# A free group and Galois theory. [closed]

Let $$F_S$$ be a free group with finite generating set $$S$$.

How to show that, there exists a Galois extension $$E$$ of the rational numbers $$\mathbb{Q}$$ such that, $$\operatorname{Gal}( E / \mathbb{Q} ) = F_S$$ ?

Thank you for the help.

Re-edit,

Let $$F_S$$ be a free profinite group which is the profinite completion of a free group with finite generating set $$S$$.

How to show that, there exists a Galois extension $$E$$ of the rational numbers $$\mathbb{Q}$$ such that, $$\operatorname{Gal}( E / \mathbb{Q} ) = F_S$$ ?

Thank you for the help.

• Don't Galois groups have to be more-or-less profinite? Oct 18, 2020 at 21:53
• Yes, it's possible. :-) Oct 18, 2020 at 21:57
• As paul garrett says, Galois groups must be profinite, and profinite groups are either finite or uncountable and so can't be countable. Do you mean to ask that the Galois group be the free profinite group on $S$? Oct 18, 2020 at 22:16
• See for example math.stackexchange.com/questions/2057432/… . Why do you believe this is true? Were you given this as an exercise? Oct 18, 2020 at 22:44
• @YoYo12: every free group on a finite (nonempty) set of generators is countable, so no such group is profinite. Oct 18, 2020 at 23:12

With "free group" replaced by "free profinite group," this is either false or an open problem. If it were known to be true it would imply that every finite group $$G$$ occurs as the Galois group of a finite Galois extension of $$\mathbb{Q}$$ and this is not known.
• Thank you. Can you explain to me why it imply that every finite group $G$ occurs as the Galois group of a finite Galois extension of $\mathbb{Q}$. Oct 18, 2020 at 22:56