# Profinite completion of finitely presented groups

Let $G$ be a finitely presented group, $\widehat{G}$ be the profinite completion of $G$, and $f: G\rightarrow \widehat{G}$ be the natural map.

My question is:

Is there an example of $G$ for which $\text{Im} f$ is not finitely presented?

Yes. Take the Baumslag-Solitar group $$G=\mathrm{BS}(2,3)=\langle t,x\mid tx^2t^{-1}=x^3\rangle$$ Then $G$ is finitely presented; the image of $G$ in its profinite completion (i.e., the largest residually finite quotient of $G$) is $\mathbf{Z}[1/6]\rtimes_{2/3}\mathbf{Z}$, which is not finitely presented. (Here $(2,3)$ can be replaced with any coprime pair $(n,m)$ with $n,m\ge 2$.)

Here's a similar example where I can provide details. Fix $n\ge 2$. Define $$H_n:\langle t,x,y|txt^{-1}=x^n,t^{-1}yt=y^n,[x,y]=1\rangle;$$ let $u$ be the group endomorphism of $H_n$ mapping $(t,x,y)\mapsto (t,x^n,y)$. It well-defined, since the triple of images satisfies the relators, and is clearly surjective (since the image of $t^{-1}xt$ is $t^{-1}x^nt=x$).

It is not injective, because $[t^{-1}xt,y]$ belongs to the kernel; to show that this element is not trivial can be obtained by observing that the presentation describes $H_n$ as an HNN-extension of $\mathbf{Z}^2=\langle x,y\mid [x,y]=1\rangle$ with an isomorphism $\langle x,y^n\rangle\to \langle x^n,y\rangle$ mapping $(x,y^n)$ to $(x^n,y)$.

As in every surjective endomorphism $u$ of a finitely generated group, all elements in $L_u=\bigcup_m\mathrm{Ker}(u^m)$ belong to the kernel of the profinite completion homomorphism. In the present case, all elements $[t^{-m}xt^m,y]$ belong to $L_u$. But in the quotient by these additional relators, the elements $t^{-m}xt^m$ generate a copy of $\mathbf{Z}[1/n]$, the elements $t^{m}yt^{-m}$ generate another one, and they commute with each other. This allows to show that the quotient (by these additional relators) is isomorphic to $$\Gamma_n=\mathbf{Z}[1/n]^2\rtimes_A\mathbf{Z},$$ where the action is by the diagonal matrix $A=\mathrm{diag}(n,n^{-1})$. The group $\Gamma_n$ is residually finite, and is easily deduced to be the largest residually finite quotient of $H_n$. Since the kernel of the quotient homomorphism $H_n\to\Gamma_n$ is the strictly increasing union of normal subgroups $\mathrm{Ker}(u^m)$, it is not finitely presented (the latter observation also follows from classical results of Bieri-Strebel).

• I am very late on this, but do you have a reference for the BS example? Commented Apr 17, 2020 at 12:40
• @frafour I should but I don't have it at the top of my head (but I can write the proof down if you like).
– YCor
Commented Apr 17, 2020 at 14:06