O. Baues and F. Gruewald in their paper "AUTOMORPHISM GROUPS OF POLYCYCLIC-BY-FINITE GROUPS AND ARITHMETIC GROUPS" have stated that a polycyclic-by-finite group has a unique maximal finite normal subgroup (see page 216, Section 1.2). I was wondering if somone could give a proof for that? Does this result still hold for finitely presented amenable groups, in particular elementary amenable ones?


**My try**: Let $G$ be a virtually polycyclic group. Since a virtually polycyclic group is finitely presented, it is countbale. So assume that $N\_1 ,N\_2 ,\ldots $ are finite normal subgroups of $G$. For each inetegr $n\geq 1$, $N\_1 N\_2 \cdots N\_n$ is a finite subgroup. On other hand, a given virtually polycyclic group has only finitely many finite subgroups up to isomorphism. Hence the ranks of finite subgroups are bounded. Thus there are only at most $n\_G$, where $n\_G$ is an integer depending only on $G$, possible different finite subgroups $N\_i$ in $G$.



Does a finitely presented amenable group contain a unique maximal finite normal subgroup?
In a group $G$, the polyfinite radical $W(G)$ is the union of all finite normal subgroups. So, $G$ has a maximal finite normal subgroup iff $W(G)$ is finite.


You're asking whether $W(G)$ is finite for every finitely presented elementary amenable group. The answer is negative. There are indeed finitely presented solvable groups in which the center has infinite torsion (the torsion part of the center is clearly contained in the polyfinite radical, which is actually the torsion part of the FC-center).


I'm not sure what are the simplest known examples, but here are ones: for a prime $p$, the group of matrices
$$\begin{pmatrix}1 & a\_{12} & a\_{13} & a\_{14}\\ 0 & t^{n\_1}(t+1)^{n\_2} & a\_{23} & a\_{24} \\ 0 & 0 & t^{n\_3}(t+1)^{n\_4} & a\_{34}\\ 0 & 0 & 0 & 1\end{pmatrix}: a\_{ij}\in\mathbf{F}\_p[t,(t+t^2)^{-1}],n\_k\in\mathbf{Z}.$$


