[GAP Forum] All factorizations of a permutation

[Max Horn]

Dear Ahmet,

I think I now see the error in my argument:

> On 23. Apr 2021, at 11:35, Max Horn <horn at 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