Examples
The table of marks for the cyclic group of order 6:
| Z6 | 1 | Z2 | Z3 | Z6 |
| Z6 / 1 | 6 | . | . | . |
| Z6 / Z2 | 3 | 3 | . | . |
| Z6 / Z3 | 2 | 0 | 2 | . |
| Z6 / Z6 | 1 | 1 | 1 | 1 |
The table of marks for the symmetric group S3 on 3 letters:
| S3 | 1 | Z2 | Z3 | S3 |
| S3 / 1 | 6 | . | . | . |
| S3 / Z2 | 3 | 1 | . | . |
| S3 / Z3 | 2 | 0 | 2 | . |
| S3 / S3 | 1 | 1 | 1 | 1 |
The dots in the two tables are all zeros, merely emphasizing the fact that the tables are lower-triangular.
(Some authors use the transpose of the table, but this is how Burnside defined it originally.)
The fact that the last row is all 1s is because is a single point. The diagonal terms are m(H, H) = | NG(H)/H |.
The ring structure of Ω(G) can be deduced from these tables: the generators of the ring (as a Z-module) are the rows of the table, and the product of two generators has mark given by the product of the marks (so component-wise multiplication of row vectors), which can then be decomposed as a linear combination of all the rows. For example, with S3,
as (3, 1, 0, 0).(2, 0, 2, 0) = (6, 0, 0, 0).
Read more about this topic: Burnside Ring
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