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Can there be an element $\sigma$ of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, other than the identity and complex conjugation, which is completely determined up to conjugacy by its action on $\sqrt[n]r$ for all $r \in \mathbb{Q}$ and all $n \in \mathbb{N}$?

I expect not, but if I am wrong, can the subgroup formed by such elements be given an alternative characterization?

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    $\begingroup$ Can’t I multiply on the right by an element of \Gal(\Qbar/\Q^{ab}({p^{1/n}: p prime, n > 1}))? Why is the identity determined up to conjugacy by its action on these elements? I think I’m being dumb, so forgive me. $\endgroup$
    – alpoge
    Commented Apr 25, 2019 at 19:50
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    $\begingroup$ When there's already an ambient group acting on a set $X$, I understand what it means to say that a particular element of that group is distinguished from others by its action on some subset $X'$; but I don't know what it means to look at an element in isolation and ask if it is, or isn't, so determined (in order to consider the set of elements that are). Could you clarify? Do you mean to consider the set of all elements of the absolute Galois group whose restriction to the set you indicate is shared with no other element? If so, then I agree with @alpoge. $\endgroup$
    – LSpice
    Commented Apr 25, 2019 at 20:07
  • $\begingroup$ (Indeed I interpreted it as: \sigma is so distinguished if and only if (\tau agrees with \sigma on that set implies \tau = \sigma) holds.) $\endgroup$
    – alpoge
    Commented Apr 25, 2019 at 20:10
  • $\begingroup$ @alpoge, right, but my question is: which $\tau$? If we allow $\tau$ to range over the whole Galois group, then, as you point out, there is no such $\sigma$. $\endgroup$
    – LSpice
    Commented Apr 25, 2019 at 20:11

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