I am new to StackExchange so apologies if my question is poorly asked or does not abide by the standards.
Referring to the 2016 edition of Tomkowicz and Wagons' book on the Banach-Tarski Paradox, looking specifically at the second part of corollary 5.12, it states:
Suppose $G$ acts on $X$ and a proper system of $m$ congruences involving $A_1,\ldots,A_r$ is given. If $G$ has $m$ independent elements such that the group $F$ that they generate acts on $X$ without non-trivial fixed points, then $X$ may be partitioned into sets $A_i$ satisfying the system. If the action of $F$ is merely locally commutative, then $X$ can be partitioned into sets $A_i$ satisfying the system.
I was able to prove that the property of local commutativity is equivalent to any orbit having at most one cycle. However, I would like to know if the statement of the second part of the corollary were still true if there were at most $2$ cycles?
The definition I am using for an orbit having at most $k$ cycles is the following:
Every finite set $X$ of vertices spans at most $|X|+(k-1)$ edges.
I have tried to split into cases where we choose an arbitrary $x$ in an orbit $\mathcal{O}$ and consider the different ways in which the cycles are positioned with respect to $x$. For example, two cases may be:
And then I would consider some arbitrary $y$ in Orb$(x)$ and show how it can be placed into set $A_i$, satisfying the containment relations but I cannot get very far. Is this even necessarily true?