# Topological groups in which all subgroups are closed

General question: does there exist a nondiscrete topological group $$G$$ such that all subgroups of $$G$$ are closed? Or, does there exist a nondiscrete topological vector space $$V$$ such that all vector subspaces of $$V$$ are closed?

I am interested specifically in topological abelian groups with linear topology, or in topological vector spaces with linear topology. A topological group $$A$$ is said to have linear topology if open subgroups form a base of neighborhoods of zero in $$A$$. A topological vector space $$V$$ is said to have linear topology if open vector subspaces form a base of neighborhoods of zero in $$V$$.

Hence my more specific question is: does there exist a nondiscrete topological abelian group $$A$$ with linear topology such that all subgroups of $$A$$ are closed? Or, does there exist a nondiscrete topological vector space $$V$$ with linear topology such that all vector subspaces of $$V$$ are closed?

Motivation: I am trying to work out the very basics of the theory of topological abelian groups/vector spaces with linear topology. In particular, I am trying to understand closed maps.

For comparison, let us start with open maps. Let $$B$$, $$C$$ be topological abelian groups with linear topologies, and let $$p\colon B\longrightarrow C$$ be an abelian group homomorphism. Then $$p$$ is an open map (as a map of topological spaces) if and only if the image of every open subgroup in $$B$$ is an open subgroup in $$C$$.

Moreover, let $$p\colon B\longrightarrow C$$ be a surjective group homomorphism between topological groups. Then the topology of $$C$$ is the quotient topology of the topology of $$B$$ (via $$p$$) if and only if $$p$$ is an open continuous map.

The latter assertion is not at all true for topological spaces in general. It is easy to come up with an example of a topological space $$Y$$ and a surjective map of sets $$p\colon Y\longrightarrow Z$$ such that, when $$Z$$ is endowed with the quotient topology of the topology of $$Y$$ (via $$p$$), the map $$p$$ is not open. But topological groups are better behaved.

Closed maps appear to be more complicated. I am interested specifically in injective closed maps. Let us start with an injective map between topological spaces $$i\colon X\longrightarrow Y$$. Then $$i$$ is a closed continuous map if and only if $$i(X)$$ is a closed subset in $$Y$$ and the the topology of $$X$$ is induced from the topology of $$Y$$ via the embedding $$i$$.

Now let $$A$$, $$B$$ be topological abelian groups with linear topologies and $$i\colon A\longrightarrow B$$ be a continuous group homomorphism. Assume that the image of any closed subgroup in $$A$$ is a closed subgroup in $$B$$. Does it follow that $$i$$ is a closed map (as a map of topological spaces), i.e., that the image of any closed subset in $$A$$ is a closed subset in $$B$$?

Suppose that we've managed to find a nondiscrete topological abelian group $$B$$ with linear topology such that all subgroups of $$B$$ are closed. Let $$A$$ denote the same abelian group as $$B$$, but endowed with the discrete topology; and let $$i$$ be the identity map. Then the image of any subgroup of $$A$$ is closed in $$B$$, but $$i$$ is not a closed map (and the topology on $$A$$ is not induced from $$B$$). So we obtain a counterexample to the previous question.

• Yes, in $\mathbf{Z}$ endowed with the profinite topology, every subgroup is closed, while it's not discrete (since the sequence $(n!)_{n\ge 0}$ tends to $0$). In this case open subgroups are indeed a basis of neighborhoods of $0$.
– YCor
Dec 2, 2020 at 17:53
• @YCor Thank you very much. I should have thought about this topological group! Is there also a topological vector space example? Dec 2, 2020 at 17:55
• I think a topological vector space over a nondiscrete field cannot have any proper open subspace.
– YCor
Dec 2, 2020 at 17:55
• Yes, when speaking about topological vector spaces with linear topology, one presumes the ground field to be discrete. Dec 2, 2020 at 17:56
• Then take any infinite-dimensional vector space over a discrete field $K$. Take the "pro-finite-codimensional topology": a basis of neighborhoods of $0$ consists of finite-codimensional vector subspaces. Then every subspace is closed, but it's not discrete since the only open subspaces are finite-codimensional.
– YCor
Dec 2, 2020 at 18:00

$$\mathbf{Z}$$ with the profinite topology has the property that every subgroup is closed. That's because every subgroup is an intersection of finite index subgroup. However it is not discrete (the profinite topology on an infinite group is never discrete).
If $$K$$ is any discrete field and $$V$$ an infinite-dimensional vector space over $$K$$, we can endow $$V$$ with the "pro-finite-dimensional" topology, for which a basis of neighborhoods of $$0$$ consists of finite-codimensional subspaces. Then every subspace is closed, but only finite-codimensional subspaces are open, so it's not discrete.