Burnside ring

Remarks

As an abelian group, $A(G)$ is free with basis the classes $[G/H]$, where $H$ runs over representatives of the conjugacy classes of subgroups of $G$. This holds because each finite $G$-set is the disjoint union of its orbits, and the orbit $G/H$ is isomorphic to $G/K$ exactly when $H$ and $K$ are conjugate. Example: for $G = \Z/2\Z$, the basis is $a = [G/G]$, the point, and $b = [G/\set{0}]$, the free orbit with two elements. The product $b \cdot b = [G \times G]$ is a free $G$-set with $4$ elements, so $b^2 = 2b$, and $A(\Z/2\Z) \cong \Z[b]/(b^2 - 2b)$. Non-example: the set of isomorphism classes of finite $G$-sets is not itself $A(G)$, since a nonempty $G$-set has no additive inverse there. For a subgroup $H$, the map $S \mapsto \abs{S^H}$, counting fixed points, extends to a ring homomorphism $A(G) \to \Z$.