Every finite Abelian group is realized as a Galois group over ℚ.

All things are difficult before they are easy, Thomas Fuller.

Recall: G is solvable if ∃ a normal tower whose subquotients are all Abelian, {id} = G₀ ⊆ G₁ ⊆ G₂ ⊆ ··· ⊆ G_l-1 ⊆ G_l = l, G_i-1 ◁ G_i, G_i/G_i-1 is Abelian.

 

Dirichlet’s theorem on arithmetic progression. Let any be two positive coprime integers a and d, there are infinitely many primes of the form a + nd, where n is also a positive integer, i.e., there are infinitely many primes that are congruent to a modulo d.

It was not long before an obvious, but yet quite challenging question was raised. Given a finite group G, define G as realizable if there exists a polynomial in ℚ[X] having G as its Galois group. The big question is, which groups are realizable?

Proposition. Every finite Abelian group is realized as a Galois group over ℚ. In other words, every finite Abelian group is the Galois group of some finite extension K over ℚ.

Proof.

First, let’s assume that G is cyclic, G ≋ ℤ/nℤ, [Dirichlet’s theorem on arithmetic progression] ∃p prime such that p ≡ 1 (mod n), i.e. n divides p -1, and consider the pth cyclotomic field ℚ(ξ_p) where ξ_p is a primitive p-th root of unity. It has a Galois group that is isomorphic to (ℤ/pℤ)*, i.e. G = Gal(ℚ(ξ_p)/ℚ) ≋ (ℤ/pℤ)* ≋ ℤ/(p-1)ℤ.

Recall: There is an isomorphism, ψ: (ℤ/nℤ)* → Gal(ℚ(ξ_n)/ℚ), a (mod n) → σ_a where σ_a(ξ_n) = ξ_n^a

p ≡ 1 (mod n) ⇒ n | p -1 ⇒ [G = Gal(ℚ(ξ_p)/ℚ) ≋ (ℤ/pℤ)*, therefore G is a cyclic group of order p-1] ∃ H ≤ G of order ^p-1⁄_n. Because it is an Abelian extension, every subgroup is Galois, H ◁ G. Let K = ℚ(ξ_p)^H be the fixed field of H (🎉 This is the Galois extension we are looking for) ⇒ K/ℚ is Galois and Gal(K/ℚ) ≋ Gal(ℚ(ξ_p)/ℚ)/H ≋ (ℤ/pℤ)*/H ≋ [H ≤ (ℤ/pℤ)* of order ^p-1⁄_n, (ℤ/pℤ)*/H is a cyclic group of order $\frac{p-1}{\frac{p-1}{n}}=n$] ℤ/nℤ ≋ G ∎

Let G be a finite Abelian group ⇒ [The Structure Theorem for Finite Abelian Groups. Every finite Abelian group is isomorphic to a direct product of cyclic groups of orders that are powers of prime numbers.] G ≋ ℤ/n₁ℤ x ℤ/n₂ℤ x ··· x ℤ/n_rℤ.

Next, using Dirichlet’s theorem on arithmetic progression, take distinct primes p₁, p₂, ···, p_r such that n_i|p_i-1, notice that there are infinitely many such primes for each i by Dirichlet’s theorem.

Consider $ \tilde{G} $ = (ℤ/p₁ℤ)* x (ℤ/p₂ℤ)* x ··· x (ℤ/p_rℤ)*, m = p₁ ··· p_r. Then, Gal(ℚ(ξ_m)/Q) ≋ (ℤ/mℤ)* ≋ $\tilde{G} $

Let’s construct a subgroup of $\tilde{G}$, say H, as follows, H ≤ $\tilde{G} $: H = H₁ x H₂ x ··· x H_r where H_i ≤ (ℤ/p_iℤ)*, |H_i| = ^p_i-1⁄_ni ⇒ $\tilde{G}$/H ≋ (ℤ/p₁ℤ)*/H₁ x (ℤ/p₂ℤ)*/H₂ x ··· x (ℤ/p_rℤ)*/H_r ≋ [|(ℤ/p_iℤ)*/H_i| = $\frac{p_i-1}{\frac{p_i-1}{n_i}}$] ℤ/n₁ℤ x ℤ/n₂ℤ x ··· x ℤ/n_rℤ ≋ G

The argument is the same as before: Gal(ℚ(ξ_m)/Q) ≋ (ℤ/mℤ)* ≋ $\tilde{G} $, Abelian, H ◁ $\tilde{G}$ ⇒ K/ℚ = ℚ(ξ_m)^H/ℚ is Galois and Gal(K/ℚ) ≋ $\tilde{G}$/H ≋ G∎

Exercise. Find a degree 4 extension of ℚ with no intermediate fields.

Proof.

Any such extension K/ℚ cannot be Galois, because in that case the Galois group is either C₄, also denoted ℤ/4ℤ or V₄ = {e, a, b, c}, both of which have subgroups of order 2 (and therefore intermediate fields), namely C₂ = ℤ/2ℤ and {e, a} (also {e, b} and {e, c}) respectively.

Computing Galois groups. Let f ∈ ℚ[x] be an irreducible quantic such that Gal(f) = A₄, e.g., x⁴ +8x +12 or x⁴ +24x +36. Let L be the splitting field of f and H = A₃ = ⟨(123)⟩ ≤ A₄, H is a cyclic subgroup of A₄ of order 3.

Let H ≤ H’ ≤ A₄ ⇒ [H = A₃ ⟨(123)⟩ is a maximal subgroup of A₄] H = H’ or H’=A₄

As it is required, K/ℚ is a degree four extension and has no non-trivial intermediate fields. For the sake of contradiction if there was a non-trivial intermediate fields, such a field, say M (ℚ ⊆ M ⊆ K=L^H) would, by the Galois correspondence lead to a subgroup of H’ (there's an inclusion-reversing bijective map or correspondence from subgroups of G and intermediate fields of K/F, so H ⊆ H'), such that H ≤ H’ ≤ A₄ ⊥

It is an open problem to determine whether every finite group appears as the Galois group for some polynomial over ℚ. It is indeed the case that every Abelian group is a Galois group over ℚ.

Lemma. Let G be a given group, let n = |G|. There is a Galois extension K/F such that Gal(K/F) ≋ S_n

Proof.

K = ℚ(t₁, t₂, ···, t_n), where t₁, t₂, ···, t_n are variables or indeterminates, that is, there are no polynomial relations over ℚ. Then, S_n acts on K by permuting the variables t_i.

We are going to define a map Φ: S_n → Gal(K/ℚ) as follows. Let σ be an arbitrary permutation, σ ∈ S_n, we can define an automorphism Φ_σ, given by permuting the subscripts of the variables t₁, t₂, ···, t_n, i.e., Φ_σ(t_i) = t_σ(i), and extending this map in the natural way to K = ℚ(t₁, t₂, ···, t_n), e.g., σ = (123), p = t₁ + 2·t₂ -4·t₃, then σ(p) = t₂ + 2·t₃ -4·t₁. The map σ → Φ_σ is an isomorphism. Identifying σ ∈ S_n with this automorphism of ℚ(t₁, t₂, ···, t_n), identifies S_n as a subgroup of Gal(K/ℚ), i.e., S_n ≤ Gal(K/ℚ)

Let F = K^S_n ⇒ [We have already demonstrated in the past that K/K^G is Galois with Gal(K/K^G)≋ G] Therefore, K/F is Galois with Gal(K/F) ≋ S_n, [K : K^G] = [K : F] = [ℚ(t₁, t₂, ···, t_n) : K^S_n] = |S_n| = n!

In fact, F = ℚ(s₁, s₂, ···, s_n). First, the reader should understand that F includes all the elementary symmetric polynomials (and obviously all rational combinations of them). In other words, the elementary symmetric functions s₁, s₂, ···, s_n are fixed under any permutation of their subscripts ⇒ ℚ(s₁, s₂, ···, s_n) ⊆ F = K^S_n where the symmetric polynomials are defined as follows:

s₁ = t₁ + t₂ + ··· + t_n
s₂ = t₁t₂ + t₁t₃ + ··· + t₂t₃ + t₂t₄ + ··· + t_n-1t_n = $\prod_{ i < j} t_it_j$
···
s_n = t₁t₂···t_n

i.e., the i^th symmetric function s_i of t₁, t₂, ···, t_n is the sum of all products of the t_j’s taken i at a time.

A rational function f(x₁, x₂, ···, x_n) is called symmetric if it is not changed by any permutations of the variables t₁, t₂, ···, t_n-

Examples:

1. The expression t₁² + t₂² + t₃² is symmetric in s₁, s₂, and s₃ because $t_1^2+t_2^2+t_3^3 =(t_1+t_2+t_3)^2-2(t_1t_2 + t_1t_3 + t_2t_3) = s_1^2 -2s_2$ ∈ ℚ(s₁, s₂)
2. In general, t₁² + t₂² + ··· + t_n² = s₁² -2s₂ ∈ ℚ(s₁, s₂, ···, s_n)
3. The expression (t₁ -t₂)² is symmetric in s₁, s₂ because (t₁ -t₂)² = (t₁ + t₂)² -4t₁t₂ = s₁² -4s₂ ∈ ℚ(s₁, s₂)

If f(t₁, ···, t_n) is a polynomial in t₁, ···, t_n, then it can be seen that f is actually a polynomial in s₁, s₂, ···, s_n.

The polynomial f(x) = xⁿ -s₁x^n-1 + ··· + (-1)ⁿs_n = [It can be proven by induction] (x -t₁)(x -t₂)···(x -t_n) ∈ F[x]. K = ℚ(t₁, t₂, ···, t_n) is the splitting field of the polynomial f of degree n over ℚ(s₁, s₂, ···, s_n) because K contains their n distinct roots, namely t₁, ···, t_n.

Therefore, we have [F(t₁, t₂, ···, t_n) : F(s₁, s₂, ···, s_n)] ≤ [Splitting field of degree n polynomial. Let f(x)∈F[x] be a polynomial of degree n. Let K be a splitting field of f(x) over F. Then [K:F] ≤ n! Hint: Automorphisms of splitting fields permute roots, these automorphisms are determined or defined by how they permute the roots ⇒ the Galos group can be embedded in S_n, and since the order of the automorphism group is the same of the degree of the extension, we get that the degree is no bigger than |S_n| ] n!

Therefore, F = K^S_n = ℚ(s₁, s₂, ···, s_n), i.e., the fixed field of the symmetric group S_n acting on the field of rational functions in n variables K = ℚ(t₁, t₂, ···, t_n) is the field of rational functions in the elementary symmetric functions ℚ(s₁, s₂, ···, s_n) and K/F, that is, ℚ(t₁, t₂, ···, t_n)/ℚ(s₁, s₂, ···, s_n) is the desired Galois extension with Galois group S_n.

Proposition. Every finite group is isomorphic to some Galois group for some finite normal extension of some field.

Proof.

By Cayley’s theorem, every finite group G is isomorphic to a subgroup of some symmetric group S_n. By the previous Lemma, there’s a Galois extension such that Gal(K/F) ≋ S_n and it is indeed constructed as follows,

Maximal subgroup of S_n. Let S_n denote the symmetric group on {1, 2, ···, n}. Let H be the subgroup {σ ∈ S_n: σ(1) = 1} -it is equivalent to {σ ∈ S_n: σ(n) = n}-. H is a maximal subgroup of S_n

Proof.

Suppose we have a subgroup H’, H ⊆ H’ ⊆ S_n. We must show that H’ = H or H’ = S_n. It suffices to show that H’ contains all transpositions (1, j), j = 2, ···, n ⇒ H’ contains S_n.

If H’ = H, then we are done🎉. Suppose, H’ ≠ H.

Let β ∈ H’, β ∉ H = {σ ∈ S_n: σ(1) = 1}. ∃k ≠ 1 such that β(k) = 1, and β(1) = r ≠ 1. Multiplying β on the left by (r, k) ∈ H -r≠1 and k≠1-, that is, (r, k)β sends 1 to k (1 → r → k) and k to 1 (k → 1) ⇒ (r, k)β = (1, k)τ for some τ that is disjoint from {1, k}, so τ ∈ H ⇒ (1, k) = (r, k)βτ^-1 [(r, k)∈ H⊆H’, β∈H’, τ∈H⊆H’ ⇒τ^-1∈H’] ∈ H’

Now conjugating with (k, j) ∈ H, j = 2,···, n, we get (k, j)(1, k)(k, j) = (1, j) ∈ H’, proving that H’ contains all transpositions of the form (1, j) ⇒ H’ = S_n

Example. Given a fixed n ≥ 1. Give an example of an extension K/F such that [K : F] = n and K/F has no non-trivial intermediate fields.

Solution. Let K/F be an extension such that Gal(K/F) ≋ S_n (previous result, Every finite group is isomorphic to some Galois group for some finite normal extension of some field) ⇒ K/L Galois

Consider H = {σ ∈ S_n σ(1) = 1} ⊆ S_n. H is a subgroup of S_n isomorphic to S_n-1 ⇒ Any intermediate field M (M ⊆ L) corresponds to a subgroup of G containing H (there's an inclusion-reversing bijective map or correspondence from subgroups of G and intermediate fields of K/F, so H ⊆ H') ⇒ L/F has the desired property because H ⊆ H’ ⊆ S_n ⇒ [S_n-1 is a maximal subgroup of S_n] H = H’ or S_n = H’. Either way, we arrive at L and F again, but no proper intermediate fields.