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Let $G$ be a finite group and $B(G)$ be its Burnside ring, i.e. formal sums of isomorphism classes of finite $G$-sets with addition given by disjoint union and multiplication given by Cartesian product. I denote by $[X]$ the class of a $G$-set $X$ in $B(G)$.

Let $X$ be a $G$-set of the form $G/H$ for some subgroup $H \subset G$. $\textit{I'm trying to prove that}$ $$\sum\limits_{n=0}^\infty [Sym^n(X)]t^n$$ is a rational function. Here $Sym^n(X) = \{ f \colon X \to \mathbb{N}\cup\{0\} : \sum\limits_{x \in X} f(x) = n\}$ (multisubsets of $X$ of cardinality $n$).

In order to do this, I'm trying to prove that there is a linear recurrence relation for $[Sym^n(X)]$ in terms of $\{[Sym^k(X)]\}_{k < n}$, where $n \gg 0$. I'm able to prove this fact in the case of $|X|=2$, because then $$[Sym^{2k}(X)] = k[X]+pt\quad \text{and} \quad [Sym^{2k+1}(X)] = 2k[X],$$ and also in the case of $|X|=3$, because then, if I'm not mistaken, $$[Sym^n(X)] = 2[X]+[Sym^{n-2}(X)]+[Sym^{n-3}(X)]-[Sym^{n-5}(X)].$$

The problem is that I do not know how to generalize these results. In the general case, I understand that $Sym^n(X) \supset Sym^1(X)$ and $Sym^{n-k}(X)$, where $|X|=k$, because the subset of functions taking the only non-zero value (equal to $n$) is $G$-isomorphic to $Sym^1(X)$, while the subset of functions $f \in Sym^n(X)$ such that $f(x) \neq 0$ for all $x \in X$ is $G$-isomorphic to $Sym^{n-k}(X)$ (one can subtract $1$ from each value). I also understand that one can stratify $Sym^n(X)$ by partitions of $n$ with length less or equal to $k$, but this stratification is not in terms of previous symmetric powers, but in terms of some arbitrary $G$-sets, so it does not help.

Any help or advice would be appreciated. I'm really stuck with that.

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1 Answer 1

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You can use the pre-$\lambda$ ring identity $$ \big(\sum_n [\mathrm{Sym}^n(X)]t^n\big)\big( \sum_k \lambda^k[X](-t)^k\big) = 1.$$ So it is enough to show that the generating series for the exterior powers is a polynomial, for any finite $G$-set $X$. There is an explicit formula for $\lambda^k[X]$ in a paper of Rökaeus, from which it in particular follows that $\lambda^k[X]=0$ if $k>|X|$.

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  • $\begingroup$ Thank you for the answer. I know that given a pre-$\lambda$ structure one can define the opposite one, but could you please explain why in this case I obtain exactly exterior powers? It seems reasonable, but I do not understand how to prove this fact. $\endgroup$ Commented Jun 6, 2022 at 9:53
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    $\begingroup$ It seems this is in fact not true. I edited the answer. $\endgroup$ Commented Jun 6, 2022 at 12:41

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