For which $n$ and $k$ are there nontrivial homomorphisms from $S_n$ to $\mathbb{Z}/k\mathbb{Z}$? The question How do I find the number of group homomorphisms from $S_3$ to $\mathbb{Z}/6\mathbb{Z}$? got me curious and thinking about the following: 
For which $n$ and $k$ are there nontrivial homomorphisms from the symmetric group $S_n$ to the cyclic group $\mathbb{Z}/k\mathbb{Z}$ of order $k$? I'm thinking it might be an open problem. Is this a problem with a known solution? If so, how can we prove it? 
Thanks! 
 A: You can proceed as follows. Let $f:S_n\to \mathbb{Z}/k\mathbb{Z}$.
We know that any two transpositions are conjugate in $S_n$, so they have conjugate images by $f$. Since $\mathbb{Z}/k\mathbb{Z}$ is abelian, it follows that two transpositions have same image.
Let $\tau$ be a fixed transposition. Then $2f(\tau)=f(\tau^2)=f(Id)=0$, so $f(\tau)$ has order $1$ or $2$. In the first case, the image of any transposition is trivial, and since they generate $S_n$, $f$ is trivial. In the second case, which may happen only if $k$ is even, a transposition is mapped to the only element of order $2$ of $\mathbb{Z}/k\mathbb{Z}$.
In this case, $f(\sigma)$ is $0$ is $\sigma$ is even, and $\overline{k/2}$ if $\sigma$ is odd. In other words, $f$ will be the composition of the signature morphism with the morphism $\theta: \{\pm 1\}\to \mathbb{Z}/k\mathbb{Z}$ sending $-1$ to $\overline{k/2}$.
Conclusion. If $k$ is odd, there is no nontrivial morphism. If $k$ is even, there is exactly one nontrivial morphism (which is the signature in disguise)
