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?
$$
begin{pmatrix}
1 & * & *\
0 & 1 & *\
0 & 0 & 1
end{pmatrix}
$$










share|cite|improve this question




























    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?
    $$
    begin{pmatrix}
    1 & * & *\
    0 & 1 & *\
    0 & 0 & 1
    end{pmatrix}
    $$










    share|cite|improve this question


























      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?
      $$
      begin{pmatrix}
      1 & * & *\
      0 & 1 & *\
      0 & 0 & 1
      end{pmatrix}
      $$










      share|cite|improve this question















      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






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      edited Dec 2 at 17:36









      YCor

      27k380132




      27k380132










      asked Dec 2 at 14:13









      Z.Zhou

      410114




      410114






















          1 Answer
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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.






          share|cite|improve this answer

















          • 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











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          1 Answer
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          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.






          share|cite|improve this answer

















          • 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















          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.






          share|cite|improve this answer

















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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














          • 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


















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