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I'm trying to solve the following exercise:

Let $n\geq 2$ be an integer. Find all the homomorphism between $S_n$ and $\mathbb{C}^*$.

I started with the simplest case: with $n=2$ i think that the function which assign $e\rightarrow 1$ and $a\rightarrow -1$ would be fine, and i think this is the only one nontrivial for $n=2$ since $-1$ is the only element of order $2$ in $\mathbb{C}^*$, am i wrong? But then i don't know how to proceed and find a general method for all $n$.

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    $\mathbb{C}^{\ast}$ is an abelian group. That tells you something about the kernel of such a homomorphism. – 2017-01-09
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    i know that the kernel of an homomorphism is a normal subgroup, and the only normal subgroup of $S_n$ should be 1, $S_n$ and $A_n$, but i don't know how to relate it with the fact that C is abelian – 2017-01-09

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Hint (see Daniel's comment): Any homomorphism $G \to A$ into an abelian group $A$ must contain the commutator subgroup $[G,G]$ in its kernel. We have $[S_n,S_n]=A_n$. Furthermore then all homomorphisms $G \to A$ into an abelian group $A$ are lifts of homomorphisms $G / [G, G] \to A$.

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    So the kernel must be $A_n$ or $S_n$, in this second case the homorphism is trivial. In the first case how can i find the function? – 2017-01-09
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    What element of $\Bbb C^{\ast}$ has order 2? – 2017-01-09
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Let $\phi$ be such an homomorphism and consider $M = \max_{g \in S_n}(\mid \phi(g)\mid)$. Knowing $\exists g$ s.t. $\mid\phi(g)\mid = M$ what can you say about $M$, considering $\mid\phi(g^k)\mid$ where $k$ is the order of $g$? Apply the same reasoning to $\min_{g \in S_n}(\mid \phi(g)\mid)$.