Let $G$ be an abelian topological group. Let $n$ be an integer and consider $G\to G$ given by sending $x\mapsto nx$.
Is this a closed map? I do not have any particular reason to think it should be, but it would make a certain step in something I am reading much easier, so I am curious about whether it is true or not.
It might be useful to assume that $G$ is Hausdorff and locally compact, but those might not be necessary.
Thanks for any help, direction, or insight.
No, not necessarily, even if $G$ is locally compact Hausdorff. For instance, let $A$ be any infinite discrete abelian group where every element has order dividing $n$ (e.g., an infinite direct sum of copies of $\mathbb{Z}/n$) and let $G=A\times \mathbb{R}$. Let $(a_k)$ be a sequence of distinct elements of $A$, and consider the set $C=\{(a_k,1/k):k\in\mathbb{N}\}\subset G$. Then $C$ is closed, since the $a_n$ are all distinct. But the image of $C$ under multiplication by $n$ is $\{(0,n/k:k\in\mathbb{N})\}$, which is not closed, because $(0,0)$ is in its closure.