How many non-trivial homomorphism are possible from $S_n$ to $\Bbb C^*$?
So I'm trying to think it with the basic properties of homomorphism.
My approach was like:
Sending all the even permutations to $1$ and odd permutations to $-1$. Here we get first non-trivial homomorphism.
Sending all the even permutations to $1$ and odd permutations to $-i$. This satisfies the basic condition to be a homomorphism. But order of $-i$ is $4$ and we may get an odd permutation of order $k$ such that $4$ does not divides $k$. So homomorphism is not defined.
Now I can't think more with examples.
One thing I have got in my hand that if $f$ be a homomorphism then $\ker(f)$ is normal in Sn. Now apparently $S_n$ has normal subgroups $\{e\}$, $A_n$ and $S_n$. For $A_n$ as $\ker(f)$ I have already got at least one homomorphism. For $S_n$ as $\ker(f)$ it's trivial. For $\{e\}$ there should be at least one but it can't possible since $S_n$ is finite where $\Bbb C^*$ is infinite.
Now I need some hits to proceed. In this regard, I was also wondering whether the statement is true or not:
If there exist a finite group $G$ and another same order finite group $H$, then for all the normal subgroups of $G$ there must exist at least one homomorphism from $G$ to $H$.
Any kind of help is appreciated. Thanks in advance!
$S_n$ is generated by its transpositions. Since $C^*$ is abelian, all conjugate permutations must map to the same element, so all transpositions must map to the same element. This has to square to $1$, so it has to be $1$ or $-1$. So there are only the two homomorphisms that you already found. (Note that this approach is more efficient in this case than the one with normal subgroups, where you would have had to treat the case of $S_4$ with its irregular $4$-element normal subgroup separately.)
In the statement you were wondering about, I assume that you mean that for each normal subgroup $N$ of $G$ there must be a homomorphism from $G$ to $H$ that has $N$ as its kernel. That can’t be true because in particular there would have to be a homomorphism from $G$ to $H$ with trivial kernel; since $G$ and $H$ have the same finite order, this would be an isomorphism; but not all groups of the same finite order are isomorphic.