Explicit computation of the Burnside ring
$begingroup$
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
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
$endgroup$
add a comment |
$begingroup$
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
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
$endgroup$
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
add a comment |
$begingroup$
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
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
$endgroup$
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
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
asked Nov 20 '18 at 7:13
user51223user51223
1,341617
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a_ia_j=p^{n-j}a_i$.
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
$endgroup$
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.
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
answered Nov 20 '18 at 7:41
S. carmeliS. carmeli
2,300519
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $a_ia_j=p^{n-j}a_i$.
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
$endgroup$
add a comment |
$begingroup$
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 $a_ia_j=p^{n-j}a_i$.
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
$endgroup$
add a comment |
$begingroup$
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 $a_ia_j=p^{n-j}a_i$.
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
$endgroup$
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 $a_ia_j=p^{n-j}a_i$.
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
answered Nov 24 '18 at 8:57
user43326user43326
1,850818
1,850818
add a comment |
add a comment |
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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