The Galois Group

Look at the field \mathbb{C}. It has a hidden symmetry you have used for years without naming it: complex conjugation, a + bi \mapsto a - bi. It scrambles \mathbb{C}, yet it leaves every real number untouched and — this is the magic — it respects both arithmetic operations: \overline{z + w} = \bar z + \bar w and \overline{zw} = \bar z\, \bar w. A shuffle of a field that preserves its addition and multiplication and fixes a chosen subfield is exactly what Galois built his entire theory on. Collect all such shuffles and you get a group — the Galois group — and it turns out to encode, in the language of symmetry, everything about how the roots of a polynomial relate to one another.

This is the idea that ties the whole module together. An algebraic question — can this polynomial be solved by radicals? is this length constructible? — is translated into a question about a finite group, where it becomes answerable. This page defines that group and shows it acting.

Automorphisms that fix the base

Fix a field extension E/F. An automorphism of E is a bijection \sigma : E \to E that is a field homomorphism — it respects the arithmetic:

\sigma(a + b) = \sigma(a) + \sigma(b), \qquad \sigma(ab) = \sigma(a)\,\sigma(b).

We want the automorphisms that leave the base field alone. Say \sigma fixes F if \sigma(a) = a for every a \in F. The collection of all automorphisms of E that fix F is the Galois group, written \mathrm{Gal}(E/F), with composition of maps as the group operation.

The central insight: automorphisms permute roots

Here is why the Galois group is so powerful. Let \alpha \in E be a root of a polynomial f with coefficients in F, say f(\alpha) = \alpha^2 - 2 = 0. Apply \sigma \in \mathrm{Gal}(E/F). Because \sigma respects arithmetic and fixes the coefficients of f (they live in F),

f(\sigma(\alpha)) = \sigma\big(f(\alpha)\big) = \sigma(0) = 0.

So \sigma(\alpha) is also a root of f! An automorphism can only send a root to another root of the same polynomial. Since \sigma is a bijection, it permutes the finite set of roots. This gives a group action of \mathrm{Gal}(E/F) on the roots, and — because an automorphism of F(\text{roots}) is pinned down by where it sends the roots — the action is faithful, embedding the Galois group into a symmetric group S_n. A subtle field extension becomes a concrete group of permutations you can list.

Three worked Galois groups

Complex over real, \mathrm{Gal}(\mathbb{C}/\mathbb{R}). An automorphism of \mathbb{C} fixing \mathbb{R} must send i (a root of x^2 + 1) to another root, so i \mapsto i or i \mapsto -i. Two automorphisms: the identity and conjugation. Hence \mathrm{Gal}(\mathbb{C}/\mathbb{R}) = \{\,\mathrm{id}, \text{conj}\,\} \cong \mathbb{Z}/2\mathbb{Z}, matching [\mathbb{C} : \mathbb{R}] = 2.

A single square root, \mathrm{Gal}(\mathbb{Q}(\sqrt 2)/\mathbb{Q}). An automorphism sends \sqrt 2 to a root of x^2 - 2, i.e. \pm\sqrt 2. So there are two: the identity, and \sigma : \sqrt 2 \mapsto -\sqrt 2 (which sends a + b\sqrt 2 \mapsto a - b\sqrt 2). The group has order 2, again equal to the degree.

Two independent square roots, \mathrm{Gal}(\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q}). Now \sqrt 2 \mapsto \pm\sqrt 2 and independently \sqrt 3 \mapsto \pm\sqrt 3. Four combinations, each a genuine automorphism — so the group has order 4, matching [\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}] = 4. Every non-identity element squares to the identity (flipping a sign twice restores it), so this is the Klein four-group:

\mathrm{Gal}(\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q}) \cong \mathbb{Z}/2 \times \mathbb{Z}/2 = V_4.

Three extensions, three groups — \mathbb{Z}/2, \mathbb{Z}/2, and V_4 — each read off simply by asking "where can the roots go?"

The commonest misconception is to imagine an automorphism can send \sqrt 2 anywhere it likes — to \sqrt 3, to 5, to some random number. It cannot. Two iron constraints bind it. First, \sigma fixes \mathbb{Q}, and second, it respects the relation (\sqrt 2)^2 = 2, so \sigma(\sqrt 2)^2 = \sigma(2) = 2 — meaning \sigma(\sqrt 2) must itself square to 2. The only such numbers are \pm\sqrt 2. An automorphism may only move a root to a conjugate root of its own minimal polynomial, nowhere else.

A second, deeper trap: for a non-normal extension the Galois group can be smaller than the degree. Take \mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}, of degree 3. Its only real cube root of 2 is \sqrt[3]{2} itself; the other two roots of x^3 - 2 are complex and do not lie in the field. An automorphism must send \sqrt[3]{2} to a root inside the field — but there is only one — so \sigma is forced to be the identity. Thus \mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}) is trivial, and here \lvert \mathrm{Gal} \rvert = 1 < 3 = [E : F]. Equality of order and degree is a privilege of Galois (normal, separable) extensions only.

Because groups are finite and computable where fields are unwieldy. The crown jewel — the Fundamental Theorem of Galois Theory — says the subfields sitting between F and E correspond exactly, and order-reversingly, to the subgroups of \mathrm{Gal}(E/F). A tangled lattice of fields becomes a tidy lattice of subgroups you can draw on a page. This dictionary is what let Galois prove that the general quintic has no formula in radicals: the relevant group is S_5, which is not "solvable," and solvability of the group is precisely solvability by radicals. A teenager's idea, still running the whole show.