Characters on symmetric groups $S_n$ and alternating groups $A_n$. For an integer $2 \leq n$ , 
let $S_n$ be the permutation on $n$ symbols and $A_n$ be the alternating group.  Also let's denote by $\mathbb{C}^*$, the multiplicative group of non-zero complex number.
Which of the following is correct? 


for every integer $2 \leq n$, 
there is a non-trivial homomorphism 
$φ:S_n \rightarrow \mathbb{C}^*$. 
for every integer $2 \leq n$, 
there is a unique non-trivial homomorphism 
$φ:S_n \rightarrow \mathbb{C}^*$. 
for every integer $3 \leq n$, 
there is a non-trivial homomorphism 
$φ:A_n \rightarrow \mathbb{C}^*$. 
for every integer $5 \leq n$, 
there is no non-trivial homomorphism 
$φ:A_n \rightarrow \mathbb{C}^*$. 


my attempt: for $5 \leq n$ $A_n$ is simple, so option $4$ is correct.
Note that every finite subgroup of $C^\times$ is cyclic.

Option 1 is  correct: Take the sign homomorphism (i.e., $\phi(g)=1$ for $g\in A_n$ and $\phi(g)=-1$ otherwise.

Option 2 is also correct as the sign homomorphism is the only non-trivial homomorphism: For $n\ge 5$, this follows because the kernel must contain the simple subgroup $A_n$. For $n\le 4$, by inspection.

Option 3 is correct because it is weaker than option 1.

Option 4 is wrong because it contradicts option 1.
Note that every finite subgroup of $C^\times$ is cyclic.

Option 1 is  correct: Take the sign homomorphism (i.e., $\phi(g)=1$ for $g\in A_n$ and $\phi(g)=-1$ otherwise.

Option 2 is also correct as the sign homomorphism is the only non-trivial homomorphism: For $n\ge 5$, this follows because the kernel must contain the simple subgroup $A_n$. For $n\le 4$, by inspection.

Option 3 is correct because it is weaker than option 1.

Option 4 is wrong because it contradicts option 1.