Some more basics on group operation Let $G$ be a group, $H \le G$ and $f \colon G \times G \rightarrow G$ the group operation. We know that $\complement_GH$ (the complement of $H$ in $G$) contains the inverse of any of its elements, so, whatever $G$ and $H$ are, $\lbrace e \rbrace \subseteq f(\complement_GH \times \complement_GH)$.
On the other hand, if we take $G=(\mathbb{Z},+)$ and  $H=2 \mathbb{Z}$, we get that $f(\complement_GH \times \complement_GH)=H$, because by summing pairwise all the odd integers we get all the even integers (and them, only).
This makes me conclude that, in general, at least the following holds: $\lbrace e \rbrace \subseteq f(\complement_GH \times \complement_GH) \cap H \subseteq H$.
I ask the following:


*

*what's the characterization of $H$ and/or $G$ to get $f(\complement_GH \times \complement_GH)=\lbrace e \rbrace$, if any?

*what's the characterization of $H$ and/or $G$ to get $f(\complement_GH \times \complement_GH)=H$?


(by "characterization of $H$ and/or $G$" I mean something like, e.g., "$H$ normal in $G$", or the like).
 A: Note that you should begin with $H<G$ instead of $H\le G$, as $H=G$ certainly makes  $f(\complement_GH \times \complement_GH)=\emptyset$.
So assume $H<G$. If we pick $a\in \complement_GH$, then for any $h\in H$, we have $ha^{-1}\notin H$ and hence $h=ha^{-1}\cdot a\in f(\complement_GH \times \complement_GH)$. This make
$$H\subseteq f(\complement_GH \times \complement_GH)\qquad \text{if }H<G. $$
Consequently, the situation in your first question occurs iff $H=\{e\}$ and the situation in your second question occurs.
For the second part, in order to obtain only $H$, the case $ha^{-1}\cdot a$ we used above must be "essentially" the only one. Indeed,
$$H= f(\complement_GH \times \complement_GH)\iff[G:H]=2.$$
Proof:

*

*If $H$ is of index 2, pick $a\in \complement_GH$ such that Then if $x,y\in\complement_GH$, we have $h_1:=xa\in H$, $h_2:=a^{-1}y\in H$ and so $xy=xaa^{-1}y=h_1h_2\in H$.


*On the other hand, if $f(\complement_GH \times \complement_GH)=H$, then we already know $H\ne G$. If $a,b\in\complement_GH$, it follows that $a^2\in H$ and $ab\in H$, hence $a^2H=abH$ and thereby $aH=bH$, i.e., there are only two cosets
Back to the first question: We have $f(\complement_GH \times \complement_GH)=\{e\}$ iff $G$ is of order $2$ and $H$ the trivial subgroup.

Generalization: If there exist  $a,b\in\complement_GH$ with $ab\notin H$, then $f(\complement_GH \times \complement_GH)=G$. Indeed, If $g\in G$, then at least one of $ag$, $b^{-1}g$ is $\notin H$ because otherwise also $ab=ag(b^{-1}g)^{-1}\in H$. We conclude  $g=a^{-1}\cdot ag=b\cdot b^{-1}g\in f(\complement_GH \times \complement_GH)$.
Such $a,b$ exist whenever $[G:H]>2$. Therefore,
$$f(\complement_GH \times \complement_GH)=\begin{cases}\emptyset&\text{if }H=G\\H&\text{if }[G:H]=2\\G&\text{otherwise}\end{cases}$$
