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\title{\large\bf Tits Alternative for Maximal Subgroups\\ of Skew Linear Groups }
\author{D. Kiani, M. Mahdavi-Hezavehi\\
 dkiani@aut.ac.ir, mahdavih@sharif.edu}

\date{}
\begin{document}
\maketitle
\begin{abstract}

Let $ D $ be a noncommutative finite dimensional $ F $-central
division algebra, and let $ N $ be a normal subgroup of $ GL_n(D)
$ with $ n \geq 1 $. Given a maximal subgroup $ M $ of $ N $, it
is proved that either $M$ contains a noncyclic free subgroup or
there exist an abelian subgroup $ A $ and a finite family
$\{K_i\}^r_1$ of fields properly containing $ F $ with $ K^*_i
\subset M $ for all $1\leq i\leq r$ such that $M/A$ is finite if
$ Char F = 0 $ and $ M/A $ is locally finite if $ Char F = p>0 $,
where $A \subseteq K^*_1 \times \cdots \times K^*_r$.

\end{abstract}
\section {Introduction}

Let $ D $ be a finite dimensional $ F $-central division algebra.
Denote by $ M_n(D) $ the $ n \times n $ matrix ring over $ D $
and $ SL_n(D) $ the commutator subgroup of the multiplicative
group $ GL_n(D) = M_n(D)^* $. Let $ N $ be a normal subgroup of $
GL_n(D) $ with $ n \geq 1 $. Given a subgroup $ M $ of $ N $, we
shall say that $ M $ is {\it maximal} in $ N $ if for any
subgroup $ H $ of $ N $ with $ M \subset H $, one concludes that
$ H = N $. Now, let $ D $ be a noncommutative division ring not
necessarily of finite dimension over its centre $ F $. The
problem of whether $ GL_n(D) $ contains a noncyclic free subgroup
seems to be posed first by Lichtman in \cite{lich}. Stronger
versions of this problem which essentially deal with the
existence of noncyclic free subgroups in normal or subnormal
subgroups of $ GL_n(D) $ have been investigated in \cite{g1} and
\cite{g2}. It is known so far that these problems have positive
answers as long as we work in a division algebra of finite
dimension over its centre. Further investigations for the
infinite dimensional case are also dealt with in those articles.
Given a maximal subgroup $ M $ of $ GL_n(D) $, it is proved in
\cite{m} that either $M$ contains a noncyclic free subgroup or
there exists a finite family $\{K_i\}^r_1$ of fields with $F^*
\subset K^*_i\subset M$ for all $1\leq i\leq r$ such that $M/A$
is finite if $ Char F = 0 $ and $ M/A $ is locally finite if $
Char F = p>0 $, where $A=K^*_1\times\dots\times K^*_r$. Now, let
$ N $ be a normal subgroup of $ GL_n(D) $ with $ n \geq 1 $, and
$ M $ be a maximal subgroup of $ N $. The aim of this note is to
investigate the structure of maximal subgroups of $ N $ with
respect to non-cyclic free subgroups sitting in $ M $. To be more
precise, let $ D $ be a noncommutative division algebra of finite
dimension over its centre $ F $. Given a maximal subgroup $ M $
of $ N $, it is proved that either $M$ contains a noncyclic free
subgroup or there exist an abelian subgroup $ A $ and a finite
family $\{K_i\}^r_1$ of fields with $F^* \subset K^*_i$ for all
$1\leq i\leq r$ such that $M/A$ is finite if $ Char F = 0 $ and $
M/A $ is locally finite if $ Char F = p>0 $, where $A \subseteq
K^*_1 \times \cdots \times K^*_r$. This, in particular,
generalizes the main result of \cite{m} concerning maximal
subgroups of $ GL_n(D) $.


\section {Notations and conventions}

Let $ D $ be an infinite $ F $-central division ring and assume that $ N $ is a non-central normal
subgroup of $ GL_n(D) $ with a maximal subgroup $M$.
Given a subgroup $ G
$ of $ GL_n(D) $, we denote by $ F[G] $ the $ F $-algebra
generated by elements of $ G $ over $ F $. We also denote by $
D^n $ the space of row $ n $-vectors over $ D $. Then $ D^n $ is
a $ D-G $ bimodule in the obvious manner. $ G $ is said to be an
{\it irreducible} (reducible) subgroup of $ GL_n(D) $ whenever $
D^n $ is irreducible (reducible) as $ D-G $ bimodule. Considering
the elements of $ D^n $ as column vectors, we may regard $ D^n $
as a $ G-D $ bimodule. It is easily shown that $ D^n $ is
irreducible (reducible) as a $ G-D $ bimodule precisely when it
has the property as $ D-G $ bimodule. We shall say that $ G $ is
{\it absolutely irreducible} if $ M_n(D) = F[G] $. For any group
$ G $ we denote its centre by $ Z(G) $. Given a subgroup $ H $ of
$ G $, $ N_G (H) $ means the {\it normalizer} of $ H $ in $ G $,
$ [G:H] $ denotes the {\it index} of $ H $ in $ G $, and $ <H,K>
$ the group generated by $ H $ and $ K $, where $ K $ is a
subgroup of $ G $. We shall say that $ H $ is {\it
soluble-by-finite} if there is a soluble normal subgroup $ K $ of
$ H $ such that $ H/K $ is finite. Let $ S $ be a subset of $
M_n(D) $, then the {\it centralizer} of $ S $ in $ M_n(D) $ is
denoted by $ C_{M_n(D)} (S) $. We shall identify the centre $ FI
$ of $ M_n(D) $ with $ F $. For each $ d \in D^* $, denote by $
A_d $ the matrix obtained from the unit matrix by replacing the $
(1,1)$-th and $ (n,n)$-th entries with $ d $ and $ d^{-1} $,
respectively. Some notations and conventions for linear groups
and skew linear groups from \cite{shir}, \cite{sup2} and
\cite{rowen1} are frequently used throughout.


\section {Maximal subgroups of normal \\subgroups in $Gl_n(D)$}

Given a division ring $D$ with center $F$, let $ N $ be a normal
subgroup of $ GL_n(D) $. This section essentially deals with the
structure of maximal subgroups $ M $ of $ N $ and how they sit in
$ N $ with respect to noncyclic free subgroups. To prove our main
result, we shall need some commutativity theorems that enable us
to understand better the structure of $ M $. To be more precise,
given an $ F $-central division ring $ D $, let $ N $ be a
non-central normal subgroup of $GL_n(D)$ with $ n \geq 1 $.
Assume that $M$ is a maximal subgroup of $N$. It is shown that
either $M$ is irreducible or there exists $P\in GL_n(D)$ such
that $P^{-1}A_dP \in M $ for any $d\in D^*$. Using this result,
it is also proved that if $ D $ is infinite, then there exists no
non-abelian maximal subgroup $M$ of $N$ such that $\mid M/M \cap
F^* \mid < \infty$. We then show that $M$ is nilpotent if and
only if $M$ is contained in the multiplicative group of a
subfield of $M_n(D)$. Finally, using above results as well as
various other results from algebraic group and skew linear group
theory, it is proved that either $M$ contains a noncyclic free
subgroup or there exist an abelian subgroup $ A $ and a finite
family $\{K_i\}^r_1$ of fields with $F^* \subset K^*_i$ for all
$1\leq i\leq r$ such that $M/A$ is finite if $ Char F = 0 $ and $
M/A $ is locally finite if $ Char F = p>0 $, where $A \subseteq
K^*_1 \times \cdots \times K^*_r$. We begin our study with the
following lemmas:

\begin{lem}\label{irreducible}
Let $D$, N, and $M$ be as above. Then either $ M $ is
irreducible or there exists $P\in GL_n(D)$ such that $P^{-1}A_dP
\in M $ for any $d\in D^*$.
\end{lem}

\begin{proof}{ If $n=1$, then $M$ is clearly irreducible. Thus, we may assume  $n\geq 2 $. If $M$ is reducible, then there exists an invertible matrix
$P$ and a natural number $0< m < n$ such that
$$PMP^{-1} \subseteq H= \left\{ \left [\begin{array}{cc}
  A & B \\
  0 & C
\end{array}\right]
\mid A\in GL_m(D),C\in GL_{n-m}(D),B\in
GL_{m\times(n-m)}(D)\right\}.$$ By maximality of $M$ in $ N $,
two cases may occur. If $H \cap N=PMP^{-1}$, since $ A_d \in H $
and $ A_d \in SL_n(D) \subset N $ (cf. \cite{artin}), we conclude
that $A_d\in PMP^{-1}$ for any $d\in D^{*}$,and so the result
follows. Otherwise, we have $H \cap N=N$. Since $SL_n(D)\subseteq
N$ we conclude that $SL_n(D)\subseteq H$ which is clearly a
contradiction.
 }\end{proof}

We observe that the above lemma holds also for a maximal subgroup
of a subnormal subgroup in $ GL_n(D) $, as it is known that for $
n \geq 2 $ subnormal subgroups of $ GL_n(D) $ are normal, and for
$ n = 1 $ the conclusion is clear.

\begin{lem} \label{division}
If $G$ is an irreducible subgroup of
$GL_n(D)$, then $C_{M_n(D)}(G)$ is a division ring.
\end{lem}
\begin{proof}{It is clear that $C_{M_n(D)}(G)$ is a ring.
Given $0\neq X\in C_{M_n(D)}(G)$, we may view $ X $ as a
transformation of the row vectors $ D^n $. By definition, $KerX$
is a $D-G$ bi-submodule of $D^{n}$. Since $G$ is irreducible we
conclude that $KerX=0~ or~ D^n$. Thus, we have $KerX=0$ because $
X \neq 0 $. By a similar argument and using the fact the $ X $
commutes with each elements of $ M_n(D) $ one may easily show
that $ImgX=D^n$, and hence $C_{M_n(D)}(G)$ is a division ring.
 }\end{proof}

\begin{lem}\label{abel}
If $ n \geq 2 $ and $M$ satisfies a group identity, then
$C_{M_n(D)}(M)$ is a field.
\end{lem}

\begin{proof}{By Lemma \ref{irreducible}, we know that $M$
is either irreducible or there exist $P\in GL_n(D)$ such that
$P^{-1}A_dP \in M $ for any $d\in D^*$. If the second case
occurs, then one may easily show that $D^*$ satisfies a group
identity and hence, by a theorem of \cite[p.304]{rowen}, we
conclude that $D$ is a field which is a contradiction. Therefore,
we may assume that $M$ is irreducible. Now, by Lemma
\ref{division}, we conclude that $C_{M_n(D)}(M)=D_1$ is a division
ring. Let $ x $ be in the derived group of $ D_1 $, i.e., $x\in
D^{'}_1$. Since $SL_n(D)\subseteq N$ we conclude that $x\in N$.
Now, by maximality of $M$, we have either $\langle x,M \rangle =M$
or $\langle x,M \rangle =N$. In the first case we have $x\in M\cap
D_1$ and so $x\in Z(M)$. In the second case we obtain $x\in
Z(N)$. Now, it is known that $ N $ as a non-central normal subgroup of 
$ GL_n(D) $ with $ n \geq 2 $ must contain $ SL_n(D) $ \cite{artin}. Therefore,
$ Z(N) \subseteq F^* $, i.e., $x\in
Z(N)\subseteq F^{*}$. Hence in any case we have $x\in F^{*}Z(M)$,
and so $D_1'\subseteq F^{*}Z(M)$. This means that $D_1'$ is
abelian and hence, by a result of Hua \cite[p. 223]{lam}, we conclude that $D_1$ is
a field. }\end{proof}

One of the consequences of the above lemma is the following
result which will be used later on.

\begin{lem}\label{solv}
If $ n \geq 2 $ and $M$ soluble, then $C_{M_n(D)}(M)$ is
a field.
\end{lem}

The next result essentially says that if $ D $ is infinite, then
there exists no non-abelian maximal subgroup $M$ of $N$ such that
$\mid M/M \cap F^* \mid < \infty$.

\begin{thm}\label{torsion}
Assume the notations of section 2. Then
there exists no non-abelian maximal subgroup $M$ of $N$ such that
$\mid M/M \cap F^* \mid < \infty$.
\end{thm}

\begin{proof}{
Assume first that $ F^* \subseteq M $, where $ M $ is a
nonabelian maximal subgroup of $ N $. If $n=1$, set $D_1=F[M]$.
Since $\mid M/F^* \mid < \infty$ it is clear that $D_1$ is a
finite dimensional division algebra. If $N \subseteq D_1$, by
Cartan-Brauer-Hua Theorem, we obtain $D_1=D$. Therefore, $D$ is a
finite dimensional division algebra. Let $x_1,\cdots,x_t$ be the
representatives for cosets of $F^*$ in $M$, i.e., $M=F^*x_1 \cup
\cdots \cup F^*x_t$. Then, we have $M= \langle x_1,\cdots,x_t
\rangle F^*$, where  $\langle x_1,\cdots,x_t \rangle$ is the
group generated by $x_1,\cdots,x_t$. Take $x \in N \setminus M$.
By maximality of $M$, we obtain $N= \langle x_1,\cdots,x_t,x
\rangle F^*$. Put $H= \langle x_1,\cdots,x_t,x \rangle$ so that
$N=HF^*$. Therefore, $N'= H$ is a normal subgroup in $D^*$. Now,
by Corollary 1 of \cite{mmy}, we conclude that $N' \subseteq
F^*$. This implies that $N$ is central which is a contradiction.
Therefore, we must have $N \nsubseteq D_1$. Now, by maximality of
$M$ we obtain $N \cap D_1=M$. It is clear that $M=D_1 \cap N $ is
a subnormal subgroup of $D_1^*$. Since $|M/F^*|< \infty $, by a
result of Herstien \cite{Herstien}, we obtain $M \subseteq
Z(D_1)$ which is a contradiction to our assumption.

We now assume that $n \geq 2$. By Lemma \ref{irreducible}, we
conclude that either $M$ is irreducible or there exists $P\in
GL_n(D)$ such that $P^{-1}A_dP \in M $ for all $d\in D^*$. If the
second case occurs, we obtain $A_d^t = aI$, where $ a\in F^* $
and $ t = [M:F^*] $. This in turn implies that $ d^{2t} = 1 $.
Therefore, by a result of \cite[p. 225]{lam}, we conclude that
$D$ is finite which is a contradiction. So we may assume the
first case, i.e., $ M $ is irreducible. By a theorem of \cite[p.
14]{shir}, we obtain that $F[M]$ is a prime ring with $[F[M]:F]<
\infty  $. If $F^*$ is finite, then so is $ M$. Now, by Theorem 4
of \cite{colloquium}, which assert that a normal subgroup of
$GL_n(D)$ does not contain any finite maximal subgroup, we arrive
at a contradiction. So, assume that $F^*$ is infinite and set
$K=C_{M_n(D)}(M)$. By Lemma \ref{abel}, we know that $K$ is a
field. Therefore, $Z(M)$ is an integral domain. Since $[F[M]:F] <
\infty $ we conclude that $Z(M)$ is field. Therefore, by a result
of \cite{rowen}, $F[M]$ is simple ring. Thus, by Centralizer
Theorem, we have $M_n(D) \otimes_F F[M]^{op} \simeq K \otimes_F
M_s(F) \simeq M_s(K)$, for some positive integer $ s $.
Therefore, $M_n(D)$ is a PI-ring and hence, by a result of
Kaplansky, $D$ is a finite dimensional division algebra over its
center. Now, let $x_1,\cdots,x_t$ be a set of representatives for
cosets of $F^*$ in $M$, i.e, $M=F^*x_1 \cup \cdots \cup F^*x_t$.
Then, we have $M= \langle x_1,\cdots,x_t \rangle F^*$, where
$\langle x_1,\cdots,x_t \rangle$ is the group generated by
$x_1,\cdots,x_t$. Take $x \in N \setminus M$. By maximality of
$M$, we obtain $N= \langle x_1,\cdots,x_t,x \rangle F^*$. Put $H=
\langle x_1,\cdots,x_t,x \rangle$ so that $N=HF^*$. This implies
that $N'= H$ is a normal subgroup of $GL_n(D)$. Now, by Corollary
1 of \cite{mmy}, we conclude that $N' \subseteq F^*$ and hence
$N$ is central, which is a contradiction and so the result follows
in this case.

Finally, if $ F^* $ is not a subset of $ M $, set $M_1=MF^*$ and
$N_1=NF^*$. It is clearly seen that $M_1$ is a maximal subgroup
of $N_1$. If $\mid M/M \cap F^* \mid $ is finite, then so is
$\mid M_1/ F^* \mid $. Therefore, this reduces to the first case
and so the result follows. }\end{proof}

As a consequence of the above theorem, setting $ N = GL_n(D) $, we
obtain a short proof of Lemma 1 in \cite{m}, i.e.,

\begin{cor} \label{alttorsion}
Let $D$ be a division ring not necessarily of finite dimension
over its center $F$. If either $n=1$ and $D$ is noncommutative or
$n > 1$ and $D$ infinite, then there exists no maximal subgroup
$M$ of  $GL_n(D)$, $n \geq 1$, containing $F^*$ such that $\mid
M/F^* \mid < \infty$ .
\end{cor}


\begin{proof}{
By Theorem \ref{torsion}, there exists no nonabelian maximal
subgroup with the stated property. If $M$ is abelian, by Lemma
\ref{abel}, we conclude that $K=C_{M_n(D)}(M)$ is a field. Now,
by maximality of $M$, we obtain $M=K^*$, and hence $K^*/F^*$ is
finite. But it is known that this is not possible unless $M=K^*$
is finite. Now, by Corollary 3.11 of \cite{kiani}, we conclude
that $D=F$ and $D$ is finite, which is a contradiction.
 }\end{proof}

The next result provides us with a criterion for $ M $ to be
nilpotent.

\begin{thm}\label{nil}
Let $D$ be a noncommutative finite dimensional $ F $-central
division algebra. Then, $M$ is nilpotent if and only if $M$ is contained in the
multiplicative group of a subfield of $M_n(D)$.
\end{thm}

\begin{proof}{ One way is clear. Assume that $ M $ is
nilpotent and $n \geq 2$. By Lemma \ref{irreducible}, either $M$
is irreducible or there exists $P\in GL_n(D)$ such that
 $P^{-1}A_dP \in M $ for any $d\in D^*$. If the second case
happens, then one may easily conclude that $D^*$ is soluble.
Therefore, by Hua's Theorem \cite[p. 223]{lam}, $D$ is commutative
which is a contradiction. Therefore, we may assume that $M$ is
irreducible. By a result of \cite[p. 14]{shir},
$F[M]$ is a prime ring. Set $K=C_{M_n(D)}(M)$. By Lemma
\ref{solv}, we know that $K$ is field. Therefore, $Z(M)$ is an
integral domain. Since $[F[M]:F] < \infty $ we conclude that
$Z(M)$ is field. Now, by a result of \cite[p. 47 ]{rowen}, $F[M]$
is a simple ring. Thus, by Artin-Wedderburn's Theorem, we obtain
$F[M] \simeq M_{n_1}(D_1)$ for some positive integer $n_1$ and a
division ring $D_1$. Since $SL_n(D)\subseteq N$, we have
$SL_{n_1}(D_1) \subseteq (GL_n(D)'=SL_n(D) \subseteq N$. If
$SL_{n_1}(D_1) \subseteq M$, then, by a theorem of
\cite[p.154]{shir}, we conclude that $D_1$ is a locally finite
field. This implies that $F $ is a locally finite. Therefore, $D$
is algebraic over a finite field and hence, by Jacobson's Theorem
\cite[p.219]{lam}, $D$ is commutative which is a contradiction.
Thus, $SL_{n_1}(D_1) \nsubseteq M$ and by maximality of $M$ in $
N $ we must have $MSL_{n_1}(D_1)=N$. Since $ SL_{n_1}(D_1)
\subseteq F[M]$ we have $N \subseteq F[M]$ and hence by
Cartan-Brauer-Hua Theorem \cite{rose}, we obtain $F[M]=M_n(D)$.
Therefore, one may easily show that $Z(M)=M \cap F^*$. Since $ D
$ is of finite dimension over $ F $, we may view $ M $ as an
irreducible nilpotent linear group. Therefore, by a theorem of
\cite[p. 57]{sup1}, we obtain $[M:Z(M)] < \infty$. Now, by Theorem
\ref{torsion}, we conclude that $M$ is abelian and so $M \subseteq
K=C_{M_n(D)(M)}$ and the result follows in this case. It remains
to show that the result also holds for $ n = 1 $. So, let $n=1$
and set $D_1=F[M]$. If $N \subseteq D^*_1$, then, by
Cartan-Brauer-Hua Theorem, we have $F[M]=D$. Therefore, we obtain
$Z(M)=M \cap F^*$. Now, by the same argument as used above, we
conclude that $M$ is abelian and so the result follows in this
case. Finally, assume that $N \nsubseteq D_1$. Thus, by
maximality of $M$, we have $N\cap D_1=M$. It is clear that $D_1
\cap N \lhd D^*_1$ and so $M\lhd D^*_1$. Now, we know that any
noncentral normal subgroup of a finite dimensional division
algebra contains a noncyclic free subgroup \cite{g2}. Therefore,
$M$ as a noncentral normal subgroup of $ D_1^* $ contains a
noncyclic free subgroup. This is in contradiction with the fact
that $ M $  is nilpotent and hence result follows. }\end{proof}

Now, as a consequence of the above result, we may easily obtain
the following corollary which is the last part of Proposition 2 in
\cite{m}.

\begin{cor}\label{altnil}
Let $D$ be a noncommutative finite dimensional $ F $-central
division algebra, and let $M$ be a maximal subgroup of $GL_n(D)$
with $n \geq 1$. Then, $M$ is nilpotent if and only if $M$ is the
multiplicative group of a maximal subfield of $M_n(D)$.
\end{cor}

\begin{proof}{ Set $ N = GL_n(D) $. By Theorem \ref{nil}, $ M \subseteq K^* \subseteq
GL_n(D) $, where $ K $ is a subfield of $ M_n(D) $. Now, by maximality of $ M $, the result follows.}
\end{proof}

Given a maximal subgroup $ M $ of $ GL_n(D) $, it is proved in
\cite{m} that either $M$ contains a noncyclic free subgroup or
there exists a finite family $\{K_i\}^r_1$ of fields with $F^*
\subset K^*_i\subset M$ for all $1\leq i\leq r$ such that $M/A$
is finite if $ Char F = 0 $ and $ M/A $ is locally finite if $
Char F = p>0 $, where $A=K^*_1\times\dots\times K^*_r$. Now, let
$ N $ be a normal subgroup of $ GL_n(D) $ with $ n \geq 1 $, and
$ M $ be a maximal subgroup of $ N $. In the next result we
essentially generalize the above mentioned result to maximal
subgroups of $ N $ as follows:

\begin{thm}\label{tits}
Let $D$ be a noncommutative finite dimensional $ F $-central
division algebra, and $N$ be a non-central normal subgroup of
$GL_n(D)$ with $n \geq 1$. Given a maximal subgroup $ M $ of $N$,
then, either $M$ contains a noncyclic free subgroup or there
exist an abelian subgroup $ A $ and a finite family $\{K_i\}^r_1$
of fields properly containing $ F $ with $ K^*_i \subset M $ for
all $1 \leq i \leq r$ such that $M/A$ is finite if $CharF=0$ and
$M/A$ is locally finite if $Charf=p>0$, where $A \subseteq K^*_1
\times \cdots \times K^*_r$.
\end{thm}

\begin{proof}{ We first consider the case $n=1$. If $ M $ is abelian,
then as in the proof of last corollary, $ M $ is the multiplicative group
of a subfield of $ D $ and so the result follows. So, we may
assume that $ M $ is nonabelian. Set $E=F[M]$. By maximality of
$M$, we have either $N \cap E^*=M$ or $N \subseteq E^*$. If the
first case occurs, we conclude that $M $ is normal in $ E^*$.
Therefore, by a result of \cite{g2}, $M$ contains a noncyclic
free subgroup. If the second case happens, by Cartan-Brauer-Hua
Theorem, we have $F[M]=D$. Assume that $M$ does
not contain a noncyclic free subgroup and consider the following cases.\\

Case 1: $CharF=0$. Since $M$ does not contain a noncyclic free
 subgroup we conclude, by Theorem 1 of \cite{j.t}, that $M$
contains a soluble normal subgroup $ T $ of finite index, i.e.,
$[M:T]< \infty$. If $ T \subset F^* $, then we obtain $ [M:F^*]<
\infty $ which contradicts Theorem \ref{torsion}. Now, $T $ as a
subgroup of $ D^*$ is a completely reducible linear group.
Therefore, by a theorem of \cite[p. 154]{sup1}, $T$ is
abelian-by-finite. Thus, $M$ contains an abelian normal subgroup
$A$, say, of finite index. If $ A \subset F^* $, then we obtain $
[M:F^*]< \infty $ which contradicts Theorem \ref{torsion}.
Therefore, $ A $ is noncentral and $A \subseteq F[A]^*=K^*_1$,
where $K_1$ is a field and so the result follows in this case.\\

Case 2: $CharF= p>0$. Since $M$ does not contain a noncyclic free
subgroup, by Tit's Theorem \cite{j.t}, we conclude that every
finitely generated subgroup of $M$ contains a soluble normal
subgroup of finite index. Therefore, by a result of Wehrfritz
\cite{wehr}, $M/Solv(M)$ is a torsion linear group, where
$Solv(M)$ is the unique maximal soluble normal subgroup obtained
by Zassenhaus-Maltsev Theorem \cite{platonov}. Therefore, by
Schur's Theorem, $M/Solv(M)$ is locally finite. Set $S=Solv(M)$.
Now, as in the above case, $ S $ contains an abelian normal
subgroup $ B $ of finite index. Therefore, $ M/B $ is locally
finite and we have $ B \subseteq F[B]=K $ which completes the
proof of this
case.\\

Now, assume that $n>1$. If $ M $ is abelian, by Theorem
\ref{nil}, the result follows. So, we may assume that $ M$ is
nonabelian. By Lemma \ref{irreducible}, we have that either $M$ is
irreducible or there exists $P\in GL_n(D)$ such that $P^{-1}A_dP
\in M$ for any $d\in D^*$. If the second case occurs, then $ M $
contains a copy of $D^*$. Now, by a result of \cite{g1}, we know
that $ D^* $ contains a noncyclic free subgroup and hence so does
$ M $. If the first case happens, by Theorem \cite[p. 9]{shir}, we
conclude that $F[M]$ is prime ring. Now, by Lemma \ref{division},
$C=C_{M_n(D)}(M)$ is a division ring. We note that $Z(F[M])
\subseteq C$. Therefore, $Z(F[M])$ is an integral domain. Since
$[Z(F[M]):F] < \infty$ we conclude that $Z(F[M])$ is field. Now,
$F[M]$ is a prime PI-ring and $Z(F[M])$ is a field. Thus, by a
result of \cite[p. 47]{rowen}, $F[M]$ is simple ring and hence by
Artin-Wedderburn's Theorem, we have $F[M] \simeq M_{n_1}(D_1)$,
for some positive integer $n_1$ and a division ring $D_1$. Since
$SL_n(D)\subseteq N$ we have $SL_{n_1}(D_1) \subseteq
(GL_n(D)'=SL_n(D) \subseteq N$. By maximality of $M$ in $ N $, we
may consider two cases.\\

Case 1: $SL_{n_1}(D_1) \subseteq M$. If $n_1 > 1$, by Theorem
\cite[p.154]{shir}, either $SL_{n_1}(D_1)$ contains a noncyclic
free subgroup, or $D_1$ is a locally finite field. In the first
case $ M $ contains a noncyclic free subgroup and so the result
follows. The second case implies that $F$ is locally finite. This
in turn asserts that $D$ is algebraic over a finite field and
hence, by Jacobson's Theorem \cite[p.219]{lam}, we conclude that
$D$ is a field which is a contradiction. If $n_1=1$, then $D_1$ is
noncommutative since otherwise $ M $ is abelian which contradicts
our assumption. Now, $D^{'}_1$ as a normal subgroup of $ D_1^* $
contain a noncyclic free subgroup. Therefore, $M$ contain
noncyclic free subgroup and so the result follows in this case.\\

Case 2: $SL_{n_1}(D_1) \nsubseteq M$. By maximality of $M$, we
have $MSL_{n_1}(D_1)=N$. Since $M $ and $ SL_{n_1}(D_1) $ are
contained in $ F[M]$ we have $N \subseteq F[M]$. Now, by
Cartan-Brauer-Hua Theorem \cite{rose}, we have $F[M]=M_n(D)$. If
$M$ does not contain a noncyclic free subgroup, we may consider
the
following subcases.\\

Subcase 1: $CharF=0$. Since $M$ does not contain a noncyclic free
 subgroup we conclude, by Theorem 1 of \cite{j.t}, that $M$
 contains a soluble normal subgroup $ T $ of finite index, i.e. $[M:T]<
 \infty$. By Theorem \cite[p. 14]{shir}, $F[T]$ is semisimple.
Since $T$ is a completely reducible linear group, by a result of
\cite[p. 154]{sup1}, $T$ is abelian-by-finite. Therefore, $M$
contains an abelian normal subgroup $A$, say, of finite index.
Now, by a theorem of \cite[p. 14]{shir}, $F[A]$ is commutative
semisimple ring. Therefore, by Artin-wedderburn's Theorem, there
exists a finite family $\{K_i\}^r_1$ of fields with $F^* \subset
K^*_i $ for all $1 \leq i \leq r$ such that $F[A]^ \simeq K_1
\times \cdots \times K_r$. Thus, $A \subseteq F[A]^* \simeq K^*_1
\times \cdots \times K^*_r$, and so the result follows in this case.\\

Subcase 2: $CharF= p>0$. Since $M$ does not contain a noncyclic
free subgroup, by Tit's Theorem \cite{j.t}, we conclude that
every finitely generated subgroup of $M$ contains a soluble
normal subgroup of finite index. Therefore, by a result of
Wehrfritz \cite{wehr}, $M/Solv(M)$ is a torsion linear group,
where $Solv(M)$ is the unique maximal soluble normal subgroup
obtained by Zassenhaus-Maltsev Theorem \cite{platonov}.
Therefore, by Schur's Theorem, $M/Solv(M)$ is locally finite. Set
$S=Solv(M)$. Now, by a result of \cite[p. 14]{shir}, $F[S]$ is
semisimple. As in Subcase 1, $S$ contains an abelian normal
subgroup $B$ ,say, of finite index. Therefore, $M/B$ is locally
finite. By Theorem \cite[p. 14]{shir} $F[B]$ is a commutative
semisimple ring. Therefore, as in Subcase 1, $B \subseteq F[B]^*
\simeq K^*_1 \times \cdots \times K^*_r$, and so the proof is
complete. }\end{proof}

As a consequence of the above theorem, we may prove the following
corollary which is the main result of \cite{m}.

\begin{cor}\label{alttits}
Let $D$ be a noncommutative finite dimensional $ F $-central
division algebra. Assume that $M$ is a maximal subgroup of
$GL_n(D)$ with $n \geq 1$. Then, either $M$ contains a noncyclic
free subgroup or there exists a finite family $\{K_i\}^r_1$ of
fields with $F^* \subset K^*_i \subset M $ for all $1 \leq i \leq
r$ such that $M/A$ is finite if $CharF=0$ and $M/A$ is locally
finite if $Charf=p>0$, where $A = K^*_1 \times \cdots \times
K^*_r$.\end{cor}

\begin{proof}{
If $M$ does not contain a noncyclic free subgroup, by Theorem
\ref{tits} and a result of \cite[p. 14]{shir}, we have $F[A]^*
\simeq K^*_1 \times \cdots \times K^*_r$. If $F[A]^* \subseteq
M$, then the result follows. If $L=F[A]^* \nsubseteq M$, by
maximality of $M$, we obtain $LM=GL_n(D)$. Therefore, $ L $ is
normal in $GL_n(D)$, i.e., $ SL_n(D) \subset L $. But this
contradicts the fact that $ SL_n(D) $ contains a noncyclic free
subgroup and so the result follows. }\end{proof}


The first author is indebted to the Institute for Studies in Theoretical Physics and Mathematics (IPM) for partial support (Grant No. 83160044). 
The second author thanks Professor Ulf Rehmann for his hospitality during 
his stay at the Bielefeld University in March 2006.  \vspace{0.2 cm}

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{\footnotesize\textsc D.~Kiani, Department of Pure Mathematics, Faculty of Mathematics and Computer Science, Amirkabir University of Technology (Tehran Polytechnic), 424, Hafez Ave., Tehran 15914, Tehran, Iran, and Institute for Studies in Theoretical Physics and Mathematics (IPM)}

{\footnotesize\textsc M.~Mahdavi-Hezavehi, Department of Mathematical Sciences, Sharif University of Technology, P. O. Box 11365-9415, Tehran, Iran, and Fakult\"at f\"ur Mathematik, Universit\"at Bielefeld, Postfach 100131, D-33501 Bielefeld, Germany}
\end{document}

