By [this answer](https://mathoverflow.net/q/304708), we know that if $K/\mathbb{Q}\_p$ is a finite extension, the centralizer of $G\_K$ in $G\_{\mathbb{Q}\_p}$ is trivial. The argument there uses that the abelinization of $G\_K$ is the pro-finite completion of $K^\times$ and that the action of $G\_{\mathbb{Q}\_p}$ is the natural one. My question is how does the situation look in the number field case? In that case (now global) class field theory gives us that
$$G\_K^{ab}\cong K^\times \backslash \mathbb{I}\_K/\mathcal{O}.$$
It seems that there ought to be a non-trivial element of $G\_\mathbb{Q}$ which fixes an element of $\mathbb{I}\_{K}$ modulo $K^\times$ and $\mathcal{O}$ not that I can give an example.... If that were the case, the centralizer would be non-trivial. Is there a good description of the centralizer or is it trivial?



Centralizer of the absolute Galois group of a number field
If $L/K$ are number fields, then the centralizer $C\_{G\_K}(G\_L)$ of $G\_L$ in $G\_K$ is trivial: As $C\_{G\_K}(G\_L)$ is normal in the normalizer $N\_{G\_K}(G\_L)$, the fixed field $E$ of $C\_{G\_K}(G\_L)$ is a Galois extension of the fixed field $F$ of $N\_{G\_K}(G\_L)$. Now $F$ lies between $K$ and $L$, in particular it is a number field. The compostium $LE$ is a finite extension of $E$ with abelian absolute Galois group. It follows from a result of Weissauer in [this](https://www.degruyter.com/document/doi/10.1515/crll.1982.334.203/pdf) paper (see also Theorem 13.9.1 in the book Field Arithmetic by Fried-Jarden), that every proper finite extension of the Galois extension $E$ of the number field $F$ is Hilbertian, in particular its absolute Galois group is not abelian. Therefore, $LE$ has no proper finite extensions, and so $\bar{\mathbb{Q}}$ is a finite extension of $E$, and therefore in fact $E=\bar{\mathbb{Q}}$.


