I am trying to find an example of the following situation.


$G$ is a t.d.l.c. (totally disconnected locally compact) $\sigma$-compact topological group in which every compact open subgroup is topologically finitely generated. $H$ is a maximal compact open subgroup and $K$ is a subgroup abstractly isomorphic to $H$ such that the intersection $H \cap K$ has countably infinite index in both $H$ and $K$.


I believe that I can show that this situation is not possible if $G$ is the group of rational points of an absolutely quasi-simple algebraic group over a non-archimedean local field $k$. 


Would welcome any useful suggestions for determining if this situation is possible.



locally topologically finitely generated t.d.l.c. group
Consider a basis of the $\mathbf{Q}$-vector space $\mathbf{Q}\_p$ given as elements of $\mathbf{Z}\_p\smallsetminus p\mathbf{Z}\_p$, written as elements $e\_n$ for integers $n\in\mathbf{Z}$ and $f\_i$ for $i$ in some uncountable index set. Define a $\mathbf{Q}$-linear automorphism $q$ of $\mathbf{Q}\_p$ by $q(e\_n)=p^ne\_n$ and $q(f\_i)=f\_i$. It fixes (pointwise) a subgroup of countable index of $\mathbf{Q}\_p$, namely the $\mathbf{Q}$-subspace $S$ spanned by the $f\_i$. Define $H=q(\mathbf{Z}\_p)$. Then by construction, $H\cap\mathbf{Z}\_p$ has infinite countable index in both $\mathbf{Z}\_p$ and $H$. 


This would be enough if you weren't requesting the compact open subgroup be maximal. This can be done as follows: define $G$ as the amalgamation of $\mathbf{Z}\_p$ with $\mathbf{Q}\_p$ along $p\mathbf{Z}\_p$. Then the left-hand $\mathbf{Z}\_p$ (call it $A$) is a maximal compact (open) subgroup of $G$, is isomorphic to $H$ as an abstract group, and $H\cap A$ has countable infinite index in both $A$ and $H$.


