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The automorphism group of the complex numbers $\mathbb{C}$ and the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ are amongst the most mysterious and worst understood objects in Galois theory and group theory.

Is there a good (= "better") description of the "intermediate group" $\mathrm{Gal}(\mathbb{C}/\overline{\mathbb{Q}})$ ?

Is this group better understood, or is it as mysterious ?

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    $\begingroup$ $\text{Gal}(\mathbb{C}/\overline{\mathbb{Q}})$ is as big and messy as $\text{Gal}(\mathbb{C}/\mathbb{Q})$, and for precisely the same reason: there's a load of algebraically independent transcendental elements in $\mathbb{C}$ which you can essentially permute as you which. Not sure if my comment is very satisfying as an answer, maybe there's more to be said (?) $\endgroup$
    – user484137
    Commented Jun 16, 2022 at 15:25
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    $\begingroup$ Does $\operatorname{Gal}(\mathbb C/\overline{\mathbb Q})$ just mean $\operatorname{Aut}(\mathbb C/\overline{\mathbb Q})$? I am not accustomed to the ‘$\operatorname{Gal}$’ notation for non-algebraic extensions, though I know there are different conventions. $\endgroup$
    – LSpice
    Commented Jun 16, 2022 at 15:48
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    $\begingroup$ This group has cardinal $2^c=2^{2^{\aleph_0}}$ $\endgroup$
    – YCor
    Commented Jun 16, 2022 at 18:31
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    $\begingroup$ But "big and messy" doesn't mean there's nothing to say. For instance one can wonder whether it's simple ($\mathrm{Aut}(\mathbf{C})=\mathrm{Aut}_{\mathbf{Q}\text{-alg}}(\mathbf{C})$ is not simple because it admits $\mathrm{Aut}(\bar{\mathbf{Q}})$ as a quotient). A natural related question is whether it has a trivial abelianization. $\endgroup$
    – YCor
    Commented Jun 16, 2022 at 18:32
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    $\begingroup$ Also one known thing is that this group is torsion-free. Indeed, the group $\mathrm{Aut}(\mathbf{C})$ has only one non-identity conjugacy class of elements of finite order, and it consists of elements of order 2 (and this also forms the only nontrivial conjugacy class of finite subgroups). These elements map nontrivially on $\mathrm{Aut}(\bar{\mathbf{Q}})$, hence the kernel $\mathrm{Aut}_{\mathbf{Q}\text{-alg}}(\mathbf{C})$ is torsion-free. $\endgroup$
    – YCor
    Commented Jun 16, 2022 at 18:39

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