Let $\text{Sym}(\omega)$ denote the set of all bijections $f:\omega\to\omega$ together with composition as group operation. Does $\text{Sym}(\omega)$ have $2^{\aleph\_0}$ pairwise non-isomorphic subgroups?



Does $\text{Sym}(\omega)$ have $2^{\aleph_0}$ pairwise non-isomorphic subgroups?
To get a continuum of a selection of different subgroups, take that many proper countable infinite subsets of primes. For each such subset S consider the abelian subgroup where an element is composed of one or more disjoint cycles each cycle of length a prime p belonging to S. As an isomorphism must map an element of finite order to another of the same order, different subsets S must give rise to non isomorphic Abelian subgroups of the symmetric group.


Gerhard "Probably Works For Direct Sum" Paseman, 2018.06.02.


