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I have this problem

Let $m \geq 2$ an integer and $\sigma \in S_n$, when do we have $f \in S_n$ such that $f^m = \sigma$

I think a half answer is that every cycle in the decomposition of $\sigma$ must be of length divisible by $m$.

Should be there a condition on $n$, like for example $\sigma =(1325)(46)$ is a square in $S_{12}$ but not in $S_6$. I am thinking that $m$ times the length of the support of $\sigma$ must be less than or equal the number of fixed elements?

Edouard
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    https://math.stackexchange.com/questions/266569/how-to-find-the-square-root-of-a-permutation may be relevant. – Rob Arthan Dec 30 '24 at 00:01
  • At first, since $m \geq 2$, I thought $n \geq 2$? But no one requires $m$ is the least iterate of $f$ giving $\sigma$... – Eric Towers Dec 30 '24 at 01:04
  • How is $\sigma$ a square in $S_{12}$? – Steve Kass Dec 30 '24 at 01:04
  • @SteveKass the square of $(1a3b2c5d)(4e6f)$ is equal to $\sigma$ where $a,b,c,d,e,f$ are distinct and invariant by $\sigma$ – Edouard Dec 30 '24 at 01:15
  • @EricTowers well, if $n=2$, $(12)$ is a $m$ power of itself if $m$ is odd, in general we always have $\sigma ^ {k \times ord(\sigma) +1} $ for all $k$. So if $m$ is minimal, we should have that condition, otherwise I should look at the gcd? – Edouard Dec 30 '24 at 01:24
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    The square of $(1a3b2c5d)(4e6f)$ is $(1325)(46)(abcd)(ef)$, which is not $\sigma$. – Steve Kass Dec 30 '24 at 01:52
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    @SteveKass Thank you very much, I did a huge mistake, so is it true to say that $\sigma$ has $m$th root iff (for fixed $k$) the number of $km-$cycles in its decomposition is divisible by $m$? so we can combine them all in one cycle in the root – Edouard Dec 30 '24 at 23:22
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    What you suggest does sound plausible, as it neatly seems to generalize the argument about square roots in the question Rob Arthan linked to in the first comment. – Steve Kass Jan 01 '25 at 00:39

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