Is residual finiteness a quasi isometry invariant for f.g. groups?Is Hopf property a quasi-isometry...
Is residual finiteness a quasi isometry invariant for f.g. groups?
Is Hopf property a quasi-isometry invariant?Residual Finiteness of Fundamental Group of Compact 3-ManifoldIs Hopf property a quasi-isometry invariant?Characterizations of product groups under quasi-isometryQuasi-isometry and left invariant orderability for groupsResidual finiteness: why do we care?Is being Noetherian a quasi-isometric invariance for f.g. groups?A generalization of residual finiteness to topological groupsKazhdan's property (T) vs. residual finitenessThe finiteness criterium $F$ under quasi-isometryWeaker version of locally extended residual finiteness
$begingroup$
A "residually finite group" is group for which the intersection of all finite index subgroups is trivial. Suppose $G$ and $G'$ are two quasi-isometric finitely generated groups. Does the residual finiteness of $G$ implies the same property for $G'$?
gr.group-theory geometric-group-theory
edited 2 hours ago
YCor
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
$endgroup$
add a comment |
$begingroup$
A "residually finite group" is group for which the intersection of all finite index subgroups is trivial. Suppose $G$ and $G'$ are two quasi-isometric finitely generated groups. Does the residual finiteness of $G$ implies the same property for $G'$?
gr.group-theory geometric-group-theory
edited 2 hours ago
YCor
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
$endgroup$
$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago
add a comment |
$begingroup$
A "residually finite group" is group for which the intersection of all finite index subgroups is trivial. Suppose $G$ and $G'$ are two quasi-isometric finitely generated groups. Does the residual finiteness of $G$ implies the same property for $G'$?
gr.group-theory geometric-group-theory
edited 2 hours ago
YCor
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
$endgroup$
A "residually finite group" is group for which the intersection of all finite index subgroups is trivial. Suppose $G$ and $G'$ are two quasi-isometric finitely generated groups. Does the residual finiteness of $G$ implies the same property for $G'$?
gr.group-theory geometric-group-theory
gr.group-theory geometric-group-theory
edited 2 hours ago
YCor
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
edited 2 hours ago
YCor
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
edited 2 hours ago
YCor
28k482135
edited 2 hours ago
YCor
28k482135
edited 2 hours ago
YCor
28k482135
28k482135
asked 3 hours ago
MostafaMostafa
2,15111133
asked 3 hours ago
MostafaMostafa
2,15111133
asked 3 hours ago
MostafaMostafa
2,15111133
2,15111133
$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago
add a comment |
$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago
$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago
$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago
add a comment |
1 Answer
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$begingroup$
No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ (which is RF) and the wreath product $Qwrmathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(mathbf{Z}/2mathbf{Z})^3wrmathbf{Z}$, and the latter shares a (non-labeled) Cayley graph with $Qwrmathbf{Z}$.
Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finite-by-RF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $Cwrmathbf{Z}$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.
answered 2 hours ago
YCorYCor
28k482135
$endgroup$
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ (which is RF) and the wreath product $Qwrmathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(mathbf{Z}/2mathbf{Z})^3wrmathbf{Z}$, and the latter shares a (non-labeled) Cayley graph with $Qwrmathbf{Z}$.
Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finite-by-RF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $Cwrmathbf{Z}$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.
answered 2 hours ago
YCorYCor
28k482135
$endgroup$
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
add a comment |
$begingroup$
No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ (which is RF) and the wreath product $Qwrmathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(mathbf{Z}/2mathbf{Z})^3wrmathbf{Z}$, and the latter shares a (non-labeled) Cayley graph with $Qwrmathbf{Z}$.
Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finite-by-RF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $Cwrmathbf{Z}$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.
answered 2 hours ago
YCorYCor
28k482135
$endgroup$
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
add a comment |
$begingroup$
No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ (which is RF) and the wreath product $Qwrmathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(mathbf{Z}/2mathbf{Z})^3wrmathbf{Z}$, and the latter shares a (non-labeled) Cayley graph with $Qwrmathbf{Z}$.
Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finite-by-RF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $Cwrmathbf{Z}$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.
answered 2 hours ago
YCorYCor
28k482135
$endgroup$
No: let $Q$ be a non-abelian group of order 8. Then the standard lamplighter groups $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ (which is RF) and the wreath product $Qwrmathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(mathbf{Z}/2mathbf{Z})wrmathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(mathbf{Z}/2mathbf{Z})^3wrmathbf{Z}$, and the latter shares a (non-labeled) Cayley graph with $Qwrmathbf{Z}$.
Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finite-by-RF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $Cwrmathbf{Z}$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.
answered 2 hours ago
YCorYCor
28k482135
edited 2 hours ago
answered 2 hours ago
YCorYCor
28k482135
answered 2 hours ago
YCorYCor
28k482135
answered 2 hours ago
YCorYCor
28k482135
28k482135
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
add a comment |
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
$begingroup$
More generally if G and A are finite groups of the same cardinality with G non-abelian and A abelian and you take the Cayley graphs of $Gwr mathbb Z$ and $Awr mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not.
$endgroup$
– Benjamin Steinberg
28 secs ago
add a comment |
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$begingroup$
Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/…
$endgroup$
– YCor
2 hours ago