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 <h...@mathematik.uni-kl.de> şunları yazdı (23 Nis 2021 12:35):
>
>
>> On 23. Apr 2021, at 10:25, Ahmet Arıkan <ari...@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
>
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