Find all group homomorphisms $A_n \rightarrow \mathbb{C}^*$ for all integers $n \geq 2$
What I have up until now:
Define $f: A_n \rightarrow \mathbb{C}*$
Then by the first isomorphism theorem, we have that :
$A_n/ \ker(f) \cong f[A_n] \subseteq \mathbb{C}^*$
Thus, as $\mathbb{C}^* $ is abelian, so is $A_n/ \ker(f)$
Hence $[A_n,A_n] \subseteq \ker(f)$
Then by using the fundamental theorem of homomorphism we can easily find all $f$ if we find all homomorphism $g: A_n \rightarrow \mathbb{C}^*$
First, we begin with $n \geq 5 $
Then we have $[A_n,A_n] =A_n$. Which means that $A_n/[A_n,A_n] = A_n/A_n \cong (\mathbb{Z}/1\mathbb{Z})$
So the $g: A_n/[A_n, A_n] \rightarrow \mathbb{C}^*$ can only be the trivial homomorphism
This means that for $n \geq 5 $ all homomorphisms $f: A_n \rightarrow \mathbb{C}^*$ are the trivial homomorphism.
Now for $n=2$ we have $[A_2 , A_2]={(1)}$. So $A_2/[A_2,A_2] = A_2 = \{ (1) \}$
So once again $g: A_n/[A_n, A_n] \rightarrow \mathbb{C}^*$ can only be the trivial homomorphism.
So, all homomorphisms $f: A_2 \rightarrow \mathbb{C}^*$ are the trivial homomorphism.
For $n=3$ we have $[A_3 , A_3]={(1)}$. So $A_3/[A_3,A_3] = A_3 = \{ (1),(1 \ 2 \ 3),(1 \ 3\ \ 2 ) \}$
And then I am not quite sure how to proceed and I also do not know how I could easily do this for $A_4$
Is what I have up until now correct? And how should I proceed further?
Best Answer
What you have so far is correct.
Notice that $A_3 \cong C_3$, so a homomorphism $A_3 \to \mathbb C^*$ is the same as a solution to $x^3 = 1$ in $\mathbb C^*$ (by choosing such an $x$, you choose the image of a generator, which determines the whole morphism).
For $A_4$, you can show $[A_4, A_4] = V_4 = \{(1), (12)(34), (13)(24), (14)(23) \}$. Since $\# A_4 = \frac{4!}{2} = 12$, we have $\# A_4/[A_4, A_4] = 12/4 = 3$, thus $A_4/[A_4, A_4] \cong C_3$. This means we are in the same situation as before, with $A_3$.