If $\alpha: \Bbb{Z}_{5} \rightarrow \text{Aut}(\Bbb{Z}_{26})$ is some group homomorphism, how do we show that $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5} \cong \Bbb{Z}_{26} \times \Bbb{Z}_{5}$ as groups? I'm pretty sure that they are isomorphic as groups, but I'm struggling to prove it.
By definition of semidirect product, $\Bbb{Z}_{26}$ is a normal subgroup of $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$. My idea is to show that $\Bbb{Z}_{5}$ is also a normal subgroup of this semidirect product, because then we can fairly easily construct the desired direct product and show that they're isomorphic as groups. But it's proving a lot harder to prove that than I thought it would be.
I've also tried brute-forcing an isomorphism, such as $\varphi: \Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5} \rightarrow \Bbb{Z}_{26} \times \Bbb{Z}_{5}$ defined by $(n, a) \mapsto (n, a)$, but that has been... less than fruitful.
This is actually a part of something bigger I'm working on, but I think understanding this might help me understand the more general case. I think the biggest issue is that the specifics of $\alpha$ are unknown.
Edit: Something I forgot to include. My last idea is to show that $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$ is cyclic, since $\Bbb{Z}_{26} \times \Bbb{Z}_{5}$ is cyclic. But without knowing what $\alpha$ is, it's pretty difficult to find a generating element of $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5}$.
Note that $|\text{Aut}(\Bbb{Z}_{26})|=\varphi(26)=12$ where $\varphi$ is the Euler phi function.
By the First Isomorphism Theorem and Lagrange's Theorem, $|\Bbb{Z}_5:\ker \alpha|$ divides $|\text{Aut}(\Bbb{Z}_{26})|=12$. This is only possible when $\ker\alpha=\Bbb{Z}_5$, that is, $\alpha$ is the trivial automorphism. Therefore $\Bbb{Z}_{26} \rtimes_{\alpha} \Bbb{Z}_{5} \cong \Bbb{Z}_{26} \times \Bbb{Z}_{5}$.