Burnside ring

Let $G$ be a finite group. Recall that by $G$-set we understand a pair $(X,\circ )$, where $X$ is a set and $\circ :G\times X\to X$ is a group action of $G$ on $X$. For short notation the pair notation will be omitted and $G$-sets will be simply denoted by capital letters.

Recall that for each subgroup $H\subseteq G$ we have canonical $G$-set $G/H=\{gH;g\in G\}$ where group action is defined as follows: for any $g,k\in G$ we have $(g,kH)⟼gkH$.

Let $X$ and $Y$ be $G$-sets. Recall that by $G$-map from $X$ to $Y$ we understand any function $F:X\to Y$ such that for any $g\in G$ and $x\in X$ we have $F(gx)=gF(x)$.

It is easy to see that family of all $G$-sets and $G$-maps forms a category (with standard comoposition). We shall denote this category by $G-𝕊$. Moreover, by $G-{𝕊}_0$ we shall denote full subcategory of $G-𝕊$ whose objects are all finite $G$-sets.

From $G$-sets $X$ and $Y$ one can construct another $G$-set in two interesting (from our point of view) ways, i.e. by taking disjoint union $X\bigsqcup Y$ with obvious group action and by taking product $X\times Y$ with group action as follows: $(g,(x,y))⟼(gx,gy).$ Moreover it is clear that when $X$ and $Y$ are finite, so are $X\bigsqcup Y$ and $X\times Y$.

Consider a finite $G$-set $X$. Then there exist a natural number $n\in \mathbb{N}$, finite family ${\{H_i\}}_{i=1}^n$ of subgroups of $G$ and an isomorphism (in $G-{𝕊}_0$ category)

$$X\simeq \coprod _{i=1}^{n}G/H_i.$$

Therefore (since $G$ is finite) family of isomorphism classes of $G-{𝕊}_0$ forms a countable set.

Denote by ${\mathrm{\Omega }}^{+}(G)=\{[X];X\in G-{𝕊}_0\}$ the set of isomorphism classes of category $G-{𝕊}_0$. Then one can turn ${\mathrm{\Omega }}^{+}(G)$ into a semiring as follows: for any finite $G$-sets $X$ and $Y$ define

$$[X]+[Y]=[X\bigsqcup Y];$$

$$[X][Y]=[X\times Y].$$

Note that here we treat the empty set as a $G$-set (with one and unique group action), therefore ${\mathrm{\Omega }}^{+}(G)$ has zero element $[\mathrm{\varnothing }]$ (the other way is to formally add the zero to ${\mathrm{\Omega }}^{+}(G)$ - this is just technical thing).

Define by $\mathrm{\Omega }(G)=K({\mathrm{\Omega }}^{+}(G))$ the Grothendieck group of $({\mathrm{\Omega }}^{+}(G),+)$. If $A$ is an abelian semigroup and $f:A\times A\to A$ is a bilinear map, then it can be uniquely extended to a bilinear map $K(f):K(A)\times K(A)\to K(A)$, therefore $\mathrm{\Omega }(G)$ can be uniquely turned into a ring from ${\mathrm{\Omega }}^{+}(G)$. This ring is called the Burnside ring of $G$.

Some properties:

$(0)$ each element of $\mathrm{\Omega }(G)$ can be expressed as a formal diffrence $[X]-[Y]$;

$(1)$ $\mathrm{\Omega }(G)$ is a commutative, unital ring, where $[G/G]$ is the unity of $\mathrm{\Omega }(G)$;

$(2)$ $\mathrm{\Omega }$ can be turned into a contravariant functor from the category of finite groups to the category of commutative, unital rings;

$(3)$ $({\mathrm{\Omega }}^{+}(G),+)$ is a cancellative semigroup, therefore it embedds into $\mathrm{\Omega }(G)$;

$(4)$ for the trivial group $E$ there is a ring isomorphism $\mathrm{\Omega }(E)\simeq \mathbb{Z}$;

$(5)$ for any group $G$ there is a ring monomorphism $\phi :\mathrm{\Omega }(G)\to {\oplus }_{i=1}^n\mathbb{Z}$ for some natural number $n\in \mathbb{N}$; this is called the characteristic embedding;

$(6)$ for any two groups $G,H$ we have: if $\mathrm{\Omega }(G)$ and $\mathrm{\Omega }(H)$ are isomorphic (as a rings), then $|G|=|H|$; generally $G$ need not be isomorphic to $H$.

Title	Burnside ring
Canonical name	BurnsideRing
Date of creation	2013-03-22 18:08:02
Last modified on	2013-03-22 18:08:02
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	10
Author	joking (16130)
Entry type	Definition
Classification	msc 16S99