For $G$ a finite group there exists a field extension $F\subset K$ s.t $G(K,F)\cong G$.

Question

Prove: For $G$ a finite group there exists a field extension $F\subset K$ s.t $G(K,F)\cong G$.

How to choose $K$ and $F$?

Thanks!

Answer 1

Let $n=|G|$.

There is a field extension $K/E$ with $\mathrm{Gal}(K,E)\cong S_n$ (for example $E=\mathbb{Q}(s_1,\ldots,s_n)$, $K=\mathbb{Q}(x_1,\ldots,x_n)$, where $s_1,\ldots,s_n$ are the symmetric polynomials in $x_1,\ldots x_n$).

By Cayley's Theorem, we know that $G$ is isomorphic to a subgroup $H$ of $S_n$, so if we let $L$ be the fixed field of $H$ in $K$, then by the Fundamental Theorem of Galois Theory we know that $K/L$ is Galois and $\mathrm{Gal}(K,L) = H\cong G$.