Finding a homomorphism between groups with a given kernel












6












$begingroup$


What is a homomorphism defined on the group of invertible upper-triangular $3times 3$ matrices whose kernel consists of matrices $begin{bmatrix} 1 & 0 & a \ 0 & 1 & 0 \ 0 &0 & 1end{bmatrix}$?



I want to use this to study the quotient group, also is there always a way to find a group homomorphism, once the kernel is given and one the group is known in general ?



What changes when the diagonal entries in the group of upper triangular matrices are all equal to one .



I prefer a correct answer as much as easy to follow explanation, because my backgroud in algebra does not go beyond "first abstract algebra course"



Thank you.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
    $endgroup$
    – Arturo Magidin
    Jan 10 at 19:00










  • $begingroup$
    @ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:05






  • 1




    $begingroup$
    For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
    $endgroup$
    – 0x539
    Jan 10 at 19:10










  • $begingroup$
    @0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:17
















6












$begingroup$


What is a homomorphism defined on the group of invertible upper-triangular $3times 3$ matrices whose kernel consists of matrices $begin{bmatrix} 1 & 0 & a \ 0 & 1 & 0 \ 0 &0 & 1end{bmatrix}$?



I want to use this to study the quotient group, also is there always a way to find a group homomorphism, once the kernel is given and one the group is known in general ?



What changes when the diagonal entries in the group of upper triangular matrices are all equal to one .



I prefer a correct answer as much as easy to follow explanation, because my backgroud in algebra does not go beyond "first abstract algebra course"



Thank you.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
    $endgroup$
    – Arturo Magidin
    Jan 10 at 19:00










  • $begingroup$
    @ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:05






  • 1




    $begingroup$
    For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
    $endgroup$
    – 0x539
    Jan 10 at 19:10










  • $begingroup$
    @0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:17














6












6








6


4



$begingroup$


What is a homomorphism defined on the group of invertible upper-triangular $3times 3$ matrices whose kernel consists of matrices $begin{bmatrix} 1 & 0 & a \ 0 & 1 & 0 \ 0 &0 & 1end{bmatrix}$?



I want to use this to study the quotient group, also is there always a way to find a group homomorphism, once the kernel is given and one the group is known in general ?



What changes when the diagonal entries in the group of upper triangular matrices are all equal to one .



I prefer a correct answer as much as easy to follow explanation, because my backgroud in algebra does not go beyond "first abstract algebra course"



Thank you.










share|cite|improve this question











$endgroup$




What is a homomorphism defined on the group of invertible upper-triangular $3times 3$ matrices whose kernel consists of matrices $begin{bmatrix} 1 & 0 & a \ 0 & 1 & 0 \ 0 &0 & 1end{bmatrix}$?



I want to use this to study the quotient group, also is there always a way to find a group homomorphism, once the kernel is given and one the group is known in general ?



What changes when the diagonal entries in the group of upper triangular matrices are all equal to one .



I prefer a correct answer as much as easy to follow explanation, because my backgroud in algebra does not go beyond "first abstract algebra course"



Thank you.







abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 18:56







Kasmir Khaan

















asked Jan 10 at 18:54









Kasmir KhaanKasmir Khaan

39722




39722








  • 1




    $begingroup$
    First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
    $endgroup$
    – Arturo Magidin
    Jan 10 at 19:00










  • $begingroup$
    @ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:05






  • 1




    $begingroup$
    For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
    $endgroup$
    – 0x539
    Jan 10 at 19:10










  • $begingroup$
    @0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:17














  • 1




    $begingroup$
    First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
    $endgroup$
    – Arturo Magidin
    Jan 10 at 19:00










  • $begingroup$
    @ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:05






  • 1




    $begingroup$
    For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
    $endgroup$
    – 0x539
    Jan 10 at 19:10










  • $begingroup$
    @0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:17








1




1




$begingroup$
First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
$endgroup$
– Arturo Magidin
Jan 10 at 19:00




$begingroup$
First, you need to specify what kind of entries your matrices have; real numbers? Integers? Integers modulo $n$? Rationals? Second: the general way is precisely to use the projection onto the quotient! But what you are implying is that you don't find the quotient to be directly accessible, so that you are actually trying to understand the quotient by finding a homomorphism into some other group that you are more comfortable with. That makes matters much more difficult, because it amounts to figuring out what the quotient "is" abstractly in order to find the homomorphism..
$endgroup$
– Arturo Magidin
Jan 10 at 19:00












$begingroup$
@ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
$endgroup$
– Kasmir Khaan
Jan 10 at 19:05




$begingroup$
@ArturoMagidin thanks for your comment ! Yes that is exactly right, I could not make sense of the quotient as it is, computed some cosets aK with no luck, ie, did not get clear picture. the entries in any Field F , sorry I should have mentioned that, but my main question was about how does one find such maps in general? for me unless I did work out something similar before or seen it somewhere , it is not easy to find a map !
$endgroup$
– Kasmir Khaan
Jan 10 at 19:05




1




1




$begingroup$
For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
$endgroup$
– 0x539
Jan 10 at 19:10




$begingroup$
For a given kernel group $G$ and a kernel $N$ (which needs to be a normal subgroup) you always have the projection $pi : G to G/N$, which has kernel $ker pi = N$.
$endgroup$
– 0x539
Jan 10 at 19:10












$begingroup$
@0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
$endgroup$
– Kasmir Khaan
Jan 10 at 19:17




$begingroup$
@0x539 thanks for your comment! yes that is the projection map, but it does not tell me much about the structure of G/N , in this example. unless am missing something very fundamental .
$endgroup$
– Kasmir Khaan
Jan 10 at 19:17










2 Answers
2






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6












$begingroup$

You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant,
$$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix} $$
This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry:
$$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix}begin{pmatrix}x'&y'&?\0&z'&u'\0&0&v'end{pmatrix}=begin{pmatrix}xx'&xy'+yz'&?\0&zz'&zu'+uv'\0&0&vv'end{pmatrix} $$





By removing all decoration, this becomes a group structure on the set $$G:={,(x,y,z,u,v,D)mid yzvD-1=0,}$$ and with multiplication rule
$$(x,y,z,u,v,D)cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$
But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.



As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:31












  • $begingroup$
    Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:49










  • $begingroup$
    You of course mean $xzvD-1=0$. :)
    $endgroup$
    – Ted Shifrin
    Jan 11 at 22:23



















1





+250







$begingroup$

I have also not done much beyond a first abstract algebra course, but this should be correct:



I am assuming you ask about homomorphisms to study quotient groups.



To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.



When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!



For the next question: "what happens when all the diagonal entries are all equal to one"



one way to see what happens in the factor group is to see the details in proving the normality of the kernel.



let $$G = begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix} in the set of invertible upper trianglular matrices$$



Then by elementary methods we calculate its inverse: $$G^-1 = begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



using matrix multiplication we check the normality of our kernel by checking if...



$$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



is also in the kernel. we get



$$begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\0&1&0\0&0&1end{pmatrix}$$



which works in any field F and we are still in the kernel.



IF the entries of the diagonal are all one then the matrix multiplication above becomes



$$begin{pmatrix}1&0&bc-be+x+c\0&1&0\0&0&1end{pmatrix}$$



let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.



The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or



$$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}a&b&ax+c\0&d&e\0&0&fend{pmatrix}$$



and of course if the entries of the diagonals are all one then it turns into



$$begin{pmatrix}1&b&c\0&1&e\0&0&1end{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}1&b&x+c\0&1&e\0&0&1end{pmatrix}$$



so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd.



AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.



when all the diagonals are 1 then for coset multiplication we get



$$begin{pmatrix}1&y&?\0&1&u\0&0&1end{pmatrix}begin{pmatrix}1&y'&?\0&1&u'\0&0&1end{pmatrix}=begin{pmatrix}1&y'+y'&?\0&1&u'+u\0&0&1end{pmatrix}$$



assuming a finite field F:
in our modified group (all diagonal entries are one)
we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.






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    2 Answers
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    2 Answers
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    6












    $begingroup$

    You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant,
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix} $$
    This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry:
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix}begin{pmatrix}x'&y'&?\0&z'&u'\0&0&v'end{pmatrix}=begin{pmatrix}xx'&xy'+yz'&?\0&zz'&zu'+uv'\0&0&vv'end{pmatrix} $$





    By removing all decoration, this becomes a group structure on the set $$G:={,(x,y,z,u,v,D)mid yzvD-1=0,}$$ and with multiplication rule
    $$(x,y,z,u,v,D)cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$
    But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.



    As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:31












    • $begingroup$
      Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:49










    • $begingroup$
      You of course mean $xzvD-1=0$. :)
      $endgroup$
      – Ted Shifrin
      Jan 11 at 22:23
















    6












    $begingroup$

    You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant,
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix} $$
    This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry:
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix}begin{pmatrix}x'&y'&?\0&z'&u'\0&0&v'end{pmatrix}=begin{pmatrix}xx'&xy'+yz'&?\0&zz'&zu'+uv'\0&0&vv'end{pmatrix} $$





    By removing all decoration, this becomes a group structure on the set $$G:={,(x,y,z,u,v,D)mid yzvD-1=0,}$$ and with multiplication rule
    $$(x,y,z,u,v,D)cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$
    But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.



    As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:31












    • $begingroup$
      Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:49










    • $begingroup$
      You of course mean $xzvD-1=0$. :)
      $endgroup$
      – Ted Shifrin
      Jan 11 at 22:23














    6












    6








    6





    $begingroup$

    You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant,
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix} $$
    This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry:
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix}begin{pmatrix}x'&y'&?\0&z'&u'\0&0&v'end{pmatrix}=begin{pmatrix}xx'&xy'+yz'&?\0&zz'&zu'+uv'\0&0&vv'end{pmatrix} $$





    By removing all decoration, this becomes a group structure on the set $$G:={,(x,y,z,u,v,D)mid yzvD-1=0,}$$ and with multiplication rule
    $$(x,y,z,u,v,D)cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$
    But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.



    As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group






    share|cite|improve this answer









    $endgroup$



    You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant,
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix} $$
    This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry:
    $$begin{pmatrix}x&y&?\0&z&u\0&0&vend{pmatrix}begin{pmatrix}x'&y'&?\0&z'&u'\0&0&v'end{pmatrix}=begin{pmatrix}xx'&xy'+yz'&?\0&zz'&zu'+uv'\0&0&vv'end{pmatrix} $$





    By removing all decoration, this becomes a group structure on the set $$G:={,(x,y,z,u,v,D)mid yzvD-1=0,}$$ and with multiplication rule
    $$(x,y,z,u,v,D)cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$
    But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.



    As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 10 at 19:18









    Hagen von EitzenHagen von Eitzen

    282k23272505




    282k23272505












    • $begingroup$
      Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:31












    • $begingroup$
      Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:49










    • $begingroup$
      You of course mean $xzvD-1=0$. :)
      $endgroup$
      – Ted Shifrin
      Jan 11 at 22:23


















    • $begingroup$
      Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:31












    • $begingroup$
      Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
      $endgroup$
      – Kasmir Khaan
      Jan 10 at 19:49










    • $begingroup$
      You of course mean $xzvD-1=0$. :)
      $endgroup$
      – Ted Shifrin
      Jan 11 at 22:23
















    $begingroup$
    Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:31






    $begingroup$
    Thank you so much ! I have to read this few times ! to understand where did that yzvD = 1 came from and other things I did not grasp yet.
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:31














    $begingroup$
    Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:49




    $begingroup$
    Can you please explain further? what does D stand for ? and why do we have that condition yzvD =1 ?
    $endgroup$
    – Kasmir Khaan
    Jan 10 at 19:49












    $begingroup$
    You of course mean $xzvD-1=0$. :)
    $endgroup$
    – Ted Shifrin
    Jan 11 at 22:23




    $begingroup$
    You of course mean $xzvD-1=0$. :)
    $endgroup$
    – Ted Shifrin
    Jan 11 at 22:23











    1





    +250







    $begingroup$

    I have also not done much beyond a first abstract algebra course, but this should be correct:



    I am assuming you ask about homomorphisms to study quotient groups.



    To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.



    When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!



    For the next question: "what happens when all the diagonal entries are all equal to one"



    one way to see what happens in the factor group is to see the details in proving the normality of the kernel.



    let $$G = begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix} in the set of invertible upper trianglular matrices$$



    Then by elementary methods we calculate its inverse: $$G^-1 = begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



    using matrix multiplication we check the normality of our kernel by checking if...



    $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



    is also in the kernel. we get



    $$begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\0&1&0\0&0&1end{pmatrix}$$



    which works in any field F and we are still in the kernel.



    IF the entries of the diagonal are all one then the matrix multiplication above becomes



    $$begin{pmatrix}1&0&bc-be+x+c\0&1&0\0&0&1end{pmatrix}$$



    let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.



    The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or



    $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}a&b&ax+c\0&d&e\0&0&fend{pmatrix}$$



    and of course if the entries of the diagonals are all one then it turns into



    $$begin{pmatrix}1&b&c\0&1&e\0&0&1end{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}1&b&x+c\0&1&e\0&0&1end{pmatrix}$$



    so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd.



    AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.



    when all the diagonals are 1 then for coset multiplication we get



    $$begin{pmatrix}1&y&?\0&1&u\0&0&1end{pmatrix}begin{pmatrix}1&y'&?\0&1&u'\0&0&1end{pmatrix}=begin{pmatrix}1&y'+y'&?\0&1&u'+u\0&0&1end{pmatrix}$$



    assuming a finite field F:
    in our modified group (all diagonal entries are one)
    we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.






    share|cite|improve this answer









    $endgroup$


















      1





      +250







      $begingroup$

      I have also not done much beyond a first abstract algebra course, but this should be correct:



      I am assuming you ask about homomorphisms to study quotient groups.



      To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.



      When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!



      For the next question: "what happens when all the diagonal entries are all equal to one"



      one way to see what happens in the factor group is to see the details in proving the normality of the kernel.



      let $$G = begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix} in the set of invertible upper trianglular matrices$$



      Then by elementary methods we calculate its inverse: $$G^-1 = begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



      using matrix multiplication we check the normality of our kernel by checking if...



      $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



      is also in the kernel. we get



      $$begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\0&1&0\0&0&1end{pmatrix}$$



      which works in any field F and we are still in the kernel.



      IF the entries of the diagonal are all one then the matrix multiplication above becomes



      $$begin{pmatrix}1&0&bc-be+x+c\0&1&0\0&0&1end{pmatrix}$$



      let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.



      The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or



      $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}a&b&ax+c\0&d&e\0&0&fend{pmatrix}$$



      and of course if the entries of the diagonals are all one then it turns into



      $$begin{pmatrix}1&b&c\0&1&e\0&0&1end{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}1&b&x+c\0&1&e\0&0&1end{pmatrix}$$



      so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd.



      AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.



      when all the diagonals are 1 then for coset multiplication we get



      $$begin{pmatrix}1&y&?\0&1&u\0&0&1end{pmatrix}begin{pmatrix}1&y'&?\0&1&u'\0&0&1end{pmatrix}=begin{pmatrix}1&y'+y'&?\0&1&u'+u\0&0&1end{pmatrix}$$



      assuming a finite field F:
      in our modified group (all diagonal entries are one)
      we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.






      share|cite|improve this answer









      $endgroup$
















        1





        +250







        1





        +250



        1




        +250



        $begingroup$

        I have also not done much beyond a first abstract algebra course, but this should be correct:



        I am assuming you ask about homomorphisms to study quotient groups.



        To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.



        When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!



        For the next question: "what happens when all the diagonal entries are all equal to one"



        one way to see what happens in the factor group is to see the details in proving the normality of the kernel.



        let $$G = begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix} in the set of invertible upper trianglular matrices$$



        Then by elementary methods we calculate its inverse: $$G^-1 = begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



        using matrix multiplication we check the normality of our kernel by checking if...



        $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



        is also in the kernel. we get



        $$begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\0&1&0\0&0&1end{pmatrix}$$



        which works in any field F and we are still in the kernel.



        IF the entries of the diagonal are all one then the matrix multiplication above becomes



        $$begin{pmatrix}1&0&bc-be+x+c\0&1&0\0&0&1end{pmatrix}$$



        let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.



        The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or



        $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}a&b&ax+c\0&d&e\0&0&fend{pmatrix}$$



        and of course if the entries of the diagonals are all one then it turns into



        $$begin{pmatrix}1&b&c\0&1&e\0&0&1end{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}1&b&x+c\0&1&e\0&0&1end{pmatrix}$$



        so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd.



        AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.



        when all the diagonals are 1 then for coset multiplication we get



        $$begin{pmatrix}1&y&?\0&1&u\0&0&1end{pmatrix}begin{pmatrix}1&y'&?\0&1&u'\0&0&1end{pmatrix}=begin{pmatrix}1&y'+y'&?\0&1&u'+u\0&0&1end{pmatrix}$$



        assuming a finite field F:
        in our modified group (all diagonal entries are one)
        we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.






        share|cite|improve this answer









        $endgroup$



        I have also not done much beyond a first abstract algebra course, but this should be correct:



        I am assuming you ask about homomorphisms to study quotient groups.



        To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.



        When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!



        For the next question: "what happens when all the diagonal entries are all equal to one"



        one way to see what happens in the factor group is to see the details in proving the normality of the kernel.



        let $$G = begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix} in the set of invertible upper trianglular matrices$$



        Then by elementary methods we calculate its inverse: $$G^-1 = begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



        using matrix multiplication we check the normality of our kernel by checking if...



        $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} begin{pmatrix}1/a&-b/ad&bc/af\0&1/d&-e/fd\0&0&1/fend{pmatrix}$$



        is also in the kernel. we get



        $$begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\0&1&0\0&0&1end{pmatrix}$$



        which works in any field F and we are still in the kernel.



        IF the entries of the diagonal are all one then the matrix multiplication above becomes



        $$begin{pmatrix}1&0&bc-be+x+c\0&1&0\0&0&1end{pmatrix}$$



        let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.



        The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or



        $$begin{pmatrix}a&b&c\0&d&e\0&0&fend{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}a&b&ax+c\0&d&e\0&0&fend{pmatrix}$$



        and of course if the entries of the diagonals are all one then it turns into



        $$begin{pmatrix}1&b&c\0&1&e\0&0&1end{pmatrix}begin{pmatrix}1&0&x\0&1&0\0&0&1end{pmatrix} = begin{pmatrix}1&b&x+c\0&1&e\0&0&1end{pmatrix}$$



        so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd.



        AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.



        when all the diagonals are 1 then for coset multiplication we get



        $$begin{pmatrix}1&y&?\0&1&u\0&0&1end{pmatrix}begin{pmatrix}1&y'&?\0&1&u'\0&0&1end{pmatrix}=begin{pmatrix}1&y'+y'&?\0&1&u'+u\0&0&1end{pmatrix}$$



        assuming a finite field F:
        in our modified group (all diagonal entries are one)
        we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 9:01









        Edcookie274Edcookie274

        38610




        38610






























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