# Burnside convolution

Let $$G$$ be a group. Say that an orbit is a nonempty transitive $$G$$-set. Let $$\Xi$$ be a set of finite orbits such that each finite orbit is isomorphic to exactly one element of $$\Xi$$.

If $$X,Y,Z\in\Xi$$, then $$X\times Y$$ is a disjoint union of orbits, and the multiplicity of $$Z$$ in $$X\times Y$$ is a well-defined nonnegative integer $$b(X,Y,Z)$$.

Say that $$G$$ satisfies Condition (C) if for all $$Z\in\Xi$$ there are only finitely many pairs $$(X,Y)\in\Xi^2$$ such that $$b(X,Y,Z)\ne0$$.

Question 1. Do all groups satisfy Condition (C)?

The motivation for introducing Condition (C) is that, if it holds, then we can define a convolution on the group $$A:=\mathbb Z^\Xi$$ of all maps from $$\Xi$$ to $$\mathbb Z$$ by $$(f*g)(Z):=\sum_{X,Y\in\Xi}f(X)\ g(Y)\ b(X,Y,Z),$$ and it is easy to see that $$(A,+,*)$$ is a commutative ring with one, which coincides with the Burnside ring of $$G$$ if $$G$$ is finite.

(We can of course take only the finitely supported fonctions in $$\mathbb Z^\Xi$$, but this was the subject of this question.

If we start with a monoid $$M$$ instead of a group $$G$$, we can generalize the above lines by replacing the notion of orbit by that of nonempty $$M$$-set which is not a disjoint union of nonempty sub-$$M$$-sets, and we can ask

Question 2. Do all monoids satisfy Condition (C)?

The monoids given as examples in this question satisfy Condition (C), but I don't even know if the additive monoid $$\mathbb N$$ does, so let me ask formally

Question 3. Does $$\mathbb N$$ satisfy Condition (C)?

Edit. In view of this answer it seems appropriate to ask a fourth question.

Say that an $$M$$-set is indecomposable if it is neither empty nor a disjoint union of nonempty sub-$$M$$-sets.

Say also that $$M$$ satisfies Condition (D) if for all $$X,Y,Z$$ such that

$$\bullet\ X$$ and $$Y$$ are two finite indecomposable $$M$$-sets,

$$\bullet\ Z$$ is a maximal indecomposable sub-$$M$$-set of the product $$X\times Y$$,

the map $$Z\to X$$ induced by the projection is surjective.

Clearly groups satisfy Condition (D).

Note that (D) implies (C), because, up to isomorphism, there are only finitely many quotients of a given finite $$M$$-set.

Question 4. Do all monoids satisfy Condition (D)?

Let me restate your question for groups. Transitive $$G$$-sets modulo isomorphisms are in bijective correspondence with the set of subgroups modulo conjugation, and finite ones correspond to finite index subgroups. For subgroups $$A,B,C$$ of $$G$$, the condition $$b(G/A,G/B,G/C)\neq 0$$ means that there exist conjugates $$A',B'$$ of $$A,B$$ such that $$C=A'\cap B'$$.

Since every finite index subgroup is contained in only finitely many subgroups, this leaves finitely many possibilities for $$A',B'$$, and hence, up to conjugation, finitely many possibilities for $$A,B$$.

So the answer (to question 1) is yes.

• Thanks! Did you take into account the fact that only finite $G$-sets are considered? – Pierre-Yves Gaillard Mar 22 at 15:03
• Thanks, it's fixed now. – YCor Mar 22 at 15:07

Edit 2. The answer all questions is Yes: see Edit 1 below. End of Edit 2.

The answer to Question 3 is Yes.

Say that an $$M$$-set is indecomposable if it is neither empty nor a disjoint union of nonempty sub-$$M$$-sets.

Say also that $$M$$ satisfies Condition (D) if for all $$X,Y,Z$$ such that

$$\bullet\ X$$ and $$Y$$ are two finite indecomposable $$M$$-sets,

$$\bullet\ Z$$ is a maximal indecomposable sub-$$M$$-set of the product $$X\times Y$$,

the map $$Z\to X$$ induced by the projection is surjective.

Clearly groups satisfy Condition (D).

Note that (D) implies (C), because, up to isomorphism, there are only finitely many quotients of a given finite $$M$$-set.

So it suffices to show that $$\mathbb N$$ satisfies (D).

For ease of notation replace the additive monoid $$\mathbb N$$ with the multiplicative monoid $$M$$ freely generated by the element $$a$$.

Say that a point of an $$M$$-set is periodic if it is a fixed point of $$a^n$$ for some $$n\ge1$$.

The proof of the following facts is left to the reader:

(a) If $$y$$ is a periodic point of an $$M$$-set $$X$$ and $$n$$ is a nonnegative integer, then $$y=a^nx$$ for some $$x\in X$$.

(b) If $$x$$ is a point of a finite $$M$$-set, then $$a^nx$$ is periodic for $$n$$ large enough.

(c) If $$x$$ and $$y$$ are two points of a finite indecomposable $$M$$-set, and if $$y$$ is periodic, then there is an $$n\in\mathbb N$$ such that $$a^nx=y$$.

Let us prove that $$M$$ (or equivalently $$\mathbb N$$) satisfies (D).

Let $$(x_0,y_0)$$ be in $$Z$$ and let $$x$$ be in $$X$$. It is enough to show that there is a $$y\in Y$$ such that $$(x,y)\in Z$$. By (b) there is an $$i\in\mathbb N$$ such that $$(x_1,y_1):=a^i(x_0,y_0)$$ is a periodic point in $$Z$$. There is, by (c), a $$j\in\mathbb N$$ such that $$a^jx=x_1$$, and, by (a), a $$y\in Y$$ such that $$a^jy=y_1$$. This implies $$a^j(x,y)=(x_1,y_1)\in Z$$, and thus $$(x,y)\in Z$$, as desired.

Edit 1. Let us prove that the answer to Question 4 is Yes. This will imply that the answer all questions is Yes.

The argument is similar to the one used above to answer Question 3, but I prefer to make this edit self-contained.

Let us fix an element $$a$$ of $$M$$. Say that a point of an $$M$$-set is periodic if it is a fixed point of $$a^n$$ for some $$n\ge1$$.

The proof of the following facts is left to the reader:

(a) If $$y$$ is a periodic point of an $$M$$-set $$X$$ and $$n$$ is a nonnegative integer, then $$y=a^nx$$ for some $$x\in X$$.

(b) If $$x$$ is a point of a finite $$M$$-set, then $$a^nx$$ is periodic for $$n$$ large enough.

In the setting of Question 4, let $$p:X\times Y\to X$$ be the projection, and assume by contradiction that $$p(Z)$$ is a proper subset of $$X$$. Then there is a tuple $$(a,x_1,x_2,y_2)$$ with $$a\in M;\ x_1,x_2\in X;\ x_1\notin p(Z);\ ax_1=x_2;\ y_2\in Y;\ (x_2,y_2)\in Z.$$ It suffices to show $$x_1\in p(Z)$$. By (b) we can pick an $$n\in\mathbb N$$ such that $$a^n(x_2,y_2)\in Z$$ is periodic. Set $$x_3:=a^nx_2=a^{n+1}x_1,\ y_3:=a^ny_2.$$ By (a) there is a $$y_1\in Y$$ such that $$a^{n+1}y_1=y_3$$, and we get $$a^{n+1}(x_1,y_1)=(x_3,y_3)\in Z,\$$ which implies $$(x_1,y_1)\in Z$$ and thus $$x_1\in p(Z)$$, contradiction.