Every finite group is isomorphic to the Galois group of some polynomial I was reading through chapter 14 of Dummit and Foote just now and I came across the sentence  "It is an open problem to determine whether every finite group appears as the Galois group for some polynomial over $\mathbb{Q}$" and I was wondering 2 things:

*

*Is it known whether each finite group is the Galois group of some polynomial (not necessarily over $\mathbb{Q}$)?


*Why is this problem so hard? (I know this question may defy a simple answer, but from the very naive perspective of someone who knows very little--me--it isn't very clear why this should be too much harder than the abelian case considering what is known about the structure of finite groups in general)
thanks :)
 A: Yes, every finite group is the Galois group of some extension. Consider the field $K = \mathbb{Q}(e_1, \dots e_n)$ and its extension $L = \mathbb{Q}(x_1, \dots x_n)$ where the $x_i$ satisfy
$$\prod (t - x_i) = \sum (-1)^i e_i t^{n-i}$$
where $e_0 = 1$, so the $e_i$ are the elementary symmetric polynomials in the $x_i$. Then $L$ is a finite Galois extension of $K$ with Galois group $S_n$ (exercise). Since every finite group $G$ embeds in some $S_n$, we get finite Galois extensions $L$ of fields $L^G$ with Galois group $G$. 
The inverse Galois problem is hard not because we don't know enough about finite groups but because we don't know enough about the Galois theory of $\mathbb{Q}$. It's equivalent to showing that every finite group is a quotient of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, and this is just hard. The abelian case is much easier because abelian quotients of the Galois group are handled abstractly by class field theory and concretely by the Kronecker-Weber theorem, but "nonabelian class field theory" is much harder, not completely understood, and very much the subject of active research. 
