Does there exist a group $G$ and a subgroup $H$ of $G$ which is not equal to $G$, such that the set-theoretic complement $G – H$ is closed under the group operation? I have tried to come up with some examples, but I can't find any. I am starting to suspect there is no such group and subgroup.
Can a group have a subgroup whose complement is closed under the group operation
group-theory
Best Answer
Let $G$ be a group and $H\subsetneq G$ a proper subgroup such that $G-H$ is closed w.r.t. the group operation. Consider $g\in G-H$. Then $g\notin H$ and hence also $g^{-1}\notin H$, and so $g^{-1}\in G-H$. It follows that $gg^{-1}=e\in G-H$, but of course $e\in H$, a contradiction.