EDIT: After talking to some experts on the subject, I have concluded that a) the answer is not obvious or well-known for locally compact groups in general, b) the answer should be 'no' and I have some idea how to construct examples, but would rather try to write them up properly somewhere. Perhaps this question should be closed? Thanks for the help anyway.


This is a fairly basic question, but I can't seem to find a clear answer.


Let $G$ be a locally compact group. Suppose that the open normal subgroups of $G$ have trivial intersection. Does it follow that every open subgroup of $G$ contains an open normal subgroup of $G$?


If so, can the locally compact condition here be weakened?


Edit: some steps towards an answer:


* The open subgroups of $G$ have trivial intersection, so $G$ is totally disconnected.
* Any compact group satisfying the conditions is profinite and in particular pro-discrete. (Profinite = compact totally disconnected.)
* A locally compact totally disconnected group has an open compact (indeed profinite) subgroup by van Dantzig's theorem; this compact subgroup is either finite (in which case $G$ is discrete) or uncountable. So any non-discrete example would need to be uncountable.
* To show every open subgroup of $G$ contains an open normal subgroup, I think it would suffice to show there is an open compact normal subgroup $K$ say. For then, given $H$ open, then $H$ contains a finite index subgroup of $K$, and so by intersecting $K$ with finitely many suitably chosen open normal subgroups we can obtain an open normal subgroup contained in $H$.



pro-discrete = locally compact and open normal subgroups have trivial intersection?
Let $K$ be an infinite profinite group and let $K\_n < K$ be a decreasing family of open subgroups with trivial intersection. Thus, $K$ acts continuously on the discrete space $X = \sqcup\_n K/K\_n$ and this in turn gives rise to a continuous action of $K$ on the free abelian group $\mathbb Z[X]$. Define $G$ to be the semidirect product $G = K \ltimes \mathbb Z[X]$.


$G$ is a locally compact group and if we denote $X\_n = \sqcup\_{m \geq n} K/K\_m \subset X$ then since $K\_n$ acts trivially on $X \setminus X\_n$ we have that $K\_n \ltimes \mathbb Z[X\_n]$ is a family of open normal subgroups with trivial intersection in $G$. Also, $K < G$ is an open subgroup but contains no nontrivial normal subgroups since the action of $K$ on $X$ is faithful.


