$f: S_n \to Z$ where $S_n$ is the set of permutations. $f(ab)=f(a)f(b)$ for all $a,b$ in $S_n$, then $f$ is identical $0, 1,$ or signature function.

Question

I need to prove the statement in the title:

$f: S_n \to Z$ where $S_n$ is the set of permutations. $f(ab)=f(a)f(b)$ for all $a,b$ in $S_n$, then $f$ is identical $0, 1,$ or signature function. Prove and explain.

I don't want an answer but a hand on the question's explanation on what should I look for. I think $S_n$ is a group with operation of composition. Should I use the fact that f is a homomorphism? What does it have to do with the signature function.

Also does f being identical to 0, 1 or signature function means that f is identical to 0 function or identity function or signature function? I am kind of confused if you can help me I would be really grateful!

Answer 1

Let $f: S_n \to \mathbb C^*$ be a group homomorphism. Then $f$ is the signature function or $f$ is trivial:

Reference: Finding all group homomorphisms of $S_n\to\mathbb{C}^*$ and $A_n\to\mathbb{C}^*$

In particular, every group homomorphism $f: S_n \to \Bbb Z$ is either trivial or the signature function. The idea is to use that $S_n$ is generated by transpositions.