Every finite group is isomorphic to some Galois group for some finite normal extension of some field.

Question

Every finite group is isomorphic to some Galois group for some finite normal extension of some field.

I'm trying to write a proof, but I know this is incorrect. Can anyone point me in the write direction?

Proposition: Every finite group is isomorphic to some Galois Group $\text{Aut}_F(K)$ for some finite normal extension K of some field F.

My hack at a proof: For every $n \in \mathbb{Z}^+$ where $n = |\text{Aut}_E K|$ there exists a normal extension $K$ of $E$ such that $\text{Aut}_E K \cong S_n$. Then consider a subgroup $H \leq \text{Aut}_E K$ where $H \cong \text{Aut}_E K$. It follows that $H$ is the Galois group of $K$ over $K_H$.

Answer 1

Your idea is right, and in fact it is the standard proof: let $G$ be a given group, and let $n=|G|$.

There is a field extension $K/E$ with $\mathrm{Aut}_E(K)\cong S_n$ (e.g., $E=\mathbb{Q}(s_1,\ldots,s_n)$, $K=\mathbb{Q}(x_1,\ldots,x_n)$, 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 $\mathrm{Aut}_{L}(K) = H\cong G$.