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\begin{document}

\author{Mikl\'{o}s Ab\'{e}rt \and Alexander Lubotzky \and L\'{a}szl\'{o} Pyber}
\title{Bounded generation and linear groups}
\date{6 September, 2001.}
\maketitle

\begin{abstract}
A group $\Gamma $ is called boundedly generated (BG) if it is the
set-theoretic product of finitely many cyclic subgroups. We show that a BG
group has only abelian by finite images in positive characteristic
representations.

We use this to reprove and generalise Rapinchuk's theorem by showing that a
BG group with the FAb property has only finitely many irreducible
representations in any given dimension over any field. We also give a
structure theorem for the profinite completion $G$ of such a group $\Gamma $.

On the other hand, we exhibit boundedly generated profinite FAb groups which
do not satisfy this structure theorem.
\end{abstract}

\section{Introduction}

A group (resp. profinite group) $\Gamma $ is said to have \emph{bounded
generation} (or finite cyclic width) if $\Gamma $ is a product of its cyclic
(resp. procyclic) subgroups $C_{1},\ldots ,C_{k}$. The smallest number $k$
for which $\Gamma $ has such a decomposition is called the \emph{cyclic width%
} of $\Gamma $ (which we denote by $cw(\Gamma )$).

Following the discovery \cite{carkell} that the groups $SL(n,\Bbb{Z})$ ($%
n\geq 3$) have bounded generation, it has been shown that many other $S$%
-arithmetic groups over number fields have this property \cite{tavgen}, and
the notion received a considerable amount of attention \cite{platrap}, \cite
{lubot}, \cite{rapinchuk2}, \cite{liepyb} and \cite{kumar}.

Recall that $\Gamma $ is said to have property FAb if for any subgroup $%
\Gamma _{0}\subseteq \Gamma $ of finite index the commutator quotient $%
\Gamma _{o}/[\Gamma _{o},\Gamma _{o}]$ is finite.

A major open problem in the area proposed by Bass \cite{bass} is whether a
residually finite FAb BG group $\Gamma $ is linear. This is indeed the case
if $\Gamma $ is residually nilpotent \cite{manseg} (in this case we don't
even have to assume property FAb).

In the current paper we make a step forward in understanding the structure
of FAb BG groups and their linear representations.

We first show that unlike in characteristic $0$, there are very few
boundedly generated groups in characteristic $p$.

\begin{theorem}
\label{linfotetel}Let $\Bbb{F}$ be a field of positive characteristic. If $%
\Gamma $ is a boundedly generated subgroup of $GL(n,\Bbb{F})$, then $\Gamma $
has an Abelian normal subgroup of finite index.
\end{theorem}

>From this we can also deduce the following result, which was proved by
Rapinchuk \cite{rapinchuk} in characteristic $0$.

\begin{corollary}
\label{vegesrep}Let $\Gamma $ be an FAb BG group and let $\Bbb{F}$ be an
arbitrary field. Then $\Gamma $ has finitely many inequivalent completely
reducible representations over $\Bbb{F}$ in any given dimension.
\end{corollary}

\medskip 

Using the corollary, we show the following.

\begin{theorem}
\label{keveslie}Let $\Gamma $ be an FAb BG group and $p$ a prime. Then $%
\Gamma $ has finitely many upper composition factors $L$ in $Lie(p)$.
\end{theorem}

Theorem \ref{keveslie} together with some results from \cite{bamapy} gives
the following structure theorem for the profinite completions of FAb BG
groups.

\begin{theorem}
\label{strucc}Let $G=\widehat{\Gamma }$ be the profinite completion of an
FAb BG group $\Gamma $. Then there exist a series 
\[
1\subseteq S\subseteq H\subseteq G
\]
\newline
of normal subgroups of $G$ and a natural number $r$ (which depends only on $%
cw(\Gamma )$) such that \newline
1) $G/H$ is finite;\newline
2) $S$ is prosolvable;\newline
3) $H/S$ is a cartesian product of finite simple groups of Lie type such
that the multiplicities and the Lie ranks of these simple groups are at most 
$r$;\newline
4) For every prime $p$ only finitely many of these simple groups are in $%
Lie(p)$.
\end{theorem}

\begin{corollary}
\label{justif}If $\Gamma $ is in addition just infinite (i.e., every
non-trivial normal subgroup has finite index), then exactly one of the
following holds:\newline
0) $\Gamma $ is finite;\newline
1) $\Gamma $ is not linear and it has a finite index subgroup $\Gamma _{0}$
such that every finite quotient of $\Gamma _{0}$ is solvable;\newline
2) $\Gamma $ is a linear group embeddable as a Zariski-dense subgroup in an $%
S$-arithmetic subgroup of a semisimple algebraic group $\mathcal{G}$ defined
over a number field $K$.
\end{corollary}

So to give a positive answer to the question of Bass for just infinite
groups one should eliminate case 1) above.

It is an interesting problem to decide whether a group $\Gamma $ as in
Corollary \ref{justif}, case 2) is in fact $S$-arithmetic. This question may
be considered as a replacement of Platonov's conjecture, to which a
counterexample has been given recently \cite{basslub}.

Finally we show that Corollary \ref{vegesrep}, Theorem \ref{keveslie} and
Theorem \ref{strucc}, part 4) do not hold for general FAb BG profinite
groups. In Section 3 we consider examples like $\prod_{p}PSL(2,2^{p})$ where 
$p$ runs over all primes and prove that these groups are BG and FAb. At the
same time we note that $\prod_{n}PSL(2,2^{n})$ where $n$ runs over all
positive integers, is not BG.

Recall that in \cite{pyber} it was shown that BG groups have subgroup growth
at most $n^{c\log n}$. We also construct BG profinite groups of this growth
type showing that the result in \cite{pyber} is best possible. We do not
know whether such a discrete group exists.

\section{Boundedly generated groups}

We start with a slight extension of a result of Shalev \cite[Proposition 5.6]
{shalev}.

\begin{proposition}
\label{analprop}Let $G$ be a $p$-adic analytic pro-$p$ group. Assume $G$ can
be embedded in $GL(n,\Bbb{K})$ for some local field $\Bbb{K}$ of positive
characteristic $\ell $. Then \newline
1) If $\ell \neq p$ then $G$ is finite. \newline
2) If $\ell =p$ then $G$ is virtually abelian.
\end{proposition}

If $\ell \neq p$, then the $p$-Sylow subgroups of any profinite subgroup of $%
GL(n,\Bbb{K})$ are finite, so $G$ must be finite.

The case $\ell =p$ is deduced from the following result of Pink 
\cite[Corollary 0.5]{pink}.

\begin{theorem}
Let $\Bbb{K}$ be a local field of characteristic $p>0$ and $L$ a compact
subgroup of $GL(n,\Bbb{K})$. Then there exist normal subgroups $%
L_{3}<L_{2}<L_{1}<L$ such that: \newline
1) $L/L_{1}$ is finite. \newline
2) $L_{1}/L_{2}$ is abelian of finite exponent. \newline
3) There exists a local field $E$ of characteristic $p$, a connected adjoint
semi-simple group $H$ over $E$ with universal covering $\pi :\tilde{H}\to H$
and an open compact subgroup $\Delta \subseteq \tilde{H}(E)$, such that $%
L_{2}/L_{3}$ is isomorphic to $\pi (\Delta )$ as a topological group.\newline
4) $L_{3}$ is a solvable subgroup of derived length at most $\log _{2}n$.
\end{theorem}

We apply Pink's theorem for $L=G$ in proving case 2 of Proposition \ref
{analprop}. As $G$ is a $p$-adic analytic pro-$p$ group, it is finitely
generated and so is $L_{1}$. Moreover $L_{1}/L_{2}$ is abelian of finite
exponent and finitely generated, hence finite and so $L_{2}$ is also of
finite index. Now, $L_{2}/L_{3}$ is a $p$-adic analytic pro-$p$ group which
is isomorphic to an open compact subgroup $\Delta $ of a semi-simple group $%
H $ over a characteristic $p$ local field $E$. This is impossible. Indeed, $%
E $ is isomorphic to $\Bbb{F}_{q}((t))$ for some $q=p^{\alpha },\;\alpha \in 
\Bbb{N}$, and $\Delta $ is commensurable to $H(\Bbb{F}_{q}[[t]])$. We claim
that the latter has infinite rank. For $n\in \Bbb{N}$, let 
\[
K(n)=Ker(H(\Bbb{F}_{q}[[t]])\to H(\Bbb{F}_{q}[[t]]/(t^{n})))\text{.} 
\]
It is easy to see that $[K(n),K(n)]\subset K(2n)$ and $K(n)^{p}\subset K(pn)$%
. Thus $K(n)/K(2n)$ is an elementary abelian $p$-group. When $n$ is going to
infinity, the rank of $K(n)/K(2n)$ is going to infinity as well. Thus $H(%
\Bbb{F}_{q}[[t]])$, and $\Delta $, cannot be of finite rank and cannot be
isomorphic to a $p$-adic analytic pro-$p$ group \cite{sms}. (One could also
deduce the same conclusion from \cite[Corollary 0.3]{pink} which shows that
if two open compact subgroups of simple algebraic groups over local fields
are isomorphic, then the fields are the same, and the algebraic groups are
isomorphic).

We can therefore deduce that $H$ is trivial and $L_{3}$ is of finite index
in $L=G$. We have concluded that $G$ is virtually solvable. We claim now
that $G$ is actually virtually abelian. Indeed, if $G$ is a solvable
subgroup of $GL(n,\Bbb{K})$, then over $\overline{\Bbb{K}}$, the algebraic
closure of $\Bbb{K}$, a finite index subgroup of $G$ is conjugate to the
group of upper triangular matrices. So we can assume that $G$ is upper
triangular, hence unipotent by abelian. But the upper unipotent subgroup of $%
GL(n,\overline{\Bbb{K}})$ is torsion. As $G$ is a $p$-adic analytic pro-$p$
group it is virtually torsion-free \cite[4.20]{sms}. The finite index
torsion-free subgroup of $G$ has trivial intersection with the unipotent
subgroup hence it must be abelian. This finishes the proof of Proposition 
\ref{analprop}.

\medskip

We deduce the following.

\begin{proposition}
\label{cicamica}Let $\Bbb{F}$ be a field of positive characteristic $p$.
Suppose that $\Gamma $ is a finitely generated subgroup of $GL(n,\Bbb{F})$
such that the pro-$p$ completion of every finite index subgroup of $\Gamma $
is $p$-adic analytic. Then $\Gamma $ is virtually abelian.
\end{proposition}

\noindent \textbf{Proof. }As $\Gamma $ is finitely generated it is actually
in $GL(n,\Bbb{A})$ where $\Bbb{A}$ is a finitely generated subring of $\Bbb{F%
}$. Now $\Bbb{A}$ can be embedded into a local field. Moreover, it can be
embedded into the ring of integers of some local field. So we can assume
that $\Gamma $ is a subgroup of $M=GL(n,\Bbb{F}_{q}[[x]])$ where $q$ is a
power of $p$ and $\Bbb{F}_{q}$ is the field of order $q$. $M$ is virtually
pro-$p$ so $\Gamma $ has a finite index subgroup $\Gamma _{0}$ such that the
closure $G$ of $\Gamma _{0}$ in $M$ is a pro-$p$ group. By our hypothesis $G$
must be $p$-adic analytic (being a quotient of a $p$-adic analytic pro-$p$
group). We can now apply Proposition \ref{analprop} to deduce that $G$,
hence $\Gamma $ is virtually abelian. $\square $

\medskip

This general criterion can be used in conjunction with the known
characterisations of $p$-adic analytic pro-$p$ groups to derive various
results.

\medskip

\noindent \textbf{Proof of Theorem \ref{linfotetel}.}

Let $\Gamma $ be a boundedly generated subgroup of $GL(n,\Bbb{F})$. Then
every finite index subgroup $\Gamma _{0}$ of $\Gamma $ is boundedly
generated \cite{tavgen}, hence the same is true for the pro-$p$ completion $%
G_{0}$ of $\Gamma _{0}$. By a result of Lazard (see \cite{sms}) every such
group $G_{0}$ is $p$-adic analytic.

Applying Proposition \ref{cicamica} we see that $\Gamma $ is virtually
abelian. $\square $

\medskip

Recall that a group $\Gamma $ is said to have polynomial index growth (PIG),
if there is a constant $c$ such that for every finite quotient $\Gamma /N$
we have $\left| \Gamma /N\right| \leq (\exp (\Gamma /N))^{c}$ (where $\exp
(G)$ denotes the exponent of the group $G$). It is clear that boundedly
generated groups have PIG. Using the results of Lazard (see \cite{sms} and 
\cite{shalev1}) together with Proposition \ref{cicamica} it follows that in
fact the conclusion of Theorem \ref{linfotetel} holds under the weaker
assumption that $\Gamma $ has PIG.

We point out another consequence of Proposition \ref{cicamica}.

Let $\Gamma $ be a finitely generated residually-finite group and let $%
s_{n}(\Gamma )$ denote the number of subgroups of $\Gamma $ of index at most 
$n$. The asymptotic growth of $s_{n}(\Gamma )$ has been a topic of intensive
research in the last decade -- see \cite{luseg} and the references therein.

In \cite{lubot} it was shown that a subgroup growth gap exists for groups $%
\Gamma $ which are linear over fields of characteristic $0$.

\begin{theorem}
Let $\Bbb{F}$ be a field of characteristic $0$ and $\Gamma $ a finitely
generated subgroup of $GL(n,\Bbb{F})$. Then one of the following holds:%
\newline
1) $\Gamma $ is a virtually solvable group of finite rank (and therefore it
has polynomial subgroup growth); \newline
2) There exists a constant $c$ such that $s_{n}(\Gamma )\ge n^{c\log n/\log
\log n}$ for every $n$.
\end{theorem}

Here we prove that a larger subgroup growth gap exists for linear groups
over fields of positive characteristic.

\begin{theorem}
Let $\Bbb{F}$ be a field of characteristic $p>0$ and $\Gamma $ a finitely
generated subgroup of $GL(n,\Bbb{F})$. The one of the following holds: 
\newline
1) $\Gamma $ is a virtually abelian group of finite rank (and therefore it
has polynomial subgroup growth);\newline
2) There exists a constant $c$ such that $s_{n}(\Gamma )\ge n^{c\log n}$ for
infinitely many $n$.
\end{theorem}

\noindent \textbf{Proof. }Suppose that $\Gamma $ is not virtually abelian.
Then by Proposition \ref{cicamica} $\Gamma $ has a subgroup $\Gamma _{0}$,
say, of index $t$ such that the pro-$p$ completion $G_{0}$ of $\Gamma _{0}$
is not $p$-adic analytic. By \cite[Corollary 2.5]{shalev1} there exists
infinitely many $k$ for which $s_{p^{k}}(G_{0})\geq p^{k^{2}/9}$. Hence for
infinitely many numbers $n$ we have 
\[
s_{n}(\Gamma )\geq (n/t)^{\log (n/t)/9\log p}\geq \frac{n^{\log n/9\log p}}{%
n^{\frac{2\log t}{9\log p}}}\text{.} 
\]
This implies our statement. $\square $

\medskip

By a recent result of Segal \cite{segal2} no such gaps exist for the
subgroup growth of arbitrary finitely generated groups.

We remark that for $\Gamma =SL(n,\Bbb{Z})$ ($n\geq 3$), and many other $S$%
-arithmetic groups $\Gamma $ we have $s_{n}(\Gamma )\leq n^{\frac{c^{\prime
}\log n}{\log \log n}}$ (\cite{lubot}). By a very recent result of Nikolov 
\cite{nikolov}, for $\Gamma =SL(n,F_{p}[t])$ ($n\geq 3$) we have $%
s_{n}(\Gamma )\leq n^{c^{\prime }\log n}$. Hence the above bounds are sharp.

We need the following result which is proved implicitly in \cite{LMS}.

\begin{lemma}
\label{fablem}Let $\Gamma $ be a finitely generated group, $n$ a positive
integer and $\Omega $ a family of fields of characteristic $p$. If $\Gamma $
can be embedded in the Cartesian product $\prod\limits_{F\in \Omega }GL(n,F)$
then $\Gamma $ is a subdirect product of finitely many linear groups of
degree $n$ over fields of characteristic $p$.
\end{lemma}

Our next result extends Rapinchuk's theorem \cite{rapinchuk}.

\medskip

\noindent \textbf{Proof of Corollary \ref{vegesrep}. }

Case 1. By Theorem \ref{linfotetel} any representation of $\Gamma $ over a
field of positive characteristic has finite image. Let $\pi $ be a complex
linear representation of $\Gamma $. Since $\Gamma $ is finitely generated,
there is a finitely generated ring $A$ with $\pi (\Gamma )\subseteq GL(n,A)$%
. For any transcendental $a\in A$, there is a ring homomorphism $\zeta
:A\rightarrow \Bbb{K}$ where $\Bbb{K}$ is a field of positive characteristic
with $\zeta (a)$ still transcendental. If for some $\gamma \in \Gamma $ we
have $tr(\pi (\Gamma ))\notin \overline{\Bbb{Q}}$ then $\zeta $ defines a
representation $\pi _{\zeta }$ over $\Bbb{K}$ with $tr(\pi _{\zeta }(\gamma
))$ transcendental. Hence $\pi _{\zeta }(\Gamma )$ is infinite, a
contradiction.

We proved that the traces of all finite dimensional complex linear
representations are algebraic numbers which is equivalent to our conclusion,
when the characteristic of $\Bbb{F}$ is $0$ (see \cite[Example 5.12 (3)]
{bass2}).

Case 2. Suppose now that $\Bbb{F}$ has characteristic $p$. Denote by $K$ the
intersection of the kernels of $\Bbb{F}$-representations of degree $n$ of
the group $\Gamma $. By Lemma \ref{fablem} $\Gamma /K$ is a subdirect
product of finitely many linear groups of degree $n$ over fields of
characteristic $p$. Each of these linear groups is abelian by finite by
Theorem \ref{linfotetel} hence the same is true for $\Gamma /K$. Since $%
\Gamma $ is an FAb group this implies that $\Gamma /K$ is finite. Hence the
number of irreducible $\Bbb{F}$-representations of degree $n$ of $\Gamma $
is bounded by the total number of irreducible $\Bbb{F}$-representations of $%
\Gamma /K$ which is at most $\left| \Gamma /K\right| $ \cite{berman}. Let $%
\rho $ be a completely reducible representation of degree $n$ of $\Gamma $.
By the above we have finitely many choices for the irreducible components $%
\rho _{1},\ldots ,\rho _{t}$ of $\rho $. The images $\rho _{i}(\Gamma )$ are
finite and $\rho (\Gamma )$ is a subgroup of their direct product hence
there are finitely many choices for $\rho $. $\square $

\medskip

\noindent \textbf{Remark.} As the above proof shows, when the field $\Bbb{F}$
has positive characteristic, we can say a bit more; in this case all
representations of a given degree factor through some finite quotient $%
\Gamma /K$ of $\Gamma $.

\medskip

Recall that an \emph{upper section} of $\Gamma $ is a group $X\cong H/K$
with $H$ and $K$ finite index subgroups of $\Gamma $, $K$ normal in $H$. If $%
H$ is subnormal in $\Gamma $ and $X$ is simple then $X$ is an \emph{upper
composition factor} of $\Gamma ,$ if both $H$ and $K$ are normal in $\Gamma
, $ then $X$ is an \emph{upper factor} of $\Gamma $.

It is proved in \cite{pyber} that the degrees of alternating upper
composition factors and the Lie ranks of Lie-type upper composition factors
of a group $\Gamma $ are bounded in terms of its cyclic width (if it is
finite). Moreover in \cite{bamapy} it is proved that if $L^{r}$ is a
non-abelian upper factor of $\Gamma $, with $L$ simple then $r$ is bounded
in terms of $cw(\Gamma )$.

To prove Theorem \ref{keveslie} we need these results, whose proofs rely on
the classification theorem of finite simple groups (CFSG) and the following
observation.

\begin{lemma}
\label{normfab}Let $\Gamma $ be a finitely generated FAb group, $H$ a normal
subgroup of index $m$ in $\Gamma $ and $H/K$ a soluble upper section of
derived length $d$. Then there is a finite bound for $\left| \Gamma
:K\right| $ which depends only on $m,d$ and $\Gamma $.
\end{lemma}

\noindent \textbf{Proof. }Denote by $I=I(m)$ be the intersection of all
normal subgroups $H$ of index $m$ in $\Gamma $. Since $\Gamma $ is finitely
generated $I$ has finite index in $\Gamma $. Let $I^{(d)}$ be the $d$-th
term of the derived series of $I$. By the FAb property $I^{(d)}$ is of
finite index. This index gives the desired bound. $\square $

\medskip

\noindent \textbf{Proof of Theorem \ref{keveslie}.}

Let $L$ be any non-abelian upper composition factor of $\Gamma $. Since $%
\Gamma $ is finitely generated, there is an upper factor $H/K$ of $\Gamma $
which is isomorphic to $L^{r}$ for some $r$. Set $C=C_{\Gamma }(H/K)$. Then $%
\Gamma /C$ is the finite group of automorphisms that $\Gamma $ induces on $%
H/K$. Hence $\Gamma /C$ contains a subgroup $B/C$ isomorphic to $L^{r}$ such
that $\Gamma /B$ has an embedding into $Out(L)wrSym(r)$. As mentioned above $%
r$ is bounded in terms of $cw(\Gamma )$. It is a well-known consequence of
CFSG that $Out(L)$ is soluble of derived length at most $3$, hence $\Gamma
/B $ has a normal subgroup of $cw(\Gamma )$-bounded index of derived length
at most $3$. By Lemma \ref{normfab} we see that $\left| \Gamma /B\right| $
is bounded in terms of $\Gamma $.

If $L$ is a simple group of rank $l$ in $Lie(p)$ then it has a faithful
representation of degree roughly $l^{2}$ over $\overline{F_{p}}$, the
algebraic closure of $F_{p}$ (see the proof of Theorem D in \cite{manseg}).
By \cite{pyber} $l$ is bounded in terms of $cw(\Gamma )$. Therefore $%
B/C\cong L^{r}$ and then also $\Gamma /C$ has a faithful representation of $%
\Gamma $-bounded degree over $\overline{F_{p}}$. By Corollary \textbf{\ref
{vegesrep} }there are only finitely many such representations of $\Gamma $
hence for given $p$ there are only finitely many choices for $L$. This
completes the proof. $\square $

\medskip

\noindent \textbf{Proof of Theorem \ref{strucc}.}

>From \cite[Theorem 2.3 and Corollary 2.4]{bamapy} one deduces that $G$ has
such a series of subgroups satisfying 1), 2) and 3). Now 4) follows from
Theorem \ref{keveslie}. $\square $

\medskip

\noindent \textbf{Proof of Corollary \ref{justif}.}

As $\Gamma $ is residually finite, it is embedded in $G=\widehat{\Gamma }$.
Now using the notations of Theorem \ref{strucc} we distinguish between
several cases.

Assume first that $S$ is of finite index in $G$. This implies that $\Gamma $
has a finite index subgroup $\Gamma _{0}$ with $\widehat{\Gamma _{0}}$
prosolvable. If $\Gamma _{0}$ itself is solvable, then by the FAb property
of $\Gamma $, $\Gamma _{0}$ and $\Gamma $ are finite and we are in Case 0).

If $\Gamma _{0}$ is not solvable, then $\Gamma $ can not be a linear group,
since every non virtually-solvable linear group has infinitely many simple
groups as upper composition factors. This follows easily from the Strong
Approximation Theorem (see \cite{weisfeiler}). We also give a direct
argument: if such a group $\Gamma $ is linear, then $\Gamma $ and also $%
\Gamma _{0}$ is residually finite-linear-group of the same degree. These
finite linear quotients of $\Gamma _{0}$ are solvable of bounded derived
length. Hence $\Gamma _{0}$ is solvable, a contradiction.

Now assume that $S$ is of infinite index in $G$ in which case $\Gamma \cap
S=1$. It follows that $\Gamma $ is separated by bounded degree
representations over finite fields, such that every characteristic occurs
only finitely many times. This implies that $\Gamma $ is linear over a field
of characteristic $0$ (see \cite{LMS}).

As $\Gamma $ is just infinite, it has a faithful specialisation into $GL(n,%
\overline{\Bbb{Q}})$ where $\overline{\Bbb{Q}}$ is the field of algebraic
numbers (see \cite[Proposition 2.2]{lubman}). As $\Gamma $ is finitely
generated, it is inside $GL(n,K)$ for some number field $K$. In fact, $%
\Gamma $ is in $\mathcal{G}(O_{S})$, where $\mathcal{G}$ is its Zariski
closure, $O$ is the ring of integers in $K$, $S$ a finite set of valuations
containing all the archimedean ones and $O_{S}=\left\{ x\in K\mid v(x)\geq 0%
\text{ for all valuations }v\notin S\right\} $. $\mathcal{G}$ can be made
semisimple since $\Gamma $ is just infinite. $\square $

\section{Profinite constructions}

In contrast to Theorem \ref{linfotetel} below we prove that the groups $%
SL(n,q)$ with $n$ fixed form a family of linear groups of dimension $n$ with 
$n$-bounded cyclic width and having no abelian normal subgroups of $n$%
-bounded index (see a forthcoming paper \cite{abpy} for a more general
result saying that if $G$ is a finite completely reducible linear group of
dimension $n$ over any field then its cyclic width is at most $1000n$). This
shows that in Theorem \ref{linfotetel}, unlike in Jordan's theorem, we
cannot bound the index of the abelian normal subgroup.

In fact, we obtain a more precise result which we use to construct examples
of profinite groups showing that Corollary \ref{vegesrep}, Theorem \ref
{keveslie} and Theorem \ref{strucc} do not hold in the profinite category.

\begin{lemma}
\label{adottrend}Let $q>3$ be a prime power and $n\geq 2$. Then $SL(n,q)$ is
the product of $10n(n-1)$ cyclic subgroups of order $q-1$.
\end{lemma}

\noindent \textbf{Proof. }We claim that the upper triangular subgroup $%
U\subseteq SL(2,q)$ is contained by the product of $5$ cyclic subgroups of
order $q-1$. We use conjugates of the diagonal subgroup $D$ (which is a
cyclic subgroup of order $q-1$). Denoting the elements of $D$ and $U$ by 
\[
d_{x}=\left[ 
\begin{array}{ll}
x^{-1} & 0 \\ 
0 & x
\end{array}
\right] \text{ and }u_{y}=\left[ 
\begin{array}{ll}
1 & y \\ 
0 & 1
\end{array}
\right] \text{ } 
\]
\newline
we have $d_{x}^{-1}u_{1}d_{x}=u_{x^{2}}$. Since $u_{x}u_{y}=u_{x+y}$ and in
any finite field every nonzero element is the sum of two squares, we have 
\[
\left\{ u_{x}\mid x\in F_{q}^{*}\right\} =\left\{
d_{x^{-1}}u_{1}d_{xy^{-1}}u_{1}d_{y}\mid x,y\in F_{q}^{*}\right\} \text{.} 
\]
Choose $t\in F_{q}^{*}$ such that $t^{2}\neq 1$ (here we need $q>3$). Set $%
a=u_{(t^{-2}-1)^{-1}}$. A simple calculation shows that $%
d_{t}^{a}d_{t^{-1}}=u_{1}$ hence $\left\{ I,u_{1}\right\} \subseteq D^{a}D$.
Therefore $U=$ $\left\{ u_{x}\mid x\in F_{q}\right\} \subseteq
DD^{a}DDD^{a}DD=DD^{a}DD^{a}D$. This proves our claim.

Now let $E_{ij}\subseteq SL(n,q)$ be the subgroup consisting of the matrices
which differ from $I$ only in the $(i,j)$-th entry ($i\neq j$). It is
well-known that $E_{ij}$ and $E_{ji}$ generate a subgroup isomorphic to $%
SL(2,q)$ and $E_{ij}$ and $E_{ji}$ correspond to the upper and lower
triangular subgroups. By the above claim every subgroup $E_{ij}$ is
contained by the product of $5$ cyclic subgroups of order $q-1$ in $SL(n,q)$.

Also let $U$ and $L$ denote the upper and lower triangular subgroups of $%
SL(n,q)$. We have $U=\prod_{i<j}E_{ij}$ and $L=\prod_{i>j}E_{ij}$ hence $U$
and $L$ are contained by the product of $5\frac{n(n-1)}{2}$ cyclic subgroups
of order $q-1$. On the other hand $LULU=SL(n,q)$ (see \cite{denvas}, the
proof of Corollary 14). This implies our statement. $\square $

\medskip

Denote by $p_{i}$ the $i$-th prime. It is easy to see that the numbers $%
2^{p_{i}}-1$ are pairwise relatively prime.

Let $G=\prod\nolimits_{i=1}^{\infty }PSL(n,2^{p_{i}})$ be the cartesian
product of the groups $PSL(n,2^{p_{i}})$ for some fixed $n\geq 2$. $G$ is a
profinite group with the usual product topology.

\begin{theorem}
\label{pslkon}1) $G$ is boundedly generated as a profinite group. \newline
2) $G$ has polynomial index growth and the FAb property as an abstract
group. \newline
3) $G$ does not contain any dense discrete boundedly generated subgroups.
\end{theorem}

\noindent \textbf{Proof. }Set $k=10n(n-1)$. By Lemma \ref{adottrend} any
group $PSL(n,2^{p_{i}})$ is the product of $k$ cyclic subgroups $%
C_{1}^{i},\ldots ,C_{k}^{i}$ of orders dividing $2^{p_{i}}-1$. For $j$ fixed
the orders of the groups $C_{j}^{i}$ are pairwise relatively prime therefore
the subgroup $C_{j}=\prod\nolimits_{i=1}^{\infty }C_{j}^{i}$ of $G$ is
procyclic. $G$ is the product of the procyclic subgroups $C_{j}$ i.e., it is
boundedly generated.

It was proved independently by Saxl and Wilson \cite{saxlwilson} and
Martinez and Zelmanov \cite{marzel} that in a finitely generated profinite
group which is a cartesian product of nonabelian finite simple groups every
finite index subgroup is open. It also follows easily, that groups of this
type are perfect (using the fact, first proved by Wilson \cite{wilson}, that
there exists a constant $r$, such that every element of a nonabelian finite
simple group is the product of $r$ commutators).

In particular every finite index normal subgroup $N$ of our $G$ is open. It
follows that $G/N$ is a product of $k$ cyclic subgroups hence $\left|
G/N\right| \leq (\exp (G/N))^{k}$ for every $N$, i.e., $G$ has polynomial
index growth.

It is also clear that every such group $N$ is also perfect. Hence if $H$ is
a finite index subgroup of $G$ then $H$ contains a perfect normal subgroup
of finite index therefore $G$ has the FAb property.

Let $\Gamma $ be a finitely generated subgroup of $G$. For $q$ arbitrary the
group $PSL(n,q)$ can be embedded into $SL(n^{2}-1,q)$. Using Lemma \ref
{fablem} we see that $\Gamma $ is a subdirect product of finitely many
linear groups over fields of characteristic $2$. If $\Gamma $ is boundedly
generated then by Theorem \ref{linfotetel} it is a subdirect product of
finitely many abelian by finite groups hence $\Gamma $ itself is abelian by
finite. Therefore $\Gamma $ can not be dense in $G$. $\square $

\medskip

In view of the above example it is natural to ask whether the profinite
group $H=\prod\nolimits_{i=1}^{\infty }PSL(n,p^{i})$ with $n\geq 2$ and $p$
fixed is boundedly generated. We give a negative answer.

\begin{proposition}
\label{szorzat}$H$ is not boundedly generated as a profinite group.
\end{proposition}

\noindent \textbf{Proof. }Suppose that $H$ is the product of $m$ procyclic
subgroups. For an arbitrary $k$ let $r_{k}=p_{1}p_{2}\cdots p_{k}$ (the
product of the first $k$ primes) and let $H_{k}=\prod\limits_{d\mid
r_{k}}PSL(n,p^{d})$. Set 
\[
o_{k}=\left| GL(n,p^{r_{k}})\right| =p^{r_{k}\frac{1}{2}n(n-1)}\prod%
\limits_{i=1}^{n}(p^{r_{k}i}-1)\text{.} 
\]
If $d\mid r_{k}$ then $GL(n,p^{d})$ can be embedded into $GL(n,p^{r_{k}})$,
which yields $\left| PSL(n,p^{d})\right| \mid o_{k}$. As a consequence,
every element of the product $H_{k}$ has order dividing $o_{k}$. Since $%
H_{k} $ is a quotient of $H$ it is a product of $m$ cyclic subgroups. Thus
using the obvious inequality $\left| PSL(n,p^{d})\right| \geq p^{d}$ we
obtain that 
\[
o_{k}^{m}\geq \left| H_{k}\right| \geq p^{\sum\limits_{d\mid
r_{k}}d}=p^{(1+p_{1})(1+p_{2})\cdots (1+p_{k})}\text{.} 
\]
\newline
Using $o_{k}\leq p^{r_{k}n^{2}}$ and taking logarithms we obtain 
\[
m\geq \frac{1}{n^{2}}\frac{(1+p_{1})(1+p_{2})\cdots (1+p_{k})}{%
p_{1}p_{2}\cdots p_{k}}\geq \frac{1}{n^{2}}\sum\limits_{i=1}^{k}\frac{1}{%
p_{i}}\text{.} 
\]

This is a contradiction since the right side can be made arbitrarily large. $%
\square $

\medskip

We remark here that our proof of Theorem \ref{linfotetel} used a 'local
method', i.e., embedding $\Gamma $ into $GL(n,\Bbb{K})$ where $\Bbb{K}$ is a
local field. One can give a different proof by 'global' considerations,
i.e., appealing to the Strong Approximation Theorem for linear groups (see 
\cite{weisfeiler} and \cite{pink2}). From the theorem it follows that $%
\Gamma $ has many nonabelian simple quotients and then a variant of
Proposition \ref{szorzat} shows that $\Gamma $ can not be boundedly
generated. We omit the details.

Note that for every $n$ the group of the form $\prod\nolimits_{p}PSL(n,p)$ ($%
p$ runs over all prime numbers) is a BG profinite group (see \cite{liepyb}
for a more general result).

\medskip

Our next result has a variety of uses in constructing boundedly generated
profinite groups.

\begin{lemma}
\label{stabprod}Let $G$ be a transitive permutation group on $n$ elements
and $H$ a pointstabiliser.\newline
1) If $G$ is $2$-transitive then it is a product of $3$ pointstabilisers.%
\newline
2) Let $N$ be the normal subgroup of $G$ generated by all pointstabilisers.
Suppose $H$ has an orbit of length $\geq \frac{n}{r}$. Then $N$ is a product
of at most $2r$ pointstabilisers.
\end{lemma}

\noindent \textbf{Proof. }If $G$ is $2$-transitive, then for any $g\in
G\setminus H$ we have $G=H\cup HgH$. Now $H$ fixes exactly one point, say $%
\alpha $, hence if $H_{1}$ is the stabiliser of $\beta \neq \alpha $ then
there is a $g\in H_{1}\setminus H$. Therefore $G=HH_{1}H$ which proves 1).

Now let $G$ be as in 2). Our conditions imply that there is a double coset $%
HgH$ of size $\geq \frac{\left| G\right| }{r}$. Hence the subset $%
X=g^{-1}HgH $ (which is a product of $2$ pointstabilisers) also has size $%
\geq \frac{\left| G\right| }{r}$.

By a result of Hamidoune \cite{hami} if a finite group $G$ is generated by a
set of size $d$, then every element of $G$ is a product of at most $2\frac{%
\left| G\right| }{d}$ elements of this set. Hence if $M$ is the subgroup of $%
N$ generated by $X$, then every element of $M$ is a product of at most $2%
\frac{\left| M\right| }{\left| X\right| }$ elements of $X$.

If $M\neq N$ then there is a pointstabiliser $H_{1}\nsubseteq M$. The set $%
MH_{1}$ is the union of at least $2$ right cosets of $M$. If $MH_{1}\neq N$
then there is a pointstabiliser $H_{2}$ such that $MH_{1}$ is properly
contained in $MH_{1}H_{2}$ which then must be the union of at least $3$
right cosets of $M$. Repeating this argument we see that $N$ is a product of
at most $2\frac{\left| M\right| }{\left| X\right| }+\left| N:M\right| -1$
pointstabilisers. But $2\frac{\left| M\right| }{\left| X\right| }+\left|
N:M\right| -1\leq 2\frac{\left| N\right| }{\left| X\right| }\leq 2r$ which
proves 2). $\square $

\medskip

Now we define a metabelian group with some interesting properties. Let $%
A=\prod\nolimits_{k=1}^{\infty }AGL(1,2^{p_{k}})$ be the cartesian product
of the finite affine groups $AGL(1,2^{p_{k}})$.

\begin{proposition}
1) $A$ is an extension of an elementary abelian $2$-group by a procyclic
group;\newline
2) $A$ is boundedly generated as a profinite group;\newline
3) $A$ has polynomial index growth as an abstract group;\newline
4) $A$ has at least $n^{\frac{1}{8}\log n-1}$ subgroups of index $n$ for
infinitely many $n$;\newline
5) $A$ does not contain any dense discrete boundedly generated subgroups.
\end{proposition}

\noindent \textbf{Proof. }By definition any group $AGL(1,2^{p_{k}})$ is an
extension of an elementary abelian group $V_{k}$ of order $2^{p_{k}}$ by a
cyclic group of order $2^{p_{k}}-1$. It is naturally a $2$-transitive
permutation group. Hence by Lemma \ref{stabprod} it is a product of $3$
cyclic subgroups of order $2^{p_{k}}-1$. 1)\ and 2) follow immediately
(using the fact that the numbers $2^{p_{k}}-1$ are relatively prime).

By a result of Segal \cite{segal} any finite index subgroup of a finitely
generated prosoluble group is open hence 2) implies 3).

For any $k$ the linear space $V_{k}$ has at least $2^{x(p_{k}-x)}$ subspaces
of codimension $x$. Each such subspace defines - in an obvious way - a
subgroup of index $(2^{p_{k}}-1)2^{x}$ in $A$. Setting $x=\left[ \frac{p_{k}%
}{3}\right] $ we obtain 4) by straightforward computation.

Finally note that $AGL(1,2^{p_{k}})$ is isomorphic to a subgroup of $%
SL(2,2^{p_{k}})$ (namely to the subgroup of all matrices with $0$ above the
main diagonal). As in the proof of Theorem \ref{pslkon} we see that any
boundedly generated subgroup $\Gamma $ of $A$ is abelian by finite hence it
can not be dense in $A$. $\square $

\medskip

There is nothing really special about the prime $2$ in the above
constructions. The group $AGL(1,p^{t})$ has a unique subgroup $A^{0}(p^{t})$
which is an extension of an elementary abelian $p$-group by a cyclic group
of order $\frac{p^{t}-1}{p-1}$. It is naturally a permutation group of
permutation rank $p$ (i.e., the cyclic pointstabiliser has $p$ orbits).
Using Lemma \ref{stabprod} one can see that $A^{0}(p^{t})$ is a product of
at most $4p$ cyclic groups of order $\frac{p^{t}-1}{p-1}$. It follows that
the profinite group $A^{0}=\prod\nolimits_{k=1}^{\infty }A^{0}(p^{p_{k}})$
is boundedly generated.

Using similar ideas one can reprove Theorem \ref{pslkon} and show that in
fact for every fixed prime $p$ the profinite groups of the form $%
\prod\nolimits_{k=1}^{\infty }PSL(n,p^{p_{k}})$ are boundedly generated.

To obtain somewhat different examples one can use the primitive affine
permutation groups of rank $3$ listed in \cite{liebeck}. For instance for $%
2^{d}=q^{6}$ the group $SL(2,q)$ acts on the nonzero elements of the group $%
C_{2}^{d}$ with $2$ orbits. Denote the corresponding affine groups of the
form $C_{2}^{d}\Join SL(2,q)$ by $A^{1}(q)$. Using Lemma \ref{stabprod} one
can see that $A^{1}(q)$ is a product of at most $40$ cyclic subgroups of
order $q-1$. Therefore the profinite group $A^{1}=\prod\nolimits_{k=1}^{%
\infty }A^{1}(2^{p_{k}})$ is boundedly generated.

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\end{document}
