Existence of a Field not containing $\sqrt{2}$, such that any finite extension is cyclic.

Question

I want to show the existence of a field $E$ not containing $\sqrt{2}$, such that any finite extension of $E$ in $\overline{\mathbb{Q}}$ is cyclic.

I think the maximal field not containing $\sqrt{2}$ should work.

I think that any finite extension of $E$ must contain $2^{\frac{1}{2^n}}$, but I am not sure about it. How can I show that any finite extension of $E$ is cyclic?

Answer 1

By taking the Galois closure if necessary, we may assume that $F/E$ is finite Galois. By construction of $E$, there is a unique minimal subextension of $F/E$, which is $E(\alpha)$. Under the Galois correspondence, it corresponds to a unique maximal subgroup $H$ of $\operatorname{Gal}(F/E)$. We have now reduced to the following group theoretic statement:

Let $G$ be a finite group with a unique maximal subgroup. Then $G$ is cyclic.

See here for a proof.

Even more is true: $G$ is cyclic of prime power order, $E(\alpha)/E$ is a $\mathbf{Z}/p\mathbf{Z}$-extension and $\operatorname{Gal}(\overline{\mathbf{Q}}/E)\cong \mathbf{Z}_p$.