Explanation behind $\text{Gal}(K/K^H)=H$. I am studying Galois correspondence theorem i.e. the fundamental theorem of Galois theory.The first part of the theorem says that given a finite Galois extension $K/F$ there exists an inclusion reversing bijection between the subgroups of $\text{Gal}(K/F)$ and the intermediate fields of $K/F$.The maps in the two directions are given by $H\mapsto K^H$ and $L\mapsto \text{Gal}(K/L)$.Now I am doing the proof of the fact that these two maps are inverses of each other.For that I need to show that $\text{Gal}(K/K^H)=H$ for a subgroup $H$ of $\text{Gal}(K/F)$.But I cannot find a proper reasoning for this.Can someone help me with this because this is the only part of the proof where I am stuck.
 A: The fundamental theorem of Galois theory claims that the maps $H\mapsto\mathbb{K}^H$ (also denoted $\operatorname{Fix}_\mathbb{K}(H)$) and $\mathbb{L}\mapsto\operatorname{Gal}(\mathbb{K}/\mathbb{L})$ form an antitone Galois connection.

*

*Let $\mathbb{L}_1<\mathbb{L}_2$, then we have $\operatorname{Gal}(\mathbb{K}/\mathbb{L}_2)<\operatorname{Gal}(\mathbb{K}/\mathbb{L}_1)$ because:
$$\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_2)
\Leftrightarrow
\forall x\in\mathbb{L}_2\colon
\varphi(x)=x
\Rightarrow
\forall x\in\mathbb{L}_1\colon
\varphi(x)=x
\Leftrightarrow
\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_1).$$

*Let $H_1<H_2$, then we have $\mathbb{K}^{H_2}<\mathbb{K}^{H_1}$ because:
$$x\in\mathbb{K}^{H_2}
\Leftrightarrow
\forall\varphi\in H_2\colon
\varphi(x)=x
\Rightarrow
\forall\varphi\in H_1\colon
\varphi(x)=x
\Leftrightarrow
x\in\mathbb{K}^{H_1}.$$

*We have $\mathbb{L}<\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}$ since:
$$x\in\mathbb{L}
\Rightarrow
\forall\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L})\colon
\varphi(x)=x
\Rightarrow
x\in\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}.$$

*We have $H<\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)$ since:
$$\varphi\in H
\Rightarrow
\forall x\in\mathbb{K}^H\colon\varphi(x)=x
\Rightarrow
\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H).$$
That is all that is necessary to establish a Galois connection, but when we take a Galois extension like $\mathbb{K}/\mathbb{F}$, where all intermediate extensions $\mathbb{K}/\mathbb{L}$ for $\mathbb{F}<\mathbb{L}<\mathbb{K}$ are also galois, we even get equalities for the last two results.


*Because of Artin's theorem and since $\mathbb{K}/\mathbb{L}$ is a Galois extension, we have:
$$[\mathbb{K}:\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}]
=|\operatorname{Gal}(\mathbb{K}/\mathbb{L})|
=[\mathbb{K}:\mathbb{L}],$$
which implies $\mathbb{K}^{\operatorname{Gal}(\mathbb{K}/\mathbb{L})}=\mathbb{L}$ together with the upper inclusion.

*Since $\mathbb{K}/\mathbb{K}^H$ is a Galois extension and because of Artin's theorem, we have:
$$|\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)|
=[\mathbb{K}:\mathbb{K}^H]
=|H|,$$
which implies $\operatorname{Gal}(\mathbb{K}/\mathbb{K}^H)=H$ together with the upper inclusion.

You can also find this calcuation in a little different way here.
