I am trying to prove that for all $n$, $S_n$ is isomorphic to a subgroup of $A_{n+2}$.
Say $S_n$ acts on $\{\alpha_1,...,\alpha_n\}$ and $A_{n+2}$ acts on $\{\alpha_1,...,\alpha_n,\alpha_{n+1},\alpha_{n+2}\}$. Let $\sigma \in S_n$ and let $\phi: S_n \to A_{n+2}$ be given by $\phi(\sigma)=\sigma$ if $\sigma$ is an even permutation, and $\phi(\sigma)=\sigma(\alpha_{n+1}\alpha_{n+2})$ if $\sigma$ is an odd permutation. This way $\phi(\sigma) \in A_{n+2}$.
Now, let $\sigma, \rho \in S_n$. Then $\phi(\sigma \rho)=\sigma \rho (\alpha_{n+1}\alpha_{n+2})$. Also, $\phi(\sigma)\phi(\rho)=\sigma(\alpha_{n+1}\alpha_{n+2})\rho(\alpha_{n+1}\alpha_{n+2})=\sigma\rho$.
So, $\phi(\sigma \rho)$ does not seem to equal $\phi(\sigma)\phi(\rho)$. But the idea is that we want $\phi(S_n) \cong S_n$, and $(\alpha_{n+1}\alpha_{n+2})$ does not permute any members of $\{\alpha_1,...,\alpha_n\}$, so $\phi(\sigma \rho)$ and $\phi(\sigma)\phi(\rho)$ are equal as permutations on this (sub)set.
This feels a little subtle to me and I'm not sure if my idea is valid or not. If it is not, can this proof still be saved? I appreciate any thoughts on this. Thanks.