First, we show that the maps are inclusion-reversing. Let $K \subset F_1 \subset F_2 \subset L$. Let $G_i = \operatorname{Gal} \left({L / F_i}\right)$. Let $\sigma \in G_2$. Then $\sigma$ is an automorphism of $L$ which fixes $F_2$. Since $F_1 \subset F_2$, it follows that $\sigma$ fixes $F_1$ and consequently $\sigma \in G_1$. Let $H_1 \subset H_2 \subset \operatorname{Gal} \left({L / K}\right)$. Let $F_i = L_{H_i}$. Let $x \in F_2$. Then $\sigma \left({x}\right) = x$ for all $\sigma\in H_2$. Since $H_1\subset H_2$, the same equality holds for each element of $H_1$ and thus $x \in F_1$. For the remainder of the proof: :let $G$ denote $\operatorname{Gal} \left({L / K}\right)$ :for any field $K \subset F \subset L$ let $G_F$ denote $\operatorname{Gal} \left({L / F}\right)$. Next, we demonstrate that the two functions described are inverses. That is: :For any intermediate field $K \subset F \subset L$: ::$F = L_{G_F}$ :For any subgroup $H \subset G$: ::$H = G_{L_H}$ {{proofread|This is why $F \subseteq L_{G_F}$.  A second option would be appreciated.}} For the first equality, $F$ is fixed by every element of ${G_F}$. By definition $L_{G_F}$ includes all elements of $L$ which are fixed by ${G_F}$. Thus we have $F \subseteq L_{G_F}$. Let $\alpha \in L_{G_F} \setminus F$. Then: :$\left[{F \left({\alpha}\right) : F}\right] > 1$ {{explain|notation $[H : K]$ -- index of subgroup?}} We can express the minimal polynomial of $\alpha$ in terms of $G_F$ as: :$\displaystyle m_\alpha \left({x}\right) = \prod_{\sigma \mathop \in G_F} \left({x - \sigma \left({\alpha}\right)}\right)^\frac 1 {\left[{L : F \left({\alpha}\right)}\right]}$ However, by our assumption, $\sigma \left({\alpha}\right) = \alpha$ for each $\sigma$. Thus: {{begin-eqn}} {{eqn | l = m_\alpha \left({x}\right)       | r = \left({x - \alpha}\right)^\frac {\left[{L : F}\right]} {\left[{L : F \left({\alpha}\right)}\right]}       | c =  }} {{eqn | r = \left({x - \alpha}\right)^{\left[{F \left({\alpha}\right) : F}\right]}       | c =  }} {{end-eqn}} Since $\left[{F \left({\alpha}\right) : F}\right] > 1$, this contradicts the separability of $L / F$. Therefore, the first equality holds. For the second equality, it is immediate that $H \subset G_{L_H}$. Suppose $H$ were a proper subset of $G_{L_H}$. By the Primitive Element Theorem, there exists an $\alpha \in L$ such that $L = L_H \left({\alpha}\right)$. Consider the polynomial: :$\displaystyle f = \prod_{\sigma \mathop \in H} \left({x - \sigma \left({\alpha}\right)}\right)$ The coefficients of $f$ are evidently elements of $L_H$ and $f$ is monic by construction. However: :$\left[{L : L_H}\right] = \operatorname{deg} \left({f}\right) = \left\vert{H}\right\vert < \left\vert{G_{L_H}}\right\vert = \left[{L : L_H}\right]$ by definition of Galois extension. This is a contradiction and it follows that $H = G_{L_H}$. Finally, we demonstrate the correspondence between normal subgroups of $G$ and the intermediate normal extensions of $K$. Suppose $K \subset F \subset L$ is an intermediate field and $F / K$ is a normal extension. We let $H = \operatorname{Gal} \left({F / K}\right)$ denote the Galois group of interest. Let $\sigma \in G$ and $\tau \in H$. We want to show that $\sigma^{-1} \tau \sigma \in H$ to conclude that $H$ is normal. We have that $F \subset L$. Thus $\sigma$ restricts to an embedding of $F$ in $\overline K$. However, since $F / K$ is a normal extension, the image of every embedding of $F$ is again $F$. Thus, $\sigma$ restricts to an automorphism of $F$. Let $x \in F$. Then $\sigma \left({x}\right) \in F$. We have that $\tau$ fixes $F$. Thus: :$\tau \left({\sigma \left({x}\right)}\right) = \sigma \left({x}\right)$ Therefore: :$\sigma^{-1} \left({\tau\left({\sigma \left({x}\right)}\right)}\right) = x$ and we conclude that: :$\sigma^{-1} \tau \sigma \in H$ Next, suppose $H$ is a normal subgroup of $G$ and $F = L_H$. Let $\tau \in H$ and $\sigma:F \mapsto \overline K$ be an embedding of $F$. By Extension of Isomorphisms, we extend $\sigma$ to $\overline \sigma$, an automorphism of $L$. Consider the composition $\hat \sigma^{-1} \tau \hat \sigma = \hat \tau \in H$ by our assumption of normality. Then: :$\hat \sigma^{-1} \tau \hat \sigma \left({x}\right) = \hat \tau \left({x}\right) = x$ which implies that: :$\tau \left({\hat \sigma \left({x}\right)}\right) = \hat \sigma \left({x}\right) \in F$ Since $x \in F$: :$\hat \sigma \left({x}\right) = \sigma \left({x}\right) \in F$ which was required to be shown. {{qed}} {{Namedfor|Évariste Galois|cat = Galois}} Category:Galois Theory 4gkocipz90bt8qlv8byfieiwqfkcgme