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\author{ Ulrich Meierfrankenfeld}
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\title{Locally Finite, Simple Groups}
\bibliographystyle{abbrv}
\begin{document}
\maketitle
\tableofcontents
\chapter{Local Properties}\l{local}
\begin{definition}\l{locally}\index{locally} Let $\cal C$ be a class of groups and $G$ a
group. Then $G$ is called a {\em locally $\cal C$-group}, provided that for each finite
subset
$X$ of $G$ there exists $Y\leq G$ with $X\subseteq Y$ and $Y\in {\cal C}$.
\end{definition}
For example a locally abelian group is
abelian as any two elements of $G$ have to lie in an abelian subgroup of $G$
and so commute. A less trivial observation is:
\begin{lemma}\l{ls=s} Let $G$ be a locally simple group. Then $G$ is simple.
\end{lemma}
\proof Let $N$ be a non-trivial normal subgroup of $G$ and $1\neq x\in N$. We
need to show that $G=N$. For this let $y \in G$. Since $G$
is locally simple there exists a simple subgroup $S$ of $G$ so that $x,y\in S$.
As
$x\in S\cap N$, $S\cap N$ is a non-trivial normal subgroup of $S$ and so
$S=S\cap N\leq N$. Thus $y\in N$. Now $y$ was arbitrary and so $G=N$\qed\bs
Another explanation why "simple" is a local property is as follows. A group
$G$ is simple if and only if $G=\langle x^G\rangle$ for all $1\neq x\in G$.
This is true if and only if $y\in \langle x^G\rangle$ for all $y\in G$. This
in turn just means that $y$ can be written as a product of finitely many
conjugates of $x$ and $x^{-1}$. But the last statement just involves finitely
many elements of $G$ and so is a local property.
\bs
The next lemma introduces a fairly general method to construct locally ${\cal
C}$-groups.
\begin{lemma}\l{clcg} Let $\cal C$ be a class of groups and let
%
$$G_1\leq G_2\leq G_3\leq\ldots\leq G_i\leq G_{i+1}\leq \ldots$$
%
be an ascending chain of ${\cal C}$-groups .
Put $G=\bigcup_{i=1}^\infty G_i$. Then $G$ is a locally ${\cal C}$-group.
\end{lemma}
\proof Let $X$ be a finite subset of $G$. As $G$ is the union of the $G_i$, for
each $x\in X$ there exists $i_x\geq 1$ with $x\in G_{i_x}$. Put
$i=\max\{i_x\mid x\in X\}$. Then $x\in G_{i_x}\leq G_i$ and so $X\subseteq G_i$.
Since $G_i$ is a $\cal C$ group, this proves that $G$ is locally a ${\cal
C}$-group.\qed \bs
For example the infinite cyclic group $(\mathbb{ Z},+)$ has lots of subgroups
isomorphic to itself and we easily obtain a chain of groups as in \ref{clcg}
with each of the $G_i$'s isomorphic to $\mathbb{ Z}$ and properly contained in
$G_{i+1}$. The resulting group $G$ is locally cyclic, but (as any possible
element of $G$ lies in one of the $G_i$'s and so generates a proper
subgroup), $G$ is not cyclic.
A more concrete example of a locally cyclic but not cyclic group is $({\cal
Q},+)$. The reader might also convince herself that our first
example of such a locally cyclic, but not cyclic group is isomorphic to a subgroup of
${\cal Q}$. For example if $|G_{i+1}/G_{i}|=2$ for all $i$, then $G$ is isomorphic to
$\{\frac{n}{2^k}\mid n\in \mathbb{ Z}, k\in {\cal N}\}$.\bs
To obtain an example of an infinite, locally finite, locally cyclic, groups,
let $p$ be prime and choose $G_i\cong C_{p^i}=\mathbb{ Z}/p^i\mathbb{ Z}$.
The resulting group $G$ is called the Pr\"ufer $p$-group and will be denoted by
$C_{p^\infty}$ or $\mathbb{ Z}_{p^\infty}$.\footnote{The reader may check
that the Pr\"ufer $p$-groups are subgroups of the multiplicative group of
complex numbers with norm equal 1}\bs
Next let $\Omega_i=\{1,2,\ldots,i\}$. Then we can identify
$\Sym(\Omega_i)$ with the subgroup of $\Sym(\Omega_{i+1})$ fixing $i+1$.
Application of \ref{clcg} yields a locally finite group $G=\bigcup_{i=1}^\infty
\Sym(\Omega_i)$. Note that we can view $G$ as a subgroup of
$\Sym(\Omega)$, where $\Omega=\{1,2,3,\ldots \}$. But $G$ is not the full
symmetric group, indeed
%
$$\ldots\ra 8 \ra 6\ra 4\ra 2\ra 1\ra 3\ra 5\ra 7\ra 9\ra\dots$$
%
is an element of infinite order in $\Sym(\Omega)$ and so cannot be contained in
$G$. Note also that the elements in $G$ fix all but finitely many
elements of $\Omega$. Conversely, if $\pi\in \Sym(\Omega)$ fixes all but
finitely many elements of $\Omega$, then $\pi\in G_i$, where $i$ is the
largest positive integer not fixed by $\pi$. For a set $\Delta$ define the
finitary symmetric group on $\Delta$ by
$$\FSym(\Delta)=\{\pi\in \Sym(\Delta)\mid |\Supp(\pi)|<\infty \}.$$
With this notation $G$ is $\FSym(\Omega)$. The groups
$\FSym(\Delta)$ are always locally finite and we have found
locally finite groups of arbitrary cardinality. $\FSym(\Delta)$ is a normal
subgroup of $\Sym(\Delta)$ and has as a normal subgroup the alternating group
$\Alt(\Omega)$, which consists of those permutations which are
product of an even number of 2-cycles. Also it might be interesting to notice that
for infinite $\Delta$, $|\FSym(\Delta)|=|\Delta|$ while $|\Sym(\Delta)|=2^{|\Delta|}$.
\bs
Next we use \ref{clcg} one more time to construct one of the most
fascinating locally finite groups: P.Hall's universal locally finite group
$U$.
For this let $n_1=3$ and put $G_1=\Sym(3)$. Let $n_2=|G_1|=n_1!$ and
$G_2=\Sym(n_2)=\Sym(G_1)$.
View $G_1$ as a subgroup of $G_2$ via the regular permutation
action. Inductively put $n_{k+1}=|G_k|=n_k!$,
$G_{k+1}=\Sym(n_{k+1})=\Sym(G_k)$ and view $G_k$ as a subgroup of $G_{k+1}$ via
the regular permutation. Put $U=\bigcup_{i=1}^\infty G_i$. Then $U$ is a locally
finite group. We will study $U$ in more details later on but for now let us
notice that as $U$ contains symmetric groups of arbitrary large finite groups,
$U$ contains an isomorphic copy of any finite group.
\chapter{ P.Hall's universal locally finite group}\l{PHall}
\begin{definition}\l{universal} A locally finite group $G$ is called universal
provided that $G$ fulfills the following property:\smallskip
(Uni)\, For all finite groups $F$ and $E$ and embeddings $\alpha: F \ra G$ and
$\beta: F \ra E$ there exists an embedding $\gamma: E\ra G$ with
$\alpha=\gamma\beta$.
\end{definition}
The goal of this chapter is to show that P.Hall's group $U$ is up to
isomorphism the unique countable, universal, locally finite group.
\begin{definition}\l{semiregular} Let $G$ be a group acting on a set $\Omega$.
Then we say that
$G$ acts semi-regularly on $\Omega$ provided that $C_G(\omega)=\{ g\in G\mid
\omega^g=\omega\}=1$ for all $\omega\in \Omega$. $G$ acts regularly on $\Omega$ if $G$
acts transitively and semi-regularly on $\Omega$.
\end{definition}
The proofs of the next three lemmas are left to the reader.
\begin{lemma}\l{esr} Let $G$ be acting on a set $\Omega$. Then the following
three statements are equivalent:
\begin{itemize}
\AI{a} $G$ acts semi-regularly on $\Omega$.
\AI{b} $G$ acts regularly on each of its orbits on $\Omega$.
\AI{c} Each orbit for $G$ on $\Omega$ is isomorphic to the action of $G$ on $G$
by right multiplication.\qed
\end{itemize}
\end{lemma}
\begin{lemma}\l{ssr} Let $G$ be a group acting semi-regularly on a set $\Omega$
and $H\leq G$. Then also $H$ acts semi-regularly on $\Omega$.\qed
\end{lemma}
\begin{lemma}\l{isr} Let $G$ be acting semi-regularly on the sets $\Omega_1$ and
$\Omega_2$. Then $\Omega_1$ and $\Omega_2$ are isomorphic as $G$-sets if and
only if $|\Omega_1|=|\Omega_2|$.\qed
\end{lemma}
\begin{proposition}\l{phu} P.Hall's group $U$ is universal.
\end{proposition}
\proof Let $F$ and $E$ be finite groups and $\alpha: F \ra U$ and $\beta: F
\ra E$ be embeddings. Also let $G_1< G_2 < \ldots < G_i< \ldots$ be the
ascending chain of finite symmetric groups used to construct $U$. Since
$\alpha(F)$ is finite, there exists $j$ with $\alpha(F)\leq G_j$. Moreover,
pick $k$, $n_k\geq |E|$ and $k\geq j$. Then $\alpha(F)\leq G_k$ and there
exists an embedding $\gamma^*: E\rightarrow G_k$. Let $i=k+1$ and $\Omega=G_k$. Then
$G_i=\Sym(\Omega)$. By construction $G_k$ acts regularly on $\Omega$ and
so by \ref{ssr} we see that both $\alpha$ and $\gamma^*\beta$ define
semi-regular actions of $F$ on $\Omega$. By \ref{isr} these actions are
isomorphic. But that just means that there exists $g\in \Sym(\Omega)=G_i\leq
U$ so that $\alpha=\mbox{\rm Inn}(g) \gamma^*\beta$. Putting
$\gamma=\mbox{\rm Inn}(g) \gamma^*$ we see that (Uni)
in Definition \ref{universal} is fulfilled.\qed
\begin{proposition}\l{uulg} Let $X$ and $Y$ be countable, universal, locally
finite groups. Then $X$ and $Y$ are isomorphic.
\end{proposition}
\proof Since $X$ and $Y$ are countable, $X=\{ x_1,x_2,x_3,\ldots\}$ and $Y=\{
y_1,y_2,y_3,\ldots\}$. Put $X_i=\langle x_1,x_2,\ldots, x_i\rangle$ and
$Y_i=\langle y_1,y_2,\ldots, y_i\rangle$ and note that $X_i$ and $Y_i$ are
finite subgroups of $X$ and $Y$, respectively. Put $X_0^*=1=Y_0^*$. By induction
we will define finite subgroups $ X_i^*\leq X$ and $Y_i\leq Y_i^*$ and
embeddings
$\phi_i: X_i^*\ra Y$ and $\psi_i: Y_i^*\ra X$ so that for all $i >0$
\begin{itemize}
\NI{1} $X_i^*=\langle \psi_{i-1}(Y^*_{i-1}), X_i\rangle$
\NI{2} $\phi_i\circ \psi_{i-1}$ is the identity on $Y_{i-1}^*$.
\NI{3} $\phi_i$ equals $\phi_{i-1}$, when restricted to $X_{i-1}^*$
\NI{4} $Y_i^*=\langle \phi_{i}(X^*_{i}), Y_i\rangle$
\NI{5} $\psi_i\circ \phi_{i}$ is the identity on $X_i^*$.
\NI{6} $\psi_i$ equals $\psi_{i-1}$, when restricted to $Y_{i-1}^*$
\end{itemize}
For $i=0$ there is nothing to do. So assume such groups and maps exist for
$i-1$. Then use 1. as the definition of $X_i^*$. Let $F=\psi_{i-1}(Y^*_{i-1})$
and $\alpha: F\ra Y^*_{i-1}\leq Y$ the inverse map of $\psi_{i-1}$. Let $\beta$
be the inclusion map from $F$ to $X_i^*.$. As $Y$ is universal there exist
$\gamma: X_i^*\ra Y$ with $\gamma\beta=\alpha$. But this just means that
$\gamma$ restricted to $F$ is $\alpha$ and so 2. holds with
$\phi_i=\gamma$. In particular, $\phi_i$ is the inverse of $\psi_{i-1}$ on
$X_{i-1}^*$. By 5. the same is true for $\phi_{i-1}$ and so 3. holds.
4., 5. and 6. are done in a similar way.
Define $\phi: X\ra Y$ by $\phi(x_i)=\phi_i(x_i)$ and $\psi:Y\ra X$ by
$\psi(y_i)=\psi_i(y_i)$. Then it is easy to check that $\psi$ and $\phi$ are
group homomorphisms which are inverse to each other.\qed
\chapter{ Kegel-covers of locally finite, simple groups}\l{Kegel}
>From now on $G$ will always be a locally finite group and ${\cal F}$ the set of
finite subgroups of $G$, i.e the set of finitely generated subgroups of $G$.
\begin{definition}\l{kegelcover}
\begin{itemize}
\AI{a} A set of pairs
$\{(H_i,M_i)| i
\in I\}$ is called a {\em sectional cover} for $G$ if, for all $i$ in $I$, $H_i\in
{\cal F} $ and $M_i$ is a normal subgroup of $H_i$, and if, for
each $H\in {\cal F}$ there exists $i$ in $I$ with $H \leq H_i$ and
$H \cap M_i = 1$. The groups $H_i/M_i$, $i \in I$, are called the {\em factors} of
the sectional cover.
\AI{b} A {\em Kegel-cover} for $G$ is a sectional cover all of whose factors are
simple groups.
\end{itemize}
\end{definition}
\begin{definition}\l{gi} $G^\infty$ is subgroup of
$G$ generated by the all the perfect finite subgroups of $G$. $G$ is called {\em
absolutely perfect} if
$G=G^\infty$.
\end{definition}
Note that $G^\infty$ is the smallest normal subgroup $N$ of $G$ so that $G/N$ is
locally solvable. Also for finite $G$, $G^\infty$ is just the last term of the derived
series of $G$ and $G$ is perfect if and only if its is absolutely perfect.
\begin{lemma}\l{psc} Let $\{(G_i,N_i)| i \in I\}$ a
sectional cover for $G$. Suppose that $I=\bigcup_{t=1}^k I_t$. Then for at least one
$1\leq t\leq k$, $\{(G_i,N_i)| i \in I_t\}$ is a sectional cover for $G$.
\end{lemma}
\proof Suppose not. Then for each $t$ there exists a finite subgroup $F_t\leq G$
so that for all $i\in I_t$ with $F_t\leq G_i$ we have $F_t\cap N_i\neq 1$. Define
$F=\langle F_t\mid 1\leq t\leq k\rangle$. By the definition of a sectional cover there
exists
$i\in I$ with $F\leq G_i$ and $F\cap N_i=1$. But then $F_t\leq G_i$ and $F_t\cap
M_i=1$. Hence $i\not\in I_t$, a contradiction to $I=\bigcup_{t=1}^k I_t$.\qed
\begin{lemma}\l{elementarykegel} Let $G$ be a locally finite, simple group
and
$\{(G_i,N_i)| i \in I\}$ a sectional cover for $G$.
\begin{itemize}
\AI{a} There exists a Kegel cover $\{(H_j,M_j)| j \in J\}$ such that
for all
$j
\in J$ there exists $i \in I$ with $N_i \leq M_j\unlhd H_j \unlhd G_i$.
\AI{b} For $i \in I$ let $M_i$ be a normal subgroup of $G_i$. Then at
least one of $\{(G_i,M_i)| i \in I\}$ and $\{(M_i, M_i \cap N_i)|
i \in I\}$ is a sectional cover for $G$.
\AI{c} Suppose $G$ is not cyclic of prime order.Then $\{(G_i^\infty, G_i^\infty
\cap N_i) | i \in I \}$ is a sectional cover for $G$.
\AI{d} Suppose $G$ is not cyclic of prime order. Let $\cal E$ be a class of groups such
for each
$i\in I$ and each non-abelian composition factor $K$ of $G_i/N_i$ one has $K \in {\cal
E}$. Then there exists a Kegel cover for $ G$ all of whose factors are in $\cal E$.
\end{itemize} \end{lemma}
\proof (a) Let $E$ be a non trivial finite subgroup of $G$ and $1 \neq e
\in E$.
As a first step we show there exists a finite subgroup $T$ of $G$
with $E\leq T$ and $E\leq \langle e^{\langle e^T\rangle }\rangle$ for all $1\neq
e\in E$. Indeed since $G$ is simple, $ E \leq \langle e^G\rangle$ and since
$G$ is locally finite, $E \leq \langle e^{F_e}\rangle$ for some finite subgroup $F_e$
of $G$. Similarly $F_e \leq \langle e^{T_e}\rangle$ for some finite subgroup $T_e$
of $G$. Then $E \leq \langle e^{\langle e^{T_e}\rangle}\rangle$. Now just let $T$ be
the finite subgroup of $G$ generated by $E$ and all the $T_e$, $1 \neq e \in
E$ and the first step is completed.
Pick $i\in I$ with $T \leq G_i$ and $T\cap N_i = 1$. Put $H_E = \langle
E^{G_i}\rangle N_i$.
Our second step is to find a maximal normal subgroup $M_E$ of
$H_E$ with $N_i\leq M_E$ and $E\cap M_E=1$. For this let ${\cal R}$ be the set of the
maximal normal subgroups of $H_E$ containing $N_i$ and
$R=\bigcap {\cal R}$. Then
$G_i$ leaves ${\cal R}$ invariant and so $R$ is a normal subgroup of $G_i$ with $R<
H_E$. Suppose that $E\leq R$. Then $H_E=\langle E^{G_i}\rangle N_i \leq R$, a
contradiction. Thus $E\not\leq R$. Thus there exists $M_E\in {\cal R}$ with
$E\not\leq M_E$. To complete the second step it remains to show that
$E\cap M_E=1$. Otherwise pick $1 \neq e \in E\cap M_E $. Then
%
$$E \leq \langle e^{\langle e^{T}\rangle}\rangle\leq\langle e^{H_E}\rangle \leq
M_E,$$
%
a contradiction.
It follows now immediately from the second step that
$\{(H_E,M_E) \mid E\in{\cal F}\}$ is a Kegel cover that fulfills (a).
(b) Assume that $\{(G_i,M_i)| i \in I\}$ is not a
sectional cover for $G$. Then there exists a $H\in {\cal F}$ such
that $H \cap M_i \neq 1$ for all $i\in I$ with $H \leq G_i$. Without loss $H
\leq G_i$ for all $i\in I$. For $1\neq h \in H$ let $I_h = \{ i \in I | h \in
M_i \}$. Then $I$ is the finite union of these $I_h$ and so by \ref{psc} there exists
$1\neq h \in H $ such that $\{(G_i,N_i)| i \in I_h\}$ is a sectional
cover for $G$. Hence we may assume that $I = I_h$. Let $E$ be any finite subgroup of
$G$. Since $G$ is $LFS$, there exists a finite subgroup $T$ in $G$ with $E\leq T$ and
$E \leq \langle h^T\rangle$. Pick $i \in I$ with $T \leq G_i$ and
$T\cap N_i = 1$. Then $E \leq \langle h^T\rangle \leq M_i$ and
$E\cap (N_i \cap M_i)\leq T\cap N_i=1$. Thus $\{(M_i, M_i \cap N_i)| i \in I\}$ is a
sectional cover for $G$.
(c) Otherwise we conclude from (b) that $\{(G_i, G_i^\infty )| i \in I \}$ is
a sectional cover for $G$. Hence by (a) $G$ has a Kegel cover all of whose
factor are of prime order. Hence every finite subgroup of $G$ is embedded into a
group of prime order and so $G$ itself has prime order.
(d) By (a) there exists a Kegel cover all of whose factors have prime order or
lie in ${\cal E}$. Thus by \ref{psc} there exists a Kegel cover all of whose factors
have prime order or there exists a Kegel cover all of factors are in ${\cal E}$. In the
first case $G$ itself has prime order and (d) is proved.\qed
\bs
We remark that there exists examples of locally finite groups which do
have a Kegel cover, but are not simple.
\begin{corollary}\l{lss} Non-cyclic, locally finite simple groups are absolutely
perfect. In particular, a locally solvable, locally finite, simple group is cyclic of
prime order.
\end{corollary}
\proof Observe that every locally finite simple groups has a sectional cover, for
example $\{ (F,1)\mid F\in {\cal F}\}$. Hence the corollary follows from
\ref{elementarykegel}c.\qed
\chapter{Serial Subgroups}\l{Serial}
The main goal of this section is to show that a subgroup of $G$ is serial if and only
if its locally subnormal.
For the next definition recall that a directed set is a partially ordered set so
that each two elements have an upper bound. For example ${\cal F}$ is a directed set
under inclusion.
\begin{definition}\l{localsystem} Let $I$ be a directed set and for $i\in I$ let
$L_i$ be a subgroup of $G$. Then $\{L_i\mid i \in I\}$ is called a {\em local system}
for $G$ with respect to $I$ provided:
\begin{itemize}
\AI{a} $L_i\leq L_j$ for all $i\leq j\in I$.
\AI{b} $\bigcup_{i\in I} L_i= G$
\end{itemize}
\end{definition}
By (b) each of the elements of $G$ lie in one of the $L_i$'s and by (a) and
since $I$ is directed each finite subgroup of $G$ lies in one of $L_i$'s.
\begin{definition}\l{series} Let $R$ be a group and $H$ a group acting on $R$.
Let
${\cal S}$ a set of $H$-invariant subgroups of $R$.
\begin{itemize}
\AI{a} ${\cal S}$ is called an {\em $H$-series} on $R$ provided that:
\begin{itemize}
\AI{a} $1\in {\cal S}$ and $R\in {\cal S}$.
\AI{b} ${\cal S}$ is totally ordered with respect to inclusion. That is, if
$D,E\in {\cal S}$ then $D\leq E$ or $E\leq D$.
\AI{c} ${\cal S}$ is closed with respect to arbitrary intersections and
unions. That is, if ${\cal D}\subseteq {\cal S}$, then $\bigcap {\cal
D}\in {\cal S}$ and $\bigcup {\cal D}\in {\cal S}$.
\AI{d} For $D\in {\cal S}$ put $D^-=\bigcup \{ E\in {\cal S}\mid E |N_1|$
and so by induction on $|G/N_1|$, $\langle N_1,N_2\rangle =\langle \langle
N_1^{N_2}\rangle, N_2\rangle$ is a subnormal subgroup of $G$.
So we may assume that $N_2$ does normalize $N_1$. Thus $\langle
N_3,N_2\rangle\leq N_G(N_1) < G$. Thus by induction on $|G|$, $\langle
N_1,N_2\rangle$ is subnormal in $\langle N_3, N_2\rangle$. Moreover by induction
on $|G/N_1|$, $\langle N_3,N_2\rangle$ is subnormal in $G$. Thus $\langle
N_1,N_2\rangle$ is subnormal in $G$ by (e).
(d) By induction on $|G|$. If $N=G$ we are done. So suppose $N\unlhd\unlhd M\lhd
G$. Then $H\leq \langle N^H\rangle\leq \langle M^H\rangle =M$. As $|M|< |G|$
and $N$ and $H$ are contained in $M$, we are done by induction.\qed
\begin{lemma}\l{clsn} Let $\{ L_i\mid i\in I\}$ be a {\em local system} for $G$ and
for $i\in I$ let $N_i$ be a subgroup of $L_i$. Suppose that
\begin{itemize}
\AI{a} For all $i\in I$, $N_i$ is locally subnormal in $L_i$.
\AI{b} For all $i\leq j\in I$, $N_i\leq N_j$.
\end{itemize}
Then $N=\bigcup_{i\in I} N_i$ is a locally subnormal subgroup of $G$.
\end{lemma}
\proof First notice that by (b) and as $I$ is directed, $N$ is a subgroup of
$G$. Let $F\in {\cal F}$. Then since $F$ and $F\cap N$ are finite there exists
$i\in I$ so that $F\leq L_i$ and $F\cap N\leq N_i$. Since $N_i$ is locally
subnormal in $L_i$, $F\cap N_i$ is subnormal in $F$. But
$$ F\cap N_i\leq F\cap N\leq F\cap N_i.$$
So $F\cap N_i=F\cap N$ and $F\cap N$ is subnormal in $F$, as desired.\qed
We are now able to generalize \ref{fsn} to infinite locally finite groups.
\begin{proposition}\l{lfsn} Let $H$ a subgroup of $G$, $N$ a locally subnormal
subgroup of $G$, $M$ a normal subgroup of $G$ and ${\cal N}$ a set of
locally subnormal subgroups of $G$. Then
\begin{itemize}
\AI{a} $H\cap N$ is locally subnormal in $H$.
\AI{b} $NM/M$ is locally subnormal in $G/M$.
\AI{c} $\bigcap {\cal N}$ and $\langle {\cal N}\rangle $ are locally subnormal
in $G$.
\AI{d} If $H\leq \langle N^H\rangle$ and $H$ is finite, then $H\leq N$.
\AI{e} If $H$ is locally subnormal in $N$, then $H$ is locally subnormal in
$G$.
\end{itemize}
\end{proposition}
\proof We will prove (d) and the second part of (c). The easy proofs for the
remaining parts are left to the reader.
(c) part 2: For $F\in {\cal F}$ put $N_F=\langle E\cap F\mid E\in {\cal N}\rangle$.
As each $E\in {\cal N}$ is locally subnormal $E\cap F$ is subnormal in
$F$ and so by \ref{fsn}c, $N_F$ is subnormal in $F$. Also $F_1\leq F_2\in
{\cal F}$ clearly implies $N_{F_1}\leq N_{F_2}$. Application of \ref{clsn} (with
$I={\cal F}$ and $L_F=F$) yields that $N=\bigcup_{F\in {\cal F}} N_F$ is locally
subnormal in $G$. To complete the proof of (c) it now suffice to show that
$N=\langle {\cal N}\rangle$. By definition of $N_F$, $ N_F\leq \langle {\cal
N}\rangle $ and so also $N\leq\langle {\cal N}\rangle$. Conversely, if $g\in
E\in {\cal N}$, then $g\in N_{\langle g\rangle}\leq N$ and so $\langle {\cal
N}\rangle\leq N$.
(d) Since $H$ is finite and $H\leq \langle N^H\rangle$, there exists a finite
subset $A$ of $N$ with $H\leq \langle A^H\rangle$. Put $F=\langle A,H\rangle$.
Then $A\subseteq F\cap N$ and so $H\leq \langle (F\cap N)^H\rangle$. As $N$ is
locally subnormal $F\cap N$ is subnormal in $F$. So by \ref{fsn}d $H\leq F\cap
N\leq N$.\qed
\begin{lemma}\l{mlsnen} Let $N$ subgroup of $G$ that is maximal with respect
to being locally subnormal. Then $N$ is normal in $G$.
\end{lemma}
\proof Suppose $N$ is not normal in $G$. Then there exists $g\in G$ with
$N\neq N^g$. By \ref{lfsn}c, $\langle N,N^g\rangle$ is locally subnormal in $G$
and thus by maximality of $\langle N,N^g\rangle=G$. In particular, $
\langle g\rangle \leq \langle N^{\langle g\rangle}\rangle$ and so by
\ref{lfsn}d, $g\in N$ a contradiction to $N\neq N^g$.\qed
\begin{theorem}\l{selsn} Let $H$ be a subgroup of $G$. Then $H$ is serial in
$G$ if and only if it is locally subnormal in $G$.
\end{theorem}
\proof Suppose first that ${\cal S}$ is a series on $G$ containing $H$. Let
$F\in {\cal F}$. Then ${\cal S}\cap F =\{E\cap F\mid E\in {\cal S}\}$ is a
subnormal series of $F$ containing $F\cap H$. So $H$ is locally subnormal.
Suppose next that $H$ is locally subnormal in $G$. Let ${\cal S}$ be a
maximal chain of locally subnormal subgroups of $G$ with $H\in{\cal S}$. That
such a maximal chain exists follows from Zorn's lemma. To complete the proof of
the theorem it suffices to show that
${\cal S}$ is a series. Clearly $1\in {\cal S}$ and $G\in {\cal S}$.
Let $E$ be a locally subnormal
subgroup of $G$ so that for each $S\in {\cal S}$, $E\leq S$ or $S\leq E$. Then
${\cal S}\cup \{E\}$ is still a chain of locally subnormal subgroups and so
the maximality of ${\cal S}$ implies $E\in{\cal S}$.
Let ${\cal D}\subseteq {\cal S}$. By \ref{lfsn}c, $\bigcup {\cal D}$ is
locally subnormal in $G$. Let $E\in {\cal S}$. If $E\leq D$ for some $D\in
{\cal D}$, then $E\leq \bigcup {\cal D}$. Otherwise $D\leq E$ for all $D\in
{\cal D}$ and so $\bigcup {\cal D}\leq E$. Hence by the previous paragraph,
$\bigcup {\cal D}\in {\cal S}$. With a similar argument $\bigcap {\cal D}\in
{\cal S}$.
Finally we need to show that $E^-\unlhd E$ for all $E\leq {\cal S}$.
As $E^- \in \cal S$, it is locally subnormal. Now let
$E^-\leq R\leq E$ so that $R$ is locally subnormal in $E$. Then $R$ is also
locally subnormal in $G$ and so by maximality of ${\cal S}$ we get $R\in {\cal
S}$. But this implies that $R=E^-$ or $R=E$. So $E^-$ is a maximal locally
subnormal subgroup of $G$. Thus by \ref{mlsnen}, $E^-\unlhd E$.
So ${\cal S}$ is indeed a series, $H$ is serial in $G$ and the theorem is
proved.\qed
\chapter{ On the Jordan H\"{o}lder Theorem for locally finite groups}
\l{sJordan}
\begin{definition}\l{djoho}\index{$H$-isomorphc} Let $H$ be a groups acting on the groups
$X$ and
$Y$.
\begin{itemize}\AI{a} $X$ and $Y$ are called {\em $H$-isomorphic} provided that there
exists an isomorphism $\alpha: X\ra Y$ which commutes with the action of $H$.
\AI{b} Let ${\cal C}$ and ${\cal D}$ be sets of groups acted upon by $H$. Then ${\cal
C}$ and ${\cal D}$ are $H$ {\em isomorphic} provided that there exists a bijection
$\beta: {\cal C}\ra {\cal D}$ so that $C$ and $\beta(C)$ are
$H$-isomorphic for all $C\in {\cal C}$.
\AI{c}\index{Jordan-H\"{o}lder} We say that the {\em Jordan-H\"{o}lder theorem holds for
$H$ on
$X$} provided that the sets of factors for any two composition series for $H$ on $X$ are
$H$-isomorphic.
\AI{d} Let ${cal E}$ be a class of groups acted upon by $H$. We say that the {\em
Jordan-H\"{o}lder theorem holds for $H$ on $X$ with respect to ${\cal E}$ } provided that
the sets of ${cal E}$-factors for any two composition series for $H$ on $X$ are
$H$-isomorphic.
\end{itemize}
\end{definition}
The following lemma will produces lots of groups for which the Jordan H\"older theorem
fails.
\begin{lemma}\l{sersub} Let ${\cal I}$ be a set of groups and $\oplus {\cal I}\leq H\leq
\prod {\cal I}$. Then $H$ has a normal series whose set of factors is isomorphic to ${\cal
I}$.
\end{lemma}
\proof Let $"\preceq"$ be any reversed well ordering ( that is any non empty subset has a
maximal element) on ${\cal I}$ such that $I$ has a minimal element. For
$I\in {\cal I}$ let $H^+_I=\{ f\in H\mid f_J=1 \text{ for all } I\prec J\in {\cal I}\}$ and
$H^-_I=\{ f\in H\mid f_J=1 \text{ for all } I\preceq J\in {\cal I}\}$. Let ${\cal S}=\{
H_I^+,H^-_I\mid I\in {\cal I}\}$. We claim that ${\cal S}$ is a series for $H$ with jumps
$(H^-_I, H^+_I)$, $I\in {\cal H}$. As $\oplus {\cal I}\leq H$, it is clear that
$H_I^+/H_I^-\cong I$, so the lemma is proved once the claim is established.
If $m$ is the minimal element of ${\cal I}$, then $1=H_m^-\in {\cal H}$ and if $M$ is the
maximal element of $I$, then $H=H^+M\in {\cal I}$.
If $J\prec I\in{\cal I}$, then $H_J^+\leq H^-_I\leq H^+_I$ and so ${\cal S}$ is totally
ordered.
Let ${\cal D}$ be a subset of ${\cal S}$. Let ${\cal R}$ be the set of all $I\in {\cal
I}$ with $H_I^{\epsilon}\in {\cal D}$ for some ${\epsilon}\in \{\pm\}$.
As ${\cal R}$ has a minimimal element, there exist a largest lower bound $r$ for
for ${\cal R}$. If $H^-_r\in {\cal R}$, then $\bigcap {\cal D}=H^-_r$. If $H^-_r\not\in
{\cal R}$ we will show that $\bigcap {\cal D}=H^+_r$. Clearly $H^+_r\leq \bigcap
{\cal D}$. Let $f\in \bigcap {\cal D}$ and let $I\in {\cal I}$ with $r\prec I$. As $r$ is
the largest lower bound of ${\cal R}$, there exists $J\in {\cal R}$ with $J\prec I$. Then
$H^\epsilon_J\in {\cal D}$ for some $\epsilon$ and so $f\in H^{\epsilon}\leq H^+_J$.
Hence $f_I=1$ and $f\in H^+_r$. So indeed $\bigcap {\cal D}=H^+_r$.
Let $R$ be the maximal
element of ${\cal R}$. If $H^+_R\in {\cal R}$, then $\bigcup {\cal D}=H^+_R$ and
if $H^+_R\not\in {\cal R}$ then $H^-_R\in {\cal R}$ and $\bigcup {\cal D}=H^-_R$.
So ${\cal S}$ is closet under union and intersection. Next let $(H^\epsilon_J, H^\mu_I)$.
be the a jump of ${\cal S}$. We need to show that $(H^\epsilon_J, H^\mu_I)=(H^-_K,
H^+_K)$ for some $K\in {\cal I}$. If $\mu=+$, then $H^\epsilon_J\leq H^-_I< H^+_I$ and
so $(H^\epsilon_J, H^\mu_I)=(H^-_I,H^+I)$. If $\epsilon=-$, then $H^-_J< H^+_J\leq
H^\mu_I$ and so $(H^\epsilon_J, H^\mu_I)=(H^-_J,H^+J)$. So suppose $\epsilon=+$ and
$\mu=-$. Then $J\prec I$. Suppose there exists $K\in {\cal I}$ with $J < K * |A|\cdot |B|$$
Since $X$ acts transitively on $\Omega_i$ each element of $\Omega_i$ lies in the same
number (say $u_i$) of conjugates of $A$. Also each conjugate of $A$ meet $\Omega_i$
in the same number of elements as $A$, that is $|A\cap \Omega_i|$. Counting the pairs
$(\omega,D)$ with $\omega\in D$ and $D\in A^X$ in two different ways we get
$|\Omega_i|\cdot u_i =|A^X| |A\cap \Omega_i|$. Thus $u_i=\frac{ |A^X| |A\cap
\Omega_i|}{|\Omega_i|}$. This means that there exist at most $u_i |\Omega_i\cap
B|=\frac{ |A^X| |A\cap \Omega_i| |B\cap \Omega_i|}{|\Omega_i|}$ conjugates of $A$ which
meet $B\cap \Omega_i$ non-trivially. Since $|B\cap \Omega_i|\leq |B|$ and $|\Omega_i|
> |A|\cdot |B|$ we have
$\frac{ |A^X| |A\cap \Omega_i| |B\cap \Omega_i|}{|\Omega_i|}
< \frac{|A^X| |A\cap \Omega_i|}{|A|}$. Summing over all $i$ we see that there exist
less than
%
$$ \sum_{i=1}^r \frac{|A^X| |A\cap \Omega_i|}{|A|}=\frac{|A^X| }{|A|}
\sum_{i=1}^r |A\cap \Omega_i|= \frac{|A^X| }{|A|} |A|=|A^X|$$
%
conjugates of $A$ that meet $B$ non-trivially. But this means that some conjugate of
$A$ meets $B$ trivially. This completes the proof in Case 2.\smallskip
{\bf General Case}\smallskip
Let $\Omega_0$ be the union of the infinite $X$-orbits and $\Omega_1$
the union of the finite $X$ orbits. Let $A_i$ and $B_i$ be the intersection of
$\Omega_i$ with $A$ and $B$, respectively. By Case 2 applied to $\Omega_1$ there
exists $x\in X$ with $A_1^x\cap B_1=\emptyset$. Let $Y=C_X(A_1^x)$. Since $A_1^x$ is
finite and $a^X$ is finite for all
$a\in A_1^x$, $Y$ has finite index in $X$. Thus the orbits for $Y$ on $\Omega_0$ are
still infinite and we can apply Case 1 to $Y$ and $\Omega_0$ to get $y\in Y$ with
$A_0^{xy}\cap B_0=\emptyset$. But $A_1^{xy}=A_1^x$ and so $A_1^{xy}\cap B_1=\emptyset$
and $A^{xy}\cap B=\emptyset$. This completes the proof of the proposition also in the
general case.\qed
We remark that the bound $|A||B|$ in the previous lemma is the best possible. To see
this, let $C$ and $D$ be any finite sets and $X=\Sym(C)\times \Sym(D)\leq Sym(C\times
D)$. Let $c\in C$, $d\in D$, $A=C\times \{d\}$ and $B=\{c\}\times D$. Then the
conjugates of $A$ under $X$ are of the form $C\times \{d^\prime\}$ with
$d^\prime\in D$, so they meet $B$ in $(c,d^\prime)$.
\begin{lemma}\l{ntnsg} Let $G$ be transitive and finitary on the infinite set
$\Omega$.
\begin{itemize}
\AI{a} $G$ is neither abelian nor finite.
\AI{b} Let $N$ be an intransitive normal subgroup of $G$. Then all orbits for $N$
on $\Omega$ are finite. Moreover $G/N$ acts transitively and finitarily on the
infinite set of orbits of $N$ on
$\Omega$. In particular, $G/N$ is neither finite nor abelian.
\end{itemize}
\end{lemma}
\proof (a) $G$ cannot be finite. If $G$ is abelian, let $g$ be an element acting
non-trivially on $\Omega$. Then $\Supp_\Omega(g)$ is a finite $G$-invariant
set, a contradiction.
(b) Since the orbits of $N$ form a system of imprimitivity, each of the orbits
is finite. As $\Omega$ is infinite, $N$ has an infinite number of orbits.
The remaining assertions now are readily verified.\qed
\begin{theorem}\l{trnope} Let $G$ be transitive and finitary on the infinite set
$\Omega$.
\begin{itemize}
\AI{a} $G^\prime$ acts transitively on $\Omega$.
\AI{b} Let $N$ be a transitive, subnormal subgroup of $G$.
Then $G^\prime\leq N$ and and $N$ is normal.
\AI{c} $G^\prime$ is perfect and has no proper subgroups of finite index.
\end{itemize}
\end{theorem}
\proof (a) Since $G/G^\prime$ is abelian, \ref{ntnsg}(b) implies that $G^\prime$ is
transitive.
(b) By induction on the defect of $N$ in $G$ we may assume that $N$ is normal in $G$.
Thus in view of \ref{ccpcp} and \ref{cp} it remains to verify that $N$ has the
conjugacy centralizer property. For this let $F\in {\cal F}$ and put
$\Delta=\Supp(F)$. As $G$ is finitary, $\Delta$ is finite. So by \ref{disjoint} applied
with $X=N$ and $A=B=\Delta$, there exists $n\in N$ with
$\Delta\cap \Delta^n=\emptyset$. Since $\Supp(F^n)=\Delta^n$, this clearly implies
$[F,F^n]=1$. Thus $N$ has the conjugacy centralizer property and (b) holds.
(c) By (a) $G^{\prime\prime}$ is transitive and so by (b) $G^\prime\leq
G^{\prime\prime}$. Thus $G^\prime$ is perfect. Let $N$ be a subgroup of finite index
in $G^\prime$. Without loss, $N$ is normal in $G^\prime$. Then $N$ is
transitive and again by \ref{ntnsg}b $G^\prime\leq N$.\qed
\begin{lemma}\l{pj} Let $X$ be a group acting primitively on a set $\Omega$ and let
$a\in \Omega$.
\begin{itemize}
\AI{a} Suppose that $X_a$ has $k$ orbits on $\Omega$ and $X_a$ has an orbit
of length $m$ on $\Omega-a$. If $k$ and $m$ are finite, then $|\Omega|\leq
\sum_{i=0}^{k-1} m^i$.
\AI{b} If $g\in X$ acts non-trivially on $\Omega$ then the number of orbits of $X_a$
on $\Omega$ is at most $\deg(g)$.
\end{itemize}
\end{lemma}
\proof (a) Let $a\neq b\in \Omega$ with $|b^{X_a}|=m$. Define a directed graph on
$\Omega$ by $\alpha\ra \beta$ if and only if $(\alpha,\beta)=(a^x,b^x)$ for some $x\in
X$. Let
$\Delta(a)$ be the set of elements in $\Omega$ that can be reached from $a$ via a
directed path. Let $\Delta_i(a)$ be the set of elements in $\Delta(a)$ that are of
distance $i$ from $a$. Then $|\Delta_0(a)|=|\{a\}|=1$, $|\Delta_1(a)|=|b^{X_a}|=m$ and
by induction $|\Delta_{i+1}(a)|\leq m |\Delta_{i}(a)|\leq m^{i+1}$. Clearly each
$\Delta_i(a)$ is a union of $X_a$-orbits and so at most $k$ of the $\Delta_i(a)$'s
are not empty. It follows that
$$ |\Delta(a)|= \sum_{i=0}^{k-1} |\Delta_i(a)|\leq \sum_{i=0}^{k-1}
m^i.$$
In particular, $\Delta(a)$ is finite. If $c\in \Delta(a)$ then clearly
$\Delta(c)\subseteq \Delta(a)$. Since $X$ is transitive, $|\Delta(a)|=|\Delta(c)|$
and, since $\Delta(a)$ is finite, we conclude $\Delta(a)=\Delta(c)$. But this implies
that $\Delta(a)$ is a block and, as $X$ is primitive, we conclude $\Delta(a)=\Omega$.
Thus (a) holds.
(b) Without loss $g$ does not fix $a$. Since $X$ is primitive, $X_a$ is by \ref{premx}
a maximal subgroup of $X$. Thus $X=\langle X_a, g\rangle$ and so $g$ does not
normalize any of the orbits of $X_a$ on $\Omega$. But then each orbit of $X_a$ meets
$\Supp(g)$ non-trivially and so there are at most $\deg(g)$ orbits.\qed
\begin{theorem}\l{jordan} Let $X$ be a primitive subgroup of $\Sym(\Omega)$ and
suppose that $X$ contains a non-trivial element of finite degree $d$ and that
$$ |\Omega|> \sum_{i=0}^{d-1} (d-1)^{2i}.$$
%
Then $\Alt(\Omega)\leq X$.
\end{theorem}
\proof Let $1\neq g\in X$ with $\deg(g)=d$. By assumption $|\Omega|> d$ and so
there exists $a\in \Omega$ with $a=a^g$. By \ref{pj}b, $X_a$ has at most $d$
orbits on $\Omega$. If $X_a$ has an orbit of length less or equal to $(d-1)^2$
on $\Omega-a$, then \ref{pj}a implies
$|\Omega|\leq \sum_{i=0}^{d-1} (d-1)^{2i}$, a contradiction.
Hence all orbits of $X_a$ on $\Omega-a$ have length larger than $(d-1)^2$. Let $t$ be
a conjugate of $g$ in $X$ with $a\neq a^t$, that is $a\in \Supp(t)$. Then we can apply
\ref{disjoint} to $X=X_a$,
$\Omega=\Omega-a$ and
$A=B=\Supp(t)-a$ to obtain $h\in X_a$ with $(\Supp(t)-a)\cap
(\Supp(t)-a)^h=\emptyset$. As $h$ fixes
$a$,
$\Supp(t)\cap \Supp(t)^h=\{a\}$. A straight-forward calculation now shows that
$[t,t^h]$ is a 3-cycle. $\Alt(\Omega)\leq X$ is now a consequence of
exercise \ref{e3cycle}\qed
\begin{corollary}\l{psio} Let $G$ be a primitive subgroup of $\FSym(\Omega)$ with
$\Omega$ infinite. Then $G=\Alt(\Omega)$ or $G=\FSym(\Omega)$.\qed
\end{corollary}
\begin{corollary}\l{sifipu} Let $G$ be an infinite, simple group of finitary permutations.
Then $G$ is an alternating group.
\end{corollary}
\proof Let $G\leq \FSym(Omega)$. Since $G$ is simple, it acts faithfully on each of its
orbits. So we may assume that $G$ is transitive. Suppose that $G$ is is totally
imprimitive. Then by \ref{dti}b, $G$ is the union of proper normal subgroups, a
contradcition to the simplicity of $G$. Hence by \ref{apoti} $G$ is almost primitive on
$\Omega$. Thus $G$ is primitive or acts primitively on some bock system ${\cal D}$. In the
latter case, the simplicity forces $G$ to be faithful on ${\cal D}$ and so in any case $G$
is a primitive group of finitary permutations. As $G$ is infinite we can apply \ref{psio}
to conclude that $G$ is an alternating or a symmetric group. As $G$ is simple, $G$ is
alternating.\qed
\begin{proposition} Let $G$ be an almost primitive subgroup of $\FSym(\Omega)$
with $\Omega$ infinite.
Let $B$ be a maximal block for $G$ on $\Omega$ and ${\cal D} := B^G$.
Put $W=G^B\wr \FSym({\cal D})$ and view $W$ as a subgroup of $\FSym(\Omega)$ with $G\leq W$. Then
$$ W^\prime\leq G\leq W.$$
\end{proposition}
\proof The goal is to show that $G$ has the centralizer property in $W$. The
proposition then follows from \ref{cp}.
Let ${\cal E}$ be any finite subset of ${\cal D}$ and $\Delta=\bigcup {\cal E}$.
\smallskip
{\em Claim:} $G^\Delta=W^\Delta$
Let $E\in {\cal E}$. Then by definition of $W$, $G^E=W^E$. Hence there exists a
finite subgroup $F$ of $N_G(E)$ with $F^E=W^E$.
Pick a finite subset ${\cal S}$ of ${\cal D}$ with
$\Supp(F)\subseteq \bigcup {\cal S}$.
As $G$ induces at least $\Alt({\cal D})$ on ${\cal D}$, there exists a $g$ in
$N_G(E)$ with ${\cal E}\cap {\cal S}^g=\{E\}$. Put $H(E)=F^g$.
Then $H(E)$ induces $W^E$ on $E$ and centralizes all
other $E^\prime\in{\cal E}$. Put $H=\langle H(E)\mid E\in {\cal E}\rangle$. Then
$H^\Delta\cong\times_{ E\in {\cal E}} W^E\cong C_W({\cal E)}^\Delta$. Moreover,
$G^{\cal E}=\Sym({\cal E})=W^{\cal E}$ and so the claim holds.
\smallskip
Next let $F$ be any finite subgroup of $W$. Since $W$ is finitary we can choose our
${\cal E}$ as above with $\Supp(F)\subseteq \Delta$. Then by the claim $F\leq
C_W(\Delta)N_G(\Delta)$. Clearly $C_W(\Delta)\leq C_W(F)$ and so $F\leq GC_W(F)$. So
$G$ indeed has the centralizer property and the proposition is proved.\qed
\chapter{Linear periodic groups in characteristic 0}
\l{LinChar0}
\begin{definition}\l{dlin}\index{degree}\index{linear} Let $X$ be a group, $n$ a positive
integer and
$K$ a field. Then $X$ is {\em linear of degree $n$} over $K$ if $X$ is isomorphic
to a subgroup of $GL_n(K)$. $X$ is {\em linear} if it is linear of some
degree over some field.
\end{definition}
\begin{lemma}\l{locirr} Let $X\leq GL_n(K)$. Then there exists a finitely
generated subgroup $H$ of $X$, such that the images of $KX$ and $KH$ in
$\End_n(K)$ are equal. In particular, $X$ and $H$ have the same submodules
in $K^n$, and $X$ is completely reducible if and only if $H$ is.
\end{lemma}
\proof The image of $KX$ in $\End_n(K)$ is a vector space of dimension at
most $n^2$ over $K$ and spanned by the images of $X$.
So it is spanned by the images of at most $n^2$ elements.
Take $H$ to be the group generated by these elements.\qed
\begin{lemma}\l{masch} Suppose that $G\leq GL_n(K)$ and $\gcd(|G|,\char K)=1$.
Then $G$ is completely reducible.
\end{lemma}
\proof By \ref{locirr} we may assume that $G$ is finite. Let $W$ be proper $KG$-subspace of
$V=K^n$. Then there exits a $K$-subspace $U$ of $V$ with $V=W\oplus U$. Define a
$K$-linear map $\pi: V\to W$ by $\pi(w+u)=w$ for all $w\in W, u\in U$. Then define
$\phi:V\to W$ by $\phi(v)=\sum_{g \in G} \pi(v^g)$. Since $\phi_W$ is just multiplication
by $|G|$ and since $\gcd(|G|,\char K)=1$, $\phi_W$ is an isomorphism. Thus $V=W\oplus \ker
\phi$. Let $h\in G$. Then $\phi(v^h)=\phi(v)$ and so $\ker \phi$ is a $KG$-module. By
induction on the dimension both $W$ and $\ker \phi$ are direct sum of irreducible
$KG$-modules. Hence $V$ is completely irreduicble as $KG$-module. \qed
\begin{proposition}\l{perloc} Periodic linear groups are locally finite.
\end{proposition}
\proof \tbc
\begin{definition}\l{duni}\index{unitary} Let $V$ be a vector space over the field $F$ and
$\sigma$ an automorphism of $F$ of order $2$. Let $f: V\times V \to K$ be a map.
\begin{itemize}
\AI{a} $f$ is called a {\em unitary form} provided that
\begin{itemize} \AI{a} $f$ is linear in the first coordinate.
\AI{b} $f$ is $\sigma$-linear in the second coordinate.
\AI{c} $f(u,w)=f(w,u)^\sigma$ for all $u,w\in V$.
\end{itemize}
$GU(V,f)$ denotes the group of invertible linear transformations
fixing $f$).
\AI{b} $f$ is called a {\em hermitian form} if $f$ is unitary with
$F=\mathbb C$ and $\sigma$ is the complex conjugation.
\AI{c} $f$ is a {\em positive definite hermitian form} provide that
$f$ is a hermitian form and $f(u,u) > 0$ for all $0\neq u\in V$.
\end{itemize}
\end{definition}
The next lemma is easily proved by induction on $\dim U$ and we will
leave the details to the reader.
\begin{lemma}\l{ortsum} Let $f$ be a positive definite hermitian form
on $V$ and $U$ a finite dimensional subspace of $V$.
Then $V=U\oplus U^\perp$, where $U^\perp=\{ v\in V\mid f(u,v)=0
\mbox{ for all } u\in U\}$.
\end{lemma}
\begin{lemma}\l{invher} Let $G$ be acting on the finite dimensional vector space $V$
over ${\mathbb C}$. Then there exists a positive definite $G$-invariant hermitian form
$V$.
\end{lemma}
\proof Suppose first that $G$ is finite. Let $f$ be any positive definite hermitian
form on $G$. For $g\in G$ define $f^g(u,v)=f(u^g,v^g)$. Then $\sum_{g\in G} f^g$
does the trick.
For the general case we may assume without loss that $G$ is irreducible. By
\ref{locirr} there exists a finite, irreducible subgroup $F$ of $G$. By the
finite case there exists such an $F$-invariant positive definite form $f$ on
$V$. We claim that $f$ is already $G$-invariant.
For this let be $F\leq H\in {\cal F}$ and $h$ be an $H$-invariant positive
definite hermitian form on $V$. Let $V^*=\Hom_{\mathbb{ C}}(V,\mathbb{C})$
be the dual of $V^*$. Define
$$\phi_f: V\to V^*, \phi_f(w)(v)=f(v,w)$$.
Then $\phi_f(w)^g(v)=\phi_f(w)(v^{g^{-1}})=f(v^{g^-1},w)=f(v,w^g)=\phi_f(w^g)(v)$.
Thus $\phi_f\in \Hom_{\mathbb C G}(V,V^*)$. By Schur's lemma and since
$\mathbb C $ is algebraically closed
$\Hom_{\mathbb C G}(V,V^*)=\mathbb C \cdot \phi_f$. Now $H$ is also an
$F$-invariant hermitian form and so $\phi_f=\lambda\cdot \phi_h$ for some
$\lambda\in \mathbb C $. Thus $f(u,v)=\lambda
h(u,v)$. Since $h$ is $H$-invariant this implies that $f$ is $H$-invariant.
This is true for any finite $H$ containing $F$ and so $f$ is indeed
$G$-invariant.\qed
\begin{definition}\l{norm} Let $f$ be a positive definite hermitian form on $V$.
For $v\in V$ put $||v||= \sqrt{f(v,v)}$. For $A\in \End_{\mathbb C} (V)$ put
$||A||=\sup\{ \frac{||Av||}{||v||}\mid 0\neq v\in V\}$.
\end{definition}
\begin{lemma} \l{trie} Let $f$ be a positive definite hermitian form on $V$.
\begin{itemize}
\AI{a} $f(v,w)+f(w,v) \leq 2 ||v||\cdot ||w||$.
\AI{b} $||v+w||\leq ||v|| + ||w||$.
\AI{c} $||A+B||\leq ||A|| + ||B||$.
\AI{d} If $A\in GU(V,f)$ then $||A||=1$ and $||A-1||\leq 2$.
\AI{e} Suppose that $A\in GU(V,f)$ and there is a $0\neq v \in V$ with
$f(Av,v)=0$. Then $||A-1||\geq \sqrt{2}$.
\AI{f} $$||AB||\leq ||A||\cdot ||B||$$
\end{itemize}
\end{lemma}
\proof Let $v,w\in V$ and put $a=||v||$, $b=||w||$ and $t= f(v,w)+f(w,v)$.
(a) Let $r\in \mathbb R$. Then
$0\leq f(v+rw,v+rw)=a^2+ r t + r^2 b^2$. Choosing $r=-\frac{a}{b}$ we conclude
that $a^2 - \frac{a}{b} t + a^2\geq 0$. Thus $t\leq 2ab$ and $a$ holds.
(b) $f(v+w,v+w)= a^2 + t + b^2\leq a^2+ 2 ab +b^2=(a+b)^2$.
(c) Follows from (b) and the definition of $||A||$.
(d) As $f$ is $A$-invariant, $||Av||=||v||$. Moreover, $||(A-1)v||=||Av-v||\leq
||Av||+||v||=2\cdot ||v||$.
(e) If $f(Av,v)=0$, then $f(Av-v,Av-v)=2f(v,v)$.
(f) follows immediately from the definition of $||A||$.\qed
\begin{lemma}\l{norcom} Let $f$ be a positive definite hermitian form on the
finite dimensional $K$-vectorspace $V$ and $A,B\in GU(V,f)$.
\begin{itemize}
\AI{a} $V$ is the orthogonal sum of the eigenspaces of $A$.
\AI{b} $$||AC||=||C|| = ||CA||$$ for all $C\in\End_{\mathbb C }(V)$.
\AI{c} $||[A,B]-1||\leq 2 \cdot ||A-1||\cdot ||B-1||$.
\AI{d} If $||B-1|| < \sqrt{2}$, $||A-1||\leq \frac{1}{2}$ and $[B,A,k]=1$,
\footnote{$[B,A,2] := [[B,A],A]$ and $[B,A,n+1] := [[B,A],A]$}
for some $k\leq 1$, then $[B,A]=1$.
\end{itemize}
\end{lemma}
\proof (a) As $\mathbb C $ is algebraically closed, there exists a non-zero
eigen-vector $v$. Then by \ref{ortsum} $V=Kv\oplus v^\perp$, and (a) follows
by induction on $\dim V$.
(b) is clear.
(c) $||[A,B]-1||=||A^{-1}B^{-1}AB-1||=|| AB-BA||=||(A-1)(B-1)-(B-1)(A-1)||\leq
\leq ||(A-1)(B-1)||+||(B-1)(A-1)||\leq 2||A-1||\cdot ||B-1||$.
(d) By (c) and induction, $||[B,A,k]-1||\leq \sqrt{2}$ for all $k$. So by induction on $k$
we may assume that $k=2$. So suppose that
$[B,A]\neq 1$. Then $A\neq A^B$. Since $A$ commutes with $[B,A]=A^{-B}A$, $A$ also commutes
with $A^B$. As $A\neq A^B$ there exists some eigenspace $D$ for $A$ on $V$
with $D\neq D^B$. Since $A^B$ commutes with $A$, $A^B$ leaves $D$ invariant and so $D$ is
the orthogonal sum of the eigenspaces for $A^B$ on $D$. Hence $D=(D\cap D^A)\oplus
(D\cap D^{B\perp})$. So there exists $0\neq v\in (D\cap D^{B\perp})$. But then
$v^B\in D^A$ and $v\perp v^B$. Thus \ref{trie}e implies $||B-1||\geq \sqrt{2}$.\qed
\begin{theorem}\l{linjor} There exists a function
$J: {\mathbb N}\to {\mathbb N}$ so that whenever $G$ is a periodic subgroup of
$GL_n(\mathbb C )$, then $G$ has an abelian normal subgroup of index
at most $J(n)$.
\end{theorem}
\proof Let $V=\mathbb C ^n$. By \ref{invher} there exists a $G$-invariant
hermitian form on $V$. Let ${\cal A}=\{g\in G\mid ||g-1|| < \frac{1}{2}\}$
and $H=\langle {\cal A}\rangle$.
\smallskip
{\bf Step 1} {\em $H$ is abelian.}
Let $a,b\in{\cal A}$ and put $a_k=[a,b,k]$. By \ref{perloc} $\langle
a,b\rangle$ is finite. Hence there exists a $k$ with $||a_k-1||$ minimal.
Suppose that $a_k\neq 1$. Then
$||a_k-1||\neq 0$ and so by \ref{norcom}c
%
$$||a_{k+1}-1||=||[a_k,b]-1||\leq 2\cdot ||a_k-1||\cdot ||b-1|| < ||a_k-1||,$$
%
a contradiction to the minimal choice of $a_k$. Thus $a_k=1$ and so $[a,b]=1$
by \ref{norcom}d.\smallskip
{\bf Step 2} {\em $|G/H|$ is bounded by a function of $n$.}\smallskip
Let ${\cal R}$ be a transversal to $H$ in $G$ and $r,s\in {\cal R}$. Since
$Hr\neq Hs$, $rs^{-1}\not\in H$ and so
$$||s-r||= ||1-rs^{-1}||\geq \frac{1}{2}.$$
It follows that the spheres of radius $\frac{1}{5}$ around the members of
${\cal R}$ are disjoint. As the elements of ${\cal R}$ have norm 1, all these
spheres are contained in a sphere of radius $\frac{6}{5}$. Hence
$|G/H|={\cal R}$ is bounded by the fraction of the volumes of spheres of
radius $\frac{6}{5}$ and $\frac{1}{5}$ in
the $2n^2$ dimensional metric space $\End_{\mathbb C} (V)$.\qed
\begin{corollary} Let $G$ be a periodic subgroup of $GL_n(K)$, where $K$
is a field of characteristic 0. Then $G$ is an abelian normal subgroup of
index less or equal to $J(n)$.
\end{corollary}
\proof We contend our self with a sketch of a proof. Suppose first that $G$ is
finite. Then it can be show that there exists an $F\leq K$ with $G\leq GL_n(F)$
and $F$ is algebraic over $\mathbb Q$. But then $F\leq \mathbb C$, and so
$G\leq GL_n(\mathbb C )$. So the theorem holds by \ref{linjor}.
In the general case we see that all finite subgroups of $G$ have an abelian normal
subgroup of index at most $J(n)$. Let $I$ be the set of finite subgroups of $G$. For $H\in I$ let
$T_H$ be the set of all abelian normal subgroups $A$ of $H$ with $|H/A|\leq J(n)$. Then $T_H$ is
not empty. Let $L\leq H\in I$ and $A\in T_H$. Then $A\cap L\unlhd L$ and $|L/A\cap L|=|LA/A|\leq
|H/A|\leq J(n)$. Thus $A\cap L\in T_L$ and we obtain a map, $\phi_{HL}: T_H \to T_L, A\to A\cap
L$. Clearly $\phi_{HH}$ is the identity function on $T_H$. Also if $M\leq L$ and $A\in T_H$, then
$M\cap A=M\cap (L\cap A)$ and so $\phi_{LM}\circ \phi_{HL}=\phi_{HM}$. So the $T_H$ and $\phi_{HL}$
form an inverse limit system. Since each $T_H$ is finite, \ref{inlifi} shows that that the
inverse limit system has an inverse limit, $(A_H)_{H\in I}$. Put $A=\bigcup_{H\in I} A_H$. We will
show that $A$ is an abelian normal subgroups of $A$ and $|G/A|\leq J(n)$.
Let $H\in I$. We claim that $A_H=A\cap H$. Clearly, $A_H\leq A\cap H$. Let $a\in A\cap H$. Then
there exists $L\in I$ with $a\in A_L$. Put $M=\$. Then $M\in I$, $A_H=A_M\cap H$ and
$A_L=A_M\cap L$. Since $a\in A_L$, $a\in A_M$ and so $a\in A_M\cap H=A_H$. So $A\cap H\leq A_H$.
Let $a,b\in A$ and $g\in G$. Put $H=\$. Then $H\in I$ and so $\{a,b\}\in A\cap
H=A_H$. Since $A_H$ is a normal abelian subgroup of $H$ we have $\\leq A_H\leq A$, $a^g\in
A_H\leq A$ and $[a,b]=1$. Thus $A$ is an abelian normal subgroup of $G$.
Let $R$ be a transversal to $A$ in $G$ and $S$ a finite subset of $R$. Put $H=\*~~$. Then $H\in
I$. Since $As\neq At$ for all $s\neq t\in S$ and since $A_H=A\cap H$, $A_Hs\neq A_Ht$ for all $s\neq
t\in S$. Thus $|S|\leq |H/A_H|\leq J(n)$. Since this holds for all finite subsets of $R$, $|R|\leq
J(n)$ and so $|G/A|\leq J(n)$.\qed
\chapter{ Inverse limits of finite sets}
In this section we will show that inverse limits of finite sets are not empty.
\begin{definition}\l{dinvlim}\index{inverse limit} Let $I$ be a directed set. Then an {\em
inverse limit system} based on $I$ is a tuple $(T_i \mid i\in I, \phi_{ij}\mid i\geq j\in
I)$ such that
\begin{itemize}
\AI{a} For $i\in I$, $T_i$ is a non-empty set.
\AI{b} For $i\geq j\in I$, $\phi_{ij}: T_i\ra T_j$ is a function.
\AI{c} For $i\in I$, $\phi_{ii}$ is the identity on $T_i$.
\AI{d} For $i\geq j\geq k$, $\phi_{jk}\circ\phi_{ij}=\phi{ik}$.
\end{itemize}
The {\em inverse limit} of a inverse limits system is the set of all tuples $(t_i\mid
i\in I)$ such that $t_i\in T_i$ and $\phi_{ij}(t_i)=t_j)$ for all $i\geq j\in I$.
\end{definition}
\begin{proposition}\l{inlifi} The inverse limit of an inverse limits system of finite
sets is not empty.
\end{proposition}
\proof Let ${\cal A}$ be the set of all tuples $(J, t_j\mid j\in J)$ such $J\subseteq
I$, $t_i\in T_i$ and \smallskip
(*) {\em For all finite subsets $F$ of $J$ and all $F\leq s\in I$
there exists $t_s\in T_s$ so that $\phi_{sf}(t_s)=t_f$ for all $f\in F$.}
\smallskip
Order ${\cal A}$ by $(J, t_j\mid j\in J) \leq (K, s_k\mid k\in K)$ if $J\subset K$
and $t_j=s_j$ for all $j\in J$. It is easy to verify that the assumptions of Zorn's
lemma are fulfilled and so there exists a maximal elements $(J, t_j\mid j\in J)$ in
${\cal A}$. If $J=I$ then (*) easily implies that $(t_j\mid j\in J)$ is in
the inverse limit. So we may assume that $J\neq I$. Pick $i\in I\setminus J$ and put
$$X=\bigcap \left\{ \phi_{si}(\bigcap_{f\in F} \phi_{sf}^{-1}(t_f)) \mid F\subseteq J,
|F|<\infty, F\leq s\in I \right\}$$
We claim that $X$ is non empty. Since $T_i$ is finite, there exists
finite subsets $F_1,F_2,\ldots F_l$ of $J$ and elements $s_1,s_2,\ldots s_l$ with
$F_k\leq s_k$
$$X=\bigcap \left\{ \phi_{s_k i}(\bigcap_{f\in F_k} \phi_{s_kf}^{-1}(t_{f})) \mid
1\leq k\leq l \right\}$$
Let $s$ be an upper bound for the $s_i's$ and $F=\bigcap_{k=1}^l F_k$. Then by (*)
there exists $t_s\in T_s$ with $\phi_{sf}(t_s)=t_f$ for all $f\in F$. Put
$x=\phi_{si}(t_s)$. Then for all $k$ and all $f\in F_k$
$x=\phi_{s_ki}(\phi_{ss_k}(t_s))$ and
$\phi_{s_kf}(\phi_{ss_k}(t_s))=\phi_{sf}(t_s)=t_{f}$. Hence $\phi_{ss_k}(t_s)\in
\bigcap_{f\in F_k} \phi_{s_kf}^{-1}(t_{f})$ and $x\in X$. Thus $X$ is indeed not empty.
Now let $t_i\in X$ and put $J^*=J\cup\{i\}$. It's clear from the definition of $X$
that $(J^*, t_j\mid j\in J^*)$ fulfills (*), a contradiction to the maximality of
$(J, t_j\mid j\in J)$.\qed
\chapter{Ultra-products and a little bit of model theory}\l{sUltra}
In this section we will prove Malcev's Theorem which asserts that a group with a local
system of linear groups of bounded degree is itself linear. The main tool in th
proof are ultra-products. The most natural setting for ultra products is model theory.
So we will use the occasion to introduce the basic definitions in model theory.
\begin{definition}\l{dmodth}\index{ first order language}\index{symbols}
\begin{itemize}
\AI{a} A {\em first-order language} is a set ${\cal FL}$ together with a partition
$({\cal FS},{\cal RS}, {\cal CS},{\cal LS})$ of $\cal FL$. The elements of ${\cal FL}$,
${\cal FS}$, ${\cal RS}$, ${\cal CS}$ and ${\cal LS}$ are referred to as {\em symbols,
function symbols, relations symbols}, {\em constant symbols} and {\em logical symbols}
Also ${\cal LS}=\{),(,\wedge,\neg,\equiv,\forall,v_1,v_2,v_3\ldots, v_n,\ldots\}$. The
$v_i$'s are called variables and $\forall$ a quantifier. The functions
symbols are assumed to be partitioned into 1-placed,2-placed, 3-placed, $\ldots$
function symbols and the relation symbols are partitioned into 1-placed, 2-placed,$
\ldots$ relation symbols. ${\cal V}$ is the set of variables.
\AI{b} A {model } ${\cal A}$ for the language ${\cal 1L}$ is a set $A$ together with a
function which assigns to each $m$-placed functions symbol $s$ an function $A^{m}\la
A$, and to each $m$-placed relation symbol an $m$-placed relation on $A$ and
to each constant symbol an element of $A$. $A$ is called the {\em universe} of the
model. The object assigned to a symbol is called the interpretation of the symbol in
${\cal A}$.
\AI{c} A {\em term }is any sequence of symbols obtain in finitely many steps by the
following procedure:
\begin{itemize}
\AI{a} A variable is a term.
\AI{b} A constant symbol is a term.
\AI{c} If $f$ is an $m$-placed function symbol and $t_1,t_2,\ldots, t_m$
are terms, then $f(t_1t_2\ldots t_m)$ is a term.
\end{itemize}
\AI{d} An {\em atomic formula} is a sequence of symbols of one of the following two
forms:
\begin{itemize}
\AI{a} $t_1\equiv t_2$, where $t_1$ and $t_2$ are terms.
\AI{b} $R(t_1t_2\ldots t_m)$, where $R$ is an $m$-placed relation
symbol and $t_1,\ldots t_m$ are terms.
\end{itemize}
\AI{e} A {\em formula} is any sequence of symbols obtain in finitely many steps by the
following procedure:
\begin{itemize}
\AI{a} An atomic formula is a formula.
\AI{b} If $\phi$ and $\psi$ are formulas, then $(\phi\wedge\psi)$ and
$(\neg \phi)$ are formulas.
\AI{c} If $v$ is a variable and $\phi$ is an formula then $(\forall v)\phi$
is a formula.
\end{itemize}
\AI{f} Let $v$ be a variable and $\phi$ a formula. If $v$ is in the sequence $\phi$,
we say that $v$ is a variable of $\phi$. If $v$ is a variable of $\phi$ and the last
occurrence of $v$ is of the form $\forall v$ then $v$ is called a bound variable of
$\phi$. A variable of $\phi$ which is not bound is called a free variable of $\phi$.
Otherwise $v$ is a bounded variable of $\phi$.
\end{itemize}
\end{definition}
\begin{definition}\l{dquant} Given formulas $\phi$ and $\psi$ and a variable $v$.
\begin{itemize}\AI{a} $\phi\vee psi)$ denotes the formula $(\neg ((\neg
\phi)\wedge(\neg \psi)))$.
\AI{b} $(\phi\la \psi)$ denotes the formula $((\neg \psi)\wedge \phi)$
\AI{c} $(\psi\leftrightarrow \psi)$ denotes the formula $((\phi\la \psi)\wedge
(\psi\la phi))$.
\AI{d} $(\exists v) \phi $ denotes the formula $(\neq (\forall v) (\neg \psi)))$.
\end{itemize}
\end{definition}
\begin{definition}\l{dsatmod} Let ${\cal 1L}$ be a first order language and ${\cal A}$
a model for ${\cal 1L}$ with universe $A$. Let $E$ be a function from
${\cal V}\ra A$.
\begin{itemize}
\AI{a} Let $t$ be a term. The the value $t\mid_E$ of $t$ at $E$ is defined as follows:
\begin{itemize}\AI{a} If $t$ is a variable, then $t\mid_E=E(t)$.
\AI{b} If $t$ is a constant, then $t\mid_E$ is the interpretation
of $t$ in ${\cal A}$.
\AI{c} If $t=F(t_1\ldots t_m)$, for a function symbol $F$ and terms
$t_1,\ldots,t_m$, then
%
$$t\mid_E=F^*(t_1\mid_E,\ldots ,t_m\mid_E).$$
%
where $F^*$ is the interpretation of $F$ in ${\cal A}$.
\end{itemize}
\AI{b} Let $\phi$ be a formula. Then the statement $E$ satisfies $\phi$
in ${\cal A}$ (written ${\cal A}\models \phi\mid_E$) is defined as follows.
\begin{itemize}
\AI{a} If $\phi$ is the atomic formula $t_1\equiv t_2$ for terms $t_1$, $t_2$
when ${\cal A}\models \phi\mid_E$ if and only if $t_1\mid E=t_2\mid E$.
\AI{b} If $\phi$ is the $R(t_1\ldots t_m)$ for a relation symbol $R$ and terms
$t_1,\ldots t_m$. Then ${\cal A}\models \phi\mid_E$ if and only if
$(t_1\mid_E,\ldots,t_m\mid_E)$ fulfills the interpretation of$R$ in ${\cal A}$.
\AI{c} If $\phi=\phi_1\wedge phi_2$, then ${\cal A}\models \phi\mid_E$ if and
only if both ${\cal A}\models \phi_1\mid_E$ and ${\cal A}\models \phi_2\mid_E$ hold.
\AI{d} If $\phi=\neg \psi$, then ${\cal A}\models \phi\mid_E$ if an only if not
${\cal A}\models \psi\mid_E$.
\AI{e} If $\phi=(\forall v) \psi$, then ${\cal A}\models \phi\mid_E$ if and only
if ${\cal A}\models \phi\mid_{E*}$, for all functions $E^*:{\cal V}\ra A$,which
agree with $E$ except maybe on $v$.
\end{itemize}
\end{itemize}
\end{definition}
The easy proof of the following lemma is left to the reader.
\begin{lemma}\l{dov} Let ${\cal 1L}$ be a first order language, ${\cal A}$ a model for
${\cal 1L}$ and $E,F: {\cal V}\ra A$.
\begin{itemize}
\AI{a} If $t$ is a term and $E$ and $F$ agree on the variables of $t$, then $t\mid
E= t\mid F$.
\AI{b} If $\phi$ is a formula and $E$ and $F$ agree on the free variable of $\phi$
then ${\cal A}\models \phi\mid_E$ if and only if ${\cal A}\models \phi\mid_F$.\qed
\end{itemize}
\end{lemma}
In view of the previous lemma we can talk about $t\mid_E$, whenever $t$ is a term and
$E$ is a function from the variables of $t$ to $A$ and about ${\cal A}\models
\phi\mid_E$ whenever
$\phi$ is a formula and $E$ a function defined on the free variables of $\phi$. In
particular, if $\phi$ has no free variable, the statement ${\cal A}\models \phi$
already makes sense.
\begin{definition}\l{dtheory} Let ${\cal FL}$ be a first order language and ${\cal A}$
a model for ${\cal A}$.
\begin{itemize}\AI{a} $\phi$ is a {\em sentence} if $\phi$ has no free variables.
AI{b} Let $\phi$ be a sentence then we say that ${\cal A}$ is a model of $\phi$ if
${\cal A}\models \phi$.
\AI{c} A {\em theory} for ${\cal A}$ is a set of sentences.
\AI{d} A model for a theory $\Sigma$ is a model for ${\cal FL}$ which is a model for
each of the sentences in $\Sigma$.
\AI{e} Two models ${\cal A}$ and ${\cal B}$ for ${\cal FL}$ are called elementary
equivalent if for each sentence $\phi$ of ${\cal FL}$, ${\cal A}\models \phi$ if and
only if ${\cal B}\models \phi$.
\end{itemize}
\end{definition}
\bs
Choose ${\cal FL}=\{ \cdot,1\}\cup {\cal LS}$, where $"\cdot"$ as 2-placed
function symbol and $"1"$ as a constant symbol. Then groups are exactly the models of
the theory with the following sentences: (where we write $a\cdot b$ for $\cdot(ab)$
\begin{itemize}
\item $(\forall v_1)(\forall v_2)(\forall v_3) v_1\cdot(v_2 \cdot v_3) \equiv
(v_1\cdot v_2)\cdot v_3$.
\item $(\forall v_1) v_1\cdot 1 \equiv v_1$
\item $(\forall v_1)(\exists v_2) v_1\cdot v_2=1$
\end{itemize}
\begin{definition}\l{dfilter} Let $I$ be a set.
\begin{itemize}
\AI{a} Let ${\cal P}(I)$ is a power set of $I$, i.e the set of all subsets of $I$.
\AI{b} A filter ${\cal D}$ on $I$ is a subset of ${\cal P(I)}$ such that
\begin{itemize}
\AI{a} $I\in {\cal D}$.
\AI{b} If $D,E\in {\cal D}$, then $D\cap E\in {\cal D}$.
\AI{c} If $D\in {\cal D}$ and $D\subseteq E\subseteq I$, then $E\in {\cal
D}$.
\end{itemize}
\AI{c} An ultra filter on $I$ is a filter ${\cal D}$ on $I$ so that for each
$D\subseteq I$ exactly one of $D$ and $I\subset D$ is in ${\cal D}$.
\end{itemize}
\end{definition}
Examples for filter on a set $I$ are:
The {\em improper filter} ${\cal P}(I)$
The {\em trivial filter} $\{I\}$.
For $D\subseteq I$ the {\em principal filter} for $D$ on $I$: $\{ E\mid D\subset
E\subseteq I\}$.
The {\em cofinite or Frechet filter }: $\{ E\subset I\mid |I\setminus E|< \infty$.
A {\em proper filter} is any filter other than the improper filter
Its an easy exercise to show that on a finite set each filter is a principal filter.
Also arbitrary intersections of filters are filters and unions of chains of filters
are filters. In particular, if
${\cal E}$ is any subset of ${\cal P}(I)$ than there exists a smallest filter
containing ${\cal E}$, namely the intersection of all the filters containing ${\cal E}$.
This filter is called the filter generated by ${\cal E}$. We say that ${\cal E}$ has
the finite intersection property if any intersection of finitely many members of
${\cal E}$ is not empty. The following proposition follows immediately from the
definitions:
\begin{lemma}\l{genfil} Let $I$ be a set and ${\cal E}\subseteq {\cal P}(I)$.
Then the filter generated by ${\cal E}$ is the filter
$$\{ D\mid \bigcap {\cal
J}\subset D\subset I \text{ for some finite subset } {\cal J} \text{ of} {\cal E}.$$
In particular, the filter generated by ${\cal E}$ is proper if and only if ${\cal E}$
has the finite intersection property.\qed
\end{lemma}
\begin{lemma} \l{ult=max} A filter is an ultra-filter if and only if its a
maximal proper filter.
\end{lemma}
\proof Clearly a ultra-filter is a maximal proper filter. So let ${\cal D}$ be maximal
proper filter and $E\subset I$ with $E\not\in {\cal E}$. We need to show that
$I\setminus E\in {\cal D}$. Since ${\cal E}$ is maximal, the filter generated by
${\cal D}+E$ is improper. By \ref{genfil} this means that ${\cal D}+E$ does not have
the finite intersection property. Since finite intersection of members of ${\cal D}$
are still in ${\cal D}$ we conclude that there exists $D\in {\cal D}$ with $D\cap
E=\empty$. Hence $D\subset I\setminus E$ and so $I\setminus E\in{\cal D}$.\qed
\begin{corollary} Every proper filter, and so also every subset of ${\cal D}$ with
the finite intersection property, lies in an ultra filter.
\end{corollary}
\proof By Zorn's lemma every filter lies in a maximal filter. By \ref{ult=max} this
maximal filter is an ultra filter.\qed
\begin{definition}\l{dultprod} Let $I$ be a set, ${\cal D}$ a filter on $I$ and for
$i\in I$ let
$A_i$ be a non-empty set. Let $A=\prod_{i\in I} A_i=\{\langle a_i|mid a_i\in A_i,i\in
I\rangle\}$ be the cartesian product of the $A_i$'s.
The equivalence relation $\equiv_{\cal D}$ on $A$ is defined by
$$\langle a_i\mid i\in I\rangle \equiv_{\cal D} \langle b_i\mid i\in I\rangle\text{ if
and only if } \{ i\in I\mid a_i=b_i\}\in {\cal D}$$
For $a\in A$ let $a_{\cal D}$ be the equivalence class of $\equiv_{\cal D}$ containing
$a$. Let $\prod_{\cal D} A_i$ be the set of equivalence classes of $\equiv_{\cal D}$.
$\prod_{\cal D} A_i$ is called the {\em reduced product} of $A_i$ modulo ${\cal D}$
A reduced product modulo ${\cal D}$ is called an ultra-product of
$A_i$ modulo ${\cal D}$ if ${\cal D}$ is an ultra filter. If $A_i=B$
for all $i\in I$, then $\prod_{\cal D} B$ is called an ultra-power of $B$.
\end{definition}
Since over-sets of filter sets are filter sets $\langle a_i\mid i\in I\rangle \equiv
\langle b_i\mid i\in I\rangle$ if and only if there exists $D\in {\cal D}$ with
$a_i=b_i$ for all $i\in {\cal D}$. This observation will be usefully in any concrete
calculation since in it might be rather difficult to determine the exact sets on
which $a_i$ and $b_i$ agree
\begin{definition} Let $I$ be a set, ${\cal FL}$ a first order language and ${\cal
A}_i,i\in I$ a model for ${\cal FL}$. Let $A_i$ be the universe of ${\cal A}_i$ and
for a non-logical symbol $s$ let $s_i^*$ be the interpretation of $s$ in ${\cal
A}_i$. Then the {\em reduced product} $\prod_{\cal D} {\cal A}_I$ is the model for
${\cal FL}$ describes as follows.
\begin{itemize}
\AI{a} The universe $A$ of $\prod_{\cal D} {\cal A}_i$ is $\prod_{\cal D} A_i$.
\AI{b} If $R$ is an $m$-placed relation symbol, when the
interpretation $R^*$ for $R$ in $\prod_{\cal D} {\cal A}_I$ is defined by
$$R^*(a^1,\ldots a_m) \text{ if and only if } \{ i\in I \mid R^*_i(a^1(i),\ldots a^m(i)
\}\in{\cal D}$$
where $a^k=\langle a^k(i)\mid i\in I\rangle_{\cal D}$
\AI{b} If $F$ is an $m$-placed function, symbol then the interpretation $F^*$ for $F$
in $\prod_{\cal D}{\cal A}_i$ is defined by
$$F^*(a^1,\ldots, a^m) = \langle R^*_i(a^1_i,\ldots,
a^m_i) \mid i\in I\rangle_{\cal D}$$
where $a^k=\langle a^k_i)\mid i\in I\rangle_{\cal D}$.
\AI{c}
If $c$ is an constant symbol then the interpretation $c^*$ for $c$
in $\prod_{\cal D}{\cal A}_i$ is defined by
$$c = \langle c*_i \rangle \mid i\in I\langle_{\cal D}$$.
\end{itemize}
\end{definition}
The reader should convince herself that this is well defined, i.e.
does not depend on the particular choice of the representatives $\langle a^k(i)\mid
i\in I\rangle$ for $a^k$.
\begin{theorem}\l{fundult} Let ${\cal D}$ be an ultra filter on $I$ and ${\cal
A}=\prod_{\cal D}{\cal A}_i$ be an ultra product of models of the first order language
${\cal FL}$. Let $E_i:{\cal V} \la A_i$ be a functions and $E=\prod_{\cal D} E_i: {\cal
V}\la A, v\la \langle E_i(v)\mid i\in I\rangle_{\cal D}$.
\begin{itemize}
\AI{a} Let $t$ be a term of ${\cal FL}$, then
$$t_E=\langle t_{E_i} \mid i\in I\rangle_{\cal D}$$
\AI{b} Let $\phi$ be a formula in ${\cal FL}$. Then
$${\cal A}\models \phi_E \text{ if and only } \{ i\mid {\cal A}_i\models
\phi\mid_{E_i}\} \in {\cal D}$$
\AI{c} Let $\Sigma$ be a finite theory. Then
${\cal A}$is a model for $\Sigma$ if and only if there exists $D\in {\cal D}$
so that for all $i\in {\cal D}$, ${\cal A}_i$ is a model for $\Sigma$.
\end{itemize}
\end{theorem}
\proof (a) This is merely an unwinding of the definition and the details are left to
the reader.
(b) As usual this is proved using the inductive definition of a formula. We will
carry out this analysis for formulas of the form $(\forall v)\phi$ and $\neq \phi$ an
d leave the remaining easier cases to the reader. So suppose that (b) holds for the
formula $\phi$.
Let $v$ be a variable. We wish to show that (b) also holds for $(\forall v) \phi$.
By definition ${\cal A}\models ((\forall v) \phi)\mid_E$
if and only if for all $\hat{E}: {\cal
V}\la {\cal A}$ such that $\hat{E}$ agrees with $E$ on $V-v$ one has
$${\cal A}\models \phi_{\hat E}.$$
As (b) holds for $\phi$ this is equivalent to
$$(*)\quad\quad \{ i\mid {\cal A}_i \models \phi\mid_{\hat E_i} \} \in {\cal D}$$
for all such $\hat{E}$ and where we may choose the ${\hat E}_i$ to agree with $E_i$
on $V-v$.
Let $D=\{ i\mid {\cal A}_i\models (\forall v \phi)\mid _{E_i}\}\in {\cal D}$.
Then for any ${\hat E}$, $D$ is a subset of the set in (*). So if $D\in {\cal D}$,
(*) holds.
Conversely suppose that $D\not\in {\cal D}$. Then by definition of $\models (\forall
v)\phi\mid_{E_i}$, for each $i\in I\setminus D$ there exists $\hat{E}_i$ so that
$\hat{E}_i$ agrees with $E_i$ on $V-v$ and we do not have $ {\cal A}_i\models
\phi\mid _{\hat E_i}$. Put $\hat{E}_d==E_d$ for $d\in D$ and put $\hat{E}=\prod_{\cal
D} \hat E_i$. For this choice of $\hat{E}$ the set in (*) is a subset of $D$ and so
as $D\not\in {\cal D}$ (*) does not hold. This completes the proof of (b) for formulas
of the form $(\forall v) \phi$.\smallskip
Next we consider the formula $\neq \phi$.
By definition ${\cal A}\models \neq\phi\mid_E$ if and only if not ${\cal A}\models
\phi_E$. As (b) holds for $\phi$, this is equivalent to
$$\{ i\mid {\cal A}_i\models \phi)\mid _{E_i}\}\not\in {\cal D}$$
As ${\cal D}$ is an ultra-filter (this is the only place there the ultra filter
assumption is used) this holds if and only if
$$\{ i\mid \text{ not } {\cal A}_i\models \phi)\mid _{E_i}\}\in {\cal D}$$
By definition of $\models \neg\phi\mid_E$ this is the same as
$$\{ i\mid {\cal A}_i\models (\neg \phi)\mid _{E_i}\}\in {\cal D}$$
(c) By (b), (c) holds a single sentence, As a ${\cal D}$ is closed under finite
intersection (c) holds for arbitrary finite theory (or more general for any theory
with a finite set of axioms).\qed\ bigskip
As an immediate consequence of \ref{fundult} we obtain that an ultra products of
groups is again a group and an ultra-product of fields is a field. Also if $n$ is a
given positive integer and each $A_i$ is a set of size at most $n$, then any
ultra-product of $A_i$ is a set of size $n$. Indeed, a set has size at most $n$ if and
only if it satisfies the sentence
$$(\exists v_1\ldots v_n)(\forall v_{n+1})(v_1\equiv v_{n+1}\vee \ldots\vee
v_n\equiv v_{n+1})$$
To be able to talk about groups,fields and vector spaces at the same time we
add an additional set $\Xi$ to a first order language ,where each
element of $\Xi$ stands for one of the structures. The universe of now is of the form
$(A_\xi|\xi\in \Xi)$ where each
$A_\xi$ is a set. The function symbols are now $(m_\xi |\mid \xi \in
\Xi)-\xi_0)$-placed (with $\sum_{\xi\in Xi} m_\xi)$ finite) where such a symbol is
interpreted by a function $\prod_{\xi\in \Xi} A_\xi^{m_\xi}\ra A_{\xi_0}$. The
relation symbols are treated similarly. The variables are of the form $v_n^\xi,
n\in \mathbb{N}, \xi\in Xi$. To evaluate terms $v_n^\xi$ is only allowed to be replaced
to element of $A_\xi$, that is the only consider functions $E:\bigcup {\cal V}
\bigcup A_\xi$ with $E(v^\xi_k)\in A_\xi$. All the assertions about first order
language we made so far are still true in this more general set-up. Indeed, it is not
difficult to see that this set can be expressed in terms of the old set up.\bs
If $V_i$ is vector space over the field $K_i$, we now obtain
$\prod_{\cal D} A_i$ is a vector space over the field $\prod_{\cal D} K_i$. Now if
each $A_i$ has dimension at most $n$, then $\prod_{\cal D} A_i$ has dimension at most
$n$ over $\prod_{\cal D} K_i$. Indeed a vector space over a field has dimension at
most $n$ if the following sentence is fulfilled:
$$(\exists v_1\ldots v_n)(\forall v_{n+1})(\exists k_1\ldots k_n)
v_{n+1}\equiv k_1\cdot v_1+k_2\cdot v_2+\ldots+k_nv_n)$$
where the $v_i$'s are vector space variables and the $k_i$'s a field variables.
If $g\in GL_K(V)$ define $\deg(g)=\deg_V(g)=\dim [V,g]=\dim V/C_V(g)$. Furthermore
$\pdeg_V(g)=\inf\{ \deg {k\cdot g} \mid o\neq k\in K\}$. The statements $\deg_V(x)\leq
n$ and $\pdeg_V(g)\leq n$ can be expressed in the first order language of groups
acting on vector-space . For example $\pdeg_(g)\leq n$ reads as
$$(\exists v_1\ldots v_n)(\exists k_{n+1})(\neg (k_{n+1}\equiv 0)\wedge ((\forall
v_{n+1})(\exists k_1\ldots k_n) k_{n+1}v_{n+1}^g \equiv k_1\cdot v_1+k_2\cdot
v_2+\ldots+k_nv_n+ v_{n+1})$$
We also have a natural embedding $\prod_{\cal D} GL_{K_i}(V_i)\ra GL_{\prod_{\cal
D} K_i}(\prod_{\cal D} V_i$. This induces an embedding $\prod_{\cal D}
PGL_{K_i}(V_i)\ra PGL_{\prod_{\cal D} K_i}(\prod_{\cal D} V_i$. Indeed, $\langle
g_i\rangle_{\cal D}$ acts as a scalar $\langle k_i\rangle{\cal D}$ on $\prod_{\cal D}
V_i$ if and only if there exists $D\in {\cal D}$ with $k_i=g_i$ for all $i\in D$. But
this is the same as saying that $\langle g_i\rangle_{\cal D}$ is in the kernel of
the map from $\prod_{\cal D} GL_{K_i}(V_i)$ to $\prod_{\cal D} PGL_{K_i}(V_i)$.
If $\pdeg{V_i}(g_i)\leq n$ for all $i\in D\in {\cal D}$, then we get
$\pdeg_{\prod V_i}(\langle g_i\rangle_{\cal D})\leq n$.\smallskip
Let $X$ be a group, $(L_i\mid i\in I)$ a local system for $X$ and for
each $i\in I$, let $\phi_i: L_i\rightarrow H_i$ be a group homomorphism. For $i\in
I\leq I^{\geq I}=\{ j\in I\mid j\geq i\}$. Since $I$ is directed, $\{ I^{\geq i}\mid
i\in I\}$ as the finite intersection property. Hence there exists an ultra-filter ${\cal
D}$ on $I$ so that $I^{\geq i}\in {\cal D}$ for all $i\in I$.
Let $H=\prod_{\cal D} H_i$. Extend $\phi_i$ to a map (but not a homomorphism)
$X\ra H_i$ by $\phi_i(x)=1$ if $x\in X\setminus L_i$. Define
$\phi: X\ra H, x\ra \langle \phi_i(x)\langle_{\cal D}$.
Let $x,y\in X$ and pick $i$ with $\langle x,y\rangle\leq L_i$. Then
$\phi_j(xy)=\phi_j(x)\phi_j(y)$ for all $j\in I^{\geq i}$. Since $I^{\geq i}\in{\cal
D}$ this implies $\phi(xy)=\phi(x)\phi(y)$. Hence $\phi$ is a homomorphism. In general
it seems to a non-trivial task to determine the kernel of $\phi$ and in general the
kernel depends on the choice of the ultra filter. But there is an easy
sufficient condition which guarantees $\phi(x)\neq 1$. Namely if there exists $i\in I$
so that $\phi_j(x)\neq 1$ for all $j \geq i$ then $\phi(x)\neq 1$. In particular if
all $\phi_i$ are one to one, then
For locally finite $X$ (or more generally for arbitrary groups if instead of finite
groups, finitely generated groups are considered) we give a similar conditions which
ensures that $\phi$ is one to one:
Suppose that $\{(L_i,M_i)|i \in I\}$ is a sectional cover and that
$\ker \phi_i=M_i$. For $i\in I$ let $I^{\la i}=\{j\in I\mid L_i\leq L_j, L_i\cap
M_j=1\}$. The definition of a sectional cover implies that $\{L^{\la i}\mid i\in I$ as
the finite intersection property. Hence we can choose an ultra-filter ${\cal D}$ with
$L^{\la i}\in {\cal D}$ for all $i\in I$. Then clearly $\phi$ is one to one.
\begin{theorem}\l{malcev} Let $X$ be a group with a local system of linear groups of
degree $n$. Then $X$ is linear of degree $n$.
\end{theorem}
\proof Let ${\cal L_i\mid i\in I}$ be the local system such that there exist
field $K_i$ and embeddings
$\phi_i\ra GL_n(K_i)$. By the preceding discussion
$X$ can be embedded into $\prod_{\cal D}\ra GL_n(K_i)$ which in turn is embedded into
$GL_n(\prod_{\cal D} K_i)$.\qed
With the same sort of argument we obtain:
\begin{theorem}\l{joncev} Let $X$ be a group and $\{ (L_i,M_i)\mid i\in
I\}$ be a sectional cover for $G$. For $i\in I$ let $V_i$ be a vector space over the
field $K_i$ and $\phi_i: L_i/M_i\ra PGL_{K_i}(V_i)$ an embedding. Furthermore let
$Y\subset X$ so that for all $y\in Y$ there exists an integer $d_y$ and a finitely
generated subgroup $F_y$ so that for all $i\in I$ with $F_y\leq L_i$ and $F_y\cap
M_i=1$ one has $\pdeg_{V_i}(\phi_i(y))\leq d_y$.
Then there exists a vector space $V$ over the field $K$ and an embedding $\phi: X\ra
PGL_K(V)$ so that $\pdeg_V(\phi(y))\leq d_y$ for all $y\in Y$.\qed
\end{theorem}
\chapter{Periodic, simple, linear groups}\l{slinear}
In this chapter we present and partly prove the classification of the periodic,
simple, linear groups.
\begin{definition}\l{dpr} A {\em projective representation} of the group $X$ is
homomorphism of $X$ into $PGL_K(V)$, where $V$ is a vector space over the field $K$.
$\dim_K V$ is called the {\em degree} of the projective representation.
\end{definition}
For example $PSL_n(K)$ has projective representation of degree $n$. This indicates
that a projective representation can be a more natural object to look at as a
linear representation. Also we remark that any group with a faithful projective
representation of finite degree is linear (but maybe of larger degree). Indeed if $V$
is a projective representation for $X$, then $X\otimes X^*$ provides a linear
representation, where $X^*$ is the dual of $X$.
\begin{definition}\l{md} Let $H$ be a finite group. Then $\mdeg(H)$ is the minimal
positive integer $d$ so that there exists a field $K$ and finite groups
$B\unlhd T\leq GL_d(K)$ such that $T\cap K\leq B$ and $M\cong H$.
\end{definition}
\begin{lemma}\l{irrext} Let $q$ be a prime, $M$ a finite $q$-group with $1\neq
M^\prime\leq \Phi(M)\leq Z(M)$. Let $|M/Z(M)|=q^n$.
\begin{itemize} \AI{a} If $Z(M)$ is cyclic, then $M$ has an abelian subgroup of rank at
least $\frac{n}{2}$.
\AI{b} Let $K$ be a field and $V$ faithful irreducible $KM$-module.
Then $V$ has dimension at least $q^{\frac{n}{2}}$.
\end{itemize} \end{lemma}
\proof (a) Let $x\in M$. Then the commutator
map induces defines an isomorphism from $M/C_M(x)$ into $[M,x]\leq Z(M)$. Since
$M/C_M(x)$ is elementary abelian and $Z(M)$ is cyclic this implies $|M/C_M(x)\leq q$.
An easy induction argument shows that $|M/C_M(A)|\leq |A/A\cap Z(M)|$ for all $A\leq
M$. Since $C_M(A)\leq A$ for any maximal abelian subgroup of $M$, (a) holds.
(b) Since $M$ is irreducible, $Z(M)$ is cyclic Also if $1\neq z\in Z(M)$, then
$C_V(z)$ is an $M$ submodule and so $C_V(z)=0$. Without loss $K$ is algebraically
closed.
Let $A$ be a maximal abelian subgroups of $M$. Then $Z(M)\leq A$ and $A$ is normal in
$M$. Let $W$ be a Wedderburn component for $A$ on $V$ and let $B=N_M(W)$. Then
$[A,B]\leq M^\prime\leq Z(M)$. As $A$ is abelian and $K$ is algebraically closed, $A$
has an eigenvector on $W$ and so acts as scalars on $W$. Thus $[A,B]$ centralizes
$W$. Put $C_{Z(M)}(W)=1$ and so $[A,B]=1$. The maximality of $A$ implies $B\leq A$.
Hence $N_M(W)=A$ and $\dim V =\dim W \cdot M/A|$. It remains to show that $|M/A|\geq
|A/Z(M)|$. For this pick $B\leq M$ with
$M=AB$ and $A\cap B=Z(M)$. Then as seen above $|M/C_M(B)|\leq |B/Z(M)|=|M/A|$. Thus
also $|A/C_A(B)|\leq |M/A|$. Put $C_A(B)\leq C_M(AB)=Z(M)$ and so $|A/Z(M)|\leq M$.
\qed
\begin{lemma}\l{minsim} Let $H$ be a non-cyclic, finite, simple group, $d=\mdeg(H)$ and
$B\unlhd T\leq GL_d(K)$ finite groups with $H\cap K\leq B$ and $T/B\cong H$. Then $T$
acts primitively on $K^d$ and $B\leq K$.
\end{lemma}
\proof Suppose that $T$ normalizes a proper subspace $W$ in $V=K^d$. By minimality of
$d$ neither $T/C_T(W)$ not $T/C_T(V/W)$ have composition factor isomorphic to $H$. But
neither does the abelian group $C_T(W)\cap C_T(V/W)$.
Hence $T$ is irreducible on $V$. Suppose next that $\Delta$ is a system of
imprimitivity for $T$ on $V$. Since $C_T(\Delta)$ is not irreducible it cannot have
composition factor isomorphic to $H$. Thus $T^\Delta$ does. But $T^\delta\leq
Sym(\Delta)\leq Sym(n)$ and $Sym(n)$ is isomorphic to a subgroup of $GL_{d-1}(K)$ via
the even permutation module, a contradiction.
So $T$ is indeed primitive on $V$.
Suppose now that $B\not K$ and pick a counter example with $|T|$ minimal.
Without loss $K$ is algebraically closed. Put $Z=T\cap K$ and let $q$ be a prime
dividing the order of $B/Z$. Let
$Q$ be a Sylow $q$-subgroup of $B$. By the Frattini argument $T=BN_T(Q)$ and so
$N_T(Q)/N_M(Q)\cong H$. Thus by minimality of $|B|$, $Q$ is normal in $T$. Let $M$ be
a subgroup of $Q$ minimal with respect to $M\not\leq Z$ and $M\unlhd T$.
As $T$ is primitive Clifford's theorem implies that $V$ is the direct sum of isomorphic
irreducible $M$-modules. If $M$ would be abelian then as $K$ is algebraically closed we
get $M\leq Z$, a contradiction.
Thus $M$ is not abelian. By minimality of $M$. $\Phi(M)\leq Z$ and $Z(M)\leq Z$.
Put $\bar{M}=M/M\cap Z$.By minimality of $T$ and as $M\leq C_T(\bar{M})$ the
latter group has no composition factor isomorphic to $H$. Hence $T/C_T(\bar{M})$
has a composition factor isomorphic to $H$. Let $\bar{M}=q^m$. Note that $\bar{M}$ is
a vector space over $GF(q)$ and so by minimality of $d$, $d\leq m$. On the other hand
by \ref{irrext}, $d\geq q^{frac{m}{2}}$. So $\frac{q^m}\leq m^2$. As $m > 1$, $q=2$
and $m\leq 4$. But then $(\Aut(M)/Inn(M))\infty \leq Sp_4(2)^\infty\cong PSL_2(9)$.
Hence $d\leq 2$, $m=2$ and $Aut(M)$ is solvable, a contradiction.
\begin{lemma}\l{boundq} Let $A$ be a elementary abelian $r$ subgroup of $X=GL_n(K)$
with $r\neq \char K$. Then $A$ has rank at most $n$ and $|N_X(A)/C_X(A) |\leq n!$.
\end{lemma}
\proof Without loss $K$ is algebraically closed. Let ${\cal E}$ be the set of
eigenspaces for $A$ on $K^n$. Then $|{\cal E}|\leq n$.
Let $E\in {\cal E}$. Then $A/C_A(E)$ is cyclic and so $A$ has rank at most $|{\cal
E}|$. Since $A$ acts as scalars on $E$ we get that $[N_X(E),A]$ centralizes $E$ and so
$[C_X({\cal E}),A]$ centralizes $K^=\bigoplus {\cal E}$. Hence $N_X(A)/C_X(A)$ is
is isomorphic to a section of $\Sym({\cal E})$ and so has order at most $|{\cal E}|!$.\qed
\begin{proposition}\l{lsls} Let $G$ be a infinite, periodic,linear, simple group and
$d$ the minimal degree of a projective representation for $G$.
\begin{itemize}
\AI{a} Let ${\cal K}$ be any Kegel cover for $G$. Let ${\cal J}$ consists of whose
$K\in {\cal K}$ with $\bar{K}$ perfect and $\mdeg(\bar{K})=d$. Then ${\cal J}$ is Kegel
cover for $G$, ${\cal K}\setminus {\cal J}$ is not a Kegel cover and $H_J$ is simple
for all $J\in {\cal J}$.
\AI{b} $G$ is the union of an ascending chain $G_1\leq G_2\leq G_3\ldots$ of finite
simple subgroups with $\mdeg(G_i)=d$ for all $i$.
\end{itemize}
\end{proposition}
\proof (a)
Without loss ${\cal K}$ has non cyclic factors.Suppose that ${\cal
K}\setminus {\cal J}$ is a Kegel-cover for $G$. Then by \ref{minsim} all the
factors of this Kegel cover have a projective representation of degree less than $d$.
By Malcev's theorem \ref{malcev} we conclude that also $G$ has a projective
representation of degree less then $d$. A contradiction.
So ${\cal K}\setminus {\cal J}$ is not a Kegel cover and thus ${\cal J}$ is a Kegel
cover. Let $J\in {\cal J}$. Since $H_J$ is has a projective representation of
degree d and $d=\mdeg{\bar J}$. Hence $H_J$ is simple by \ref{minsim}
(b) By exercise \ref{eplc} $G$ is countable. Thus (b) follows from (a).\qed
In order to proceed we proudly present:\bs
\centerline{\bf \huge The classification of finite simple groups}\bs
Any finite simple groups is isomorphic to one of the following groups where $p$
is a prime, $q=p^k$ and $n,k$ are positive integers.
\begin{itemize}
\item $C_p$
\item $Alt(n)$, $n\geq 5$
\item $A_{n}(q)=PSL_{n+1}(q)$, $(n,q)\neq (1,2),(1,3)$.
\item $\mbox{}^2A_{n}(q^2)=PSU_{n+1}(q^2)$, $n\leq 3$, $(n,q)\neq (3,2)$
\item $B_n(q)=P\Omega_{2n+1}(q)$, $n\geq 3$, $p$ odd.
\item $\mbox{}^2B_2(q)=Sz(q)$, $p=2$, $k>1$ odd.
\item $C_n(q)=PSp_{2n}(q)$, $n\geq 2$, $(n,q)\neq (4,2)$.
\item $D_n(q)=P\Omega_{2n}^{+}(q)$, $n\geq 4$
\item $\mbox{}^2D_n(q^2)=P\Omega_{2n}^{-}(q)$, $n\geq 4$.
\item $\mbox{}^3D_4(q^3)$
\item $E_6(q),E_7(q),E_8(q)$
\item $\mbox{}^2E_6(q^2)$
\item $F_4(q)$
\item $\mbox{}^2 F_4(q)$, $p=2$, $k>1$ odd, $\mbox{}^2 F_4(q)^\prime$
\item $G_2(q)$, $q\neq 2$.
\item $\mbox{}^2G_2(q)$, $p=3, k>1$.
\item $M_{11},M_{12},M_{22},M_{23},M_{24},J_1,J_2,J_3,J_4, HS, McL, Suz, Ly ,He,Ru,$
$O'N, Co_3,C0_2,C0_1,Fi_{22}, Fi_{23}, Fi_{24}^\prime, F_3, F_5, F_3,F_1$
\item $DDELNPPRST$
\end{itemize}
All of these groups are simple. And they are pairwise non-isomorphic except for
$Alt(5)\cong PSL_2(4)\cong PSL_2(5)$, $PSL_2(9)\cong Alt(6)$, $PSL_3(2)\cong
PSL_2(9)$ and $PSp_4(3)\cong PSU_4(2^2)$.\medskip
\begin{proposition}\l{lslt} Let $G$ be a infinite, periodic,simple linear group.
\begin{itemize}\AI{a} There
exists a prime $p$, an positive integer $n$, a Lie-type $X$ and an ascending chain
$G_1< G_2< G_3< \ldots $ of finite simple subgroups of $G$ such that
$G=\bigcap_{i=1}^\infty G_i$ and $G_i\cong X_n(p^{k_n})$. Moreover, any linear
representation of $G$ has to be over a field in characteristic $p$.
\AI{b} If $G$ is a linear representation over the field $K$, then $\char K=p$.
\end{itemize}
\end{proposition}
\proof Let $G_1\leq G_2\leq ldots$ be as \ref{lsls}. Using
\ref{psc} we will select a subsequence which fulfills (a). (b) will be proved along
the way. First of all we may assume that none. of the $G_i$'s is one of the sporadic
groups. Also suppose that $K$ is a field so that $G$ has a faithful projective
representation of degree $d$ over $K$. Let $p=\char K$.
Let $r$ be any prime with $r\neq p$ and let $L$ be any simple section of one of
the $G_i$'s. By \ref{minsim} $L$ has faithful projective
representation of degree $d$. $L$ is linear of degree
$d^2$ and so $L$ has no elementary abelian subgroups of rank larger than $d^2$. In
particular there is bound on the size of the alternating
sections. Hence only finitely many of the $G_i$'s can be alternating groups. So we
may assume that $G_i\cong X(i)_{n_i}(q_i)$,
$q_i=p_i^{k_i}$.
We will use the following facts about groups of Lie type without proof:
\begin{itemize}
\NI{1} $X_{n}(q)$ has a section isomorphic to $Alt(n)$
\NI{2} $X_{n}(q)$ has an elementary abelian subgroup $A$ with $|A|=q$ and
$|N(A)/C(A)|\geq \frac{q-1}{2}$
\end{itemize}
>From 1. we conclude that there are only finitely many distinct $n_i$'s. Also there
are only finitely many $X(i)$'s and so by
\ref{psc} we may assume that for all $i$ and some $X$ and $n$, $X(i)=X$ and $n_i=n$.
By 2. and \ref{boundq} there is a bound on the $q_i$'s with $p_i\neq \char K$. Thus,
$p=\char K$ is a prime and $p_i=p$ for all but finitely many $i$.\qed
\begin{lemma}\l{bglv} Let $K$ be a field of characteristic $p\geq 0$, $V$ a finite
dimensional vector space over
$K$, ${\cal B}$ a basis for $V$, $G=GL_V(K)$ and $S=SL_H(V)$. Let $W$ be a $K$-subspace
of
$V$ and put $Q_W=C_G(W)\cap C_G(V/W)$
\begin{itemize}
\AI{a} $Q_W\leq SL_K(W)$ and $Q_W\cap K=1$.
\AI{b} $N_G(W)/Q_W\cong GL_K(W)\times GL_K(W)$,
$N_S(W)/W\cong \{(a,b)|a\in GL_K(W), b\in GL_K(V/W), \det(a) \det(b)=1\}$.
\AI{c} $Q_W$ is an elementary abelian $p$-group (where an elementary abelian
$0$-group is defines to be a torsion free abelian group). As
${\mathbb Z}N_G(W)/Q_W$-module $Q_W$ is isomorphic to $\Hom_K(V/W,W)$.
\AI{d} If $W$ is a proper subspace of $V$, then $C_G(Q_W)=Q_WK^\#$ and
$C_{PSL_K(V)}(Q_W)=Q_WK^\#/K^\#$
\AI{e} $SL_K(V)=\langle Q_{Kb}\mid b\in {\cal B}\rangle$.
\AI{f} $Q_B$ acts transitively on the points of $V\setminus B$.
\AI{g} Let $V=A\oplus B$ with $\dim A=1$. Then $SL_K(V)=\langle Q_A, Q_B\rangle$.
\AI{h} Suppose that $\dim_K V\geq 3$ or $|K|\geq 4$. Then every proper normal
subgroup of $SL_K(V)$ is contained in $K$ and $PSL_K(V)$ is simple.
\end{itemize}
\end{lemma}
\proof (a) Obvious.
(b) There is an obvious homomorphism from $\phi: N_G(W) \rightarrow GL_K(W)\times
GL_K((V/W)$ and $Q_W$ is the kernel of this homomorphism. Also if $V=W\oplus U$,then
$GL_K(W)\times GL_K(U)\leq GL_K(V)$, and $\phi$ restricted to this subgroup is an
isomorphism. Hence $\phi$ is onto. This proves the first part of (b). The second is
obvious.
(c) For $q\in Q_W$ define $q^*\in \Hom(V/W,W)$ by $q*(v+W)=[v,q]$. As $q$
centralizes $V/W$, $[v,q]$ indeed lies in $W$ and since $q$ centralizes $W$ this well
defined. Conversely if $\alpha\in \Hom(V/W,W)$ define $\alpha^*\in Q_W$ be
$\alpha^*(v)=v+\alpha(v+W)$.
(d) Let $g\in C_G(Q_W)$. Let $H$ be a hyper-plane (that is a subspace of codimension 1)
of $V/W$ and
$U$ a 1-space in $W$. Pick $\alpha\in \Hom(V/W,W)$ with $\ker \alpha =H$ and $\Im
\alpha =U$. Then $\ker \alpha^g==H^g$ and $\Im \alpha^g= U^g$. Since
$\alpha^*=\alpha^g$ we also have $\alpha=\alpha^g$ and so $H^g=H$ and $U^g=U$. Since
this is true for all such $H$ and $U$ it is easy to see that
$g$ acts as an scalar on $W$ and on $U$. $\alpha g=g\alpha$ implies that these scalars
are the same. Thus the first part of (d) holds. The second follows from the first and
(a).
(e) Let $n=\dim K V$. If $n=1$, there is nothing to prove. So suppose $n\geq 2$. Let
$H=\langle Q_{Kb}\mid b\in {\cal B}$. Let $a\in{\cal B}$, ${\cal B}_0={\cal B}-a$,
$H_0=\langle Q_{Ka}\cap C_G(a)\mid a\in {\cal B}_0\rangle$ and $V_0=K\langle {\cal
B}_0\rangle$. Let $b\in {\cal B}$. By induction on $n$, $H_0=SL_K(V_0)$.
\smallskip
{\em \bf Step 1} If $v+Ka=w+Ka$, then $v^q=w$ for some $q\in Q_{Ka}$.
\smallskip
Pick $\alpha\in \Hom_{V/Kb,Kb}$ with $\alpha(v+Ka)=w-v$. Then
$v^{\alpha^*}=w$.
\smallskip
{\em \bf Step 2} $H$ acts transitively on $V^\#$.
\smallskip
By step 1 every 1-space in $V\setminus K_a$ is conjugate under
$Q_{Ka}$ to am element of
$V_O$. Since $H_0$ acts transitively one the 1-spaces of $V_0$, we conclude that
$Q_{Ka}H_0$ acts transitively on the 1-spaces different from $1$. Moreover, $Q_{Kb}$
does not fix $Ka$ and so $H$ is transitive one the 1-spaces of $V$. It remains to show
that for all $\lambda\in K^\#$, $a$ and $\lambda \cdot a$ are conjugate under $H$.
By step 1 (sometimes applies to $b$ in place of $a$),
$a$ and $a+b$ are conjugate under $Q_{Kb}$, $a+b$ and $\lambda\cdot a+b$ are conjugate
under $Q_{Ka}$, and $\lambda \cdot a+b$ and $\lambda \cdot a$ are conjugate under
$Q_{Kb}$.
\smallskip
{\em \bf Step 3} $Q_{Ka}$ acts transitively on the complements to $Ka$ in $V$.
\smallskip
Let $W$ be any complement to $Ka$ in $V$.
Let $\beta(v_0+Ka)$ be the projection of $v_0$ onto $Ka$ via $V= Ka\oplus W$. Then
$\beta\in \Hom_{V/Kb,Kb}$ and $v_O+\beta(v_0)\in W$ for all $v_0\in V_0$. Thus
$V_0^{\beta^*}=W$ and Step 3 is established.
Let $s\in SL_V(K)$. By Step 2 and Step 3 there exists $h\in H$ so that $sh$
centralizes $a$ and normalizes $V_0$. Thus $sh\in SL_K(V_0)\leq H_0\leq H$ and $s\in
H$.
(f) This is just the dual statement to Step 3 in (e).
(g) By (f) $V=\langle A^{Q_B}\rangle$ and so $A^{Q_B}$ contains a basis. Thus by (e)
$\langle Q_A^{Q_B}\rangle=SL_K(V)$.
(g) Let $N$ be a normal subgroup of $S$. Let $A$ be a 1-space and $M=N_S(A)$.
Clearly $SL_K(V)$ acts double transitive on the $1$-spaces and so $M$ is a maximal
subgroup of $G$. If $N\leq M$, then as $N$ is normal in $S$ and $S$ is transitive, $S$
normalizes all the $1$-spaces in $V$. Thus $N\leq K$. So suppose $N\not\leq M$.
Since $M$ normalizes $Q_{A}$ we conclude that $MQ_{A}\unlhd S$. Thus $S=NQ_A$ and
$[Q_A,M]\leq N$. Suppose that $[Q_A,M]\neq Q_A$. Since $C_G(A)$ acts irreducibly on
$Q_A^\#$ we conclude that $[Q_A,M]=1$. Thus by (d) $M\leq Q_AK$. This implies $\dim
V=2$ and $|K|\leq 3$, a contradiction to the assumptions.
Hence $Q_A=[Q_A,M]\leq N$ and $S=N$.\qed
\begin{lemma}\l{mdegsl} Let $K$ be a field in positive characteristic $p$. Let $n>1$ an
integer. Let $\phi: PSL_n(K)\la PGL_F(W)$ be a embedding
with $\char K=\char F$ and $\dim_F W\leq n$. Then $\dim F W=n$. Moreover if $A$ is
point or a hyper-plane in $V$, then $\phi(Q_A)\leq Q_B$ for a unique point or hyper-plane
$B$ in $W$.
\end{lemma}
\proof By induction on $n$. Let $p=\char F$. Since $Q_A$ is a $p$-group, $Q_A$
centralizes proper subspace $B$ in $W$. If $n=2$, $B$ has to be a point and we are
done. So suppose $n> 2$. Note that $N_{SL_n(K)}(A)/Q$ has a subgroup
isomorphic to $SL_{n-1}(K)$. Induction and the simplicity of
$PSL_{n-1}(K)$ (respectively the structure of $PSL_2(2)$ and $PSL_2(3)$ implies
either $B$ or $W/B$ has dimension at least $n-1$. Thus $B$ is a point or a hyper-plane
and $Q_A\leq Q_B$.\qed
\begin{theorem}\l{pslslt} Let $G$ and $G_1\leq G_2\leq\ldots$ be as in
\ref{lslt}(a) with $X=A$. Let $F$ be the subfield of $\bar{{\mathbb F}}_p$ generated by
the ${\mathbb F}_{p^{k_i}}$. Then $G\cong A_n(F)$.
\end{theorem}
\proof Let $d=n=1$. Let $F_i={\mathbb
F}_{p^{k_i}}$ and $V_i=F_i^d$. Let $V_\infty$ and $F_\infty$ be a suitable
ultra product of the $V_i$ and $F_i$, respectively. Then $V_\infty$ is a
projective $G$-module of degree $d$ over $F_\infty$. Put $G_\infty=G$. Let $Z_i=\{
k\in F_i \mid k^d=1\}$. Then for $i<\infty$, $G_i\cong PSL_{K_i}(V_i)\cong
SL_{K_i}(V_i)/Z_i$.
$|Z_i|\leq d$ and so passing to a subsequence we may assume $|Z_i|=z$ for all $1\leq
i< \infty$ and some $z$. Then also $|Z_\infty|=z$.
Let $V_1=A_1\oplus B_1$ for a point $A_1$ and a hyper-plane
$B_1$. Then for $1\leq i\leq \infty$, $Q_{A_1}\leq Q_{A_i}$ for some point or
hyper-plane $A_i$ in $V_i$. Replacing $V_i$ by its dual if necessary we may
assume that
$A_i$ is a point. Also let $B_i$ be the point or hyper-plane of $V_i$ with
$Q_{B_1}\leq Q_{B_i}$. Then $G_i=\langle Q_{A_1},Q_{B_1}\rangle$ normalizes $A_i+B_i$
and so $V_i=A_i\oplus B_i$. Note also that if $i\leq j\leq \infty$, then
$Q_{A_i}\leq Q_{A_j}$. Fix $1\leq i <\infty$. Let
$M_i=N_{G_i}(A_i)$ and $M_i^*$ the inverse image of $M_i\in
SL_{F_\infty}(V_\infty)$. Let $K_i$ be the subfield of
$F_\infty$ generated by the scalars induced by $M_i^*$ on $A_\infty$. The
$p$-elements in $M_i^*$ centralize $A_\infty$. Also $M_i/O^{p^\prime}(M_i)\cong
F_i^\#/Z_i$, $M_i^*/Z_\infty\cong M_i$ and $Z_i\cong Z_\infty$. Thus
$K_i$ is isomorphic to a subfield of $F_i$. Let $O\neq a\in A_\infty$. Let $W_i$ be
the $K_iG_i$ submodule of $V_\infty$ spanned by $a$. Also let $t_1,t_2,\ldots t_d\in
G_1$ so that $V_1=\oplus_{i=1^d} A_1^{t_i}$ and let $a_j=a^{t_j}$
\smallskip
{\em \bf Claim} For all $i<\infty$, $W_i=K_ia\oplus (W_i\cap B_\infty)$
\smallskip
Let $g\in G_i$. If $A_i^gB_i$, then $Q_{B_i}\leq M_i^g$ and so $A_i^g\leq
C_{W_i}(Q_{A_i})\leq B_\infty$. If $A_i^g\not\leq B_i$, then by \ref{bglv}f,
$A_i^g=A_i^q$ for some
$q\in Q_{B_i}$. Hence $g=mq$ for some $m\in M_i^*$. By definition of $K_i$, $m$
normalizes $K_ia$ and so $K_ia^g=K_ia^q\leq K_ia[K_i,q]\leq K_ia(B_\infty\cap W_i)$.
Thus the claim is proved. It follows that $W_i\cap A_\infty=K_ia$ and so
$[W_i,Q_{A_i}]\leq K_ia$. Hence $G_i=\langle Q_{A_i}^{t_j}\mid 1\leq j\leq
d$ normalizes $\oplus_{j=1}^d K_ia_j$. Hence $a_1,a_2,\ldots a_d$ is a $K_i$-basis
for $W_i$. $W_i$ is a faithful projective $G_i$ module and so $|PSL_d(K_i)|\geq
|PSL_d(K_i)$. Since $K_i$ is a isomorphic to a subfield of $F_i$ this implies that
$K_i$ and $F_i$ are isomorphic. Moreover $G_i$ maps isomorphically onto
$PSL_{K_i}(W_i)$.
As $G=\bigcup G_i$ we have $W_\infty=\bigcup{i=1}^\infty W_i$, $K_\infty=\bigcup
K_i\cong F$ and $G=\bigcap PSL_{K_i}(W_i)=PSL_{K_\infty}(W_\infty)\cong PSL_d(F)$.\qed
The next theorem (or in some cases parts of it) was proved independently by
Belyeav \cite{Be}, Borovik \cite{Bo}, Hartley and Shute \cite{HS} and Thomas
\cite{Th}. We refer the reader to whose papers for a proof.
\begin{theorem}\l{BBHST} Let $G$ be an infinite, periodic,linear, simple groups. Then
$G$ is a groups of Lie type over an infinite,locally finite field. More precisely
if $G_1\leq G_2\leq\ldots$ are as \ref{lslt}(a) and $F$ is the subfield of
$\bar{{\mathbb F}}_p$ generated by the ${\mathbb F}_{p^{k_i}}$. Then $G\cong
X_n(F)$.\qed
\end{theorem}
\chapter{ The maximal subgroups of $SL_2(K)$}
\l{masusl}
In this section $p$ is a prime and $K$ is locally finite field in characterictic $p$, that
is $K$ is a subfield of the algebraic closure of ${\mathbb F}_p$. Let $V$ be a
2-dimensional vector space over $K$ , $L=\SL_K(V)$,
$Z=Z(L)=\{\pm 1\}$ and $\Bar{L}=L/Z(L)=\PSL_K(V)$. The goal of this section is
to determine the maximal subgroup of $L$.
\begin{lemma}\l{ffson} Let $m=\sup\{ i\mid {\mathbb
F}_{p^{2^i}}\leq K\}$.
\begin{itemize}
\AI{a} $F$ has a field extension of degree 2 if and only if $m<\infty$.
\AI{b} Let $F$ be a field extension fo degree $2$ of $K$. Then
\begin{itemize}
\AI{a} $F:K$ is Galois.
\AI{b} $F$ is unique up to $K$-isomorphism. Namely $F$ is the splitting field if
$x^{p^{2^{m+1}}}-1$ over $K$
\AI{c} Let $\sigma$ the non-trivial $K$-automorphism of $F$. Then $K=\{ ff^\sigma\mid f\in
F\}$.
\end{itemize}
\end{itemize}
\end{lemma}
\proof Let $E$ the algebraic closure of $K$. We start with describing the subfields of
$E$. Let $q$ be a power of
$p$. Then $(ab)^q=a^qb^q$ and $(a+b)^q=a^q+b^q$ for all $a,b\in E$. Hence the $q$ roots of
the polynomial $x^q-x$ for a subfield ${\mathbb F}_q$ of $E$. Conversely, if $T$ is a
subfield of order $q$ in $E$, then as $(T^\#,\cdot)$ is a groups of order $q-1$,
$t^{q-1}=1$ for all $o\neq t\in T$ and so $t^q-t=0$ for $t\in T$. Thus ${\mathbb F}_ q$
is a unique subfield of order $q$ in $E$. Let $T$ be any subfield and for a prime
$r$ let $m_r(T)=\sup \{ r^i \mid {\mathbb F}_{p^{r^i}}\leq T\}$. Let $N_K$ be the
formal product $\prod_r m_T(r)$,there the product is taken of all the primes. For two such
formal we can build the product, the greates comman divisor and leat commom multiple in
an obviuos way. Given a formal product $N= \prod m_r$ let ${\mathbb
F}_{p^N}=\bigcup_{n\in {\mathbb N}}
\{ {\mathbb F}_{p^n}\mid n \text {divides} N\}$. Then ${\mathbb F}_{p^N}$ is a subfield of
$E$ with $N_{ {\mathbb F}_{p^N}}=N$ and $T={\mathbb F}_{p^{N_T}}$. Hence we obtain a one
to one correspondence between subfield of $E$ and the formal products. Given two
formal products $N$, $M$. Then ${\mathbb F}_{p^N}\cap {\mathbb F}_{p^M}={\mathbb
F}_{p^{\gcd(N,M)}}$, The subfield of $E$ generated by ${\mathbb F}_{p^N}$ and $ {\mathbb
F}_{p^M}$ is ${\mathbb F}_{p^{\lcm(N,M)}}$. Also ${\mathbb F}_{p^N}\leq {\mathbb
F}_{p^M}$ if and only if $N$ divides $M$, in which case the degree of this extension is
$M/N$.
\medskip
By the preceeding dicussion it remains to prove (bc) Since $F$ is the union
of its finite subfields and since all finite subfields are invariant under $\sigma$ we may
assume that $F$ is finite. Let $q=|K|$. Then $x\to x^q$ is non trivial $K$-autmorphism
of $F$ and so $\sigma(f)=f^q$ for all $f\in F$. Since $\sigma$ has order two,
$ff^\sigma$ is fixed by $\sigma$ and so lies in $K$. The map $f\to ff^\sigma=f^{q+1}$
induces a homomorphism form the multiplicative group of $F$ to the multiplicative
group of $K$. Since $F$ has at most $q+1$ $(q+1)$'th roots of unity, the cernel of
this map has order at most $q+1$. As $|F^\#|=q^2-1=(q-1)(q+1)$ and $|K^\#|=q-1$, this
map must be onto.\qed
\begin{lemma}\l{absl} Let $A$ be an abelian subgroup of $L$. Then $C_L(A)$ is is the unique
maximal abelian subgroup of $L$ containing $A$, $N_L(A)$ is solvable of derived length
two and one of the following holds:
\begin{itemize}
\NI{1}
\begin{itemize}\AI{a}
$\Bar{A}$ is an elementary abelian $p$-group.
\AI{b} $KA$ has exactly one proper submodule $W$ in $V$.
\AI{c} $C_L(M)=Z(L)U$ where $U=C_L(W)\cong (K,+)$ is a maximal unipotent subgroup of $L$.
With respect to some of basis $V$, $U=\{ \left(\begin{array}{cc} 1 & 0\\ k
&1\end{array}\right)\mid k\in K\}$.
\AI{d} $N_L(A)=N_L(W)$, $N_L(W)/U\cong (K^\#,\cdot)$ and with respect to some basis of $V$,
$N_L(A)=\{
\left( \begin{array}{cc} \lambda & 0 \\ k & \lambda^{-1} \end{array} \right) \mid k\in
K,\lambda\in K^\# \}$.
\end{itemize}
\NI{2}
\begin{itemize}
\AI{a} $A$ is a locally cyclic $p^\prime$ group.
\AI{b} $KA$ has excactly two proper submodules $W_1,W_2$ on $V$
\AI{c} $C_L(A)=N_L(W_1)\cap N_L(W_2)\cong (K^\#,\cdot)$. With respect to some basis
$V$, $C_L(A)=\{ \left(\begin{array}{cc} k& 0\\ 0& k^{-1}\end{array} \right)\mid k\in
K^\#\}$.
\AI{d} $N_L(A)=N_L(\{ W_1,W_2\})$, $|N_L(A)/C_L(A) |=2$ and $N_L(A)/C_L(A)$ inverts
$C_L(A)$.
\end{itemize}
\NI{3}
\begin{itemize}
\AI{a} $A$ is a locally cyclic $p^\prime$-group.
\AI{b} $A$ acts irreducible on $V$.
\AI{c} Let $F=\End_{KA}(V)$. Then $F$ is a field extension $K$ of degree
$2$. Let $1\neq \sigma\in \Gal(F,K)$, then $C_L(A)=F\cap L=\{ f\in F\mid
ff^\sigma=1\}$.
\AI{d} $N_L(A)=N_L(F)$, $|N_L(A)/C_L(A) |=2$ and $N_L(A)/C_L(A)$ inverts
$C_L(A)$.
\end{itemize}
\end{itemize}
\end{lemma}
\proof
{\em Case 1} $A$ is not a $p^\prime$ group.
\medskip
Let $1\neq b\in B$ be an element of order $p$. Then $(b-1)^p=b^p-1=0$. Hence there exists
$v_2\in V$ with $v_1=v_2^{(b-1) }\neq neq 0$ and $v_1(b-1)=0$. Then $W=C_V(b)=Kv_1$ is the
only proper $K\angle b\rangle$-invarinant subspace of $V$. As $A$ is abelian, $A$
normalizes $W$ and so 1b holds. Hence also $N_L(A)$ leaves $W$ invariant and 1d holds. A
simple calculation shows that $C_L(b)=UZ(L)$ where $U$ is as in 1c. Since $UZ(L)$ is
abelian, $UZ(L)=C_L(A)$ and so 1c holds. Also $\Bar{A}\leq \Bar{U}$ and so 1a holds.
\medskip
{\em Case 2} $A$ is a $p^\prime$ group and $A$ is not irreducible.
Let $W_1$ be a proper $KA$-submodule in $V$. Let $a\in A\setminus Z(L)$. Then $a$ acts as
a scalar $\lambda$ on $W_1$. Since $\det a=1$, $a$ acts as $\lambda^{-1}$ on $V/W_1$.
Suppose that $\lambda=\lambda^{-1}$. Then $\lambda=\pm 1$. Put then $a^p=\pm 1$ and as $A$
is a $p$-group, $a=\pm 1$ a contradcition to $a\not\in Z(L)$. So $\lambda\neq 1$. Let
$W_2$ be the eigenspace of $a$ corresponding to the eigen vector $\lambda^{-1}$. Then
$W_1$ and $W_2$ are the only proper $K\langle a\rangle$ submodules in $V$ and so 2b
holds. Is is now easy to check that also 2c and 2d hold. Finally finite subgroups the
multiplicative groups of field are cyclic and so also 2a holds.
\medskip
{\em Case 3} $A$ is irreducible( and so also a $p^\prime$ group)
\medskip
By Schur lemma $F$ is a division ring. Clearly $K\leq F$ and as $A$ is abelian $A\leq F$
and so $F\neq A$. Since $V$ is a vector space over $F$ we get $2=\dim_KV= \dim_FV\cdot
\dim_KF$. Hence $F$ as degree $2$ over $K$ and $V$ is 1-dimensional over $F$. In particular
we may and do identify the $K$-space $V$ with the $K$-spave $F$. Let
$f\in F\setminus K$. Then $1,f$ is a $K$ basis for $V$. Let $f^2=-k-lf$ for some $k,l\in
K$. So $m_f(x)=x^2+lx+k$ is the minimal polynomial for $f$ over $K$ and the matrix for
$f$ with respect to the basis $1,f$ is $\left(\begin{array}{cc}0 &1\\ -k &
-l\end{array} \right)$. Thus $\det f=k$. As $f\in SL(V)$ if and only if $k=1$.
In particular for $a\in A\setminus Z(L)$ we see that $a$ and $a^{-1}$ are the roots of
$m_a(x)$, $F$ contains all roots of $m_a(x)$ and $m_a(x)$ does
not have a double root. Thus $F:K$ is a Galois extension.
Let $\sigma$ be the non trivial
element of $Gal(F:K)$. Then $f,f^\sigma$ are the roots of $m_f(x)$ and so $k=ff^\sigma$.
Hence $f\in L$ if and only if $ff^\sigma=1$. Note that $\sigma\in GL_K(F)=GL_K(V)$. With
rspect to the basis $a,a^{-1}$ we see that $\det \sigma=-1$. So in order to show that
$N_L(F)\not\leq C_L(F)$ we need find $f\in F$ with $\det f=-1$, for then $f\sigma\in
N_L(F)\setminus C_L(F)$. This amounts to finding $f\in F$ with $ff^\sigma-1$. But such an
$f$ exists by \ref{ffson}. Since $C_L(A)\leq F$, 3a holds and thus 3. holds in this case.
\qed
\begin{lemma}\l{solmn1} Let $M\leq L$. Then one of the following holds.
\begin{itemize}
\AI{a} $M$ has an abelian normal subgroup $A$ with $A\not\leq Z(L)$ and the structure of
$M\leq N_L(A)$ is described by \ref{absl}. In particular, $M$ is solvable of derived
length at most 2.
\AI{b} $p$ is odd and there exists $Q\unlhd M$ so that $Q$ is a quaternion group of order
eigth. Moreover $N_{\bar L}(Q)\cong Alt(4)$ or $\Sym(4)$.
\AI{c} $\bar M^\infty$ is locally "finite,simple" and so simple.
\end{itemize}
\end{lemma}
\proof Suppose first that $M$ has a solvable normal subgroup $Q$ with $Q\not\leq Z(L)$.
Choose $Q$ of minimal derived length. If $Q$ is abelian, then (a) holds. So suppose that
$Q$ is not abelian. Then by minimal choice $Z(Q)=Q^\prime=Z(L)$ and $p$ is odd. By
\ref{XY} $Q/C_Q(Q)\leq 2^2=4$. If $|Q/Z(Q)|\leq 3$, $Q/Z(Q)$ is cyclic and so $Q$ is
abelian. So $|Q/Z(Q)|=4$ and $|Q|=8$. By \ref{absl}, $C_2\times C_2$ is nor a subgroup of
$L$ and so $Q$ is a quaternion group. It is easy to see that $C_L(Q)=Z(L)$ and so $N_{\bar
L}(Q)$ is ismorphic to a subgroup of $\Aut(Q)$. Since $\Aut(Q)\cong \Sym(4)$ it rmains to
verify that $3$ divides $N_L(Q)$. We leave this as an exercise to the reader.
Note that we have probed that all solvable subgroups of $L$ have derived length at most 4
and so all locally solvable subgroups of $L$ are solvable. So we may assume that
$\LSol(M)\leq Z(L)$. Let $R$ be non trivial normal finite perfect subgroup of $M$ and $T$
a normal subgroup of $R$ minimal with respect to $T\not\leq Z(L)$. Let $S$ be some Sylow
subgroup of $T$ with $S\not\leq Z(L)$. Then by Frattini, $R=TN_R(S)$. Since $N_L(S)$ is
solvable and $R$ is perfect, $R=T$. We conclude that $\Bar R$ is simple.\qed
\begin{lemma}\l{BurTho} Let $H$ be a finite group.
\begin{itemize}\AI{a} Let $r$ a prime and $S$ a Sylow $r$-subgroup of $G$. If $S$ is cyclic
and $N_H(S)\leq C_H(S)$, then $H$ has a normal $p$-complement, that is $H=NS$ for some
normal $p^\prime$ subgroups $N$ of $H$.
\AI{b} Let $T< S\leq G$ with $S$ a Sylow $2$-subgroup of $H$. Let $t$ be an involution in
$H$ and suppose that $|S/T|=2$. Then either $t$ is conjugate in $H$ to an element of $T$
or $H$ has a normal subgroup of index two which does not contain $t$.
\end{itemize}
\end{lemma}
(a) Since $S$ is cyclic $S$ has a faithful $1$ dimensional module over ${\mathbb C}$.
Let $W$ be the ${\mathbb C}$ module induced from $U$. For $h\in H$ let $\sign(h)\in
\{1,-1\}$ be the sign of $h$ as a permutation on $H/S$ and let $\det h$ the determinant of
$h$ as a linear map map on $W$.
If $h$ is a $r^\prime$ element then no non-trivial power
of $h$ normalizes a conjugate of $W$ under $H$ and so as ${\mathbb C}\langle
h\rangle$-module,
$W$ is isomorphic to the permutation module on $H/S$. Thus $\det h=\sign h$.
So suppose now that $s\in S$ with $|h|=p$. Let $\lambda$ be scalar induced by $s$ on
$U$. Let $g\in H$ so that $s$ fixes $Sg$. Then both $s\in S\cap S^g$ and so both $S$ and
$S^g$ are Sylow $p$-subgroups of $C_H(s)$. Hence $S^g=S^c$ for some $c\in C_H(s)$. Thus
$gc^{-1}\in N_H(S)=C_H(S)\leq C_H(s)$ and so $g\in C_H(s)$. In particular $s$ acts as
$\lambda$ on $U^g$. Let $m=|C_H(s)/S|$. We conclude that $\det s=\lambda^m \sign h$.
Define $\phi(h)=\det(h)\sign (h)$. Then $\phi$ is a homorphism form $H$ into the
multiplicative group of ${\cal C}$. Also by the above $\phi(h)=1$ for all $p^\prime$
elements and $\phi(s)=\lambda^m\neq 1$ as $\lambda$ is a $p$-root of unity and $m$ is
coprime to $p$. Hence $H=\ker \phi S$ and $\ker \phi \cap S=1$.
(b) Consider ther action of $H$ on $\Omega=H/T$ by right
multiplication. Suppose first that $H$ has a fixed point $Th$ on $\Omega$. Then $Tht=Th$
and so
$t^{h_1}\in T$. So suppose that $t$ acts fixed point freely on $\Omega$. Then $t$ as
$\frac{|\Omega|}{2}=\frac{|H|}{|S|}$ orbits on $\Omega$. Thus $t$ induces an odd
permutaion on $\Omega$. Thus the normal subgroup of $H$ consisting of all the elements
inducing an even permutation on $\Omega$ has index two and does no contain $\Omega$.\qed
\begin{lemma}\l{ssl} Let $M\leq L$ so that $\bar{M}$ is non-abelian, finite simple simple.
The one of the following holds.
\begin{itemize}
\NI{1} There exists a finite subfield $K_M$ of $K$ and $2$-dimensional $K_M$ subspace $V_M$
with $KV_M=V$ and $M=SL_{K_M}(V_M)$.
\NI{2} $p$ is odd and $\bar{M}\cong \Alt(5)$.
\end{itemize}
\end{lemma}
\proof Let $A_1,A_2,\ldots A_r$ be representatives for the conjucagcy classes of maximal
abelian $p^\prime$ subgroups of $M$. Also let $\bar{A_0}$ be a Sylow $p$ subgroup of
$\bar{M}$.
\begin{equ}\l{ssl-1} Every non trivial element of $\bar{M}$ is contained in exactly
one conjugate of exactly on of the $\bar A_i$'s, $0\leq i\leq r$.
\end{equ}
Let $x\in M\setminus Z(L)$ and put $A=\langle x\rangle$. Then $A$ is abelain. If $A$ is
not a $p^\prime$ group then by \ref{absl} $\bar A$ is a $p$-group and so $A$ is contained
in a conjugate of $A_O$. If $A$ is a $p^\prime$ group, then $A$ is contained a maximal
abelian $p$-prime subgroup of $M$, which by definition is a conjugate of one of the
$A_i$'s, $1\leq i\leq r$. Finally if $A\leq A_i^x$ for some $x\in M$, then $A_i^x\leq
C_M(A)$ which by \ref{absl} is abelian and either (modulo $ Z(L)$ a $p$ or a $p^\prime$
group. Thus $A_i^x=C_M(A)$ is unique.
\medskip
Let $a_i=|\bar A_i|$, $m=|\bar M| $ and $r_i=|N_L(A_i)/A_i|$. Let $e=\gcd(2,p-1)$
\begin{equ}\l{ssl-2} $1-\frac{1}{m}=\sum_{i=0}^r \frac{1}{r_i}(1-\frac{1}{a_i})$
\end{equ}
Fist note $|N_{\bar L}(A_i)|=r_ia_i$ and so $A_i$ has $|L/N_L(A_i)|=\frac{m}{r_ia_i}$
conjugates. Each of these conjugates contains $a_i-1$ non-trivial elements and so by
\ref{ssl-1} and as $\bar{M}$ as $m-1$ non-trivial elements:
$$ m-1=\sum_{i=0}^r \frac{m}{r_ia_i}(a_i-1).$$
Dividing boths sides by $m$ gives \ref{ssl-2}.
\begin{equ}\l{ssl-2.5} $a_i\neq 1$ for at least two $0\leq i\leq r$
\end{equ}
Suppose $a_i\neq 1$ for exactly one $i$. Then as $a_ir_i\leq m$, \ref{ssl-2}
$$1-\frac{1}{a_ir_i}\leq
1-\frac{1}{m}=\frac{1}{r_i}(1-\frac{1}{a_i})=\frac{1}{r_i}-\frac{1}{a_i}{r_i}\leq
1-\frac{1}{a_ir_i}$$
Hence all inequalites are inequalities and we conclude, $a_ir_i=m$, $r_i=1$ and $a_i=m$.
Thus $\bar M=\bar A_i$ is abelian a contradiction.
\begin{equ}\l{ssl-3}\begin{itemize}
\AI{a} Let $1\leq i\leq r$, then $r_i\leq 2$.
\AI{b} If $a_0\neq 1$, then $r_0=a_j$ for some $1\leq j\leq r$.\end{itemize}
\end{equ}
(a) follows directly from \ref{absl}. For (b) suppose that $a_0\neq 1$. Suppose that
$r_0=1$. If $a_0\leq 3$, then $\bar{A_0}$ is a cyclic Sylow $p$-subgroup of $\Bar{M}$.
Since $r_0=1$, $N_M(A_0)=A_0$ is abelian, a contradiction to \ref{BurTho}. Thus $a_0\geq
4$ and so $\frac{1}{r_0}(1-\frac{1}{a_0}\geq \frac{3}{4}$. By \ref{ssl-2.5} $r\geq 1$ and
by (a)
$\frac{1}{r_1}(1-\frac{1}{a_1}\geq \frac{1}{2}(1\frac{1}{2})=\frac{1}{4}$.
Thus \ref{ssl-2} provides the contradiction
$$ 1-\frac{1}{m}\geq \frac{3}{4}+\frac{1}{4}=1$$
Hence $r_0\neq 1$. Let $r$ be a prime divisor of $r_0$ and $T$ be a Sylow $r$ subgroup of
$N_M(A_0)$. Let $A=C_M(T)$. Then be \ref{ssl-2}, $A=A_i^g$ for some $i$ and $g$ in $M$.
By \ref{absl}(1) there exists a unique 1-space $Y$ invariant under $A_0$. In particular
also $T$ normalises $Y$ and by \ref{absl}(2), $T$ fixes exactly two $1$-spaces and both of
these are also invarinat under $A$. Thus $A$ normalizes $Y$ and so also $A_0$. Thus
$A=N_{N_M(A_0)}(T)$. Since $N_M(A_O)/A_0$ is cyclic, $A_0T$ is normal in $N_M(A_0)$ and
so by a Frattini argument, $N_M(A_0)=A_0A$. Thus $r_0=|\bar A|=a_i$.
\begin{equ}\l{ssl-4} Let $r$ be a primes divisor of $\bar{M}$.
\begin{itemize}
\AI{a} $r$ divides exactly one of the $a_i$'s.
\AI{b} Let $S_r$ be a Sylow $r$ subgroup of $M$. Then exactly one of the following holds:
\begin{itemize}
\NI{1} $r$ is odd or $p=2$, and $S_r\leq A_i^g$ for some $g\in M$.
\NI{2} $r=2$, $p$ is odd and for some $g\in M$, $S_r\cap A_i^g$ is a Sylow
$r$-subgroup of $A_i^g$ and $|S_r/S_r\cap A_i^g|=2$.
\end{itemize}
\AI{c} $r_i=2$ for all $i\geq 1$.
\end{itemize}
\end{equ}
Let $\bar{a}$ be an element of order $r$ in $\bar M$. Then $a\in A_i^g$ for some $i$
and some $g\in M$. Without loss $a\in A_i$. Then $r$ divides $a_i$.
Let $T_r$ be the Sylow $r$-subgroup of $A_i$. Let $t\in
N_M(T_r)$. Then $T_r=T_r^g\leq A_i\cap A_i^g$ and so by \ref{ssl-1},
$A_i=A_i^g$. Thus $N_M(T_r)=N_M(A_i)$.
Supppose first that $r$ is odd or $p=2$. Without loss $T_r\leq S_r$. By \ref{ssl-3},
$r$ does not divide $r_i$. So $T_r$ is a Sylow $p$-subgroup of $N_M(T_r)$ and hence
$N_{S_r}(T_r)=T_r$. Hence
$T_r=S_r$. So (b) holds. Assume that $r$ also divides $a_j$. Then also $A_j$ contains a
Sylow $r$-subgroup of $M$ and so \ref{ssl-2} implies $i=j$. Thus (a) holds. Suppose
$i\geq 1$ and $r_i=1$, then $N_M(T_r)=A_i$ is abelian. Since $T_r$ is a cyclic Sylow $r$
subbgroup of $M$, this contradicts \ref{BurTho}. So (c) holds for this $i$.
Suppose next that $r=2$ and $p$ is odd. Without loss $\bar{a}\leq Z(\bar S)$. Then
$T_r=C_{S_r}(a)$. $S_r$ acts on the coset $\bar{a}=aZ(L)$ and $aZ(L)$ contains only two
elements. Thus $|S_r/T_r|=2$ and $S_r\leq N_M(T_r)$. If $S_r=T_r$ we obtain the same
contradiction as at the end of the last paragraph. Thus $S_r\neq T_r$ and $r_i=2$. So (b)
and (c) hold. Assume that $r=2$ also divides $a_j$. Then $\bar{A_j}$ contains an
involution $\bar{b}$. As $T_r$ has indes two on $S_r$, \ref{BurTho} implies that $b^g\in
T_r\leq A_i$ for some $g$. Thus $b^g\in A_j^g\cap A_i$ and so by \ref{ssl-2}, $j=i$. So
also (a) holds and all parts of \ref{ssl-4} are proved.
\begin{equ}\l{ssl-5} $m=e \prod_{i=0}^r a_i$.
\end{equ}
Clearly $m=\prod |\bar S_r|$, where the product is taken over all prime divisors $r$ of
$m$ and $S_r$ is a Sylow $rp$-subgroup of $M$. Thus \ref{ssl-5} follows from
\ref{ssl-4}(b).
\begin{equ}\l{ssl-6} One of the following holds
\begin{itemize}
\AI{a} $r=3$
\AI{b} $r=2$ and $a_0\neq 1$.
\end{itemize}
\end{equ}
Since $r_i\leq 2$ for all $i\geq 1$ \ref{ssl-2} implies
$$1-\frac{1}{m}\geq r\frac{1}{2}\frac{1}{2}$$
hence $1< \frac{r}{4}$ and $r<4$.
Suppose that neither (a) nor (b) hold. Then $r\leq 2$ and $a_0=1$, or $r\leq
1$. So by \ref{ssl-2.5} there exists exactly two $a_i$'s ( say
$i$ and $j$) with $a_i\neq 1$. Then $r_i\geq 2$ and $r_j\geq 2$ and so by \ref{ssl-2}
$$1-\frac{1}{m}=\frac{1}{r_i}(1-\frac{1}{a_i}+\frac{1}{r_j}(1-\frac{1}{a_j}\leq
\frac{1}{2}(1-\frac{1}{a_i}+1-\frac{1}{a_j})$$
Thus
$$\frac{1}{a_i}+\frac{1}{a_j}\leq \frac{1}{m}$$
By \ref{ssl-5}, $m=ea_ia_j$ so multiplication with $m$ implies
$$ e(a_j+a_i)\leq 1$$
A contradiction, as $e,a_i$ and $a_j$ ar all larger or equal to $a$.
\begin{equ}\l{ssl-7} Suppose that $r=3$. Then $r_0=1$, $p>5$,
$\{a_1,a_2,a_3\}=\{2,3,5\}$ and $m=60$. In particular, $\bar{M}\cong \Alt(5)$.
\end{equ}
Since $r=3$, \ref{ssl-2} implies:
$$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}=1+\frac{2}{m}+\frac{2}{r_0}(1-\frac{1}{a_0})>
1$$
Suppose that $\{a_1,a_2,a_3\}\neq \{2,3,5\}$. Then since the $a_i$'s are relatively prime
$$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\leq
\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{21+14+6}{42}=\frac{41}{42} <1$$
a contradiction.
Thus $\{a_1,a_2,a_3\}=\{2,3,5\}$. Since $p$ does not divide any $a_i$ for
$i\geq 1$,
$p> 5$. In particualar, $p$ is odd and $m=2\cdot 2\cdot 3\cdot 5\cdot a_0$ and so $m=60
a_0$. Also
$$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}
=\frac{15+10+6}{30}=\frac{31}{30}$$
$$\frac{1}{60}=\frac{1}{60a_0}+frac{1}{r_0}(1-\frac{1}{a_0})$$
But this clearly implies, $a_0=1$ and so $m=60$.
\begin{equ}\l{ssl-8} Suppose that $r=2$ and $a_0\neq 1$. Put $q=a_0$ and choose notation
so that $r_0=a_1$. Then one of the follwing holds:
\begin{itemize}
\AI{a} $ea_1=q-1, ea_2=q+1$, $Z(M)=Z(L)$ and $|M|=q(q+1)(q-1)$.
\AI{b} $q=p=3$, $a_1=2$, $a_2=5$ and $m=60$. In particular, $\bar{M}\cong Alt(5)$.
\end{itemize}
\end{equ}
Without loss $A_1\leq N_M(A_0$. By \ref{ssl-1} $C_{\bar M}\leq \bar{A}_0$ for all $1\neq
a\in \bar A_0$ and so all orbits of $\bar A_1$ on $\bar A_0^\#$ have length $a_1$. Thus
$$ a_1 \text{ divides} q-1\eqno{(1)}$$
By \ref{ssl-2}
$$1-\frac{1}{eqa_1a_2}=\frac{1}{a_1}(1-\frac{1}{q})+\frac{1}{2}(1-\frac{1}{a_1})
+\frac{1}{2}((1-\frac{1}{a_2})$$
Thus
$$\frac{1}{2a_2}-\frac{1}{eqa_1a_2}=\frac{1}{2a_1}-\frac{1}{a_1q}$$
Put $d=\frac{2}{e}=\gcd(2,p)$ and multiply the previous equation with $2a_1a_2q$. We
obtain:
$$qa_1-d=qa_2-2a_2=(q-2)a_2\eqno{(2)}$$
Thus $q-2$ divides $qa_1-d$ and so also $qa_1-d-a_1(q-2)= 2a_1-d$. Since $a_1>1$ and
$d\leq 2$, $2a_2-d>0$ and thus $q-2\leq 2a_1-d$ and so
$$ a_1\geq \frac{q-1}{2}+ \frac{d-1}{2}\eqno{(3)}$$
Suppose first that $p=2$. Then $d=2$ and (3) and (1) imply that $a_1=q-1$ and by (2),
$a_2=\frac{q(q-1)-2}{q-2}=\frac{(q+1)(q-2)}{q-2}=q+1$. Thus \ref{ssl-8} (a) holds in this
case.
Soppose next that $p$ is odd. Then $d=1$ by (3) and (1), $a_1=\frac{q-1}{2}$ or
$a_1=q-1$. If $a_1=\frac{q-1}{2}$ then by (2)
$a_2=\frac{q \frac{q-1}{2}-1}{q-2}=\frac{1}{2}\frac{q(q-2)}{q-2}=\frac{q+1}{2}$ and again
\ref{ssl-8} holds. So suppose $a_1=\frac{q-1}{2}$. Thus by (2) $a_2=q+1+\frac{1}
{q-2}$. This implies $q=3$, $a_1=2$ and $a_2=5$. Hence \ref{ssl-8} is proved.
\begin{equ}\l{ssl-9} Suppose that \ref{ssl-8}(a) holds. Then there exists a subfield $K$
of $F$ and a $2$-dimensional $F$ subspace $W$ of $V$ with $V=KW$ such that $M=SL_2(W)$,
where $SL_F(W)$ is (in the canonical way) viewed as a subgroup of $SL_K(V)$.
\end{equ}
Let $U$ be the 1-dimensional subspace normalizes by $A_0$ and choose notation so that
$N_M(A_0)=N_M(U)=A_0A_1$. For $a\in A_1$ let $\lambda(a)\in K$ be defined by
$u^a=\lambda(a) u$ for all $u\in U$. Let $F=\lambda(a)\mid a\in A_1\cup\{ 0\}$.
Since $|A_1|=q-1$ and $\lambda$ is a homorphism, $F^\#$ is cyclic of order $q$. Thus
$f^q=f$ for all $f\in F$ and as $q$ is a power of $p$, $F\cong {\mathbb F}_q$ is a
subfield of $U$. Let $w\in N_M(A_1)\setminus A_1$. Then $N_M(A_O)\cap N_M(A_0^w)=
N_M(U)\cap N_M(U^w)=A_1$. In particular, $|A_0^{wA_0}|=|A_0|=q$. On the otherhand, $
|A_O^M|=|M/N_M(A_0)|=|q(q-1)(q+1)/q(q-1)|=q+1$ and so $A_0^{wA_0}\cup \{A_0\}= A_O^M$.
Hence $\{1\}\cup wA_0$ is a transversal to $A_OA_1$.Thus
$$ M= A_0A_1\cup A_0A_1wA_1\eqno{(4)}$$
Let $Y$ be a $1$-dimensional $F$ subspace of $U$. Let $W=\langle Y^M\rangle$. Then
$W$ as an $FM$-submodule of $V$. Since $Y$ is $A_0A_1$ invariant,(4) implies
$W=\langle Y+ Y^{wA_0}\rangle= Y+Y^w+[Y^w,A_0]$
Since $[V,A_0]\leq U$ and $Y\leq U$ we conclude that $W=(W\cap U)+Y^w$. Thus $W\cap
U^w=(W\cap U\cap U^w)+Y^w=Y^w$ and so $W\cap U=Y$ and $W=Y+Y^w$. Hence $W$ is 2-dimensioal
over $F$ and $M\leq SL_2(W)$. As $|SL_2(W)=q(q-1)(q+1)=|M|$ we conclude $M=SL_2(W)$ and
\l{ssl-9} is proved.\qed
\begin{lemma}\l{altsl} Let $p=3$ or $p>5$ and $M\leq L$ with $\bar{M}\cong\Alt(5)$. Then
$M=N_L(M)$.
\end{lemma}
\proof Note that $Z(L)\leq M$. Let $R=N_M(L)$ and let $T$ be the kernel of the actions of
$R$ on the five Sylow $2$-subgroups of $M$. If $T\not
Z(L)$ we conclude that $N_L(T)/T$is solvable, a contradcition to $M\leq N_L(T)$ and
$M\cap T=Z(L)$. Thus $T=Z(L)$ and so $\bar R$ is isomorphic to a subgroup of $\Sym(5)$.
Thus $R=M$ or $\bar{R}\cong \Sym(5)$. Let $A$ be a Sylow $5$-subgroup of $M$. Then $A$ is
abelian and as $p\neq 5$ we conclude from \ref{absl} that $|N_R(A)/C_R(A)|\leq 2$. This
excludes the case $\bar R\cong \Sym(5)$. Thus $R=M$.\qed
\begin{lemma}\l{issl} Let $M\leq SL_2(V)$ so that $\bar M$ is infinite simple. Then
there exists an infinite subfield $F$ of $K$ and a 2-dimensional $F$-subspace $W$ of $V$
with
$M=SL_2(W)$ and $V=KW$.
\end{lemma}
\proof By \ref{solmn1} $\bar{M}$ is locally " finite ,simple". As $M$ is countbale we
conclude that $M$ is an ascending union of finite subgroups
$$M_1< M_2< M_3< M_4<\ldots M_k<$$
such that $\bar{M_k}$ is simple. Without loss $\bar M_k$ is non-abelian and has order
larger than 60. Thus by \ref{ssl} $M_k=SL_2(F_k)$ for some finite subfield $F_k$ of $K$.
Let $k\leq j$. Then $M_k\leq M_j$ and so by \ref{ssl} applied to $K=M_j$, $F_k$ is a
subfield of $F_j$. Let $F=\bigcup_{k=1}^\infty F_k$. Then $F$ is an infinite subfield of
$K$. Let $a$ be an element of order $p$ in $M_1$ and $0\neq\in u\in C_V(a)$. Put
$W_k=\langle F_ku^{M_k}\rangle$. Then as seen in the proof of \ref{ssl} \ref{ssl-p}, $W_k$
is $2$-dimensional over $F_k$ and $M_k=SL_{F_k}(W_k)$. Put $W=\langle
Fu^M=\bigcup_{k=1}^\infty W_k$. Then $W$ is $2$ dimensional over $F$ and
$M=\bigcup_{k=1}^\infty M_k=\bigcup_{k=1}^\infty SL_{F_k}(W_k)=\bigcup_{k=1}^\infty
SL_F(W)$.\qed
\begin{lemma}\l{nslf} Let $F$ be a subfield of $K$ and $W$ a 2-dimensional subspace $F$
subspace of $V$ with $V=KW$. Let $M=SL_K(W)$
\begin{itemize}
\AI{a}$N_L(M)=GL_K(W)F^\#\cap L= \{ \lambda b\mid b\in GL_F(W),\lambda \in F^\#,
\lambda^2\det b=1\}.$
\AI{b} Put $F^{\frac{1}{2}}=\{ k\in K\mid k^2\in F\}$. Then
$$|N_L(M)/M|=|F^{\frac{1}{2}\#}/F^\#|\leq 2$$
\end{itemize}
\end{lemma}
\proof Let $M=SL_F(W)$ and $a\in GL_K(V)$. Then $M^a=SL_F(W^a)$. Hence $M=M^a$ if and only
of $M$ normalizes $W^a$. Let $u$ be a non-trivial unipotent element in $M$, $0\neq w\in
C_W(u)$ and $0\neq y\in C_{W^a}(u)$. Then $y=\lambda w$ for some $\lambda\in K$. Then
$W^a$ and $\lambda W$ are both 2-dimensional $FM$-submodules of $V$ and containe
$y=\lambda w$. Thus $W^{a}=\lambda W$. Hence $a\lambda^{-1}\in N_{GL_K(V)}(W)=GL_F(W)$ and
$a=b\lambda$ for some $b\in GL_K(W)$. Moreover $a\in L$ if and only if $1=\det a=
\lambda^2 \det b$. Thus (a) holds.
To prove (b) define $\phi: N_L(M)\to F^{\frac{1}{2}\#}/F^\#$, $\lambda b\to \lambda$. We
claim that $\phi$ is a well defined homomorphism, is onto and has kernel $L$. Indeed,
suppose that $\lambda b\mu c$. Then $\lambda\mu^{-1}=cb^{-1}\in GL_F(W)\cap K^\#=F^\#$
and so $\lambda F^\#=\mu F^\#$. Also as $\lambda^2\det b=1$, $\lambda\in F^{\frac{1}{2}}$.
So $\phi$ is well defined. Clearly $\phi$ is a homomorphism. Let $\lambda\in
F^{\frac{1}{2}\#}$. Then there exists $b\in GL_F(W)$ with $\det b=\lambda^{-2}$ and
so $\lambda b\in N_L(M)$ and $\phi$ is onto. If $\lambda b$ is in the kernel of
$\phi$ if and only if $\lambda\in F$. And that is the case if and only if $\lambda b\in
GL_F(W)\cap L=SL_F(W)$. So the kernel of $\phi$ is $M$. Note that $F^{\frac{1}{2}\#}$ isa
multiplicative subgroup of $K^\#$ and so is locally cyclic. It folloes that
$F^{\frac{1}{2}\#}/F$ is a locally cyclic group, in which the square of every element is
trivial. Thus $F^{\frac{1}{2}\#}/F$ has order at most two, and the proof of (b) is
completed.\qed
\begin{theorem}[Dickson's List]\l{masusl} Let $M$ be a maximal subgroup of $L$. Then one of
the follwing holds.
\begin{itemize}
\AI{a} $M=N_L(U)$, $U$ a $1$-dimensional $K$ subspace of $W$.
\AI{b} $M=N_L(\{U_1,U_2\})$, $U_1,U_2$ distinct $1$-dimensional $K$-subspaces of $V$.
\Al{c} $M=N_L(F)$, where $F$ is field extension of degree teo of $K$.
\Al{d} $M=N_L(SL_F(W))$, where $F$ is a subfield of $K$, and $W$ is a $2$-dimensional
$F$-subspace of $V$ with $V=FW$.
\Al{e} $p$ is odd and $M=N_L(Q)$, where $Q\leq L$ with $Q\cong Q_8$.
\Al{f} $p=3$ or $p> 5$, and $\bar M\cong \Alt(5)$.
\end{itemize}
\end{theorem}
\proof This merely summarizes the results of this chapte. We also refer the reader to the
lemmas in this section for a more detailed descriptions of structure of $M$.\qed
\chapter{ The generalized Fitting subgroup of periodic, linear group}
This section is devoted to the general structure of periodic linear groups. Our aim
goal is to show that modulo the unipotent radical, the generalized Fitting subgroup
of a periodic linear group has finite index. As an application we show that
Jordan Hoelder Theorem holds for countable, periodic, linear groups.
\begin{definition}\l{dgf} Let $X$ be a group.
\AI{a} $X$ is called {\em quasi simple} if $X$ is perfect, $X\neq 1$ and $X/Z(X)$ is
simple.
\AI{b} $X$ is {\em (absolutely semisimple} if $X$ is a direct sum of
(absolutely) simple groups.
\AI{c} A {\em component} for $X$ is a quasi simple, subnormal subgroup of $X$.
\AI{d} The {\em layer} $\E(X)$ is the subgroup of $X$ generated by the components of
$X$.
\AI{e} The {\em Hirsch-Plotkin radical} $\LN(X)$ is the subgroup of $X$ generated by
all the locally nilpotent normal subgroup of $X$.
\AI{f} The generalized Fitting subgroup $\F^*(X)$ of $X$ is the group $\LN(X)\E(X)$
\end{definition}
We remark that the Hirsch-Plotkin radical is always locally nilpotent so
$\LN(X)$ is the largest locally nilpotent normal subgroups of $X$. A proof
for this non-trivial fact can be found in \cite{Ro}. On the other hand for
locally finite groups this is readily verified and we leave the details to
the reader.
\begin{lemma}\l{comp} Let $X$ be group.
\AI{a} Let $Y$ be a subnormal subgroup of $X$ and $K$ a component. Then $K\leq Y$ or
$[Y,K]=1$.
\AI{b} Distinct components of $X$ commute. In particular $E(X)/Z(E(G)\cong \bigoplus
\{K/Z(K)\mid K \text{ a component of } X$.
\end{lemma}
\proof Note first that (a) implies (b). Indeed let $K_1$ and $K_2$ be components of
$X$ with $[K_1,K_2]\neq 1$. Then by (a) $K_1\leq K_2$ and $K_2\leq K_1$.
We prove (a) by induction on the defects of $K$ and $Y$. If
$K=G$, then either $Y\leq Z(K)$ or $YZ(K)/Z(K)$ is a subnormal subgroup of the simple
group $K/Z(K)$. The latter implies $K=YZ(K)$ Thus $K=K^\prime\leq Y$. So we may assume
that $K\unlhd\unlhd L\lhd X$
If $Y=G$ we are done. So $Y\unlhd\unlhd N\lhd G$. If $K\leq N$, we are done by
induction. Hence $[K,N]\leq L\cap N$ and $K\not\leq L\cap N$. Hence by induction
$[L\cap N,K]=1$. Thus $[N,K,K]=1$ and the three subgroup lemma implies, $[N,K]=1$
and so $[Y,K]=1$.
\begin{proposition}\l{ff*g} Let $G$ be a finite group. Then $C_G(F^*(G)\leq F^*(G)$.
\end{proposition}
\proof Let $D=C_G(F^*(G))$. Clearly a component of $D$ is a component of $G$ and
$F(D)\leq F(G)$. Thus $F^*(D)\leq F^*(G)\leq C_G(D)$ and $F^*(D)=Z(D)$. If $D=Z(D)$ we
are done. So let $E/Z(D)$ be a minimal subnormal subgroup of $D$. Then $E/Z(D)$ is
simple and so either $E/Z(D)$ is abelian or perfect. In the first case $E$ is
nilpotent and in the second case $E=E^\prime Z(D)$ and $E^\prime$ is a component of
$G$. Hence $E\leq F^*(D)\leq Z(D)$, a contradiction.\qed
\begin{lemma}\l{siass} Let ${\cal S}$ be a set of perfect, simple
groups and $X=\oplus {\cal S}$.
\begin{itemize}\AI{a} If $N\unlhd\unlhd X$, then $N=\oplus {\cal N}$ for some ${\cal
N}\subset {\cal S}$.
\AI{b}If $N$ is serial in $X$ and $X$ is absolutely semisimple, then
$N=\oplus {\cal N}$ for some ${\cal N}\subset {\cal S}$.
\end{itemize}
\end{lemma}
\proof Let ${\cal C}$ be a (subnormal) series on $X$ with $N\in {\cal S}$. Let
$S\in{\cal S}$ and $C\in {\cal S}$. Then $C\cap S$ is a subnormal (serial) subgroup
of $S$ and as $S$ is (absolutely) simple. $C\cap S =1$ or $C\cap S=S$. In the latter
case $S\leq S$. Let $T_S$ the intersection of the elements of ${\cal S}$ containing
$S$ and $B_S$ the union of the elements which meet $S$ trivially. Then $(B_S,T_S)$ is a
jump, $T_S\leq B_S$, $B_S\cap S=1$ and since $SB_S/B_S$ is a normal subgroup of
$T_S/B_S$, $T_S=SB_S$. It $T_S=T_{S^*}$ for some other $S^*\in {\cal S}$ we conclude
$[T_S,T_S]\leq [S,S^*]B_S\leq B_S$,a contradiction. Let $1\neq n\in N$ and write
$s=s_1\cdot\ldots
\cdot s_n$ with $1\neq s_i\in S_i$ and distinct $S_1,\ldots S_n\in {\cal S}$.
To complete the proof of the lemma it suffices to show that $S_i\leq N$ for all
$i$. Without loss $T_{S_i}\leq B_{S_n}$ for all $i 1$, i.e. $\Delta_i\leq\Supp(t)$. Then
$\frac{1}{2}|\Delta_i|\leq |\Delta_i|-1$ for all $1\leq i\leq r$ and so
$$|\Delta|-s=\sum_{i=1}^s |\Delta_i|-1=\sum_{i=1}^r |\Delta_i|-1\geq \sum_{i=1}^r
\frac{1}{2}|\Delta_i|=\frac{1}{2}|Supp_\Delta(t)|.\qed$$
\begin{lemma}\l{Ffd} Let $F\in {\cal F}$. Then there exists a finite dimensional
$KF$-submodule
$U$ in $V$ and $C\leq C_V(F)$ with $V=U\oplus C$. In particular,$F$ is a linear group.
\end{lemma}
\proof Let $F=\langle I\rangle$ for some finite subset $I$ of $F$. Then
$C_V(F)=\bigcap_{i\in I} C_V(i)$ and $[V,F]=\sum_{i\in I} [V,i]$. Hence as $X$ is
finitary, $V/C_V(F)$ and $[V,F]$ are finite dimensional. Thus there exists a finite
dimensiona $K$-subspace $U$ of $V$ with $[V,F]\leq U$ and $V=U+C_V(F)$. Since $[V,F]\leq
U$, $U$ is $F$-invariant. Let $C$ be a complement to $C_U(F)$ in $C_V(F)$. Then $V=U\oplus
C$ and the lemma is proved.\qed
\begin{lemma}\l{lufu} Suppose that $X$ acts locally unipotently on $V$ and $V$ is finite
dimnsional. Then $X$ acts unipotently on $V$. In particular, if $V\neq O$, then $V\neq
[V,X]$
\end{lemma}
\proof Without loss $V\neq 0$. Let $F\in {\cal
F}$. Then $F$ acts unipotently on $V$ and so $[V,F]\neq V$. Pick $F\in {\cal F}$ so that
$[V,F]$ has maximal $K$-dimension. Then for all $F\leq F^*\in {\cal F}$,
$[V,F]=[V,F^*]$ and so $[V,X]=[V,F]\neq V$. That $X$ is unipotent on $V$ now follows by
induction on $\dim_K V$.\qed
\begin{lemma}\l{unpcen}
Let ${\cal D}$ be an $KX$-composition series on $V$. Then ${\cal LU}(X)$ is the largest
subgroup of $X$ acting trivially on all of the factors of ${\cal D}$.
\end{lemma}
\proof Let $W$ be factor of ${\cal D}$ and $N$ a normal subgroup of $X$ acting locally
unipotent on $W$. Suppose that $[W,N]\neq O$. Then as $X$ is irreducible on $V$,
$W=[W,N]$. Let $0\neq w\in W$. Then $w\in W=\langle \langle w^X\rangle,N]$ and so there
exists $F\in {\cal F}$ with $[w\in [\langle w^F\rangle,F\cap N]$. Put $U\langle
w^F\rangle$. Then $w\in [U,F\cap N]$ ans since $F$ normalizes $[U,F\cap N]$, $U=[U,F\cap
N]$. But $F\cap N$ acts locally unipotent on $U$ and we obtain a contradiction to
\ref{lufu}. Hence $N$ centralizes $W$.
Conversely let $N$ be a subgroup of $X$ which centralizes all the factor of ${\cal D}$. Let
$U$ be a finite dimensional subspace of $V$ and $F$ a finitely generated subgroup of $N$.
Put $W=U[V,F]$. Then $W$ is finite dimensional and $F$ invariant. Moreover $F$ centralizes
all factors of the series ${\cal D}\cap W$ and so $F$ avts unipotently on $W$. Thus $N$
acts locally unipotently on $V$\qed
\begin{lemma}\l{sbbd} Let $H\leq X$ so that $d=\dim_K [V,H]$ is finite and put $L=\langle
H^X\rangle$.
\begin{itemize}
\AI{a} Let ${\cal C}$ be a series for $KX$ on $V$. Then at most $d$ factors of $L$ are
non centralized by $L$.
\AI{b}
There exists a $KX$ series ${\cal D}$ on $V$ so that
\begin{itemize}
\AI{a} $|{\cal D}|\leq 2d+2.$
\AI{b} If $W$ is a factor of ${\cal D}$, then either $L$ act trivially on $W$ or $W$ is
irreducible as $KX$-module.
\end{itemize}
\end{itemize}
\end{lemma}
\proof Let $E=[V,H]$ and let ${\cal C}$ be any series for $KX$ on $V$. Then
${\cal C}\cap E$ is a finite series
$$1=E_0 ~~