If we restrict the character $\chi:G\to S_{o(\chi)}$, then is this map surjective?

Question

If $\chi$ is a character of a finite abelian group $G$, then each element $\chi(g)$ is a $o(\chi)-$root of unity in $\mathbb{C}$. In fact

$\chi(g)^{o(\chi)}=(\chi^{o(\chi)})(g)=1$

The question is the following:

If we restrict the character $\chi:G\to S_{o(\chi)}$, then is this map surjective?

I think no because otherwise $G/\ker(\chi)$ would be a cyclic group in general. Is there a counter-example?

Answer 1

First note that $G/\ker(\chi)$ has to be cyclic, since the image of $\chi: G\to \mathbb{C}^\times$ is a finite subgroup of $\mathbb{C}^\times$, and for any field $K$, all finite subgroups of $K^\times$ are cyclic. You can also see that subgroups of $S_n$ are cyclic, so in any case the image of $\chi$ is definitely cyclic.

But actually it is fairly easy to see that the image of $\chi$ is exactly $S_n$ (I am not a fan of this notation, by the way, but I will keep consistent with the question) where $n$ is the order of $\chi$ in the group of characters $\hat{G}$.

Indeed, suppose $\chi:G\to S_n$ is not surjective. Then its image is a subgroup of $S_n$, so it is $S_m$ for some $m$ which is a proper divisor of $n$. But then for all $g$, we have $\chi(g)^m=1$, so actually $\chi^m=1$, which is a contradiction.