Problem from Dixon&Mortimer textbook: (allegedly) sufficient conditions for a transporter between two finite subsets to be non-empty I should perhaps begin by apologizing for asking a not very profound question, but there is something that bothers me about exercise 1.5.19 from Permutation groups by Dixon&Mortimer. May I restate the hypothesis in a language closer to my personal preference. For any action of a group $G$ on a set $A$ and any subset $X\subseteq A$ define its fixator as $$\mathrm{Fix}_{G}\ X=\bigcap_{x \in X} \mathrm{Stab}_{G}\ x$$
when $X \neq \emptyset$ and set in particular $\mathrm{Fix}_{G}\ \emptyset=G$.
Assume now the following for a given transitive action of $G$ on $A$:


*

*$B, C \subseteq A$ are finite such that $|B| \leqslant |C|$

*$\mathrm{Fix}_{G}\ B$ and $\mathrm{Fix}_{G}\ C$ act transitively
on the complementary subsets $A \setminus B$ respectively $A\setminus C$.


Then the exercise would have it that there exists an operator $\lambda \in G$ such that $\lambda B \subseteq C$. The exercise also asks whether the alleged result remains valid for infinite subsets $B$ and $C$. 
Although I can't come up right away with a counterexample, these hypotheses of mere transitivity seem too weak to afford the nonemptiness of the transporter of $B$ to $C$. Expressing the conditions of the problem in terms of a certain point stabilizer and intersections of some of its conjugates (which is what the fixators would amount to) also doesn't seem too useful. There will exist of course only finitely many injections from $B$ to $C$, however why should one of them act like one of the operators of $G$ on $B$?  
I can't completely deny the possibility of there being some trickery of maneuvering $B$ into $C$, but at any rate I can't see how. If anyone could enlighten me, I would be grateful.
 A: I prefer to use Dixon&Mortimer's notation $G_{(X)}$ for what you call the fixator of $X$ in $G$. And like D&M, I very much prefer to use right actions, so I will use $B^g$ where you write $gB$.
Choose $g \in G$ such that $|B^g \cap C|$ is as large as possible, and then replace $B$ with $B^g$. If $B \cap C = B$ then we are done so suppose not. So there exists $b \in B \setminus C$. Since $|C| \ge |B|$ and $B,C$ are finite, there exists $c \in C \setminus B$.
If $A \ne B \cap C$, then there exists $a \in A \setminus (B \cap C)$, and we can find $g \in G_{(C)}$ with $b^g = a$ and $h \in G_{(B)}$ with $a^h=c$, so $b^{gh} = c$, and since $(B \cap C)^{gh} = B \cap C$, we now have $(B \cap C) \cup \{c\ \} \in B^{gh} \cap C$, contrary to the maximality of $|B^g \cap C|$.
It remains to deal with the case when $A = B \cup C$, in which canse $A$ is the disjoint union of $B \cap C$, $B \setminus C$ and $C \setminus B$ (so in particular $A$ is finite). By transitivity of $G$ on $A$ (which we haven't used yet), there exsists $g \in G$ with $c^g = b$. Since $c \not\in B$, $g$ can map at most $|B \setminus C| - 1$ elements of $B$ into $B \setminus C = A \setminus C$. So $g$ must map at least $|B| - |B \setminus C| + 1 = |B \cap C| + 1$ elements of $B$ into $C$. Hence $|B^g \cap C| > |B \cap C|$ again contradicting the maximality of $|B \cap C|$.
