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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 $H$.

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2 Answers

up vote 2 down vote accepted

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.

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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.

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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. –  Ian Agol Oct 19 '10 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. –  Mark Sapir Oct 19 '10 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 $H=\mathbb{Z}<\mathbb{Z}[\frac{1}{7}]$? –  Tom Church Oct 19 '10 at 4:54
    
@Tom Church: what topology are you putting on this group? –  Kevin Buzzard Oct 19 '10 at 6:59
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@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. –  Colin Reid Oct 19 '10 at 10:41
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