Using the first isomorphism theorem to find group homomorphisms During class we learned about the fiest isomorphism theorem, we were also told that it can be applied to find (the number of) group homomorphisms between two groups. 
Now as an example I'm trying to find group homomorphisms between $A_n$ and $C*$
So I think that I have to find a smart finction $g$ which takes $A_n/Ker(f)$ to C* 
However how do I go about it?
 A: $\newcommand{\C}{\mathbb{C}}$I assume $\C^{\star}$ is the multiplicative group of non-zero complex numbers.
Note that $\C^{\star}$ is an abelian group. So $A_{n}/\ker(f)$ must be abelian, For which values of $n$ is it the case that $A_{n}$ has proper normal subgroups with abelian quotient?
A: The kernel is a normal subgroup, so homomorphisms are determined by normal subgroups. First assume $n\neq 3, 4$. In this case you may know $A_n$ is simple, so any map is either trivial or injective. Suppose you have an injective map. Then $A_n$ is isomorphic to a subgroup of $\mathbb{C}^*$. You should check that isomorphism preserves commutativity. From here you can find a contradiction...
If $n=4$, $A_n$ has one non-trivial normal subgroup namely $V$, the Klein four group. By first isomorphism theorem you can show $A_4/V \simeq \mathbb{Z}/3\mathbb{Z}$. So essentially you're determining maps from $\mathbb{Z}/3\mathbb{Z}$ to $\mathbb{C}^*$. A hint for this: if $\varphi$ is your isomorphism, $g \in G$, then $\varphi(g^3) = \varphi(g)^3$. 
For $n=3$, $A_n$ is abelian of order $3$, so the problem is similar to above. 
