Is the absolute Galois Group of $\Bbb Q$ countable? Is $\text{Gal} (\overline{\Bbb Q}/\Bbb Q)$ countable or uncountable? It seems like it should be countable (because the algebraic closure of $\Bbb Q$ is countable and there are countably many permutations of the irrational algebraic numbers, and a countable union of countable sets is countable. However, I've seen references that seem to imply it is uncountable. What is the answer, and why/how?

Let $I\subseteq \Bbb N$ be any subset and let $K_I=\Bbb Q(\{\sqrt{p_i}\}_{i\in I})$ where $p_i$ is the $i^{th}$ prime. Then there are precisely $2^{\Bbb N}$ in fact the Galois group of the compositum of this extension is exactly isomorphic to
$$\prod_{i\in\Bbb N}\Bbb Z/2\Bbb Z$$
and this of course indicates there are uncountably many elements in $\text{Gal}(\overline{\Bbb Q}/\Bbb Q)$
You can also do a simple cardinality argument using inverse limits, but that technology is a bit stronger.
