Suppose $F$ is a subfield of $\mathbb{R}$ and let $K=F(\sqrt[n]{a})$, then any Galois extension of $F$ contained in $K$ must have degree less than 3.
How do i go about proving this claim? I have considered the galois extension $K(\zeta_n)$ over $F$, but don't know how to proceed. If I can somehow show that all the normal subgroups of $Gal(K(\zeta_n)/F)$ have order 1 or 2, then I will be done, but again don't see how that might be true.