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I am trying to do the following exercise:

Let $C_{n,k}$ denote the set of $k$-cycles of $S_n$. Show that $|C_{n,k}|=\binom{n}{k}(k-1)!$.

Approach and Definitions:

Here are the Definition I know: Let $\pi \in S_n$ be a permutation. \pi is a $k$ cycle if

  1. $k$ is the smallest number s.t. $\pi^k(i)=i$
  2. $\pi$ fixes each $j$ not in $\{i,\pi(i),...,\pi^{k-1}(i)\}$.

The Approach: I don't know how to count the $k$-cycles. I know that a $k$ can be represented as $\{i,\pi(i),...,\pi^{k-1}(i)\}$. But also as $\{\pi(i),...,\pi^{k-1}(i),i\}$ or as $\{\pi^2(i),...,\pi^{k-1}(i),i,\pi(i)\}$,..., $\{\pi^{k-1}(i),...,\pi(i),i\}$. This would give $k$ ways to represent this one cycle. (But the qeustions wasn't how many ways can it be represented, but how many c-cylces are there.)

So I am still clueless on how to proceed.

NTc5
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1 Answers1

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A $k$-cycle is a $k$ tuple with elements from $\{1,...,n\}$ that "shift" between those $k$ elements. You can take those $k$ elements and rename them, for simplicity, to $1,...,k$. Then you can ask yourself how many permutations of $S_k$ are there, as we know $k!$. But every $k$-cycle can be represented in $k$ ways, as you already elaborated in your questions. So we get $(k-1)!$. But we have not considered the k numbers from $\{1,...,n\}$ that we chose. There are $\binom{n}{k}$ ways to choose them. Thus, the final result is $\binom{n}{k}(k-1)!$.

Philip
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