9

Let $\sigma=(1,2,3,\dots,n)$ be an odd $n-$cycle in $S_n$ (so $n$ is even). It is known that the size of its conjugacy class is $|cl_{S_n}(\sigma)|=(n-1)!$. I am interested in the size of the subset $S=cl_{cl_{S_n}(\sigma)}(\sigma)$, that is, the set of all the $n-$cycles that we can obtain by conjugating $\sigma$ only with elements in its conjugacy class. In particular, I would like to show that $S$ contains at least $|cl_{S_n}(\sigma)|/2=(n-1)!/2$ elements, which happens to seem true in the numerical experiments I run.

Approaches I tried:

  1. Construct enough elements by hand. I found this trick: If we conjugate $\sigma=(1,2,3,\dots,n)$ by $(1, a_1, 3, a_3, 5, a_5, \dots, 4, a_4, 2, a_2, n)$ for a free choice of the $a_i$ in the set of the remaining half symbols, we obtain a permutation sending $(1,a_1,a_2,a_3,a_4,\dots,a_{n/2-1},\dots)$. This way it is possible to produce explicitly some mutually different elements, but still not enough.

  2. By elementary group theory, two elements $\alpha,\beta\in S$ are such that $\alpha\sigma\alpha^{-1}=\beta\sigma\beta^{-1}$ if and only if $\beta=\alpha\sigma^k$ for some even power $k$ of $\sigma$. This way we obtain that $S$ contains at least $\frac{(n-1)!}{n/2-1}$ elements, still too few, unfortunately, due to the denominator depending on $n$ (but observe that this case solves the case $n=6$).

Notes:

  1. The fact that we are choosing as $\sigma$ that particular odd cycle is just a simplification of the original problem. This claim still seems to be true for any odd permutation $\sigma\in S_n$ for every $n$ big enough to contain $\sigma$.

I feel that this problem is not impossible, and just a clever simple counting trick is needed. Thanks in advance for any suggestions or comments.

fspa
  • 569
  • 3
  • 13
  • Thanks for your comment Derek. The number $n$ must be even for $\sigma$ to be an odd permutation, so unfortunately I knew already that for odd values of $n$ counterexamples exist. – fspa Apr 25 '24 at 09:18
  • 1
    I started to write an answer, but currently it looks as a comment: "Let $n,m$ be any natural numbers, $\sigma=(x_1 x_2\dots x_m)\in S_n$ be any cycle, and $\tau\in S_n$ be any permutation. It is easy to check that $\tau^{-1}\sigma\tau=(\tau(x_1) \tau(x_2)\dots \tau(x_m))$. In particular, if $\sigma=(1 2\dots n)$ then $\tau^{-1}\sigma\tau=(\tau(1) \tau(2)\dots \tau(n))$. If $\tau$ is conjugated with $\sigma$, that is if $\tau$ a cycle of $n$ elements then ..." I shall try to continue later. – Alex Ravsky Apr 28 '24 at 05:13
  • 1
    Thanks for your comment and your approach, Alex. I am curious to see where it will lead! :) – fspa Apr 29 '24 at 13:24

0 Answers0