# 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 ⊆ ··· ⊆ Gl-1 ⊆ Gl = l, Gi-1 ◁ Gi, Gi/Gi-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(ξp) where ξp is a primitive p-th root of unity. It has a Galois group that is isomorphic to (ℤ/pℤ)*, i.e. G = Gal(ℚ(ξp)/ℚ) ≋ (ℤ/pℤ)* ≋ ℤ/(p-1)ℤ.

Recall: There is an isomorphism, ψ: (ℤ/nℤ)* → Gal(ℚ(ξn)/ℚ), a (mod n) → σa where σan) = ξna

p ≡ 1 (mod n) ⇒ n | p -1 ⇒ [G = Gal(ℚ(ξp)/ℚ) ≋ (ℤ/pℤ)*, therefore G is a cyclic group of order p-1] ∃ H ≤ G of order p-1n. Because it is an Abelian extension, every subgroup is Galois, H ◁ G. Let K = ℚ(ξp)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(ξp)/ℚ)/H ≋ (ℤ/pℤ)*/H ≋ [H ≤ (ℤ/pℤ)* of order p-1n, (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/nrℤ.

Next, using Dirichlet’s theorem on arithmetic progression, take distinct primes p1, p2, ···, pr such that ni|pi-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/prℤ)*, m = p1 ··· pr. Then, Gal(ℚ(ξm)/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 Hr where Hi ≤ (ℤ/piℤ)*, |Hi| = pi-1ni ⇒ $\tilde{G}$/H ≋ (ℤ/p1ℤ)*/H1 x (ℤ/p2ℤ)*/H2 x ··· x (ℤ/prℤ)*/Hr ≋ [|(ℤ/piℤ)*/Hi| = $\frac{p_i-1}{\frac{p_i-1}{n_i}}$] ℤ/n1ℤ x ℤ/n2ℤ x ··· x ℤ/nrℤ ≋ G

The argument is the same as before: Gal(ℚ(ξm)/Q) ≋ (ℤ/mℤ)* ≋ $\tilde{G}$, Abelian, H ◁ $\tilde{G}$ ⇒ K/ℚ = ℚ(ξm)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=LH) 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) ≋ Sn

Proof.

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

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

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

In fact, F = ℚ(s1, s2, ···, sn). 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, ···, sn are fixed under any permutation of their subscripts ⇒ ℚ(s1, s2, ···, sn) ⊆ F = KSn where the symmetric polynomials are defined as follows:

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

i.e., the ith symmetric function si of t1, t2, ···, tn is the sum of all products of the tj’s taken i at a time.

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

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 + ··· + tn2 = s12 -2s2 ∈ ℚ(s1, s2, ···, sn)
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, ···, tn) is a polynomial in t1, ···, tn, then it can be seen that f is actually a polynomial in s1, s2, ···, sn.

The polynomial f(x) = xn -s1xn-1 + ··· + (-1)nsn = [It can be proven by induction] (x -t1)(x -t2)···(x -tn) ∈ F[x]. K = ℚ(t1, t2, ···, tn) is the splitting field of the polynomial f of degree n over ℚ(s1, s2, ···, sn) because K contains their n distinct roots, namely t1, ···, tn.

Therefore, we have [F(t1, t2, ···, tn) : F(s1, s2, ···, sn)] ≤ [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 Sn, 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 |Sn| ] n!

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

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 Sn. By the previous Lemma, there’s a Galois extension such that Gal(K/F) ≋ Sn and it is indeed constructed as follows,

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

Proof.

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

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

Let β ∈ H’, β ∉ H = {σ ∈ Sn: 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’ = Sn

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) ≋ Sn (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 ∈ Sn σ(1) = 1} ⊆ Sn. H is a subgroup of Sn isomorphic to Sn-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’ ⊆ Sn ⇒ [Sn-1 is a maximal subgroup of Sn] H = H’ or Sn = H’. Either way, we arrive at L and F again, but no proper intermediate fields.

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