Question: Let $\sigma \in S_9$ satisfy $o(\sigma)=5$. How many $k \in [9]$ are there such that $\sigma(k)=k$?
I struggle with the combinatorics in this question.
I can see the permutation requires a 5-cycle which makes the cycle structure $[5,1,1,1,1]$ so the answer is ${9\choose5}\cdot4! $ ?
${9\choose5}$ represents the number of ways to choose the elements in the 5-cycle and $4!$ is the number of ways to choose the 4 elements s.t. $\sigma(k)=k$
Is this the correct way to think about the problem?
Thanks.