TITLE: Every finite group is the Galois group of a field extension
QUESTION [11 upvotes]: How can I show that every finite group is the Galois group of an extension $K/F$ where $F$ is itself a finite extension of $\mathbb Q$? 

I know  the following:

Every finite group is contained in $S_p$ for a large enough prime $p$.
Every irreducible polynomial in $\mathbb Q[x]$ of degree $p$ having exactly $p-2$ real roots has a Galois group $S_p$ over $\mathbb Q$.
For any $n$ there is an irreducible polynomial in $\mathbb Q[x]$ of degree $n$ having exactly $n-2$ real roots.

Does this have something to do with the inverse Galois problem?

REPLY [11 votes]: Well this is "a" inverse Galois problem in some sense but significantly easier than "the" inverse Galois problem.
Your three results basically solve the problem already.
Take your finite group $G$ and embed it in $S_p$ for some prime $p$ (via (1)). Take some irreducible polynomial $f$ over $\mathbb{Q}$ with exactly $p-2$ real roots (via (3)). Let $K$ be the splitting field of $f$ (over $\mathbb{Q}$) then $K$ has Galois group $S_p$ (via (2)). Set $F$ the fixed field of $G$ (considered as a subgroup of $S_p$). Then by the main theorem of Galois theory $G$ is the Galois group of $K/F$.