Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\cal F$ denote the group of all finitely-supported permutations of $\mathbb N$. Say that a finite subgroup $G$ of $\cal F$ is singular if $G$ acts transitively on $\lbrace 1,2,3 \rbrace$ but no cyclic subgroup of $G$ acts transitively on $\lbrace 1,2,3 \rbrace$ (this is equivalent to saying that some element in $G$ sends $1$ to $2$, another sends $1$ to $3$ but no element of $G$ has all of $1,2$ and $3$ in a single orbit).

The Klein group (products of disjoint transpositions on $\lbrace 1,2,3,4 \rbrace$) is in example of such a subgroup.

Question 1 : are there other simple examples of minimal singular subgroups ? Is there a parametric description of all of them up to isomorphism ?

Question 2 : Denote by ${\cal F}(i \to j)$ the set of all permutations in $\cal F$ sending $i$ to $j$. Say that a permutation $s\in {\cal F}(1 \to 2)$ and a permutation $t\in {\cal F}(1 \to 3)$ are related iff the subgroup generated by $s$ and $t$ is a minimal singular subgroup of $\cal F$. Given $s$, let $R(s)$ denoted the set of all $t$'s such that $s$ and $t$ are related. Does $R(s)$ admit a simple description ?

Of course, any answer to question 2 automatically provides an answer to question 1.

share|improve this question

1 Answer 1

A simple example for Q1 generalizing your example for the Klein group:

Let $G$ be any non-cyclic finite group generated by two elements $g$ and $h$. Take the regular action of $G$ on itself, i.e., let $G$ act on itself by right rsp. left multiplication. Identify $G$ (as set acted upon) with a subset of $\mathbb N$ where the $1$ of the group is identified with $1 \in \mathbb N$, $g$ is identified with $2$ and $h$ with $3$.

As the orbits of the cyclic subgroups of $G$ containing $1$ are just the cyclic subgroups, none of them containing both $g$ and $h$, $G$ considered as subgroup (the regular action is faithful) of $\cal F$ is minimal singular.

Probably the condition minimal singular is too weak to be helpful. By the way, "finitely-supported" permutations are often called finitary.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.