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$p$-completion is pro-$p$ free

Why is the free pro-c-group on an infinite set not the pro-c-completion of the free group?Group presentations - againBasis for the completion of a free moduleNontrivial examples of pro-$p$ groupsClosure in profinite topologyIf a subset of a free group $F$ is Nielsen reduced, then it is a basis of $F$. Is the converse statement true?Is pronilpotent completion of free group projective?Equivalent basis of free groupsAbelianization of free profinite groupExtension of surface group by cyclic is residually finite

0

$begingroup$

Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, mathbb{F}_p) neq 0 implies H^2(widehat{G_p},mathbb{F}_p) neq 0$. There seems to be no need for a basis of $widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.

abstract-algebra group-theory free-groups group-presentation profinite-groups

share|cite|improve this question

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  web.ma.utexas.edu/users/areid/StAndrews3.pdf This paper seems to pose this as an open problem for the profinite case.
  $endgroup$
  – Henrique Augusto Souza
  2 days ago
  

  
  
  
  


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  $begingroup$
  It's clearly false, just take $G=mathbf{Z}/qmathbf{Z}$ for $qge 2$ coprime to $p$. You should at least assume that $G$ is residually-$p$.
  $endgroup$
  – YCor
  2 days ago
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  If it's true (assuming residually-$p$), it will certainly not be an easy argument. I'm rather inclined to believe that it's false (although it might be hard to disprove), i.e. construct a counterexample.
  $endgroup$
  – YCor
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YCor Yes, that is indeed a counterexample... However, in this case we have $widehat{G_p} = {0}$! While the trivial group is indeed free, it is a trivial type of free group. What if we suppose that $widehat{G_p}$ is a free pro-$p$ group of rank $geq 1$? Also, Corollary 4.13 of that paper establishes conditions of when is this valid for the profinite completion, but it doesn't seem to work in general either.
  $endgroup$
  – Henrique Augusto Souza
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's trivial to convert my example to another one yielding a nontrivial free group.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, mathbb{F}_p) neq 0 implies H^2(widehat{G_p},mathbb{F}_p) neq 0$. There seems to be no need for a basis of $widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.

abstract-algebra group-theory free-groups group-presentation profinite-groups

share|cite|improve this question

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  web.ma.utexas.edu/users/areid/StAndrews3.pdf This paper seems to pose this as an open problem for the profinite case.
  $endgroup$
  – Henrique Augusto Souza
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's clearly false, just take $G=mathbf{Z}/qmathbf{Z}$ for $qge 2$ coprime to $p$. You should at least assume that $G$ is residually-$p$.
  $endgroup$
  – YCor
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If it's true (assuming residually-$p$), it will certainly not be an easy argument. I'm rather inclined to believe that it's false (although it might be hard to disprove), i.e. construct a counterexample.
  $endgroup$
  – YCor
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YCor Yes, that is indeed a counterexample... However, in this case we have $widehat{G_p} = {0}$! While the trivial group is indeed free, it is a trivial type of free group. What if we suppose that $widehat{G_p}$ is a free pro-$p$ group of rank $geq 1$? Also, Corollary 4.13 of that paper establishes conditions of when is this valid for the profinite completion, but it doesn't seem to work in general either.
  $endgroup$
  – Henrique Augusto Souza
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's trivial to convert my example to another one yielding a nontrivial free group.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, mathbb{F}_p) neq 0 implies H^2(widehat{G_p},mathbb{F}_p) neq 0$. There seems to be no need for a basis of $widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.

abstract-algebra group-theory free-groups group-presentation profinite-groups

share|cite|improve this question

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

$endgroup$

share|cite|improve this question

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

share|cite|improve this question

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514

asked 2 days ago

Henrique Augusto SouzaHenrique Augusto Souza

1,313514