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

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.

The question (and comment) you refer to is a sketch of how to solve this problem. The key fact is the following: $C^* = \{ z \in \mathbb{C} \mid z \neq 0 \}$ is an abelian group under the operation of multiplication of complex numbers.

The relevant theorem is the following: if $G$ is a group and $A$ is an abelian group, then any homomorphism $f:G \to A$ must factor through the abelianization of $G$. This means that there is a homomorphism $f_1:G/[G,G] \to A$ such that $f = f_1 \circ \pi$, where $\pi: G \to G/[G,G]$ is the natural map to the abelianization, which we take to be defined as quotient of $G$ by its commutator subgroup $[G,G]$.

In your case, it is well-known that if $G = S_n$ and $n > 2$, then $G/[G,G] \cong C_2 = \{-1,1\}$, the cyclic group of order two. Any homomorphism $f: S_n \to C^*$ will be determined by a homomorphism $f_1:C_2 \to C^*$.

So, this reduces your question to the following: What are all homomorphisms $f_1:C_2 \to C^*$?