Linear representation of the free metabelian / 2-step nilpotent profinite groups on 2 generators
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Let G be the free profinite group on 2 generators, $A=G/[G,[G,G]],B=G/[[G,G],[G,G]]$, then what is the structure of the groups $A$ and $B$?
I heard that $A$ is isomorphic to the group of such ($3times 3$ below) matrices with entries in $hat{mathbb{Z}}$, is this right and why?
$$
begin{pmatrix}
1 & * & *\
0 & 1 & *\
0 & 0 & 1
end{pmatrix}
$$
gr.group-theory profinite-groups
edited Dec 2 at 17:36
YCor
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
add a comment |
up vote
5
down vote
favorite
Let G be the free profinite group on 2 generators, $A=G/[G,[G,G]],B=G/[[G,G],[G,G]]$, then what is the structure of the groups $A$ and $B$?
I heard that $A$ is isomorphic to the group of such ($3times 3$ below) matrices with entries in $hat{mathbb{Z}}$, is this right and why?
$$
begin{pmatrix}
1 & * & *\
0 & 1 & *\
0 & 0 & 1
end{pmatrix}
$$
gr.group-theory profinite-groups
edited Dec 2 at 17:36
YCor
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let G be the free profinite group on 2 generators, $A=G/[G,[G,G]],B=G/[[G,G],[G,G]]$, then what is the structure of the groups $A$ and $B$?
I heard that $A$ is isomorphic to the group of such ($3times 3$ below) matrices with entries in $hat{mathbb{Z}}$, is this right and why?
$$
begin{pmatrix}
1 & * & *\
0 & 1 & *\
0 & 0 & 1
end{pmatrix}
$$
gr.group-theory profinite-groups
edited Dec 2 at 17:36
YCor
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
Let G be the free profinite group on 2 generators, $A=G/[G,[G,G]],B=G/[[G,G],[G,G]]$, then what is the structure of the groups $A$ and $B$?
I heard that $A$ is isomorphic to the group of such ($3times 3$ below) matrices with entries in $hat{mathbb{Z}}$, is this right and why?
$$
begin{pmatrix}
1 & * & *\
0 & 1 & *\
0 & 0 & 1
end{pmatrix}
$$
gr.group-theory profinite-groups
gr.group-theory profinite-groups
edited Dec 2 at 17:36
YCor
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
edited Dec 2 at 17:36
YCor
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
edited Dec 2 at 17:36
YCor
27k380132
edited Dec 2 at 17:36
YCor
27k380132
edited Dec 2 at 17:36
YCor
27k380132
27k380132
asked Dec 2 at 14:13
Z.Zhou
430114
asked Dec 2 at 14:13
Z.Zhou
430114
asked Dec 2 at 14:13
Z.Zhou
430114
430114
add a comment |
add a comment |
1 Answer
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oldest
votes
up vote
7
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The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $widehat{mathbb Z}^X$, the free pro-abelian group on $X$. Then we can consider the edge set of its Cayley graph $E=widehat{mathbb Z}^Xtimes X$, which is a profinite space with the product topology. Let $H$ be the free pro-abelian group on the profinite space $E$. Then $widehat{mathbb Z}^X$ acts continuously on $E$ via the usual action on its Cayley graph, i.e., via left multiplication in the first coordinate and this extends to a continuous action on $H$ by automorphisms. Form the semidirect product $Hrtimes widehat{mathbb Z}^X$. Then your group $B$ embeds in $Hrtimes widehat{Z}^X$ in the following way. Send $xin X$ to the pair $((1,x),x)$ where $(1,x)$ should be thought of as the edge from $1$ to $x$ labeled by $x$ in the Cayley graph and the second $x$ is the corresponding generator of $widehat{mathbb Z}^X$. This extends to an embedding of $G$.
Your question about $A$ boils down to whether the $3times 3$ Heisenberg group has the congruence subgroup property, which I leave to more knowledgeable people than I.
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $widehat{mathbb Z}^X$, the free pro-abelian group on $X$. Then we can consider the edge set of its Cayley graph $E=widehat{mathbb Z}^Xtimes X$, which is a profinite space with the product topology. Let $H$ be the free pro-abelian group on the profinite space $E$. Then $widehat{mathbb Z}^X$ acts continuously on $E$ via the usual action on its Cayley graph, i.e., via left multiplication in the first coordinate and this extends to a continuous action on $H$ by automorphisms. Form the semidirect product $Hrtimes widehat{mathbb Z}^X$. Then your group $B$ embeds in $Hrtimes widehat{Z}^X$ in the following way. Send $xin X$ to the pair $((1,x),x)$ where $(1,x)$ should be thought of as the edge from $1$ to $x$ labeled by $x$ in the Cayley graph and the second $x$ is the corresponding generator of $widehat{mathbb Z}^X$. This extends to an embedding of $G$.
Your question about $A$ boils down to whether the $3times 3$ Heisenberg group has the congruence subgroup property, which I leave to more knowledgeable people than I.
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
add a comment |
up vote
7
down vote
accepted
The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $widehat{mathbb Z}^X$, the free pro-abelian group on $X$. Then we can consider the edge set of its Cayley graph $E=widehat{mathbb Z}^Xtimes X$, which is a profinite space with the product topology. Let $H$ be the free pro-abelian group on the profinite space $E$. Then $widehat{mathbb Z}^X$ acts continuously on $E$ via the usual action on its Cayley graph, i.e., via left multiplication in the first coordinate and this extends to a continuous action on $H$ by automorphisms. Form the semidirect product $Hrtimes widehat{mathbb Z}^X$. Then your group $B$ embeds in $Hrtimes widehat{Z}^X$ in the following way. Send $xin X$ to the pair $((1,x),x)$ where $(1,x)$ should be thought of as the edge from $1$ to $x$ labeled by $x$ in the Cayley graph and the second $x$ is the corresponding generator of $widehat{mathbb Z}^X$. This extends to an embedding of $G$.
Your question about $A$ boils down to whether the $3times 3$ Heisenberg group has the congruence subgroup property, which I leave to more knowledgeable people than I.
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $widehat{mathbb Z}^X$, the free pro-abelian group on $X$. Then we can consider the edge set of its Cayley graph $E=widehat{mathbb Z}^Xtimes X$, which is a profinite space with the product topology. Let $H$ be the free pro-abelian group on the profinite space $E$. Then $widehat{mathbb Z}^X$ acts continuously on $E$ via the usual action on its Cayley graph, i.e., via left multiplication in the first coordinate and this extends to a continuous action on $H$ by automorphisms. Form the semidirect product $Hrtimes widehat{mathbb Z}^X$. Then your group $B$ embeds in $Hrtimes widehat{Z}^X$ in the following way. Send $xin X$ to the pair $((1,x),x)$ where $(1,x)$ should be thought of as the edge from $1$ to $x$ labeled by $x$ in the Cayley graph and the second $x$ is the corresponding generator of $widehat{mathbb Z}^X$. This extends to an embedding of $G$.
Your question about $A$ boils down to whether the $3times 3$ Heisenberg group has the congruence subgroup property, which I leave to more knowledgeable people than I.
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
The group $B$, the free pro-metabelian group, has the following description, due to Jorge Almeida. I’ll do it for an arbitrary finite set $|X|$ of cadinality at least $2$. Consider $widehat{mathbb Z}^X$, the free pro-abelian group on $X$. Then we can consider the edge set of its Cayley graph $E=widehat{mathbb Z}^Xtimes X$, which is a profinite space with the product topology. Let $H$ be the free pro-abelian group on the profinite space $E$. Then $widehat{mathbb Z}^X$ acts continuously on $E$ via the usual action on its Cayley graph, i.e., via left multiplication in the first coordinate and this extends to a continuous action on $H$ by automorphisms. Form the semidirect product $Hrtimes widehat{mathbb Z}^X$. Then your group $B$ embeds in $Hrtimes widehat{Z}^X$ in the following way. Send $xin X$ to the pair $((1,x),x)$ where $(1,x)$ should be thought of as the edge from $1$ to $x$ labeled by $x$ in the Cayley graph and the second $x$ is the corresponding generator of $widehat{mathbb Z}^X$. This extends to an embedding of $G$.
Your question about $A$ boils down to whether the $3times 3$ Heisenberg group has the congruence subgroup property, which I leave to more knowledgeable people than I.
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
answered Dec 2 at 15:40
Benjamin Steinberg
22.9k265124
22.9k265124
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
add a comment |
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
3
3
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Yes it's very easy to show by hand that the $3times 3$-Heisenberg group has the congruence subgroup property. More generally this holds in general, in the sense that for every unipotent $mathbf{Q}$-subgroup $U$ of $mathrm{GL}_d$, every finite index subgroup of $mathrm{GL}_d(mathbf{Z})cap G(mathbf{Q})$ has the congruence subgroup property.
– YCor
Dec 2 at 17:39
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
Thanks @YCor I suspected as much.
– Benjamin Steinberg
Dec 2 at 17:46
add a comment |
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