In (2) of Example 1505, we have seen that $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)$ is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$, so $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)/ \mathbb{Q}$ is Galois. By (5) of Example 2106, we know that $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}$ with $$\begin{aligned} \sigma_1 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, &\qquad \sigma_2 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ \sqrt{-3} &\longmapsto \sqrt{-3}, &\qquad \sqrt{-3} &\longmapsto -\sqrt{-3}. \end{aligned}$$ The nontrivial proper subgroups of $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right)$ are $\left< \sigma_{1} \right>, \left< \sigma_{2} \right>, \left< \sigma_{1}\sigma_{2} \right>, \left< \sigma_{1}^{2}\sigma_{2} \right>$, and $\left< \sigma_{1} \right>$ is normal of order $3$, $\left< \sigma_{2} \right>, \left< \sigma_{1} \sigma_{2} \right>, \left< \sigma_{1}^{2} \sigma_{2} \right>$ are not normal and of order $2$. $$\begin{aligned} \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1 \rangle} = \mathbb{Q}(\sqrt{-3}), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_2 \rangle} = \mathbb{Q}(\sqrt[3]{2}), \\ \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2 \rangle} = \mathbb{Q}\left(\sqrt[3]{2}\frac{-1-\sqrt{-3}}{2}\right), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2^2 \rangle} = \left(\sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}\right), \end{aligned}$$ and these are all the proper nontrivial sub-extensions of $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$.

(D&F p.576) Since every subgroup of the Klein-4 group is normal, all the subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$ are Galois extensions of $\mathbb{Q}$.

(D&F Example p.577-581)

Consider $\mathbb{F}_{p^{n}}$, the unique subfield of $\overline{\mathbb{F}}_{p}$ of cardinality $p^{n}$.

By Example 22016, $$ \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) = \left< \operatorname{Fr} \right> \cong \mathbb{Z} / n\mathbb{Z}. $$ The subgroups of $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$ are $\left< \operatorname{Fr}^{d} \right> \cong d\mathbb{Z} / n\mathbb{Z}, d \mid n, d \geq 1$. (They are all normal subgroups since $\mathbb{Z} / n\mathbb{Z}$ is abelian.) $$ \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } = \{ \alpha \in \mathbb{F}_{p^{n}} \mid \alpha^{p^{d}} = \alpha \} = \mathbb{F}_{p^{d}}. $$ By the Fundamental Theorem, we have $$ \mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) = \mathrm{Gal}\left( \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } / \mathbb{F}_{p} \right) \cong \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) / \left< \operatorname{Fr}^{d} \right> = \left< \operatorname{Fr} \right> / \left< \operatorname{Fr}^{d} \right> \cong \mathbb{Z} / d\mathbb{Z}, $$ compatible with what we have proved in Example 22016: $\mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / d\mathbb{Z}$.