"Galois subfields of real radical extensions are at most quadratic" [closed]

Question

I was looking at the following page that attempts to prove that any Galois extension of a subfield $F$ of $\mathbb R$ contained in $F(\sqrt[n]{a})$ for some real $a$ with a real $n$th root must have degree at most $2$. http://planetmath.org/node/40163

I'm not sure about a step in their proof, where they conclude from $L'=F'(\sqrt[n]{\beta})$ that $L=F(\sqrt[n]{\beta})$. Is this step actually valid? If so, why? If not, how can the proof be repaired?

Answer 1

Let $N$ be a normal closure of $K$, so $N$ is $K$ adjoin an $n$th root of unity. Then you know the Galois group of $N$ over $K$, and you know the Galois group of $N$ over $F$, and the group of $N$ over $L$ has to be a normal subgroup of the group of $N$ over $F$, containing the group of $N$ over $K$. A little group theory should finish things off.