Mixing
What is a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$?
up vote
0
down vote
favorite
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.
abstract-algebra group-theory group-presentation
asked 2 days ago
mnewman
214
add a comment |
up vote
0
down vote
favorite
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.
abstract-algebra group-theory group-presentation
asked 2 days ago
mnewman
214
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
2 days ago
@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
2 days ago
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
2 days ago
1
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
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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.
abstract-algebra group-theory group-presentation
asked 2 days ago
mnewman
214
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.
abstract-algebra group-theory group-presentation
abstract-algebra group-theory group-presentation
asked 2 days ago
mnewman
214
asked 2 days ago
mnewman
214
edited 2 days ago
asked 2 days ago
mnewman
214
asked 2 days ago
mnewman
214
asked 2 days ago
mnewman
214
214
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
2 days ago
@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
2 days ago
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
2 days ago
1
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
2 days ago
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
2 days ago
@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
2 days ago
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
2 days ago
1
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
2 days ago
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
2 days ago
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
2 days ago
@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
2 days ago
@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
2 days ago
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
2 days ago
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
2 days ago
1
1
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
2 days ago
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
2 days ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005503%2fwhat-is-a-presentation-of-the-upper-triangular-subgroup-of-gl2-mathbb-q%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 days ago
@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
2 days ago
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
2 days ago
1
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
2 days ago