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 $K_H$.


1 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 $\mathrm{Aut}_{L}(K) = H\cong G$.

  • $\begingroup$ 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 $\endgroup$
    – FShrike
    Aug 12, 2022 at 16:54
  • 1
    $\begingroup$ @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 $x_i$. $\endgroup$ Aug 12, 2022 at 17:03
  • $\begingroup$ 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 $\endgroup$
    – FShrike
    Aug 12, 2022 at 17:13

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