Lemma 1   Let $K$ be an extension field of $F$ contained in $L$ . Then $L$ is Galois over $K$ , and $\Gal(L/K)$ is a subgroup of $G$ .

Proof. Note that $L/F$ is normal and separable because it is a Galois extension; it remains to prove that $L/K$ is also normal and separable. Since $L$ is normal and finite over $F$ , it is the splitting field of a polynomial $f\in F[X]$ over $F$ . Now $L$ is also the splitting field of $f$ over $K$ (because $F\subset K\subset L$ ), so $L/K$ is normal.

To see that $L/K$ is also separable, suppose $\alpha\in L$ , and let $f^\alpha_F\in F[X]$ be its minimal polynomial over $F$ . Then the minimal polynomial $f^\alpha_K$ of $\alpha$ over $K$ divides $f^\alpha_F$ , which has no double roots in its splitting field by the separability of $L/F$ . Therefore $f^\alpha_K$ has no double roots in its splitting field for any $\alpha\in L$ , so $L$ is separable over $K$ .

The assertion that $\Gal(L/K)$ is a subgroup of $G$ is clear from the fact that $K\supset F$ .

Lemma 2   The function $\phi$ from the set of extension fields of $F$ contained in $L$ to the set of subgroups of $G$ defined by $$ \phi(K)=\Gal(L/K) $$ is an inclusion-reversing bijection. The inverse is given by $$ \phi^{-1}(H)=L^H, $$ where $L^H$ is the fixed field of $H$ in $L$ .

Proof. The definition of $\phi$ makes sense because of Lemma 1. The identities $$ \phi^{-1}\circ\phi(K)=K\quad\hbox{and}\quad \phi\circ\phi^{-1}(H)=H $$ for all subgroups $H\subset G$ and all fields $K$ with $F\subset K\subset L$ follow from the properties of the Galois group. The fixed field of $\Gal(L/K)$ is precisely $K$ ; on the other hand, since $L^H$ is the fixed field of $H$ in $L$ , $H$ is the Galois group of $L/L^H$ .

For extensions $K$ and $K'$ of $F$ with $F\subset K\subset K'\subset L$ , we have $$ \sigma\in\Gal(L/K')\iff\sigma\in\Gal(L/K), $$ so $\phi(K)\supset\phi(K')$ . This shows that $\phi$ is inclusion-reversing.

Lemma 3   Let $H$ be a subgroup of $G$ . Then the following are equivalent:

1. $L^H$ is normal over $F$ .
2. $\sigma\left(L^H\right)=L^H$ for all $\sigma\in G$ .
3. $\sigma H\sigma^{-1}=H$ for all $\sigma\in G$ .

In particular, $L^H$ is normal over $F$ if and only if $H$ is a normal subgroup of $G$ .

Proof. $1\Rightarrow2$ : Since for all $\sigma\in G$ and $\alpha\in L^H$ , $\sigma(\alpha)$ is a zero of the minimal polynomial of $\alpha$ over $F$ , we have $\sigma(\alpha)\in L^H$ by the normality of $L^H/F$ .

$2\Rightarrow3$ : For all $\sigma\in G,\tau\in H$ the equality $$ \sigma\tau\sigma^{-1}(x)=\sigma\sigma^{-1}(x)=x $$ holds for all $x\in L^H$ (from the assumption it follows that $\sigma^{-1}(x)\in L^H$ , which is fixed by $\tau$ ). This implies that $$ \sigma\tau\sigma^{-1}\in\Gal(L/L^H)=H $$ for all $\sigma\in G,\tau\in H$ .

$3\Rightarrow1$ : Let $\alpha\in L^H$ , and let $f$ be the minimal polynomial of $\alpha$ over $F$ . Since $L/F$ is normal, $f$ splits into linear factors in $L[X]$ . Suppose $\alpha'\in L$ is another zero of $f$ , and let $\sigma\in G$ be such that $\sigma(\alpha')=\alpha$ (such a $\sigma$ always exists). By assumption, for all $\tau\in H$ we have $\tau':=\sigma\tau\sigma^{-1}\in H$ , so that $$ \tau(\alpha')=\sigma^{-1}\tau'\sigma(\alpha') =\sigma^{-1}\tau'(\alpha)=\sigma^{-1}(\alpha)=\alpha'. $$ This shows that $\alpha'$ lies in $L^H$ as well, so $f$ splits in $L^H[X]$ . We conclude that $L^H$ is normal over $F$ .

Lemma 4   Let $H$ be a normal subgroup of $G$ . Then $L^H$ is a Galois extension of $F$ , and the homomorphism \begin{eqnarray*} r\colon G&\to&\Gal(L^H/F) \\ \sigma&\mapsto&\sigma\vert_{L^H} \end{eqnarray*}induces a natural identification $$ \Gal(L^H/F)\cong G/H. $$

Proof. By Lemma 3, $L^H$ is normal over $F$ , and because a subextension of a separable extension is separable, $L^H/F$ is a Galois extension.

The map $r$ is well-defined by the implication $1\Rightarrow2$ from Lemma 3. It is surjective since every automorphism of $L^H$ that fixes $F$ can be extended to an automorphism of $L$ (if $L\ne L^H$ , for example, we can choose an $\alpha\in L\setminus L^H$ such that $L=L^H(\alpha)$ using the primitive element theorem, and we can extend $\sigma\in\Gal(L^H/F)$ to $L$ by putting $\sigma(\alpha)=\alpha$ ). The kernel of $r$ is clearly equal to $H$ , so the first isomorphism theorem gives the claimed identification.