Is there an algorithmic way to compute the quotient of two subgroups of $mathbb Z^n$?












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In algebraic topology one sometimes has to compute the homology groups via CW complexes, reducing the problem to the calculation of $Z/B$ where $B subseteq Z subseteq mathbb{Z}^n $. It is difficult enough for me sometimes if $B$ is complicated.



For instance the isomorphism $mathbb Z oplus mathbb Z / left<(2,0),(-1,2)right> cong mathbb Z / 4 mathbb Z $ is not trivial at all.



Is there a general algorithm to determine $Z/B$, as both of $Z$ and $B$ are represented as the subgroups generated by several elements of $mathbb Z^n $? Thanks.






abstract-algebra homology-cohomology abelian-groups







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    $begingroup$


    In algebraic topology one sometimes has to compute the homology groups via CW complexes, reducing the problem to the calculation of $Z/B$ where $B subseteq Z subseteq mathbb{Z}^n $. It is difficult enough for me sometimes if $B$ is complicated.



    For instance the isomorphism $mathbb Z oplus mathbb Z / left<(2,0),(-1,2)right> cong mathbb Z / 4 mathbb Z $ is not trivial at all.



    Is there a general algorithm to determine $Z/B$, as both of $Z$ and $B$ are represented as the subgroups generated by several elements of $mathbb Z^n $? Thanks.






    abstract-algebra homology-cohomology abelian-groups







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      In algebraic topology one sometimes has to compute the homology groups via CW complexes, reducing the problem to the calculation of $Z/B$ where $B subseteq Z subseteq mathbb{Z}^n $. It is difficult enough for me sometimes if $B$ is complicated.



      For instance the isomorphism $mathbb Z oplus mathbb Z / left<(2,0),(-1,2)right> cong mathbb Z / 4 mathbb Z $ is not trivial at all.



      Is there a general algorithm to determine $Z/B$, as both of $Z$ and $B$ are represented as the subgroups generated by several elements of $mathbb Z^n $? Thanks.






      abstract-algebra homology-cohomology abelian-groups







      share|cite|improve this question











      $endgroup$




      In algebraic topology one sometimes has to compute the homology groups via CW complexes, reducing the problem to the calculation of $Z/B$ where $B subseteq Z subseteq mathbb{Z}^n $. It is difficult enough for me sometimes if $B$ is complicated.



      For instance the isomorphism $mathbb Z oplus mathbb Z / left<(2,0),(-1,2)right> cong mathbb Z / 4 mathbb Z $ is not trivial at all.



      Is there a general algorithm to determine $Z/B$, as both of $Z$ and $B$ are represented as the subgroups generated by several elements of $mathbb Z^n $? Thanks.






      abstract-algebra homology-cohomology abelian-groups




      abstract-algebra homology-cohomology abelian-groups






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 12 at 8:38









      user26857

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      asked Jan 10 at 14:39









      DromedaDromeda

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      373






















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          $begingroup$

          Yes. For instance, you can use Smith normal form to do this. This amounts to doing Gaussian elimination over the integers.



          https://en.wikipedia.org/wiki/Smith_normal_form



          In somewhat more detail, any subgroup of $mathbf{Z}^n$ can be thought of as the image of a $mathbf{Z}$-linear (that is, additive) map $mathbf{Z}^m to mathbf{Z}^n$ for some integer $m$, which in turn may be represented by a matrix $A$ with integer entries. Performing elementary row operations on $A$ amounts to gradually changing the chosen basis of the target $mathbf{Z}^n$, and performing elementary column operations on $A$ amounts to gradually changing the chosen basis of the domain $mathbf{Z}^m$. The relevant fact is therefore that you can use elementary row and column operations to reduce $A$ to a diagonal matrix of the form
          $$A=left(begin{matrix} a_1 & 0 & 0 & cdots \ 0 & a_2 & 0 & cdots \ 0 & 0 & a_3 & cdots \ vdots & vdots & vdots end{matrix} right)$$ in which $a_i$ divides $a_{i+1}$ for all $i$.



          This is a computationally effective form of the fundamental theorem of abelian groups, and shows that the cokernel of $A$ is isomorphic to $$mathrm{coker}(A) cong mathbf{Z}/a_1 mathbf{Z} oplus mathbf{Z}/a_2 mathbf{Z} oplus cdots.$$



          In your chosen example, the matrix $A$ is
          $$A=left(begin{matrix} -1 & 2 \ 2 & 0 end{matrix} right)$$ and by first doing the row operation replace row two by row two plus twice row one and then the column operation replace column two by column two plus twice column one reduces it to
          $$A=left(begin{matrix} -1 & 0 \ 0 & 4 end{matrix} right),$$ showing that the cokernel is $mathbf{Z}/4 mathbf{Z}$ as claimed.






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            $begingroup$

            Yes. For instance, you can use Smith normal form to do this. This amounts to doing Gaussian elimination over the integers.



            https://en.wikipedia.org/wiki/Smith_normal_form



            In somewhat more detail, any subgroup of $mathbf{Z}^n$ can be thought of as the image of a $mathbf{Z}$-linear (that is, additive) map $mathbf{Z}^m to mathbf{Z}^n$ for some integer $m$, which in turn may be represented by a matrix $A$ with integer entries. Performing elementary row operations on $A$ amounts to gradually changing the chosen basis of the target $mathbf{Z}^n$, and performing elementary column operations on $A$ amounts to gradually changing the chosen basis of the domain $mathbf{Z}^m$. The relevant fact is therefore that you can use elementary row and column operations to reduce $A$ to a diagonal matrix of the form
            $$A=left(begin{matrix} a_1 & 0 & 0 & cdots \ 0 & a_2 & 0 & cdots \ 0 & 0 & a_3 & cdots \ vdots & vdots & vdots end{matrix} right)$$ in which $a_i$ divides $a_{i+1}$ for all $i$.



            This is a computationally effective form of the fundamental theorem of abelian groups, and shows that the cokernel of $A$ is isomorphic to $$mathrm{coker}(A) cong mathbf{Z}/a_1 mathbf{Z} oplus mathbf{Z}/a_2 mathbf{Z} oplus cdots.$$



            In your chosen example, the matrix $A$ is
            $$A=left(begin{matrix} -1 & 2 \ 2 & 0 end{matrix} right)$$ and by first doing the row operation replace row two by row two plus twice row one and then the column operation replace column two by column two plus twice column one reduces it to
            $$A=left(begin{matrix} -1 & 0 \ 0 & 4 end{matrix} right),$$ showing that the cokernel is $mathbf{Z}/4 mathbf{Z}$ as claimed.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Yes. For instance, you can use Smith normal form to do this. This amounts to doing Gaussian elimination over the integers.



              https://en.wikipedia.org/wiki/Smith_normal_form



              In somewhat more detail, any subgroup of $mathbf{Z}^n$ can be thought of as the image of a $mathbf{Z}$-linear (that is, additive) map $mathbf{Z}^m to mathbf{Z}^n$ for some integer $m$, which in turn may be represented by a matrix $A$ with integer entries. Performing elementary row operations on $A$ amounts to gradually changing the chosen basis of the target $mathbf{Z}^n$, and performing elementary column operations on $A$ amounts to gradually changing the chosen basis of the domain $mathbf{Z}^m$. The relevant fact is therefore that you can use elementary row and column operations to reduce $A$ to a diagonal matrix of the form
              $$A=left(begin{matrix} a_1 & 0 & 0 & cdots \ 0 & a_2 & 0 & cdots \ 0 & 0 & a_3 & cdots \ vdots & vdots & vdots end{matrix} right)$$ in which $a_i$ divides $a_{i+1}$ for all $i$.



              This is a computationally effective form of the fundamental theorem of abelian groups, and shows that the cokernel of $A$ is isomorphic to $$mathrm{coker}(A) cong mathbf{Z}/a_1 mathbf{Z} oplus mathbf{Z}/a_2 mathbf{Z} oplus cdots.$$



              In your chosen example, the matrix $A$ is
              $$A=left(begin{matrix} -1 & 2 \ 2 & 0 end{matrix} right)$$ and by first doing the row operation replace row two by row two plus twice row one and then the column operation replace column two by column two plus twice column one reduces it to
              $$A=left(begin{matrix} -1 & 0 \ 0 & 4 end{matrix} right),$$ showing that the cokernel is $mathbf{Z}/4 mathbf{Z}$ as claimed.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Yes. For instance, you can use Smith normal form to do this. This amounts to doing Gaussian elimination over the integers.



                https://en.wikipedia.org/wiki/Smith_normal_form



                In somewhat more detail, any subgroup of $mathbf{Z}^n$ can be thought of as the image of a $mathbf{Z}$-linear (that is, additive) map $mathbf{Z}^m to mathbf{Z}^n$ for some integer $m$, which in turn may be represented by a matrix $A$ with integer entries. Performing elementary row operations on $A$ amounts to gradually changing the chosen basis of the target $mathbf{Z}^n$, and performing elementary column operations on $A$ amounts to gradually changing the chosen basis of the domain $mathbf{Z}^m$. The relevant fact is therefore that you can use elementary row and column operations to reduce $A$ to a diagonal matrix of the form
                $$A=left(begin{matrix} a_1 & 0 & 0 & cdots \ 0 & a_2 & 0 & cdots \ 0 & 0 & a_3 & cdots \ vdots & vdots & vdots end{matrix} right)$$ in which $a_i$ divides $a_{i+1}$ for all $i$.



                This is a computationally effective form of the fundamental theorem of abelian groups, and shows that the cokernel of $A$ is isomorphic to $$mathrm{coker}(A) cong mathbf{Z}/a_1 mathbf{Z} oplus mathbf{Z}/a_2 mathbf{Z} oplus cdots.$$



                In your chosen example, the matrix $A$ is
                $$A=left(begin{matrix} -1 & 2 \ 2 & 0 end{matrix} right)$$ and by first doing the row operation replace row two by row two plus twice row one and then the column operation replace column two by column two plus twice column one reduces it to
                $$A=left(begin{matrix} -1 & 0 \ 0 & 4 end{matrix} right),$$ showing that the cokernel is $mathbf{Z}/4 mathbf{Z}$ as claimed.






                share|cite|improve this answer











                $endgroup$



                Yes. For instance, you can use Smith normal form to do this. This amounts to doing Gaussian elimination over the integers.



                https://en.wikipedia.org/wiki/Smith_normal_form



                In somewhat more detail, any subgroup of $mathbf{Z}^n$ can be thought of as the image of a $mathbf{Z}$-linear (that is, additive) map $mathbf{Z}^m to mathbf{Z}^n$ for some integer $m$, which in turn may be represented by a matrix $A$ with integer entries. Performing elementary row operations on $A$ amounts to gradually changing the chosen basis of the target $mathbf{Z}^n$, and performing elementary column operations on $A$ amounts to gradually changing the chosen basis of the domain $mathbf{Z}^m$. The relevant fact is therefore that you can use elementary row and column operations to reduce $A$ to a diagonal matrix of the form
                $$A=left(begin{matrix} a_1 & 0 & 0 & cdots \ 0 & a_2 & 0 & cdots \ 0 & 0 & a_3 & cdots \ vdots & vdots & vdots end{matrix} right)$$ in which $a_i$ divides $a_{i+1}$ for all $i$.



                This is a computationally effective form of the fundamental theorem of abelian groups, and shows that the cokernel of $A$ is isomorphic to $$mathrm{coker}(A) cong mathbf{Z}/a_1 mathbf{Z} oplus mathbf{Z}/a_2 mathbf{Z} oplus cdots.$$



                In your chosen example, the matrix $A$ is
                $$A=left(begin{matrix} -1 & 2 \ 2 & 0 end{matrix} right)$$ and by first doing the row operation replace row two by row two plus twice row one and then the column operation replace column two by column two plus twice column one reduces it to
                $$A=left(begin{matrix} -1 & 0 \ 0 & 4 end{matrix} right),$$ showing that the cokernel is $mathbf{Z}/4 mathbf{Z}$ as claimed.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 14:57

























                answered Jan 10 at 14:51









                StephenStephen

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                10.5k12339






























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