embedding of abelian topological group. Let $A$ be an abelian topological group and let $n$ be a natural number, such that the homomorphism of topological group 
$i: A\rightarrow A$, $a\mapsto n.a$ is injective. Is it a trivial fact that $i$ is an embedding of the underlying topological spaces ?

Let $A=\mathbb{Z}$, with the topology given by embedding it in the product $\prod_n \mathbb{Z}/(2\cdot 3^n\mathbb{Z})$ by the quotient map on each coordinate.  (So, we have the $3$-adic topology restricted to $2\mathbb{Z}$, but $2\mathbb{Z}$ is open in $\mathbb{Z}$.)  Then multiplication by $2$ is injective on $A$, but it is not an embedding, because $2\mathbb{Z}$ is open in $A$ but its image $4\mathbb{Z}$ is not.
