TITLE: For every integer $n \geq 2$ , there is a unique non-trivial homomorphism $f: {S_n} \to \mathbb{C^*}$.
QUESTION [5 upvotes]: For every integer $n \geq 2$ , there is a unique non-trivial homomorphism  $f: {S_n} \to \mathbb{C^*}$. 
How can I prove this statement? I know there will be a homomorphism as $S_n \over A_n$ is isomorphic to ${(1 ,-1).}$ But how can I prove that there is no other homomorphism than that?
Can anyone please help me?

REPLY [4 votes]: I think there's a proof that uses fewer advanced facts than the ones that have been posted. 
$S_n$ is generated by a 2-cycle $a=(1\ 2)$ and an $n$-cycle $b=(1\ 2\ \cdots\ n)$. Also, $ab=(1\ 3\ 4\ \cdots\ n)$, an $n-1$-cycle. 
Now let $f$ be a homomorphism. Then $1=f(1)=f(a^2)=(f(a))^2$, so $f(a)=1$ or $f(a)=-1$. Similarly, $(f(b))^n=1$, and $(f(a))^{n-1}(f(b))^{n-1}=1$. It follows that $f(b)=(f(a))^{n-1}$. So there are two choices for $f(a)$, and once you've made that choice, that determines the value of $f(b)$ – and once you have $f(a)$ and $f(b)$, you have $f$. Hence, there are only the two homomorphisms.