Condition for existence of a sort of complement subgroup Reading the wikipedia page for "Complement (group theory)" I am not convinced that the "complement of a subgroup" is what I am looking for, hence the title.
Let $G$ be a group, and let $H$ be a proper nontrivial subgroup of $G$. Let $K$ be the set of elements in $G$ not in $H$, also including the identity. In general it is not true that $K$ is a subgroup of $G$. A simple example using $S_3$ is the observation that $(132)(23)(123) = (12)$, and that $\{e,(12)\}$ is a subgroup of $S_3$. We see from this example that even if $H$ is abelian, it is not true in general.
I am looking for a condition, if there is one, under which it is true that $K$ actually is a subgroup.
 A: “Let $K$ be the set of elements in $G$ not in $H$, also including the identity.” — that's a contradiction, as the identity is always in $H$. I'll assume you mean $K = (G\setminus H)\cup \{e\}$ where $e$ is the identity.
$K$ cannot be a subgroup.
Proof: Take $h\in H$ and $k\in K$, both not the identity. Then $g:=hk\in K$ (because if it were in $H$, then $k=gh^{-1}$ would be in $H$, too, and everything not in $H$ is in $K$). Now if $K$ were a group, it would follow that $gk^{-1}\in K$. But that's a contradiction, as $gk^{-1}=h\in H$, and the only element both in $H$ and $K$ is, by construction, the identity. But we assumed that $h$ is not the identity. $\square$
A: Suppose $h\in H$ and $k\in K$. Where is $hk$? If $H$ and $K$ are both subgroups, with $H\cup K=G$ and $H\cap K=\{e\}$, then we have a problem. Do you see why?
A: In general, if $H$ is a proper subgroup of a group $G$, then $\langle G\backslash H \rangle = G$, that is, the set-theoretic complement generates the whole group. This follows from the fact that a group cannot be the union of two proper subgroups (I leave the proof to you). 
