All Questions
7 questions
21
votes
1
answer
564
views
Partitions of ${\rm Sym}(\mathbb{N})$ induced by convergent, but not absolutely convergent series
Let $(a_n) \subset \mathbb{R}$ be a sequence such that the series
$\sum_{n=1}^\infty a_n$ converges, but does not converge absolutely.
Then there is a partition of the symmetric group ${\rm Sym}(\...
13
votes
1
answer
1k
views
When taking the fixed points commutes with taking the orbits
Let $G$ and $H$ be groups, both acting on a set $X$ on the left, in such a way that the two actions commute. (Equivalently, let $G \times H$ act on $X$.)
The set $\text{Fix}_H(X)$ of $H$-fixed ...
6
votes
0
answers
245
views
A group action on another group action quotient: how to best describe the resulting structure and does it have a name?
Suppose I have an action $\alpha:G\times X\to X$ of a group $G$ on a set $X$ and, on top of that, an action $\beta:H\times(X/G)\to(X/G)$ of another group $H$ on the set of $G$-orbits.
Is there a nice ...
5
votes
1
answer
889
views
A generalized Burnside's lemma
Let $G$ be a finite group acting on a set $X$, and let $S\subseteq G$ be a union of conjugacy classes. Then I believe I can prove:
$$ \sum_{[x]\in X/G} \frac{|G_x \cap S|}{|G_x|} = \sum_{g\in S} \...
5
votes
1
answer
316
views
Connected permutation groups and wreath product
Let $G$ and $H$ be subgroups of the symmetric groups $\mathfrak S_m$ and $\mathfrak S_n$. Assume that $n>1$ and that $H$ is a 'connected' permutation group, that is, there is no non-trivial $H$-...
4
votes
1
answer
225
views
Integer-valued polynomials from Pólya counting
Let finite group $G$ act on a finite set $X$ and hence on colorings $Y^X$, where $Y=\{1,2,\ldots,k\}$ is a set of colors. The Burnside-Pólya-Redfield-etc. counting theorem says that the number of ...
0
votes
0
answers
186
views
Subset of reals associated to pairs of matrices in $\mathrm{SL}(2,\mathbb{R})$
Let $\Gamma$ be a subgroup of $\mathrm{SL}(2,\mathbb{R})$. I would like to ask if there is any research on the following set:
$$\Gamma*\Gamma:=\bigg\{\dfrac{(a+b)(a'+b')}{(c+d)(c'+d')}\bigg|\begin{...