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{\bf PARTITIONS WHICH ARE $p$- AND $q$-CORE}
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{\bf J.-C. Puchta}\\
{\smallit Mathematical Institute, 24-29 St. Giles', Oxford OX1 3LB,
United Kingdom}\\
{\tt puchta@maths.ox.ac.uk}\\
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\centerline{\smallit Received:  6/19/01, Revised:  9/1/01,
 Accepted:  9/6/01,
Published:  9/12/01}
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\centerline{\bf Abstract}

\noindent
Let $p$ and $q$ be distinct primes, $n$ an integer with
$n>p^2q^2$. Then there is no partition of $n$ which is at the same
time $p$- and $q$-core. Hence there is no irreducible representation
of $S_n$ which is of $p$- and $q$-defect zero at the same time.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY
 \smalltt 1 (2001),
\#A06\hfill}

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Let $n$ be an integer. Then there is a natural bijection between the set of
partitions of $n$ and the irreducible representations of the symmetric group on
$n$ letters $S_n$. A representation of a finite group $G$ with character $\chi$
is called of $p$-defect zero, if $|G|_p\big|\chi(1)$. In the case of the
symmetric group this is known to be equivalent to the statement that the
corresponding partition has no hook-number divisible by $p$, in this case the
partition is called a $p$-core partition. Granville and Ono \cite{GO}
proved that for any $t\geq 7$ and any $n$ there is a $t$-core partition
of $n$, thus for every $p\geq 7$ there is an irreducible representation
of $S_n$ with
$p$-defect zero, an easier proof was given by Kiming \cite{Kim}.

In a recent paper Navarro and Willems \cite{NW} asked for relations
between the $p$- and the $q$-blocks of representations. In this note
we will show that the property of having defect zero exclude each
other, if $n$ is large enough compared to $p$ and $q$. More precisely
we will prove the following theorem. 

\noindent
{\bf Theorem 1.}
Let $p$ and $q$ be primes, $n$ an integer with $n>p^2q^2$. Then there is no
irreducible representation of $S_n$ with $p$- and $q$-defect zero.

By the correspondence between irreducible representations of the $S_n$ and
partitions of $n$ this will follow from the following statement.

\noindent
{\bf Theorem 2.}
Let $s$ and $t$ be relatively prime integers, $n$ an integer with $n>s^2t^2$.
Then there is no partition of $n$ which is at the same time $s$- and $t$-core.

Especially, the number of partitions which are simultaneously $s$- and
$t$-core is finite. J. Kohles Anderson \cite{JKA} proved a more precise
version of this statement: The number of 
\newpage
partitions with this property
is in fact equal to $\frac{1}{s+t}{{s+t}\choose t}$. However, the
proof we give here seems to be simpler then the one given by her.

I would like to thank the referee for making me aware of \cite{JKA}.

The proof will use the description of $t$-core partitions introduced by Garvan,
Kim and Stanton \cite{GKS}.

For the sequel we choose an arbitrary partition $n=\lambda_1+\ldots+\lambda_k$
of $n$ and assume that it is $t$-core and $s$-core at the same time. We thus
have to show that $n<s^2t^2$.

Consider the diagram of the partition, i.e. the set of cells whose first
row consists of $\lambda_1$ cells $(1, 1), (1, 2),\ldots, (1, \lambda_1)$, the
second of $\lambda_2$ cells and so on.
Label a cell $(i, j)$ with $j-i\pmod{st}$, cells in column 0 are labeled in
the same way. A cell at the end of a row is called exposed. Now divide the
diagram into regions $S_k$, such that a cell belongs to $S_k$ if and only if
$s(k-1)\leq j-i<sk$, in the same way $T_k$ denotes the cells with
$t(k-1)\leq j-i<tk$. Now by \cite{GKS}, paragraph 2, we know that if the
partition is $s$-core, and there is an exposed cell labeled with $i$ in the
region $S_k$, then there is an exposed cell labeled with
$\tilde{i}\equiv i\pmod{s}$ in every region $S_l$ with $l\leq k$. Especially,
there is some sequence $k_\nu, 0\leq\nu\leq l, k_0=1$, such that
$\lambda_{k_\nu}\equiv\lambda_1 - (k_\nu-1)\pmod{s}$,
$(k_{\nu+1}-k_\nu)<\lambda_{k_\nu}-\lambda_{k_{\nu+1}}< 2s-(k_{\nu+1}-k_\nu)$
and $\lambda_{k_l}< s$, i.e. $\lambda_{k_\nu}=\lambda_1 - \nu s+k_\nu$. Assume
that $l<t$. Since $\lambda_{k_\nu}\leq\lambda_{k_{\nu+1}}$, we have $k_{\nu+1}
\leq k_\nu+s$, thus the partition under consideration consists of at
most $ls<st$ summands, each being $st$ at most, thus we have $n\leq s^2t^2$.

Now if $l>t$, then the labels of the exposed cells in the rows $k_\nu$ run
through a complete remainder system $\pmod{t}$, since $s$ and $t$ are
coprime, the remainders of $\lambda_{k_\nu}-k_\nu = \lambda_1-\nu s, \,
0\leq \nu<t$ are therefore all different. However, by \cite{GKS} we know
that if the partition is
$t$-core, and there is an exposed cell in region $T_k$ with the label $i$, then
there is no exposed cell with a label $\overline{i}\equiv t-i-1\pmod{t}$ in
any region $T_l$ with $l\geq 1-k$. If $\lambda_1$ is in region $T_k$, then
$\lambda_{k_{t-1}}$ is in region $T_l$ with $l\geq k-s$, thus $k-s<1-k$, i.e.
$k\leq s/2$. By the definition of $T_k$ we have $\lambda_1< t(s/2+1)\leq st$.

Since
the property of being a $t$-core partition is unchanged under conjugation, by
the same reasoning we get that there are less than $st$ summands, thus we
obtain $n<s^2t^2$ again.

Thus in any case the assumption that our partition is at the same time $s$-core
and $t$-core leads to the estimate $n<s^2t^2$ which proves our theorem.

\newpage

\begin{thebibliography}{9}
\bibitem{GKS} F. Garvan, D. Kim, D. Stanton, {\em Cranks and $t$-cores},
Invent. Math. 101, No.1, 1-17 (1990)
\bibitem{GO} A. Granville, K. Ono, {\em Defect zero $p$-block for finite simple
groups}, Trans. Am. Math. Soc. 348, No.1, 331-347 (1996)
\bibitem{JKA} J. Kohles Anderson, {\em Partitions which are
simultaneously $t_1$- and $t_2$-core}, to appear in Discrete Mathematics
\bibitem{Kim} I. Kiming, {\em A note on a theorem of A. Granville and K.
Ono}, J. Number Theory 60, No.1, 97-102 (1996)
\bibitem{NW} G. Navarro, W. Willems, {\em When is a $p$-block a $q$-block?}, 
Proc. Am. Math. Soc. 125, No.6, 1589-1591 (1997)
\end{thebibliography}
Mathematics Subject Classification: 05A17, 11P83, 20C30

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