Exercise 4.3.35 (Number of conjugacy classes of elements of order $p$ in $S_n$)

Question

Let $p$ be a prime. Find a formula for the number of conjugacy classes of elements of order $p$ in $S_n$ (using the greatest integer function).

Richard Ganaye · Accepted Answer

\begin{proof} 
Since $p$ is a prime, the elements of order $p$ in $S_n$ are the products of $k$ cycles of length $p$, where $0 <  k p \leq n$, so 
$$1\leq k \leq \left \lfloor \frac{n}{p} \right \rfloor.$$

All the products of $k$ cycles of length $p$ form a conjugacy class, with cycle type $(1,1,\ldots,1, p, p \ldots, p)$, with $k$ occurrences of $p$, and $n - kp$ occurrences of $1$. Hence the number $N$ of conjugacy classes of elements of order $p$ in $S_n$ is
$$N = \left \lfloor \frac{n}{p} \right \rfloor.$$
\end{proof}

Example: In exercise 1.3.7, there are $2 = \left \lfloor \frac{4}{2} \right \rfloor$ conjugacy classes of elements order $2$ in $S_4$: the conjugacy class of $(1\ 2)$ and the conjugacy class of $(1\ 2)(3\ 4)$. In exercise 1.3.6, there is only $1 =\left  \lfloor \frac{4}{4} \right \rfloor$ class of conjugacy of elements of order $4$ in $S_4$, the class of $(1\ 2\ 3\ 4)$.

By Exercise 33, the number of elements in the class of conjugacy of a product of $k$ cycles of length $p$ in $S_n$ is $\frac{n!}{(n-kp)!\, k!\,  p^k}$.

Exercise 4.3.35 (Number of conjugacy classes of elements of order $p$ in $S_n$)

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Exercise 4.3.35 (Number of conjugacy classes of elements of order $p$ in $S_n$)

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