# Two versions of the Murnaghan-Nakayama rule

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?

• 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? – Mark Wildon Mar 27 at 13:21