The fundamental theorem of Galois theory

Theorem

The fundamental theorem of Galois theory

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Statement

Let $L/K$ be a finite Galois extension (i.e. $L/K$ is normal and separable), and write $G = \operatorname{Gal}(L/K)$. For a subfield $K \subseteq M \subseteq L$ write $\Phi(M) = \operatorname{Gal}(L/M) = \{\sigma \in G : \sigma(x) = x \ \forall x \in M\}$, and for a subgroup $H \leq G$ write $\Psi(H) = L^{H} = \{x \in L : \sigma(x) = x \ \forall \sigma \in H\}$, the fixed field of $H$. Then:

(1) $\Phi$ and $\Psi$ are mutually inverse, inclusion-reversing bijections between the set of intermediate fields $\{M : K \subseteq M \subseteq L\}$ and the set of subgroups $\{H : H \leq G\}$. (2) For each intermediate $M$, $[L:M] = |\Phi(M)|$ and $[M:K] = [G : \Phi(M)]$. (3) $M/K$ is a normal (equivalently Galois) extension if and only if $\Phi(M)$ is a normal subgroup of $G$, and in that case $\operatorname{Gal}(M/K) \cong G/\Phi(M)$ via $\sigma \mapsto \sigma|_M$.

Why it matters

This theorem converts a continuous-looking problem — classifying all fields sitting between $K$ and $L$ — into a finite, purely group-theoretic problem: classifying subgroups of $G$. It is the mechanism by which "solvability by radicals" of a polynomial gets translated into "solvability" of the Galois group, and it is the template every later Galois-type correspondence (étale fundamental groups, covering space theory, differential Galois theory) is modelled on.

Beyond its use in the insolvability of the quintic, the correspondence is the standard tool for computing splitting fields, deciding constructibility questions (doubling the cube, trisecting angles, constructing regular $n$-gons), and organising ramification theory in algebraic number theory.

Hypotheses

$L/K$ is algebraic and finite. Without finiteness the group $G$ need not determine the field lattice via fixed points alone: for infinite algebraic Galois extensions (e.g. $\overline{\mathbb{Q}}/\mathbb{Q}$) the naive statement fails — one must topologise $G$ (Krull topology) and restrict the correspondence to closed subgroups. Dropping finiteness silently, as many first attempts do, produces a bijection claim that is simply false: there are subgroups of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ (using the axiom of choice) with no corresponding intermediate field.

$L/K$ is normal. Take $K = \mathbb{Q}$, $L = \mathbb{Q}(\sqrt[3]{2})$ (real cube root), so $[L:K]=3$ but $L$ does not contain the other roots of $x^3-2$. Here $\operatorname{Aut}(L/K) = \{1\}$ is trivial, so counting automorphisms gives no information distinguishing $K$ from $L$: the "correspondence" would have one group but three intermediate fields $K, L$, and (inside a splitting field) genuinely different subfields, so no bijection can exist with $G$ this small.

$L/K$ is separable. In characteristic $p$, take $K = \mathbb{F}_p(t)$ and $L = \mathbb{F}_p(t^{1/p})$, so $L = K(\alpha)$ with $\alpha^p = t$. This extension is normal (it is the splitting field of $(x^p - t) = (x-\alpha)^p$) but purely inseparable, so $\operatorname{Aut}(L/K) = \{1\}$ even though $[L:K] = p \gt 1$. Again the group is too small to see the field $L$ between $K$ and itself.

Only closed intermediate objects are used (automatic here since everything is finite). This is not a separate hypothesis in the finite case — every subgroup of a finite group is closed in the discrete topology, and every subfield of a finite extension is automatically an "allowed" object. It is flagged because the infinite generalisation (Krull's theorem) has to add this closedness condition explicitly, and forgetting it is the commonest way to misquote the theorem in the infinite setting.

Proof

\text{Fix a splitting field description: } L \text{ is the splitting field over } K \text{ of a separable polynomial } f \in K[x].

Equivalent characterisation of finite Galois extensions: normal + separable $\iff$ splitting field of a separable polynomial. This is the standing hypothesis, used repeatedly to transport normality/separability to subextensions. A

\text{For every intermediate field } M,\ L/M \text{ is again Galois, with } \operatorname{Gal}(L/M) \leq G \text{ a subgroup.}

$L$ is still the splitting field of $f$ over $M$ (the roots of $f$ that generate $L$ over $K$ also generate it over the larger base $M$), and $f$ is separable over $M$ since separability is inherited by considering $f$ over any base containing $K$. Any $K$-automorphism of $L$ fixing $M$ pointwise is in particular a $K$-automorphism, giving $\Phi(M) \leq G$. A

\textbf{Artin's Lemma. } \text{If } H \leq \operatorname{Aut}(L) \text{ is a finite group of field automorphisms and } L^H \text{ its fixed field, then } [L:L^H] = |H| \text{ and } \operatorname{Gal}(L/L^H) = H.

Cited lemma (E. Artin), proved via Dedekind's theorem on linear independence of characters: distinct automorphisms $\sigma_1,\dots,\sigma_n \in H$, viewed as characters $L^\times \to L^\times$, are linearly independent over $L$; consequently no non-trivial $L^H$-linear relation $\sum a_i \sigma_i = 0$ can hold, which forces $[L:L^H] \geq |H|$. Combined with the elementary bound $[L:L^H] \leq |H|$ (obtained by showing any $|H|+1$ elements of $L$ are $L^H$-linearly dependent, using that each element satisfies a degree-$\le|H|$ polynomial $\prod_{\sigma \in H}(x - \sigma(\alpha))$ with coefficients in $L^H$), this gives equality $[L:L^H]=|H|$; since $H \subseteq \operatorname{Gal}(L/L^H)$ trivially and $|\operatorname{Gal}(L/L^H)| \leq [L:L^H] = |H|$ always, we get $\operatorname{Gal}(L/L^H) = H$ exactly. C

\Psi(\Phi(M)) = L^{\operatorname{Gal}(L/M)} = M \quad \text{for every intermediate } M.

By Step 2, $L/M$ is Galois, and by definition a finite extension $L/M$ is Galois precisely when $M = L^{\operatorname{Gal}(L/M)}$ (this is the defining property of "Galois", equivalent to $|\operatorname{Gal}(L/M)| = [L:M]$, which itself follows from Artin's Lemma applied with $H=\operatorname{Gal}(L/M)$ once one checks $M \subseteq L^{\operatorname{Gal}(L/M)}$ trivially and the degree count in Step 3 forces equality). So $\Psi \circ \Phi = \mathrm{id}$. B

\Phi(\Psi(H)) = \operatorname{Gal}(L/L^H) = H \quad \text{for every subgroup } H \leq G.

Direct application of Artin's Lemma (Step 3) to the finite group $H \leq \operatorname{Aut}(L)$ — note $H \leq G = \operatorname{Gal}(L/K)$ already consists of field automorphisms of $L$ fixing $K$, so the Lemma applies verbatim and gives $\operatorname{Gal}(L/L^H) = H$ exactly, not merely a group containing $H$. So $\Phi \circ \Psi = \mathrm{id}$. Together with Step 4, $\Phi$ and $\Psi$ are mutually inverse bijections. B

M_1 \subseteq M_2 \iff \operatorname{Gal}(L/M_2) \subseteq \operatorname{Gal}(L/M_1).

($\Rightarrow$) An automorphism fixing the larger field $M_2$ pointwise certainly fixes the smaller field $M_1$ pointwise, since $M_1 \subseteq M_2$. ($\Leftarrow$) Apply $\Psi$ to both sides and use Steps 4–5 ($\Phi,\Psi$ mutually inverse and order-preserving in this direction) to recover $M_1 = \Psi\Phi(M_1) \subseteq \Psi\Phi(M_2) = M_2$. Hence the bijection is inclusion-reversing. A

[L:M] = |\operatorname{Gal}(L/M)|, \qquad [M:K] = [G : \operatorname{Gal}(L/M)].

The first equality is Artin's Lemma (Step 3) applied with $H = \Phi(M)$, using $M = L^H$ from Step 4. The second follows from the multiplicativity of degrees in the tower $K \subseteq M \subseteq L$, $[L:K] = [L:M][M:K]$, together with $[L:K]=|G|$ (Step 3 with $H=G$, $L^G=K$ by Step 4 applied to $M=K$) and Lagrange's theorem $|G| = |\operatorname{Gal}(L/M)| \cdot [G:\operatorname{Gal}(L/M)]$: dividing gives $[M:K] = [G:\operatorname{Gal}(L/M)]$. A

\text{For } \sigma \in G,\ \operatorname{Gal}(L/\sigma(M)) = \sigma \, \operatorname{Gal}(L/M) \, \sigma^{-1}.

Direct computation: $\tau$ fixes $\sigma(M)$ pointwise $\iff$ $\sigma^{-1}\tau\sigma$ fixes $M$ pointwise (substitute $x=\sigma(m)$ and unwind), i.e. $\tau \in \operatorname{Gal}(L/\sigma(M)) \iff \sigma^{-1}\tau\sigma \in \operatorname{Gal}(L/M) \iff \tau \in \sigma\operatorname{Gal}(L/M)\sigma^{-1}$. This uses only that $\sigma$ is a bijection of $L$ fixing $K$, so $\sigma(M)$ is again an intermediate field between $K$ and $L$. B

M/K \text{ normal} \iff \sigma(M) = M \ \forall \sigma \in G \iff H := \operatorname{Gal}(L/M) \trianglelefteq G.

First equivalence: since $L/K$ is normal, $L$ contains every $K$-conjugate of every element of $M$; every $K$-embedding of $M$ into $L$ extends (by the isomorphism-extension theorem for algebraic extensions) to a $K$-automorphism of $L$, i.e. to some $\sigma \in G$, and $M/K$ is normal precisely when every such embedding has image inside $M$ itself, i.e. $\sigma(M) \subseteq M$, hence $\sigma(M)=M$ by degree/finiteness, for all $\sigma \in G$. Second equivalence: by Step 8, $\sigma(M)=M$ for all $\sigma$ is equivalent (applying the injective, inclusion-reversing correspondence $\Phi$ from Steps 4–6) to $\sigma H \sigma^{-1} = H$ for all $\sigma \in G$, which is the definition of $H \trianglelefteq G$. C

\text{If } H \trianglelefteq G, \text{ the restriction map } \rho: G \to \operatorname{Gal}(M/K),\ \sigma \mapsto \sigma|_M \text{ induces } G/H \xrightarrow{\ \cong\ } \operatorname{Gal}(M/K).

By Step 9, $\sigma(M)=M$ for every $\sigma \in G$, so $\sigma|_M$ is indeed a $K$-automorphism of $M$ and $\rho$ is a well-defined group homomorphism. Its kernel is exactly $\{\sigma \in G : \sigma|_M = \mathrm{id}_M\} = \operatorname{Gal}(L/M) = H$. Surjectivity: as in Step 9, every element of $\operatorname{Gal}(M/K)$ extends to some $\sigma \in G$ (isomorphism-extension theorem), so $\rho$ is onto. By the first isomorphism theorem, $G/H \cong \operatorname{Gal}(M/K)$; this also independently confirms $|G/H| = |\operatorname{Gal}(M/K)| = [M:K]$, consistent with Step 7. B

Result

\{\, K \subseteq M \subseteq L \,\} \;\; \longleftrightarrow \;\; \{\, H \leq G \,\}, \qquad M \mapsto \operatorname{Gal}(L/M), \quad L^H \mathrel{\reflectbox{$\mapsto$}} H

Reading. Every field strictly between $K$ and $L$ corresponds to exactly one subgroup of $G = \operatorname{Gal}(L/K)$, with bigger fields matched to smaller groups. The size of a subgroup tells you the degree of $L$ over the matching field, and the index tells you the degree of the matching field over $K$. Normal subgroups pick out exactly the intermediate fields that are themselves Galois over $K$, and for those the quotient group is literally the Galois group of the sub-extension.

Scope. Applies precisely to finite Galois extensions $L/K$ (normal + separable); this covers all finite extensions in characteristic $0$ or over finite fields that arise as splitting fields of separable polynomials, and in particular every splitting field of a separable polynomial over $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{F}_q$. Does not directly apply to non-normal or inseparable extensions (see "Fails without"), and the infinite version requires the Krull topology restriction to closed subgroups.

Corollaries & converses

There are only finitely many intermediate fields between $K$ and $L$ when $L/K$ is finite Galois, since $G$ is finite and has finitely many subgroups.
$M/K$ is Galois for every intermediate $M$ exactly when $G$ is abelian (every subgroup is then normal) — this is why "abelian extension" and "extension with every intermediate field itself Galois" coincide inside a fixed Galois closure.
$G$ is simple and non-abelian $\iff$ the only intermediate fields normal over $K$ are $K$ and $L$ themselves — the group-theoretic phenomenon behind the insolvability of the general quintic traces directly to $A_5$ being simple.
The correspondence is a genuine order-reversing lattice isomorphism: it sends the compositum of two intermediate fields to the intersection of the corresponding subgroups, and the intersection of two intermediate fields to the subgroup generated by the two corresponding subgroups.
Converse direction: the theorem as stated is already an "if and only if" for normality (Step 9) and is a bijection (not merely an implication), so there is no separate converse to check — every subgroup does arise as $\operatorname{Gal}(L/M)$ for a genuine intermediate field $M$, by Artin's Lemma (Step 5), and this uses finiteness essentially.

Fails without

Dropping normality: $K=\mathbb{Q} \subset L=\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt[3]{2},\omega)$ (the last being the actual splitting field of $x^3-2$). Working instead with the non-normal $L=\mathbb{Q}(\sqrt[3]{2})$ over $K=\mathbb{Q}$: $\operatorname{Aut}(L/K)$ is trivial (the only real embedding of $L$ fixing $\mathbb{Q}$ is the identity, since the other cube roots of $2$ are non-real), so the "Galois group" has one subgroup but $L/K$ has intermediate structure invisible to it — the correspondence collapses.
Dropping separability: $K=\mathbb{F}_p(t)$, $L = K(\alpha)$ with $\alpha^p=t$. Here $[L:K]=p$ but $\operatorname{Aut}(L/K)=\{1\}$ (any $K$-automorphism must send $\alpha$ to a root of $x^p-t=(x-\alpha)^p$, forcing it to $\alpha$), so again one trivial group tries to correspond to the two distinct fields $K \subsetneq L$; no bijection with the field lattice is possible.
Dropping finiteness (infinite algebraic extensions): for $L=\overline{\mathbb{Q}}$, $K=\mathbb{Q}$, the abstract group $G=\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ has (by Zorn's-lemma arguments) subgroups of index $2^{\aleph_0}$ that are not closed in the Krull topology and to which no intermediate field corresponds at all; the naive bijection statement is false and must be replaced by a bijection with only the *closed* subgroups.

Common errors

Matching field degree to subgroup index and subgroup order the wrong way round — writing $[M:K]=|\operatorname{Gal}(L/M)|$ instead of $[L:M]=|\operatorname{Gal}(L/M)|$; the correspondence is order-reversing, so bigger fields go with smaller groups.
Concluding $M/K$ is Galois whenever $M$ corresponds to some subgroup $H$, forgetting the extra requirement that $H$ be normal in $G$ — every intermediate field corresponds to a subgroup, but only the normal ones give a Galois sub-extension.
Confusing $\operatorname{Gal}(M/K)$ with $H = \operatorname{Gal}(L/M)$ itself when $M/K$ is Galois, rather than with the quotient $G/H$.
Applying the theorem to a field $L$ that is merely a splitting field of *some* polynomial without checking separability, silently assuming characteristic $0$ reasoning transfers unchanged to positive characteristic.
Trying to run the finite-case bijection verbatim for an infinite Galois extension, ignoring the need to restrict to closed subgroups of the profinite Galois group.

Discussion

The theorem, in essentially its modern form, crystallised from Galois's original (1832) insight that the "group of permutations" of the roots of a polynomial that respect all algebraic relations among them governs solvability by radicals; Dedekind and later Artin reformulated the theory in terms of automorphisms of fields rather than permutations of roots, which is the formulation given here and is what makes the fixed-field construction $L^H$ available as the inverse map.

Structurally, the correspondence is best understood as an instance of a much more general phenomenon: a group $G$ acting on a set (here, on $L$ by field automorphisms, and via that, on the lattice of intermediate structures) produces an order-reversing Galois connection between "sub-objects fixed pointwise" and "subgroups". The same skeleton reappears as the correspondence between subgroups of $\pi_1(X)$ and covering spaces of $X$, and between closed subgroup schemes and quotients in algebraic geometry — the name "Galois theory" has been exported to all of these by analogy.

Artin's Lemma (Step 3) is the load-bearing technical result: it is what upgrades the a priori inequality $|\operatorname{Aut}(L/K)| \leq [L:K]$ (true for any finite extension) to an equality precisely when $L/K$ is Galois, and it does so via linear independence of characters — a fact about the multiplicative group $L^\times$ that has nothing to do with field addition, which is why the same trick reproves normal basis existence and underlies Hilbert's Theorem 90.

Common misconception: students sometimes read the theorem as saying subgroups of $G$ are "the same as" intermediate fields, as if the correspondence were an identification rather than a bijection between two different kinds of objects; the correct statement keeps $M$ and $H=\operatorname{Gal}(L/M)$ as different objects in different categories (a field versus a group) that determine each other uniquely once $L/K$ is fixed — nothing is literally equal except degrees, orders, and indices.

Worked examples

\text{Let } L = \mathbb{Q}(\sqrt2,\sqrt3), \ K=\mathbb{Q}. \text{ Find every intermediate field.}

$L$ is the splitting field over $\mathbb{Q}$ of $(x^2-2)(x^2-3)$, which is separable in characteristic $0$; so $L/K$ is Galois. A

[L:K] = 4, \qquad G = \operatorname{Gal}(L/K) = \{1,\sigma,\tau,\sigma\tau\} \cong \mathbb{Z}/2 \times \mathbb{Z}/2,

where $\sigma:\sqrt2\mapsto-\sqrt2,\ \sqrt3\mapsto\sqrt3$ and $\tau:\sqrt2\mapsto\sqrt2,\ \sqrt3\mapsto-\sqrt3$. Degree $4$ since $\sqrt3 \notin \mathbb{Q}(\sqrt2)$ (else $\sqrt6=\sqrt2\sqrt3 \in \mathbb{Q}(\sqrt2)$, contradicting a direct norm computation), so $[L:K]=[L:\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2):K]=2\cdot2=4=|G|$ as required by the theorem (Step 7), confirming $L/K$ is indeed Galois with this $G$. A

\text{Subgroups of } G: \ \{1\},\ \langle\sigma\rangle,\ \langle\tau\rangle,\ \langle\sigma\tau\rangle,\ G.

$G\cong (\mathbb{Z}/2)^2$ has exactly these five subgroups (Lagrange restricts subgroup order to $1,2,4$; the three order-$2$ subgroups are generated by the three non-identity elements). A

L^{\{1\}}=L,\quad L^{\langle\sigma\rangle}=\mathbb{Q}(\sqrt3),\quad L^{\langle\tau\rangle}=\mathbb{Q}(\sqrt2),\quad L^{\langle\sigma\tau\rangle}=\mathbb{Q}(\sqrt6),\quad L^{G}=K.

Each fixed field is computed directly (e.g. $\sigma$ fixes exactly the elements built from $\sqrt3$ alone, so fixes $\mathbb{Q}(\sqrt3)$; $\sigma\tau$ sends $\sqrt6=\sqrt2\sqrt3 \mapsto (-\sqrt2)(-\sqrt3)=\sqrt6$, fixing $\mathbb{Q}(\sqrt6)$) and each has degree $[L:L^H]=|H|$ over $L$ matching Step 3 of the theorem, e.g. $[L:\mathbb{Q}(\sqrt3)]=2=|\langle\sigma\rangle|$. B

\text{Every intermediate field of } \mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q} \text{ is } \mathbb{Q},\ \mathbb{Q}(\sqrt2),\ \mathbb{Q}(\sqrt3),\ \mathbb{Q}(\sqrt6),\ \mathbb{Q}(\sqrt2,\sqrt3) \text{ — exactly five, matching the five subgroups.}

Reading. Since $G$ is abelian every subgroup is normal, so all five intermediate fields are themselves Galois over $\mathbb{Q}$ — consistent with each being a splitting field of a separable quadratic-type polynomial.

\text{Let } L \text{ be the splitting field of } x^4-2 \text{ over } K=\mathbb{Q}.

Roots are $\pm\alpha,\pm i\alpha$ with $\alpha=\sqrt[4]{2}\in\mathbb{R}_{\gt0}$, so $L=\mathbb{Q}(\alpha,i)$; $x^4-2$ is separable (its derivative $4x^3$ shares no root with it) so $L/K$ is Galois. A

[L:K]=8,\qquad G=\operatorname{Gal}(L/K)\cong D_4 = \langle r,s \mid r^4=s^2=1,\ srs=r^{-1}\rangle,

with $r:\alpha\mapsto i\alpha,\ i\mapsto i$ (order $4$) and $s:\alpha\mapsto\alpha,\ i\mapsto-i$ (complex conjugation restricted to $L$, order $2$); $[L:K]=[L:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):K] = 2\cdot4=8=|G|$, so by Step 7 of the theorem $G$ has order exactly $8$ and, checking the relations on generators, is dihedral of order $8$. B

H=\langle r \rangle \trianglelefteq G, \quad [G:H]=2, \quad L^{H} = \mathbb{Q}(i).

Every index-$2$ subgroup of any group is automatically normal (its two cosets are the fibres of the sign map $G\to G/H\cong\mathbb{Z}/2$). Elements of $\langle r\rangle$ fix $i$ (check on the generator $r$) and $[L:\mathbb{Q}(i)]=4=|\langle r\rangle|$, so by uniqueness of the fixed field of a given order (Steps 4–5) $L^{H}=\mathbb{Q}(i)$ exactly. B

\text{By the theorem's normal-subgroup clause (Step 9–10): } \mathbb{Q}(i)/\mathbb{Q} \text{ is Galois with } \operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q}) \cong G/H \cong \mathbb{Z}/2.

$H=\langle r\rangle$ normal in $G$ (Step 3) triggers the theorem's guarantee that the corresponding field $\mathbb{Q}(i)$ is Galois over $\mathbb{Q}$, with Galois group the quotient $G/H$, of order $[G:H]=2$ — matching the direct, elementary fact that $\mathbb{Q}(i)/\mathbb{Q}$ is the splitting field of the separable polynomial $x^2+1$. C

H=\langle r\rangle \trianglelefteq D_4 \;\longleftrightarrow\; M=\mathbb{Q}(i),\qquad \operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q}) \cong D_4/\langle r \rangle \cong \mathbb{Z}/2

Reading. Instead of separately proving $\mathbb{Q}(i)/\mathbb{Q}$ is Galois of degree $2$, the correspondence reads this off automatically from a single group-theoretic fact — that $\langle r \rangle$ is normal in $D_4$.

Problems

Let $L=\mathbb{Q}(\zeta_5)$ where $\zeta_5=e^{2\pi i/5}$, $K=\mathbb{Q}$. Given $G=\operatorname{Gal}(L/K)\cong \mathbb{Z}/4$ (cyclic, generated by $\sigma:\zeta_5\mapsto\zeta_5^2$), list all intermediate fields and their degrees over $K$.
Solution
$G\cong\mathbb{Z}/4$ has exactly three subgroups: $\{1\}$, $\langle\sigma^2\rangle$ (order $2$), and $G$ itself (order $4$), since a cyclic group of order $4$ has a unique subgroup of each divisor order $1,2,4$. By the theorem this gives exactly three intermediate fields: $L^{\{1\}}=L=\mathbb{Q}(\zeta_5)$ of degree $4$ over $K$; $L^{G}=K=\mathbb{Q}$ of degree $1$; and $L^{\langle\sigma^2\rangle}$, a field of degree $[G:\langle\sigma^2\rangle]=2$ over $K$. Since $\sigma^2:\zeta_5\mapsto\zeta_5^4=\bar\zeta_5$ is complex conjugation restricted to $L$, the fixed field is the maximal real subfield $\mathbb{Q}(\zeta_5+\zeta_5^{-1}) = \mathbb{Q}\!\left(\frac{\sqrt5-1}{2}\right)$, i.e. $\mathbb{Q}(\sqrt5)$, of degree $2$. All three are automatically Galois over $\mathbb{Q}$ since $G$ is abelian, so every subgroup is normal.
Explain, using the theorem, why $D_4$ (from Worked Example 2) having a subgroup of order $2$ that is not normal must correspond to an intermediate field that is not Galois over $\mathbb{Q}$, and exhibit one.
Solution
Take $H'=\langle s \rangle=\{1,s\}$ with $s:\alpha\mapsto\alpha,\ i\mapsto -i$ from Example 2. Using the dihedral relation $srs=r^{-1}$, i.e. $sr=r^{-1}s$, compute the conjugate $rsr^{-1}$: it satisfies $r(sr^{-1})=r(rs)=r^2s$ (using $sr^{-1}=rs$, which follows from $sr=r^{-1}s$ by taking inverses of both sides and rearranging), so $rsr^{-1}=r^2s \neq s$. Hence $rH'r^{-1} = \langle r^2 s\rangle \neq H' = \langle s \rangle$, so $H'$ is not normal in $G$. By the theorem's Step 9, $M=L^{H'}$ is then not normal, hence not Galois, over $\mathbb{Q}$. Concretely $L^{H'}=\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt[4]2)$, which indeed fails to be normal over $\mathbb{Q}$: it contains the real root $\sqrt[4]2$ of $x^4-2$ but not the non-real root $i\sqrt[4]2$, so it is not a splitting field, confirming the group-theoretic prediction without any further computation.
Suppose $L/K$ is finite Galois with $|G|=p$ a prime. Use the theorem to show there are no non-trivial intermediate fields, and identify the only two.
Solution
By Lagrange, a group of prime order $p$ has only the trivial subgroup $\{1\}$ and itself as subgroups (any non-trivial proper subgroup would have order dividing $p$ and strictly between $1$ and $p$, impossible). By the bijection, the only intermediate fields are $L^{\{1\}}=L$ and $L^{G}=K$; there is nothing strictly between $K$ and $L$. This matches the direct fact that $[L:K]=p$ is prime, so no intermediate degree can divide it non-trivially.
Let $L/K$ be finite Galois with group $G=S_3$ (order $6$). $S_3$ has subgroups of orders $1,2,2,2,3,6$ (three conjugate order-$2$ subgroups, one order-$3$ subgroup, which is normal). Describe the shape of the lattice of intermediate fields, and say which intermediate fields are Galois over $K$.
Solution
By the bijection there are six intermediate fields, order-reversed against the six subgroups: $K=L^{S_3}$ (degree $1$); a unique field $M_0=L^{A_3}$ of degree $[S_3:A_3]=2$ over $K$ (here $A_3\cong\mathbb{Z}/3$ is the unique, hence normal, subgroup of order $3$); three fields $M_1,M_2,M_3$ of degree $[S_3:\langle\text{transposition}\rangle]=3$ over $K$, one for each of the three order-$2$ subgroups; and $L$ itself (degree $6$). Since $A_3\trianglelefteq S_3$, the theorem says $M_0/K$ is Galois, with $\operatorname{Gal}(M_0/K)\cong S_3/A_3\cong\mathbb{Z}/2$. The three order-$2$ subgroups are conjugate to each other (all transpositions are conjugate in $S_3$) but each is non-normal (index $3$, and $S_3$ is non-abelian so normality is not automatic), so $M_1,M_2,M_3$ are each non-Galois over $K$ — they are permuted among themselves by the action of $G$, consistent with Step 8 of the proof, since conjugate subgroups correspond to $G$-translates of the same field.
A student claims: "Since $\operatorname{Gal}(L/M)$ is a subgroup of $G$ for every intermediate $M$, and every subgroup of an abelian group is normal, it follows that every finite Galois extension with intermediate field $M$ has $M/K$ Galois." Find the error and give a specific extension $L/K$ with non-abelian $G$ where the claim's conclusion fails, referencing an earlier example.
Solution
The error is treating "abelian" as a hypothesis that always holds, rather than as a special property some Galois groups have and others don't. The claim's if-clause ("every subgroup of an abelian group is normal") is true, but it only applies to those $L/K$ whose Galois group actually is abelian (Problem 1's $\mathbb{Z}/4$ example, or Worked Example 1's $(\mathbb{Z}/2)^2$); it says nothing about extensions like Worked Example 2 ($L=\mathbb{Q}(\sqrt[4]2,i)$, $G\cong D_4$, non-abelian) or Problem 4 ($G\cong S_3$, non-abelian), where non-normal subgroups exist and, as shown there, produce genuinely non-Galois intermediate fields such as $\mathbb{Q}(\sqrt[4]2)$ or $M_1$ in the $S_3$ example. So the claim is false in general; it is a special feature of abelian Galois groups, not a consequence of the Fundamental Theorem itself.