I am trying to solve an exercise in D. Robinson's book A Course in the Theory of Groups, which asks me to show that if $G$ is polycyclic and residually finite p-group for infinitely many prime p, then $G$ is nilpotent and finitely generated torsion-free.
How do I show the nilpotent part?
The hint given is that first assume $G$ is not nilpotent then show that there exists some $i$ such that $\gamma _i (G)/\gamma _{i+1} (G) $ is finite.
I asked my lecturer and he told me to start with finding an abelian normal subgroup $A$ such that $G/A$ is nilpotent and torsion-free. Does such a subgroup always exist and why?
Here is an outline of a proof. The fact that $G$ is polycyclic implies that there must in any case exist an $i$ with $\gamma_i(G)/\gamma_{i+1}(G)$ finite, and if $G$ is not nilpotent we have $\gamma_{i+1}(G) \ne 1$.
Now if $s$ is the exponent of $\gamma_i(G)/\gamma_{i+1}(G)$, then you can show by induction on $j$ that the exponent of $\gamma_j(G)/\gamma_{j+1}(G)$ must divide $s$ for all $j \ge i$.
Choose a prime $p$ such that $G$ is residually a finite $p$-group and $p$ does not divide $s$. Choose some $g$ with $1 \ne g \in \gamma_i(G)$, and let $N$ be a normal subgroup of $G$ with $g \not\in N$ such that $G/N$ is a finite $p$-group.
Then $\overline{G}:=G/(\gamma_i(G) \cap N)$ is nilpotent, and $\gamma_i(G)/(\gamma_i(G) \cap N)$ is a nontrivial $p$-group, so $p$ divides the exponent of $\gamma_j(\overline{G})/\gamma_{j+1}(\overline{G})$ for some $j \ge i$. Hence the same is true for $G$, contradiction.