$H=\{\sigma \in S_n :\sigma(3)=3\}$
where $S_n$ is the symmetric group of degree $n$. We need to find the order of the subgroup $H$. My book has the answer 2.
But my intuition says that by fixing $3$, we can permute the rest of the $n-1$ symbols, thus we should have $(n-1)!$ as the order of $H$.
Is that the correct way to think?