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 $\text{id_\omega}\in \text{Sym}(\omega)$?