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
edited Nov 30 at 2:48
Shaun
8,205113578
asked Nov 19 at 20:46
mnewman
264
add a comment |
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
edited Nov 30 at 2:48
Shaun
8,205113578
asked Nov 19 at 20:46
mnewman
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
add a comment |
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
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
edited Nov 30 at 2:48
Shaun
8,205113578
asked Nov 19 at 20:46
mnewman
264
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
edited Nov 30 at 2:48
Shaun
8,205113578
asked Nov 19 at 20:46
mnewman
264
edited Nov 30 at 2:48
Shaun
8,205113578
asked Nov 19 at 20:46
mnewman
264
edited Nov 30 at 2:48
Shaun
8,205113578
edited Nov 30 at 2:48
Shaun
8,205113578
edited Nov 30 at 2:48
Shaun
8,205113578
8,205113578
asked Nov 19 at 20:46
mnewman
264
asked Nov 19 at 20:46
mnewman
264
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
add a comment |
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
add a comment |
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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