advanced33 min studyLesson 160 in Mathematics

Solvable Groups and Solvability by Radicals

Translate radical formulas into solvable Galois groups, and distinguish the general quintic obstruction from special solvable examples.

On this page

What you’ll understand

  • Test solvability through normal series or derived subgroups
  • State the characteristic-zero radical criterion with the splitting-field group
  • Explain why some quintics are solvable even though the general quintic is not

A radical formula uses arithmetic and repeated extraction of integer roots. Can every root of a polynomial be placed inside a field obtained by finitely many such operations? In characteristic zero, Galois theory answers yes exactly when the splitting field has a solvable Galois group. Solvable has a precise group-theoretic meaning; it does not mean that a numerical algorithm can approximate the roots.

Peeling a group into abelian layers

A finite group is solvable when it admits a chain from the identity to the whole group such that each subgroup is normal in the next and every successive quotient is abelian. Equivalently, repeatedly take the subgroup generated by commutators; the derived series must eventually reach the identity. A commutator measures failure to commute, so this process asks whether all noncommutativity can be removed in finitely many abelian layers.

Abelian groups are solvable, but solvable groups need not be abelian. The symmetric group S3 has the normal subgroup A3, giving quotients C3 and C2. Its noncommuting permutations therefore still admit a two-layer reduction. By contrast, A5 is nonabelian and simple, so its commutator subgroup is the whole group: it cannot descend through a nontrivial abelian quotient. Consequently S5, which contains A5 as its derived subgroup, is not solvable.

G(0)=G,G(j+1)=[G(j),G(j)],G solvable⟺G(r)={1} for some rG^{(0)}=G,\qquad G^{(j+1)}=[G^{(j)},G^{(j)}],\qquad G\ \text{solvable}\Longleftrightarrow G^{(r)}=\{1\}\ \text{for some }r
The derived series gives a test equivalent to a normal series with abelian factors.↔ Scroll wide equations horizontally.

Why extracting roots creates group restrictions

A radical tower adjoins elements whose positive integer powers already belong to the preceding field. Individual steps need not be normal: adjoining the real cube root of two omits its complex companions. After adjoining suitable roots of unity, however, automorphisms of a normal radical step act by multiplying a chosen root by a root of unity. Those actions produce abelian, and after refinement cyclic, layers. Taking normal closures and organizing the layers is essential to the theorem.

In characteristic zero, a polynomial is solvable by radicals if and only if the Galois group of its splitting field is solvable. The converse constructs radical extensions after supplying roots of unity and resolving the solvable group into cyclic layers. This is an existence criterion for containment in a radical extension. It does not require the original splitting field itself to be presented by a radical tower without enlarging it, and it does not promise a short or aesthetically simple formula.

char⁡K=0:f solvable by radicals⟺Gal⁡(L/K) solvable,L=splitting field of f\operatorname{char}K=0:\quad f\ \text{solvable by radicals}\Longleftrightarrow\operatorname{Gal}(L/K)\ \text{solvable},\quad L=\text{splitting field of }f
The criterion concerns the Galois group of the full splitting field, with radical solvability understood inside a suitable extension.↔ Scroll wide equations horizontally.

The quintic obstruction has a precise scope

The generic degree-n polynomial has Galois group Sn over its rational-function coefficient field. For n at least five this group is not solvable, proving that no universal radical formula handles arbitrary coefficients. This does not say that every polynomial of degree five lacks radical solutions. The polynomial x to the fifth minus two plainly has roots built from a fifth root of two and fifth roots of unity; its Galois group is a solvable semidirect product of orders five and four.

To establish impossibility for a particular polynomial, determine enough of its Galois group to prove nonsolvability. Irreducibility alone does not suffice. Conversely, numerical approximation remains available for nonsolvable groups: radical expressibility restricts permitted exact operations, not accurate decimal computation. Keeping these distinctions separate prevents an algebraic impossibility theorem from becoming a mistaken claim of general computational impossibility.

A nonabelian cubic group that is still solvable

For x cubed minus two over Q, identify the splitting field, determine its Galois group, and explain the radical criterion through a normal series.

  1. Let a be the real cube root of two and zeta a primitive cube root of unity. The three roots are a, zeta a, and zeta squared a, so L = Q(a, zeta). Eisenstein at two gives [Q(a):Q] = 3. Since Q(a) is real and zeta is nonreal with quadratic minimal polynomial, adjoining zeta contributes degree two, giving [L:Q] = 6.

  2. The extension is a characteristic-zero splitting field, so its Galois group has six elements and acts faithfully on three roots. Therefore it is S3. The automorphism r sending a to zeta a while fixing zeta has order three; complex conjugation s fixes a and inverts zeta. Direct substitution gives s r s inverse = r inverse.

  3. The subgroup generated by r is A3, normal of order three. Its quotient in S3 has order two. The chain identity inside A3 inside S3 has abelian factors C3 and C2, establishing solvability. The roots are indeed radical expressions using a = cube root of two and zeta = (minus one plus square root of minus three)/2.

{1}◃A3◃S3,A3≅C3,S3/A3≅C2\{1\}\triangleleft A_3\triangleleft S_3,\qquad A_3\cong C_3,\qquad S_3/A_3\cong C_2
Abelian successive quotients, rather than commutativity of the entire group, are the decisive property.↔ Scroll wide equations horizontally.
The result

The splitting-field group is nonabelian S3 but solvable. Its normal series explains why noncommuting root symmetries remain compatible with radical expressions for all three roots.

Explore the model

Interactive lab

Interactive lab · vectors

Inspect a cyclic rotation before adding reflection

Set magnitude to one and compare angles 0, 120, and minus 120 degrees (equivalent to 240). Predict a return to the starting vector after three 120-degree increments. Then compare 120 and minus 120 degrees as a reflection pair. These planar operations illustrate the relations behind S3, but the lab does not compute a derived series, splitting field, or radical expression.

Inspect a cyclic rotation before adding reflectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forCyclic layers can sit inside a noncommutative group; the actual solvability conclusion requires a normal-series proof.

Check your understanding

Which conclusion follows from the nonsolvability of the generic quintic Galois group S5?

Try it yourself

Explain why the dihedral group of order eight is solvable, and state the consequence for a characteristic-zero polynomial whose splitting-field Galois group is that group.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the subgroup generated by a quarter-turn rotation. It has order four and is normal because reflections invert rotations.

Compare with the solution

Write D4 = <r,s> with r to the fourth = s squared = identity and srs inverse = r inverse. The rotation subgroup <r> is normal and cyclic of order four, and the quotient D4/<r> is cyclic of order two. Thus identity inside <r> inside D4 is a normal series with abelian factors, proving solvability. The characteristic-zero Galois criterion then guarantees that every root of the polynomial belongs to some radical extension of its base field. This is an existence conclusion; it does not by itself display the radicals.

Take it with you

  • Solvable groups can be nonabelian when their successive normal quotients are abelian
  • Radical solvability in characteristic zero is equivalent to solvability of the splitting-field group
  • The general quintic obstruction leaves special radical-solvable quintics and numerical methods available

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.