Here is my question:
Given that $a$ and $b$ are relatively prime positive integers, and $\alpha$ is some group homomorphism $\alpha: \Bbb{Z}_{b} \rightarrow \text{Aut}(\Bbb{Z}_{a})$, where $\text{Aut}(\Bbb{Z}_{a})$ is the automorphism group of $\Bbb{Z}_{a}$, how do I show that $\Bbb{Z}_{a} \rtimes_{\alpha} \Bbb{Z}_{b} \cong \Bbb{Z}_{a} \times \Bbb{Z}_{b}$ as groups?
Here is my thought process so far: It follows easily that $\Bbb{Z}_{a}$ (or an isomorphic copy thereof) is a normal subgroup of $\Bbb{Z}_{a} \rtimes_{\alpha} \Bbb{Z}_{b}$, basically by construction of a semidirect product. Similarly, $\Bbb{Z}_{a} \cap \Bbb{Z}_{b} = \{0\}$ as subgroups of $\Bbb{Z}_{a} \rtimes_{\alpha} \Bbb{Z}_{b}$.
If I can show that $\Bbb{Z}_{b}$ is also a normal subgroup of $\Bbb{Z}_{a} \rtimes_{\alpha} \Bbb{Z}_{b}$, then the desired isomorphism follows. One way to show this is to show that $\Bbb{Z}_{a} \rtimes_{\alpha} \Bbb{Z}_{b} \cong \Bbb{Z}_{b} \rtimes_{\beta} \Bbb{Z}_{a}$ for some group homomorphism $\beta: \Bbb{Z}_{a} \rightarrow \text{Aut}(\Bbb{Z}_{b})$, because then $\Bbb{Z}_{b}$ is normal in both of those semidirect products. But this has been very difficult, since we don't know what $\alpha$ is.
I have also tried brute-forcing this by trying to construct an isomorphism between the two groups, but that hasn't gotten me anywhere.