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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?

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    $\begingroup$ Indeed. For each proper countable infinite subset of primes, consider the abelian group which has finite cycles of order each prime. Gerhard "That Should Be Enough Examples" Paseman, 2018.06.02. $\endgroup$ Jun 3, 2018 at 6:10
  • $\begingroup$ Oh - nice, thanks! Can you post this as an answer so we can close this thread? $\endgroup$ Jun 3, 2018 at 6:12
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    $\begingroup$ Just out of curiosity, can we show that it actually have more than $2^{\aleph_0}$? $\endgroup$ Jun 3, 2018 at 6:14
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    $\begingroup$ @i707107 this is the right question. $\endgroup$
    – YCor
    Jun 3, 2018 at 6:48
  • $\begingroup$ That $Sym(\omega)$ has $c$ non-isomorphic subgroups is trivial since there are uncountably many non-isomorphic countable groups, and highly a duplicate (since the construction of $c$ non-isomorphic countable groups appears at many places). $\endgroup$
    – YCor
    Jun 9, 2018 at 7:45

2 Answers 2

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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.

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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 $\mathrm{Sym}(\omega)$.

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    $\begingroup$ Another approach: it's immediate that ultrafilter stabilizers are pairwise distinct: this is historically probably the first family of $2^c$ subgroups detected in $Sym(\omega)$ (it's observed in 1956 by Rudin; F. Richman in 1966 shows that these are maximal subgroups). It's likely that ($\ast$) the stabilizers of ultrafilters $\eta_1,\eta_2$ are isomorphic if and only if $\eta_1$ and $\eta_2$ are in the same orbit under $Sym(\omega)$. If so, this would imply the existence of $2^c$ non-isomorphic subgroups among these stabilizers. I don't know if anybody checked ($\ast$). $\endgroup$
    – YCor
    Jun 3, 2018 at 8:50

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