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

Recall: G is solvable if ∃ a normal tower whose subquotients are all Abelian, {id} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ 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

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/n_{1}ℤ x ℤ/n_{2}ℤ x ··· x ℤ/n_{r}ℤ.

Next, using Dirichlet’s theorem on arithmetic progression, take distinct primes p_{1}, p_{2}, ···, 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/p_{1}ℤ)* x (ℤ/p_{2}ℤ)* x ··· x (ℤ/p_{r}ℤ)*, m = p_{1} ··· 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 = H_{1} x H_{2} x ··· x H_{r} where H_{i} ≤ (ℤ/p_{i}ℤ)*, |H_{i}| = ^{pi-1}⁄_{ni} ⇒ $\tilde{G}$/H ≋ (ℤ/p_{1}ℤ)*/H_{1} x (ℤ/p_{2}ℤ)*/H_{2} x ··· x (ℤ/p_{r}ℤ)*/H_{r} ≋ [|(ℤ/p_{i}ℤ)*/H_{i}| = $\frac{p_i-1}{\frac{p_i-1}{n_i}}$] ℤ/n_{1}ℤ x ℤ/n_{2}ℤ x ··· x ℤ/n_{r}ℤ ≋ G

The argument is the same as before: **Gal(ℚ(ξ _{m})/Q)** ≋ (ℤ/mℤ)* ≋

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 C_{4}, also denoted ℤ/4ℤ or V_{4} = {e, a, b, c}, **both of which have subgroups of order 2** (and therefore intermediate fields), namely C_{2} = 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) = A_{4}, e.g., x^{4} +8x +12 or x^{4} +24x +36. Let L be the splitting field of f and H = A_{3} = ⟨(123)⟩ ≤ A_{4}, H is a cyclic subgroup of A_{4} of order 3.

Let H ≤ H’ ≤ A_{4} ⇒ [H = A_{3} ⟨(123)⟩ is a maximal subgroup of A_{4}] H = H’ or H’=A_{4}

**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’ ≤ A_{4} ⊥

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(t_{1}, t_{2}, ···, t_{n}), where t_{1}, t_{2}, ···, 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** t_{1}, t_{2}, ···, t_{n}, i.e., Φ_{σ}(t_{i}) = t_{σ(i)}, and *extending this map in the natural way* to K = ℚ(t_{1}, t_{2}, ···, t_{n}), e.g., σ = (123), p = t_{1} + 2·t_{2} -4·t_{3}, then σ(p) = t_{2} + 2·t_{3} -4·t_{1}. The map σ → Φ_{σ} is an isomorphism. Identifying σ ∈ S_{n} with this automorphism of ℚ(t_{1}, t_{2}, ···, t_{n}), identifies S_{n} as a subgroup of Gal(K/ℚ), i.e., **S _{n} ≤ Gal(K/ℚ)**

Let **F = K ^{Sn}** ⇒ [We have already demonstrated in the past that K/K

In fact, F = ℚ(s_{1}, s_{2}, ···, 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 s_{1}, s_{2}, ···, s_{n} are fixed under any permutation of their subscripts ⇒ **ℚ(s _{1}, s_{2}, ···, s_{n}) ⊆ F** = K

s_{1} = t_{1} + t_{2} + ··· + t_{n}

s_{2} = t_{1}t_{2} + t_{1}t_{3} + ··· + t_{2}t_{3} + t_{2}t_{4} + ··· + t_{n-1}t_{n} = $\prod_{ i < j} t_it_j$

···

s_{n} = t_{1}t_{2}···t_{n}

i.e., the i^{th} symmetric function s_{i} of t_{1}, t_{2}, ···, t_{n} is the sum of all products of the t_{j}’s taken i at a time.

A rational function f(x_{1}, x_{2}, ···, x_{n}) is called symmetric if it is not changed by any permutations of the variables t_{1}, t_{2}, ···, t_{n}-

Examples:

- The expression t
_{1}^{2}+ t_{2}^{2}+ t_{3}^{2}is symmetric in s_{1}, s_{2}, and s_{3}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(s_{1}, s_{2}) - In general, t
_{1}^{2}+ t_{2}^{2}+ ··· + t_{n}^{2}= s_{1}^{2}-2s_{2}∈ ℚ(s_{1}, s_{2}, ···, s_{n}) - The expression (t
_{1}-t_{2})^{2}is symmetric in s_{1}, s_{2}because (t_{1}-t_{2})^{2}= (t_{1}+ t_{2})^{2}-4t_{1}t_{2}= s_{1}^{2}-4s_{2}∈ ℚ(s_{1}, s_{2})

If f(t_{1}, ···, t_{n}) is a polynomial in t_{1}, ···, t_{n}, then it can be seen that f is actually a polynomial in s_{1}, s_{2}, ···, s_{n}.

The polynomial f(x) = x^{n} -s_{1}x^{n-1} + ··· + (-1)^{n}s_{n} = [It can be proven by induction] (x -t_{1})(x -t_{2})···(x -t_{n}) ∈ F[x]. **K = ℚ(t _{1}, t_{2}, ···, t_{n}) is the splitting field of the polynomial f of degree n over ℚ(s_{1}, s_{2}, ···, s_{n})** because K contains their n distinct roots, namely t

Therefore, we have [F(t_{1}, t_{2}, ···, t_{n}) : F(s_{1}, s_{2}, ···, 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^{Sn} = ℚ(s_{1}, s_{2}, ···, 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(t_{1}, t_{2}, ···, t_{n}) is the field of rational functions in the elementary symmetric functions ℚ(s_{1}, s_{2}, ···, s_{n}) and **K/F, that is, ℚ(t _{1}, t_{2}, ···, t_{n})/ℚ(s_{1}, s_{2}, ···, 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

**Maximal subgroup of S _{n}**. Let S

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.