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If $G$ is any group, then by $\text{Sub}(G)$ we denote the collection of all subgroups, ordered by $\subseteq$. If $(P,\leq)$ is a partially ordered set we let $\text{Max}(P)$ and the set of maximal elements.

Let $\frak{S}$ be the group of all bijections $f:\omega\to\omega$ together with composition.

Is there a member of $\text{Sub}(\frak{S}) \setminus \{\frak{S}\}$ that is not contained in some member of $\text{Max}\big(\text{Sub}(\frak{S})\setminus \{\frak{S}\}\big)$?

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  • $\begingroup$ @andreasthom sorry got the question totally wrong -- please see the revised one $\endgroup$
    – Dominic van der Zypen
    Commented Jun 9, 2018 at 13:50
  • $\begingroup$ Are you asking whether there's an element of $P$ which is not upper bounded by a maximal element of $P$, where $P$ is some poset of subgroups? $\endgroup$
    – Jalex Stark
    Commented Jun 9, 2018 at 14:15
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    $\begingroup$ Jalex: My guess is that he is asking if ``there's an element of P which is not upper bounded by a maximal element of P, where P is THE poset of subgroups''. $\endgroup$
    – Péter Komjáth
    Commented Jun 9, 2018 at 14:44
  • $\begingroup$ That's correct @peterkomjath. I want P to be the poset of proper subgroups -- that is, all the subgroups of $\text{Sym}(\omega)$ except $\text{Sym}(\omega)$ itself. $\endgroup$
    – Dominic van der Zypen
    Commented Jun 9, 2018 at 15:26

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As was written by Andreas Thom in his comment, Google gives an extensive literature on this subject. For example, look at the papers https://link.springer.com/chapter/10.1007/978-94-011-2080-7_18 or https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/jlms/s2-42.1.85 of Macpherson and his coauthors.

The question posed by Dominic van der Zypen was considered by Baumgartner, Shelah and Thomas who constructed a consistent example of a proper subgroup of $Sym(\omega)$ that is not contained in a proper maximal subgroup of $Sym(\omega)$.

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