# When is the profinite completion of a centerless group itself centerless?

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.

• 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. – YCor Jun 9 at 11:02
• @YCor Thanks for the good answer! I went ahead and posted the question here on MathOverflow as well. – Santana Afton Jun 9 at 15:35

## 1 Answer

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.

• 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. – Santana Afton Jun 4 at 22:19
• I meant $\rtimes$, I fixed the typo. I used that $A$ has trivial profinite completion too. – YCor Jun 5 at 4:38