Let $G$ be a finite group, and $B(G)$ be its Burnside ring. Show that each ring homorphism $\varphi:B(G)\to\mathbb{Z}$ is the mark of some $H\le G$, i.e. it maps to an equivalent class of finite $G$-set to the number of elements fixed by $H$.
I am working on Alperin's Groups and Representations. I think it should be an easy exercise, but I cannot prove it. I considered representatives of conjugate classes of $G$: $G_1=1,G_2,\ldots, G_n=G$, and correspondent coset spaces $G/G_i$. But I am not sure how to recover $H$ by $\varphi$.
Thanks for your help.
(Sorry if my answer is a bit messy, I learned about Burnside rings reading your question and got interested in solving the exercise, so my notations may very well differ with the usual ones.)
First, let $H$ be a subgroup of $G$. We are going to try to recover $H$ from the data of its mark $\varphi_H$.
We put $E$ the set of subgroups of $G$, and for every $K\leqslant G$ we write $\pi_K:G/H\to E$ defined by $\pi_K(gK)=gKg^{-1}$ (which is well-defined).
Let $K$ be a subgroup of $G$ ; we compute $\varphi_H(G/K)$. If $g\in G$, then for any $h\in H$, $h\cdot gK=gK$ iff $h\in gKg^{-1}$. So $gK\in G/K$ is fixed by all of $H$ iff $H\subset gKg^{-1}$. So $\varphi_H(G/K)$ is the number of $x\in G/K$ such that $H\subset \pi_K(x)$.
In particular, $\varphi_H(G/K)\neq 0$ iff $H$ is conjugated to a subgroup of $K$. Thus we can recover $H$ (or rather its conjugacy class, of course) : there is exactly one conjugacy class of subgroups of $G$ such that any $K$ in this class has minimal cardinality for the property $\varphi_H(G/K)\neq 0$.
Now let $\varphi:B(G)\to \mathbb{Z}$ be any ring morphism. Let $H$ be a subgroup with minimal cardinality such that $\varphi(G/H)\neq 0$ (which must exist since at worst $\varphi(G/G)=1$). We will show that $\varphi = \varphi_H$.
Let $K$ be any subgroup of $G$, and put the following equivalence relation on $G/K$ : $gK\sim g'K$ iff $H\cap (g'Kg^{-1})\neq \emptyset$. Choose $(x_i)_{i\in I}$ a set of representatives for the relation $\sim$.
Then in $B(G)$ we have $G/H\cdot G/K = \sum_{i\in I} G/N_i$ with $N_i = H\cap \pi_K(x_i)$ (because each orbit of $G/H\times G/K$ must contain at least one element of the form $x_g=(H,gK)$, whose stabilizer is $H\cap (gKg^{-1})$, and $x_g$ is in the same orbit as $x_{g'}$ iff $g\sim g'$).
Now if $H\not\subset \pi_K(x_i)$, then $|N_i|<|H|$ so by minimality of $|H|$, $\varphi(G/N_i)=0$. And if $H\subset \pi_K(x_i)$, then $N_i=H$ so in particular $\varphi(G/N_i)=\varphi(G/H)$. Thus we get that $\varphi(G/H\cdot G/K)=n_K\varphi(G/H)$ where $n_K$ is the number of $i\in I$ such that $H\subset \pi_K(x_i)$. Clearly then $n_K = \varphi(G/K)$ since $\varphi(G/H)\neq 0$.
It just remains to show that $n_K = \varphi_H(G/K)$. But $\varphi_H(G/K)$ counts the $x\in G/K$ such that $H\subset \pi_K(x)$, and $n_K$ counts those $x$ only modulo $\sim$. So we have to show that if $x,y\in G/K$ are such that $H\subset \pi_K(x)$ and $H\subset \pi_K(y)$, then $x\sim y$ only if $x=y$.
Suppose $x=gK$ and $y=g'K$ satisfy that. Since $x\sim y$, we have $h=g'kg^{-1}$ for some $h\in H$ with $k\in K$. Now since $H\subset gKg^{-1}$, $h=gk'g^{-1}$ for some $k'\in K$. Then $g'=gk'k^{-1}$ and $gK=g'K$, ie $x=y$.
In the end, we do have $\varphi = \varphi_H$.