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It is well known that for every positive cardinal $\kappa$, every group of cardinality $\kappa$ can be embedded into $\text{Sym}(\kappa)$, the group of bijections on $\kappa$ with composition as group operation.

Is there an infinite cardinal $\kappa$ and a group $U$ of cardinality $\kappa$ such that every group of cardinality $\kappa$ is isomorphic to a subgroup of $U$?

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    $\begingroup$ Just for context, the answer is no for $\kappa<2^{\aleph_0}$, using that there exists continuum many non-isomorphic finitely generated groups, and that each countable group embeds in one of them. $\endgroup$
    – YCor
    Commented Oct 8, 2019 at 7:18
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    $\begingroup$ For $\kappa=2^{\aleph_0}$, I think this is an open question. Actually, concerning the group $G=\mathrm{Sym}(\aleph_0)/\mathrm{fin}$, the question whether every group of continuum cardinal embeds into $G$ is open, and is also open in ZFC+CH; in (ZFC + $2^{\aleph_0}=\aleph_2$) it consistently has a negative answer. $\endgroup$
    – YCor
    Commented Oct 8, 2019 at 7:24
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    $\begingroup$ Jónsson (1956) proved that under GCH, the answer is yes for every uncountable $\kappa$, see Theorem 3.1 therein. However to get the result for a single $\kappa$ probably does not require GCH, see notably Theorem 2.9 (unfortunately I can't find a definition for $\mathfrak{n}_\mu$ used there, so I'm not sure of meaning of the assumption). $\endgroup$
    – YCor
    Commented Oct 8, 2019 at 7:47
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    $\begingroup$ Quoting from page 3 of Banach spaces and groups - order properties and universal models by Shelah and Usvyatsov: "Note that every AEC with amalgamation has a universal model in every regular $\lambda$ satisfying $\lambda=\lambda^{<\lambda}$ or $\lambda=\mu^+$ and $2^{<\mu}\leq\lambda$." So it seems one could take $\lambda=\beth_\omega^+$. $\endgroup$
    – Gabe Conant
    Commented Oct 8, 2019 at 8:09
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    $\begingroup$ @AsafKaragila Yes, I am, it's no contradiction. Continuum many pairwise non-isomorphic 2-generated groups. And since you talk about formulas, continuum pairwise non-elementary equivalent 2-generated groups. Precisely for every set $J$ of primes there exists a 2-generated group $G$ in which for $p$ prime, the formula $\exists g\neq 1:g^p=1$ holds in $G$ iff $p\in J$. $\endgroup$
    – YCor
    Commented Oct 8, 2019 at 14:37

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