# There are uncountably many subgroups of $\text{Gal}( \bar{\mathbb{Q}}/\mathbb{Q})$ of index $2$.

I am reading infinite Galois theory, and the motivation for introducing the Krull topology seems to be that this is the way in which we can solve the problem that the Galois correspondence fails. This is the counterexample given:

If $$G=\text{Gal}( \bar{\mathbb{Q}}/\mathbb{Q})$$ there are uncountably many subgroups of $$G$$ with index $$2$$, while the number of subfields of $$\bar{\mathbb{Q}}$$ of degree $$2$$ over $$\mathbb{Q}$$ is countable, thus there cannot be a bijection.

I understand why the number of subfields of $$\bar{\mathbb{Q}}$$ of degree $$2$$ over $$\mathbb{Q}$$ is countable (these are of the form $$\mathbb{Q}(\alpha)$$ with $$\alpha$$ a root of an irreducible polynomial of degree $$2$$, and there are countably many possible $$\alpha$$) but I don't get why there are uncountably many subgroups of $$G$$ with index $$2$$.

Any hints or help will be appreciated.

• I think your question is handled here. I won't vote to close as a duplicate right away, because my vote would be immediately binding, and I want to give others the chance to raise their objections. Apr 30 at 18:46
• @JyrkiLahtonen not quite, they didn't take $\bar{\mathbb{Q}}$, but a smaller extension. This will give also a counterexample in the same way, but in my case this argument does not work. Apr 30 at 19:36
• If $L$ stands for the compositum of all the quadratic extensions (like in the other thread, then $L/\Bbb{Q}$ is Galois, and won't $Gal(L/\Bbb{Q})$ be a quotient of the absolute Galois group. So the index two subgroups of the former give rise to index two subgroups of the latter? Mind you, I'm not overly confident about my understanding of infinite Galois theory, but I think the argument there definitely helps here as well. Waiting for others to comment also. Apr 30 at 19:46