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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.

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  • $\begingroup$ What does t.d.l.c. stand for? I'm guessing l.c. is "locally compact". $\endgroup$
    – Todd Trimble
    Oct 31, 2014 at 11:44

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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 $H$.

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