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First of all, I saw that question Number of Elements of order $p$ in $S_{p}$ and additionally ask this question. Please inform me if there is such question in the site then we can close question.

Numbers of elements of order $p$ in $Sym(n)$ where $Sym(n)$ denotes symetric group of order $n!$ .

I thought it is better to start with $p$-cycles as their order is $p$. And order is depends on $n$ and $p$ so we should divide in a cases.

Any comments, suggestions would be appreciated. Thank you.

Arturo Magidin
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Fuat Ray
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1 Answers1

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If the OP meant $p$ is prime then $S_n$ acts by conjugation on the elements of order $p$.

The orbits are the $o_k,k\le n/p$ where $o_k$ is the set of elements with $k$ ($p$-cycles) and $n-kp$ ($1$-cycles).

The stabilizer of an element of $o_k$ has $(n-kp)! p^k k!$ elements.

So $o_k$ contains $\frac{n!}{(n-kp)! p^k k!}$ elements and there are $$\sum_{k=1}^{n/p} \frac{n!}{(n-kp)! p^k k!}$$ elements of order $p$.

reuns
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