# Every finite group is the Galois group of a field extension

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:

1. Every finite group is contained in $S_p$ for a large enough prime $p$.

2. Every irreducible polynomial in $\mathbb Q[x]$ of degree $p$ having exactly $p-2$ real roots has a Galois group $S_p$ over $\mathbb Q$.

3. 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?

Well this is "a" inverse Galois problem in some sense but significantly easier than "the" inverse Galois problem.

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 $K/F$.

• How is this different from the inverse Galois problem? Just set $F=\mathbb{Q}$? I know there is something I'm missing here, but what is it? Jul 7, 2019 at 22:47
• You can't set $F=\mathbb{Q}$ because then the Galois group of $K/F$ is $S_p$ and not $G$. $F$ has to be the fixed field of $G$... Jul 8, 2019 at 14:23
• Aaah, I see, you get to choose who the field $F$ is Jul 8, 2019 at 15:24