# Exercise of permutation groups

Suppose that the group $G$ acts transitively on $\Omega$ and $\Gamma$ and $\Delta$ are finite subsets of $\Omega$ with $|\Gamma| < |\Delta|$. If $G_{(\Gamma)}$ and $G_{(\Delta)}$ act transitively on $\Omega \setminus \Gamma$ and $\Omega \setminus \Delta$, respectively, show that ${\Gamma}^x \subset \Delta$ for some $x \in G$. Does the result remain true if $\Gamma$ and $\Delta$ are infinite?

I think if $\Gamma \subset \Delta$ is trivial. I use induction on the members in intersection. But it doesn't work. I try to solve it but I stuck. I don't have any idea how to solve it. Any kind of suggestion is appreciated. Thanks to everyone for the help

• What's $G_{(\Gamma)}$? The setwise stabiliser of $\Gamma$? The pointwise stabiliser? Something else? – Lord Shark the Unknown Apr 28 '18 at 4:07
• @Lord shark the unknown The pointwise stabiliser. – N math Apr 28 '18 at 4:09

You need to induct on $|\Gamma\setminus\Delta|$ rather than on $|\Gamma\cap\Delta|$.
• @c monsour I also induct on $|\Gamma\setminus \Delta|$ but i can't to solve it. Can you explain more? – N math Apr 28 '18 at 4:17
• Use $G_{(\Delta)}$ to move a point of $\Gamma\setminus\Delta$ to a point in $G\setminus(\Gamma\cup\Delta)$, and then use $G_{(\Gamma)}$ to map it into $\Delta$. This maps one more element of $\Gamma$ into $\Delta$ without disturbing the rest. You do need a separate argument for the starting point in the case you have $\Omega=\Gamma\cup\Delta$. I leave that as an exercise. – C Monsour Apr 28 '18 at 4:34
• @C Monsour Thank you for your answer. For the case $|\Gamma \setminus \Delta| =1$ I proved as you said but alittle more complicated. But I was not sure this way is true because I got stuck solving in the following. Now, I try to continue and solve it. Thank you :) – N math Apr 29 '18 at 4:35
For the infinite case, consider the group of all permutations of an uncountable set $\Omega$ with finite support (i.e. permutations that move only finitely many points).
• So the result remains true if $\Gamma$ and $\Delta$ are infinite as long as $\Gamma\setminus\Delta$ is finite, but not otherwise. – C Monsour Apr 28 '18 at 11:50
• Yes I think that's right. If $\Gamma \setminus \Delta$ is finite, then your inductive proof still works. But otherwise we would need to move infinitely many points to map $\Gamma$ into $\Delta$. – Derek Holt Apr 28 '18 at 12:37