A perfect dictionary
For a Galois extension K/F, the fundamental theorem of Galois theory gives a one-to-one correspondence: each field E sitting between F and K matches exactly one subgroup H of the Galois group — namely the automorphisms that fix every element of E. The correspondence reverses inclusions: bigger fields ↔ smaller subgroups.
And the bookkeeping is exact: [K : E] = |H| while [E : F] equals the index of H in the group. By Lagrange's theorem these multiply to |Gal(K/F)|, mirroring the tower law for degrees. Geometry of fields becomes arithmetic of groups — that is the magic.
Solvable by radicals = solvable group
To “solve by radicals” means to reach the roots from F using only +, −, ×, ÷ and nth roots — exactly the quadratic-formula style of answer. Each radical you adjoin builds one rung of a tower of fields. Translating through the dictionary, that tower becomes a chain of subgroups, each normal in the next with abelian quotients. A group with such a chain is called solvable.
This is the heart of the matter: a polynomial is solvable by radicals if and only if its Galois group is a solvable group. The entire question of formulas has been converted into a question about the internal structure of one finite group.
The quintic falls
Degree 2, 3, 4 equations are solvable because their Galois groups (subgroups of S₂, S₃, S₄) are all solvable — which is why the classical quadratic, cubic, and quartic formulas exist. But a “generic” fifth-degree polynomial has Galois group S₅, the full symmetric group on 5 letters.
Why S₅ is NOT solvable:
S₅ ⊃ A₅ (the alternating group, 60 elements)
A solvable chain would need abelian quotients all the way down.
But A₅ is SIMPLE (no normal subgroups except {e} and A₅ itself)
and A₅ is NOT abelian.
So the chain S₅ ⊃ A₅ ⊃ ... jams: there is nowhere to go.
⇒ S₅ is not solvable
⇒ the general quintic has NO formula in radicals.This is not “we haven't found the formula yet.” It is a proof that no such formula can exist, for any finite combination of radicals. The specific polynomial x⁵ − x − 1 over Q has Galois group S₅ and genuinely cannot be solved in radicals. That is the unsolvability of the quintic — perhaps the most beautiful “impossible” theorem in algebra.