Here are the things I did:
Claim: Let $p$ be prime. $\sigma \in S_n$ with $|\sigma|=p$ if and only if $\sigma$ is product of $p$-cycles.
$\rightarrow$ Supose $|\sigma|=p$ and $\tau$ is a $k$-cycle on the decomposition of $\sigma$ with $k\leq p$, then $\tau^p=id$, so $k$ divides $p$ and if we exclude the $1$-cycles we get that $k=p$, by the other hand, if $k>p$, $\sigma^p=id$ and $\tau^p \neq id$, absurd.
$\leftarrow$ Clear.
Obs: The number of $p$-cycles in $S_n$ is $\dfrac{n!}{p(n-p)!}$.
Now, let $\sigma \in S_n$ of order $p$, $\sigma=\tau_1\tau_2\cdots\tau_r$ with $1\leq r \leq\dfrac{n!}{p(n-p)!}$ where each $\tau_i$ is a $p$-cycle, then the type of $\sigma$ is $(1^{n-r}\; p^r)$, and by the Orbit-Stabilizer Theorem we get:
$[S_n:C_{S_n}(\sigma)]=|O(\sigma)|=\dfrac{n!}{1^{n-r}(n-r)!p^rr!}=\dfrac{n!}{(n-r)!p^rr!}$.
This is a Dummit & Foote exercise, sec. 4.3, exercise 35. But the say to use the greatest integer function, which I did not use, so didn't. Have I calculated the number I was supposed to calculate? Or is there another way to do it? Either way, could some one explain me the right path?
Thanks.
Best Answer
All permutations of cycle type $$(a_1\,a_2\,\ldots\,a_p) $$ are conjugates of each other. Likewise, all permutations of cycle type $$(a_1\,a_2\,\ldots\,a_p)(b_1\,b_2\,\ldots\,b_p) $$ are conjugates of each other. More generally, for every $k$ with $1\le k\le \frac np$, we have one conjugacy class comprised of those permutatons consisting of $k$ $p$-cycles. The number of such $k$ is $\lfloor \frac kp\rfloor$.