Theorem Let F be a field of characteristic 0, containing all roots of 1. Let f(x) ∈ F(x) be an irreducible polynomial of prime degree p. Suppose f(x) is solvable by radicals over F. Let K be the splitting field of f(x). Then [K:F] divides p (p - 1).

Proof Converting it to a statement about groups, let G be Gal(K/F). G is a solvable subgroup of the symmetry group S_p on p elements, with order(G) divisible by p; we want to show that the order of G divides p (p - 1).

If p = 2 then order(G) = 1 or 2, so it's true. So assume p > 2. Let x be a p-cycle in G, <x> the subgroup generated by x.

Lemma 1: Let H be a subgroup of G which properly contains <x>. Then, <x> is properly contained in N_H, the normalizer of <x> in H.

Proof: Suppose on the contrary that N_H = <x>. Then the order of class(<x>) = order(H) / p. So there are order(H) (p - 1)/ p elements of order p in H. So there are order(H)/p - 1 non-identity elements not of order p in H.

Choose generators x₁, ..., x_n, where n = order(H)/p, for the p-groups in H. Pick α_j, j ∈ {2, ..., n} such that z_j = x_j^α_j x₁ fixes 1. The z_j's are order(H)/p - 1 distinct non-identity elements not of order p.

But more can easily be found.   x₂^α₂ x₁ doesn't fix everything; suppose it doesn't fix  i. Choose β₂ such that x₂^β₂ x₁ fixes  i. x₂^β₂ x₁ is distinct from all the z_j's.

So N_H must properly contain <x>.

Lemma 2: Let 1 = H_n < H_n-1 < ... < H₁ = G be a minimal solvable chain for G ( i.e. H_i-1 / H_i is abelian). Then <x> ≤ H_n-1.

Proof:  Proof by induction on i. Suppose that <x> < H_i. By Lemma 1, <x> is properly contained in N_i, the normalizer of <x> in H_i. Let y ∈ N_i - <x>.   y x y^-1 x^-1 ∈ <x>, but y x y^-1 x^-1 ≠ 1 since the centralizer of <x> in S_p is <x>. So x ∈ <y x y^-1 x^-1>. So x is in the commutator subgroup of H_i. Since H_i/H_i+1 is abelian, <x> ≤ H_i+1. This is why I think of this proof as shaking a tree. You shake <x> down the chain of subgroups.

Lemma 3: G is contained in N, the normalizer of <x> in S_p.

Proof: Let 1 = H₀ < H₁ < ... < H_n = G be a solvable chain for G.   The proof is by induction on the index of the H's. By Lemma 2, <x> ≤ H₁, and since H₁ is abelian, H₁ ≤ N.

Suppose that H_i ≤ N.  Pick h ∈ H_i+1.  Since H_i is normal in H_i+1,  h <x> h^-1 ≤ H_i. But then h <x> h^-1 = <x>, since there is only one p-group in H_i. So <x> is normal in H_i+1. So the lemma is proved.

It only remains to show that the order of N is p (p - 1). But that's easy: there are (p - 1)!  p-cycles in S_p; each p-group has p-1  p-cycles, so there are (p - 2)!  p-groups in S_p. So the order of N is p (p - 1).

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