Complement of a proper subgroup generates the group ($⟨G\setminus H⟩=G(H<G)$)

For a proper subgroup $H$ of any group $G$, its complement is a generating set of a group. Choose $x\notin H$. For every $h\in H$, also $xh\notin H$, and $h=x^{-1}(xh)$ lies in the subgroup generated by the complement. That subgroup consequently contains both $H$ and its complement. No finiteness assumption is needed.