I have always see the following Murnaghan-Nakayama rule for a partition $\lambda$ and a permutation $\sigma \in \mathfrak{S}_n$ of cycle structure $(\sigma_1, ..., \sigma_n)$: $$ \chi_{\lambda}(\sigma) = \displaystyle \sum_{\xi \in R(\lambda, \sigma_1)} (-1)^{ht(\xi)} \chi_{\lambda \setminus \xi}(\sigma \setminus \sigma_1). $$ where $R(\lambda, \sigma_1)$ are the skew-hooks of $\lambda$ of length $\sigma_1$.

However, in the proof I am reading, they use a somewhat different rule. Let $\sigma \in \mathfrak{S}_{n+m}$ be a permutation that fixes (at least) $m$ elements and $\tau$ be a permutation of the $m$ elements fixed by $\sigma$. Then, $$ \chi_{\lambda}(\sigma \tau) = \displaystyle \sum_{\xi \in R(\lambda, \sigma_1)} (-1)^{ht(\xi)} \chi_{\lambda \setminus \xi}(\sigma). $$ The equivalence between the two formulas is clear if $|\tau| \geq \sigma_1$, but I don't see why this is true when $|\tau| < \sigma_1$. Is it trivial? Am I missing something, or it is a non-trivial corolary?

Thanks in advance!

  • 3
    $\begingroup$ Your first version of the Murnaghan–Nakayama rule does not require any particular order on the cycles. You can take $\sigma_1$ to be the shortest or the longest cycle, or anything in between. For instance $\chi^{(4,1,1)}$ vanishes on a $5$-cycle, and this is most easily seen by taking $\sigma_1 = 5$ rather than $\sigma_1 = 1$. Does this answer your question? $\endgroup$ – Mark Wildon Mar 27 at 13:21
  • $\begingroup$ Yes, I thought I had to take the biggest part! Thank you! $\endgroup$ – eti902 Mar 27 at 13:46

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