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

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 $K_H$.

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 $\mathrm{Aut}_{L}(K) = H\cong G$.
• Excuse me. This relies on $K/E$ being Galois. It is easy to see that $S_n$ is a subgroup of the automorphism group of $K$ over $E$, but to complete the picture, everything falls into place if I can see that $[K:E]=n!$. However, I'm not finding this very clear Aug 12, 2022 at 16:54
• @FShrike: work the other way around. $K$ is Galois over the fixed field of any subgroup of its automorphism group, with Galois group that subgroup. Now consider the subgroup given by permuting the $x_i$. Aug 12, 2022 at 17:03
• Ah, thank you. I appreciate the response, that is a much better way to look at it. My attempts at an induction on $n$ were going haywire Aug 12, 2022 at 17:13