Number of homomorphisms from a finite group to an infinite group

Question

I am starting to self-study abstract algebra, and came across, in my humble opinion, a fairly difficult question. More specifically, the question I was trying to answer is along the lines of:

"Let $G$ a finite group, and $Hom(G,\Bbb C^\times)$ the set of all group homomorphisms $\phi:G\rightarrow\Bbb C^\times$, where $\Bbb C^\times$ is the multiplicative group of the non-zero complex numbers. Prove that $Hom(G,\Bbb C^\times)$ is finite."

I found many answers for the case where both groups were finite, but didn't have much success in this case.

I was trying to think about $G$ having a finite number of generators might lead to some brute-force technique, or maybe induction on the order of $G$, but no success up to now. I suspect there is something very basic I'm forgetting, so please go easy on me if that's the case.

I would be glad if you could give some insights about finding all group homomorphisms between groups too. What kind of facts are normally used to prove such a thing?

Thanks in advance.

Answer 1

For any $g \in G$, $g^{|G|} = 1$, so for any homomorphism $\phi \in \text{Hom}(G, \mathbb{C}^ \times)$, $\phi(g)^{|G|} = 1$. This means that $\phi$ is a map from the finite set $G$ to the finite set of $|G|$-th roots of unity in $\mathbb{C}$. There are only finitely many maps between two finite sets, hence $|\text{Hom}(G, \mathbb{C}^ \times)|$ is finite.

Answer 2

The circle group has only a single subgroup of any given finite order. If $f: G \to H$ is a surjective homomorphism, the order of $H$ divides the order of $G$. Therefore there are only a finite number of subsets of $\Bbb{C}^{\times}$ that a homomorphism can map $G$ to, and since there are only a finite number of functions from $G$ to any one of those sets the total number of homomorphisms from $G$ to $\Bbb{C}^{\times}$ is finite.