# Does a cocompact subgroup of a topological group contain a cocompact normal subgroup?

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.

• Uniform lattices in $SL_3(\mathbb{R})$? – user6976 Mar 16 at 15:40
• @NajibIdrissi Put $G=H=\mathbb{R}$ and $K=\{0\}$ then $G/K$ is not a continuous image of $G/H$. Right? – Ali Taghavi Mar 16 at 15:47
• @MarkSapir may I ask you to ellaborate your comment or write a complete answer? Thank you. – Ali Taghavi Mar 16 at 15:51
• @AliTaghavi a non-compact simple connected Lie group $G$ has no proper normal cocompact subgroup. But it has cocompact lattices, and also has connected cocompact proper subgroups (e.g., upper triangular matrices in $\mathrm{SL}_{n\ge 2}$ of $\mathbf{R}$ or $\mathbf{C}$). – YCor Mar 16 at 15:55
• The situation now that the title and the question are answered in comments, while the "remark" asks a more complicated and distinct question. I'd suggest to either (a) ask the "remark" in separate question (in which case I could answer this one with a cw answer to make the question settled) or (b) change the current "remark" into the main question (changing the title in particular) giving the original question and its easy examples only as context. I think (a) is a better solution (since these are quite drastic changes). – YCor Mar 19 at 20:22

For instance, non-compact simple connected Lie group $$G$$ has no proper normal cocompact subgroup. But it has cocompact lattices, and also has connected cocompact proper subgroups (e.g., upper triangular matrices in $$\mathrm{SL}_n(\mathbf{R})$$ or $$\mathrm{SL}_n(\mathbf{C})$$, for $$n\ge 2$$).