Automorphisms that fix the floor
Start concretely. Given a field extension L/K, the Galois group in the loose sense is Aut(L/K), the group of field automorphisms of L that fix every element of K. Each such σ permutes the roots of any polynomial with coefficients in K, because applying σ to p(α)=0 gives p(σα)=0. So an automorphism cannot send a root to a non-root: the Galois group lives entirely inside the symmetries of root-sets. That single observation is the engine of the whole subject.
But Aut(L/K) can be embarrassingly small. Take K=Q and L=Q(∛2). Any σ must send ∛2 to a root of x³−2, and the other two roots are complex while L sits inside R. So σ has nowhere to send ∛2 except ∛2 itself: Aut(L/K) is trivial even though [L:K]=3. The group is too small because L is not normal — it does not contain all the roots of the minimal polynomial it started one.
Counting by embeddings
Here is the clean way to count automorphisms. Fix an algebraic closure of K and count K-embeddings of L into it. For a simple extension K(α), an embedding is determined by where it sends α, and α can go to any root of its minimal polynomial m. So the number of embeddings equals the number of distinct roots of m. If m has degree n and no repeated roots, you get exactly n embeddings.
Building up a tower one generator at a time, the number of embeddings multiplies — it is itself multiplicative across a tower, mirroring the tower law for degrees. The count is called the separable degree [L:K]_s, and it is always at most [L:K]. The two agree precisely when every minimal polynomial along the way has distinct roots, i.e. when the extension is separable.
Count Aut(L/Q) for L = Q(sqrt 2, sqrt 3). Degrees: [L:Q] = [L:Q(sqrt 2)] * [Q(sqrt 2):Q] = 2 * 2 = 4. Embeddings of L into C, by choosing images of the generators: sqrt 2 -> +sqrt 2 or -sqrt 2 (roots of x^2 - 2) sqrt 3 -> +sqrt 3 or -sqrt 3 (roots of x^2 - 3) All 2 * 2 = 4 combinations land inside L itself (L is normal), so every embedding is an automorphism. => |Aut(L/Q)| = 4 = [L:Q]. L/Q is Galois. The four maps, by sign pattern (s2, s3): e = (+, +) identity a = (-, +) flips sqrt 2 b = (+, -) flips sqrt 3 ab = (-, -) flips both Each has order 2, and the group is Z/2Z x Z/2Z (Klein four).
The two conditions, side by side
Normal fixes the first failure: L/K is normal iff L is a splitting field over K, so every embedding into the closure has image L — every embedding *is* an automorphism. Separable fixes the second: it forces [L:K]_s = [L:K]. Put them together and |Aut(L/K)| = [L:K]_s = [L:K]. That equality, |Gal(L/K)| = [L:K], is the operational definition of Galois we will lean on for the rest of the track.