To find a counterexample disproving a generalization of a theorem in the theory of scale functions on locally compact totally disconnected groups, initiated by George Willis, I am looking for a group with certain properties. Alternatively I am looking for arguments why such a group cannot exist. The precise question is:
Is there an example of a totally disconnected locally compact topological group $P$ such that
(1) $P$ is algebraically a (semi)direct product of subgroups $G$ and $H$
(2) $G$ and $H$ are not closed in $P$, but locally compact in the topology inherited from $P$,
(3) $K\cap G$ and $K\cap H$ are not compact for any compact open subgroup $K\subset P$,
(4) $P$ is not compact itself?
Usual construction methods of totally disconnected locally compact groups such as direct limits of discrete groups seem to fail to produce such an example. In particular (3) seems hard to achieve. So maybe an example which occurs naturally in some other context has a better chance to work than building one with bottom up methods.
I hope I didn't overlook any obvious argument contradicting the existence of the desired group.
Best Answer
This question just got bumped to the front page by the Mathoverflow bot. But the question was already answered in the comments by BCnrd. Following advice from this Meta article, I'm going to copy BCnrd's answer here so the OP can accept this answer and this question won't get bumped up again. My response is CW so I can't gain any reputation.
The answer to the question is "no." Here's why:
First, note that every totally disconnected topological group is Hausdorff. Next any locally compact subspace of a Hausdorff space is locally closed, hence open in its closure (see e.g. Munkres Topology for a proof of this). If $P$ was a totally disconnected topological group and $G$ is a locally compact subgroup then $G$ is open in its closure $\overline{G} \subset P$, which in turn is a closed subgroup of P. But open subgroups of topological groups are closed, so $G$ is closed in $\overline{G}$ and hence $G=\overline{G}$. This proves (2) is impossible.