Dear Ahmet,
I think I now see the error in my argument:
> On 23. Apr 2021, at 11:35, Max Horn <h...@mathematik.uni-kl.de> wrote:
>
[...]
> For suppose G and X are as desired. Then G must be non-trivial and hence
> there is a point a \in \Omega which G moves. Let
>
> X' := { \pi \in X | a^\pi \neq a }
>
> Then this set still must be infinite, for if it was finite, then
> X'':=X\setminus X' would be an infinite subset of X but the group it
> generates fixes a and hence cannot be G.
>
> So we may replace X by X'. Now we can repeat this process for any point moved
> by G (of which there must be infinitely many).
>
> In the end, the set X only contains permutations with infinite support.
In this last step, I was hasty. The problem is that one has to repeat the
reduction an infinite number of times... And then one may end up with a finite
or even empty subset. Indeed, suppose wlog that G fixes no points in \Omega.
Then for each a\in\Omega we can define
X_a := { \pi \in X | a^\pi \neq a }
as above. My last step essentially claimed that we can switch to
\bigcap_{a\in\Omega} X_a
but this intersection then could very well be empty.
Cheers
Max
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