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  "body": "\n## Statement\n\nLet $L/K$ be a finite [[Galois extension|0DS2F7GK05]] with [[Galois group|0XVR72M51M]] $G = \\operatorname{Gal}(L/K)$.\n\n1. The maps $H \\mapsto L^H$ and $E \\mapsto \\operatorname{Gal}(L/E)$ are inverse bijections between the subgroups $H$ of $G$ and the [[intermediate fields|02NBKVQFRQ]] $E$ of $L/K$.\n2. These bijections reverse inclusions: $H_1 \\subseteq H_2$ if and only if $L^{H_1} \\supseteq L^{H_2}$.\n3. For every subgroup $H$, $[L : L^H] = |H|$ and $[L^H : K] = |G|/|H|$.\n4. The extension $L^H/K$ is normal if and only if $H$ is a normal subgroup of $G$. Then $\\operatorname{Gal}(L^H/K) \\cong G/H$.\n\nHere $L^H$ is the [[fixed field|0KMJ6919WS]] of $H$.\n\n## Remarks\n\nFor $L = \\mathbb{Q}(\\sqrt{2}, \\sqrt{3})$ and $K = \\mathbb{Q}$, the group $G$ has order $4$ and three subgroups of order $2$. They match the three intermediate fields $\\mathbb{Q}(\\sqrt{2})$, $\\mathbb{Q}(\\sqrt{3})$ and $\\mathbb{Q}(\\sqrt{6})$.\nThe top field $L$ matches the trivial subgroup, and the base field $K$ matches $G$.\nA finite Galois extension has only finitely many intermediate fields, because a finite group has finitely many subgroups.\nThe theorem is used to prove that the general equation of degree $5$ is not solvable by radicals.\n\n## History\n\nÉvariste Galois proved the theorem in his development of Galois theory.\n\n## Explanations\n\n### L1 human\n\nThe theorem turns questions about fields into questions about groups.\nEach intermediate field matches one subgroup of the Galois group, and the bigger the field, the smaller the subgroup.\nFinite groups are often easier to study than fields.\n",
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