Characterize the natural numbers $n$ such that there is a surjective group homomorphism from $S_n$ to $S_{n-1}$. This has always been one of my most favorite exercises from Group Theory, and I was surprised to see that this hasn't been asked before. To repeat:

Characterize the natural numbers $n$ such that there is a surjective group homomorphism from $S_n$ to $S_{n-1}$.

I have a solution which I will post in a couple days (if someone doesn't recreate it), but I am more interested in seeing how other people would approach this problem. I am very interested in alternate proofs of this characterization.
 A: What I liked most about this problem is that it doesn't lend itself to be easily solved by counting (it might, but I haven't seen such a proof). Thus, we are forced to travel different paths to attack the problem from which we might learn something else interesting in its own right.
For example, GA316's observation that $A_n$ is the only normal subgroup of $S_n$ for $n\geq 5$ is something you might discover while trying to solve this problem. The proof of that lemma essentially relies on the simplicity of $A_n$ for $n\geq 5$ and has been addressed on this site here.
My proof of this problem also relies on the simplicity of $A_n$ for $n\geq 5$, but I went down a different road to solve it. Namely, I proved the following lemmas:

Lemma 1: $A_n$ is the only subgroup of $S_n$ with index 2.
Lemma 2: Let $G$ and $H$ be two groups, and let $\phi:G\rightarrow H$ be a surjective group homomorphism. Let $K$ be a subgroup of $G$ with finite index $n$. Then $\phi(K)$ is a subgroup of $H$ with finite index, and its index divides $n$.

I'll leave those lemmas to the interested reader (I know at least the first is on this site).
Now I'll show that there is no surjective homomorphism from $S_n$ to $S_{n-1}$ for $n\geq 5$. Let $n\geq 5$, and suppose to the contrary that $\phi:S_n\rightarrow S_{n-1}$ were a surjective homomorphism. Consider the subgroup $\phi(A_n)$ in $S_{n-1}$. Its index must divide $2$ by the lemma. Hence its index is either $1$ or $2$. If its index were $1$, then $\phi$ restricts to a surjective homomorphism from $A_n$ to $S_{n-1}$. But $A_n$ is simple and only has trivial normal subgroups. Thus the only quotients of $A_n$ have order $1$ or $n!/2$. But $S_{n-1}$ has order $(n-1)!$, and for $n\geq 5$ we have that $n!/2> (n-1)!$. So $\phi(A_n)$ cannot have index $1$, thus it must have index $2$. By the lemma, this means that $\phi(A_n)=A_{n-1}$. But this means that $\phi$ restricts to a surjective homomorphism from $A_n$ to $A_{n-1}$. But $A_n$ is still simple and we reach another contradiction in cardinalities. And we reached our final contradiction.
A: Hint : For $n \ge 5$, $A_n$ is the only normal subgroup of $S_n$
