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 $n\geqslant 1$. 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.

The Bernside ring is generated by the classes of transitive $G$-sets. The multiplication table is then given by a Mackey-like formula: $$[G/H]\times [G/K]=\sum_{H\backslash G/K} [G/H \cap gKg^{-1}]$$. This follows from the classical fact that $H\backslash G/K\cong G \backslash (G/H\times G/K)$ For example, in the case $G=\mathbb{Z}/p^n $ the generators are $a_k = [G/(\mathbb{Z}/p^k)]$ for $k\le n$. The relations are then $$a_i a_j = a_i p ^{n-j}$$ is $j\ge i$. 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.

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

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 $a_ia_j=p^{n-j}a_i$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.