Two maps that undo each other
Fix a Galois extension L/K with group G = Gal(L/K). There are two natural maps. Up: a subgroup H ⊆ G goes to its fixed field L^H = { x ∈ L : σx = x for all σ ∈ H }, an intermediate field K ⊆ L^H ⊆ L. Down: an intermediate field F goes to Gal(L/F), the automorphisms fixing F pointwise, a subgroup of G. The fundamental theorem says these two are mutually inverse bijections.
Both maps reverse inclusions — a bigger subgroup fixes fewer elements, so a bigger H gives a smaller L^H. This makes the correspondence an order-reversing bijection between two lattices, an instance of a Galois connection that happens to be a perfect duality. The bottom of one lattice (the trivial subgroup, fixing everything: L^{1}=L) matches the top of the other, and vice versa.
Why it is a bijection: Artin
The hard direction is that you never lose information passing to the fixed field. Artin's theorem does the heavy lifting: if H is a finite group of automorphisms of L, then L over its fixed field L^H is Galois with Gal(L/L^H) = H, and crucially [L : L^H] = |H|. So the subgroup H is reconstructed exactly as Gal(L/L^H). Going the other way, |Gal(L/F)| = [L:F] because L/F is Galois too, and L^{Gal(L/F)} = F. The two round-trips both close.
Normal matches normal
The last clause is the prettiest. An intermediate field F is normal over K exactly when its subgroup H = Gal(L/F) is a normal subgroup of G, and then restriction gives Gal(F/K) = G/H. The word ‘normal’ matching in two senses is not a coincidence: σ(F)=F for every σ ∈ G is the field-theoretic statement, and σHσ⁻¹ = H is its translation into the group. The quotient G/H is born as the Galois group of the bottom piece.
L = Q(sqrt 2, sqrt 3), G = Gal(L/Q) = {e, a, b, ab} = Z/2Z x Z/2Z.
a: sqrt2 -> -sqrt2, sqrt3 -> +sqrt3
b: sqrt2 -> +sqrt2, sqrt3 -> -sqrt3
Subgroups of G (all normal, since G is abelian) and their fixed fields:
{e} <-> L = Q(sqrt2, sqrt3) [L:F] = 4
{e, a} <-> Q(sqrt3) fixed by a
{e, b} <-> Q(sqrt2) fixed by b
{e, ab} <-> Q(sqrt6) ab fixes sqrt2*sqrt3
G <-> Q [F:Q] = 1
Check the index/degree mirror for F = Q(sqrt2), H = {e, b}:
[L:F] = |H| = 2 [F:Q] = [G:H] = 4/2 = 2. OK.
Every subgroup is normal in G, so every one of the three
quadratic subfields is normal over Q, with Gal(F/Q) = G/H = Z/2Z.