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

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.

• If something is not in $H$, its inverse can't be in $H$ either. What happens if you multiply an element with its inverse?
– D_S
Commented Jul 20 at 22:17

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.

• Anyone care to explain the downvotes? Commented Jul 22 at 22:35

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).

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.

$$\dagger:$$ Hint:

It fails the two-step subgroup test.

• What's with the downvotes? Commented Jul 22 at 21:13