# $\text{Max}\big(\text{Sub}(\text{Sym}(\omega))\setminus \{\text{Sym}(\omega)\}\big)$

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 $\text{Max}\big(\text{Sub}(\frak{S})\setminus \{\frak{S}\}\big)$?

• @andreasthom sorry got the question totally wrong -- please see the revised one – Dominic van der Zypen Jun 9 '18 at 13:50
• 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? – Jalex Stark Jun 9 '18 at 14:15
• 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''. – Péter Komjáth Jun 9 '18 at 14:44
• 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. – Dominic van der Zypen Jun 9 '18 at 15:26

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 $Sym(\omega)$.