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.

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