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This is an homework question:

Determine the homomorphisms $S_n \to \mathbb{C}^*$ for $n \geq 2$.

If $f: S_n \to \mathbb{C}^*$ is an homomorphism and since $\mathbb{C}^*$ is abelian, then here exist an homomorphism:

$$h: S_n/ [S_n,S_n]=A_n \to \mathbb{C}^*$$

where $f = h \circ \varphi$. Since $S_n/ [S_n,S_n]= S_n / A_n$, then there exist an homomorphism:

$$h: S_n / A_n \to \mathbb{C}^*$$

So this implies that $h(A_n)=1$ and $h(S_n-A_n)= \pm 1$. So $f$ should suffice: $f(\sigma)=h(\sigma A_n)=h(A_n)1$ for $\sigma \in A_n$ and $f(\sigma)=h(\sigma A_n)=h(S_n - A_n) = 1$ for $\sigma \in (S_n -A_n)$.

My questions are: a) Is this correct ? b) Are there other ways to see this ?

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b) Yes. You can use the normal subgroups of $S_n$ because $\ker f$ is one of them. – user26857 May 26 '13 at 14:27

Looks quite good. Assuming that you allow the homomorphism to map elements of $S_n\setminus A_n$ either all to $+1$ or all to $-1$. This is implied in some parts of your answer but not all of them. In other words, for the sake of completeness you could explain, why both cases occur.

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