TITLE: Why is this extension of Galois?
QUESTION [7 upvotes]: Let $F$ be a subextension of $\mathbb{C}$ maximal with respect to not containing $\sqrt2$. Let $K/F$ be a finite extension with $K\subset\mathbb{C}$. Then $K/F$ is of Galois and $[K:F]$ is a power of a prime. Could you help me solve this exercise?
Following the idea of Olivier:
Let $L$ be the normal closure of $K/F$. If $K\neq F$, then $\sqrt2\in L$ and so $2|[L:F]=|\mathrm{Gal}(L/F)|=G$, let $H<G$ be a 2-Sylow. I want to prove $L^H=F$. If $L^H/F$ is proper then $\sqrt2\in L^H$, as in the answer of Olivier we prove that every element of $G$ fixes $\sqrt2$ and so $\sqrt2\in L^G=F$ contradiction, so $L^H=F$ and so $G=H$, so $[L:F]$ is a power of 2 and so it is $[K:F]$. What I'm missing is how to prove that $L=K$, i.e. that $K/F$ is normal.

REPLY [3 votes]: Let $L/F$ be the normal closure of $K/F$. Let $G=Gal(L/F)$, and let $P$ be a Sylow 2-subgroup of $G$. Then $M=\mathrm{Inv}(P)$ is an odd degree extension field of $F$, so it cannot contain $\sqrt2$ either, and we conclude by maximality that $M=F$ and that $G$ is a 2-group.
Let $H=Gal(L/K)$. As you observed, if we can show that $H$ is a normal subgroup of $G$, then we are done. Let $G_0=Gal(L/F[\sqrt2])$. As $F[\sqrt2]$ is contained in all the finite extensions of $F$, we see that $G_0$ must be the unique maximal subgroup of $G$. 
The only abelian $2$-groups with a unique maximal subgroup are the cyclic ones. Using this we can prove the following.
Claim. A 2-group $G$ with a unique maximal subgroup $G_0$ is abelian.
Proof. We know that $G$ is nilpotent by virtue of being a 2-group. Consider the lower central series: $G_1=[G,G]$, $G_2=[G,G_1]$.
The derived subgroup of the quotient $\overline{G}=G/G_2$ is then obviously $\tilde{G}=[\overline{G},\overline{G}]=G_1/G_2$. Because $[\tilde{G},\overline{G}]$ is trivial, we have $\tilde{G}\subseteq Z(\overline{G})$. 
The maximal subgroups of the abelian group $G/[G,G]$ are in 1-1 correspondence of maximal subgroups of $G$ containing $[G,G]$. As $G_0$ is the only such subgroup of $G$, we must have that $G_0/[G,G]$ is the only maximal subgroup of $G/[G,G]$. Therefore $G/[G,G]$ must be cyclic. We are now in a position to apply
the well-known lemma: 
$$G/Z(G)\ \text{cyclic}\implies G\ \text{abelian}$$
to the group $\overline{G}$. Therefore $[\overline{G},\overline{G}]$ is trivial, and we thus have $G_2=G_1$. The lower central series of a nilpotent group can stall only at the trivial subgroup, so we have shown that $G_1=\{1_G\}$. Q.E.D.
Of course, this settles the original question as all the subgroups of an abelian group are normal. At this point we can also conclude that $G$ must be cyclic.