0) \leq N_H(U)$, in particular $S$ normalizes $U$.
Let $P_1 \in {\cal P}_M(S)$ such that $P_1 \not\leq \w C$. By (1.7)
$Q \not\leq O_p(P_1)$, and so by (1.3)(b) $P_1 = (P_1)^0S$ and $(P_1)^0S_0\leq
M_0$. Since $O_p(M)\leq O_p((P_1)^0S_0)$, $(P_1)^0S_0$ has
characteristic $p$, whence (1.3)(a) and the uniqueness of $U$ give
$$(P_1)^0S_0 = \langle U,(P_1)^0S_0\cap \w C\rangle.$$
Since $P_1$ is a minimal parabolic subgroup not contained in $\w C$ we get
that $P_1 = US$; in particular $P = US$, and (a)(ii) follows.
From $O_p(U)\leq O_p(P)$ and (1.2)(b) we get $Y_P\leq Y_M$, thus $Y_P\leq
Y$ and (++) yields $Y_P=Y$. Now (2.1) gives $P^* = UO_p(P)\leq M_0$,
whence (a)(i) and $P\cap M_0=P^*$ follow.
Note that $M_0C_M(Y_M)$ is a normal subgroup of $M$. It follows that
$$M \cap \w C = C_M(Y_M)(M_0\cap \w C)N_{M\cap \w C}(S_0)\leq (M_0 \cap \w
C)N_M(Z_0),$$
so (+) and (1.3)(a) yield $M \cap \w C \leq N_M(Z_0)$. On the other
hand by
$Q$-Uniqueness $C_M(Z_0) \leq \w C$,
so by (1.6) $Q$ is the unique conjugate of $Q$ in $C_M(Z_0)$. Hence
$N_M(Z_0) \leq N_M(Q) = M \cap \w C$.
By the Structure Theorem $C_S(Y_M)=O_p(M_0)\in Syl_p(C_{M_0}(Y_M))$,
whence by (1.2)(e) $Y_{M_0}=\O _1 (Z(C_S(Y_M)))\leq Y_M$. This gives
$Z_0\leq Y_M$ and thus $Z_0=C_{Y_M}(S_0)$. From (++) it follows
that $Z_0\leq Y=Y\cap Z(O_p(P))$, therefore $S\cap P^*=O_p(P)S_0$
yields $Z_0=\O _1(Z(S\cap P^*))$. This shows (a)(iii).
Assume now that case (b) of the Structure Theorem holds. Let $P_1$ and
$P_1^*$ be
as given there and set $S_0 := P_1^*\cap S$ and $Z_0 := \O_1(Z(S_0))$. Then
$P_1 = M^0S$ and by (2.1) $P_1^* = U(P_1)$; moreover, by (1.3)(c) and (1.7)
${\cal P}_M(S) = {\cal P}_{P_1}(S) \cup {\cal P}_{M\cap\w C}(S)$. The
maximality of $P$ gives $P = P_1$ and $P^* = P_1^*$, and (b)(i)holds.
Since
$P^*C_M(Y_M)$ is normal in $M$ we get as above
$$M \cap \w C = C_M(Y_M)(P^* \cap \w C)N_{M\cap \w C}(S_0).$$
As $P$
is a minimal parabolic subgroup, the structure of $P^*$ and its action on
$Y_P$ show that $N_P(Z_0)$ is the unique maximal subgroup containing $S$. It
follows that $P^* \cap \w C \leq N_P(Z_0)$ and thus $M \cap \w C \leq
N_M(Z_0)$.
This is (b)(ii).
Let $P \leq L \in {\cal M}_H(S)$ and $L << \w L \in {\cal L}_H^*(S)$. Then $L
= (L\cap \w L)C_L(Y_L)$ and thus
$$P^0 \leq L^0 = (L\cap \w L)^0 \leq \w L^0.$$
It follows that $P = P^0S \leq \w L$, and we are allowed to apply the
Structure Theorem to $\w L$.
If case (a) of the Structure Theorem holds for
$\w L$, then by case (a) of the Local P! Theorem
$P \cap \w L_0 =P^*=U(P).$
But then $Q \leq P^*$, a contradiction.
If case (b) of the Structure Theorem holds for $\w L$, then the maximality of
$P$ gives $Y_P = Y_{\w L}$ and thus $Y_{\w L}=Y_M$; in particular $M = \w L$. This shows (b)(iii).
\bigskip
{\bf Notation.} We fix $M$, $P$ and $P^*$ as in the Local
P!-Theorem. (Observe that in case (b) of the Local $P!$-Theorem the definition of $P^*$ differs from that
given in the $P!$-Theorem. But it will be shown in section 4 that this
case does not occur.) Furthermore, we set $\o P := P/C_P(Y_P)$, $S_0 :=S \cap P^*$ and $Z_0 := \O_1(Z(S_0))$.
Recall that $P$ satisfies the hypotheses of (2.1) -- (2.5) and if $B(S) \not
\leq O_p(P)$ also those of (2.6) -- (2.9). Later in the course of the amalgam
method we will apply these Lemmata not only to $P$ but also to conjugates of $P$.
\bigskip
{\bf (3.1)} $P$ admits the decompositions
$$\eqalign{({\cal D}_1)&\quad \o P^* = K_1 \times \cdots
\times K_r,
\quad K_i \cong SL_2(p^m),\hbox{ and}\cr
({\cal D}_2)&\quad Y_P = V_1 \times \cdots \times V_r,\quad
V_i\hbox{ a natural $SL_2(p^m)$-module for $K_i$}.\cr}$$
Moreover, $[Y_P,Q\cap S_0] = Z_0$ and either $S_0 = B(S)$ or $B(S) \leq O_p(P)$.
\bigskip
Proof. The decompositions ${\cal D}_1$ and ${\cal D}_2$ are from the Local
P!-Theorem. Assume that $B(S) \not\leq O_p(P)$. Since $P^* = U(P)$
(2.6), (2.1) and (2.8) show that $S_0 = B(S)$.
\bigskip
{\bf Remark.} The next result, Theorem $2$, establishes part (a) and (c) of the
$P!$-Theorem if case (a) of the Local $P!$-Theorem holds. We then
embark on the proof of the main result of this section, Theorem $3$,
where we show that $Z_0$ is normal in $\w C$. This establishes part
(b) of the $P!$-Theorem in all cases. It then remains to treat case
(b) of the Local $P!$-Theorem. This is done in the next section, where
the $F!$-Theorem eliminates this case.
\bigskip
{\bf Theorem 2.} Assume Hypothesis I. Then either ${\cal P}_H(S) = {\cal
P}_P(S) \cup {\cal P}_{\w C}(S)$, or the following hold:
(a) $Z_0$ is normal in $\w C$.
(b) $Q = B(S) = S_0$.
(c) $\w P$ is of type $L_3$ for every $\w P \in \P_H(S) \setminus
\P_{\w C}(S)$.
\bigskip
Proof. Assume first that $P$ is of type
$L_3$. Then by (2.2) $Y_P \in {\cal A}(S)$, $B(S) = S_0$, and for every $A \in
{\cal A}(S)$ either
$$S_0 = AY_P\hbox{ or }A = Y_P.$$
Moreover, $Y_P \leq Q$ by (1.2)(b) and Hypothesis I. It follows that also
$J(S) = S_0 \leq Q$ since $Y_P$ is not normal in $\w C$. But then $J(S) =
J(Q)$ and $Z_0 = \Omega_1(Z(S_0)) = \O_1(Z(J(S)))$ is normal in $\w
C$. On the other hand, $N_P(S_0)$ is transitive on $Z_0$ and by (1.7)
contained in $\w C$, so $Z_0 \leq
Z(Q)$ and $Q \leq S_0$. We conclude that $Q = S_0$; in particular
$N_H(B(S)) = \w C$.
Let $\w P \in \P_H(S)\setminus \P_{\w C}(S)$. Then $\w P \not\leq
N_H(\O_1(Z(B(S))))$, and Corollary 1 shows that also $\w P$ is of type
$L_3$. Hence, Theorem $2$ holds if $P$ is of type $L_3$.
We may assume now:
(1) $P$ is not of type $L_3$ and ${\cal P}_H(S) \ne {\cal
P}_P(S) \cup {\cal P}_{\w C}(S)$.
By (1) there exists $\w P \in \P_H(S)$ such that
(2) $\w P \not \leq P$ and $\w P \not\leq \w C$.
Assume that $O_p(\langle P,\w P\rangle) \ne 1$. Then there exists $L \in
\o \L_H(S)$ such that $\langle P,\w P \rangle : = R \leq L$. Since $P = P^0S$
and $\w P = \w P^0S$, we also get $R \leq L^0S$. Now (1.4) shows that there
exists $\w M \in \L_H^*(S)$ such that $R \leq \w M$. The Local
P!-Theorem applied to $\w M$, together with the maximal choice of $P$, gives $\w P \leq P$, which contradicts (2). We have
shown:
(3) $O_p(\langle P,\w P\rangle) = 1$.
We now apply Theorem 1. Then (3) shows that the cases (a) and (b) of Theorem 1
do not hold. Assume that $B(S)$ is not normal in $P$, so by (1.3) also
$\O_1(Z(B(S)))$ is not normal in $P$.
Hence by Corollary 1 $\,O_p(P) = Y_P$ and $P^*/Y_P \cong SL_2(p^m)$,
and Corollary 1 and (1) show that $N_H(B(S)) \leq N_H(Y_P)$. On the
other hand, as above, $Y_P \leq Q$ implies $B(S) = S_0 = Q$ since
$Y_P$ is not normal in $\w C$. Hence $\w C = N_H(B(S))\leq N_H(Y_P)$,
and $Y_P$ is normal in $\w C$, a contradiction. We have shown:
(4) $P \leq N_H(B(S))$.
By (3) and (4) $\O_1(Z(B(S)))$ is not normal in $\w P$. Hence again (3) and Corollary 1 show
that $\w P$ is of type $L_3$. In particular $p \ne 2$, and there exists an
involution $t \in N_{\w P}(S)$ such that $[S,t] = Y_{\w P}$. Since $Y_{\w P}
\leq B(S)$ and $Y_{\w P} = O_p(\w P)$ we get $Y_P \leq \O_1(Z(B(S)) \leq Y_{\w P}$. Hence $Y_P =
[Y_P,t]$, and $t$ inverts $Y_P$. This shows that $[t,P] \leq C_H(Y_P) \cap
N_H(O_p(P)) =:X$, and $P^0$ normalizes $\langle t \rangle X$. Since
$$[\langle t \rangle X,Q] \leq Q \cap \langle t \rangle X \leq C_S(Y_P) =
O_p(P)$$
we conclude that $[t,P^0] \leq O_p(P)$ and thus also $[t,P] \leq
O_p(P)$. Hence, $P$ normalizes $\langle t \rangle O_p(P)$ and thus also
$O^p(\langle t\rangle O_p(P)) = \langle t \rangle Y_{\w P}$. It follows that
$P$ normalizes $Y_{\w P}$, which contradicts (3). This completes the
proof of Theorem 2.
\bigskip
{\bf (3.2)} Suppose that $O^p(\o P) \leq \langle \o
x,\o A\rangle$, where $x$ is a $p$-element in $P$ and $A$ a normal subgroup of $S$ in
$Q$. Then $O^p(P) \leq \langle x,A\rangle$.
\bigskip
Proof. Let $P_0 = \langle x,A\rangle$ and $P_1 = O^p(P)$. Note that $P_1
\not\leq C_P(Y_P)$ by our choice of $P$, so $P_1 \leq \langle A^P\rangle$ by
(1.3)(b). Note further that
$[C_P(Y_P),A] \leq O_p(P)$ since $A \leq Q$ and that $P_1 \leq
P_0C_P(Y_P)$. It follows that
$$P_1 \leq \langle A^P \rangle = \langle A^{P_1}\rangle \leq \langle
A^{P_0}\rangle O_p(P).$$
Since $\langle A^{P_0} \rangle$ is normal in $P_0O_p(P)$ we get that
$$P_1 = O^p(\langle A^P\rangle) = O^p(P_0O_p(P)) = O^p(P_0).$$
\bigskip
{\bf Hypothesis II.} Assume Hypothesis I and ${\cal P}_H(S) =
{\cal P}_P(S) \cup {\cal P}_{\w
C}(S)$. Further assume that there exists $\w P \in {\cal P}_{\w
C}(S)$ such that
$(P,\w P)$ is an amalgam and $N_{\w P}(Z_0)$ is a maximal
subgroup of $\w P$.
\bigskip
Our goal, which we will achieve in (3.9), is to prove that no group $H$
satisfies Hypothesis II.
\bigskip
{\bf (3.3)} Assume Hypothesis II. Let $x \in \w P$ and
$O_p(P) \leq N_{\w P}(Z_0^x)$.
Then $x \in N_{\w P}(Z_0)$.
\bigskip
Proof. Assume first that $J(S) \leq O_p(\w P)$. Then $J(S)$ is normal in
$\w P$ and thus not normal in $P$ since $(P,\w P)$ is an amalgam. Hence, by
(3.1) $S_0 = B(S)$ and $Z_0 = \O_1(Z(J(S)))$. But then $Z_0$ is
normal in $\w P$, a contradiction. Thus, $J(S)$ is not normal in $\w P$. Since
$\w P$ is minimal parabolic we get that $N_{\w P}(J(S)) \leq N_{\w P}(Z_0)$
and that $N_{\w P}(Z_0)$ is self-normalizing.
Assume now that $x \not\in N_{\w P}(Z_0)$ but $O_p(P) \leq N_{\w P}(Z_0^x)$,
so $N_{\w P}(Z_0) \ne N_{\w P}(Z_0^x)$. We choose $x$ in addition such that
$|T|$ is maximal, where
$$O_p(P) \leq T \in Syl_p(N_{\w P}(Z_0) \cap N_{\w P}(Z_0^x)).$$
Note that $O_p(\w P) \leq T\cap S$.
After conjugation in $N_{\w P}(O_p(P))$ we may assume that $T_1 := N_T(O_p(P))
\leq S$, so $T_1 = T\cap S$. Note that $T \not\in Syl_p(\w P)$ since $\w P$ is minimal parabolic;
in particular $T$ is not a Sylow $p$-subgroup of $N_{\w P}(Z_0^x)$. Hence, the
maximality of $T$ yields
(1) $ N_{\w P}(T) \not\leq N_{\w P}(Z_0)$.
From (1) and $N_{\w P}(J(S)) \leq N_{\w P}(Z_0)$ we get:
(2) $J(S) \ne J(T)$ and $J(S) \not \leq T$.
In particular $J(S) \not \leq O_p(P)$, and (3.1) and (2.7) yield
(3) $S_0 = B(S) = J(S)O_p(P)$ and $[\O_1(Z(J_0)),J(S)] = Z_0 = \O_1(Z(J(S)))$,
where $J_0 := J(O_p(P))$.
Assume that $J(T_1) \ne J_0$. Since $Q \leq T_1$ the $Q$-transitivity and
(2.1) imply
$$S_0 = J(T_1)O_p(P) \leq T.$$
This contradicts (2) and (3). We have
shown:
(4) $J_0 = J(T_1)$.
Since $(P,\w P)$ is an amalgam and $O_p(\w P) \leq T_1$ we get from (4) $J_0
\not \leq O_p(\w P)$ and thus $N_{\w P}(J_0) \leq N_{\w P}(Z_0)$.
Set $T_2 := N_T(J_0)$ and note that $J_0 \ne J(T_2)$ by (1). There exists $y \in N_{\w P}(J_0)$ such that $T_2 \leq S^y$. From (3) we get
$$[\O_1(Z(J_0)),J(S)^y] = Z_0,$$
in particular $J(T_2) \leq N_H(Y_P)$ since $Y_P \leq \O_1(Z(J_0))$.
Hence also $T_3 := \langle O_p(P),J(T_2) \rangle \leq N_H(Y_P)$, and $O_p(P) =
C_{T_3}(Y_P)$ is normal in $T_3$ since $O_p(P) \in Syl_p(N_H(Y_P))$. It
follows that $T_3 \leq T_1$ and thus by (4) $J_0 = J(T_2)$, a contradiction.
\bigskip
{\bf (3.4)} Assume Hypothesis II. Let $V = \langle Y_P^{\w
P}\rangle$. Then $V$ is abelian.
\bigskip
Proof. Set $V_0= \langle Z_0^{\w P}\rangle$. By Hypothesis I and (1.2)(b) $Y_P \leq Q$ and thus
$V \leq O_p(\w P) \leq S$. Assume that $V$ is not abelian. Then there exists $x \in \w P$ such
that $A := Y_P^x \not\leq O_p(P)$. Then (2.1) and the $Q$-invariance
of $A$ show that $[V,Y_P] = [A,Y_P] = Z_0$ and $AO_p(P)= VO_p(P) =
S_0$. Moreover $V_0 \leq Z(V) \leq O_p(P)$, and $O^{p^\prime}(C_{\w P}(V_0)) \leq O_p(\w P)$ since $Z_0$ is not normal in $\w P$.
There exists $y \in P$ such that $\langle V,V^y\rangle C_{P^*}(Y_P) =
P^*$. Since $V$ is contained in $Q$ and normal in $S\,$ (3.2) implies
$O^p(P)\leq \langle V,V^y\rangle$. Hence
$Z(P) = 1$ gives $Z(\langle V,V^y\rangle) = 1$.
Note that $V_0 \leq O_p(P) \leq S^y$ and thus
$$[V_0,V_0^y] \leq V_0 \cap V_0^y \leq Z(\langle V,V^{y}\rangle)
= 1.$$
Let $z^\prime,z \in \w P$ such that for $A_1 = Y_P^z$ and $A_2 =
Y_P^{z^\prime}$
$$[A_1,A_2] = Z_0^z \ne Z_0.$$
It follows that $U := \langle A_1,A_2 \rangle \leq O_p(P)$.
In addition $V_0^y \leq O^{p^\prime}(C_{\w P}(V_0)) \leq C_{O_p(\w P)}(Z_0^z)$
and thus $[V_0^y,U] \leq V_0^y \cap V_0 = 1$. Hence $U \leq
C_{O_p(\w P^y)}(V_0^y)$ and thus $[A_1,A_2,V^y] = 1$. It
follows
that $Z_0^z$ centralizes $V^y$ and
$$Z_0^z \leq Z(\langle V,V^{y}\rangle) = 1,$$
a contradiction.
\bigskip
{\bf Notation.} From now through (3.9) we will apply the
amalgam method to the amalgam $(P,\w P)$. With one exception we will use the
standard terminology (see [DS], [KS] and the proof of Theorem 1). In particular we
choose $\a,\be,\b \in
\Gamma$ so that $(\a,\b)$ is a critical pair and so that $\{G_\a,G_\be\} = \{P,\w P\}$. The
exception to standard notation is the definition of $Z_\delta$. For $\delta \in \Gamma$ we
define
$$Z_\delta := Y_{G_\delta}.$$
In addition, we define for $g \in G$, $\delta = \a^g$ and $\lambda
=\be^g$
$$\eqalign{&Z_\lambda^* = C_{Z_\delta}(O^p(G_\lambda)),\ \w
Q_\lambda =
Q^g,\ Z(\delta,\lambda) = Z_0^g,\ \w C_\lambda = \w C^g,\cr
&V_\lambda^* = \langle x^h \mid h \in G_\lambda,\ x \in
Z_\delta\hbox{ and } [x,S^g] \leq Z_\lambda^*\rangle.\cr}$$
Note that $Z_\lambda^*$ is normal in $G_\lambda$ and thus
$[V_\lambda^*,Q_\lambda] \leq Z_\lambda^*$. Note further that
$$V_\lambda^* = \langle (Z_\delta \cap V_\lambda^*)^{G_\lambda}\rangle.$$
\bigskip
{\bf (3.5)} Assume Hypothesis II. Then $Z = Y_{\w P}$ and $\w P =
G_\be$.
\bigskip
Proof. Clearly $Z = Y_{\w P}$ implies $\w P = G_\be$. Thus, we
may assume that
$Z \ne Y_{\w P}$. Then by (1.3)(b) $C_S(Y_{\w P}) = O_p(\w P)$ and
$[Z_\a,Z_\b] \ne 1$. Let $1\ne x \in [Z_\a,Z_\b]$.
Assume that $G_\a = \w P$. Then $Z_\a \leq Y_{\w C} \leq Z(Q)$ by (1.2)(b)
and $C_H(x) \leq \w C$ by $Q$-Uniqueness. Since $Z_\a \not\leq Q_\b$ we get
$G_\b\not\leq \w C$. It follows that $G_\b$ is
conjugate to $P$.
Hence, after switching to another critical pair we may assume
that $G_\a = P$. (3.4) shows that $b > 2$.
Let $\a -1\in \Delta(\a)$ such that $\langle G_{\a-1}\cap
G_\a,Z_\b\rangle =G_\a$ and set $A := Z_{\b-1}(Z_\b \cap Q_\a)$. Since $b > 2$
we have
$$(*)\quad [Z_{\a-1},A,Z_\b] = 1.$$
Assume first that $[Z_{\a-1},A] =: R \ne 1$. As above $C_H(R) \leq \w
C_{\a-1}$ since $G_{\a-1}$ is conjugate to $\w P$. Hence ($*$) gives
$$\langle G_{\a-1},Z_\b\rangle = \langle G_{\a-1},G_\a \rangle \leq \w
C_{\a-1},$$
a contradiction.
Assume now that $[Z_{\a-1},A] = 1$. Then $Z_{\a-1} \leq G_\b$ and
$$Z_\b \cap Q_\a = C_{Z_\b}(Z_\a) \leq C_{Z_\b}(Z_{\a-1}),$$
while (2.1) gives
$$|Z_\a/C_{Z_\a}(Z_\b)| = |Z_\b/C_{Z_\b}(Z_\a)|.$$
It follows that
$$(**)\quad |Z_\b/C_{Z_\b}(Z_\a Z_{\a-1})| = |Z_\b/C_{Z_\b}(Z_\a)| =
|Z_\a/C_{Z_\a}(Z_\b)| \leq |Z_\a Z_{\a-1}/C_{Z_\a Z_{\a-1}}(Z_\b)|.$$
According to (2.1)(e), this time applied to $G_\b$, equality holds in ($**$), so $Z_{\a-1} \leq
Z_\a Q_\b$ and $[Z_{\a-1},Z_\b] \leq [Z_\a,Z_\b] \leq Z_\a$. Hence,
$Z_{\a-1}Z_\a$ and thus also $[Z_{\a-1},Q_\a]$ is normal in $G_\a$. Now the
irreducibility of $Z_\a$ and (1.2)(e) yield $Z_{\a-1} \leq Z_\a$. But then $Q_\a \leq
Q_{\a-1}$ and thus also $Q_\a \leq Q_\be$. Since $Z_\b \leq Q_\be$ (2.1) and
(3.1)
give $S_0 \leq Q_\be$ and $S_0 = B(S)$. Hence, $Z_0$ is normal in $\w P$,
which contradicts Hypothesis II.
\bigskip
{\bf (3.6)} Assume Hypothesis II. Then $[Z_\a,Z_\b] = 1$.
\bigskip
Proof. Asssume that $[Z_\a,Z_\b] \ne 1$. From (3.5) we get
that $\w P = G_\be$ and $Z_\be = Z$. In particular $b$ is even, and $G_\b$ is
conjugate to $G_\a $. Moreover, (3.4) gives:
(1) $V_\be$ is an elementary abelian subgroup of $Q_\be$, and $b \geq 4$.
Pick $\b +1 \in \Delta(\b)$ such that $Z(\b,\b+1) \ne Z(\b,\b-
1)$. The $\w Q_{\b+1}$-transitivity shows that $O^p(G_\b) \leq \langle Z_\a,\w
Q_{\b+1}\rangle C_{G_\b}(Z_\b)$. So (3.2) yields $O^p(G_\b) \leq \langle Z_\a,\w Q_{\b+1}\rangle$.
(2) $Z_\a \cap V_\be^* \leq Z(\a,\be)$.
Note that $S_0 = Q_\a\langle Z_\b^{\w Q_\be} \rangle$ by (2.1) and
$Q$-transitivity since $Z_\b \in \U(P)$,
so $[Z_\be^*,S_0] = 1$. Hence $Z_\be^* \leq Z(\a,\be)$. Moreover
$$D:=[Z_\a \cap V_\be^*,S_0] \leq [Z_\a \cap V_\be^*,Q_\a Q_\be] \leq
[V_\be^*,Q_\be] \leq Z_\be^*.$$
Note that $D$ is $Q$-invariant. Hence, the action of $S_0$ on $Z_\a$ and the
$Q$-transitivity either give $D =1$, or $D = Z(\a,\be)$. The first case
implies (2). In the second case $Z(\a,\be) = Z_\be^*$ is normal in $G_\be$,
which contradicts Hypothesis II.
(3) $V_{\b+1}^* \leq Q_{\a+2}$.
This follows from (2) since $Z_{\a+2}$ centralizes $Z(\mu,\b+1)$
for all $\mu \in \Delta(\b+1)$.
(4) Let $A \leq V_{\b+1}^*$ such that $O^p(G_\b) \leq N_G(AZ_\b
)$. Then $A \leq Z(\b,\b+1)$.
Since $\langle A^{G_\b \cap G_{\b+1}}\rangle$ satisfies the hypothesis of (4)
we may assume that $A$ is $(G_\b\cap G_{\b+1})$-invariant; i.e. $AZ_\b$ is
normal in $G_\b$. Then also $Y := [AZ_\b,Q_\b]$ is normal in $G_\b$ and $Y
\leq V_{\b+1}^*$.
If $Y = 1$, then (1.2)(e) shows that $A \leq \O_1(Z(Q_\b)) =
Z_\b$ since $G_\b$ is conjugate to $P$. Now (2) yields $A \leq Z(\b,\b+1)$. If $Y \ne 1$, then the
irreducibility of $Z_\b$ gives $Z_\b \leq Y$, which contradicts (2).
(5) $V_{\b+1}^* \not\leq G_\a$.
Assume that $V_{\b+1}^* \leq G_\a$. As $b > 2$ and thus $V_{\b+1}^* \leq Q_\b$, (2.4) gives
$$[Z_\a,V_{\b+1}^*] \leq [Z_\a,Z_\b][Z_\a \cap Q_\b,V_{\b+1}^*] \leq Z_\b
V_{\b+1}^*,$$
so $Z_\b V_{\b+1}^*$ is normal in $\langle Z_\a,G_\b \cap G_{\b+1}\rangle = G_\b$. Now (4) shows that $V_{\b+1}^* =
Z(\b,\b+1)$, which contradicts Hypothesis II.
By (1.3)(b) and Hypothesis II $Q_\be$ is the unique Sylow p-subgroup of
$\cap_{\rho\in \Delta(\be)}N_{G_\be}(Z(\rho,\be))$. Hence, by (5) there
exists $\rho \in \Delta(\be)$ such that $V_{\b+1}^* \not\leq
N_{G_\be}(Z(\rho,\be))$. Note that by (3) and (3.3) also $\langle
Q_\rho^{V_{\b+1}^*} \rangle \not\leq N_{G_\be}(Z(\rho,\be))$.
(6) $Z_\rho \leq Q_\b$.
Assume that $Z_\rho \not\leq Q_\b$. Then $(\rho,\b)$ is a
critical pair, and $Z(\rho,\be)
= \langle [Z_\rho,Z_\b]^{\w Q_\be}\rangle $ centralizes $\langle
Q_\rho^{V_{\b+1}^*} \rangle$, a contradiction.
(7) Set $R := [Z_\rho,V_{\b+1}^*]$. Then $|R| < |Z(\rho,\be)|$.
Note that by (3) and (6) $R \leq V_{\b+1}^*\cap V_\be$ and by (1)
$[R,Z_\a] = 1$. Since $[V_{\b+1}^*,Q_{\b+1}] \leq Z_{\b+1}^* \leq Z_\b$ we get
that $RZ_\b$ is normalized by $\langle Z_\a,Q_{\b+1}\rangle$ anf thus by
$O^p(G_\b)$. Now (4) shows that $R \leq Z(\b,\b+1)$; and
equality does not hold since $Z_\a$ centralizes $R$ but not
$Z(\b,\b+1)$.
We now derive a final contradiction. Let $t \in
V_{\b+1}^*\setminus N_{G_\be}(Z(\rho,\be))$, $U = \langle
Q_\rho,t\rangle$ and $Y_0 =
C_{Z_\rho}(t)$. Note that
$$|Z_\rho/Y_0| \leq |[Z_\rho,t]|\leq |[Z_\rho,V_{\b+1}^*]|,$$
so by (7) $|Z_\rho/Y_0| < |Z(\rho,\be)|$. On the other hand, by (3.1) $|Z_\rho|
= |Z(\rho,\be)|^2$ and so $|Y_0| > |Z(\rho,\be)|$.
Set $U_0:= \langle Q_\rho^U\rangle$ and $Y_1 = C_{Z_\rho}(U_0)$.
By (3.3) $U_0 \not\leq N_{G_\be}(Z(\rho,\be))$. Since
$Y_0 \leq Y_1$ we also have $|Y_1| > |Z(\rho,\be)|$. Moreover, $Y_1$ and $U_0$
are $Q_\be$-invariant.
Let $x \in G_\be$ such that $\a^x = \rho$. As seen above $S_0^x \leq Q_\be
Q_\rho$, so $Y_1$ is $S_0^x$-invariant. Moreover, since $|Y_1| >
|Z(\rho,\be)|$ we also have $[Y_1,S_0^x] \ne 1$. Now (3.1), applied to $P^x\ (= G_\rho)$, and the
$Q$-transitivity yield
$$Z(\rho,\be) = \langle [Y_1,S_0^x]^Q\rangle \leq Y_1.$$
This contradicts $U_0 \not\leq N_{G_\be}(Z(\rho,\be))$.
\bigskip
{\bf (3.7)} Assume Hypothesis II. Let $A \leq \w P$ and $Y_0
:= [Y_P,A\cap P]$.
Suppose that $A \not\leq
N_{\w P}(Z_0)$ and $[Y_0,A] = 1$. Then either $Y_0 = 1$, or the following hold:
(a) $p = 2$ and $\o P \cong S_3 \ wr\ C_2$ or $S_5$.
(b) $|A\cap P/A\cap O_2(P)| = 2$, $|Y_0| = |Z_0| = 4$ and $ C_{P^*}(Y_0) = O_2(P)$.
\bigskip
Proof. Set $A_0 := A \cap P$, $U := \langle O_p(P),A\rangle$,
$U_0 := \langle O_p(P)^U\rangle$ and $Y_1 := C_{Y_P}(U_0)$. Note that
(1) $Y_0 \leq Y_1$, and $U_0$ is $Q$-invariant.
Hence $Y_1$ is the largest $Q$-invariant subgroup of $Y_P$ centralized by
$U_0$. By (3.3) $U_0 \not\leq N_{\w P}(Z_0)$ and thus
(2) $Z_0 \not \leq Y_1$.
From now on we assume that $Y_0 \ne 1$ and use the notation of (3.1); in addition we set $q := p^m$ and $R_i := [V_i,A_0]$, $i = 1,...,r$. It is
convenient to treat the following two cases separately:
($*$) There exists $i \in \{1,...,r\}$ such that $1\ne R_i \leq V_i$.
($**$) $R_i \not\leq V_i$ for all $i \in \{1,...,r\}$ with $R_i \ne 1$.
Case $(*)$: We have $A_0 \leq N_H(V_i)$ and thus $\o A_0 \leq N_{\o P
}(K_i)$. If $\o A_0 \leq K_iC_{\o P}(V_i)$, then $R_i = Z_0
\cap V_i \leq Y_0$, and (1) and the $Q$-transitivity give $\langle R_i^Q
\rangle = Z_0 \leq Y_1$, which contradicts (2). Hence, by (2.5)(e)
$|A_0/C_{A_0}(V_i)| = 2 = p$.
Assume that $r > 1$. Then there exists $x \in Q$ such that $K_i^{\o x} = K_j
\ne K_i$ and
$$[K_i\cap \o S,\o x]C_{\o P}(V_i) = (K_i \cap \o S)C_{\o P}(V_i).$$
It follows that
$$[R_i,K_i\cap \o S] = [R_i,[K_i\cap \o S,\o x]] \leq [R_i,\o Q],$$
so by (1) $[R_i,K_i \cap \o S] = Z_0 \cap V_i \leq Y_1$. Now as above the
$Q$-transitivity yields $Z_0 \leq Y_1$, which contradicts (2). Hence $r = 1$.
Thus $|A_0/A_0\cap O_2(P)| = 2$; moreover $|Y_P/C_{Y_P}(A_0)| = q$ and
$C_{Y_P}(A_0) = Y_0$ since $\o A_0$ acts as a field
automorphism on $\o P^*$.
We have proved:
(3) In case ($*$) $\ r = 1$, $p=2$, $C_{P^*}(Y_0) = O_2(P)$,
$|A_0/A_0 \cap O_2(P)| = 2$ and $|Y_P/Y_0| = q$.
Case ($**$): Fix $i \in \{1,..,r\}$ such that $R_i \ne 1$. Then $A_0 \not\leq
N_P(V_i)$ since $R_i \not\leq V_i$, and from (2.5)(e) we get that
$|A_0/C_{A_0}(V_i)| = 2 (=p)$ and there exists $j \ne i$ such that $\langle
V_i^{A_0}\rangle = V_i \times V_j$. Note that
$$V_iV_j = V_i(Y_1 \cap V_iV_j) = V_j(Y_1 \cap V_iV_j).$$
Assume that $r > 2$. Then by the $Q$-transitivity there exists $x \in Q$ such
that $V_i^x \not\in \{V_i,V_j\}$. In particular, there exists $\o b \in
(K_i \times K_i^{\o x}) \cap \o Q$ such that
$$V_i\cap Z_0 = [V_i,b] \leq [V_iV_j,b] = [V_jY_1,b] = [Y_1,b].$$
As above, (1) and the $Q$-transitivity give $Z_0 \leq Y_1$, which contradicts
(2). We have shown that $r = 2$, so $N_{A_0}(V_i) = C_{A_0}(V_i)$ implies $|A_0/A_0 \cap O_2(P)| = 2$.
For every $c \in P^* \setminus O_2(P)$ we have $[Y_0,c] \ne 1$ since $Y_P =
Y_0V_i$ for $i = 1,2$. It follows that $C_{P^*}(Y_0) = O_2(P)$. Moreover $V_i
\cap Y_0 = 1$ implies $|Y_0| = |V_i| = |Y_P/Y_0| = q^2$. We have shown:
(4) In case ($**$) $r = 2 = p$, $C_{P^*}(Y_0) = O_2(P)$, $|A_0/A_0 \cap
O_2(P)| = 2$ and $|Y_P/Y_0| = q^2$.
Assume that case (a) of the Local P!-Theorem holds for $P$. Then $r = 1$, $QO_2(P) = S_0$ and
$[y,Q] = Z_0$ for every $y \in Y_P\setminus Z_0$.
As $Y_0 \not\leq Z_0$ by (3), this gives $Z_0 \leq Y_1$, which contradicts (2). We have shown:
(5) Case (b) of the Local P!-Theorem holds for $P$; in particular ${\cal M}_H(P) = \{M\}$.
As a trivial consequence of (5) we get:
(6) $N_H(J(O_2(P))) \leq M$.
Let $O_2(P) \leq T \in Syl_2(U_0)$ and $T_0 = N_T(J(O_2(P)))$. Note that $T_0
\leq M$ by (6). By (3.1) $J(S) \leq S_0$ and by (2.1)(e)
$${\cal A}(O_2(P)) \subseteq {\cal A}(S),$$
so $J(T_0) \leq S_0^x$ for some $x \in M$. According to (5)
$P^*C_M(Y_P)$ is normal in $M$, hence $J(T_0) \leq
P^*C_M(Y_P)$. Now by (1), (3) and (4) imply
$$J(T_0) \leq C_M(Y_0) \cap P^*C_M(Y_P) = C_{P^*}(Y_0)C_M(Y_P) =
O_2(P)C_M(Y_P) = C_M(Y_P).$$
Since $O_2(P)$ is a Sylow 2-subgroup of $C_M(Y_P)$ we conclude that $J(T_0) =
J(O_2(P))$ and thus also $J(T) = J(O_2(P))$; in particular $T =
N_T(J(O_2(P))) = T_0 \leq M$. In
addition, (3.3) implies $T \leq N_{\w P}(Z_0)$ and (5) implies $Y_P = Y_M$. We have shown:
(7) $J(T) = J(O_2(P))$, and $T$ normalizes $Y_P$ and $Z_0$.
According to (5), (6), (7) and (b)(ii) of the Local P!-Theorem $N_{U_0}(T) \leq M \cap \w C \leq
N_M(Z_0)$. Since $U_0 \not \leq N_{\w P}(Z_0)$ there exists $F \in {\cal
P}_{U_0}(T)$ such that $F \not\leq N_H(Z_0)$; see (1.3)(a). As $O_2(\w P) \leq N_H(U_0)$ we
get $[U_0,O_2(\w P)] \leq O_2(U_0)$; in particular, $F$ is
$O_2(\w P)$-invariant and $[F,O_2(\w P)] \leq O_2(F)$. In addition, (3.3) and (7) show $O_2(P) \not \leq O_2(F)$ and
thus by (1.3)(c)
(8) $O^2(F) = [O^2(F),O_2(P)] \leq \langle O_2(P)^F\rangle$.
Set $W = \langle
Y_P^F\rangle$. Clearly $[W,O^2(F)] \ne 1$ since by (7) $O^2(F)
\not\leq N_H(Z_0)$. Moreover, (3.4) shows that $W$ is
elementary abelian.
Assume that $O_2(P) \cap O_2(F)$ is normal in $F$. Then by (8)
$$[O^2(F),O_2(\w P)] \leq [\langle O_2(P)^F\rangle,O_2(\w P)] \leq O_2(P) \cap
O_2(F)$$
and $W = \langle Y_P^{O^2(F)}\rangle \leq
Z(O_2(P) \cap O_2(F))$ since $Y_P \leq Z(O_2(P) \cap O_2(\w P))$ by Hypothesis I and (1.2)(b). The $P \times Q$-Lemma implies that
$[C_W(O_2(\w P)),O^2(F)] \ne 1$; in particular $[Y_{\w P},O^2(\w
P)] \ne 1$, which contradicts (3.5). We have shown:
(9) $O_2(P) \cap O_2(F)$ is not normal in $F$.
Note that $F \not\leq M$ since $M\cap \w C \leq N_M(Z_0)$, so by (6) and (7) $J(O_2(P)) = J(T) \not \leq O_2(F)$. Assume that there exists only
one non-central $F$-chief factor (in a given $F$-chief series) of $W$. As by (9)
$$[O^2(F),O_2(F)] \not\leq
O_2(F)\cap O_2(P)\hbox{ and }C_{O_2(F)}(W) \leq O_2(F) \cap O_2(P),$$
we get $[O^2(F),O_2(F),W] \ne 1$. Thus by
[Ste2, 3.3] there exists $B \leq O_2(F)$ such that
$$[Y_P,B,B] = 1 \ne [Y_P,B]\hbox{ and }|[Y_P,B]| \leq |B/C_B(Y_P)|.$$
The structure of $P$ given in (3.1) shows that $B
\leq P^*$. But then (1), (3) and (4) imply $B \leq C_{P^*}(Y_0) = O_2(P) =
C_{P^*}(Y_P)$, a contradiction.
We have shown that there are at least two non-central $F$-chief factors in
$W$. Let $B_1 \in {\cal A}(O_2(P))$ with $B_1 \not\leq O_2(F)$. From (2.1) we get that
$$|B_1/C_{B_1}(W^*)| \leq |W^*/C_{W^*}(B_1)|$$
for all non-central $F$-chief factors $W^*$ of $W$.
We now apply the $qrc$-Lemma [Ste2, 3.1(c)] to $F$ and $B_1$ and get
$(q-1)(rc-1)\leq 1$ (where $q$, $r$ and $c$ are the parameters defined in [Ste]). Since by
[Cher] $r \geq 1$ it follows that $q \leq 2$.
Hence, there exists $B
\leq O_2(F)$ such that
$$(+)\quad |B/C_B(Y_P)|^2 \geq |Y_P/C_{Y_P}(B)|.$$
Again by (3) and (4) $C_{P^*}(Y_0) = O_2(P)$ and thus $B \cap P^* \leq
O_2(P)$.
As above, we now treat the two cases ($*$) and ($**$) separately. It remains to prove the isomorphism type of $\o P$.
Assume case ($*$). Then $\o B$ induces a field automorphism of order 2 on $\o
P^*$. Hence $(+)$ gives $|Y_P| = 4^2$ and $\o P \cong S_5$.
Assume case ($**$). Then $Y_P = Y_0V_i$, $i = 1,2$, and again $|\o B| = 2$ and
$|Y_P| = 4^2$, so $\o P \cong S_3\ wr\ S_2$.
\bigskip
{\bf L-Lemma.} Let $X \in {\cal P}_H(S)$ and $A \leq S$ such that
$A \not \leq
O_p(X)$, and let $M$ be the unique maximal subgroup of $X$
containing $S$. Then
there exists a subgroup $O_p(X) \leq L \leq X$ with $A \leq L$ satisfying:
(i) $AO_p(L)$ is contained in a unique maximal subgroup $L_0$ of
$L$,
and $L_0 = L \cap M^g$ for some $g \in X$.
(ii) $L = \langle A,A^x \rangle O_p(L)$ for every $x \in L
\setminus
L_0$.
(iii) $L$ is not contained in any $X$-conjugate of $M$.
\bigskip
Proof. For $U \leq X$ set
$$U^* := \langle A^g \mid g \in X, A^g \leq U \rangle.$$
Note that $N_X(U) \leq N_X(U^*)$; in particular $N_X(S^*) \leq M$.
Choose $Y$ among all $X$-conjugates of $M$ such that
$Y \ne M$ and for $T \in Syl_p(Y \cap M)$
$$|T^*| \hbox{ is maximal}.$$
Without loss of generality we may assume that $T \leq S$. Let $h \in X$ such
that $T \leq S^h \leq Y$ and set $N := N_X(T^*)$
and $S_1 := S \cap N$. Then $T \ne S^h$ since $Y\ne M$, so also $T <
N_{S^h}(T) \leq N \cap S^h$. As $T \in Syl_p(Y\cap M)$ this gives $N \not\leq
M$. Since $N_X(S^*) \leq M$ this implies that $T^* \ne S^*$ and thus also $T^*
\ne S_1^*$. Hence, there exists a conjugate $B = A^g$, $g \in X$, such that $B \leq S_1$ and $B \not \leq T$. Choose $z \in
N \setminus M$ such that $L_1 := \langle B,z \rangle T^*$ is
minimal, and set $L:= L_1^{g^{-1}}O_p(X)$.
Since $T^* \ne BT^* = (BT^*)^*$ the maximality of $T^*$ shows that $M$ is the
unique conjugate containing $BT^*$. In particular, (iii) holds since
$L_1\not\leq M$. Moreover, the minimality of $L_1$ gives (i). Let $x \in
L_1\setminus M$. Then $M^x$ is the unique conjugate of $M$ containing
$B^xT^*$ and $M \ne M^x$, so $B^x \not \leq M$ and $\langle B,B^x\rangle T^* =
L_1$. This gives (ii).
\bigskip
{\bf (3.8)} Assume Hypothesis II. Let $A \leq S$ such that $[V_\be,A,A] =
1$ and $A \not \leq Q_\be$. Then there exist $\tau \in \Delta(\be)$, $T \in
Syl_p(G_\be\cap G_\tau)$ and $L
\leq G_\be$ such that for $L(\tau) := N_L(Z(\tau,\be))$, $W := \langle Z_\tau^L\rangle$ and $W^* := \langle v^h
\mid v \in Z_\tau,\ h \in L,\ [v,T] \leq Z_\be^*\rangle$ the following hold:
(a) $Q_\be \leq A O_p(L)\leq T\cap L \in Syl_p(L(\tau))$, and $L(\tau)$ is a
maximal subgroup of $L$.
(b) $L = \langle y,A^x\rangle O_p(L)$ for every $x \in L$ and every $y \in
L\setminus L(\tau)^x$.
(c) $[W^*,O^p(L)] \ne 1$ and $[W,O^p(L)] \not \leq W^*$.
(d) Let $U$ be a non-central $L$-chief factor of $W$. Then $C_U(A) = C_U(a)$
for every $a \in A\setminus O_p(L)$, and $|U/C_U(A)| \geq |A/A\cap O_p(L)|$.
\bigskip
Proof. According to (3.1), (3.4), (3.5) and (3.6) $b \geq 3$ and $\b \in \be^G$; in
particular $Q_\tau \not \leq Q_\be$ for all $\tau \in \Delta(\be)$ since $Z_\a \leq Q_{\b-1}$ and $Z_\a \not \leq Q_\b$. We apply the L-Lemma with $G_\be$ in place of $X$. Then there exists
$Q_\be \leq L\leq G_\be$ and $\tau \in \Delta(\be)$ such that
(i) $L(\tau)$ is the unique maximal subgroup of $L$ containing $AO_p(L)$, and
$AO_p(L) \leq T\cap L \in Syl_p(L(\tau))$ for some $T \in Syl_p(G_\be\cap G_\tau)$.
(ii) $L = \langle A,A^x\rangle O_p(L)$ for every $x \in L\setminus L(\tau)$.
(iii) $\langle L,T_0\rangle = G_\be$ for every $T_0 \in Syl_p(G_\be)$.
Claim (a) follows directly from (i).
Let $y$ and $x$ be as in (b). Then $y^\prime :=y^{x^{-1}} \in L
\setminus L(\tau)$ and by (ii)
$$L = \langle A,A^{ y^\prime}\rangle O_p(L) = \langle A, y^\prime \rangle
O_p(L).$$
This implies (b).
For the proof of (c) assume first that $[W^*,O^p(L)] = 1$. Then $W^* \leq
Z_\tau$ and $[W^*,T] \leq Z_\be^*
\leq W^*$ since $L = O^p(L)(T\cap L)$. By (iii) $W^*$ is normal
in $\langle L,T\rangle = G_\be$. But this implies that $W^* =
Z_\be^* = Z_\tau$, a contradiction.
Assume now that $[W,O^p(L)]\leq W^*$. Then $W = W^*Z_\tau$ and
$$Z_\be^*[W,\w Q_\be] = Z_\be^*[Z_\tau,\w Q_\be] \leq Z_\tau.$$
Hence $Z_\be^* [Z_\tau,\w Q_\be]$ is normal in $\langle T,L\rangle = G_\be$.
On the other hand $Q_\tau \not\leq Q_\be$ and thus by (1.3)(b) $[Z_\tau,\w Q_\be]\leq
Z_\be^*$. Let $g \in G_\be$ such that $\tau = \a^g$. Then the action of $P^g$ on $Z_\tau$, as
described in (3.1), shows that
$$[Z_\tau,\w Q_\be \cap S_0^g] = Z(\tau,\be) \leq Z_\be^*,$$
which contradicts Hypothesis II. Hence, (c) is proved.
Note that $L$ is minimal parabolic (with respect to $T\cap L$ and
$L(\tau)$). Hence by (1.3)(b) $C_{T\cap L}(U) = O_p(L)$ for every non-central
$L$-chief factor $U$ in $W$. (2.1)(e) shows that
$$|U/C_U(A)| \geq |A/A\cap O_p(L)|.$$
Let $a \in A\setminus O_p(L)$. Then by (1.3)(b) there exists $x \in L$ such that $a
\not\in L(\tau)^x$. By (b) $L = \langle a,A^x\rangle O_p(L)$ and thus, together
with the quadratic action of $A$ on $U$,
$$U = [U,a] \times [U,A^x] = C_U(a) \times C_U(A^x);$$
in particular $C_U(a) = [U,a] \leq C_U(A)$ and equality holds. This is (d).
\bigskip
{\bf (3.9)} No group satisfies Hypothesis II.
\bigskip
Proof. Assume Hypothesis II. By (3.1), (3.4), (3.5) and (3.6) $[Z_\a,Z_\b] = 1$ and $b\geq
3$. In particular, $\b\in \be^G$ and $V_\be$ acts quadratically on $V_\b$,
and vice versa. We apply (3.8) with $(G_\b,V_\be)$ in place of $(G_\be,A)$ and choose the notation
$\tau$, $L$, $T$, $W$, $W^*$ as there.
(1) $Z_\mu \not\leq G_\rho$ for every $\rho \in \Delta(\be)$ and $\mu \in
\tau^L$ such that $Z_\rho \not\leq L(\mu)$.
Assume that there exist $\rho \in \Delta(\be)$ and $\mu \in \tau^L$ such that
$Z_\rho \not\leq L(\mu)$ but $Z_\mu \leq G_\rho$.
Let $x \in L$ such that $\mu = \tau^x$. Then, with the notation of (3.1)
applied to $G_\rho$, there exists
a submodule $V_i \leq Z_\rho$ such that $V_i \not\leq L(\mu)$. By (3.8)(b)
$\langle V_i,V_\be^x\rangle O_p(L) = L$. On the other
hand $Z_\mu \leq G_\rho$, and (3.1) together with the quadratic
action of $Z_\mu$ on $Z_\rho$ gives
either
$$[V_i,Z_\mu \cap W^*] = 1\hbox{ or }[V_i,Z_\mu] = [V_i,Z_\mu \cap W^*].$$
In the first case $Z_\mu \cap W^*$ is normal in $L$. Hence $W^* = Z_\mu\cap
W^*$, and by (1.3)(b) $[W^*,O^p(L)] = 1$ since $V_i \not \leq O_p(L)$. In the second case
$[W,O^p(L)] \leq W^*$ since $O^p(L) \leq \langle V_i^L\rangle$, so both cases contradict (3.8)(c), and (1) is proved.
In particular, (1) together with $V_\be \not\leq O_p(L)$ gives $W \not\leq
Q_\be$. Hence, we are allowed to apply (3.8) to $(G_\be,W)$ in place of
$(G_\be,A)$. Again we use the notation of (3.8), but this time indicated
by $\,\,\w{}\,\,$ to distinguish from the above notation, so $\w \tau$, $\w
L$, $\w T$, $\w W$, $\w W^* $ are given as there. With the same argument as above we get
(2) $Z_{\w \mu} \not\leq G_{\w \rho}$ for every $\w \rho \in \Delta(\b)$ and
$\w\mu\in \tau^{\w L}$ such that $Z_{\w \rho} \not\leq \w L(\w \mu)$.
As above, (2) implies $\w W \not\leq O_p(L)$.
We now choose $\mu \in \tau^L$ and $\w \mu \in \w\tau^{\w L}$ such that $\w W
\not\leq L(\mu)$ and $W \not\leq \w L(\w \mu)$. From (1) and (2) we get that
$Z_{\w \mu} \not \leq O_p(L)$ and $Z_\mu \not\leq O_p(\w
L)$.
Moreover, we may assume that $|W/W \cap O_p(\w L)| \leq |\w W/\w W\cap O_p(L)|$,
since the other case follows by the same argument with the roles of $W$ and $\w W$ reversed.
From (3.8)(c) we get that there exist two non-central $L$-
chief factors $U_1$ and $U_2$ in $W$. As $Z_{\w \mu}\not\leq O_p(L)$ (3.8)(d)
implies that $C_{U_i}(V_\be) = C_{U_i}(Z_{\w \mu})$, so, again by (3.8)(d),
$$|\w W/\w W\cap O_p(L)| \leq |V_\be/V_\be \cap O_p(L)| \leq
|U_i/C_{U_i}(V_\be)| = |U_i/C_{U_i}(Z_{\w \mu})|.$$
Hence
$$\eqalign{|\w W/\w W\cap O_p(L)|^2&\leq
|U_1/C_{U_1}(Z_{\w\mu})||U_2/C_{U_2}(Z_{\w\mu})|\leq |W/C_W(Z_{\w \mu})|\cr
&\leq |W/W\cap Q_{\w \mu}| \leq |W/W \cap O_p(\w L)||W\cap G_{\w \mu}/W \cap Q_{\w \mu}|\cr
&\leq |\w W/\w W\cap O_p(L)||W\cap G_{\w \mu}/W \cap Q_{\w \mu}|.\cr}$$
On the other hand by (3.7), applied to $G_{\w \mu}$ with $A = W$, we get
$|W\cap G_{\w \mu}/W \cap Q_{\w \mu}| \leq 2$. It
follows that
(3) $|W/W\cap O_p(\w L)| = |\w W/\w W\cap O_p(L)| = 2 = p$ and $|Z_\mu| =
|Z_{\w\mu}| = 16$.
(4) $|W/C_W(Z_{\w
\mu})| = |\w W/C_{\w W}(Z_\mu)| = 4$.
As a consequence we get from (3)
(5) $Z_{\w \mu}\not\leq L(\mu)$ and $Z_\mu \not\leq \w L(\w \mu)$.
Next we prove:
(6) $L/C_L(W) \cong \w L/C_{\w L}(\w W) \cong S_3$.
Let $t \in Z_{\w \mu}\setminus O_2(L)$ and $x \in L$ such that $\mu =
\tau^x$. Then by (3) and (5) $L = \langle
t,t^x\rangle O_2(L)$ and thus $O^2(L) \leq \langle t^L\rangle$. Hence, (3.8)(c) gives $W^* \not\leq
C_W(t)$ and $W^*C_W(t) \ne W$, and (6) follows for $L$ from $|W/C_W(Z_{\w
\mu})| = 4$. A similar argument gives the claim for $\w L$.
Set $W_0 := W$ and $W_i := [W_{i-1},\w Q_\b]$ for $i\geq 1$, and note that
$W_i = \langle (W_i \cap Z_\mu)^L\rangle$.
(7) Assume that $(W_i \cap Z_\mu)W_{i+1} = W_i$. Then $W_i \leq Z_\mu$.
Note that $W_{i+1} = [W_i,\w Q_\b] \leq [Z_\mu W_{i+1},\w Q_\b] \leq Z_\mu W_{i+2}$. It
follows that $W_i = (W_i\cap Z_\mu)W_k$ for all $k \geq i+1$ and thus $W_i
\leq Z_\mu$.
(8) $[Z_{\w \mu},Z_\mu\cap O_2(\w L)] \ne 1$.
Let $A_1 := Z_\mu\cap O_2(\w L)$,and assume that $[Z_{\w \mu},A_1] = 1$. By
(6) $L(\mu) = (L(\mu) \cap G_\mu)C_L(W)$. Suppose that $Z_\mu =
A_1(Z_\mu \cap W_1)$. Then $W = Z_\mu W_1$ and by (7) $W = Z_\mu$. But then
$Z_\mu$ is normal in $\langle L,G_\b \cap G_\mu \rangle = G_\b$, a
contradiction. We have shown that $Z_\mu \cap W_1 \leq A_1$. It follows that
$Z_\mu\cap W_1$ is centralized by $Z_{\w\mu}$ and thus normalized by $L$, so $W_1 \leq A_1$ and
$[W_1,O^2(L)] = 1$. In particular $[Z_\mu,\w Q_\b]$ is normalized by $L$ and
centralized by $O^2(L)$. Hence, by the L-Lemma (iii) it is also normalized by
$G_\b$ and centralized by $O^2(G_\b)$. Since $Z(\mu,\b) \leq [Z_\mu,Q_\b]$ we
get that $Z(\mu,\b)$ is normal in $G_\b$, a contradiction to Hypothesis II.
(9) $R := [Z_{\w\mu} \cap O_2(L),Z_\mu \cap O_2(\w L)] \ne 1$, and $R$ is
centralized by a Sylow $2$-subgroup of $G_{\w \mu}$ and $G_\mu$.
Let $A := Z_\mu$ and $A_0 := A \cap G_{\w \mu}$. By (8) $Y_0 := [Z_{\w \mu},A_0]
\ne 1$, and by (5) $A$ and $G_{\w\mu}$ satisfy the hypothesis of (3.7). Then (3.7) shows that $|Y_0| = 4$ and $|A_0/A_0\cap Q_{\w \mu}| = 2$; in
particular $A_0 = A \cap O_2(\w L)$. Moreover, (3.7) gives $|Z_{\w \mu}/C_{Z_{\w
\mu}}(A_0)| = 4$ and thus $R \ne 1$ since $|Z_{\w\mu}/Z_{\w\mu}\cap O_2(L)| =
2$.
The action of $G_{\w \mu}$ on $Z_{\w \mu}$ also shows that all elements
of $Y_0$ are centralized by a Syolw 2-subgroup of $G_{\w \mu}$. This and the
symmetric argument in $G_\mu$ yields the additional claim of (9).
We now derive a final contradiction. According to (9) there exist $y \in G_{\w
\mu}$ and $z \in G_\mu$ such that $R = Z_\be^y = Z_\b^z$. Then by (1.6) $\w
C_\be^y = \w C_\b^z$ and thus $\w Q_\be^y = \w Q_\b^z$. On the other hand,
Hypothesis I and (1.2)(b) yield $Z_{\w \mu} \leq \w Q_\be^y$, so $Z_{\w \mu}
\leq \w Q_\b^z \leq G_\mu$, which contradicts (2) and (5).
\bigskip
{\bf Theorem 3.} Assume Hypothesis I. Then $Z_0$ is normal in $\w
C$.
\bigskip
Proof. Assume that $Z_0$ is not normal in $\w C$. By
the definition of $\w C$ $N_H(S)\leq \w
C$. Hence, $N_H(S)$ acts on $\P_H(S)\setminus \P_{\w C}(S)$, and Theorem 2
implies that $N_H(S) \leq N_H(P)$ and thus also $N_H(S)
\leq N_H(P^*)$ since $P^* = U(P)$. It follows that $N_H(S) \leq N_H(S_0) \leq N_H(Z_0)$.
According to (1.3)(a) there exists $\w P \in
{\cal P}_{\w C}(S)$ such that $Z_0$ is not normal in $\w P$. We
choose $|\w
P|$ minimal with this property. If $(P,\w P)$ is an amalgam, then
$(P,\w P)$
satisfies Hypothesis II, which is impossible by (3.9).
Thus, $(P,\w P)$ is not an amalgam, and there exists $L \in \L_H(S)$ such that
$\langle P,\w P\rangle \leq L$. Let $L << \w M \in \L_H^*(S)$. Then by (1.2)
$Y_L \leq Y_{\w M}$ and by (1.4) $P^0 \leq L^0
\leq \w M^0 \leq \w M$.
We now apply the Local P!-Theorem to $\w M$.
Assume that also $\w P \leq \w M$. Then
$\w P\leq \w M\cap \w C \leq N_{\w M}(Z_0)$, a contradiction. Thus, we have
$\w P \not\leq \w M$.
Assume first that case (a) of the
Local P!-Theorem holds. Then $Q \leq S_0$, so $Z_0\leq Z(Q)$ and thus also $W := \langle Z_0^{\w
P}\rangle \leq Z(Q)$. Note that
$$Z_0 \leq Y_P = [Y_P,P^0] \leq [Y_L,L^0]\hbox{ and }W \leq [Y_L,L^0]$$
by (1.2). It follows that $W \leq [Y_{\w M},\w M_0]$ since $Y_L \leq Y_{\w M}$
and $L^0 \leq \w M^0$. In case (a)
$[Y_{\w M},\w M_0]$ is a natural $SL_n(p^m)$- or $Sp_{2n}(p^m)^\prime$-module. In particular,
$C_{[Y_{\w M},\w M_0]}(Q) = Z_0$ and so $Z_0 = W$ and $\w P \leq N_H(Z_0)$, a
contradiction.
Assume finally that case (b) of the Local P!-Theorem holds for $\w M$. Then $P^0
= L^0 = \w M^0$ and $\w P \leq N_H(\w M^0) = \w M$, which contradicts $\w P
\not\leq \w M$.
\bigskip
{\bf Corollary 2.} Assume Hypothesis I and $p = 2$. Then ${\cal
P}_H(S) =
\{P\}\cup {\cal P}_{\w C}(S)$.
\bigskip
Proof. We apply Theorem 2. Then
${\cal P}_H(S) = {\cal P}_P(S) \cup {\cal P}_{\w C}(S)$, and
the structure of $P$, see (3.1), implies
${\cal P}_P(S) = \{P\}\cup {\cal P}_{N_P(Z_0)}(S)$. Now
Theorem 3 yields the assertion.
\bigskip
{\bf Corollary 3.} Assume Hypothesis I. Suppose that case (b) of the Local
P!-Theorem holds for $P\leq M\in \L_H^*(S)$. Then the following holds:
(a) $p = 2$ and $\M_H(P) = \{M\}$,
(b) $\o P^* = K_1 \times \cdots \times K_r$, $K_i \cong SL_2(2)$,
(c) $Y_P = V_1\times \cdots \times V_r$, where $V_i = [Y_P,K_i]$ is a natural
$SL_2(2)$-module for $K_i$,
(d) $r \geq 4$.
(e) $Q$ is transitive on $K_1,...,K_r$.
\bigskip
Proof. We are in case (b) of the Structure Theorem. According to Theorem 3 $Z_0$ is normal in $\w C$. Hence
$$(*)\quad [N_P(Z_0),Q] \leq O_p(N_P(Z_0)).$$
We apply (3.1). Then either $\o P^*\cong SL_2(p^m)$, or the $Q$-transitivity
and ($*$) show that $N_{K_i}(Z_0)$ is a $p$-group and $r \geq 2$.
In the first case $Y_P$ is a natural $SL_2(p^m)$-module for
$P^*$. Thus, $Y_P$ is an $F$-vector space for $F:= End_{\o P^*}(Y_P)$,
and $P$ induces semi-linear transformations on $Y_P$.
As $N_{P^*}(Z_0)$ is irreducible
on $Z_0$, we get from ($*$) that $[Z_0,Q] = 1$, so $Q$ centralizes a
$1$-dimensional $F$-subspace of $Y_P$. Hence, $Q$
induces $F$-linear transformations on $Y_P$, and $Q \leq
P^*$. But this contradicts case (b) of the Structure Theorem.
In the second case (a) -- (c) and (e) are clear. For the proof of (d)
note that $Q$-transitivity yields $r = 2$ or (d). Assume $r = 2$, so
$P/C_P(Y_P) \cong O_4^+(2)$ and $|Z_0| = 4$. Hence, Theorem 3 shows
that $\w C/C_{\w C}(Z_0)$ is a subgroup of $S_3$. If all involutions
in $Z_0$ are conjugate in $\w C$, then $Q$-Uniqueness implies that $P
\leq \w C$, which is not the case. It follows that $\w C =C_{\w
C}(Z_0)S$, in particular $C_{\w C}(Z_0)\not\leq M$. We conclude that
$C_H(x) \not\leq M$ for all $1\ne x \in Y_P$. Now Theorem 3 of [MSS2] shows that $Y_M \not\leq Q$, a contradiction.
\vfill\eject
{\bf 4. F-Uniqueness}
\bigskip
In this section we treat the exceptional case described in Corollary 3, so in
this section we assume:
\bigskip
{\bf Hypothesis III.} Hypothesis I and case (b) of the Local P!-Theorem holds
for $P \leq M\in \L_H^*(S)$; in particular $\M_H(P) = \{M\}$.
{\bf Notation.} We use the notation given in Corollary 3 (and (3.1)). Set
$$F := C_{\w C}(Z_0)\hbox{ and }\O := \{K_1,...,K_r\}.$$
Recall that by Theorem 3 $F$ is normal in $\w C$, and by Corollary 3
($*$) $p = 2$, $K_i \cong SL_2(2)$, $r \geq 4$, and $Q$ is transitive on $\O$.
We will use these facts without further reference.
\bigskip
\bigskip
{\bf (4.1)} $P^* \cap \w C = S_0C_{P^*}(Y_P)$ and $\w C = C$.
\bigskip
Proof. Assume that $U := C_{P^*}(Y_P)S_0 < P^* \cap \w C$. Then by Corollary 3
(b) $K_i \leq [\o S_0,\o{P^*\cap \w C}]\o S_0$ for some $i$, and the $Q$-transitivity
yields $P^* \leq \w C$, which is not the case.
Let $Z^* = \langle Z^{\w C}\rangle$. By Theorem 3 $Z^* \leq Z_0\cap Z(Q)$, and
by $Q$-uniqueness $C_{P^*}(z)
\leq P^* \cap \w C = S_0C_{P^*}(Y_P)$ for all $1\ne z \in Z^*$. Now Corollary
3 (c) yields $|Z^*| = 2$, so $C = \w C$.
\bigskip
{\bf (4.2)} $N_H(B(S)) \leq M$.
\bigskip
Proof. It suffices to show that $P$ and $N_H(B(S))$ are contained in a
$2$-local subgroup of $H$ since ${\cal M}_H(P) = \{M\}$. Assume that this is
not the case; i.e. $O_2(\langle P,N_H(B(S))\rangle = 1$. Then $B(S)$ is not normal in $P$ and by (3.1) $B(S) = S_0$.
Hence, $N_H(B(S)) =N_H(S_0) \leq N_H(Z_0) = \w C$. For every $i = 1,...,r$ we choose $X_i \leq P^*$
minimal with respect to
$$B(S) \leq X_i\hbox{ and }\o X_i = K_i\o{B(S)}.$$
Then $X_i \in \P_H(B(S))$ and $\langle X_i,S\rangle = P$. Moreover $V_i =
[Y_{X_i},O^2(X_i)] = [Y_P,O^2(X_i)]$ since $Y_{X_i} \leq \O_1(Z(O_2(P))) = Y_P$.
Suppose that there is a non-trivial characteristic subgroup $A$ of $B(S)$,
which is normal in $X_1$. Then $\langle S,X_1,N_H(B(S))\rangle = \langle
P,N_H(B(S))\rangle \leq N_H(A)$, which contradicts $O_2(\langle P,N_H(B(S))\rangle = 1$.
Hence, no non-trivial characteristic subgroup of $B(S)$ is normal in $X_1$. Now
[Ste1] gives $[O_2(X_1),O^2(X_1)] = V_1 \leq Y_P$. Hence also $[O_2(P),O^2(P)] \leq Y_P$, and $Z(P)
= 1$ yields
$$Y_P = O_2(P) = V_1\times \cdots \times V_r.$$
Since $Q$ is transitive on $\{V_1,...,V_r\}$ and $N_H(B(S))$ does not
normalize $Y_P$ there exists $t \in N_H(B(S))$ such that $R := [V_1,V_1^t] \ne
1$. It follows that also $[V_1^t,V_1^{t^2}] \ne 1$, so
$$R^t = [V_1^t,V_1^{t^2}] = [V_1^t,V_1] = R$$
since $\langle V_1,V_1^{t^2}\rangle \leq B(S) \leq N_P(V_1^t)$.
As $t \in
\w C$ and $Y_P$ is normal in $Q$ the $Q$-transitivity gives
$$(*)\quad S_0 = Y_PY_P^t\hbox{ and } Y_P \cap Y_P^t = Z_0.$$
Let $U = N_H(R)$ and $W = O_2(U)$. Then $\langle t,X_2,...,X_r\rangle \leq U$,
and $V_i \cap W$ is $X_i$-invariant for every $i \geq 2$.
It follows that either there exists an $i \geq 2$ such that $V_i \leq W$, or
$V_i \cap W = 1$ for every $i \geq 2$. The
first case gives $V_i^t \leq W$ and so $V_i^t \leq O_2(X_2\cdots X_r)$. On the
other hand, by ($*$) $[Y_P,V_i^t] \ne 1$, so we get that $[V_i^t,V_1] =
R$. But this implies that $R \leq V_i^t$ and $R = R^t \leq V_i$, which is
impossible since $V_1 \cap V_i = 1$ for $i > 1$.
We have shown that $V_i \cap W = 1$ for $i > 1$. It follows that $[S_0\cap
W,O^2(X_2)] = 1$. Since $S_0 \cap W$ is normalized by $X_2$ and $W$ we get
$[(S_0 \cap X_2)^x,W] \leq S_0\cap W$ for every $x \in X_2$. Hence $[W,O^2(X_2)]\leq S_0 \cap W$ and $[W,O^2(X_2),O^2(X_2)] = 1$. But then $U$ is not of characteristic
2, a contradiction.
\bigskip
{\bf (4.3)} Let $S_0 \leq T$, $T$ a $2$-subgroup of
$H$. Then $S_0$ is normal in $N_H(T)$ and $N_H(T) \leq M \cap
\w C$.
\bigskip
Proof. Note that $N_H(T) \leq N_H(B(S)) \leq M$ by (3.1) and (4.2).
Moreover, by the Structure Theorem, case (b), $Y_P = Y_M$ and
$P^*C_M(Y_M)$ is normal in $M$, so $T \cap P^*C_M(Y_M) = S_0$.
Hence Theorem 3
gives $N_M(T) \leq N_M(S_0)\leq M \cap \w C$.
\bigskip
{\bf (4.4)} Let $\w L \in \L_H(S)$. Then either $\w L \leq \w C$, or $P \leq \w L \leq M$ and $F \not\leq
\w L$.
\bigskip
Proof. Assume that $\w L\not\leq \w C$. Then (1.3)(a) and the Corollaries 2 and 3 show that
$P \leq \w L \leq M$. If in addition $F \leq M$, then the Frattini argument and
(4.3)
imply that $\w C = FN_H(S_0) \leq M$, a contradiction.
\bigskip
{\bf (4.5)} Suppose that $S_0 \leq T \leq S$ such that $|S/T| = 2$ and $S =
TQ$. Let $T \leq L \leq H$ and $O_2(L) \ne 1$. Then one of the
following holds:
(a) $L \leq M$.
(b) $L \leq \w C$.
(c) $L \in \L_H(T)$.
Proof. Let $U = N_H(O_2(L))$ and $T \leq T_0 \in Syl_2(U)$. By (4.3) $T_0 \leq M \cap \w C$
and thus either $T = T_0$ or $T_0 \in Syl_2(\w C)$ and $Q \leq
T_0$. In the second case $T_0 = TQ = S$, and (4.4) yields $L \leq U \leq M$
or $L\leq U \leq \w C$. In the first case $U \in \L_H(T)$ and thus
also $L \in \L_H(T)$.
\bigskip
{\bf Notation.} From now on we fix a maximal subgroup $T$ of $S$ containing
$N_S(K_1)$. Recall that $B(S) \leq S_0 \leq T$.
Let $Q_0:= T \cap Q$ and
$$\L_0(T) := \{U \in \L_H(T) \mid U \not\leq \w C \hbox{ and }U \cap\w C\not\leq M\}.$$
By $\L_0(T)_*$ we denote the set of minimal elements of $\L_0(T)$.
{\bf (4.6)} Let $\o P^* := P^*/C_{P^*}(Y_P)$ and $1\ne K\leq O^2(\o P^*)$. Suppose
that $K$ is $Q_0$-invariant. Then $K = O^2(\o P^*)$ or $K = \times_{X
\in \O_i}X^\prime$ for some $T$-orbit $\O_i$ of $\O$; in particular $[K,\o
Q_0] \ne 1$.
\bigskip
Proof. Since $K \ne 1$ there exist $K_i \in \O$ and $t \in S_0\cap K_i$
such that $[K,\o t] = K_i^\prime$. Let $q \in Q$ such that $K_i^q \ne
K_i$, and let $q_0 := [t,q]$ and $R:=[K,\o q_0]$. Then $q_0 \in S_0 \cap Q
\leq Q_0$ and $R \leq (K_i\times K_i^q)\cap K$ with $[R,\o t]
= K_i^\prime$.
Since $r > 2$ there exists $x \in Q$ such that $K_i^x
\not\in\{K_i,K_i^q\}$. Let $x_0 = [t,x]$. Then as above $x_0 \in Q_0\cap S_0$, while
$\o x_0C_{\o S_0}(K_i\times K_i^q) = \o t C_{\o S_0}(K_i\times K_i^q)$. It follows that $[R,\o x_0] = K_i^\prime
\leq K$.
We have shown that $K_i^\prime \leq K$ for every $K_i \in \O$ such that
$[K,K_i] \ne 1$. Now the action of $Q_0$ on $K$ and $\O$ gives the desired
structure of $K$. Moreover, $r > 2$ implies that $[K,\o Q_0] \ne 1$.
\bigskip
{\bf (4.7)} $|S/T| = 2$, $S = TQ$, and $T$ has two orbits $\O_1$ and $\O_2$ on
$\O$ such that for $Z_i := C_{\O_1(Z(T))}(\O_i)$ the following hold:
(a) $|\O_i| = {r\over 2}$ and $|Z_i| = 2$, $i = 1,2$, and
(b) $\O_1(Z(T)) = Z_1 \times Z_2$.
\bigskip
Proof. This is a direct consequence of the choice of $T$.
\bigskip
{\bf (4.8)} $\L_0(T) \ne \emptyset$.
\bigskip
Proof. Let $L := C_H(Z_1)$, $Z_1$ as in (4.7). Then $L \not\leq \w C$,
and by (4.4) $L \cap \w C \not \leq M$ since $F \leq L \cap \w C$. Now
(4.5) shows that $L \in \L_0(T)$.
\bigskip
{\bf (4.9)} Let $L \in \L_0(T)$. Then $O_2(\langle O^2(P^*),L \cap \w C\rangle) = 1$.
\bigskip
Proof. Let $L_0 := \langle O^2(P^*),L\cap \w C\rangle$ and assume that
$O_2(L_0) \ne 1$. Let $t \in Q\setminus T$. Then
$T\langle t \rangle = S$ since $T$ has index 2 in $S$. Moreover, $[t,L
\cap \w C] \leq Q_0 \leq O_2(L\cap \w C)$. It follows that $t$
normalizes $L_0$. Hence $S \leq L_0\langle t \rangle$ and $1\ne O_2(L_0)
\leq O_2(L_0\langle t\rangle)$. This contradicts
(4.4) since $L_0 \not\leq M$ as $L\cap \w C \not\leq M$
and $L_0\not\leq \w C$ as $O^2(P^*) \not\leq \w C$.
\bigskip
{\bf Theorem 4.} Suppose that $L \in \L_0(T)$. Then
$$\P_{L}(T) = \P_{L\cap M}(T) \cup \P_{L\cap \w C}(T).$$
\bigskip
Proof. Assume that there exists $X \in \P_{L}(T)$ such that $X \not\leq M$
and $X \not\leq \w C$. By (4.2) and (1.3)(b) neither $B(S)$ nor $\O_1(Z(T))$ is normal in
$X$. Hence, (2.9) implies that there exists a minimal parabolic
subgroup $X_0$ of characteristic $2$ in $X$ such that
$X_0$
satisfies (2.9)(a) -- (e) (in place of $L_i$); in particular $X = \langle
T,X_0\rangle$, $O_2(X)B(S) \in Syl_2(X_0)$ and $X_0/C_{X_0}(Y_{X_0}) \cong
SL_2(2^k)$. We choose $X^* \leq X_0$ minimal with respect to
$$B(S) \leq X^*\hbox{ and } X_0 = X^*C_{X_0}(Y_{X_0}).$$
Then $X^*$ is a minimal parabolic subgroup and $X = \langle
X^*,T\rangle$. Moreover $B(S) \in Syl_2(X^*)$ by (2.7) applied to $X^*$.
Assume that there exists a non-trivial characteristic subgroup $A$ of
$B(S)$ which is normal in $X^*$. As $A$ is also characteristic in $S$
we get
$$(*)\quad X = \langle T,X^*\rangle \leq N_H(A) \hbox{ and }S \leq N_H(A).$$
Hence by (4.4) $N_H(A)\leq \w C$ or $N_H(A) \leq M$, which contradicts
$X \leq N_H(A)$.
Thus, no non-trivial characteristic subgroup of $B(S)$ is normal in $X^*$. As
$X^*$ is a minimal parabolic subgroup the hypothesis of
[Ste1] is satisfied. We get $[O^2(X^*),O_2(X^*)] =
[Y_{X^*},O^2(X^*)]$ and $Y_{X^*}/C_{Y_{X^*}}(X^*)$ is a natural
$SL_2(2^k)$-module for $X^*/C_{X^*}(Y_{X^*})$, so $[O^2(X^*),O_2(X^*)]
\leq Y_X$. Since $[O_2(X),B(S)] \leq B(S)\cap O_2(X) \leq O_2(X^*)$ we also
get
$$[O_2(X),O^2(X^*)] \leq Y_X\hbox{ and }[O_2(X),O^2(X)] \leq Y_X.$$
As in the proof of (4.9) pick $t \in Q\setminus T$. Then
$$(**)\quad [L\cap \w C,t] \leq Q\cap T\leq O_2(L\cap
\w C).$$
Assume first that $Y_X^t \leq O_2(X)$. Then
$$S \leq \langle X,t\rangle \leq N_H(Y_XY_X^t) \in
\L_H(S),$$
and by (4.4) $N_H(Y_XY_X^t)\leq M$ or $N_H(Y_XY_X^t)\leq \w C$. But
this contradicts $X \leq N_H(Y_XY_X^t)$.
We have shown that $Y_X^t \not \leq O_2(X)$.
As $|Y_X/C_{Y_X}(Y_X^t)| = |Y_X^t/C_{Y_X^t}(Y_X)|$ we get $Y_X^t \in {\cal
U}(X)$ (for the definition see section 2). Since $Y_X^t$ is normal in $T$ we
conclude with (2.1) that $Y_X^tO_2(X) = B(S)O_2(X)$. In addition, (2.1) shows
that $B(S)C_X(Y_X)/C_X(Y_X)$ is self-centralizing in $X/C_X(Y_X)$. It follows
that $O_2(X^t) \leq Y_X^tO_2(X)$. Hence, for $D := O_2(X) \cap O_2(X^t)$ we
get $O_2(X^t) = Y_X^tD$ and similarly $O_2(X) = Y_XD$. This gives
$$\Phi(O_2(X^t)) = \Phi(D) = \Phi(O_2(X));$$
in particular $\langle X,S\rangle \leq N_H(\Phi(D))$. Now as above (4.4)
implies that $\Phi(D) = 1$, so $O_2(X) = Y_X$ and $B(S) = Y_XY_X^t$.
The action of
$T$ on $B(S)$ shows that $Y_X$ and $Y_X^t$ are the only maximal
$T$-invariant elementary abelian normal subgroups of $B(S)$; in particular
$Y_X = Y_{L}$, and by ($**$) $L \cap \w C$ normalizes
$B(S)$. Now (4.2) yields $L \cap \w C \leq M$, which
contradicts $L \in \L_0(T)$.
\bigskip
{\bf (4.10)} Let $L \in \L_0(T)_*$
and $N$ be a normal subgroup of $L$ that is minimal with respect to $N
\not\leq \w C$. Then $N = [N,Q_0] = O^2(L)$.
\bigskip
Proof. As $N(L\cap \w C) \in \L_H(T)$ the minimality of $L$ yields $L =
N(L\cap \w C)$. Hence $N_0 := [N,Q_0]$ is a normal subgroup of
$L$. Assume that $N \ne N_0$. The the minimal choice of $N$ gives $N_0
\leq \w C$, so $N_0Q_0$ is a normal subgroup of $L$ in $\w C$. It
follows that $Q_0 \leq O_2(N_0Q_0) \leq O_2(L)$. But then $[Q,O_2(L)]
\leq Q_0\leq O_2(L)$ and $S = TQ \leq N_H(O_2(L))$, so (4.4) implies
that $L \leq \w C$ or $L \leq M$. This contradicts the definition
of $\L_0(T)$.
We have shown that $N = N_0$. The minimality of $N$ also gives that $N
= O^2(N)$. Thus, it remains to prove that $L = NT$. Assume now that $L \ne NT$.
By Theorem 4
$$\P_{NT}(T) \subseteq \P_M(T) \cup \P_{\w C}(T).$$
Since $NT \not\leq \w C$ the minimality of
$L$ shows that $NT \cap \w C \leq M$. Thus $\P_{NT}(T) \subseteq
\P_M(T)$. As by (4.3) also $N_L(T) \leq M$ we conclude from (1.3)(a)
that $NT \leq M$.
Now $N = [N,Q_0] \leq P$, and $N = O^2(N)$ implies $N \leq
O^2(P^*)$. Since $N$ is also $S_0$-invariant we get from (4.1) that
$[Z,N]$ is normal in $P^*$. On the other hand by (4.6) $[Z,N] = [Z,L]$, so $[Z,L]$ is normalized by $L$ and
$P^*$. But this contradicts (4.9).
\bigskip
{\bf Corollary 4.} Let $L \in \L_0(T)_*$. There exists a unique $P_1 \in \P_L(T)$ such that $P_1 \not \leq \w C$. Moreover, the
following hold:
(a) $Q_0 \not \leq O_2(P_1)$,
(b) $O^2(P_1) \leq O^2(P^*)$, and
(c) $O^2(P_1)C_{P^*}(Y_P)/C_{P^*}(Y_P) = K_1^\prime \times \cdots \times
K_s^\prime$, where $\{K_1,...,K_s\}$ is a $T$-orbit of $\O$.
\bigskip
Proof. By (4.3) $N_L(T) \leq \w C$, so by (1.3)(a) there exists $P_1\in
\P_L(T)$ such that $P_1 \not\leq \w C$.
Now Theorem 4 gives $P_1 \leq M$ and again by (4.3) $S_0 \not \leq O_2(P_1)$. Since
$P^*C_M(Y_P)$ is normal in $M$ we get from (1.3)(c) that
$O^2(P_1) = [O^2(P_1),S_0] \leq P^*C_M(Y_P)$.
Let $\o M := M/C_M(Y_P)$. Note that $O^2(\o P_1) \ne 1$ and by (4.1) (a) and
(c) hold. By (a) and (1.3)(c)
$O^2(P_1) = [O^2(P_1),Q_0] \leq [M,Q] \leq M^0 \leq P$,
so also (b) holds.
Let $P_0$ be another minimal parabolic in $\P_L(T)$, which is not in $\w
C$. Then (a) -- (c) hold for $P_0$ in place of $P_1$. By
(4.6) either
$$O^2(P_0)O^2(P_1)C_{P^*}(Y_P) = O^2(P^*)C_{P^*}(Y_P)\hbox{ or }
O^2(P_0)C_{P^*}(Y_P) = O^2(P_1)C_{P^*}(Y_P).$$
Note that $[C_{P^*}(Y_P),Q_0] \leq O_2(P) \leq T$ and $Q_0$ is normal in
$S$. Hence, in the first case (1.3)(c) implies that $O^2(P^*)
= [O^2(P^*),Q_0] \leq O^2(P_0)O^2(P_1)O_2(P^*) \leq L$, which
contradicts (4.9). In the second case we conclude that $O^2(P_0)O_2(P) = O^2(P_1)O_2(P)$ and thus
$O^2(O^2(P_0)O_2(P)) = O^2(P_0) = O^2(P_1)$. Hence $P_0 = P_1$.
\bigskip
{\bf (4.11)} Let $X$ be a finite group and $V$ a faithful
$GF(2)X$-module, and let $S \in Syl_2(X)$ and $V_0 = C_V(S)$. Suppose that
$F^*(X)$ is simple, $V = \langle V_0^X\rangle \ne V_0$, and
($*$) there exists an elementary abelian subgroup $1\ne A \leq S$ such that
$|V/C_V(A)| \leq |A|$.
\noindent Then there exists a minimal parabolic subgroup $P_1$
containing $S$ such that $P_1 \not \leq C_X(V_0)$ and $(P_1\cap F^*(X))/O_2(P_1\cap
F^*(X)) \cong
SL_2(2^k)$ or $S_\ell$.
\bigskip
Proof. A theorem of Gasch\"utz (see for example [Hu, I.17.4]), applied to the
semidirect product of $V$ with $X$, shows that $V = C_V(X)[V,X]$. Hence, there
exists a $X$-submodule $W$ such that $\o V := V/W$ is a faithful irreducible
$X$-module. Moreover, property ($*$) implies that$|\o V/C_{\o V}(A)| \leq |A|$. Thus, the F-Module Theorem
for $\cal K$-groups, see [GM1] and [GM2], gives the conclusion.
\bigskip
{\bf F!-Theorem.} No group satisfies the hypothesis of this section.
\bigskip
Proof. We will derive a contradiction using the previous results of
this chapter. According to (4.8) there exists $L \in \L_0(T)_*$. We
fix the following additional notation:
$$C_L = L \cap \w C,\,V = \langle Z^L \rangle,\,\o L = L/C_L(V).$$
As in Corollary 4 let $P_1$ be the unique element of $\P_L(T)$ with
$P_1 \not\leq \w C$. Then
(1) $O^2(P_1) \leq O^2(P^*)$ and $O^2(P_1)C_{P^*}(Y_P)/C_{P^*}(Y_P) =
K_1^\prime\times \cdots \times K_s^\prime$,
\noindent where $\O_1:=\{K_1,...,K_s\}$ is one of the two $T$-orbits
of $\O$. From (1.3)(b) and (1) we get
$$O^2(P_1) \cap C_{P^*}(Y_P) = O_2(O^2(P_1)) \geq O^2(P_1) \cap C_L(V),$$
in particular
($*$) $O^2(\o P_1)/O_2(O^2(\o P_1)) = K_1^\prime\times \cdots \times
K_s^\prime$.
As in (4.10) let $N$ be a normal subgroup of $L$ that is minimal
with respect to $N \not\leq C_L$. Then by (4.10)
(2) $N = [N,Q_0] = O^2(L)$.
Moreover, since by (4.1) every normal subgroup of $L$ in $C_L$
centralizes $V$ we get
(3) $\o N$ is a minimal normal subgroup of $\o L$, and $O_2(\o L) = 1$.
Next we show:
(4) $C_L(V) \leq M$, in particular $L \ne (L\cap M)C_L(V)$.
Assume that $C_L(V) \not\leq M$. Then the minimality of $L$ yields $L
= C_L(V)P_1$. It follows from (2) that
$$N = N \cap (O^2(P_1)C_L(V)) = O^2(P_1)(N\cap C_L(V))$$
and
$$L = NT = [N,Q_0]T = O^2(P_1)T = P_1 \leq M,$$
which contradicts the choice of $L$ in $\L_0(T)$.
(5) $\o N \cap \o T \ne 1$; in particular $\o N$ is not abelian.
Assume that $\o N \cap \o T = 1$. For every prime $q$ the Frattini
argument gives a $\o Y_q \in Syl_q(\o N)$ such that $\o T \leq N_{\o
U}(\o Y_q)$ and $\o N = \langle \o Y_q\mid q \in \pi(\o N)\rangle$.
Let $Y_q$ be the inverse image of $\o Y_q$ in $L$.
From (1), ($*$) and (1.3)(a) we get that $Y_q \leq C_L$ for every
$q\ne 3$. Hence $\o N = \o Y_3C_{\o N}(\o Q_0)$, so by (2)
$$\o N = [\o N,\o Q_0] = [\o Y_3C_{\o N}(\o Q_0),\o Q_0] = [\o Y_3,\o
Q_0] \leq \o Y_3.$$
Now (3) shows that $\o N$ is elementary abelian, moreover $\o N =
O^2(\o P_1)$. Thus (4) gives $L \leq M$, a contradiction. Hence, (5) is
proved.
Let $\O_2$ be the $T$-orbit of $\O$ different from $\O_1 =
\{K_1,...,K_s\}$. Then by (4.7)
$$\O_1(Z(T)) = Z_1 \times Z_2,\, Z_i := C_{Y_P}(\O_i),$$
and $P_1 \leq L_1 := C_L(Z_2)$.
Assume that $L_1 \cap \w C \leq M$. Then $L \cap F \leq L_1 \cap \w C
\leq M$ since $\O_1(Z(T)) \leq Z_0$. Now (4.3) and the Frattini
argument imply $C_L \leq N_{C_L}(S_0)(L \cap F) \leq M$, which
contradicts the choice of $L \in \L_0(T)$. Thus $L_1 \cap \w C
\not\leq M$, and the minimality of $L$ yields:
(6) $Z_2 \leq Z(L)$, in particular $O^2(P^*) \not\leq L$.
Next we show:
(7) $\o N$ is simple.
According to (3) and (5) there exist subgroups $C_L(V) \leq N_i \leq
NC_L(V)$, $i = 1,...,k$ such that $\o N = \o N_1 \times \cdots \times
\o N_k$, and $\o N_1,...,\o N_k$ are simple groups conjugate under $\o
T$.
Assume first that $\o N_i
\cap \o C_L \leq \o T$, $i = 1,...,k$. The projection $\o C_i$ of $\o N\cap \o
C_L$ in $\o N_i$ is a subgroup of $\o N_i$ that normalizes $\o N_i \cap \o
T$. Hence by (5) $\o C_L(\o C_1\times \cdots \times \o C_k)$
is a proper subgroup of $\o L$, and the minimality of $L$ implies that
$\o C_i \leq \o C_L \cap \o N_i$, so $\o N \cap \o T = \o N\cap \o C_L$.
Now (4) yields $C_L \leq M$, which contradicts the choice of $L \in
\L_0(T)$.
Assume now that there exists a component $\o N_1$ such that $\o N_1
\cap \o C_L$ is not a $2$-group. Then $O^2(\o N_1 \cap \o C_L) =
O^2((\o N_1\cap \o C_L)O_2(\o N\cap \o C_L)) \ne 1$ and
$$[\o N_1\cap \o C_L,\o Q_0] \leq O_2(\o C_L) \cap \o N \leq O_2(\o N \cap \o
C_L),$$
so $\o Q_0$ normalizes $O^2(\o N\cap \o C_L)$ and thus also $\o N_1$.
It follows:
($**$) $\o Q_0$ normalizes every component of $\o N$.
\noindent Among all $T$-invariant subgroups $U \leq N$ satisfying
(i) $\o U = \o U_1 \times \cdots \times \o U_k$, $U_i \leq N_i$, and
(ii) $O^2(P_1) \leq U$
\noindent we choose $U$ to be minimal. Then $\o{U \cap N_i}$ is the
projection of $O^2(\o P_1)$ into $\o N_i$. From ($*$) and (3) we
conclude that $UT \ne L$. The minimality
of $L$ implies that $UT\cap \w C \leq M$ and thus by (1.3)(a) and
(4.3) $UT \leq M$. On the other hand the minimality of
$U$ yields $U = [U,Q_0] = O^2(U)$. It follows that $U$ is a
$Q_0$-invariant subgroup of $O^2(P^*)$. Now (4.6) and (6) show that
$$\o U = [\o U,\o Q_0] = O^2(\o P_1) = \o U_i \times \cdots \times \o
U_k.$$
By ($**$) $\o U_1$ is $Q_0$-invariant. Hence,
another application of (4.6) shows that $O^2(\o P_1) \leq \o N_1$. As
$O^2(\o P_1)$ is $\o T$-invariant, also $\o N_1$ is. Since the groups
$N_1,...,N_k$ are conjugate under $T$ we conclude that $k = 1$.
(6) $J(S) \not \leq C_L(V)$.
Assume that $J(S) \leq C_L(V)$. Then $V \leq \O_1(Z(J(S)))$ and thus also
$B(S) \leq C_L(V)$. Now the Frattini argument and (4.2) yield $L =
N_L(B(S))C_L(V) = (L\cap M)C_L(V)$, which contradicts (4).
We now derive a final contradiction. According to (8) there exists $A \in {\cal A}(S)$ such that $\o A \ne
1$. Hence, the maximality of $A$ implies that $|V/C_V(\o A)| \leq |\o A|$, so
by (7) we can apply (4.11) to $\o L$. Thus, there exists $C_L(V)T \leq P_0 \leq L$ such
that $\o P_0$ is a minimal parabolic subgroup of $\o L$, $\o P_0
\not\leq C_{\o L}(V_0)$, where $V_0 := C_V(T) =\O_1(Z(T))$, and
$$(***)\quad (\o P_0\cap \o N)/O_2(\o P_0\cap \o N) \cong SL_2(2^k) \hbox{ or }S_\ell.$$
Since by (6) $V_0 = Z_2 \times Z \leq Z(L)Z$ we get $C_L = C_L(V_0)$, $P_0
\not\leq C_L$ and $P_1 \leq P_0$. Now ($*$) and ($***$) show that $s =
1$ and $r = 2$, which contradicts $r \geq 4$.
\bigskip
{\bf The proof of the $P!$-Theorem and the Corollary.} Let $P \leq M \in \L_H^*(S)$. Then the
$F!$-Theorem and Corollary 3 show that case (a) of the Local $P!$-Theorem
and case (a) of the Structure Theorem hold for $M$. The $P!$-Theorem now
follows from Theorem 2 and Theorem 3.
For the proof of the Corollary let $L \in Loc_H(P)$. We may assume that
$C_H(Y_L) \leq L$. By (1.5) there exists $M \in \L_H^*(S)$ such that
$$P = P^0S \leq L^0S \leq M.$$
Hence, $M$ satisfies case (a) of the Structure Theorem. In particular, we get
from the structure of $M/C_M(Y_M)$ and its action on $Y_M$:
(i) $(L\cap M_0)/C_{L\cap M_0}(Y_L) \cong SL_k(p^m)$ or $Sp_{2k}(p^m)$, and
$[Y_L,L\cap M_0]$ is the corresponding natural module.
(ii) $L_0 = (L\cap M_0)C_S(Y_L)$ and $C_{L_0}(Y_L) = C_S(Y_L)C_{L_0}(Y_M)$.
\noindent This gives claim (a) of the Corollary.
Assume that $C_{L_0}(Y_L) \ne O_p(L_0)$. Then $C_{L_0}(Y_M) \ne O_p(M_0)$, and
we get $M_0/O_2(M_0) \cong Sp_4(2)^\prime$ (and $p = 2$). But then $L_0 = M_0$
since otherwise $L^0/O_2(L^0) \cong SL_2(2)$ and $C_{L_0}(Y_L) = O_2(L_0)$.
\vfill \eject
{\bf References}
\bigskip
[Bau] B. Baumann, \"Uber endliche Gruppen mit einer zu $L_2(2^n)$ isomorphen
Faktorgruppe, Proc. AMS 74 (1979), 215 -- 222.
[Cher] A. Chermak, Quadratic modules of degree less than 2 for finite groups
of intrinsic rank 1, Preprint (1997).
[DS] A. Delgado, B. Stellmacher, Weak (B,N)-pairs of rank 2, in
``Groups and Graphs: New results and Methods'', Birkh\"auser, 1985.
[Glau] G. Glauberman, Weakly closed elements of Sylow subgroups, MZ 107
(1968), 1 -- 20.
[GM1] R. Guralnick, G. Malle, Classification of 2F-modules, I, to
appear J. Alg.
[GM2] R. Guralnick, G. Malle, Classification of 2F-modules, II, in preparation.
[Hu] B. Huppert, Endliche Gruppen I, Springer 1967.
[KS] H. Kurzweil, B. Stellmacher, Theorie der endlichen Gruppen,
Springer-Verlag, 1998.
[MSS1] U. Meierfrankenfeld, G. Stroth, B. Stellmacher, Finite groups
of local characteristic $p$, an overview, Proc. Durham Conference (2001).
[MSS2] U. Meierfrankenfeld, G. Stroth, B. Stellmacher, The Structure Theorem,
Preprint 2002.
[Ste1] B. Stellmacher, Pushing Up, Arch. Math. 46 (1986), 8 -- 17.
[Ste2] B. Stellmacher, On the $2$-local structure of finite groups, in "Groups,
Combinatorics and Geometry" (Cambridge University Press, 1992), 159 -- 182.
\end
%%% Local Variables:
%%% mode: plain-tex
%%% TeX-master: "publication"
%%% End:
%%% Local Variables:
%%% mode: latex
%%% TeX-master: "pub"
%%% End: