3
$\begingroup$

Having a centerless profinite completion leads to some nice properties. For example, given a short exact sequence

$$1\to A\to B\to C\to 1$$

where $A$ is finitely generated and $\hat{A}$ has trivial center, we have an exact sequence

$$1\to\hat{A}\to\hat{B}\to\hat{C}\to 1.$$

When does a centerless group $G$ have a centerless profinite completion $\hat{G}$? Does this change if $G$ is finitely generated and/or residually finite?

I know that if $G$ is residually finite, then we have an injection $G\to \hat{G}$ with dense image, and so $Z(\hat{G})$ would have to live solely within $(\hat{G}\setminus G)\cup\{e\}$. This seems unlikely, but I don't see the proof.

$\endgroup$
  • 1
    $\begingroup$ The case when $G$ is assumed both residually finite and finitely generated sounds tricker (although there should be examples); I'd suggest to post it separately on MathOverflow. $\endgroup$ – YCor Jun 9 at 11:02
  • $\begingroup$ @YCor Thanks for the good answer! I went ahead and posted the question here on MathOverflow as well. $\endgroup$ – Santana Afton Jun 9 at 15:35
3
$\begingroup$

Here's a finitely generated group with trivial center whose profinite completion is isomorphic to the profinite completion of $\mathbf{Z}$ (hence with nontrivial center).

Namely, the subgroup of the group $\mathfrak{S}(\mathbf{Z})$ generated by the alternating group $A$ (even finitely supported permutations) and the shift $s:n\mapsto n+1$. It is actually generated by $s$ and by the 3-cycle $(012)$. This is a semidirect product $A\rtimes\langle s\rangle$. It has trivial center since the centralizer of the alternating subgroup in the whole symmetric group is trivial. Since $A$ is infinite simple, its image in the profinite completion is trivial, whence the claim.

I'm not sure now about a residually finite example.


Here's a residually finite example (not finitely generated). Consider the Baumslag-Solitar group $\mathbf{Z}_{(p)}\rtimes_{p+1}\mathbf{Z}$, where (to simplify) $p$ is prime and the action is by multiplication by $p+1$. It is easy to see that the profinite completion is $$\mathbf{Z}_p\rtimes_{p+1}\hat{\mathbf{Z}};$$ the action is not faithful because it factors through $\mathbf{Z}_p\subset \hat{\mathbf{Z}}$. So all the kernel of the action is central.

$\endgroup$
  • $\begingroup$ Hm, so you’re saying that if $H = A\ltimes\langle s\rangle$ then $\hat{H} = \hat{\mathbb{Z}}$? Strange. I’d be surprised if finite generation and residual finiteness isn’t enough, but I’ve been surprised before. $\endgroup$ – Santana Afton Jun 4 at 22:19
  • $\begingroup$ I meant $\rtimes$, I fixed the typo. I used that $A$ has trivial profinite completion too. $\endgroup$ – YCor Jun 5 at 4:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.