$1$-dimensional representations of $\operatorname{GL}_2(\mathbb{F}_q)$. I have some questions about $1$-dimensional representations of $G=\operatorname{GL}_2(\mathbb{F}_q)$. I need to show that there are $q-1$ $1$-dimensional representations of $\operatorname{GL}_2(\mathbb{F}_q)$. I am able to show that $[G,G] = \operatorname{SL}_2(\mathbb{F}_q)$. It follows that $f: G/[G,G] \cong \mathbb{F}_q^*$ given by $f(g) = \det(g)$ is an isomorphism. Therefore for each homomorphism $\xi: \mathbb{F}_q^* \to \mathbb{C}^*$, we have a homomorphism $G \to \mathbb{C}^*$ given by $G \to G/[G,G] \to \mathbb{F}_q^* \to \mathbb{C}^*$. My question is
(1) How could we know that every homomorphism $G \to \mathbb{C}^*$ is given in this way?
(2) How to show that there are $q-1$ $1$-dimensional representations of $G=\operatorname{GL}_2(\mathbb{F}_q)$?
Thank you very much.
 A: Since $\mathbb{C}^*$ is an abelian group, it follows that $[G,G] \leq \operatorname{Ker}(\theta)$ for any homomorphism $\theta: G \rightarrow \mathbb{C}^*$. Thus $\theta$ induces a well defined homomorphism $\hat{\theta}: G / [G,G] \rightarrow \mathbb{C^*}$ by $x[G,G] \mapsto \theta(x)$.
You can check that $\theta \mapsto \hat{\theta}$ defines a bijection between one-dimensional representations of $G$ and $G/[G,G]$. Injectivity is easy, and surjectivity follows from what you have done in your question.
For any finite abelian group $A$, it is possible to prove that the number of one-dimensional representations of $A$ is equal to the order of $A$. Hence for a finite group $G$, the number of one-dimensional representations of $G$ is equal to the order of $G/[G,G]$.
As Jyrki noted in the comments, you should note that $\operatorname{GL}_2(\mathbb{F}_q)$ has $q-1$ one-dimensional representations only when $q$ is odd. The problem with the $q = 2$ case is that then $[G,G]$ is not $\operatorname{SL}_2(\mathbb{F}_q)$.
