Q: If $ H = \{ \sigma \in S_n : \sigma(n) = n \} $ is a subgroup of $S_n$, then show that $H \simeq S_{n-1}$.
I know any group is isomorphic to a subgroup of the symmetric group. But I don't know how to proceed.
Q: If $ H = \{ \sigma \in S_n : \sigma(n) = n \} $ is a subgroup of $S_n$, then show that $H \simeq S_{n-1}$.
I know any group is isomorphic to a subgroup of the symmetric group. But I don't know how to proceed.
Following Tara B's suggestion I'm posting my comment as an answer:
Think of $S_n$ as the group of bijections from $\{1,\dots,n\}$ to itself with composition as the group multiplication. What does a bijection fixing $n$ do? It must give a bijection from $\{1,\dots,n-1\}$ to itself. Conversely, every bijection of $\{1,\dots,n-1\}$ gives a bijection of $\{1,\dots,n\}$ fixing $n$, so...