Number of $k$-cycles in $S_n$

Question

I've computed that the number of $k$-cycles in $S_n$ is $\frac{n!}{(n-k)!k}$ and wiki seems to agree with me. Now, we know that in $S_n$ the number of $k$-cycles is also equal to the cardinality of the conjugacy class of a $k$-cycle (two elements are conjugate if and only if they have the same cycle type, and two $k$-cycles have the same cycle type). But, according to this formula, the conjugacy class of a $k$-cycle has $\frac{n!}{k}$ elements. I found that formula on Dummit&Foote, too.

Where am I wrong?

Answer 1

Those formulas give the same thing, but you seem to have been forgetting the $a_j!$ in the denominator for $a_1$. The partition for a $k$-cycle is

$$\lambda =(\underbrace{k}_{1\text{ time}}\quad \underbrace{1\,\ldots \, 1}_{n-k\text{ times}})$$

so that

$$\begin{cases}a_k = 1 \\ a_1=n-k \\ a_j=0 & j\ne 1, k\end{cases}.$$

Then the formula gives

$$n!\bigg/\left(\prod_{j=1}^nj^{a_j}\cdot a_j!\right)$$

Since $0!=1$ and $m^0=0$ for all $m\ne 0$ we have this reduces to

$${n!\over 1^{n-k}\cdot(n-k)!\cdot (k)^1\cdot 1!}={n!\over k(n-k)!}$$