# pro-discrete = locally compact and open normal subgroups have trivial intersection?

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 $H$.

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.

Take a residually finite group $G$ and a subgroup $H$ such that no finite number of conjugates of $H$ intersect trivially, but all conjugates have trivial intersection. Now declare that subgroup and all finite index subgroups of $G$ open. $G$ becomes a locally compact non-discrete group and $H$ is open and does not contain normal non-trivial subgroups.

• You have to be careful here - when you generate the topology this way, you have to make sure that the induced topology is not discrete, because then the trivial subgroup is an open subgroup. Oct 19, 2010 at 2:42
• I assume that this is the purpose of the condition that no finite number of conjugates of $H$ have trivial intersection. But I do not know if it is enough.
– user6976
Oct 19, 2010 at 2:54
• Doesn't $\text{BS}(1,7)$, the semidirect product of $\mathbb{Z}[\frac{1}{7}]$ with $\mathbb{Z}=\langle t\rangle$ by $t\frac{p}{7^k}t^{-1}=\frac{7p}{7^k}$, provide an example, taking $H=\mathbb{Z}<\mathbb{Z}[\frac{1}{7}]$? Oct 19, 2010 at 4:54
• @Tom Church: what topology are you putting on this group? Oct 19, 2010 at 6:59
• @Tom Church, unknown: a countable group is totally disconnected locally compact if and only if it is discrete, so I don't think there are any interesting countable examples for this problem. Oct 19, 2010 at 10:41