Is there an absolute Galois group that is not a subgroup of $\hat{\mathbb{Z}}$ and that has one and only one closed non-open subgroup?



Absolute Galois group with unique closed non-open subgroup
Eventually this is purely a fact of profinite groups.


**Proposition.** *Let $G$ be an infinite profinite group in which every nontrivial closed subgroup is open. Then $G$ is isomorphic to $\mathbf{Z}\_p$ for some prime $p$.*


Proof. First, the case when $G$ is abelian. Then its Pontryagin dual is a discrete torsion abelian group, in which every proper subgroup is finite. It is standard (and easy) that this characterizes the Prüfer groups $C\_{p^\infty}$ for $p$ prime, whose Pontryagin duals are the $\mathbf{Z}\_p$.


Next, the general case. Let $x$ be a nontrivial element of $G$, and $A\_x=\overline{\langle x\rangle}$ the closure of the subgroup generated by $x$. Then $A\_x$ is open, so has finite index, hence is infinite. Hence this is an abelian group with the required condition, so is isomorphic to $\mathbf{Z}\_p$.


Hence $G$ has a normal open subgroup $N$ isomorphic to $\mathbf{Z}\_p$; choose $N$ maximal with this property. Then $G/N$ is a $p$-group: indeed, otherwise, it has an element of prime order $\ell\neq p$; lift it to an element $x\in G$: then $A\_x$ cannot be isomorphic to $\mathbf{Z}\_p$ (since it surjects onto $\mathbf{Z}/\ell\mathbf{Z}$) and cannot be isomorphic to $\mathbf{Z}\_q$ for any prime $q\neq p$ (since it intersects $\mathbf{Z}\_p$ in a subgroup of finite index), contradiction with the abelian case.


Assuming by contradiction $N\neq G$, so the nontrivial $p$-group $G/N$ has a central subgroup of order $p$; let $H$ be its inverse image in $G$. So $N$ is normal of index $p$ in $H$. If $N$ is central in $H$, then since $H/N$ is cyclic, $H$ is abelian, hence isomorphic to $\mathbf{Z}\_p$ by the abelian case, contradicting the maximality of $N$. Fix $x$ not centralizing $H$. Since $A\_x$ is infinite, $x^p$ is a nontrivial element of $N$. Hence the action by conjugation of $x$ on $N\simeq\mathbf{Z}\_p$ has a nonzero fixed point. But any automorphism of the topological group $\mathbf{Z}\_p$ is given by $x\mapsto sx$ for $s\in\mathbf{Z}\_p^\times$, which fixes only zero. This yields a contradiction again. So $N=G$, hence $G\simeq\mathbf{Z}\_p$.


