Explicit computation of the Burnside ring












6












$begingroup$


I would like to see explicit computations of the Burnside ring $A(G)$ when $G$ is a small Abelian group, such as $G=mathbb{Z}/2,mathbb{Z}/2^n,mathbb{Z}/p^n$ where $p$ is an odd prime and $ngeqslant 1$. Here, by explicit I mean in terms of generators and relations. I know that there there is a certain map with finite cokernel. But, I don't see any generators in these descriptions. Surely, for something like $mathbb{Z}/2$ it must be well known!



I would be very grateful for any reference. Here, $mathbb{Z}/k$ is the cyclic group of order $k$.



I particular, I wonder if there is a ``canonical'' presentation of this ring?!?



I would be very grateful for any references.






reference-request at.algebraic-topology gr.group-theory finite-groups







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$endgroup$








  • 3




    $begingroup$
    The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
    $endgroup$
    – Drew Heard
    Nov 20 '18 at 7:55






  • 1




    $begingroup$
    GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
    $endgroup$
    – Dima Pasechnik
    Nov 24 '18 at 16:10
















6












$begingroup$


I would like to see explicit computations of the Burnside ring $A(G)$ when $G$ is a small Abelian group, such as $G=mathbb{Z}/2,mathbb{Z}/2^n,mathbb{Z}/p^n$ where $p$ is an odd prime and $ngeqslant 1$. Here, by explicit I mean in terms of generators and relations. I know that there there is a certain map with finite cokernel. But, I don't see any generators in these descriptions. Surely, for something like $mathbb{Z}/2$ it must be well known!



I would be very grateful for any reference. Here, $mathbb{Z}/k$ is the cyclic group of order $k$.



I particular, I wonder if there is a ``canonical'' presentation of this ring?!?



I would be very grateful for any references.






reference-request at.algebraic-topology gr.group-theory finite-groups







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
    $endgroup$
    – Drew Heard
    Nov 20 '18 at 7:55






  • 1




    $begingroup$
    GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
    $endgroup$
    – Dima Pasechnik
    Nov 24 '18 at 16:10














6












6








6





$begingroup$


I would like to see explicit computations of the Burnside ring $A(G)$ when $G$ is a small Abelian group, such as $G=mathbb{Z}/2,mathbb{Z}/2^n,mathbb{Z}/p^n$ where $p$ is an odd prime and $ngeqslant 1$. Here, by explicit I mean in terms of generators and relations. I know that there there is a certain map with finite cokernel. But, I don't see any generators in these descriptions. Surely, for something like $mathbb{Z}/2$ it must be well known!



I would be very grateful for any reference. Here, $mathbb{Z}/k$ is the cyclic group of order $k$.



I particular, I wonder if there is a ``canonical'' presentation of this ring?!?



I would be very grateful for any references.






reference-request at.algebraic-topology gr.group-theory finite-groups







share|cite|improve this question









$endgroup$




I would like to see explicit computations of the Burnside ring $A(G)$ when $G$ is a small Abelian group, such as $G=mathbb{Z}/2,mathbb{Z}/2^n,mathbb{Z}/p^n$ where $p$ is an odd prime and $ngeqslant 1$. Here, by explicit I mean in terms of generators and relations. I know that there there is a certain map with finite cokernel. But, I don't see any generators in these descriptions. Surely, for something like $mathbb{Z}/2$ it must be well known!



I would be very grateful for any reference. Here, $mathbb{Z}/k$ is the cyclic group of order $k$.



I particular, I wonder if there is a ``canonical'' presentation of this ring?!?



I would be very grateful for any references.






reference-request at.algebraic-topology gr.group-theory finite-groups




reference-request at.algebraic-topology gr.group-theory finite-groups






share|cite|improve this question













share|cite|improve this question











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asked Nov 20 '18 at 7:13









user51223user51223

1,341617




1,341617








  • 3




    $begingroup$
    The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
    $endgroup$
    – Drew Heard
    Nov 20 '18 at 7:55






  • 1




    $begingroup$
    GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
    $endgroup$
    – Dima Pasechnik
    Nov 24 '18 at 16:10














  • 3




    $begingroup$
    The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
    $endgroup$
    – Drew Heard
    Nov 20 '18 at 7:55






  • 1




    $begingroup$
    GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
    $endgroup$
    – Dima Pasechnik
    Nov 24 '18 at 16:10








3




3




$begingroup$
The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
$endgroup$
– Drew Heard
Nov 20 '18 at 7:55




$begingroup$
The table of marks is a useful way to do this: en.wikipedia.org/wiki/Burnside_ring#Marks
$endgroup$
– Drew Heard
Nov 20 '18 at 7:55




1




1




$begingroup$
GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
$endgroup$
– Dima Pasechnik
Nov 24 '18 at 16:10




$begingroup$
GAP knows tables of marks for quite a few groups, see gap-system.org/Manuals/doc/ref/chap70.html
$endgroup$
– Dima Pasechnik
Nov 24 '18 at 16:10










2 Answers
2






active

oldest

votes


















7












$begingroup$

The Bernside ring is generated by the classes of transitive $G$-sets.
The multiplication table is then given by a Mackey-like formula:
$$[G/H]times [G/K]=sum_{Hbackslash G/K} [G/H cap gKg^{-1}]$$.
This follows from the classical fact that $Hbackslash G/Kcong G backslash (G/Htimes G/K)$
For example, in the case $G=mathbb{Z}/p^n
$ the generators are $a_k = [G/(mathbb{Z}/p^k)]$ for $kle n$. The relations are then
$$a_i a_j = a_i p ^{n-j}$$ is $jge i$. This follows directly from the formula for the product and the fact that all the subgroups of an abelian group are normal, so double cosets are just cosets for the sum.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
    $endgroup$
    – user51223
    Nov 20 '18 at 7:51





















1












$begingroup$

Although the answer by S. camell is perfect, personally I find the use of Mackey-like formula here is an over-kill.



Let's use the same notation. As a set $a_ia_j$ is just $Z/p^{n-i}times Z/p^{n-j}$,
and $Z/p^n$ is acting diagonally on this. Clearly $p^i$ acts trivially on this and $p^{i-1}$ don't fix any element, so all orbits are isomorphic to $a_i$. By counting elements of the set we see that $a_ia_j=p^{n-j}a_i$.






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






    active

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    active

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    7












    $begingroup$

    The Bernside ring is generated by the classes of transitive $G$-sets.
    The multiplication table is then given by a Mackey-like formula:
    $$[G/H]times [G/K]=sum_{Hbackslash G/K} [G/H cap gKg^{-1}]$$.
    This follows from the classical fact that $Hbackslash G/Kcong G backslash (G/Htimes G/K)$
    For example, in the case $G=mathbb{Z}/p^n
    $     the generators are $a_k = [G/(mathbb{Z}/p^k)]$ for $kle n$. The relations are then
    $$a_i a_j = a_i p ^{n-j}$$ is $jge i$. This follows directly from the formula for the product and the fact that all the subgroups of an abelian group are normal, so double cosets are just cosets for the sum.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
      $endgroup$
      – user51223
      Nov 20 '18 at 7:51


















    7












    $begingroup$

    The Bernside ring is generated by the classes of transitive $G$-sets.
    The multiplication table is then given by a Mackey-like formula:
    $$[G/H]times [G/K]=sum_{Hbackslash G/K} [G/H cap gKg^{-1}]$$.
    This follows from the classical fact that $Hbackslash G/Kcong G backslash (G/Htimes G/K)$
    For example, in the case $G=mathbb{Z}/p^n
    $     the generators are $a_k = [G/(mathbb{Z}/p^k)]$ for $kle n$. The relations are then
    $$a_i a_j = a_i p ^{n-j}$$ is $jge i$. This follows directly from the formula for the product and the fact that all the subgroups of an abelian group are normal, so double cosets are just cosets for the sum.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
      $endgroup$
      – user51223
      Nov 20 '18 at 7:51
















    7












    7








    7





    $begingroup$

    The Bernside ring is generated by the classes of transitive $G$-sets.
    The multiplication table is then given by a Mackey-like formula:
    $$[G/H]times [G/K]=sum_{Hbackslash G/K} [G/H cap gKg^{-1}]$$.
    This follows from the classical fact that $Hbackslash G/Kcong G backslash (G/Htimes G/K)$
    For example, in the case $G=mathbb{Z}/p^n
    $     the generators are $a_k = [G/(mathbb{Z}/p^k)]$ for $kle n$. The relations are then
    $$a_i a_j = a_i p ^{n-j}$$ is $jge i$. This follows directly from the formula for the product and the fact that all the subgroups of an abelian group are normal, so double cosets are just cosets for the sum.






    share|cite|improve this answer









    $endgroup$



    The Bernside ring is generated by the classes of transitive $G$-sets.
    The multiplication table is then given by a Mackey-like formula:
    $$[G/H]times [G/K]=sum_{Hbackslash G/K} [G/H cap gKg^{-1}]$$.
    This follows from the classical fact that $Hbackslash G/Kcong G backslash (G/Htimes G/K)$
    For example, in the case $G=mathbb{Z}/p^n
    $     the generators are $a_k = [G/(mathbb{Z}/p^k)]$ for $kle n$. The relations are then
    $$a_i a_j = a_i p ^{n-j}$$ is $jge i$. This follows directly from the formula for the product and the fact that all the subgroups of an abelian group are normal, so double cosets are just cosets for the sum.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 20 '18 at 7:41









    S. carmeliS. carmeli

    2,300519




    2,300519












    • $begingroup$
      Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
      $endgroup$
      – user51223
      Nov 20 '18 at 7:51




















    • $begingroup$
      Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
      $endgroup$
      – user51223
      Nov 20 '18 at 7:51


















    $begingroup$
    Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
    $endgroup$
    – user51223
    Nov 20 '18 at 7:51






    $begingroup$
    Does this work when $p=2$? And is this an integral description so that $$A(G)simeqmathbb{Z}[a_itextrm{ with }ileqslant n:a_ia_j=a_ip^{n-j}textrm{ whenever }jgeqslant i]$$. And I presume there is no grading associated to the generators, right? And would this be a ``canonical presentations''?
    $endgroup$
    – user51223
    Nov 20 '18 at 7:51













    1












    $begingroup$

    Although the answer by S. camell is perfect, personally I find the use of Mackey-like formula here is an over-kill.



    Let's use the same notation. As a set $a_ia_j$ is just $Z/p^{n-i}times Z/p^{n-j}$,
    and $Z/p^n$ is acting diagonally on this. Clearly $p^i$ acts trivially on this and $p^{i-1}$ don't fix any element, so all orbits are isomorphic to $a_i$. By counting elements of the set we see that $a_ia_j=p^{n-j}a_i$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Although the answer by S. camell is perfect, personally I find the use of Mackey-like formula here is an over-kill.



      Let's use the same notation. As a set $a_ia_j$ is just $Z/p^{n-i}times Z/p^{n-j}$,
      and $Z/p^n$ is acting diagonally on this. Clearly $p^i$ acts trivially on this and $p^{i-1}$ don't fix any element, so all orbits are isomorphic to $a_i$. By counting elements of the set we see that $a_ia_j=p^{n-j}a_i$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Although the answer by S. camell is perfect, personally I find the use of Mackey-like formula here is an over-kill.



        Let's use the same notation. As a set $a_ia_j$ is just $Z/p^{n-i}times Z/p^{n-j}$,
        and $Z/p^n$ is acting diagonally on this. Clearly $p^i$ acts trivially on this and $p^{i-1}$ don't fix any element, so all orbits are isomorphic to $a_i$. By counting elements of the set we see that $a_ia_j=p^{n-j}a_i$.






        share|cite|improve this answer









        $endgroup$



        Although the answer by S. camell is perfect, personally I find the use of Mackey-like formula here is an over-kill.



        Let's use the same notation. As a set $a_ia_j$ is just $Z/p^{n-i}times Z/p^{n-j}$,
        and $Z/p^n$ is acting diagonally on this. Clearly $p^i$ acts trivially on this and $p^{i-1}$ don't fix any element, so all orbits are isomorphic to $a_i$. By counting elements of the set we see that $a_ia_j=p^{n-j}a_i$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 8:57









        user43326user43326

        1,850818




        1,850818






























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