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
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There exist a non trivial homomorphism from $S_n$ to $C^*$
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There exist a unique non trivial homomorphism from $S_n$ to $C^*$
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For $n \geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$
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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)?
Best Answer
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 $n \le 4$.