Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exists a normal subgroup $K$ of $G$, $K\subset H$, with $|G/K|<\infty$.

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarily normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ such that $G/K$ is a compact topological group?

Remark: One can consider a similar problem in the context of rings, algebras, Lie algebras and Banach or $C^*$ algebras: Classification of all $A$ with such a structure with the property that every finite codimensional subalgebra $B$ contains an ideal $I$ in $A$ with $A/I$ being finite-dimensional (or in the ring case finite).

Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exists a normal subgroup $K$ of $G$, $K\subset H$, with $|G/K|<\infty$.

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarily normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ such that $G/K$ is a compact topological group?

Remark: One can consider a similar problem in the context of rings and algebras: see this follow-up question.

Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exist a normal subgroup $K$ of $G$ $K\subset H$ with $|G/K|<\infty$

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarilly normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ and $G/K$ is a compact topological group.

Remark: One can consider a similar problem in the context of Rings, algebra, Lie algebra and Banach or $C^*$ algebras: Classification of all $A$ with such an structure with the property that every finite codimensional subalgebra $B$ contains an ideal $I$ in $A$ with $A/I$ is finite dimensional.(in the ring case finite)

Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exists a normal subgroup $K$ of $G$, $K\subset H$, with $|G/K|<\infty$.

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarily normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ such that $G/K$ is a compact topological group?

Remark: One can consider a similar problem in the context of rings, algebras, Lie algebras and Banach or $C^*$ algebras: Classification of all $A$ with such a structure with the property that every finite codimensional subalgebra $B$ contains an ideal $I$ in $A$ with $A/I$ being finite-dimensional (or in the ring case finite).

Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exist a normal subgroup $K$ of $G$ $K\subset H$ with $|G/K|<\infty$

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarilly normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ and $G/K$ is a compact topological group.

Motivation: It is obvious that for a finite index subgroup $H$ of a group $G$, there exist a normal subgroup $K$ of $G$ $K\subset H$ with $|G/K|<\infty$

Our question: Let $G$ be a topological group and $H$ be a closed but not necessarilly normal subgroup of $G$ such that the quotient topological space $G/H$ is a compact space. Does there exist a closed subgroup $K\subset H$ which is normal in $G$ and $G/K$ is a compact topological group.

Remark: One can consider a similar problem in the context of Rings, algebra, Lie algebra and Banach or $C^*$ algebras: Classification of all $A$ with such an structure with the property that every finite codimensional subalgebra $B$ contains an ideal $I$ in $A$ with $A/I$ is finite dimensional.(in the ring case finite)