Find the minimal Galois extension $L$ of $\mathbb Q$ containing $\mathbb Q(\sqrt[4]{5})$. Describe the structure of $Gal(L/\mathbb Q)$.
I think $L$ is a splitting field of $X^4-5$ over $\mathbb Q$. I think the structure is a cyclic group generated by a single element.
A Galois extension of $\mathbb{Q}$ containing $\mathbb{Q}(\sqrt[4]{5})$ must be normal, so the minimal polynomial $f(X) = X^4-5 \in \mathbb{Q}[X]$ (irreducible by Eisenstein with 5) must be decomposable into linear factors.
Roots of $f$ in $\mathbb{C}$ are
$$\alpha_1 = \sqrt[4]{5}, \alpha_2 = -\sqrt[4]{5}, \alpha_3 = i\sqrt[4]{5}, \alpha_4 = -i\sqrt[4]{5}$$
So $L = \mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4) = \mathbb{Q}(i, \sqrt[4]{5})$.
As $\mathbb{Q}(\sqrt[4]{5}) \subset \mathbb{R}$ and $i \notin \mathbb{R}$, the degree of the extension is $$[L:\mathbb{Q}] = [\mathbb{Q}(i):\mathbb{Q}] \cdot [\mathbb{Q}(\sqrt[4]{5}):\mathbb{Q}] = 2 \cdot 4 = 8$$
$L/\mathbb{Q}$ is a Galois extension because it is normal as splitting field and separable. So we have $\# Gal(L/\mathbb{Q}) = 8$.
A $\sigma \in Gal(L/\mathbb{Q})$ is uniquely defined by the images of $i$ and $\sqrt[4]{5}$. As we can see from the degree of the extension, $X^2 +1$ is irreducible over $L/\mathbb{Q}(\sqrt[4]{5})$, so there is $\tau \in Gal(L/\mathbb{Q}(\sqrt[4]{5}))$ with $\tau(i) =\ ^+_-i$. Furthermore, $X^4 - 5$ is irreducible over $L/\mathbb{Q}(i)$, so there is $\phi \in Gal(L/\mathbb{Q}(i))$ with $\phi(\sqrt[4]5) =\ ^+_-\sqrt[4]5$ resp. $\phi(\sqrt[4]5) =\ ^+_-i\sqrt[4]5$.
So combining $\phi$ and $\tau$ gives us $\sigma = \tau \circ \phi \in Gal(L/\mathbb{Q})$. As we can see, there are 8 different combinations of $\tau \circ \phi$. As the Galois group is of order 8, these are already all $\mathbb{Q}$-automorphisms.
To conclude, as $f$ is separable and without multiple irreducible factors, $Gal(L/\mathbb{Q})$ operates on $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ and induces a monomorphism
$$Gal(L/\mathbb{Q}) \hookrightarrow S_4$$ $$\phi \mapsto \sigma \in S_4, \phi(\alpha_j) = \alpha_{\sigma(j)}$$
This subgroup in $S_4$ is easy to find, as we know what all the automorphisms in the Galois group look like.