Does $\text{Sym}(\omega)$ have $2^{\aleph_0}$ pairwise non-isomorphic subgroups? 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?
 A: It has $2^{2^{\aleph_0}}=2^c$ pairwise non-isomorphic subgroups (this is the maximum we could expect since this is the number of subsets).
To prove this, first start with a family of groups $(G_i)_{i\in c}$ such that


*

*all $G_i$ are countable, torsion-free, with trivial center, and directly indecomposable

*the $G_i$ are pairwise non-isomorphic.


The existence of such a family (even with $G_i$ finitely generated) is standard (just start with $H_i$ torsion-free countable, and define $G_i=H_i\ast H_i$ which will satisfy in addition the other two conditions).
For $I\subset c$, define $G_I=\bigoplus_{i\in I}G_i$, where $\bigoplus$ denotes the restricted direct product.
The $G_I$ are pairwise non-isomorphic, because being center-free and indecomposable ensures the uniqueness of the restricted direct product decomposition.
However, $G_I$ does not embed into $\mathrm{Sym}(\omega)$, so this is not the example yet.
Embed $G_I$ into $\mathrm{Sym}(\omega)/\mathrm{Sym}_{\mathrm{fin}}(\omega)$. This is possible as follows: consider $c$ infinite countable subsets $(A_i)$ of $\omega$, with pairwise finite intersection (the classical trick is to identify $\omega$ with $\mathbf{Q}$ and choose for every real the image of a sequence converging to it). Embed $G_i$ into $\mathrm{Sym}(A_i)$; the intersection property implies that this extends to a homomorphism $f$ of $G=\bigoplus_{i\in c}G_i$ into $\mathrm{Sym}(\omega)/\mathrm{Sym}_{\mathrm{fin}}(\omega)$ (quotient by the group of finitely supported permutations). Then $f$ is injective (because each $G_i$ is torsion-free). Now, for each $I$, let $H_I$ be the inverse image of $f(G_I)$ in $\mathrm{Sym}(\omega)$. Then the set of torsion elements in $H_I$ is $\mathrm{Sym}_{\mathrm{fin}}(\omega)$ and the quotient is isomorphic to $G_I$; in particular, the $H_I$ are pairwise non-isomorphic when $I$ ranges over subsets of $c$.

Edit (to complement Gerhard's answer, since his examples are abelian):
There are also $2^c$ non-isomorphic abelian subgroups in $\mathrm{Sym}(\omega)$. Indeed, there exist $2^c$ non-isomorphic torsion-free abelian groups of cardinal $c$ (see this MO answer). Each such group has its injective hull isomorphic to $\mathbf{Q}^{(c)}\simeq\mathbf{Q}^{\omega}$ and therefore embeds into $\mathrm{Sym}(\omega)$.
A: 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.
