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\newtheorem*{conA}{Conjecture A}
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\begin{document}


\title[Brauer's Problem 1]{Complex group algebras of finite
groups:\linebreak[1]
Brauer's Problem 1}

\author{Alexander Moret\'o}
\address{Departament d'\`Algebra,
Universitat de Val\`encia,
46100 Burjassot, Val\`encia, SPAIN}
\email{Alexander.Moreto@uv.es}
\thanks{Research supported by the Basque Government, the Spanish
Ministerio de Ciencia y Tecnolog\'{\i}a, grant BFM2001-0180, and
the FEDER}

\subjclass{Primary 20C15}
\date{October 12, 2004}
\commby{David J. Benson}

\begin{abstract}
 Brauer's Problem 1 asks the following: what are
the possible complex group algebras of finite groups? It seems
that with the present knowledge of representation theory it is not
possible to settle this question. The goal of this paper is to
announce a partial solution to this problem. We conjecture that if
the complex group algebra of a finite group does not have more
than a fixed number $m$ of isomorphic summands, then its dimension
is bounded in terms of $m$. We prove that this is true for every
finite group if it is true for the symmetric groups.
\end{abstract}

\maketitle

\section{Introduction}

Let $G$ be a finite group and $\Irr(G)=\{\chi_1,\dots,\chi_k\}$
the set of irreducible characters of $G$. Put $\chi_i(1)=n_i$.
Following B.~Huppert, we say that $(n_1,\dots,n_k)$ is the {\bf
degree pattern} of $G$. In recent years much information has been
obtained on the possible sets of character degrees of finite
groups, especially in the solvable case (even though a complete
classification of such sets seems to be very far away). However,
as pointed out by Huppert, almost nothing is known about Brauer's
Problem 1 (see \cite{bra}), which asks the following: What are the
possible degree patterns of finite groups?  As is well known, if
$(n_1,\dots,n_k)$ is the degree pattern of $G$, the complex group
algebra of $G$ is $\C G=\bigoplus_{i=1}^k \M_{n_i}(\C)$.
 So knowing the possible degree patterns of finite groups is
equivalent to knowing the possible isomorphism types of complex
group algebras.

Even though, in our opinion, with the present knowledge of
representation theory it is impossible to settle Brauer's
Problem 1, we think that it is possible to obtain significant
restrictions on the structure of the complex group algebras. The
goal of this note is to provide the first such restriction. For
the sake of discussion, we state the following.

\begin{conA}
The $\C$-dimension of the complex group algebra of any finite group 
$G$ is
bounded in terms of the maximum number of isomorphic summands in
the decomposition $\C G=\bigoplus_{i=1}^k \M_{n_i}(\C)$.
\end{conA}

In other words, Conjecture A says that the order of a finite group
is bounded in terms of the largest multiplicity of its irreducible
character degrees. Our main results are the following. As usual,
we say that a quantity is $(a_1,\dots,a_l)$-bounded if it is
bounded by some real-valued function that depends on
$a_1,\dots,a_l$.

\begin{thmB}
Conjecture A holds for every finite group if it holds for the
symmetric groups.
\end{thmB}


\begin{thmC}
Let $G$ be a finite group and assume that $G$ does not contain an
alternating group bigger than $\Alt(t)$ as a composition factor. If the
largest multiplicity of a character degree is $m$, then the order of 
$G$ is
$(m,t)$-bounded.
\end{thmC}


Unfortunately, we have been unable to prove that Conjecture A
holds for the symmetric groups. This seems to be a difficult
number-theoretic problem. Note that an immediate consequence of
Theorem C is that Conjecture A holds for solvable groups.
(Actually, we prove this result in our way toward a proof of
Theorems B and C.)

In the next section we will outline the proof of Theorems B and C.
Full details will appear elsewhere. In Section 3, we discuss
Conjecture A for symmetric groups.

I thank the referee for a number of useful comments that have
improved the readability of this paper.

\section{Proof of Theorems B and C}

The first step in the proof of Theorems B and C is the proof that
Conjecture A holds for solvable groups. First, we note that the
hypothesis of this conjecture is inherited by quotients. In the
key lemma, we show that the number of primes that divide the order
of a solvable group that satisfies the hypothesis is bounded. It
is an application of results and ideas from \cite{far}.


\begin{lem}
\label{far} Let $G$ be a  solvable group with at most $m$
irreducible characters of each degree and let $p$ be a prime
divisor of $|G|$. Then $p$ is $m$-bounded.
\end{lem}

Actually, in this lemma we are just using that for any character
$\chi\in\Irr(G)$, the field $\Q(\chi)$ is an extension of $\Q$ of
degree $\leq m$. This follows from the fact that all the Galois
conjugate characters have the same degree.

By Gaschutz's theorem, if $G$ is a solvable group, then $G/F(G)$
acts faithfully and completely reducibly on $F(G)/\Phi(G)$. This
is the context in which we apply the following lemma. We say that
a finite module $V$ for a group $G$ has mixed characteristic if
$V$ is an abelian group all of whose Sylow subgroups are
elementary abelian groups. For such a $G$-module $V$, $r(G,V)$
stands for the number of orbits in $V$ under the action of $G$.
For a solvable group $G$, we write $\dl(G)$ to denote the derived
length of $G$.

\begin{lem}
\label{kel}
 Let $G$ be solvable and let $V$ be a finite faithful
completely reducible $G$-module (possibly in mixed
characteristic). Then there exist constants $C_1$ and $C_2$ such
that
$$
\dl(G)\leq C_1\log\log r(G,V)+C_2
$$
if $r(G,V)>1$, and $\dl(G)\leq C_2$ if $r(G,V)=1$.
\end{lem}

\begin{proof}
This immediate consequence of Theorem 2.4 of \cite{kel} appears as
Theorem 7.2 in \cite{mosa}.
\end{proof}

Finally, we need the following result, which is the nilpotent case
of Conjecture~A.

\begin{lem}
\label{jai}
Let $G$ be a nilpotent group and assume that the
largest multiplicity of a character degree is $m$. Then $|G|$ is
$m$-bounded.
\end{lem}

\begin{proof}
This is Corollary 1.12 of \cite{jai}.
\end{proof}

With these ingredients, the idea of the proof of Conjecture A in
the solvable case is the following. We consider the solvable
groups $G$ with at most $m$ irreducible characters of any given degree.
We want to see that $|G|$ is $m$-bounded. First, we look at  the
solvable groups with fixed derived length $k$. An easy inductive
argument shows that for these groups the order of $G$ is
$(m,k)$-bounded. Thus, if we want to find a counterexample to
Conjecture A for solvable groups, we need to consider groups of
arbitrarily large derived length. Indeed, by Clifford's theory and
Lemma \ref{jai}, we need to consider groups $G$ such that
$\dl(G/F(G))$ is arbitrarily large. Now applying Lemma \ref{kel}
to the action of $G/F(G)$ on $F(G)/\Phi(G)$, we deduce that the
number of orbits in this action is doubly exponential in
$\dl(G/F(G))$. The inductive argument mentioned above shows that,
in fact,
$$
|G:F(G)|\leq m^{4^{\dl(G/F(G))}}.
$$
This together with Lemma \ref{far} leads to the fact that the
number of divisors of $|G:F(G)|$ is bounded by a function of the
form
$$
|G:F(G)|\leq 4^{\dl(G/F(G))}\cdot f(m)
$$
for some function $f$ that only depends on $m$. Since the number
of orbits of $G/F(G)$ on $F(G)/\Phi(G)$,  and hence on
$\Irr(F(G)/\Phi(G))$,  is doubly exponential in $\dl(G/F(G))$, we
deduce that the number of irreducible characters of $G$ whose
degree divides $|G:F(G)|$ is doubly exponential in $\dl(G/F(G))$.
We conclude that, since $\dl(G/F(G))$ is arbitrarily large, some
of the irreducible character degrees occur with multiplicity
larger than $m$. This contradiction concludes the proof of
Conjecture A for solvable groups.


The next step of the proof of Conjecture A is for the simple
groups of Lie type. We will present a proof, due to G. Malle, that
gives pretty good explicit bounds. An independent proof has been
found in \cite{lish}.  We will give the proof in a series of
lemmas. The idea is to find many Galois conjugate characters.


\begin{lem}
\label{1} Let $s\in\GL(n,\overline{\F}_q)$ be semisimple. Then $s$
is conjugate to at most $n!$ of its powers. If moreover all
eigenvalues of $s$ are powers of one among them, then $s$ is
conjugate to at most $n$ of its powers.
\end{lem}


\begin{lem}
\label{2} The group $\GL(n,q)$ contains a semisimple element $s$
of order $q^n-1$ all of whose eigenvalues are powers of one among
them.
\end{lem}



\begin{lem}
\label{3} Suppose $G=\SL(n,q), \SU(n,q), \Sp(2n,q), \SO(2n+1,q),
\SO^+(2n,q),\\ \SO^-(2n,q)$. Then $G$ contains a semisimple
element of order $(q^n-1)/(q-1)$, \mbox{$q^{[n/2]}-1$}, $q^n-1, q^n-1, q^n-1,
q^{n-1}-1$ conjugate to at most $n, n, 2n, 2n+1, 2n, 2n$ of its
powers.
\end{lem}


\begin{cor}
\label{4} Suppose $G=\PSL(n,q), \PSU(n,q), \PSp(2n,q), \PSO(2n+1,q),
\break
\PSO^+(2n,q), \PSO^-(2n,q)$. Then $G$ contains a semisimple
element of order at least $(q^n-1)/n(q-1), (q^{[n/2]}-1)/n,
(q^n-1)/2, (q^n-1)/2, (q^n-1)/2, (q^{n-1}-1)/2$ conjugate to at
most $n, n, 2n, 2n+1, 2n, 2n$ of its powers.
\end{cor}

These results follow from easy linear algebra arguments. With them
and  Deligne-Lusztig theory, we can prove the main result for the
classical groups of Lie type.

\begin{teo}
\label{5} Suppose $G=\PSL(n,q), \PSU(n,q), \PSp(2n,q), \PSO(2n+1,q),\break
\PSO^+(2n,q),  \PSO^-(2n,q)$. Then the largest multiplicity of the
character degrees of $G$ is at least $\varphi(q^n-1)/n^2(q-1),
\varphi(q^{[n/2]}-1)/n^2, \varphi(q^n-1)/4n$,
\mbox{$\varphi(q^n-1)/2(2n+1)$}, $\varphi(q^n-1)/4n, \varphi(q^{n-1}-1)/4n$.
\end{teo}


\begin{teo}
\label{6} Conjecture A holds for simple groups of Lie type.
\end{teo}

\begin{proof}[Sketch of proof]
For classical groups, this follows easily from Theorem \ref{5}
using the fact that for any $\varepsilon>0$,
$\varphi(k)/k^{1-\varepsilon}\to\infty$ as $k\to\infty$ (see
Theorem 327 of \cite{hawr}).


For exceptional groups, we can use, for example, that
$(P)\SL(2,q)\leq G(q)$ for $G(q)$ of exceptional type different
from $^2B_2(q)$ (see the Dynkin diagram) and $G(q)\leq\GL(a,q)$ for
some small $a$. For instance, $(P)\SL(2,q)\leq
E_8(q)\leq\GL(248,q)$. It is easy to see  using the first part of
Lemma \ref{1}  that $E_8(q)$ contains a semisimple element of
order $(q-1)/2$ conjugate to at most $(248)!$ of its powers. Now,
use Deligne-Lusztig theory.

Finally, we consider the groups $^2B_2(q)$. It is well known that
they have $(q-2)/2$ characters of degree $q^2+1$ (see Theorem
XI.5.10 of \cite{huba}). The result follows.
\end{proof}



Now, we can begin to work toward the proof of the general case of
Theorems B and C. We need some more lemmas. The first one is a
non-trivial number-theoretic result. Given an integer $n$, we
write $d(n)$ to denote the number of divisors of $n$.

\begin{lem}\label{div}
\begin{enumerate}
\item If $\varepsilon>0$, then $d(n)<2^{(1+\varepsilon)\log
n/\log\log n}$ for all $n>n_0(\varepsilon)$. 
\item
$$
\lim_{n\to\infty}\frac{d(n!)}{2^\frac{c\log n!}{(\log\log n!)^2}}=1,
$$
where $c$ is some constant.
\end{enumerate}
\end{lem}

\begin{proof}
The first part  is Theorem 317 of \cite{hawr}. The second part is in
\cite{erd}.
\end{proof}

We need one more lemma on the characters of the simple groups.

\begin{lem}
\label{ext}
Let $S$ be a finite simple group. Then there exists a non-principal 
irreducible character of $S$ that extends to $\Aut(S)$.
\end{lem}

In the proof of Theorem C, the following result is also necessary.

\begin{teo}
\label{pyba}
Let $G$ be a permutation group on a set $\Omega$ of cardinality $k$, and
assume that $G$ does not contain any alternating group bigger than
$\Alt(t)$ as a composition factor. Then the number of orbits of $G$ on 
the
power set $\PP(\Omega)$ is at least $a^{k/t}$ where $a>1$ is some 
constant.
\end{teo}

\begin{proof}
This appears in \cite{bapy}.
\end{proof}

Now we will sketch the proof of Theorem C.

\begin{proof}[Sketch of proof of Theorem C]
Using the solvable case and Clifford's theory, we may assume that
$F(G)=1$. Hence, $G$ is isomorphic to a subgroup of the
automorphism group of its socle, so it suffices to bound the
cardinality of the socle of~$G$.

We have to bound the number of times that a given simple group $S$
appears as a direct factor of the socle and also the order of each
of these simple groups. We will just do the first part.

Let $N$ be a normal subgroup of $G$ isomorphic to the direct
product of $k$ copies of a non-abelian simple group $S$. We will
bound $k$ in terms of $m$. We have that $G/C_G(N)$ embeds into
$\Aut(N)\cong \Aut(S)\wr S_k$. Put $H=G/C_G(N)$ and view this
group as a subgroup of $\Aut(S)\wr S_k$. Let $B=H\cap\Aut(S)^k$
and note that $H/B$ is a
 permutation group on $k$ letters.
Fix a non-linear character $\varphi\in\Irr(S)$ that extends to
$\Aut(S)$ (it exists by Lemma \ref{ext}). Note that the hypotheses
of Theorem \ref{pyba} hold. It follows from this result that for
some $s$, the number of orbits of $H/B$ on the subsets of
cardinality $s$ is at least $a^{k/t}/(k+1)$. Considering the
characters of $B$ that extend products of $s$ copies of $\varphi$
and $k-s$ copies of the principal character of $S$, we deduce that
there are at least $a^{k/t}/(k+1)$ $H$-orbits of characters of $B$
of the same degree. Using Corollary 11.29 of \cite{isa}, it
follows that $H$ (and hence $G$) has at least $a^{k/t}/(k+1)d(k!)$
characters of the same degree. Using part (ii) of Lemma \ref{div}
and Stirling's formula, one can see that this quotient goes to
infinity as $k$ goes to infinity. It follows that $k$ has to be
bounded in terms of $m$, as desired.
\end{proof}




Once we have proved Theorem C, Theorem B will be an immediate
consequence of the following lemma. Given a group $G$, we write
$m(G)$ to denote the largest multiplicity of the character degrees
of $G$.

\begin{lem}
\label{mu} Assume that  $\Alt(t)$ is a composition factor of a
finite group $G$ for some $t>6$. Then
$$m(G)\geq m(\Alt(t))/4.
$$
\end{lem}

In the proof of this lemma, we are using that any invariant
character of the base group of a wreath product extends to the
whole group.



\section{Symmetric groups}

The goal of this section is to discuss Conjecture A for symmetric
groups. As we will see, this is a combinatorial problem. It is
well known that the number of irreducible characters of $S_n$ is
the number of partitions of $n$. A. Young found a description of
the irreducible characters in terms of the partitions of $n$.

Given a partition of $n$, $\mu=(a_1,\dots,a_t)$, with $a_1\geq
a_2\geq\dots\geq a_t$, the Young diagram associated to $\mu$ is an
array of $n$ nodes with $a_i$ nodes in the $i$th row. We assign
numbers to the rows and columns and coordinates to the nodes. The
{\it hook number} $H(i,j)$ of the node $(i,j)$ is the number of
nodes to the right and below the node $(i,j)$, including the node
$(i,j)$. The degree of the character $\chi_{\mu}$ associated to
the partition $\mu$ is given by the {\it hook length formula}
$$
\chi_{\mu}(1)=\frac{n!}{\prod_{i, j}H(i,j)}.
$$
This description of the degrees was obtained by J. Frame, G. de B.
Robinson and R. Thrall in \cite{frt}.

Therefore, Conjecture A can be restated as follows. Given an
integer $n$, let $m(n)$ be the largest number of partitions of $n$
with the same product of hook numbers.

\begin{conA'}
$m(n)\to\infty$ as $n\to\infty$.
\end{conA'}

Computer calculations suggest that this is the case, but it seems
to be difficult to find a proof. Just to find a proof of the
weaker result $\limsup m(n)=\infty$ would be interesting.





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\end{document}