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?
Best Answer
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
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.