TITLE: Finding all group homomorphisms of $S_n\to\mathbb{C}^*$ and $A_n\to\mathbb{C}^*$
QUESTION [7 upvotes]: I need to find all group homomorphisms of $S_n\to\mathbb{C}^*$ (for $n\ge2$), and also all group homomorphisms of $A_n\to\mathbb{C}^*$ (for $n\ge2$).
Someone on this site, two years ago, seems to have already asked the same question (Finding homomorphisms of $S_n$ → C* and $A_n$ → C*). I understand what the questioner says there, but I don't understand it when he says:

Therefore there is exactly one $g:\mathbb{Z}/2\mathbb{Z}\to\mathbb{C}^*$, so all elements of $a\in\mathbb{C}^*$ give us $\operatorname{ord}(a)\mid g[\mathbb{Z}/2\mathbb{Z}]$, so $a$ can only be $-1$ or $1$. Now we know that $f:S_n\to\{-1,1\}$ …

What is he saying there? And how to go on from there? (I do not understand the answer given either …)
I hope that someone here can help me further. (And if I'm not allowed to ask this question because is has already been asked two years ago, please tell me what to do instead).
Thank you in advance!

REPLY [13 votes]: Here is another take for the first part.
Let $f: S_n \to \mathbb C^*$ be a group homomorphism.
Let $\tau$ be a transposition. Then $\tau^2=1$ implies $f(\tau)=\pm 1$.
Since every permutation $\sigma$ is a product of transpositions, we have $f(\sigma)=\pm 1$. Therefore, the image of $f$ has size at most $2$ and so the index of $\ker f$ in $S_n$ is at most $2$. Thus, the only possibilities for $\ker f$ are $S_n$ and $A_n$. Therefore, $f$ is trivial or the sign homomorphism.