Try
The nonabelian group S₃ has a normal subgroup A₃ of order three. Its quotient has order two, and both A₃ and S₃/A₃ are abelian. Thus a noncommutative group can be assembled from abelian pieces. We will make that statement precise before using it in field theory.
Permutation products act right to left. Compare xy and yx for x=(12), y=(23). Their discrepancy is measured by the commutator [x,y]=xyx⁻¹y⁻¹. Here it equals (132), a nonidentity three-cycle. Check its action on each of 1,2,3. In general [x,y]=1 exactly when xy=yx.
Predict
If we force every commutator to become the identity in a quotient, will the quotient be abelian? Conversely, must the kernel of every map to an abelian group contain every commutator? Predict both directions before proving them. The collection of individual commutators need not itself be a subgroup, so we use the subgroup they generate.
Experiment
Let G′ be the subgroup generated by all [x,y], called the derived or commutator subgroup. Define G⁽⁰⁾=G and G⁽ᵏ⁺¹⁾=(G⁽ᵏ⁾)′. A group is solvable when this derived series reaches the identity after finitely many steps.
| Group | First derived subgroup | Next derived subgroup | Conclusion |
|---|
| C₅ | 1 | 1 | Abelian, hence solvable |
| S₃ | A₃ | 1 | Nonabelian but solvable |
| A₅ | A₅ | A₅ | Not solvable; proof below |
For S₃, the sign map shows G′⊆A₃ because its target is abelian. The displayed commutator generates A₃, so equality holds. Since A₃ is cyclic, its derived subgroup is trivial. “Solvable” is a structural property; it does not mean that every calculation in the group is easy.
Understand
The universal abelian quotient
Conjugation sends [x,y] to [gxg⁻¹,gyg⁻¹], so it preserves the generated subgroup G′. Thus G′ is normal. In G/G′ every commutator is trivial, so any two cosets commute. If N is normal and G/N is abelian, each [x,y] maps to identity, so G′⊆N. Consequently G′ is the smallest normal subgroup with abelian quotient. Every homomorphism from G to an abelian group factors through G/G′: define the factor on gG′ by the original image of g; kernel containment proves representative independence.
Subgroups, quotients and extensions
For H≤G, induction gives H⁽ᵏ⁾⊆G⁽ᵏ⁾ because the commutator formula is unchanged in a subgroup. Hence subgroups of solvable groups are solvable. If π:G→Q is surjective, π(G′)=Q′: images of commutators are commutators, and every commutator of Q has lifts. Induction gives π(G⁽ᵏ⁾)=Q⁽ᵏ⁾, so quotients of solvable groups are solvable.
Suppose N is normal, N⁽ᵃ⁾=1, and (G/N)⁽ᵇ⁾=1. The quotient identity gives G⁽ᵇ⁾⊆N. Taking another a derived steps yields G⁽ᵃ⁺ᵇ⁾⊆N⁽ᵃ⁾=1. Thus an extension of a solvable group by a solvable normal subgroup is solvable. Conversely, if G is solvable, both N and G/N are solvable by the preceding arguments.
Equivalently, G is solvable if it has a finite chain G=G₀⊇G₁⊇⋯⊇Gₘ=1 with Gᵢ₊₁ normal in Gᵢ and each quotient Gᵢ/Gᵢ₊₁ abelian. A terminating derived series supplies such a chain. In the reverse direction, Gᵢ′⊆Gᵢ₊₁, so induction gives G⁽ⁱ⁾⊆Gᵢ and eventually G⁽ᵐ⁾=1. Each subgroup need only be normal in its predecessor, not necessarily in all of G.
S₄ still has abelian pieces
Let V₄ consist of identity and (12)(34), (13)(24), (14)(23). Multiplying distinct nonidentity elements gives the third; each is its own inverse, so V₄ is abelian. Conjugation permutes these three double transpositions, making V₄ normal in A₄ (indeed in S₄). The chain S₄⊇A₄⊇V₄⊇1 has quotients of orders two, three and four. The first two are cyclic by their prime orders; the last is V₄ and abelian. This proves S₄ solvable without claiming S₄ itself is abelian or that every quotient of order four is cyclic.
A₅ has no proper nontrivial normal subgroup
Even permutations of five letters have types: identity; three-cycle; product of two disjoint transpositions; or five-cycle. Their counts are 1,20,15,24. For three-cycles choose three letters and one of two cyclic orders, giving 10·2=20. For double transpositions choose the fixed letter and one of three pairings of the others, giving 5·3=15. There are 4!=24 five-cycles.
We need A₅ conjugacy classes, not only cycle types. A three-cycle has centralizer of size six in S₅: its powers and a possible swap of the two unused letters. Only the three powers are even, so its A₅ class size is 60/3=20. A double transposition has S₅ centralizer of size eight: one may exchange its two two-letter orbits and independently swap within either. An odd commuting transposition pairs the even and odd centralizer elements, so its A₅ centralizer has size four and its class size is 15. Finally, the centralizer of a five-cycle consists of its five powers: a commuting permutation is determined by the image of one letter along the cycle. All powers are even, giving A₅ class size 12. The 24 five-cycles therefore form two classes of size 12.
A normal subgroup must be a union of whole conjugacy classes and contain identity. Its possible orders are 1 plus a subset sum of 20,15,12,12. None except 1 and 60 divides 60; check the finite list before applying Lagrange. Thus A₅ is simple. It is nonabelian: for x=(123), y=(345), xy sends 3 to 4 whereas yx sends 3 to 1. Its derived subgroup is therefore nontrivial and normal, hence equals A₅ by simplicity. The derived series never shrinks. A₅ is not solvable, and neither is S₅, since it contains the nonsolvable subgroup A₅.
Apply
Repair two claims: “a nonabelian group is not solvable” and “every finite group is solvable because it has a finite composition series.” S₃ disproves the first. The second omits the requirement that the successive pieces be abelian; A₅ supplies a simple nonabelian piece.
For a changed example, prove S₃×C₅ is solvable. Compute a commutator coordinatewise and show its derived subgroup is A₃×1, then its next derived subgroup is trivial. Explain why a quotient map onto A₅ would instead prove a group nonsolvable.
Check the product and quotient reasoning
[(g,a),(h,b)]=([g,h],1), because C₅ is abelian. These commutators generate exactly A₃×1, not merely a subgroup of it: lift the generating three-cycle commutator from S₃. The next derived subgroup is trivial. In contrast, any quotient of a solvable group is solvable, so a surjection onto A₅ is impossible for a solvable domain. The kernel does not need to be trivial for this obstruction.
Continue
These results are prerequisites for the relation between radical towers and Galois groups. They do not alone prove a polynomial is solvable by radicals or identify its Galois group. Continue to solvability by radicals for the roots-of-unity lemmas, radical extensions and both directions of that theorem. Return to group proof synthesis for targeted revision.
