Let $\sigma=(1,2,...,n)$ be a long cycle in the permutation group $S_n$. Let $\lambda\vdash n$ be a certain partition.

My question is: when $\pi$ runs over the group $S_n$, how often will the comutator $\pi\sigma\pi^{-1}\sigma^{-1}$ have cycle type $\lambda$?

I have observed that cycle type $\lambda$ only happens if its length $\ell(\lambda)$ is of the same parity as $n$. How to prove this?

  • $\begingroup$ what is $l(\lambda)$? $\endgroup$
    – Asinomás
    Mar 14, 2018 at 22:13
  • $\begingroup$ the length - number of parts $\endgroup$
    – Marcel
    Mar 14, 2018 at 23:41
  • $\begingroup$ Commutators don't play nice with cycle types, so I'd expect the first question to either be intractable or deep. $\endgroup$
    – anon
    Mar 15, 2018 at 0:00

1 Answer 1


As $\pi$ runs through the group $S_n$, the permutation $\tau=\pi\sigma\pi^{-1}$ runs through the $n$-cycles. There are $n!$ possible $\pi$'s and $(n-1)!$ cycles, so each cycle appears $n$ times.

Given $\tau$, you want that $\tau\sigma$ has cycle type $\lambda$. This means that $\sigma=\tau^{-1}\theta$ with $\tau$ a $n$-cycle and $\theta$ has cycle type $\lambda$.

This is a factorization problem. The solution, from the celebrated Frobenius formula which you can find, for example, in "Factoring N-Cycles and Counting Maps of Given Genus", by Goupil and Schaeffer, is

$$\frac{|C_\lambda|}{n}\sum_\nu \frac{\chi_\nu(\lambda)}{d_\nu},$$ where $\nu$ is a hook partition $\nu=(1^r,n-r)$ and $d_\nu={n-1\choose r}$.

The solution to your original question is, therefore, given by $$ \frac{|C_\lambda|}{(n-1)!}\sum_{r=0}^{n-1} r!(n-1-r)!\chi_{(1^r,n-r)}(\lambda).$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .