[GAP Forum] All factorizations of a permutation

[Ahmet Arıkan]

Thank you Max,

I think I can clarify more as follows:

My initial question “Is there a perfect p-group G contained FSym(IN*) which satisfies my condition (G has a generating subset X such that every infinite subset of X also generates G—-I call such groups “infinitely supported”). Here IN* is the set of non-zero natural numbers and FSym(IN*) is the group of all finitary permutations of IN* as usual. It can be found some examples for non-perfect or simple cases, but such cases are not under my consideration for now.

To know the existence of such perfect groups shall provide a very important point of view for some other crucial problems. In my opinion, we need to start with some (useful?) elements to construct such an X using GAP. So I need the distinct factorizations of elements to construct X (I guess!).

Best wishes,
Ahmet Arikan



——————————-
Gazi University, 
Gazi Education Faculty, 
Department of Mathematics Education,
Ankara, 
Turkey

Max Horn <horn at mathematik.uni-kl.de> şunları yazdı (23 Nis 2021 12:35):

> 
> 
>> On 23. Apr 2021, at 10:25, Ahmet Arıkan <arikan at gazi.edu.tr> wrote:
>> 
>> Hi Chris, thank you for the reply. 
>> 
>> My main problem is to construct a group G in Sym(\Omega) satisfying the following property: 
> 
> What is \Omega? An arbitrary infinite set?
> 
>> 
>> G has a generating subset X such that every infinite subset of X also generates G.
>> 
>> So to construct such a group, we may start with an element x (say x=(1,3,4) ) to contruct X. Then we need to find suitable factorizations like (1,3,4)=(1,2,3,4)*(2,3) ( or multiple factorizations)  and  continue to construct X={(1,3,4), (1,2,3,4),(2,3),...}. This is just an explanation of why I want to find suitable factorizations of permutations.
> 
> I don't see at all why you "may start" wich such elements. To the contrary, I think looking at such elements won't help at all.
> 
> 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. Moreover, we can of course restrict it to be countably infinite.
> 
> If I am not mistaken, here is an example for a group and set as described: G=\Sym(\ZZ) together with 
> 
> X := \{ \pi_p | p is a prime \}
> 
> and
> 
> \pi_p: ZZ\to\ZZ, x \mapsto x + p
> 
> It satisfies the even strong property that any subset of X of size at least 2 still generates G.
> 
>> We do not know yet if such a perfect locally finite (p-) group G exists.
> 
> Why do you think such a group must be perfect? Or are you asking whether such a group *can* be perfect? The example I gave of course is abelian and not locally finite. So perhaps there are more requirements in your question that are missing?
> 
> 
> 
> 
> Cheers
> Max
>