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:
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The open subgroups of $G$ have trivial intersection, so $G$ is totally disconnected.
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Any compact group satisfying the conditions is profinite and in particular pro-discrete. (Profinite = compact totally disconnected.)
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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.
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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 $H$.
Best Answer
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.