# 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.

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.

1. 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: $$\varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_2) \Leftrightarrow \forall x\in\mathbb{L}_2\colon \varphi(x)=x \Rightarrow \forall x\in\mathbb{L}_1\colon \varphi(x)=x \Leftrightarrow \varphi\in\operatorname{Gal}(\mathbb{K}/\mathbb{L}_1).$$
2. Let $$H_1, then we have $$\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}.$$
3. We have $$\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})}.$$
4. We have $$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.

1. 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.
2. 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.