Let $F/\mathbb{Q}$ be a degree 4 extension, NOT Galois. Prove that the Galois closure of $F$ has Galois group either $S_4, A_4$ or $D_8$. The question is as the title states. So if $F=\mathbb{Q}(\alpha)$ for some alpha that satisfies a degree 4 polynomial $p(x)$, then we are looking for the splitting field of $p(x)$? I'm not sure what to work with, given the information...

You have a good start: by the primitive element theorem, we know that $K=\mathbb{Q}(\alpha)$ for some $\alpha$ whose minimal polynomial $p\in\mathbb{Q}[x]$ is of degree $4$. Let's denote the splitting field of $p$ by $L$.
What do we know about the extension $L/\mathbb{Q}$?


*

*it is a Galois extension, since it is a splitting field, and characteristic $0$ implies separable

*each element of the Galois group $G=\mathrm{Gal}(L/\mathbb{Q})$ is determined by how it permutes the four roots of $p$, and therefore $G$ is isomorphic to a subgroup of $S_4$

*$L$ strictly contains $K$ and $[K:\mathbb{Q}]=4$, and $G\subseteq S_4$ implies $[L:\mathbb{Q}]$ divides $24$, so we must have $[L:\mathbb{Q}]\in\{8,12,24\}$
Looking at the subgroups of $S_4$ (groupprops link), the only possibilities are $D_8$, $A_4$, and $S_4$.
