Problem of Galois Extension $\Bbb K$ is a non-Galois extension of $\Bbb Q$ and $[K:\Bbb Q]=4$. If $\Bbb F$ is the Galois closure of $\Bbb K$ then show that $Gal(\Bbb F/\Bbb Q)$ is either $S_4, A_4$ or $D_8$ with order 8. Further, I want to prove that $Gal(\Bbb F/\Bbb Q) = D_8$ iff $\Bbb K$ contains a quadratic extension of $\Bbb Q$.
 A: The Galois group $G_{F/\Bbb Q}$ must be a subgroup of $S_4$, because $F$ is the splitting field of a polynomial of degree $4$ (the minimal polynomial of any element generating $K/\Bbb Q$). The subgroup $H$ of $G_{F/\Bbb Q}$ with fixed field $K$ must have index $4$. Therefore $4\mid\# G_{F/\Bbb Q}\mid 24$ and so the only values for $\#G_{F/\Bbb Q}$ are among $4,8,12,24$. Furthermore $K/\Bbb Q$ is not normal so $H\le G_{F/\Bbb Q}$ is not normal, and this excludes the value $4$ and means $G_{F/\Bbb Q}$ must be nonabelian.
Now this turns into a group theory problem. Between the quaternion and dihedral groups of order eight show the first doesn't have an index four non-normal subgroup. Argue the alternating group is the only subgroup of the symmetric group of index two (this takes care of the order $12$).
Back to Galois theory, we can say $K$ contains a quadratic number field iff $H$ is contained in an index two subgroup (corresponding to said field) of $G_{F/\Bbb Q}$. Check which of the groups still on the table allow this possibility. (Disclaimer: I haven't checked myself yet, so it'll be an adventure.) I would use the groupprops wiki or other resources to find a lattice of subgroups of $S_4$ for aid.
