Profinite completion of finitely presented groups

Profinite completion of finitely presented groups



Let $G$ be a finitely presented group, $widehatG$ be the profinite completion of $G$, and $f: Grightarrow widehatG$ be the natural map.



My question is:



Is there an example of $G$ for which $textIm f$ is not finitely presented?




1 Answer
1



Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid tx^2t^-1=x^3rangle$$
Then $G$ is finitely presented; the image of $G$ in its profinite completion (i.e., the largest residually finite quotient of $G$) is $mathbfZ[1/6]rtimes_2/3mathbfZ$, which is not finitely presented. (Here $(2,3)$ can be replaced with any coprime pair $(n,m)$ with $n,mge 2$.)



Here's a similar example where I can provide details. Fix $nge 2$. Define
$$H_n:langle t,x,y|txt^-1=x^n,t^-1yt=y^n,[x,y]=1rangle;$$
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 $t^-1xt$ is $t^-1x^nt=x$).



It is not injective, because $[t^-1xt,y]$ 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 $mathbfZ^2=langle x,ymid [x,y]=1rangle$ with an isomorphism $langle x,y^nrangleto langle x^n,yrangle$ mapping $(x,y^n)$ to $(x^n,y)$.



As in every surjective endomorphism $u$ of a finitely generated group, all elements in $L_u=bigcup_mmathrmKer(u^m)$ belong to the kernel of the profinite completion homomorphism. In the present case, all elements $[t^-mxt^m,y]$ belong to $L_u$. But in the quotient by these additional relators, the elements $t^-mxt^m$ generate a copy of $mathbfZ[1/n]$, the elements $t^myt^-m$ generate another one, and they commute with each other. This allows to show that the quotient (by these additional relators) is isomorphic to $$Gamma_n=mathbfZ[1/n]^2rtimes_AmathbfZ,$$ where the action is by the diagonal matrix $A=mathrmdiag(n,n^-1)$. 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 $H_ntoGamma_n$ is the strictly increasing union of normal subgroups $mathrmKer(u^m)$, it is not finitely presented (the latter observation also follows from classical results of Bieri-Strebel).



Thanks for contributing an answer to MathOverflow!



But avoid



Use MathJax to format equations. MathJax reference.



To learn more, see our tips on writing great answers.



Required, but never shown



Required, but never shown




By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

Crossroads (UK TV series)

ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế