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In the following questions we are dealing with subgroups of SYM($\aleph_0$) (the group of permutations of a countable set) with each non identity element having infinite support.
1. Are there two isomorphic subgroups (of SYM($\aleph_0$)) which are not conjugate?
2. Are there subgroups as above, but with transitive natural action on $\aleph_0$?
3. Are there locally finite subgroups as above, with transitive natural action on $\aleph_0$?

Thanks, this is my first time posting here.

Horse
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1 Answers1

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For the first question. Let us consider the morphism :

$$f_3:\mathbb{N}\rightarrow \mathbb{N} $$

$$3n\mapsto 3n $$

$$3n+1\mapsto 3n+2$$

$$3n+2\mapsto 3n+1$$

This is clearly a bijection of order $2$ whose support is $(3\mathbb{N})^c$ which is infinite.

On the other hand could also consider :

$$f_2:\mathbb{N}\rightarrow \mathbb{N} $$

$$2n\mapsto 2n+1 $$

$$2n+1\mapsto 2n$$

This is clearly a bijection of order $2$ whose support is $\mathbb{N}$.

Now the subgroup $G$ generated by $f_2$ and $H$ by $f_3$ are both isomorphic to $\mathbb{Z}_2$, nevertheless they cannot be conjugate (otherwise the set of fixed points would be empty for $f_3$ because it is for $f_2$).

Clément Guérin
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  • Thank you. I am trying to see if I can use your observation about fixed points to get an answer in my particular scenario, which is a case of "3.". – Horse Mar 25 '15 at 19:54
  • For the transitive case you ask something more difficult... But if you find a subgroup which acts transitively on $\mathbb{N}$ which is small (say like isomorphic to $\mathbb{Z}$) and which commutes to both $f_3$ and $f_2$ above we can use the same kind of idea, i think, to have the '2' case. – Clément Guérin Mar 25 '15 at 21:14
  • By the way, locally finite means...? – Clément Guérin Mar 25 '15 at 21:15
  • Clément, see here. – Jyrki Lahtonen Mar 25 '15 at 21:23