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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 $H$.

Edit 2: The preprint 'Abstract Commensurators of Profinite Groups' by Barnea, Ershov and Weigel contains the following assertion (paraphrased):

The intersection of all open normal subgroups of a totally disconnected, locally compact group $L$ is trivial if and only if $L$ is pro-discrete, that is, $L$ is the inverse limit of discrete groups.

So it appears the authors think the answer to my first question is 'yes', but no explanation is given. This leads me to think it is a well-known/'obvious' result, but it's not one I've ever seen proved explicitly.

4 added 364 characters in body

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 $H$.

Edit 2: The preprint 'Abstract Commensurators of Profinite Groups' by Barnea, Ershov and Weigel contains the following assertion (paraphrased):

The intersection of all open normal subgroups of a totally disconnected, locally compact group $L$ is trivial if and only if $L$ is pro-discrete, that is, $L$ is the inverse limit of discrete groups.

So it appears the authors think the answer to my first question is 'yes', but no explanation is given. This leads me to think it is a well-known/'obvious' result, but it's not one I've ever seen proved explicitly.

3 added 568 characters in body

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 $H$.

Edit 2: The preprint 'Abstract Commensurators of Profinite Groups' by Barnea, Ershov and Weigel contains the following assertion (paraphrased):

The intersection of all open normal subgroups of a totally disconnected, locally compact group $L$ is trivial if and only if $L$ is pro-discrete, that is, $L$ is the inverse limit of discrete groups.

So it appears the authors think the answer to my first question is 'yes', but no explanation is given. This leads me to think it is a well-known/'obvious' result, but it's not one I've ever seen proved explicitly.