For any group $G$, we let $\text{Sub}(G)$ be the complete lattice of subgroups of $G$. Let $\text{Sym}(\omega)$ be the group of all bijections $f:\omega\to\omega$. 


What is an element of $U\in\text{Sub}(\text{Sym}(\omega))$ such that for all $V\in \text{Sub}(\text{Sym}(\omega))$ such that the subgroup generated by $U\cup V$ is $\text{Sub}(\text{Sym}(\omega))$, we have that $U\cap V$ contains more elements than just the neutral element $\text{id\_\omega}\in \text{Sym}(\omega)$?



Complements in $\text{Sub}(\text{Sym}(\omega))$
I understand the question as: "what is a subgroup $U\le G$ such that ($\*$) for every $V\le G$ such that $V\cap U=\{1\}$ we have $\langle U,V\rangle\neq G$"? I also understand "what is a subgroup $U$" as an awkward way to ask about the existence of such a subgroup.


Then the subgroup of finitely supported permutations satisfies ($\*$). Indeed, first observe that for a normal subgroup, Property ($\*$) is equivalent to the failure of a splitting. That the subgroup of finitely supported permutations does not split is an old result of W. Scott (1964).


