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} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ G_{r} = G such that (i) G_{i} is a normal subgroup of G_{i+1}, **G _{i} ◁ G_{i+1}** ∀i = 0, 1,… r-1, and (ii)

Examples

**G is Abelian ⇒ G is solvable**: {id} ⊆ G- S
_{n}is solvable ↭ n = 1, 2, 3, or 4.

S_{1}, S_{2} Abelian ⇒ S_{1}, S_{2} solvable

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

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

- V
_{4}= {(), (12)(34), (13)(24), (14)(23)}. V_{4}/{e} ≋ V_{4}≋ ℤ/2ℤxℤ/2ℤ (D_{2}) that is Abelian (all groups of order 4 are Abelian). - |A
_{4}/V_{4}| = |A_{4}|/|V_{4}| = 12/4 = 3 so that A_{4}/V_{4}≋ C_{3}≋ ℤ/3ℤ which is Abelian, too. - |S
_{4}/A_{4}| = 24/12 = 2, so that S_{4}/A_{4}≋ C_{2}≋ ℤ/2ℤ that is also Abelian. Fact:**Let G be a group, H ≤ G such that |G : H| = 2 ⇒ H ◁ G**, e.g., A_{4}◁ S_{4}.

- If G is solvable, and H ≤ G (H is a subgroup of G), then H is solvable.

Proof

Since G is solvable, there is a normal tower whose subquotients are all Abelian, {id} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ G_{r} = G.

Consider the sequence H ∩ {id} = {id} ⊆ H ∩ G_{1} ⊆ H ∩ G_{2} ⊆ ··· ⊆ H ∩ G_{r} = H.

- This is a normal series,
**H ∩ G**[Recall that N ◁ G ↭ gng_{i}◁ H ∩ G_{i+1}^{-1}∈ N ∀g ∈G, ∀ n ∈ N] Let g_{i+1}∈ H ∩ G_{i+1}, h_{i}∈ H ∩ G_{i}, g_{i+1}h_{i}g_{i+1}^{-1}∈ H because g_{i+1}∈ H, h_{i}∈ H and H ≤ G (it is closed under the group operation). Besides, since G_{i}◁ G_{i+1}, then g_{i+1}h_{i}g_{i+1}^{-1}∈ G_{i}⇒ g_{i+1}h_{i}g_{i+1}^{-1}∈ H ∩ G_{i}. - Futhermore, H ∩ G
_{i}≤ G_{i}(H ∩ G_{i}is going to be our “H” in the 2nd isomorphism), G_{i-1}◁ G_{i}(G_{i-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∩G_{i})G_{i-1}≤ G_{i}and G_{i-1}◁ G_{i}, 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 G_{i}/G_{i-1}is Abelian ⇒ [A subgroup of an Abelian group is itself Abelian] $\frac{H ∩ G_i}{H ∩ G_{i-1}}$ is Abelian∎

- Let N be a normal subgroup of a group G. If G is solvable, then G/N is solvable.

Proof.

{id} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ G_{r} = G

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

{id} = Φ(G_{0}) ⊆ Φ(G_{1}) ⊆ Φ(G_{2}) ⊆ ··· ⊆ Φ(G_{r}) = G/N. Claim: this sequence satisfies the conditions needed to prove that G/N is indeed solvable.

**Φ(G _{i}) ◁ Φ(G_{i+1})**. [Recall that N ◁ G ↭ gng

Then, xyx^{-1} = Φ(x’)Φ(y’)Φ(x’)^{-1} = [Φ is an homomorphism] Φ(x’y’x’^{-1}) ∈ Φ(G_{i}) because x’y’x’^{-1} ∈ G_{i} since G_{i} ◁ G_{i+1} ⇒ Φ(G_{i}) ◁ Φ(G_{i+1})

**Φ(G _{i+1})/Φ(G_{i}) is Abelian.** Consider the homomorphism G

- Let N be a normal subgroup of G. If N and G/N are solvable, then G is solvable.

Proof.

Suppose that N has a chain {id} = N_{0} ⊆ N_{1} ⊆ N_{2} ⊆ ··· ⊆ N_{m} = N and G/N has a chain {id} = U_{0} ⊆ U_{1} ⊆ U_{2} ⊆ ··· ⊆ U_{n} = G/N.

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

Consider the following series, {id} = N_{0} ⊆ N_{1} ⊆ N_{2} ⊆ ··· ⊆ N_{m} = N = Φ^{-1}(U_{0}) ⊆ Φ^{-1}(U_{1}) ⊆ Φ^{-1}(U_{2}) ⊆ ··· ⊆ Φ^{-1}(U_{n}) = 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 N_{i} 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}(U_{i+1}) → U_{i+1}/U_{i}, given by following or composing Φ (Φ: Φ^{-1}(U_{i+1}) → U_{i+1}) by the quotient homomorphism (U_{i+1} → U_{i+1}/U_{i}). Its kernel is Φ^{-1}(U_{i}), and therefore [Let Φ: G → H be a group homomorphism, then Ker(Φ) ◁ G] **Φ ^{-1}(U_{i}) ◁ Φ^{-1}(U_{i+1})**. The image is U

Alternatively, you may consider that Φ^{-1}(U_{i}) is a subgroup of G containing N (by the Correspondence theorem), say U’_{i} with U’_{i+1}/U’_{i} ≋ U_{i+1}/U_{i} 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. S_{n} is not solvable for n ≥ 5

Proof.

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

However, A_{n} is not Abelian ⇒ **there exist a non-trivial normal tower whose subquotients are all Abelian**, i.e., ∃ {id} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ G_{r} = A_{n} and r ≥ 2 (A_{n} is not Abelian) ⇒ G_{r-1} is not trivial (G_{r-1} ≠ {e}), G_{r-1} ≠ A_{n}, G_{r-1} ◁ A_{n}. Therefore, A_{n} contains a nontrivial, proper normal subgroup, namely G_{r-1} ⊥ An 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 - α_{1})···(x - α_{n}) 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} = G_{0} ⊆ G_{1} ⊆ G_{2} ⊆ ··· ⊆ G_{r} = G, G_{i} ◁ G_{i+1} ∀i = 0, 1,… r-1, and G_{i+1}/G_{i} is Abelian

Let’s apply Galois fundamental theorem to G_{r} = G (K/F is Galois):

F = F_{0} = K^{Gr} ⊆ [G_{r-1} ◁ G_{r} = G ⇒ **K ^{Gr-1}/K^{Gr}(= F) is Galois** with Galois group

K^{Gr-1} ⊆ K^{Gr-2}

Apply the Fundamental Galois Theorem to K/K^{Gr-1}, K/K^{Gr-1} is Galois, and the Galois group is G_{r-1}, G_{r-2} ◁ G_{r-1} ⇒ **K ^{Gr-2}/K^{Gr-1} is Galois** (+Galois+) with Galois group

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

F = F_{0} = K^{Gr} ⊆ K^{Gr-1} ⊆ K^{Gr-2} ⊆ K^{r-3} ⊆ ··· ⊆ K^{G2} ⊆ K^{G1} ⊆ K^{G0} = 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 = L_{0} ⊆ L_{1} ⊆ ··· ⊆ L_{n} such that α ∈ L_{n} and each L_{i}/L_{i-1} is Abelian, i.e., Galois and the Galois group is Abelian.

⇐ )

Suppose f is solvable ⇒ Every root of f, say α_{1}, α_{2}, ···, α_{n} live in a splitting field, say K = F(α_{1}, α_{2}, ···, α_{n}), is solvable, too ⇒ Each α_{i} is contained in a radical extension L_{i} of F.

Claim: we can find an extension M, L ⊆ M such that M/F is Galois and F = M_{0} ⊆_{Abelian} M_{1} ⊆_{Abelian} ⊆ M_{2} ⊆ _{Abelian} ··· ⊆_{Abelian} M_{l} = M

Let L be the composite of L_{1}, L_{2}··· L_{n} ⇒ [Recall L_{1}/F, L_{2}/F radical extensions ⇒ the composite L_{1}L_{2}/F is radical, too.] L/F is radical, so ∃ a tower of fields

F = F_{0} ⊆_{s.r.} F_{1} ⊆_{s.r.} ⊆ F_{2} ⊆_{s.r.} ··· ⊆_{s.r.} F_{m} = L, L ⊇ K = F(α_{1}, α_{2}, ···, α_{n})

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 = F_{0} ⊆_{s.r.} F_{1} ⊆_{s.r.} ⊆ F_{2} ⊆_{s.r.} ··· ⊆_{s.r.} F_{m} = 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, F_{i-1} ⊆_{s.r.} F_{i} ⇒ F_{i} = F_{i-1}(α_{i}) where α_{i}^{di} ∈ F_{i-1}. Let ξd_{1}, ξd_{2}, ···, ξd_{m} be the primitive d_{1}-nth, d_{2}-nth, ··· d_{m}-nth roots of unity respectively, and F’ = F(ξd_{1}, ξd_{2}, ···, ξd_{m}). 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), ξ^{n} = 1 ∈ F].

If you adjoined ξd_{1}, the primitive d_{1}-nth root of unity, F(ξd_{1}) contains all the d_{1}-nth roots of unity, and therefore F’ is the splitting field of (x^{d1}-1)(x^{d2}-1)···(x^{dm}-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 L_{1}/F, L_{2}/F radical extensions ⇒ the composite L_{1}L_{2}/F is radical.L_{1}/F, L_{2}/F Galois extensions ⇒ the composite L_{1}L_{2}/F is Galois.] M = F’L = F(ξ_{d1}, ξ_{d2}, ··· ξ_{dm}, α_{1}, α_{2}, ··· α_{m}) is radical and Galois

Notice that we have a tower
F ⊆_{Abelian} F(ξ_{d1}) [… 🚀]

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

[… 🚀] ⊆_{Abelian} F(ξ_{d1}, ξ_{d2}) ⊆ ··· ⊆_{Abelian} F(ξ_{d1}, ξ_{d2}, ··· ξ_{dm}) = F’⊆_{Abelian} F’(α_{1}) ⊆_{Abelian} F’(α_{1}, α_{2}) ⊆_{Abelian} ··· ⊆_{Abelian} F’(α_{1}, α_{2}, ··· α_{m}) = M ⊇ L ⊇ K

F’ = F(ξ_{d1}, ξ_{d2}, ··· ξ_{dm}) ⊆ F’(α_{1}) is Kummer, F’ contains the primitive d_{1}-nth root of unity, and it is a radical extension ⇒ Kummer ⇒ Cyclic ⇒ Abelian.

So let’s rename it, F = M_{0} ⊆_{Abelian} M_{1} ⊆_{Abelian} ··· ⊆_{Abelian} M_{l} = M ⊇ L ⊇ K

**Claim: Gal(M/F) is solvable**.

M_{1}/M_{0} 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 M_{1} is Galois over M_{0} = F, then **Gal(M/M _{1}) ◁ Gal(M/F) and Gal(M/F)/Gal(M/M_{1}) ≋ Gal(M_{1}/F)**. Futhermore, in our particular case, Gal(M

Gal(M/F) ⊇_{Abelian} Gal(M/M_{1}) ⊇_{Abelian} Gal(M/M_{2})

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.

We can use the same reasoning as before, M_{2}/M_{1} Abelian ⇒ Galois ⇒ **Gal(M/M _{2}) ◁ Gal(M/M_{1}) and Gal(M/M_{1})/Gal(M/M_{2}) ≋ Gal(M_{2}/M_{1})**. Futhermore, Gal(M

Gal(M/F) ⊇_{Abelian} Gal(M/M_{1}) ⊇_{Abelian} Gal(M/M_{2}) ⊇_{Abelian} Gal(M/M_{3}) ⊇ ·· · ⊇_{Abelian} Gal(M/M_{l-1}) ⊇_{Abelian} Gal(M/M_{l}) = {id}, and therefore **Gal(M/F) is solvable**

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 S_{n.}

Therefore, the Galois group of f (deg(f) = n ≤ 4) is a subgroup of S_{1}, S_{2}, S_{3} or S_{4}, 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.∎

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.