Finite normal subgroups of polycyclic-by-finite groups Let $G$ be a polycyclic-by-finite and $H<G$ a finite index subgroup. Let $F(G)$ be the unique maximal finite normal subgroup of $G$. Then $F(G)\cap H$ is finite and normal in $H$. 
Question 1. Suppose $F(H)$ is non-trivial. Is $F(G)\cap H$ non-trivial?
Question 2. Is $F(G)\cap H=F(H)$? 
 A: Yes (to both questions). It's essentially the contents of Proposition 2.7 in my commability Indiana UMJ paper (arxiv link), specified to (a particular instance of) discrete groups.
Here's a proof (easier in this case since we have discrete groups). In a group $G$ let $W(G)$ be its polyfinite radical, namely the subgroup generated by finite normal subgroups. 
We need to know the following ($\ast$) : $W(G)$ is the set of torsion elements in $FC(G)$, the union of finite conjugacy classes of $G$.

Let $G$ be a group and $H$ a finite index subgroup. Then $W(G)\cap H=W(H)$.

Proof: $\subset$ is clear. Conversely, let $F$ be a finite normal subgroup of $H$. Then clearly $F$ is contained in $FC(G)$, and also consists of torsion elements. Given the previous fact, it follows that $F$ is contained in $W(G)$. $\Box$
($\ast$) is an easy consequence of the fact that in an FC-group (group in which all conjugacy classes are finite), the set of torsion elements is closed under product. The latter fact is a bit harder (but well-documented).
