Galois Groups — Mathematics Knowledge System

Overview

The Galois group Gal(L/K) of a field extension L/K is the group of all field automorphisms of L that fix every element of K, under composition. When L is the splitting field of a polynomial f over K, Gal(L/K) acts faithfully by permuting the roots of f, so it can always be realized as a subgroup of a symmetric group S_n on the n roots — this concrete permutation picture is what makes Galois groups computable in practice. Identifying which subgroup of S_n a given polynomial's Galois group actually is (cyclic, dihedral, or the full S_n) is one of the central computational problems of Galois theory, since the group's structure directly answers questions about solvability by radicals and the shape of the subfield lattice.

Intuition

Every automorphism in a Galois group must send a root of f to another root of the same minimal polynomial — it can't send √2 to 3, say, because automorphisms preserve algebraic relations, and 3 doesn't satisfy x²−2=0. So each automorphism is completely determined by how it permutes the finitely many roots, which is exactly why Gal(L/K) sits naturally inside S_n. Not every permutation of the roots is realized, though — only those consistent with all algebraic relations among the roots (like α₁+α₂+α₃=0 from Vieta's formulas) survive as actual automorphisms, so the Galois group is typically a proper subgroup of S_n that encodes precisely which root-swaps are 'legal'. The size of the group equals the degree of the splitting field, giving a completely computable link between field-theoretic degree and group-theoretic order.

Formal Definition

Definition

Let L/K be a finite Galois extension (normal and separable) and let f(x) ∈ K[x] have splitting field L with distinct roots α₁,...,αₙ.

\operatorname{Gal}(L/K) = \{ \sigma : L \to L \mid \sigma \text{ field automorphism},\ \sigma|_K = \operatorname{id}_K \}

Galois group as automorphisms fixing K

\sigma(\alpha_i) \in \{\alpha_1,\ldots,\alpha_n\} \ \forall \sigma \in \operatorname{Gal}(L/K)

Automorphisms permute the roots of f

\operatorname{Gal}(L/K) \hookrightarrow S_n

Galois group embeds as a subgroup of the symmetric group on the roots

|\operatorname{Gal}(L/K)| = [L:K]

Order equals the degree of the splitting field extension

Properties

Transitive action on roots (irreducible f)

\text{If } f \text{ is irreducible over } K, \text{ then } \operatorname{Gal}(L/K) \text{ acts transitively on the roots of } f.

Cyclic Galois group for prime-degree cyclotomic-like cases

\operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \cong (\mathbb{Z}/p\mathbb{Z})^{\times} \quad \text{(cyclic of order } p-1\text{) for } \zeta_p \text{ a primitive } p\text{th root of unity}.

Galois group of a generic (Galois) cubic

\text{For an irreducible cubic with Galois splitting field, } \operatorname{Gal}(L/K) \text{ is either } \mathbb{Z}_3 \text{ (if the discriminant is a square in } K\text{) or } S_3 \text{ (otherwise).}

Subgroup–subfield correspondence

\text{Subgroups } H \leq \operatorname{Gal}(L/K) \text{ correspond bijectively (order-reversing) to intermediate fields } K \subseteq E \subseteq L, \text{ via } E = L^H.

Worked Examples

The roots of x³−2 are ∛2, ω∛2, ω²∛2 where ω = e^{2πi/3} is a primitive cube root of unity; the splitting field is L = ℚ(∛2, ω).
$L = \mathbb{Q}(\sqrt[3]{2}, \omega)$
As computed via the Tower Law in splitting-fields, [L:ℚ] = 3 × 2 = 6, so |Gal(L/ℚ)| = 6.
$[L:\mathbb{Q}] = 6 \implies |\operatorname{Gal}(L/\mathbb{Q})| = 6$
Automorphisms are determined by where ∛2 goes (any of the 3 roots) and where ω goes (ω or ω², since ω satisfies x²+x+1); this gives all 6 combinations, matching |S₃|=6.
$\sigma(\sqrt[3]{2}) \in \{\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}\},\quad \sigma(\omega)\in\{\omega,\omega^2\}$
Since the Galois group has order 6 and acts as all permutations of the 3 roots (it is non-abelian, as swapping roots and applying complex conjugation don't commute), it must be the full symmetric group S₃.
$\operatorname{Gal}(L/\mathbb{Q}) \cong S_3$

Answer: Gal(ℚ(∛2, ω)/ℚ) ≅ S₃, the full symmetric group on the 3 roots of x³−2, of order 6.

Practice Problems

Difficulty 6/10

What is |Gal(L/ℚ)| for L the splitting field of x⁴−2 over ℚ, given [L:ℚ] = 8, and why must Gal(L/ℚ) embed in S₄?

Difficulty 7/10

Explain why the Galois group of an irreducible quadratic x²+bx+c over a field K (char ≠ 2) is always either trivial or ℤ₂, and identify which case occurs.

Difficulty 8/10

Why is the Galois group of the splitting field of a 'generic' (random-coefficient) degree-n polynomial over ℚ expected to be the full symmetric group Sₙ?

Common Mistakes

Common Mistake

Assuming the Galois group of every degree-n polynomial is the full Sₙ.

Sₙ is only an upper bound; extra algebraic relations among the roots (as with x⁴−2, whose group is D₄, order 8, not S₄'s 24) can make the actual Galois group a proper subgroup.

Common Mistake

Thinking any permutation of the roots extends to an automorphism.

Only permutations respecting every algebraic relation among the roots (forced by the coefficients via Vieta's formulas, or by other identities) extend to genuine field automorphisms.

Quiz

Summary

Gal(L/K) is the group of field automorphisms of L fixing K, and it always embeds in Sₙ by permuting the n roots of a defining polynomial.
|Gal(L/K)| = [L:K] exactly, giving a direct numerical bridge between group order and field-extension degree.
Gal(ℚ(∛2,ω)/ℚ) ≅ S₃ (order 6) and Gal(ℚ(ζ₅)/ℚ) ≅ ℤ₄ illustrate that the actual group can be the full symmetric group or a much smaller cyclic group depending on hidden root relations.
The subgroup–subfield (Galois) correspondence lets every question about intermediate fields be answered by studying subgroups of the Galois group instead.
Identifying the exact Galois group (not just an upper bound of Sₙ) is the central computational problem linking a polynomial's structure to its solvability.

References

WebsiteWikipedia — Galois group
WebsiteWikipedia — Galois theory