It is known that the only elementary abelian $2$-groups (finite and nonfinite) in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ are in fact finite and cyclic – that is to say, they are of order $2$.


(Here, $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$.)


Question: Does one need the Axiom of Choice to acquire this property ?



Involutions in the absolute Galois group (and the Axiom of Choice)
No, you shouldn't need any choice for this, and it should still be true if you replace $\overline{\mathbb{Q}}$ with any other algebraic closure of $\mathbb{Q}$. Let $K$ be a field (which in our application will be $\overline{\mathbb{Q}}$) and let $G$ be a finite group of automorphisms of $K$. Then $K/K^G$ is always a degree $\#(G)$ extension. (See, e.g., [Milne - Fields and Galois theory](https://www.jmilne.org/math/CourseNotes/FT.pdf) Corollary 2.14 and Theorem 3.4.)


The [Artin–Schreier theorem](https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf) says that, if $K/F$ is a finite degree extension and $K$ is algebraically closed, then $[K:F] \leq 2$.


Combining the two statements, if $G$ is a finite subgroup of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, then $\#(G) \leq 2$.


There is no use of choice in the proof of either of the big tools that I'm using.


