I'm asked to find the Galois group of irreducible polynomials of the form $X^4-k$ and $X^6-k$ for $k\in\mathbb{Q}_{>0}$.
I think I got the solution for $f=X^4-k$ but I am not entirely sure if it is correct. For starters, we have $\Omega^f=\mathbb{Q}(\sqrt[4]{k},i)$ and we have $[\Omega^f:\mathbb{Q}]=8$ because of the Tower Law and the fact that $\mathbb{Q}(\sqrt[4]{k})$ is a real field. Therefore, Gal($f$) must be a transitive subgroup of order 8 of $S_4$. The only subgroup of $S_4$ that satisfies such conditions is $D_4$, hence Gal($f$)=$D_4$. Would this be correct? I know in this case Gal($f$) can also be found identifying automorphisms directly, but I think this way is harder.
Now, for $g=X^6-k$, $\Omega^g=\mathbb{Q}(\sqrt[6]{k},\xi_6)$ and $\xi_6=e^{i\pi/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i\implies\Omega^g=\mathbb{Q}(\sqrt[6]{k},\sqrt{3}i)\implies[\Omega^g:\mathbb{Q}]=12$ for the same reasons. Then Gal($g$) is a transitive subgroup of order $12$ of $S_6$, but I think these are much harder to characterize. How should I proceed? I have just learned about semidirect products, so it might have something to do with that.
Any hints?
This is a special case of the classical question of the Galois Group of irreducible polynomials $X^n-k.$ See e.g. here for the general answer (see in particular the reply by Dietrich Burde): The Galois Group is $\mathbb Z/n \rtimes (\mathbb Z/n)^\times$ if and only if $k$ is odd, or $n$ is even and $\sqrt k \notin ℚ(_)$. For $n=4, n=6,$ the only exceptional cases which would need special treatment are $n=6$ and $k=-3a^2$, $a \in \mathbb Q $. In this case, $(\sqrt[6]{k})^3=$ $\sqrt k = i\sqrt 3a \in \mathbb Q(\zeta_6)$, hence $\mathbb Q(\sqrt[6]{k}) \cap \mathbb Q(\zeta_6) \ne \mathbb Q$ and the degree of the splitting field would not be equal to 12 anymore. For $n=4$ one always obtains $\mathbb Z/4 \rtimes (\mathbb Z/4)^\times$ which is isomorphic to $D_4$, as stated by the OP.