Let us call a group $(G,\cdot)$ finitarily complete if $G$ is countable, and every finite group is isomorphic to a subgroup of $(G,\cdot)$.
Is there a collection of $2^{\aleph_0}$ pairwise non-isomorphic, finitarily complete groups?
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asked Nov 16, 2023 at 12:46
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2$\begingroup$ Given every countable group $G$, there are continuum many f.g. groups containing it. Indeed, take a f.g. group $H$ containing it, and a continuum family $L_i$ of f.g. groups. Then $(H\times L_i)$ includes continuum many f.g. groups. (The map $i\mapsto$ isomorphism class of $H\times L_i$ is countable-to-one.) $\endgroup$– YCorCommented Nov 16, 2023 at 13:56
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1$\begingroup$ A variant just for the question: for each subset $J$ of $\mathbf{N}_{\ge 2}$, consider the group $G_J=\ast_{n\in J}S_n$. Then $S_n$ is isomorphic to a free factor of $G_J$ iff $n\in J$, so $G_J$ determines $J$ (and $G_J$ contains all finite groups as soon as $J$ is infinite). The same argument with restricted direct product (excluding $n=2$) works since the decomposition into indecomposable factors works well for centerless groups. $\endgroup$– YCorCommented Nov 16, 2023 at 14:32
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2 Answers
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Yes. Consider groups of the form $\operatorname{FSym}(\mathbf N) \times G$, where $G$ is any countable group. Obviously any such group contains a copy of every finite group. I am not sure whether $\operatorname{FSym}(\mathbf N) \times G \cong \operatorname{FSym}(\mathbf N) \times H$ implies $G \cong H$ -- probably -- but this is certainly true if $G$ and $H$ are abelian, since in this case $G$ is the centre of $\operatorname{FSym}(\mathbf N) \times G$, and there are $2^{\aleph_0}$ isomorphism classes of countable abelian groups.
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$\begingroup$ For the number of abelian groups, see math.stackexchange.com/q/119642/23805. $\endgroup$ Commented Nov 16, 2023 at 13:31
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2$\begingroup$ To be self-contained: for any subset $S$ of the set of prime numbers let $A_S$ be the additive group of the ring $\mathbf{Z}[S^{-1}]$. Then $p\in S$ iff $pA_S=A_S$, so $A_S$ determines $S$. $\endgroup$– YCorCommented Nov 16, 2023 at 14:01
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Here's another solution. Observe that if $G_1,G_2,G_3,\dots $ is an infinite sequence of simple groups, pairwise non-isomorphic, and $G$ is the direct limit of the finite products of these, then the $G_j$ are the only simple groups that admit a surjective homomorphism from $G$. Now choose $G_j$ to be the alternating group $A_{n_j}$ for any sequence $5\le n_1<n_2<n_3<\dots$.
answered Nov 16, 2023 at 18:13
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$\begingroup$ Ah, I see that YCor indicated essentially the same solution in a comment. $\endgroup$ Commented Nov 16, 2023 at 18:15