$\overline{H}=\bigsqcup_{x\in \bar{H} \setminus H} Hx$?

Question

Is it true that if I have a topological $G$ and I consider a subgroup $H$ then $\overline{H}=\bigsqcup_{x\in \bar{H} \setminus H} Hx$? (here we have the disjoint union).

Answer 1

No, this is never true. The sets on the right hand side are not disjoint, nor is their union equal to the closure of $H$ (this is easy to see, since it does not contain $H$).

What is true is that $\overline H=\bigcup_{x\in \overline H} Hx$, since $\overline H$ is a group containing $H$, and also (for the same reason), $\overline H=\bigsqcup_{Hx\in H\backslash \overline H} Hx=\bigsqcup_{xH\in \overline H/H} xH$.

Answer 2

I guess $G$ is a topological group. The required equality never holds, because its right-hand side contains no elements of $H$. The set $\overline{H}$ is a semigroup containing $H$ so it is a union of some disjoin family of right cosets $Hx$. For any $x,y\in\overline{H}$ we have $Hx=Hy$ iff $xy^{-1}\in H$, otherwise $Hx$ and $Hy$ are disjoint.