A kind of a reverse " Inverse Galois Problem " by fixing the above field and finding a base field

Question

Let $K/\mathbb Q$ be an infinite extension field and such that for any extension $F/K$ such that $F$ is algebraically closed , $[F:K]$ is infinite . Then is it true that for any finite group $G$ , there exists an intermediate field $\mathbb Q \subseteq L \subseteq K$ such that $K/L$ is a finite Galois extension with $G \cong Gal(K/L)$ ? If not true for all finite groups , is it at least true for finite abelian or cyclic groups ?

Answer 1

No, let’s take the maximal abelian extension of $\Bbb Q$ for our $K$. The Galois group is abelian, of course, and $K$ is not so big that the field of all algebraic numbers is finite over it. The Galois group (over $\Bbb Q$ is also, in this case, the group of all automorphisms of $K$, and it doesn’t have any nonabelian subgroups. So $K\supset L$ is an abelian extension for every subfield $L$ of $K$.

For your question about the weaker situation, take a prime $p$, and the “$p$-gamma extension” of $\Bbb Q$. This is an extension $K_\Gamma$ for which $\text{Gal}^{K_\Gamma}_\Bbb Q\cong\Bbb Z_p$, the additive group of $p$-adic integers. But the only nontrivial subgroups of $\Bbb Z_p$ are infinite, so your conjecture fails in an extreme way here.