The maximal subfield of $\mathbb C$ not containing $\sqrt2$ Related:
Does a maximal subfield of $\mathbb C$ not containing $\sqrt{2}$ have index $2$?
He said,
"...fixed field is an extension of $K$ which doesn't contain $\sqrt{2}$, and thus must be $K$ itself. Thus, the Galois group is cyclic..."
Why is that so?
Could you give a reason or hint?
Thank you!

This boils down to the group theoretic proposition that any 2-group with unique subgroup of index $2$ is cyclic. A proof is here:Abelian $p$-group with unique subgroup of index $p$ 
To see the reduction let $L$ be any alg. ext. of $K$ and $E$ its normal closure.
Let $G$ be the Galois group of $E/K$ then $G$ is a $2$-group since if the $2$-sylow subgroup has index, which must be odd, greater than $1$ then you get an odd extension of $K$. Further any extension contains $K(\sqrt{2})$ and so there is a unique subgroup of index $2$. It then follows that $G$ is cyclic and all intermediate fields are also Galois and cyclic.  
