  Let $G$ be an abelian group. We prove that a group $G$ admits a Hausdorff
group topology $\tau$ such that the von Neumann radical $\mathbf{n}(G, \tau)$
of $(G, \tau)$ is non-trivial and finite iff $G$ has a non-trivial finite
subgroup. If $G$ is a topological group, then $\mathbf{n} (\mathbf{n} (G))
\not= \mathbf{n} (G)$ if and only if $\mathbf{n} (G)$ is not dually embedded.
In particular, $\mathbf{n} (\mathbf{n} (\mathbb{Z},\tau)) = \mathbf{n}
(\mathbb{Z},\tau)$ for any Hausdorff group topology $\tau$ on $\mathbb{Z}$.
