Cyclotomic: the group is (Z/nZ)^×
Let ζ be a primitive nth root of unity and consider the cyclotomic extension Q(ζ)/Q. Any automorphism σ sends ζ to another primitive nth root, so σ(ζ)=ζ^a for some a coprime to n. The map σ ↦ a (mod n) is an injective homomorphism into the unit group (Z/nZ)^×, and it is onto because the nth cyclotomic polynomial Φ_n is irreducible over Q. Hence Gal(Q(ζ)/Q) ≅ (Z/nZ)^×, an abelian group of order φ(n).
n = 7: Phi_7(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1, irreducible.
[Q(zeta_7):Q] = phi(7) = 6.
Gal = (Z/7Z)^x = {1,2,3,4,5,6} under multiplication mod 7.
This group is cyclic of order 6, generated by 3:
3^1=3, 3^2=2, 3^3=6, 3^4=4, 3^5=5, 3^6=1 (mod 7).
Subgroup lattice => intermediate field lattice (FTGT):
unique subgroup of order 2 = {1,6} <-> the unique deg-3 subfield
unique subgroup of order 3 = {1,2,4} <-> the unique deg-2 subfield
The degree-2 subfield is Q(sqrt(-7)) (a Gauss-sum fact).
This is how one PROVES quadratic subfields of cyclotomic fields
exist: read them off the cyclic group (Z/7Z)^x.Kummer: cyclic extensions from nth roots
Now fix a base field K that already contains the nth roots of unity (so we are working ‘downstream’ of the cyclotomic step). Adjoin an nth root: L = K(α) with α^n = a ∈ K. This is a Kummer extension, and it is cyclic of degree dividing n. The reason is direct: any σ ∈ Gal(L/K) sends α to another nth root of a, which must be ζ^k α for some root of unity ζ in K. The assignment σ ↦ ζ^{k} is an injection into the group μ_n of nth roots of unity — a cyclic group — so Gal(L/K) is cyclic.
This is the exact converse used in guide 3: when the base has enough roots of unity, *cyclic of degree n is the same as adjoining an nth root*. Each cyclic layer of a radical tower is a Kummer extension. So cyclotomic and Kummer together manufacture exactly the abelian-by-pieces extensions that radicals can reach — the structural reason solvable groups and radical solvability coincide.
Finite fields: Frobenius does everything
Finite fields give the cleanest Galois theory of all. There is one field F_q with q=p^n elements for each prime power, and F_{p^n}/F_p is Galois. Its group is generated by the Frobenius map φ: x ↦ x^p, which fixes F_p (Fermat: a^p=a) and has order exactly n. So the Galois group of a finite field is cyclic, Gal(F_{p^n}/F_p) ≅ Z/nZ, generated by φ.
F_{2^6} / F_2, Frobenius phi(x) = x^2, ord(phi) = 6.
Gal = <phi> = Z/6Z.
FTGT for a cyclic group Z/6Z: subgroups <-> divisors of 6.
subgroup order fixed field degree over F_2
<phi> = whole 6 F_2 1
<phi^2> 3 F_4 = F_{2^2} 2
<phi^3> 2 F_8 = F_{2^3} 3
<phi^6> = {1} 1 F_{2^6} 6
So F_{2^d} sits inside F_{2^6} iff d | 6.
d in {1,2,3,6}: yes. d = 4: NO, since 4 does not divide 6.
The subfield lattice of finite fields is literally the
divisibility lattice of the exponents -- a one-line consequence
of the Galois group being cyclic.