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Galois Groups as Permutation Groups, and the Inverse Problem

Read the Galois group off the roots: it acts faithfully and transitively on them, the discriminant detects the alternating subgroup, and resolvents pin down the group for cubics and quartics. Then meet the inverse Galois problem — which groups actually occur over Q.

The action on the roots

Let f be a separable polynomial of degree n over K with roots α_1,…,α_n in its splitting field L. Every σ ∈ Gal(L/K) permutes the roots, and because L is generated by them, σ is *determined* by that permutation. So Gal(L/K) embeds as a subgroup of the symmetric group S_n — this is the Galois group as a permutation group. Two structural facts: the action is faithful (no nonidentity σ fixes all roots) and, if f is irreducible, transitive (Gal can move any root to any other, since all roots share one minimal polynomial).

The discriminant detects A_n

Form δ = ∏_{i<j}(α_i − α_j). A transposition flips one factor's sign, so a permutation acts on δ by its sign: even permutations fix δ, odd ones negate it. Therefore Gal ⊆ A_n exactly when δ is fixed by the whole group, i.e. when δ ∈ K. The square D = δ² is the discriminant and always lies in K. So the test is clean: Gal ⊆ A_n iff D is a perfect square in K.

Cubic  f(x) = x^3 + px + q  (irreducible, separable, char != 2,3).
Discriminant:  D = -4p^3 - 27q^2.

Gal is transitive in S_3, so |Gal| is 3 or 6 => Gal = A_3 or S_3.
The discriminant decides which:

  D is a square in K   <=>  Gal = A_3 = Z/3Z   (cyclic, order 3)
  D is NOT a square    <=>  Gal = S_3          (order 6)

Examples over Q:
  x^3 - 3x + 1:  D = -4(-3)^3 - 27(1)^2 = 108 - 27 = 81 = 9^2.
      square => Gal = A_3 = Z/3Z.  (A 'cyclic cubic'.)
  x^3 - x - 1:   D = -4(-1)^3 - 27(-1)^2 = 4 - 27 = -23.
      not a square => Gal = S_3.

For a quartic one uses the RESOLVENT CUBIC instead:
  its splitting behavior over K separates the five transitive
  subgroups of S_4  (S_4, A_4, D_4, Z/4Z, V_4).
Cubics: the discriminant alone separates A_3 from S_3; quartics need a resolvent.

For quartics the discriminant is not enough — five transitive subgroups of S_4 must be told apart. The tool is the resolvent cubic, an auxiliary cubic whose roots are built symmetrically from pairs of the quartic's roots. How its roots lie in K (all, one, or none rational) combined with the discriminant test pins down which of S_4, A_4, D_4, C_4, V_4 you have. This resolvent method is the practical workhorse for low-degree Galois groups.

Which groups occur? The inverse problem

We have spent the track computing Gal(L/K) from a given extension. Turn it around: given a finite group G, is there an extension of Q with Gal ≅ G? This is the inverse Galois problem, and it is open in general — honestly, no one knows whether every finite group occurs over Q. What is known is substantial: every abelian group occurs (slice up a suitable cyclotomic field, using guide 4), every solvable group occurs (Shafarevich), the symmetric and alternating groups occur, and most finite simple groups have been realized one family at a time.

Step back and see the whole track in one breath: the Galois extension packages an extension as a group action, the correspondence makes fields and subgroups two views of one lattice, solvability of the group is exactly radical solvability of the equation, the three abelian families (cyclotomic, Kummer, finite) are where everything is explicitly computable, and the inverse problem asks the only question we cannot yet fully answer — which groups the construction can produce. That arc, from one extension to all of them, is graduate Galois theory.