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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

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  • $\begingroup$ What's $G_{(\Gamma)}$? The setwise stabiliser of $\Gamma$? The pointwise stabiliser? Something else? $\endgroup$ – Lord Shark the Unknown Apr 28 '18 at 4:07
  • $\begingroup$ @Lord shark the unknown The pointwise stabiliser. $\endgroup$ – N math Apr 28 '18 at 4:09
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You need to induct on $|\Gamma\setminus\Delta|$ rather than on $|\Gamma\cap\Delta|$.

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    $\begingroup$ @c monsour I also induct on $|\Gamma\setminus \Delta|$ but i can't to solve it. Can you explain more? $\endgroup$ – N math Apr 28 '18 at 4:17
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    $\begingroup$ 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. $\endgroup$ – C Monsour Apr 28 '18 at 4:34
  • $\begingroup$ @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 :) $\endgroup$ – N math Apr 29 '18 at 4:35
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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).

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    $\begingroup$ So the result remains true if $\Gamma$ and $\Delta$ are infinite as long as $\Gamma\setminus\Delta$ is finite, but not otherwise. $\endgroup$ – C Monsour Apr 28 '18 at 11:50
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    $\begingroup$ 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 $\Delta$. $\endgroup$ – Derek Holt Apr 28 '18 at 12:37
  • $\begingroup$ @Derek Holt Thanks for your answer. $\endgroup$ – N math Apr 29 '18 at 4:39

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