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

Question

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.

Answer 1

Hint : For $n \ge 5$, $A_n$ is the only normal subgroup of $S_n$

Answer 2

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.

Answer 3

I don't know whether my answer is correct, but I wanna share about it

similar to the construction of $Π_i$ in example 2.5.13 in Algebra, Artin, we can't construct them without contradiction when $n$ is odd

so $n$ need to be even.

let $a$ be a prime divisor of $n$, and $n=ka$

we have $\frac{C_{n}^{a}C_{n-a}^{a}···C_{a}^{a}}{A_{k}^{k}}=n-1$, that is, ${\frac{n!}{(a!)^{k}·k!}}=n-1$

when $n=2,4$, the equation is satisfied

and when $n≥6$, ${\frac{n!}{(a!)^{k}·k!}}＞＞n-1$

Answer 4

A bit different answer than the OP's:

Suppose $\varphi:S_n\to S_{n-1}$ is indeed a group epimorphism and let's also observe $\varphi(A_n).$

By definition, $\varphi':A_n\to\varphi(A_n), \varphi'(\sigma):=\varphi(\sigma)$ is surjective. Since $A_n$ is simple, every homomorphism from $A_n$ is either trivial or injective. If $\varphi'$ were injective, it would be bijective (since it is surjective) and therefore, $\varphi(A_n)$ would be a subgroup of $S_{n-1}$ of order $\frac{n!}2,$ a contradiction because $\frac{n!}2>\frac{n!}n=(n-1)!$ for $n\ge 5$ in particular.

Hence, $\varphi(A_n)=\varphi'(A_n)=\{e_{S_{n-1}}\},$ which implies $A_n\subseteq\ker\varphi.$ However, $A_n$ is maximal in $S_n,$ so, either $\ker\varphi=A_n$ or $\ker\varphi=S_n,$ the latter of which is impossible since $\varphi(S_n)=S_{n-1}\ne\{e_{S_{n-1}}\}.$ Therefore $\ker\varphi=A_n$.

By the First Isomorphism Theorem, $S_n/A_n=S_n/\ker\varphi\cong\operatorname{Im}\varphi=S_{n-1},$ yet another contradiction since $[S_n:A_n]=2,$ while $|S_{n-1}|=(n-1)!>2,$ for $n\ge 5$ in particular.