Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.8
Let $G = S_n$ and fix $i \in \{ 1, 2, \ldots, n \}$. Prove that $\mathsf{stab}_G(i)$ is a subgroup of $G$, and find $|\mathsf{stab}_G(i)|$.
Solution: Note that $\mathsf{stab}_G(i)$ is not empty since $1 \in \mathsf{stab}_G(i)$. Now suppose $\sigma$, $\tau \in \mathsf{stab}_G(i)$; we have $$(\sigma \circ \tau^{-1})(i) = \sigma(\tau^{-1}(i)) = \sigma(i) = i,$$ so that $\sigma \circ \tau^{-1} \in \mathsf{stab}_G(i)$. By the subgroup criterion, $\mathsf{stab}_G(i) \leq G$.
Now every permutation that fixes $i$ is a permutation of the remaining $n-1$ elements of the set $\{ 1, 2, \ldots, n \}$. There are $(n-1)!$ such permutations. Thus $|\mathsf{stab}_G(i)| = (n-1)!$.
