In a paper I am reading at the moment (Hrushovski Martin, elimination of imaginaries in $Q\_p$), in some proof they use the following fact (at least this would be enough to get their proof going, but maybe we have more hypothesis) :


If we have a group $G$ acting on a set $X$ containing elements $(e\_i\mid i<\omega)$ such that all $g\in G$ fixes all the $e\_i$ but a finite number, is it true that all but a finite number of $e\_i$ are fixed by $G$?


The authors say it is a consquence of Neumann's lemma (my guess is the one saying that a group covered by a finite number of cosets of subgroups is covered by the ones of finite index) but I have been trying to figure it out for some time already and cannot. If anyone has any idea of how that might work, a little help would be very welcome.




---


Some days later : Thanks a lot for the counter-examples, they really helped me understand.


If anyone is interested, what I really needed was that there are at most a finite number of $e\_i$ with an infinite orbit under $G$. This is false in general (see the counter-example given in the answer) but Newmann's lemma does imply that if there are an infinity of elements with an infinite orbit then for all $N$, we can find a $g$ and $N$ elements that are not fixed by $g$. And as it happens that my $G$ is in fact the automorphism group of a very saturated model (sorry, some model theory had to barge in) by compacity I can find an automorphism that moves an infinity of $e\_i$.


I hope any of this made sense.



A problem with Neumann's lemma.
Let $S\_\infty=\bigcup\_{n=1}^\infty S\_n$ be the infinite symmetric group consisting of permutations of $\mathbb{N}$ fixing all but finitely many elements. Then $\{e\_i=i\}$ satisfy your conditions but no element $i\in \mathbb{N}$ is fixed by $S\_\infty$.


(Sorry, I see that Andreas Blass already gave this counterexample. I've made this answer community wiki.)


