Let $K\_1$ be a perfect field. Let $K\_2/K\_1$ and $K\_3/K\_2$ be quadratic extensions. Let $K\_4/K\_3$ be the Galois closure of $K\_3$ over $K\_1$. Is it true that either $K\_3 = K\_4$ or $K\_4/K\_3$ is quadratic such that the Galois group of $K\_4$ over $K\_1$ is isomorphic to $\mathcal{D}\_4$, the dihedral group with 8 elements?



Quadratic extension of quadratic extension
Yes. Let $a\_1$ be a generator of $K\_3$ over $K\_1$ and let $a\_2$, $a\_3$, and $a\_4$ be the other three roots of the polynomial of $a\_1$. Then say that $a\_2$ is the other root of the polynomial of $a\_1$ over $K\_2$. Then The Galois group of $K\_4/K\_1$ acts on the partitioned set $\{\{a\_1,a\_2\},\{a\_3,a\_4\}\}$, and is therefore a transitive subgroup of $D\_4$, either $D\_4$ itself or $C\_4$ or $C\_2 \times C\_2$.


I think that the following is a more general fact that can be proven in the same way. If $K\_2/K\_1$ is separable and its Galois closure has Galois group $G$, and if $K\_3/K\_2$ is separable and its Galois closure has Galois group $H$, then the Galois group of the Galois closure of $K\_3/K\_1$ is a subgroup of a wreath product of $G$ and $H$. If you take a longer chain of extensions you get an iterated wreath product.


