I'm having a little trouble proving that any two subgroups of a totally disconnected, locally compact topological group are commensurable, so am looking for a push in the right direction!
I have so far a proof that shows that some subgroup $H$ of the group $G$ is commensurated by the group $G$, but this relies on the fact that I can conjugate $H$ by some $g \in G$ and then use the intersection $H \cap gHg^{-1}$ -- at which point it's trivial to just examine the cosets of this group to see that the index is finite (thanks to a finite subcover) and thus commensurability is proven.
At this point, I'm considering using the properties of the index function; specifically that $[H : H \cap K] \leq [G : K]$; since $[G : K]$ is finite (by the same argument as above), $[H : H \cap K]$ must be too. Is this the right direction to be headed, or is there something fundamental that I'm missing that makes this glaringly wrong?
It seems that any compact open subgroups $G_1$ and $G_2$ of any (Hausdorff) topological group $G$ are commensurable, because the intersection $G_1\cap G_2$ is a compact open subgroup of the group $G$, therefore $G_1\cap G_2$ is a compact open subgroup of both $G_1$ and $G_2$. Thus both indexes $[G_1:G_1\cap G_2]$ and $[G_2:G_1\cap G_2]$ are finite.