TITLE: Are all elements of order $2$ in $\mathrm{Gal}(\overline{\Bbb Q} / \Bbb Q)$ conjugate?
QUESTION [6 upvotes]: Let $G$ be the absolute Galois group of $\Bbb Q$. Is it true that any two elements of order $2$ in $G$ are conjugate (in $G$) ?
I've seen this question, but the answer only shows that any element of finite order in $G$ has order $≤2$. Apparently, Artin answered positively to my question, but I found no reference for this result. Any help would be appreciated.

REPLY [5 votes]: Yes, this is true.  If $\sigma\in G$ has order $2$, let $R$ be the fixed field of $\sigma$.  Then $R$ is a real-closed field which contains $\mathbb{Q}$ and is contained in $\overline{\mathbb{Q}}$, so it is a real closure of the ordered field $\mathbb{Q}$ for the ordering on $\mathbb{Q}$ we obtain by restricting the ordering of $R$.  But there is only one way to make $\mathbb{Q}$ an ordered field, so this ordering is just the usual ordering on $\mathbb{Q}$.
So, $R$ is a real closure of $\mathbb{Q}$ with its usual ordering.  But the real closure of an ordered field is unique up to unique isomorphism.  So if $\sigma'\in G$ is any other element of order $2$ and $R'$ is its fixed field, then $R\cong R'$.  This isomorphism extends to an automorphism $\tau$ of $\overline{\mathbb{Q}}$, and $\sigma$ and $\sigma'$ are then conjugate via $\tau$.