Homomorphisms from nonabelian simple groups into $\mathbb{C}^\times$

Question

Let be $G$ finite nonabelian simple group. My question is, why there is only one homomorphism $G\to\mathbb{C}^\times$. (In that case $g\mapsto1$ ofc.) I can only show that nontrivial homomorphism cannot be injective, but I don't know how to use the simplicity.

Answer 1

You can avoid abelianizations. Let $\theta: G \to \mathbb{C}^\times$ be such a homomorphism. The kernel must be a normal subgroup, so by simplicity it is either trivial ($\theta$ is 1-1) or $G$ ($\theta$ is the constant homomorphism). In the first case the non-abelian group $G$ embeds in the abelian group $\mathbb{C}^\times$, an impossibility. Hence the second case is the only one.

Answer 2

Since $\mathbb C^\times$ is abelian, any homomorphism $G \rightarrow \mathbb C^\times$ factors through abelianization $G \rightarrow G/G'$. Since $G'$ is a normal subgroup of $G$ and $G$ is simple, there are two choices: $G'=G$ or $G'=1$. Since $G$ is nonabelian, $G' \neq 1$. Thus, $G'=G$ and a homomorphism $G \rightarrow \mathbb C^\times$ factors as $G \rightarrow G/G' \cong 1 \rightarrow \mathbb C^\times$ and can only be the identity morphism.

In a bit simpler way, observe that the image of a simple group $G$ is either $1$ or $G$ itself. In our case, it cannot be $G$ since subgroups of abelian groups are abelian, and $G$ is not. Therefore, the image is $1$ and the homomorphism is trivial.

Note that the same argument applies for any abelian group placed instead of $\mathbb C^\times$.