The most significant discovery of Galois is that under some hypotheses, there is a one-to-one correspondence between

1. subgroups of the Galois group Gal(E/F)

2. subfields M of E such that F ⊆ M.

The correspondence goes as follows:

To each intermediate subfield M, associate the group Gal(E/M) of all M-automorphisms of E:

G = Gal : {intermediate fields} → {subgroups of Gal(E/F)}

M → G(M) = Gal(E/M)

To each subgroup H of Gal(E/F), associate the fixed subfield F(H):

F : {subgroups of Gal(E/F )} → {intermediate fields}

H → F(H)

We will prove that, under the right hypotheses, we actually have a bijection (namely G is the inverse of F). For example.

*Let E/F be a finite Galois extension with Galois group G.*

**Theorem:**- The map F is a bijection from subgroups to intermediate fields, with inverse G.
- Consider the intermediate field K = F(H) which is fixed by H, and σ ∈ G.Then the intermediate fieldσK = {σ(x), x∈K}
is fixed by σHσ

^{−1}, namely σK = F(σHσ^{−1})1. We first consider the composition of maps H → F(H) → GF(H).*Proof:*We need to prove that GF(H) = H. Take σ in H, then σ fixes F(H) by definition and σ ∈ Gal(E/F(H)) = G(F(H)), showing that

H ⊆ GF(H).

To prove equality, we need to rule out the strict inclusion. If H were a proper subgroup of G(F(H)), by the above proposition the fixed field F(H) of H should properly contain the fixed field of GF(H) which is F(H) itself, a contradiction, showing that

H = GF(H)

Now consider the reverse composition of maps K → G(K) → FG(K)

This time we need to prove that K = FG(K). But FG(K) = fixed field by Gal(E/K) which is exactly K by the above proposition (its first point). It is enough to compute F(σHσ

^{−1}) and show that it is actually equal toσK = σF(H).

F(σHσ

^{−1}) = {x ∈ E, στσ^{−1}(x) = x ∀ τ ∈ H} = {x ∈ E, τσ^{−1}(x)=σ^{−1}(x) ∀ τ ∈ H}= {x ∈ E, σ

^{−1}(x) ∈ F(H)}= {x ∈ E, x ∈ σ(F(H))} = σ(F(H))

We now look at subextensions of the finite Galois extension E/F and ask about their respective Galois group.

Let E/F be a finite Galois extension with Galois group G. Let K be an intermediate subfield, fixed by the subgroup H.**Theorem:**1. The extension E/K is Galois.

2. The extension K/F is normal if and only if H is a normal subgroup of G.

3. If H is a normal subgroup of G, then

Gal(K/F ) ≃ G/H = Gal(E/F )/Gal(E/K).

4. Whether K/F is normal or not, we have

[K : F] = [G : H]

**Proof:**That E/K is Galois is immediate from the fact that a subextension E/K/F inherits normality and separability from E/F.

First note that σ is an F-monomorphism of K into E if and only if σ is the restriction to K of an element of G: if σ is an F -monomorphism of K into E, it can be extended to an F-monomorphism of E into itself thanks to the normality of E. Conversely, if τ is an F-automorphism of E, then σ = τ|K is surely a F-monomorphism of K into E.

Now, this time by a characterization of a normal extension, we have

K/F normal ⇐⇒ σ(K) = K ∀ σ ∈ G

Since K = F(H), we just rewrite

K/F normal ⇐⇒ σ(F(H)) = F(H) ∀ σ ∈ G.

Now by the above theorem, we know that σ(F(H)) = F(σHσ−1), and we have

K/F normal ⇐⇒ F(σHσ−1) = F(H) for all σ ∈ G

We now use again the above theorem that tells us that F is invertible, with inverse G, to get the conclusion:

K/F normal ⇐⇒ σHσ−1 =H ∀ σ ∈ G

To prove this isomorphism, we will use the 1st isomorphism Theorem for groups. Consider the group homomorphism

Gal(E/F)→Gal(K/F), σ →σ|K.

This map is surjective and its kernel is given by

Ker={σ, σ|K =1}=H =Gal(E/K).

Applying the first isomorphism Theorem for groups, we get

Gal(K/F ) ≃ Gal(E/F )/Gal(E/K)

Finally, by multiplicativity of the degrees:

[E :F]=[E :K][K :F]

Since E/F and E/K are Galois, we can rewrite |G| = |H|[K : F]. We conclude by Lagrange Theorem:

[G:H]=|G|/|H|=[K :F]