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Does a (nice) centerless group always have a centerless profinite completion?
Exotic automorphisms of the fundamental group of a curve?Free subgroups vs law Open subgroups of free pro-C groupsInteresting commensurated subgroups of countable groupsProfinite groups, completions, and Schreier's formulaDense subgroups in subgroups of profinite groupsWhen is first group cohomology isomorphic to conjugacy classes of sections?What is the probability of generating a given procyclic subgroup in $mathrmGal(barK/K)$?Profinite closure of characteristic subgroupMaximal subgroups of infinite index and profinite completion
$begingroup$
This is an extension of a question I asked here on Math.SE
Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $hatG$ also has trivial center?
In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact.
gr.group-theory profinite-groups
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
$endgroup$
add a comment |
$begingroup$
This is an extension of a question I asked here on Math.SE
Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $hatG$ also has trivial center?
In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact.
gr.group-theory profinite-groups
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
$endgroup$
add a comment |
$begingroup$
This is an extension of a question I asked here on Math.SE
Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $hatG$ also has trivial center?
In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact.
gr.group-theory profinite-groups
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
$endgroup$
This is an extension of a question I asked here on Math.SE
Assume that I have a finitely generated residually finite centerless group $G$. Is it true that the profinite completion $hatG$ also has trivial center?
In the linked question, user YCor was able to show that this fails in general if you do not assume either finite generation or residually finite. However, the result happens to be true if $G$ is a surface group. I’d like to know if this is a phenomenon specific to surface groups, or if this is a more general fact.
gr.group-theory profinite-groups
gr.group-theory profinite-groups
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
edited Jun 9 at 18:18
Santana Afton
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
asked Jun 9 at 15:33
Santana AftonSantana Afton
1609 bronze badges
1609 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$widehat mathrmSL_n(mathbb Z)= prod _q ; mathrmprime mathrmSL_n(mathbb Z_q)supset widehat Gamma supset prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
$endgroup$
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$widehat mathrmSL_n(mathbb Z)= prod _q ; mathrmprime mathrmSL_n(mathbb Z_q)supset widehat Gamma supset prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
$endgroup$
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
add a comment |
$begingroup$
The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$widehat mathrmSL_n(mathbb Z)= prod _q ; mathrmprime mathrmSL_n(mathbb Z_q)supset widehat Gamma supset prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
$endgroup$
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
add a comment |
$begingroup$
The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$widehat mathrmSL_n(mathbb Z)= prod _q ; mathrmprime mathrmSL_n(mathbb Z_q)supset widehat Gamma supset prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
$endgroup$
The answer is No in general. Let $ngeq 3$ be odd (it is not necessary that $n$ be odd) and suppose $G=mathrmSL_n(mathbb Z)$. There exists a subgroup $Gamma subset mathrmSL_n(mathbb Z)$ of finite index which is torsion-free and centreless (the centre can only be $pm 1$ and because $n$ is odd the centre can only be trivial). However, $mathrmSL_n(mathbb Z)$ has the congruence subgroup property which means that we have the following inclusion of the profinite completions:
$$widehat mathrmSL_n(mathbb Z)= prod _q ; mathrmprime mathrmSL_n(mathbb Z_q)supset widehat Gamma supset prod _pin S U_p times prod _ ell notin S mathrmSL_n(mathbb Z_ell),$$
where $S$ is a finite set of primes, $U_p$ is an open subgroup of finite index in $mathrmSL_n(mathbb Z_p)$, and $ell$ runs through primes in the complement of $S$. Since for infinitely many $ell$ (for example, all $ell$ with $ellequiv 1 ; (mathrmmod;n)$), the group $mathrmSL_n(mathbb Z_ell)$ has $n$-th roots of unity in the centre, it follows that the profinite completion of $Gamma $ is not centreless.
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
edited Jun 15 at 21:28
YCor
29.9k4 gold badges89 silver badges144 bronze badges
29.9k4 gold badges89 silver badges144 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
answered Jun 9 at 15:54
VenkataramanaVenkataramana
9,5991 gold badge32 silver badges54 bronze badges
9,5991 gold badge32 silver badges54 bronze badges
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
add a comment |
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
1
1
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
$begingroup$
Great. It seems to work for every finite index subgroup of $mathrmSL_n(mathbfZ)$ when $n$ is odd, including $mathrmSL_n(mathbfZ)$ itself (and for all centerless finite index subgroups, for arbitrary $nge 3$).
$endgroup$
– YCor
Jun 9 at 16:56
1
1
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
$begingroup$
@YCor: thanks for the comment. Yes, torsion free is not assumed by the OP, so it works for every finite index subgroup if $ngeq 3$ is odd.
$endgroup$
– Venkataramana
Jun 10 at 1:17
add a comment |
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