Complement of a proper subgroup generates the group
= Complement of a proper subgroup generates the group
{title2=$\langle G\setminus H\rangle=G\quad(H<G)$}
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.