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Let $\lambda \vdash nk$. Let $n^k$ denote the partiton with $k$ parts of size $n$. We can compute $\chi^\lambda(n^k)$ by using the Murnaghan-Nakayama rule, as a signed sum over border-strip tableaux, (shape $\lambda$ and border strips of size $n$).

Now, by checking several cases on the computer, it seems like each term in the (Murnaghan-Nakayama) sum has the same sign. Combinatorially, it reduces to proving the following:

Define the height of a border-strip to be the maximal row index that it touches, minus the minimal row index. The height of a border-strip decomposition of the diagram $\lambda$ is the sum of the heights of the border-strips. Show that the parity of this sum is independent of the actual decomposition (it only depends on $\lambda$ and $n$).

I am looking for a reference of this (I do think I can in fact prove a stronger statement, but I have not worked out the details yet).

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This is corollary 10 in "A bijection proving orthogonality of the characters of $S_n$", by Dennis E. White.

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  • $\begingroup$ Perfect, this solves my problem! $\endgroup$ Dec 18 '17 at 19:54

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