What is a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$?











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I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by



$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.






group-theory group-presentation combinatorial-group-theory







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  • 2




    Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
    – Daniel Schepler
    Nov 19 at 21:15










  • @DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
    – mnewman
    Nov 19 at 21:23












  • Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
    – Daniel Schepler
    Nov 19 at 21:35








  • 2




    Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
    – reuns
    Nov 19 at 21:36

















up vote
1
down vote

favorite
1












I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by



$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.






group-theory group-presentation combinatorial-group-theory







share|cite|improve this question




















  • 2




    Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
    – Daniel Schepler
    Nov 19 at 21:15










  • @DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
    – mnewman
    Nov 19 at 21:23












  • Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
    – Daniel Schepler
    Nov 19 at 21:35








  • 2




    Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
    – reuns
    Nov 19 at 21:36















up vote
1
down vote

favorite
1









up vote
1
down vote

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1






1





I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by



$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.






group-theory group-presentation combinatorial-group-theory







share|cite|improve this question















I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by



$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.






group-theory group-presentation combinatorial-group-theory




group-theory group-presentation combinatorial-group-theory






share|cite|improve this question















share|cite|improve this question













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edited Nov 30 at 2:48









Shaun

8,205113578




8,205113578










asked Nov 19 at 20:46









mnewman

264




264








  • 2




    Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
    – Daniel Schepler
    Nov 19 at 21:15










  • @DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
    – mnewman
    Nov 19 at 21:23












  • Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
    – Daniel Schepler
    Nov 19 at 21:35








  • 2




    Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
    – reuns
    Nov 19 at 21:36
















  • 2




    Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
    – Daniel Schepler
    Nov 19 at 21:15










  • @DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
    – mnewman
    Nov 19 at 21:23












  • Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
    – Daniel Schepler
    Nov 19 at 21:35








  • 2




    Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
    – reuns
    Nov 19 at 21:36










2




2




Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15




Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15












@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23






@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23














Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35






Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35






2




2




Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36






Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36

















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