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Solvable groups. Galois' theorem.

Seeing much, suffering much, and studying much are the three pillars of learning, Benjamin Disraeli

Definition. Let G be a group. We say that G is solvable or G is a solvable group if it has a normal tower whose subquotients are all Abelian, that is, there exists a finite series or sequence of subgroups {id} = G0 ⊆ G1 ⊆ G2 ⊆ ··· ⊆ Gr = G such that (i) Gi is a normal subgroup of Gi+1, Gi ◁ Gi+1 ∀i = 0, 1,… r-1, and (ii) Gi+1/Gi is Abelian ∀i = 0, 1,… r-1.

  Examples

S1, S2 Abelian ⇒ S1, S2 solvable

S3 solvable: {()} ◁ [A3, the alternating group, is a normal subgroup of S3, A3/{()} ≋ ℤ/3ℤ] A3 {(), (123), (132)} ◁ [S3/A3 ≋ ℤ/2ℤ is also Abelian] S3

S4 is solvable. The only normal subgroups in S4 are the trivial group, the Klein 4-group V4, A4, and S4 itself. Therefore, we have a normal chain or series, namely {e} ◁ V4 ◁ A4 ◁ S4.

  1. V4 = {(), (12)(34), (13)(24), (14)(23)}. V4/{e} ≋ V4 ≋ ℤ/2ℤxℤ/2ℤ (D2) that is Abelian (all groups of order 4 are Abelian).
  2. |A4/V4| = |A4|/|V4| = 12/4 = 3 so that A4/V4 ≋ C3 ≋ ℤ/3ℤ which is Abelian, too.
  3. |S4/A4| = 24/12 = 2, so that S4/A4 ≋ C2 ≋ ℤ/2ℤ that is also Abelian. Fact: Let G be a group, H ≤ G such that |G : H| = 2 ⇒ H ◁ G, e.g., A4 ◁ S4.

Properties

Proof

Since G is solvable, there is a normal tower whose subquotients are all Abelian, {id} = G0 ⊆ G1 ⊆ G2 ⊆ ··· ⊆ Gr = G.

Consider the sequence H ∩ {id} = {id} ⊆ H ∩ G1 ⊆ H ∩ G2 ⊆ ··· ⊆ H ∩ Gr = H.

  1. This is a normal series, H ∩ Gi ◁ H ∩ Gi+1 [Recall that N ◁ G ↭ gng-1∈ N ∀g ∈G, ∀ n ∈ N] Let gi+1∈ H ∩ Gi+1, hi∈ H ∩ Gi, gi+1higi+1-1 ∈ H because gi+1∈ H, hi∈ H and H ≤ G (it is closed under the group operation). Besides, since Gi ◁ Gi+1, then gi+1higi+1-1 ∈ Gi ⇒ gi+1higi+1-1 ∈ H ∩ Gi.
  2. Futhermore, H ∩ Gi ≤ Gi (H ∩ Gi is going to be our “H” in the 2nd isomorphism), Gi-1 ◁ Gi (Gi-1 is going to be our “N” in the 2nd isomorphism) ⇒ [2nd Isomorphism Theorem. HN/N ≋ H/H ∩ N] $\frac{(H ∩ G_i)G_{i-1}}{G_{i-1}}≋\frac{H ∩ G_i}{(H ∩ G_i)∩G_{i-1}}=\frac{H ∩ G_i}{H ∩ G_{i-1}}$ ⇒ [(H∩Gi)Gi-1 ≤ Gi and Gi-1◁ Gi, Correspondence theorem: G group, H ≤ G, N ◁ G. Then, H/N ≤ G/N] $\frac{(H ∩ G_i)G_{i-1}}{G_{i-1}}≤\frac{G_i}{G_{i-1}}$, but recall that we have Abelian quotients, so Gi/Gi-1 is Abelian ⇒ [A subgroup of an Abelian group is itself Abelian] $\frac{H ∩ G_i}{H ∩ G_{i-1}}$ is Abelian∎

Proof.

{id} = G0 ⊆ G1 ⊆ G2 ⊆ ··· ⊆ Gr = G

Let Φ: G → G/N be the quotient homomorphism which sends g ∈ G to the coset gN (g → gN).

{id} = Φ(G0) ⊆ Φ(G1) ⊆ Φ(G2) ⊆ ··· ⊆ Φ(Gr) = G/N. Claim: this sequence satisfies the conditions needed to prove that G/N is indeed solvable.

Φ(Gi) ◁ Φ(Gi+1). [Recall that N ◁ G ↭ gng-1∈ N ∀g ∈G, ∀ n ∈ N] Let y ∈ Φ(Gi) and x ∈ Φ(Gi+1) ⇒ ∃x’ ∈ Gi+1, y’ ∈ Gi : y = Φ(y’), x = Φ(x’).

Then, xyx-1 = Φ(x’)Φ(y’)Φ(x’)-1 = [Φ is an homomorphism] Φ(x’y’x’-1) ∈ Φ(Gi) because x’y’x’-1 ∈ Gi since Gi ◁ Gi+1 ⇒ Φ(Gi) ◁ Φ(Gi+1)

Φ(Gi+1)/Φ(Gi) is Abelian. Consider the homomorphism Gi+1 → Φ(Gi+1)/Φ(Gi), this is obtained by after applying Φ: Gi+1 → Φ(Gi+1), following it (composing it) with the quotient homomorphism, Φ(Gi+1) → Φ(Gi+1)/Φ(Gi). This is a surjective homomorphism whose kernel is Gi ⇒ By the first isomorphism theorem, Gi+1/Gi (Abelian) ≋ Φ(Gi+1)/Φ(Gi) ⇒ Φ(Gi+1)/Φ(Gi) is Abelian∎

Proof.

Suppose that N has a chain {id} = N0 ⊆ N1 ⊆ N2 ⊆ ··· ⊆ Nm = N and G/N has a chain {id} = U0 ⊆ U1 ⊆ U2 ⊆ ··· ⊆ Un = G/N.

Let Φ: G → G/N be the quotient homomorphism, and consider the sequence,

Consider the following series, {id} = N0 ⊆ N1 ⊆ N2 ⊆ ··· ⊆ Nm = N = Φ-1(U0) ⊆ Φ-1(U1) ⊆ Φ-1(U2) ⊆ ··· ⊆ Φ-1(Un) = G.

Basically, we are using the fact that by the fourth isomorphism or the Correspondence theorem there is a bijection from the set of all subgroups of G containing N, onto the set of all subgroups of the quotient group G/N. In other words, the structure of the subgroups of G/N is exactly the same as the structure of the subgroups of G containing N, with N collapsed to the identity element.

The solvability conditions are satisfied for the Ni subgroups follows from the fact that they arise from the sequence of subgroups that shows that N is indeed solvable.

Consider the surjective homomorphisms Φ-1(Ui+1) → Ui+1/Ui, given by following or composing Φ (Φ: Φ-1(Ui+1) → Ui+1) by the quotient homomorphism (Ui+1 → Ui+1/Ui). Its kernel is Φ-1(Ui), and therefore [Let Φ: G → H be a group homomorphism, then Ker(Φ) ◁ G] Φ-1(Ui) ◁ Φ-1(Ui+1). The image is Ui+1/Ui, so by the first isomorphism theorem, Φ-1(Ui+1)/Φ-1(Ui) ≋ Ui+1/Ui and by assumption, Ui+1/Ui is Abelian, therefore Φ-1(Ui+1)/Φ-1(Ui) is Abelian

Alternatively, you may consider that Φ-1(Ui) is a subgroup of G containing N (by the Correspondence theorem), say U’i with U’i+1/U’i ≋ Ui+1/Ui and subgroups that contain a normal subgroup (e.g., N) are normal subgroups, too.

Proposition. Let G and H be groups. If G is a solvable group and Φ: G → H is a group homomorphism from G to H, then Φ(G) is a solvable. The Homomorphic Image of a Solvable Group is Solvable, Mathonline

Theorem. Sn is not solvable for n ≥ 5

Proof.

Let’s suppose for the sake of contradiction that Sn is solvable ⇒ [G is solvable, H ≤ G subgroup ⇒ H is solvable.] Consider the alternating group on n letters An, then An is solvable.

However, An is not Abelian ⇒ there exist a non-trivial normal tower whose subquotients are all Abelian, i.e., ∃ {id} = G0 ⊆ G1 ⊆ G2 ⊆ ··· ⊆ Gr = An and r ≥ 2 (An is not Abelian) ⇒ Gr-1 is not trivial (Gr-1 ≠ {e}), Gr-1 ≠ An, Gr-1 ◁ An. Therefore, An contains a nontrivial, proper normal subgroup, namely Gr-1An is simple for n ≥ 5.

Recall. Let F be a field, F ⊆ ℂ, f ∈ F[x], deg(f) = n ≥ 1. If there exists an extension field K/F such that f factors as a product of linear polynomials (x - α1)···(x - αn) over the field K, i.e., K is the splitting field of f over F then Gal(K/F) is called the Galois group of the polynomial f over F.

Galois’ theorem. Let F be a field, char(F) = 0, f ∈ F[x]. If K is a splitting field of f over F, let G = Gal(f) = Gal(K/F). Then, G is solvable ↭ f is solvable (by radicals) over F

Proof.

⇒) Notice that K/F is Galois (K is a splitting field of f over F and char(F) = 0). Suppose G = Gal(K/F) is a solvable group. We have a chain, ∃ {id} = G0 ⊆ G1 ⊆ G2 ⊆ ··· ⊆ Gr = G, Gi ◁ Gi+1 ∀i = 0, 1,… r-1, and Gi+1/Gi is Abelian

Let’s apply Galois fundamental theorem to Gr = G (K/F is Galois):

F = F0 = KGr ⊆ [Gr-1 ◁ Gr = G ⇒ KGr-1/KGr(= F) is Galois with Galois group Gr/Gr-1 being Abelian because there are part of G’s normal tower whose subquotients are all Abelian] KGr-1

KGr-1 ⊆ KGr-2

K · K K F · ^ ^ · G G = r r - + - G F 2 G 1 a o a l l o = o i i s K s ^ + G G r r / G r - 1 = G G G a r / l - G o 1 r i - s 1

Apply the Fundamental Galois Theorem to K/KGr-1, K/KGr-1 is Galois, and the Galois group is Gr-1, Gr-2 ◁ Gr-1KGr-2/KGr-1 is Galois (+Galois+) with Galois group Gr-1/Gr-2 Abelian because there are part of G’s normal tower whose subquotients are all Abelian

And using the same argument, we move on building the tower of normal fields… then

F = F0 = KGr ⊆ KGr-1 ⊆ KGr-2 ⊆ Kr-3 ⊆ ··· ⊆ KG2 ⊆ KG1 ⊆ KG0 = K ⇒ there is a tower of fields, K contains all the roots, and each intermediate extension is Abelian ⇒ f is solvable.

Recall: α is solvable over F ↭ There exists a tower of fields: F = L0 ⊆ L1 ⊆ ··· ⊆ Ln such that α ∈ Ln and each Li/Li-1 is Abelian, i.e., Galois and the Galois group is Abelian.

⇐ )

Suppose f is solvable ⇒ Every root of f, say α1, α2, ···, αn live in a splitting field, say K = F(α1, α2, ···, αn), is solvable, too ⇒ Each αi is contained in a radical extension Li of F.

Claim: we can find an extension M, L ⊆ M such that M/F is Galois and F = M0Abelian M1Abelian ⊆ M2Abelian ··· ⊆Abelian Ml = M

Let L be the composite of L1, L2··· Ln ⇒ [Recall L1/F, L2/F radical extensions ⇒ the composite L1L2/F is radical, too.] L/F is radical, so ∃ a tower of fields

F = F0s.r. F1s.r. ⊆ F2s.r. ··· ⊆s.r. Fm = L, L ⊇ K = F(α1, α2, ···, αn)

L/F is radical ⇒ [Recall Let L/F be a radical extension. Then, there exist a M/L extension such that M/F is both Galois and radical.] ∃ M/F extension such that M/F is Galois and radical.

We could assume, by extending L if needed, and without losing any generality that L/F is both Galois and radical,

F = F0s.r. F1s.r. ⊆ F2s.r. ··· ⊆s.r. Fm = L. In general, these extensions may not be Abelian 😞 so let’s attach roots of unity as we have done it previously in other proofs.

∀i, Fi-1s.r. Fi ⇒ Fi = Fi-1i) where αidi ∈ Fi-1. Let ξd1, ξd2, ···, ξdm be the primitive d1-nth, d2-nth, ··· dm-nth roots of unity respectively, and F’ = F(ξd1, ξd2, ···, ξdm). Then, F’/F is obviously a radical extension, just consider the following tower of extensions and recall that any cyclotomic extension is simple radical [F ⊆ F(ξ), ξn = 1 ∈ F].

F | F F | F | F ' ( ( ( = s ξ s s ξ s ξ s F . d . . d . d . ( r 1 r r 1 r 1 r ξ . , . . , . ) . d 1 ξ ξ , d d 2 2 ξ , ) d · 2 · , · · , · ξ · d , m ξ - d 1 m ) )

If you adjoined ξd1, the primitive d1-nth root of unity, F(ξd1) contains all the d1-nth roots of unity, and therefore F’ is the splitting field of (xd1-1)(xd2-1)···(xdm-1) ⇒ [By assumption, char(F) = 0 ⇒ normality implies separability] F’/F Galois.

L/F is radical and Galois, F’/F is radical and Galois, too. Take the composite F’L ⇒ [Recall L1/F, L2/F radical extensions ⇒ the composite L1L2/F is radical.L1/F, L2/F Galois extensions ⇒ the composite L1L2/F is Galois.] M = F’L = F(ξd1, ξd2, ··· ξdm, α1, α2, ··· αm) is radical and Galois

Notice that we have a tower F ⊆Abelian F(ξd1) [… 🚀]

Recall. Cyclotomic extensions are always Abelian, the mapping Φ: Gal(ℚ(ξ)/ℚ) → (ℤ/ℤn)*, σ → aσ mod n = [aσ], σ(ξ) = ξaσ is an injective or one-to-one homomorphism. Since (ℤ/ℤn)* is Abelian, Gal(ℚ(ξn)/ℚ) is Abelian, too.

[… 🚀] ⊆Abelian F(ξd1, ξd2) ⊆ ··· ⊆Abelian F(ξd1, ξd2, ··· ξdm) = F’⊆Abelian F’(α1) ⊆Abelian F’(α1, α2) ⊆Abelian ··· ⊆Abelian F’(α1, α2, ··· αm) = M ⊇ L ⊇ K

F’ = F(ξd1, ξd2, ··· ξdm) ⊆ F’(α1) is Kummer, F’ contains the primitive d1-nth root of unity, and it is a radical extension ⇒ Kummer ⇒ Cyclic ⇒ Abelian.

So let’s rename it, F = M0Abelian M1Abelian ··· ⊆Abelian Ml = M ⊇ L ⊇ K

Claim: Gal(M/F) is solvable.

M1/M0 is Abelian ⇒ Galois ⇒ By the fundamental Galois Theorem [An intermediate fieldL (F ⊆ L ⊆ K) is Galois over F ↭ Gal(K/L) is a normal subgroup of G, Gal(K/L) ◁ G In this case, G/Gal(K/L)≋Gal(L/F)]. Since the intermediate field M1 is Galois over M0 = F, then Gal(M/M1) ◁ Gal(M/F) and Gal(M/F)/Gal(M/M1) ≋ Gal(M1/F). Futhermore, in our particular case, Gal(M1/F) is Abelian, so the quotient Gal(M/F)/Gal(M/M1) is Abelian, too.

M M M M M | M l l l 2 1 o - - A = 1 2 b = e M l F i a n G a l o i s G a l o i s

Gal(M/F) ⊇Abelian Gal(M/M1) ⊇Abelian Gal(M/M2)

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.

M M M M | M M l l l 2 1 o - - A = 1 2 b = e M l F i a n G a l o i s G a l o i s

We can use the same reasoning as before, M2/M1 Abelian ⇒ Galois ⇒ Gal(M/M2) ◁ Gal(M/M1) and Gal(M/M1)/Gal(M/M2) ≋ Gal(M2/M1). Futhermore, Gal(M2/M1) is Abelian, so the quotient Gal(M/M1)/Gal(M/M2) is Abelian, too.

Gal(M/F) ⊇Abelian Gal(M/M1) ⊇Abelian Gal(M/M2) ⊇Abelian Gal(M/M3) ⊇ ·· · ⊇Abelian Gal(M/Ml-1) ⊇Abelian Gal(M/Ml) = {id}, and therefore Gal(M/F) is solvable

M K | F = G a F l ' o L i s = F ( ξ d 1 , ξ d 2 , · · · ξ d m , α 1 , α 2 , · · · α m )

Then, by the Fundamental Theorem of Galois, Gal(M/K) ◁ Gal(M/F), Gal(K/F) ≋ Gal(M/F)/Gal(M/K).

Gal(M/F) is solvable ⇒ [Gal(K/F) ≋ Gal(M/F)/Gal(M/K), so Gal(M/F)→Gal(K/F) is surjective group homomorphism, the map image of a solvable group, is solvable, too] Gal(K/F) = [K is the splitting field of f] Gal(f) is solvable

Corollary: Let f(x) be the general nth degree polynomial in F[x] of deg(f) = n ≤ 4, then f is solvable (by radicals).

Proof. (An alternative proof)

Recall. Let f be an irreducible polynomial of degree n over the field, and let G be the Galois group of f. The Galois group of f is a subgroup of Sn.

Therefore, the Galois group of f (deg(f) = n ≤ 4) is a subgroup of S1, S2, S3 or S4, and all of these are solvable ⇒ [If G is solvable, and H ≤ G (H is a subgroup of G), then H is solvable] G is solvable ⇒ [Galois’ theorem. Let F be a field, char(F) = 0, f ∈ F[x]. If K is a splitting field of f over F, let G = Gal(f) = Gal(K/F). Then, G is solvable ↭ f is solvable (by radicals) over F] f is solvable (by radicals) over F.∎

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
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