Abstract Algebra
Solvable Groups
In 1824, Niels Abel proved that no formula exists for solving degree-5 equations. Galois explained WHY: the symmetry group of such an equation is not solvable. This connection between group theory and algebra is one of the greatest conceptual breakthroughs in mathematical history.
- Numerical root-finding algorithms (Newton's method, bisection) are necessary precisely because no analytic formula exists for polynomials of degree ≥ 5
- In Galois theory over p-adic fields: solvability of Galois groups is linked to solvability of Diophantine equations - an active research area
Предварительные знания
The Commutator and Derived Series
The **commutator subgroup** (derived subgroup) G' = [G,G] is the subgroup generated by all commutators [g,h] = ghg⁻¹h⁻¹. The commutator measures how much g and h 'fail to commute': [g,h] = e if and only if gh = hg. G/G' is the largest abelian quotient of G (i.e., G' is the smallest normal subgroup with abelian quotient). **Derived series:** G = G⁽⁰⁾ ⊇ G⁽¹⁾ = G' ⊇ G⁽²⁾ = (G')' ⊇ ..., where G⁽ⁿ⁺¹⁾ = [G⁽ⁿ⁾, G⁽ⁿ⁾].
**G/G' is the maximal abelian quotient.** Every homomorphism G → A (A abelian) factors through G/G'. This makes G' a 'measure of non-abelianness'. If G is abelian, G' = {e} and G/G' = G. The more 'complex' G is, the 'larger' G' and the 'smaller' G/G'.
The **lower central series** (another tool): G = G₀ ⊇ G₁ = [G,G₀] ⊇ G₂ = [G,G₁] ⊇ ..., where Gₙ₊₁ = [G, Gₙ]. Groups where this series terminates at {e} are called **nilpotent** (a stronger condition than solvability). p-groups are nilpotent → solvable.
What is the commutator subgroup [S₃, S₃] of S₃?
Solvable Groups: Definition
**Definition:** A group G is **solvable** if its derived series terminates at the identity: G = G⁽⁰⁾ ⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇ ... ⊇ G⁽ⁿ⁾ = {e}. Equivalently: G has a **subnormal series** G = G₀ ⊇ G₁ ⊇ ... ⊇ Gₖ = {e} with each Gᵢ₊₁ ◁ Gᵢ and each quotient Gᵢ/Gᵢ₊₁ abelian. Intuition: a solvable group is 'built' from abelian 'bricks' via extensions.
**Feit-Thompson theorem (1963):** Every finite group of odd order is solvable. The proof is 255 pages! This means: the only finite simple non-abelian groups have even order. Corollary: every finite simple non-abelian group contains an element of order 2 (an 'embedded' Z₂).
**The Jordan-Hölder theorem.** A subnormal series G = G₀ ⊇ G₁ ⊇ ... ⊇ Gₖ = {e} with simple quotients Gᵢ/Gᵢ₊₁ is called a **composition series**. The Jordan-Hölder theorem: all composition series of G have the same length and the same factors (up to reordering). This is the 'prime factorization' for groups.
Is the group D₅ (dihedral group of order 10, symmetries of a regular pentagon) solvable?
The Abel-Ruffini Theorem and Galois's Formula
**Central theorem of Galois theory:** The equation f(x) = 0 (with coefficients in F) is **solvable by radicals** if and only if the Galois group Gal(f) is solvable. **Abel-Ruffini theorem:** The 'generic' equation of degree n ≥ 5 has no radical formula. Proof: the Galois group of the generic degree-n polynomial is Sₙ. For n ≥ 5: S₅ ⊇ A₅ ⊇ A₅ ⊇ ... - the derived series does not terminate. Sₙ is not solvable → no formula.
**Why radicals ↔ solvable groups?** Adjoining an nth root √[n]{a} to a field F corresponds to a **cyclic extension** (Galois extension with group Zₙ). A radical formula is a tower of such extensions. A tower of cyclic extensions → solvable group (extensions of solvable by solvable are solvable, and cyclic groups are abelian → solvable). Conversely: Galois proved that solvability of the group guarantees the existence of such a tower.
**Degree-5 equations solvable by radicals.** The equation x⁵ − 5x + 12 = 0 has Galois group D₅ (order 10, solvable) and is solvable by radicals. The formula is complex but finite. Solvability depends on the specific polynomial, not just its degree.
Degree-5 equations are never solvable by radicals
The Abel-Ruffini theorem says there is no GENERAL FORMULA for all degree-5 equations. Specific equations may be solvable if their Galois group is solvable.
Example: x⁵ − 1 = 0 (roots of unity) is solved by x = e^{2πik/5} = cos(2πk/5) + i·sin(2πk/5) - these are radicals. The Galois group of this equation is Z₄, which is solvable.
An equation f(x) = 0 has Galois group A₄ (order 12). Is it solvable by radicals?
Key Ideas
- Commutator subgroup G' = [G,G]: generated by all commutators; G/G' is the maximal abelian quotient
- Derived series: G⁽⁰⁾ ⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇ ..., where G⁽ⁿ⁺¹⁾ = [G⁽ⁿ⁾,G⁽ⁿ⁾]
- G solvable ⟺ derived series terminates at {e} ⟺ subnormal series with abelian quotients exists
- All p-groups, S₂, S₃, S₄ are solvable; A₅, S₅, Sₙ (n≥5) are not
- Galois theorem: f solvable by radicals ⟺ Gal(f) is solvable
- Abel-Ruffini: generic polynomial of degree n≥5 has no radical formula (Gal = Sₙ is not solvable)
Further Paths
Solvable groups are the 'safe world' of algebra: their structure is completely understood. Beyond them lie finite simple groups, the Monster, and the 10,000-page classification theorem.
- Galois Theory: Introduction — Solvability of the Galois group ↔ solvability by radicals - the central theorem connecting both lessons
- Group Actions — The proof of the Abel-Ruffini theorem uses the Galois group's action on roots and the fixed-field theorem
Вопросы для размышления
- Prove that A₄ is solvable by finding an explicit subnormal series with abelian quotients. Show that A₄ is not nilpotent.
- Why does the Feit-Thompson theorem (all groups of odd order are solvable) have enormous implications for the classification of finite simple groups?
- Find a specific degree-5 equation that is solvable by radicals and explain why its Galois group is solvable.