Can a group have a subgroup whose complement is closed under the group operation?

Question

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.

Answer 1

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.

Answer 2

Maybe good to know (and a bit more than you are asking for, except if for example $G$ is finite): if $H$ is a proper subgroup of a group $G$, then $\langle G-H\rangle=G$, that is, the complement of $H$ always generates the whole group. This follows from the fact that a group is never the union of two proper subgroups (one can find many proofs of this fact here on Math StackExhange, see for example here).

Answer 3

Let $k\in G\setminus H$. Suppose $k^{-1}\in H$. Then $H\not\le G$. (Why?${}^\dagger$) Thus $k^{-1}\in G\setminus H$. But then $e=kk^{-1}\in G\setminus H$, a contradiction.

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$\dagger:$ Hint:

It fails the two-step subgroup test.