Profinite completion of finitely presented groups

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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 a counterexample for $G$ such that $textIm f$ is not finitely presented?

gr.group-theory abstract-algebra profinite-groups

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asked 1 hour ago

Bruno

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up vote
4
down vote

favorite

1

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 a counterexample for $G$ such that $textIm f$ is not finitely presented?

gr.group-theory abstract-algebra profinite-groups

share|cite|improve this question

asked 1 hour ago

Bruno

233

New contributor

Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

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 a counterexample for $G$ such that $textIm f$ is not finitely presented?

gr.group-theory abstract-algebra profinite-groups

share|cite|improve this question

asked 1 hour ago

Bruno

233

New contributor

Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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 a counterexample for $G$ such that $textIm f$ is not finitely presented?

gr.group-theory abstract-algebra profinite-groups

gr.group-theory abstract-algebra profinite-groups

share|cite|improve this question

asked 1 hour ago

Bruno

233

New contributor

Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 1 hour ago

Bruno

233

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Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

Bruno

233

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Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Bruno

233

asked 1 hour ago

Bruno

233

233

New contributor

Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Bruno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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1 Answer
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Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid txt^-1rangle$$
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'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 endomorphism $u$ 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).

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edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the example!
  – Bruno
  1 hour ago
  

  
  
  
  

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid txt^-1rangle$$
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'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 endomorphism $u$ 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).

share|cite|improve this answer

edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the example!
  – Bruno
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid txt^-1rangle$$
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'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 endomorphism $u$ 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).

share|cite|improve this answer

edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the example!
  – Bruno
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid txt^-1rangle$$
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'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 endomorphism $u$ 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).

share|cite|improve this answer

edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

Yes. Take the Baumslag-Solitar group
$$G=mathrmBS(2,3)=langle t,xmid txt^-1rangle$$
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'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 endomorphism $u$ 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).

share|cite|improve this answer

edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

share|cite|improve this answer

share|cite|improve this answer

edited 59 mins ago

edited 59 mins ago

edited 59 mins ago

answered 1 hour ago

YCor

24.9k274117

answered 1 hour ago

YCor

24.9k274117

answered 1 hour ago

YCor

24.9k274117

24.9k274117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the example!
  – Bruno
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the example!
  – Bruno
  1 hour ago
  

  
  
  
  

Thanks for the example!
– Bruno
1 hour ago

Thanks for the example!
– Bruno
1 hour ago

add a comment | 

Bruno is a new contributor. Be nice, and check out our Code of Conduct.

 

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