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I apologize if this question is basic in some sense. I was looking for an example of a non-proper HNN-extension and I found this.

In the comments, markvs mentioned the Baumslag-Solitar group $B(2,3)$. We can show that this is non-Hopfian using Britton's lemma so that is not residually finite. So, its natural try to use this to get a non-proper example. We know also that $B(2,3)$ is an HNN-extension of $\Bbb{Z}$ with respect to the natural $f: \langle a^2 \rangle \to \langle a^3 \rangle$. Now, how to proceed now? I mean, there is a way to decompose $\widehat{B(2,3)}$ as an HNN-extension of $\widehat{\Bbb{Z}}$ or something?

Intuitively, makes sense to me to do $B(2,3) = \operatorname{HNN}(\Bbb{Z})$ then $\widehat{B(2,3)} = \operatorname{HNN}(\Bbb{\widehat{Z}})$, but I don't know if it works.

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    $\begingroup$ In general it is certainly NOT true that graph-of-groups decompositions of a group lead to corresponding decompositions of the profinite completion. For instance, there are amalgams of free groups $F_1*_M F_2$ without any finite quotients at all. (I think the first such example is due to Meenaxi Bhattacharjee.) $\endgroup$
    – HJRW
    Mar 24 at 13:12

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The largest residually finite quotient of $\mathrm{BS}(2,3)$ is the semidirect product $\mathbf{Z}\ltimes_{2/3}\mathbf{Z}[1/6]$.

So, if I'm correct, its profinite completion is $$\widehat{\mathbf{Z}}\ltimes_{2/3}\widehat{\mathbf{Z}[1/6]}.$$

Each time, the $2/3$ means that the generator $1$ acts by multiplication by $2/3$. Denoting by $P$ the set of primes, note that $\widehat{\mathbf{Z}}=\prod_{p\in P}\mathbf{Z}_p$ and $\widehat{\mathbf{Z}[1/6]}=\prod_{p\in P-\{2,3\}}\mathbf{Z}_p$. The group of automorphisms of the latter is also profinite, namely $\prod_{p\in P-\{2,3\}}\mathbf{Z}_p^\times$, so indeed the homomorphism $\mathbf{Z}\to \widehat{\mathbf{Z}[1/6]}^\times$ mapping $1$ to multiplication by $2/3$ extends uniquely to a continuous homomorphism $\widehat{\mathbf{Z}}\to \widehat{\mathbf{Z}[1/6]}^\times$.

A similar description should hold for all $\mathrm{BS}(m,n)$ whenever $\gcd(m,n)=1$. (Otherwise it's quite different since then the profinite completion is not solvable.)

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  • $\begingroup$ What role does the solvability of the profinite completion play here? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Mar 24 at 10:54
  • $\begingroup$ @Carl-FredrikNybergBrodda in these cases we get an easy description of the profinite completion as semidirect product of two procyclic groups. (While in other cases, the profinite completion contains a nonabelian free subgroup.) $\endgroup$
    – YCor
    Mar 24 at 10:58
  • $\begingroup$ OK, thanks, I suppose the part I don't quite see is why the profinite completion is not solvable in the case of $\operatorname{gcd}(m,n)=1$, i.e. (as you point out) where the non-abelian free subgroup comes from. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Mar 24 at 11:10
  • $\begingroup$ @Carl-FredrikNybergBrodda if $p\ge 2$ divides both $m,n$, the Baumslag-Solitar group $\langle t,x:tx^m=x^nt\rangle$ admits as a quotient $\langle t,x:x^p=1\rangle$ which is the free product $\mathbf{Z}\ast (\mathbf{Z}/p\mathbf{Z})$, and is residually finite with a non-abelian free subgroup (of finite index). $\endgroup$
    – YCor
    Mar 24 at 13:33

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