Find infinitely many homomorphism from $GL_2(\mathbb Q)$ to $\mathbb Q^*$ Find infinitely many homomorphism from $GL_2(\mathbb Q)$ to $\mathbb Q^*$. Here $\mathbb Q^*$ means the multiplicative group of nonzero rational numbers.
My attempt: An example is the determinant function, which is a homomorphism satisfying the statement given. But I don't see how can I make any progress by knowing this fact and generate an infinite number of homomorphisms. I thought it'll be a good idea to find all the normal subgroups of $GL_2(\Bbb Q)$, but I can't think of any other than the $SL_2(\Bbb Q)$.
Help will be really appreciated. Thanks.
 A: Take the determinant homomorphism followed by the group homomorphisms
$$
f\colon \Bbb Q^*\rightarrow \Bbb Q^*, t\mapsto t^n
$$
for every $n\ge 1$.
A: You can compose the determinant map with any homomorphism from $\Bbb Q^\times$ to $\Bbb Q^\times$ to get infinitely many of them.

As a side note, it is possible to show that every homomorphism $\operatorname{GL}_2(\Bbb Q) \rightarrow \Bbb Q^\times$ factors through the determinant map.
This is because the commutator subgroup of $\operatorname{GL}_2(\Bbb Q)$ is equal to $\operatorname{SL}_2(\Bbb Q)$ (exercise!). Since $\Bbb Q^\times$ is commutative, any homomorphism from $\operatorname{GL}_2(\Bbb Q)$ to $\Bbb Q^\times$ must factor through the quotient $\operatorname{GL}_2(\Bbb Q)/\operatorname{SL}_2(\Bbb Q)$. The result then follows from the following short exact sequence:
$$1 \rightarrow \operatorname{SL}_2(\Bbb Q)\rightarrow\operatorname{GL}_2(\Bbb Q)\rightarrow \Bbb Q^\times \rightarrow 1$$ where the map $\operatorname{GL}_2(\Bbb Q)\rightarrow \Bbb Q^\times$ is the determinant map.
This result is also true with $2$ replaced with any $n \geq 2$ and with $\Bbb Q$ replaced with any field of characteristic $0$.
