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Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known? And if so is there a description on how the order can be found?


My initial thoughts is that because $\bar{\Bbb Q}$ is countable and for an $\alpha_n\in\bar{\Bbb Q}\backslash\Bbb Q$ with $n\in\Bbb N_0$ and $\alpha_0=1$ we can say that any algebraic number $\beta\in\bar{\Bbb Q}$ can be written as $$\beta=\sum_{k\ge0}\alpha_kp_{k+1}=p_1+\alpha_1p_2+\alpha_2p_3+\cdots$$ Now there are automorphisms on $\bar{\Bbb Q}$ that I know are in a set $A\subset \operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ and $A=\{\imath, *_1, *_2,...\}$ where $*_n$ is conjugation of the $n^\text{th}$ term. So I am hoping that there is a $B=\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)\backslash A$ which is also countable or uncountable. Which means the problem is no just determining the cardinality of $B$.

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    $\begingroup$ Infinite profinite groups are all uncountable. $\endgroup$ Jun 26 '14 at 17:00
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It has the cardinality of the continuum.

To see this, note that first it's faithfully described by its action on the roots of all irreducible polynomials over $\mathbb{Q}$, of which there are countably many; this means that it injects into a product over countably many symmetric groups, which has the cardinality of the continuum. So it has at most the cardinality of the continuum.

Second, we can build up an element of the absolute Galois group by picking an arbitrary element of a Galois group of some Galois extension $\mathbb{Q} \to K_1$, then extending it in an arbitrary fashion to a Galois extension $\mathbb{Q} \to K_1 \to K_2$, and so forth; at every step we can arrange that there are at least two choices to make, and we make countably many choices, so the Galois group has at least the cardinality of the continuum.

(An overkill way to do the second step: the absolute Galois group is a profinite group, hence in particular compact Hausdorff. Such a group is either finite or uncountable due to the existence of Haar measure, since if it were countable then it could not have a Haar measure of finite total measure.)

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    $\begingroup$ You don't need Haar measure. Use the more basic fact that any infinite compact space is uncountable. (I first saw that in Munkres, for his topological proof of the uncountability of $[0,1]$.) $\endgroup$
    – KCd
    Jun 26 '14 at 17:50
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    $\begingroup$ @KCd: I think I know the result you're alluding to, but it has more hypotheses than that (e.g. the one-point compactification of $\mathbb{N}$ is compact Hausdorff but countable). You're probably thinking of en.wikipedia.org/wiki/Finite_intersection_property#Applications, right? $\endgroup$ Jun 26 '14 at 18:11
  • $\begingroup$ Whoops, you're right that the link is what I intended. I left off a condition about isolated points. The correct statement is that a nonempty compact Hausdorff space without isolated points is uncountable. See Munkres Theorem 27.7 p. 176 or the link in Qiaochu's comment. $\endgroup$
    – KCd
    Jun 26 '14 at 19:27
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Here is a different take on Qiaochu's proof.

$\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ surjects onto $\mathrm{Gal}(\mathbb{Q}(\sqrt{p} : p \text{ prime}) / \mathbb{Q}) \cong \prod_{p \text{ prime}} \mathbb{Z}/2\mathbb{Z}$, so that $\# \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \geq c$.

In the other direction, $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) = \lim_{E / \mathrm{Q} \text{ finite Galois}} \mathrm{Gal}(E/\mathbb{Q})$. Since there are only countably many finite Galois extensions and each associated Galois group is finite, it follows $\# \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \leq {\aleph_0}^{\aleph_0} = c$.

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  • $\begingroup$ Is the product supposed to be $\Bbb Z/2\Bbb Z$ or $\Bbb Z / p\Bbb Z$? $\endgroup$ Oct 29 '14 at 1:16
  • $\begingroup$ $\mathbb{Z}/2\mathbb{Z}$ is correct. Remember that for each prime $p$ we have $\mathrm{Gal}(\mathbb{Q}(\sqrt{p})/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ Oct 29 '14 at 7:30

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