Example. $\mathbb{F}_{p^{n}}$ [2305D]

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: $\mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / d\mathbb{Z}$.