Embedding of torsion free nilpotent group into Unitriangular matrix group $UT_n(mathbbZ)$
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I have given the presentation $ G = langle a_1,a_2 | [[a_i,a_j],a_k],forall i,j,kin1,2, a_1^2 = a_2^2 rangle $ of a torsionfree, nilpotent group which looks pretty similar to the Heisenberg group $ H = langle x,y | [[x,y],x] = [[x,y],y] = 1 rangle $ except for the relation $ a_1^2 = a_2^2$. This additional relation is already confusing to work with. How can two elements satisfy this relation without being inverse or equal to each other?
Second I figured out $G$ can be written as $G = langle a_1,a_2 | [a_1,a_2]=[a_2,a_1], a_1^2=a_2^2 rangle $ using both relations. Now I'm stuck with embedding $G$ into the group of unitriangular matrices $ UT_n(mathbbZ)$. Basically because the second relation doesn't make sense to me. Here the Heisenberg group is embedded Group presentation for discrete Heisenberg group.
Unfortunatly I am not able to adapt. Maybe there is also a way to picture the group elements of $G$ in a different way, I'm not exactly sure how the words are generated. I'm not asking for a full solution but every hint in the right direction is highly appreciated.
Thanks a lot for taking the time to check out the question!
abstract-algebra group-theory geometric-group-theory group-presentation
asked Aug 27 at 16:37
Arwid
61
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I have given the presentation $ G = langle a_1,a_2 | [[a_i,a_j],a_k],forall i,j,kin1,2, a_1^2 = a_2^2 rangle $ of a torsionfree, nilpotent group which looks pretty similar to the Heisenberg group $ H = langle x,y | [[x,y],x] = [[x,y],y] = 1 rangle $ except for the relation $ a_1^2 = a_2^2$. This additional relation is already confusing to work with. How can two elements satisfy this relation without being inverse or equal to each other?
Second I figured out $G$ can be written as $G = langle a_1,a_2 | [a_1,a_2]=[a_2,a_1], a_1^2=a_2^2 rangle $ using both relations. Now I'm stuck with embedding $G$ into the group of unitriangular matrices $ UT_n(mathbbZ)$. Basically because the second relation doesn't make sense to me. Here the Heisenberg group is embedded Group presentation for discrete Heisenberg group.
Unfortunatly I am not able to adapt. Maybe there is also a way to picture the group elements of $G$ in a different way, I'm not exactly sure how the words are generated. I'm not asking for a full solution but every hint in the right direction is highly appreciated.
Thanks a lot for taking the time to check out the question!
abstract-algebra group-theory geometric-group-theory group-presentation
asked Aug 27 at 16:37
Arwid
61
The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
I have given the presentation $ G = langle a_1,a_2 | [[a_i,a_j],a_k],forall i,j,kin1,2, a_1^2 = a_2^2 rangle $ of a torsionfree, nilpotent group which looks pretty similar to the Heisenberg group $ H = langle x,y | [[x,y],x] = [[x,y],y] = 1 rangle $ except for the relation $ a_1^2 = a_2^2$. This additional relation is already confusing to work with. How can two elements satisfy this relation without being inverse or equal to each other?
Second I figured out $G$ can be written as $G = langle a_1,a_2 | [a_1,a_2]=[a_2,a_1], a_1^2=a_2^2 rangle $ using both relations. Now I'm stuck with embedding $G$ into the group of unitriangular matrices $ UT_n(mathbbZ)$. Basically because the second relation doesn't make sense to me. Here the Heisenberg group is embedded Group presentation for discrete Heisenberg group.
Unfortunatly I am not able to adapt. Maybe there is also a way to picture the group elements of $G$ in a different way, I'm not exactly sure how the words are generated. I'm not asking for a full solution but every hint in the right direction is highly appreciated.
Thanks a lot for taking the time to check out the question!
abstract-algebra group-theory geometric-group-theory group-presentation
asked Aug 27 at 16:37
Arwid
61
I have given the presentation $ G = langle a_1,a_2 | [[a_i,a_j],a_k],forall i,j,kin1,2, a_1^2 = a_2^2 rangle $ of a torsionfree, nilpotent group which looks pretty similar to the Heisenberg group $ H = langle x,y | [[x,y],x] = [[x,y],y] = 1 rangle $ except for the relation $ a_1^2 = a_2^2$. This additional relation is already confusing to work with. How can two elements satisfy this relation without being inverse or equal to each other?
Second I figured out $G$ can be written as $G = langle a_1,a_2 | [a_1,a_2]=[a_2,a_1], a_1^2=a_2^2 rangle $ using both relations. Now I'm stuck with embedding $G$ into the group of unitriangular matrices $ UT_n(mathbbZ)$. Basically because the second relation doesn't make sense to me. Here the Heisenberg group is embedded Group presentation for discrete Heisenberg group.
Unfortunatly I am not able to adapt. Maybe there is also a way to picture the group elements of $G$ in a different way, I'm not exactly sure how the words are generated. I'm not asking for a full solution but every hint in the right direction is highly appreciated.
Thanks a lot for taking the time to check out the question!
abstract-algebra group-theory geometric-group-theory group-presentation
asked Aug 27 at 16:37
Arwid
61
asked Aug 27 at 16:37
Arwid
61
asked Aug 27 at 16:37
Arwid
61
asked Aug 27 at 16:37
Arwid
61
61
The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56
add a comment |Â
The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56
The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56
add a comment |Â
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The element $[a_1,a_2]$ has order $2$ (that follows from your relation $[a_1,a_2]=[a_2,a_1]$) , so $G$ is not torsion-free, and so it cannot be embedded into any $rm UT_n(mathbb Z)$.
– Derek Holt
Aug 27 at 18:10
Yes, there are several maximal torsion-free subgroups, $langle a_1 rangle$, $langle a_2 rangle$, $langle a_1[a_1,a_2] rangle$, etc, and they are all infinite cyclic.
– Derek Holt
Aug 27 at 19:56
I checked and the transformation is indeed correct and the group itself not torsionfree. Am I right that the biggest torsion free subgroup is $G_1 = langle a_1,a_2 | [[a_i,a_j]a_k] , i=j=k , a_1^2 = a_2^2 rangle = langle a_1,a_2 | a_1 = a_2 rangle = langle a_1 rangle = mathbbZ $? This than could be embedded in $ UT_n(mathbbZ)$ through $G_1 rightarrow UT_n(mathbbZ), a_1 beginbmatrix1&a_1\0&1endbmatrix$.
– Arwid
Aug 27 at 19:56