Finding A Subgroup of $A_n$ Isomorphic to $S_{n-2}$ I am asked to show there exists a subgroup of $A_n$ that is isomorphic to $S_{n-2}$ for $n \ge 3$. Here are some of my thoughts:
Certainly there exists a homomorphism $f : A_n \rightarrow S_n$---namely, take $f$ to be the inclusion map, $f(\sigma) = \sigma$. Because of this, $f^{-1}(S_n)$ will be a subgroup of $A_n$, and I suspect I can extend $f$ in someway to an isomormphism by defining $\varphi : S_{n-2} \rightarrow f^{-1}(S_n)$
I am not really sure how to move forward, but does this seem to be moving in the right direction? It seems that I want to take all even permutation $S_{n-2}$ and map them to themselves, and then take every odd permutation and "correct" it to become an even permutation, but I am not sure why this is right--or if it is even remotely correct. Can $\varphi$ map an odd permutation to an even permutation and still be an isomorphism? It doesn't seem that it could.

If $\sigma\in S_{n-2}$ is even, map it to itself. Otherwise, map it to $\sigma\circ(n-1,n)$.
