Suppose that $K/F$ is a normal extension of fields of finite degree and consider $G' \leq \mbox{Gal}(K,F)$. If $\mathcal{F} = K_{G'}$ is the fixed field of $G'$, then show that for any $\tau \in \mbox{Gal}(K,F)$, $K_{\tau G' \tau^{-1}} = \tau(\mathcal{F})$.

Here's my attempt. Suppose $a \in \mathcal{F}$ and $g' \in G'$. Then $(\tau g' \tau^{-1})(\tau a) = \tau (g' a) = \tau a$, since $g'$ fixes $a \in \mathcal{F}$. Hence, $\tau G' \tau^{-1}$ fixes $\tau(\mathcal{F})$. So if I can show that the group fixing $\tau(\mathcal{F})$ has order equal to the degree of $K$ over $\tau({\mathcal{F}})$, which is the same as the degree of $K$ over $\mathcal{F}$, and I then show that the degree of $K$ over $\mathcal{F}$ is the order of $G'$, then the proof is complete. If $K/F$ is a Galois extension, then all of the above holds, but I'm not sure that it holds in the case that $K/F$ is a normal extension of fields of finite degree.


1 Answer 1


Some of your notation is unusual: $G'$ usually denotes the derived group of $G$ while the fixed field of $H$ is usually denoted $K^H$ rather than $K_H$.

I'll use standard notation: let $G=\textrm{Aut}(K/F)$, $H$ be a subgroup of $G$ and $K^H$ be its fixed field. Let $\tau\in G$. Then $a\in K^{\tau H\tau^{-1}}$ iff $\tau\sigma\tau^{-1}(a)=a$ for all $\sigma\in H$ iff $\sigma\tau^{-1}(a)=\tau^{-1}(a)$ for all $\sigma\in H$ iff $\tau^{-1}(a)\in K^H$ iff $a\in\tau(K^H)$. Therefore $K^{\tau H\tau^{-1}}=\tau(K^H)$.

None of this depends on $K/F$ being a Galois extension. Here $G=\textrm{Aut}(K/F)$ is the group of $F$-automorphisms of $K$. The extension is Galois iff $F=K^G$, but that's not necessary for the argument above.

  • $\begingroup$ Does your proof make use of the assumption that $K/F$ is a normal extension? $\endgroup$
    – user95224
    Commented Feb 23, 2018 at 5:24
  • 1
    $\begingroup$ @user95224 On reading my proof, I find no assumptions made whatsoever about the extension $K/F$. $\endgroup$ Commented Feb 23, 2018 at 5:38

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