Galois group of $x^n-p\in\Bbb Q[x]$ 
Compute the Galois group of $x^n-p$ over $\Bbb Q$. Here, $n\in\Bbb N$ and $p$ is a prime number.

Let $\eta$ be the primitive $n$th root of unity. Then $\Bbb Q(\sqrt[n]{p},\eta)$ is a splitting field of $x^n-p$ over $\Bbb Q$. Let $H$ be a fixed field of $\Bbb Q(\eta)$ under the Galois correspondence. Since $\Bbb Q(\eta)/\Bbb Q$ is Galois, $H$ is normal in $G=\operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q)$. Hence from the s.e.s. which splits
$$0\to H = \operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q(\eta))\to G = \operatorname{Gal}(\Bbb Q(\sqrt[n]{p},\eta)/\Bbb Q)\xrightarrow{\pi} G/H = \operatorname{Gal}(\Bbb Q(\eta)/\Bbb Q)\to 0$$
since we can define an inverse $p:G/H\to G$ by extension $\sigma\in \operatorname{Gal} (\Bbb Q(\eta)/\Bbb Q)$ by $\sigma(\sqrt[n]{p}) =\sqrt[n]{p}$, we can conclude that $G\simeq H\rtimes (G/H)$.
It's a naive generalization of this answer which may require more conditions. Does the proof make sense? What extra conditions are necessary to make this proof is valid?
 A: In the end it depends what you allow yourself to know about semi-direct products, in particular what equivalent conditions there are to achieve the structure of a semi-direct product. If you say a split exact sequence defines a semi-direct product then your proof is perfectly valid and correct.
There are, however, different, more explicit approaches possible too. For example, one can show that for a group $G$ with subgroups $N,H$ such that

*

*$N\triangleleft G$

*$NH=G$

*$N\cap H=\{1\}$
we have $G=N\rtimes H$ with $H$ acting on $N$ by conjugation (the general $H\to\operatorname{Aut}(N)$ being yet another equivalent definition for semi-direct products).
For $n=p$ this gives a particular simple proof. The conditions are this case readily checked using the subgroups generated by $\sigma$ and $\tau$, with
$$
\sigma\colon\begin{cases}\sqrt[p]{p}&\mapsto\sqrt[p]{p}\zeta_p\\\zeta_p&\mapsto\zeta_p\end{cases}\quad\text{and}\quad\tau\colon\begin{cases}\sqrt[p]{p}&\mapsto\sqrt[p]{p}\\\zeta_p&\mapsto\zeta_p^c\end{cases}
$$
and $c$ a generator of $(\mathbb Z/p\mathbb Z)^\ast$. It might be possible to generalize this strategy for general $n$ but you might have to make a more elaborate argument.
