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 Sp 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 NH, the normalizer of <x> in H.

Proof: Suppose on the contrary that NH = <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, ..., xn, where n = order(H)/p, for the p-groups in H. Pick αj, j ∈ {2, ..., n} such that zj = xjαj x1 fixes 1. The zj's are order(H)/p - 1 distinct non-identity elements not of order p.

But more can easily be found.   x2α2 x1 doesn't fix everything; suppose it doesn't fix  i. Choose β2 such that x2β2 x1 fixes  i. x2β2 x1 is distinct from all the zj's.

So NH must properly contain <x>.

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

Proof:  Proof by induction on i. Suppose that <x> < Hi. By Lemma 1, <x> is properly contained in Ni, the normalizer of <x> in Hi. Let y ∈ Ni - <x>.   y x y-1 x-1 ∈ <x>, but y x y-1 x-1 ≠ 1 since the centralizer of <x> in Sp is <x>. So x ∈ <y x y-1 x-1>. So x is in the commutator subgroup of Hi. Since Hi/Hi+1 is abelian, <x> ≤ Hi+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 Sp.

Proof: Let 1 = H0 < H1 < ... < Hn = 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 Hi ≤ N.  Pick h ∈ Hi+1.  Since Hi is normal in Hi+1,  h <x> h-1 ≤ Hi. But then h <x> h-1 = <x>, since there is only one p-group in Hi. So <x> is normal in Hi+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 Sp; each p-group has p-1  p-cycles, so there are (p - 2)!  p-groups in Sp. So the order of N is p (p - 1).

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