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  "body": "\n## Proof\n\nThe main tool is [[Artin's theorem on fixed fields|X7R15C00ET]]. We prove the four parts in order.\n\n1. Let $L/K$ be a finite [[Galois extension|0DS2F7GK05]] and $G = \\operatorname{Gal}(L/K)$. By [[the order of the Galois group|0BDJBH8B4Z]], $G$ is finite with $|G| = [L : K]$.\n2. Let $H$ be a [[subgroup|0N0PXCTYA7]] of $G$. Its [[fixed field|0KMJ6919WS]] $L^H$ is a [[subfield|07QY5RPFBR]] of $L$. Each element of $H$ fixes $K$, so $K \\subseteq L^H$. Thus $L^H$ is an [[intermediate field|02NBKVQFRQ]].\n3. By Artin's theorem applied to the finite group $H$, we get $[L : L^H] = |H|$ and $\\operatorname{Gal}(L/L^H) = H$.\n4. Let $E$ be an intermediate field. By [[the characterizations of Galois extensions|P75CXH2Q2B]], $L$ is a [[splitting field|02NZZ1J0FH]] over $K$ of a [[separable polynomial|03J8BS7FMV]] $f \\in K[x]$.\n5. Write $f = c(x - a_1) \\cdots (x - a_n)$ in $L[x]$, with $L = K(a_1, \\dots, a_n)$. Then $L \\supseteq E(a_1, \\dots, a_n) \\supseteq K(a_1, \\dots, a_n) = L$, so $L = E(a_1, \\dots, a_n)$.\n6. Also $f \\in E[x]$ is separable over $E$, because the identity $u f + v f' = 1$ with $u, v \\in K[x]$ also holds in $E[x]$. So $L$ is a splitting field of a separable polynomial over $E$.\n7. The extension $L/E$ is finite, since $[L : K] = [L : E][E : K]$ by the [[tower law|0A7KJ68XHT]]. By the characterizations again, $L/E$ is Galois.\n8. Every automorphism of $L$ fixing $E$ fixes $K$. So $\\operatorname{Gal}(L/E)$ is a subset of $G$. It contains the identity and is closed under composition and inverses, so it is a subgroup of $G$.\n9. Let $H = \\operatorname{Gal}(L/E)$ and $E' = L^H$. Every element of $H$ fixes $E$, so $E \\subseteq E'$.\n10. Since $L/E$ is Galois, $|H| = [L : E]$ by step 1 applied to $L/E$. By step 3, $[L : E'] = |H|$.\n11. By the tower law, $[L : E] = [L : E'][E' : E]$. Step 10 gives $[E' : E] = 1$. So $E'$ is a $1$-dimensional $E$-space that contains $E$, and $E' = E$.\n12. Steps 3 and 11 show that $\\operatorname{Gal}(L/L^H) = H$ and $L^{\\operatorname{Gal}(L/E)} = E$. So the two maps are inverse bijections. This proves part 1.\n13. For part 2, let $H_1 \\subseteq H_2$ be subgroups. An element fixed by all of $H_2$ is fixed by all of $H_1$. So $L^{H_1} \\supseteq L^{H_2}$.\n14. Conversely, suppose $L^{H_1} \\supseteq L^{H_2}$. An automorphism fixing $L^{H_1}$ fixes $L^{H_2}$. So $\\operatorname{Gal}(L/L^{H_1}) \\subseteq \\operatorname{Gal}(L/L^{H_2})$, that is, $H_1 \\subseteq H_2$ by step 3. This proves part 2.\n15. For part 3, step 3 gives $[L : L^H] = |H|$. The tower law gives $[L : K] = [L : L^H][L^H : K]$. With step 1, $[L^H : K] = |G| / |H|$.\n16. For part 4, fix a subgroup $H$ and let $E = L^H$. Suppose first that $E/K$ is [[normal|0CBRFQ7DVZ]]. Let $\\sigma \\in G$ and $a \\in E$.\n17. Let $q$ be the [[minimal polynomial|0GK1WB9R98]] of $a$ over $K$. It is [[irreducible|0NFEG48RDR]] and has the root $a \\in E$. By normality, $q$ [[splits|0SHMC00RJV]] over $E$ as $q = (x - e_1) \\cdots (x - e_m)$ with $e_i \\in E$.\n18. Since $\\sigma$ fixes the coefficients of $q$, we get $q(\\sigma(a)) = \\sigma(q(a)) = 0$. So $\\prod_i (\\sigma(a) - e_i) = 0$ in the field $L$, and $\\sigma(a) = e_i$ for some $i$. Hence $\\sigma(E) \\subseteq E$ for every $\\sigma \\in G$.\n19. Let $\\tau \\in H$, $\\sigma \\in G$ and $a \\in E$. By step 18 for $\\sigma^{-1}$, $\\sigma^{-1}(a) \\in E$, so $\\tau(\\sigma^{-1}(a)) = \\sigma^{-1}(a)$. Thus $\\sigma \\tau \\sigma^{-1}(a) = a$.\n20. So $\\sigma \\tau \\sigma^{-1}$ fixes $E$ and lies in $\\operatorname{Gal}(L/E) = H$, by step 3. Hence $H$ is a [[normal subgroup|0W77CVBCHQ]] of $G$.\n21. Now suppose that $H$ is normal in $G$. Let $\\sigma \\in G$, $a \\in E$ and $\\tau \\in H$. Then $\\sigma^{-1} \\tau \\sigma \\in H$, so $\\tau(\\sigma(a)) = \\sigma(\\sigma^{-1} \\tau \\sigma(a)) = \\sigma(a)$.\n22. So $\\sigma(a) \\in L^H = E$, that is, $\\sigma(E) \\subseteq E$ for every $\\sigma \\in G$. Applying this to $\\sigma^{-1}$ gives $E \\subseteq \\sigma(E)$. So $\\sigma(E) = E$.\n23. Hence the restriction $\\sigma|_E$ is an automorphism of $E$ fixing $K$. The map $r \\colon G \\to \\operatorname{Aut}(E)$, $\\sigma \\mapsto \\sigma|_E$, is a [[group homomorphism|0HV9Y7WG5R]], since restriction respects composition.\n24. Its kernel is the set of $\\sigma \\in G$ that fix $E$ pointwise. This is $\\operatorname{Gal}(L/E) = H$, by step 3.\n25. The image $r(G)$ is a finite group of automorphisms of $E$. An element $a \\in E$ fixed by all of $r(G)$ lies in $L^G$. By step 12 with $E = K$, $L^G = L^{\\operatorname{Gal}(L/K)} = K$. So $E^{r(G)} = K$.\n26. By Artin's theorem applied to $E$ and $r(G)$, the extension $E/K$ is Galois, so it is normal. This proves that $E/K$ is normal if and only if $H$ is normal in $G$.\n27. Artin's theorem also gives $\\operatorname{Gal}(E/K) = r(G)$. By the [[first isomorphism theorem|0CXXJNG5B6]] and step 24, the [[quotient group|0YYWCNTPWR]] $G/H$ is isomorphic to $r(G) = \\operatorname{Gal}(L^H/K)$.\n28. In both directions of part 4, $H$ is normal, so steps 21 to 27 apply. This proves part 4 and the theorem.\n\n## Idea\n\nArtin's theorem says that a finite group $H$ of automorphisms is exactly the group of symmetries of $L$ over its fixed field, and that $[L : L^H] = |H|$. This gives one round trip. For the other, $L$ stays Galois over every intermediate field, and a degree count shows that the fixed field of $\\operatorname{Gal}(L/E)$ cannot be larger than $E$. Normality matches because $\\sigma$ moves $L^H$ to $L^{\\sigma H \\sigma^{-1}}$, so $L^H$ is stable under $G$ exactly when $H$ is.\n",
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