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4 votes
0 answers
245 views

Dessins d'enfants and the absolute Galois group

If I am not mistaken, Alexander Grothendieck introduced dessins d'enfants as they are known today, in order to better understand the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}...
THC's user avatar
  • 4,547
7 votes
1 answer
603 views

Interesting (combinatorial) actions of the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$

I have read many times that it is crucial to understand the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) =: L$, in so far that some have stated (was it Richard Taylor ?) that ...
THC's user avatar
  • 4,547
1 vote
2 answers
1k views

Is the absolute Galois group the same as the automorphism group? [closed]

Is the absolute Galois group $\mathrm{Gal}(\overline{\mathbf{Q}}|\mathbf{Q})$ the same as the group $\mathrm{Aut}_{\mathbf{Q}}(\overline{\mathbf{Q}})$ the automorphism group in the category of $\...
Ofra's user avatar
  • 1,613
9 votes
2 answers
1k views

Is it known if the absolute Galois group is "divisible"?

The definitions of a divisible group that I have seen all seem to assume abelian is an a priori property of the group. My question is as to whether or not it is known that--given a non-torsion element ...
Adam Hughes's user avatar
  • 1,049
2 votes
2 answers
1k views

Place stabilizers for the absolute Galois Group

Fix an algebraic closure, $\overline{\mathbb{Q}}$ for the rationals and consider the set, $B_p$, of all places of $\overline{\mathbb{Q}}$ over a fixed (possibly infinite) prime, $p$, of $\mathbb{Q}$. ...
Adam Hughes's user avatar
  • 1,049
5 votes
2 answers
1k views

non-continuous inverse Galois problem

Let $G=Gal(\bar{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group over $\mathbf{Q}$. Q1: Is it possible to find a (necessarily non-closed) normal subgroup $K\leq G$ such that $G/K$ is free of ...
Hugo Chapdelaine's user avatar