I have a question that how to find all homomorphisms from
dihedral group $D_{2n}=\langle s,r : s^2=r^n=1 , srs=r^{-1} \rangle$ to the multiplicative group $\mathbb C^\times$?
alternating group $A_4$ to the multiplicative group $\mathbb C^\times$?
I came up with a single approach to deal with the problems.
Note that $D_{2n}$ which is generated by $r, s$ so in order to specify a homomorphism from $D_{2n}$ to $\mathbb{C}^*$, we only need to say what happens to each of these elements. Assume that $\phi:D_{2n}\rightarrow\mathbb{C}^*$ is a homomorphism, then $\phi(r)^n=1$, so $\phi(r)= \exp(2k\pi i/n)$, $k=0,\ldots,n-1$, and similarly $\phi(s)=\pm 1$. Thus we have $2n$ options for homomorphism $\phi$.
We know that $A_4=\langle (123), (234) \rangle$ and each of generator has order $3$. Using the similar arguments above we deduce that there are $9$ possible homomorphisms.
Do the above solutions look fine? Please light up my mind by your comments.
As you figured out, the first part is not quite so simple. Neither is the second part.
Let $\sigma$ and $\tau$ be the two generators of $A_4$ that you mentioned. In particular, note that $\sigma\tau$ has order $2,$ and so if $f:A_4\to\Bbb C^\times$ is a homomorphism, then $f(\sigma)f(\tau)=f(\sigma\tau)\in\{1,-1\}.$ Given the possibilities for $f(\sigma)$ and $f(\tau),$ we conclude that $f(\sigma\tau)\ne-1,$ and so $f(\sigma)$ and $f(\tau)$ must be multiplicative inverses. This leaves $3$ possible homomorphisms $A_4\to\Bbb C^\times$. Checking the two nontrivial possibilities is straightforward.