__Theorem__ Let F be a field of characteristic 0, containing all
roots of 1. Let

__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

If

__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>) =

Choose generators _{1}, ..., x_{n},_{j}, _{j} =
x_{j}^{αj} x_{1}_{j}'s are

But more can easily be found.
_{2}^{α2} x_{1}_{2} such that
_{2}^{β2} x_{1} fixes
i.
_{2}^{β2} x_{1} _{j}'s.

So N_{H} must properly contain <x>.

__Lemma 2__: Let _{n} < H_{n-1} <
... < H_{1} = G_{i-1} / H_{i} is abelian). Then _{n-1}.

__Proof:__ Proof by induction on i. Suppose that
_{i}._{i}, the normalizer of <x> in
H_{i}. Let _{i} - <x>.
^{-1} x^{-1} ∈ <x>,^{-1} x^{-1} ≠ 1_{p} is <x>.
So ^{-1} x^{-1}>._{i}. Since
H_{i}/H_{i+1} is abelian, _{i+1}

__Lemma 3:__ G is contained in N, the normalizer of <x> in
S_{p}.

__Proof:__ Let _{0} < H_{1} < ...
< H_{n} = G be a solvable chain for G. The proof is by
induction on the index of the H's. By Lemma 2, _{1},_{1} is abelian,
_{1} ≤ N.

Suppose that _{i} ≤ N._{i+1}._{i} is normal in
H_{i+1}, ^{-1} ≤
H_{i}.^{-1} =
<x>_{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};
each p-group has _{p}. So the order of N is

back to math page