Problem about group homomorphism of symmetric and alternating group? For $n \geq 2$,Let $S_n$ and $A_n$ be symmetric and Alternating group on $n$ letters. Let $C^*$ be multiplicative group of complex numbers. Then


*

*There exist a non trivial homomorphism from $S_n$ to $C^*$

*There exist a unique non trivial homomorphism from $S_n$ to $C^*$

*For $n \geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$

*For $n \geq 5$ there in no non trivial homomorphism from $\phi: A_n \ \to C^*$
As- 
(1) option is correct as map even cycle to $1$ and odd cycle to $-1$, we get homomorphism.
(4) Correct as if there exist such homomorphism, then if Ker$\phi = 0$, then $A_n$ becomes embedded in $C^*$ which is not possible as it is non abelian. If $Ker \phi$ is non trivial, then this situation is also not possible as $A_n$ is simple for $n \geq 5$. So (3) is incorrect.
What about option (2)?
 A: You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' \le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n \cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n \ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = \{1, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n \ge 5$, you should be looking at $n \le 4$.
A: This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
