(Permutations) For $n ≥ 2$,there are at least n subgroups of Sn of order $(n − 1)!$

Question

For $n ≥ 2$, show that there are at least n subgroups of Sn of order $(n − 1)!$

Answer 1

For $k \in \{1,..,n\}$, let $$H_k=\{ \sigma \in S_n | \sigma(k)=k \}$$ Then the subsets $H_1, .. H_n$ are subgroups of order $(n-1)!$ that are mutually distinct (I leave that to you to check).