# 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!

• For $n\ge 5$, the only normal subgroup of $S_n$ is $A_n$, so $k=2$. Then one can figure out $n<5$ by force. Commented Dec 13, 2019 at 6:40
• @ElliotG: technically, one would have $2$ divides $k$, since the sign morphism onto $\mathbb{Z}/2\mathbb{Z}$ can be composed with the injection of $\mathbb{Z}/2\mathbb{Z}$ into any cyclic group of even order. Commented Dec 13, 2019 at 6:42
• Since $\mathbb C^*$ contains a copy of every cyclic group, see also math.stackexchange.com/a/3626782/589
– lhf
Commented Jun 19, 2020 at 0:11

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)

• Hi ! This is a refreshing approach. Your conclusion holds for any choice of $n \geq 1$, correct ? Commented Dec 13, 2019 at 7:31
• Yup! Say $n\geq 2$ to have at least one transposition. Commented Dec 13, 2019 at 7:34
• Note there is nothing original in this approach. This is the same approach used when trying to determine morphisms $S_n\to \mathbb{C}^\times$ (yielding the signature). In general, it is natural to try to guess the image of a set of generators. The fact you wan to exploit also conjugacy classes here is because the target is abelian, so two conjugate elements will have the same image. Commented Dec 13, 2019 at 7:49