\documentclass[12pt,a4paper]{amsart} \usepackage[cp850]{inputenc} \usepackage{latexsym, amsfonts, amssymb,amsmath} \usepackage[english]{babel} \usepackage{latexsym} \usepackage{amscd} \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \renewcommand{\P}{{\mathcal P}} \newcommand{\D}{{\mathcal D}} \renewcommand{\O}{{\mathcal O}} \newcommand{\X}{{\mathcal X}} \newcommand{\Y}{{\mathcal Y}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\N}{{\mathbb N}} \newcommand{\epps}{{\varepsilon}} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \textwidth13.2cm \begin{document} \begin{center} {\Large\bf On character tables related to \\[5pt] the alternating groups}\\[.1in] \bigskip \bigskip { \large Christine Bessenrodt} \\[5pt] Institut f\"ur Mathematik, Uni\-ver\-si\-t\"at Hannover\\ D-30167 Hannover, Germany \\ Email: bessen@math.uni-hannover.de \\[3ex] {\large J\o rn B. Olsson\footnote{Partially supported by The Danish National Research Council.} } \\[5pt] Matematisk Afdeling, University of Copenhagen \\ Copenhagen, Denmark\\ Email: olsson@math.ku.dk \\[4ex] {\small {\bf Abstract} \medskip \parbox{12cm}{ There is a simple formula for the absolute value of the determinant of the character table of the symmetric group~$S_n$. It equals $a_{\mathcal{P}}$, the product of all parts of all partitions of~$n$ (see \cite[Corollary 6.5]{James}). In this paper we calculate the absolute values of the determinants of certain submatrices of the character table $\mathcal{X}$ of the alternating group $A_n$, including that of $\mathcal{X}$ itself (Section~2). We also study explicitly the powers of 2 occurring in these determinants using generating functions (Section~3). }} \end{center} %First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire %--first part \thispagestyle{myheadings} \font\rms=cmr8 \font\its=cmti8 \font\bfs=cmbx8 \markright{\its S\'eminaire Lotharingien de Combinatoire \bfs 52 \rms (2004), Article~B52c\hfill} \def\thepage{} % % \medskip \section{Preliminaries} We fix a positive integer $n$. We will use the same notation as in \cite{bos}, which we recall here. If $\mu=(\mu_1,\mu_2, \ldots)$ is a partition of $n$ we write $\mu \in \P$ and then $z_{\mu}$ denotes the order of the centralizer of an element of (conjugacy) type $\mu$ in $S_n$. Suppose $\mu=(1^{m_1(\mu)}, 2^{m_2(\mu)},\ldots)$, is written in exponential notation. Then we may factor $z_\mu= a_{\mu}b_{\mu}$, where $$a_{\mu}=\prod_{i \ge 1}i^{m_i(\mu)}, \hspace{.1 in} b_{\mu}=\prod_{i \ge 1}m_i(\mu)!$$ Whenever $\mathcal{Q} \subseteq \mathcal{P}$ we define $$a_{\mathcal{Q}}=\prod_{\mu \in \mathcal{Q}}a_{\mu}, \hspace{.1 in} b_{\mathcal{Q}}=\prod_{\mu \in \mathcal{Q}}b_{\mu} \: .$$ \bigskip We consider the alternating group $A_n$. We let $\P^+$ denote the even partitions in $\P$, ${\mathcal O}$ the partitions into odd parts, and ${\mathcal D}$ the partitions into distinct parts. The conjugacy classes in $A_n$ are of two types. The classes labelled by partitions $\mu \in \P^+\setminus (\O \cap \D)$ are the non-split classes, which contain all $S_n$-permutations of this type; we denote a representative by $\sigma_{\mu}$ and note that the corresponding centralizer is then of order $z_{\mu}'=z_{\mu}/2$. For the partitions $\mu \in \D \cap \O$, the corresponding $S_n$-class splits into two conjugacy classes in $A_n$, for which we denote representatives by $\sigma_{\mu}^+$ and $\sigma_{\mu}^-$; their centralizers are of order $z_{\mu}'=z_{\mu}$. We briefly recall some information on the irreducible $A_n$-characters (see \cite[sect.~2.5]{JK}). Let $\mu$ be a partition of $n$. For $\mu \neq \tilde{\mu}$, i.e., $\mu$ non-symmetric, $[\mu]\downarrow_{A_n}=[\tilde{\mu}]\downarrow_{A_n}$ is irreducible. Let $\{\mu\}=\{\tilde{\mu}\}$ denote this irreducible character of $A_n$.\\ For $\mu=\tilde{\mu}$, i.e., $\mu$ symmetric, $[\mu]\downarrow_{A_n} = \{\mu\}_{+} + \{\mu\}_{-}$ is a sum of two distinct irreducible $A_n$-characters (which are conjugate in $S_n$).\\ This gives all the irreducible complex characters of $A_n$, i.e., $$ \mbox{Irr}(A_n) = \{\{\mu\}_{\pm} \mid \mu \vdash n, \mu=\tilde{\mu}\} \cup \{\{\mu\} \mid \mu \vdash n, \mu \neq \tilde{\mu}\} \: . $$ The characters $\{\mu\}_{\pm}$, for symmetric $\mu$, usually have non-rational values on the corresponding ``critical'' classes of cycle type $h(\mu)=(h_{1}, \ldots, h_{l})$, where $h_{1}, \ldots , h_{l}$ are the principal hook lengths in $\mu$; note that $h(\mu) \in \D \cap \O$, so the corresponding $S_n$-class splits. Then we have $[\mu](\sigma_{h(\mu)}) = (-1)^{\frac{n-l}{2}}=:\epps_{\mu}$ and $$ \{\mu\}_{+}(\sigma_{h(\mu)}^{\pm}) = \frac12 \left( \epps_{\mu} \pm \sqrt{\epps_{\mu}\prod_{i=1}^l h_i} \right) $$ $$ \{\mu\}_{-}(\sigma_{h(\mu)}^{\pm}) = \frac12 \left( \epps_{\mu} \mp \sqrt{\epps_{\mu}\prod_{i=1}^l h_i} \right) $$ All other irreducible $A_n$-characters have the same value on these two classes. %First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire %--restoring the headers and pagenumbering \pagenumbering{arabic} \addtocounter{page}{1} \markboth{\SMALL }{\SMALL } % % \medskip For later use, we want to recall the Jacobi minor theorem (see \cite[p.\ 21]{gantmacher}).\\ Let $A=(a_{ij})$ be an $n\times n$ matrix. Let $M_v$ be a $v$-rowed minor of the determinant $\det A$, corresponding to the rows $i_1, \ldots , i_v$ and the columns $k_1, \ldots , k_v$. Then we take the $(n-v)$-rowed complementary minor for $A$ by deleting all the rows and columns chosen for $M_v$ before, and define the signed complementary minor $M^{(v)}$ to $M_v$ by multiplying this complementary minor by the sign $\pm 1$, depending on $\sum_{j=1}^v i_j + \sum_{j=1}^v k_j$ being even or odd, respectively. (Note that for principal minors the sign is always $+$.) Let $A'=(A_{ij})$ be the $n \times n$-matrix of cofactors $A_{ij}$ for $A$, i.e., the adjoint matrix to $A$. Let $M_v$ and $M'_v$ be corresponding $v$-rowed minors of $A$ and $A'$, respectively, then $$M'_v = (\det A)^{v-1} M^{(v)} \: .$$ \medskip \section{Determinants of submatrices of the character table of $A_n$} We observe that by the Murnaghan-Nakayama formula we have for any symmetric partition $\mu$ and any $\nu \in \D \cap \O$: $$\{\mu\}_{\pm}(\sigma_{\nu}^{\pm}) =0 \; \textrm{for all } \nu > h(\mu) \; \textrm{(in lexicographic order)}$$ Hence, if we order the $k$ (say) partitions in $\D \cap \O$ in decreasing lexicographic order, and the $k$ symmetric partitions according to their principal hook lengths, then the corresponding $2k \times 2k$ part of the character table of $A_n$ is almost an upper triangular matrix, except that we have $2 \times 2$ blocks along the diagonal. We call this matrix $\X_s$. Knowing the entries of these diagonal blocks explicitly, we can easily compute their determinant and hence the (absolute value of the) determinant of this submatrix of the character table. A $2\times 2$ block corresponding to the characters $\{\mu\}_{\pm}$ on the classes $\sigma_{h(\mu)}^{\pm}$ gives a contribution of absolute value $$|\epps_{\mu}\sqrt{\epps_{\mu}\prod_i h_i}|=\sqrt{\prod_i h_i}=\sqrt{a_{h(\mu)}} \: ,$$ where $h(\mu)=(h_1, h_2 , \ldots)$. Hence the absolute value of the determinant of the whole submatrix is given by: \begin{proposition} $$|\det \X_s|= \prod_{\nu \in \D \cap \O} \sqrt{a_{\nu}} = \sqrt{a_{\D \cap \O} } \: .$$ \end{proposition} \medskip We can also easily determine the (absolute value of the) determinant for the whole character table $\X$ of $A_n$. By character orthogonality, we know that $\bar{\X}^t \X$ is a diagonal matrix with the centralizer orders as its diagonal entries. Set $\P^{(+)}=\P^+ \setminus (\D \cap \O)$. Hence we have $$\begin{array}{rcl} |\det \X|^2 & = & \left(\prod_{\mu \in \P^{(+)}} \frac{z_{\mu}}{2}\right) \left(\prod_{\mu \in \D \cap \O} z_{\mu}^2\right)\\[8pt] & = & 2^{-|\P^{(+)}|} z_{\P^+} z_{\D \cap \O} = 2^{-|\P^{(+)}|} a_{\P^+} b_{\P^+} a_{\D \cap \O} \\ \end{array}$$ Now we have $b_{\P^+}=2^{e^+} a_{\P^+}$, for some integer $e^+ \in \Z$. (This is not hard to prove by a combinatorial argument, see Lemma~\ref{Lemma3}.) Hence we obtain \begin{proposition}\label{X} $$\begin{array}{rcl} |\det \X|^2 & = & 2^{e^+-|\P^{(+)}|} a_{\P^+}^2 a_{\D \cap \O} \: . \end{array}$$ \end{proposition} In the next section we will see that $e^+=e^+(n) \in \N$, and that there is a nice generating function for the numbers $e^+(n)$ (Proposition~\ref{Proposition4}). In particular, an explicit formula for $e^+(n)$ is given by $$e^+(n)=\sum_{i=1}^{[n/2]} \tau(i) p'(n-2i) \: , $$ where $\tau(i)$ is the number of divisors of~$i$, and $p'(j)= |\D(j)\cap \O(j)|$. \medskip We are interested in determining the determinant of the integral part of the character table of $A_n$ corresponding to the non-symmetric partitions and the non-split conjugacy classes; let us call this matrix $\X_u$ (with some ordering of rows and columns chosen). (Note that this is also a submatrix of the character table of $S_n$.) This part of the character table of $A_n$ is complementary to the submatrix we have considered above, and we want to compute its determinant by employing Jacobi's theorem. \begin{theorem}\label{det(Xu)} The determinant of the matrix $\X_u$ has absolute value $$|\det \X_u| = 2^{(e^+-|\P^{(+)}|)/2} a_{\P^{(+)}} \: . $$ \end {theorem} \begin{proof} We assume that the rows of the character table $\X$ of $A_n$ are labelled such that the rows corresponding to the symmetric partitions come first, and that the columns are labelled such that the $v=|\P^{(+)}|$ partitions in $\P^{(+)}$ come first. Let $\Delta$ be the diagonal matrix with the centralizer orders $z_{\mu}'$ as its diagonal entries, and let $\Delta^{(+)}$ be the diagonal submatrix corresponding to the partitions $\mu \in \P^{(+)}$. As we have $\bar{\X}^t \cdot \X = \Delta$ , we know that the adjoint matrix to $\X$ is $$\X'=(\det \X) \Delta^{-1} \bar{\X}^t\: .$$ We now want to apply Jacobi's minor theorem as it is stated in Section~1. We take the $v$-rowed minor $M_v$ corresponding to the upper left square part in $\X$, i.e., $M_v=\det \X_u$. The corresponding minor of $\X'$ is then the determinant of $$(\det \X) (\Delta^{(+)})^{-1} \X_u$$ (remember that $\X_u$ is integral). The signed complementary minor to $M_v$ in $\X$ is then just $\det \X_s$. By Jacobi's theorem we know that $$(\det \X)^v (\prod_{\mu\in\P^{(+)}} z_{\mu}' ) ^{-1} \det \X_u = (\det \X)^{v-1} \det \X_s \: .$$ Hence $$\det \X_u = (\det \X)^{-1} 2^{-v} a_{\P^{(+)}} b_{\P^{(+)}} \det \X_s \: , $$ and thus $$\begin{array}{rcl} |\det \X_u| & = & 2^{-(e^+-v)/2} (a_{\P^{+}} \sqrt{a_{\D \cap \O}})^{-1} 2^{-v} a_{\P^{(+)}} b_{\P^{+}} \sqrt{a_{\D \cap\O}}\\ & = & 2^{(e^+-v)/2} a_{\P^{(+)}} \end{array} $$ where we have used the relation $b_{\P^+}=2^{e^+}a_{\P^+}$. \end{proof} \medskip \section{Powers of 2} We compute the generating functions for the powers of 2 occurring in the determinants of the previous section. Let $P(q), P^+(q), P^-(q)$ be the generating function for the number of partitions (resp. even/odd partitions) of $n$. The following is well-known: \medskip \begin{lemma}\label{Lemma1} $P^+(q)-P^-(q)=\Delta(q)$, where $$\Delta(q)=\prod_{k \ge 0}(1+q^{2k+1})\quad ( \; = \frac{P(q)P(q^4)}{P(q^2)^2})$$ is the generating function for the number of partitions of $n$ into distinct odd parts. \end{lemma} Indeed, using that in $P(q)=\prod_{k \ge 1} \frac{1}{1-q^k}$ the factor $\frac{1}{1-q^k}$ accounts for the parts equal to $k$ we see that $$ P^+(q)-P^-(q)=\prod_{k \ge 1} \frac{1}{1+(-q)^k} \: .$$ Substituting $q \rightarrow -q$ in the Euler identity $ \prod_{k \ge 1}(1+q^k) = \prod_{k \ge 0} \frac{1}{1-q^{2k+1}}$ and inverting we get $$\prod_{k \ge 1} \frac{1}{1+(-q)^k}=\prod_{k \ge 0} (1+q^{2k+1}) \: ,$$ proving the Lemma. $\Box$ \medskip We assume in the following always that $\delta=+$ or $-$ is a sign. \medskip \begin{corollary}\label{Corollary2} We have $$P^{\delta}(q)=\frac{P(q)+\delta \Delta(q)}{2} \; .$$ \end{corollary} We let $\mathcal{P}^{\delta}(n)$ be the set of partitions of $n$ with sign $\delta$. Then define $$a^{\delta}(n)=a_{\P^{\delta}(n)}= \prod_{\mu \in \mathcal{P}^{\delta}(n)}a_{\mu}, \hspace{.1 in} b^{\delta}(n)=b_{\P^{\delta}(n)}= \prod_{\mu \in \mathcal{P}^{\delta}(n)}b_{\mu}\; .$$ We factor each $i \in \mathbb{N}$ as a product $i=i_2i'$, where $i_2$ is a power of 2 and $i'$ is odd and consider two involutory bijections $\iota, \iota'$ on the set $$ \mathcal{T}(n)=\{(\mu,d,k)| \mu \in \mathcal{P}(n), m_d(\mu) \ge k \} \: .$$ Here $$\iota: (\mu,d,k) \mapsto (\hat{\mu},k,d)$$ where $\hat{\mu}$ is obtained from $\mu$ by replacing $k$ parts equal to $d$ by $d$ parts equal to $k$ and leaving all other parts unchanged and $$\iota': (\mu,d,k) \mapsto (\tilde{\mu},d_2k',k_2d')$$ where $\tilde{\mu}$ is obtained from $\mu$ by replacing $k$ parts equal to $d$ by $k_2d'$ parts equal to $d_2k'$ and leaving all other parts unchanged. Let $$\mathcal{T}^{\delta}_{d,k}(n)=\{\mu \in \mathcal{P}^{\delta}(n)| m_d(\mu) \ge k \}\: .$$ Then \beq |\mathcal{T}^{\delta}_{d,k}(n)|=p^{(-1)^{(d-1)k}\delta}(n-dk) \eeq Indeed removing $k$ parts equal to $d$ from a partition $\mu$ with sign $\delta$ gives you a partition with sign $(-1)^{(d-1)k}\delta$ and of cardinality $|\mu|-dk$. Note that this means that the partitions $\mu, \hat{\mu}$ in the definition of $\iota$ have different signs if and only if $(d-1)k$ and $d(k-1)$ have different parities, ie. if and only if $d$ and $k$ have different parities. Moreover the partitions $\mu, \tilde{\mu}$ in the definition of $\iota'$ have the same sign. Thus $\iota$ induces a bijection between $\mathcal{T}^{\delta}_{d,k}(n)$ and $\mathcal{T}^{\delta}_{k,d}(n)$ if $d,k$ have the same parity and between $\mathcal{T}^{\delta}_{d,k}(n)$ and $\mathcal{T}^{-\delta}_{k,d}(n)$ if $d,k$ have different parities. Moreover the bijection $\iota'$ shows that \beq a^{\delta}(n)'= b^{\delta}(n)', \eeq and hence \medskip \begin{lemma}\label{Lemma3} $b^{\delta}(n)/ a^{\delta}(n)=2^{e^{\delta}(n)}$ for some integer $e^{\delta}(n)$. \end{lemma} \medskip The power of 2 in $a^{\delta}(n)$ is $$x^{\delta}(n)= \underset{d \: \text{even}}{\prod_{d,k} } d_2^{|\mathcal{T}^{\delta}_{d,k}(n)| } $$ and the power of 2 in $b^{\delta}(n)$ is $$y^{\delta}(n)= \underset{ k \: \text{even}}{\prod_{d,k }} k_2^{|\mathcal{T}^{\delta}_{d,k}(n)| }$$ Let $x_o^{\delta}(n),x_e^{\delta}(n) $ be the product of the factors in $x^{\delta}(n)$, where $k$ is odd/even and correspondingly $y_o^{\delta}(n),y_e^{\delta}(n) $ be the product of the factors in $y^{\delta}(n)$, where $d$ is odd/even. Using the map $\iota$ we see that $$x_e^{\delta}(n)=y_e^{\delta}(n), ~~x_o^{\delta}(n)=y_o^{-\delta}(n) $$ Thus the power of 2 in $b^{\delta}(n)/ a^{\delta}(n)$ is $x_o^{-\delta}(n)/x_o^{\delta}(n)$. \\ Suppose that $x_o^{\delta}(n)=2^{f_o^{\delta}(n)}$ and $x_e^{\delta}(n)=2^{f_e^{\delta}(n)}$. Then $e^{\delta}(n)=f_o^{-\delta}(n)-f_o^{\delta}(n)$. \\ We have (since $\nu_2(d)=0$, when $d$ is odd) $$f_o^{\delta}(n)= \underset{ k \: \text{odd}}{\sum_{d,k}} \nu_2(d)|\mathcal{T}^{\delta}_{d,k}(n)|= \underset{ k \: \text{odd}}{\sum_{d,k }} \nu_2(d)p^{-\delta}(n-dk) \: .$$ Let $\tau_o(n)$ the number of odd divisors of $n$. Note that $\tau_o(n)\nu_2(n)$ equals the number $\tau_e(n)$ of {\it even} divisors of $n$. We then get (substituting $dk=t$ in the above sum and noting that then $\nu_2(d)=\nu_2(t)$ ) $$ f_o^{\delta}(n)= \sum_{t=1}^{n}\tau_o(t)\nu_2(t)p^{-\delta}(n-t)= \sum_{t=1}^{n}\tau_e(t)p^{-\delta}(n-t) \: . $$ Let $T(q)=\sum_{t \ge 1}\frac{q^t}{1-q^t}$ be the generating function for $\tau(n)$. Then $T(q^2)$ is the generating function for the number $\tau_e(n)$ of even divisors of $n$. If $F_o^{\delta}(q)$ is the generating function for $f_o^{\delta}(n)$ we obtain \beq F_o^{\delta}(q)=T(q^2)P^{-\delta}(q). \eeq Using Lemmas~\ref{Lemma1} and~\ref{Lemma3} we deduce \begin{proposition}\label{Proposition4} The generating function for $e^{\delta}(n)$ is $$E^{\delta}(n)= F_o^{-\delta}(q)-F_o^{\delta}(q)=\delta T(q^2)\Delta(q) \: .$$ \end{proposition} \begin{remark} {\rm This Proposition was also proved in \cite{mueller} in a different way. Our approach was partially inspired by an unpublished note of John Graham.\\ Note that the proposition shows that $e^+=e^+(n)$ is always a {\em positive} integer. } \end{remark} \medskip Let us also consider $F_e^{\delta}(q)$. We have $$f_e^{\delta}(n)=\sum_{\{d,k| k~ even \}}\nu_2(d)|\mathcal{T}^{\delta}_{d,k}(n)|= \sum_{\{d,k| k~ even \}}\nu_2(d)p^{\delta}(n-dk) \: .$$ We substitute $dk=2t$ in the above and obtain $$f_e^{\delta}(n)=\sum_{t \ge 1} \tau^*(t)p^{\delta}(n-2t) \: ,$$ where $\displaystyle \tau^*(t)=\sum_{d \mid t}\nu_2(d)$. We have $$\tau^*(t)=\binom{\nu_2(t)+1}{2} \prod_{p~odd}(\nu_p(t)+1) \: .$$ Thus if $T^*(q)$ is the generating function for $\tau^*(t)$ then $$F_e^{\delta}(q)=T^*(q^2)P^{\delta}(q) \: .$$ It is easily seen that $$T^*(q)=\sum_{j \ge 1} T(q^{2^j}) \: .$$ \medskip \begin{proposition}\label{Proposition5} The exponent of 2 in $a^{\delta}(n)$ has the generating function $$F_e^{\delta}(q)+F_o^{\delta}(q)= T^*(q^2)P^{\delta}(q)+T(q^2)P^{-\delta}(q) \: .$$ \end{proposition} \bigskip In Theorem~\ref{det(Xu)} we have seen that $|\det \X_u| = 2^{(e-|\P^{(+)}|)/2} a_{\P^{(+)}}$.\\ By Proposition~\ref{Proposition4}, $e=e^+(n)$ has generating function $E^+(q)=\Delta(q)T(q^2)$. \\ Moreover $|\P^{(+)}(n)|$ has generating function $P^+(q)-\Delta(q)=P^-(q)$ (Lemma \ref{Lemma1}). Clearly, $ a_{P^{(+)}}(n)$ is divided by the same power of 2 as $a^+(n)$, as the removed partitions have only odd parts. The generating function for the corresponding exponent is given by Proposition~\ref{Proposition5}. Hence the exponent of 2 in $\det \X_u$ has the generating function $$G(q)=\frac{1}{2}\left(T(q^2) \Delta(q)-P^-(q)\right)+T^*(q^2)P^+(q)+T(q^2)P^-(q),$$ and this then yields \begin{theorem} The exponent of 2 in $\det \X_u$ has the generating function $$G(q)= \frac{1}{2}\left(T(q^2)P(q)-P^-(q)\right)+T^*(q^2)P^+(q) \: .$$ \end{theorem} \medskip According to MAPLE the first values of the coefficients of $G(q)$ are the following for $n=2, \ldots ,14$: 0 0 2 2 4 6 15 19 30 43 70 94 138. \medskip Let us finally remark that the Propositions~\ref{Proposition4} and~\ref{Proposition5} also allow to compute the generating function for the exponent of 2 in $|\det(\X)|$, using Proposition~\ref{X}. \begin{thebibliography}{99} \bibitem{BesOls} C.\ Bessenrodt, J.~B.\ Olsson, A note on Cartan matrices for symmetric groups, Archiv d.\ Math.\ {\bf 81} (2003), 497--504. \bibitem{bos} C.\ Bessenrodt, J.~B.\ Olsson, R.~P.\ Stanley, Properties of some character tables related to the symmetric groups, preprint 2003; to appear in: J.\ Algebraic Combinatorics. \bibitem{gantmacher} F.~R.\ Gantmacher, \emph{The Theory of Matrices}, vol.\ 1, Chelsea, New York, 1960. \bibitem{James} G.\ James, The representation theory of the symmetric groups, Lecture notes in mathematics {682}, Springer--Verlag, 1978. \bibitem{JK} G.~James, A.~Kerber, \emph{The Representation Theory of the Symmetric Group}, Addison--Wesley, 1981. \bibitem{mueller} J.\ M\"uller, On a remarkable partition identity, J. Comb. Th. A {\bf 101} (2003), 271--280. \bibitem{OlsReg} J.~B.~Olsson, Regular character tables of symmetric groups, The Electronic Journal of Combinatorics {\bf 10} (2003), Article~\#N3. \end{thebibliography} \end{document}