We know that if $G$ is a finite subgroup of $\operatorname{Aut}(K)$, with $F$ the fixed field of $G$, then $G=\operatorname{Gal}(K/F)$.
Does this condition hold if $|G|=\infty$?
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1I believe the answer is no. Take $K=\mathbb{Q}(\ldots,u_{-2},u_{-1},u_0,u_1,u_2,\ldots)$, where the $u_i$, $i\in\mathbb{Z}$, are all transcendental over $\mathbb{Q}$. Let $G=\langle g\rangle\cong\mathbb{Z}$ act on $K$ by $g.u_i=u_{i+1}$. I believe $K^G=\mathbb{Q}$, though $\mathrm{Aut}(K)$ is substantially bigger than $G$. – David Hill Sep 18 '15 at 19:47
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2A similar example would be to take $K=\mathbb{C}(x,y)$ and let $GL_2(\mathbb{C})$ act naturally on $x,y$. Then the fixed field is just $\mathbb{C}$, but $\mathrm{Aut} (K/\mathbb{C})$ is much larger that the general linear group. – Mohan Sep 18 '15 at 19:50
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Yet another example would be the case where $K$ is an algebraic closure of $\Bbb{F}_p$, and $G$ is the group generated by the Frobenius automorphism $\phi:x\mapsto x^p$. The fixed field of $G$ is then the prime field $\Bbb{F}_p$. But $Gal(K/F)$ is much larger. In this case the closure of $G$ would be the full Galois group. – Jyrki Lahtonen Sep 18 '15 at 20:05
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Suppose if the extension K/F even turns out to be galois even then the group Gal(K/F) must be a profinite group. If the group G is countable then it cannot be galois so i guess this must be a counterexample. For example one can consider a countable subgroup of automorphism of complex numbers. So perhaps a better question could be is it possible to have profinite completion of G to be galois group of K/K^G. Also consider looking at http://math.stackexchange.com/questions/188882/is-every-group-a-galois-group. – random123 Sep 18 '15 at 21:27
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Yet another example, similar to that of @Mohan, but I think easier: Take the integers $\Bbb Z$, acting on $\Bbb C(x)$ so as to send (for $n\in\Bbb Z$) a rational function $f(x)$ to $f(x+n)$. The fixed rational functions are periodic, and the only such are the constants. – Lubin Sep 20 '15 at 21:11