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Let $L/K$ be a field extension, and let $K \subseteq A \subseteq L$ an intermediate field. Define $$ A^{FG} := \operatorname{Gal}(L/A) = \{ \phi \in \operatorname{Aut}(L) : \forall x \in A: \phi(x) = x \} $$ and let $U \subseteq \operatorname{Gal}(L/K)$ be a subgroup of the Galois group, then $$ U^{FK} := \operatorname{Fix}_L(U) := \{ l \in L : \forall \alpha \in U: \alpha(l) = l \}. $$ Two subgroups $U_1, U_2$ of a group $G$ are called conjugate, if $gU_1g^{-1} = U_2$ for some $g \in G$, two intermediate fields $K \subseteq A_1, A_2 \subseteq L$ are called conjugate, if $\beta(A_1) = A_2$ for some $\beta \in \operatorname{Gal}(L/K)$.

Now I want to show:

i) If $U_1, U_2$ are two conjugate subgroups of the Galois Group, then the intermediate fields $U_1^{FK}, U_2^{FK}$ are also conjugate.

ii) Conversely, if $A_1, A_2$ are two conjugate intermediate fields, then $A_1^{FG}, A_2^{FG}$ are conjugate too.

The proof of ii) is simple I guess, let $\beta(A_1) = A_2$, then $$ \operatorname{Gal}(L/A_1) = \beta^{-1} \operatorname{Gal}(L/A_2) \beta $$ So if $\psi = \beta \phi \beta^{-1}$ with $\phi \in \operatorname{Gal}(L/A_2)$, then $\psi \in \operatorname{Gal}(L/A_1)$, cause if $x \in A_1$ then $$ \psi(x) = \beta^{-1}(\phi(\beta(x))) = \beta^{-1}(\beta(x)) = x $$ because $\beta(x) \in A_2$ and $\phi(x) = x$ for alle $x \in A_2$, hence $\psi$ leaves the elements of $A_1$ fixed, hence $\psi \in \operatorname{Gal}(L/A_1)$, the other inclusion follows likewise.

But for i) I am stuck. Let $U_1, U_2$ be two conjugate subgroups, i.e. $\beta U_1 \beta^{-1} = U_2$ for some $\beta \in \operatorname{Gal}(L/K)$. I want to show $\beta(U_1^{FK}) = U_2^{FK}$, which shows that $U_1^{FK}, U_2^{FK}$ are conjugate. So let $x \in U_1^{FK}$, which means that $\alpha(x) = x$ for all $\alpha \in U_1$, consider $\beta(x)$, now I have to show that $\beta(x) \in U_2^{FK}$, but here I am stuck, cause with the conjugacy of the subgroups, it just follows that for $\alpha \in U_2$ $$ \alpha(\beta(x)) = \beta(\alpha(x) $$ but then, I could not infer $\alpha(x) = x$, cause $x \in U_1^{FK}$, and $\alpha \in U_2$... any hints?

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    $\begingroup$ Your proof of i) implies ii) by the main Thm of Galois theory. $\endgroup$ – Martin Brandenburg Jun 16 '13 at 8:07
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Consider $\beta(x)$. You want to show that $\beta(x) \in {U_2}^{FK}$ and so you need to show that for each $\sigma \in U_2$, $\sigma(\beta(x))=\beta(x)$. So pick some $\sigma \in U_2$. Well $U_2=\beta U_1 \beta^{-1}$ and so $\sigma=\beta \circ \alpha \circ \beta^{-1}$ for some $\alpha \in U_1$. Thus $\sigma(\beta(x))=\beta(\alpha(\beta^{-1}(\beta(x))=\beta(\alpha(x))=\beta(x)$.

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