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.

  • $\begingroup$ i still have a little trouble working out the induction on $j$ to show that the exponent divide $s$. Would you be able to elaborate? $\endgroup$
    – T.Kent
    Mar 17 '16 at 14:47
  • $\begingroup$ This follows from the fact that the map $\kappa:G/\gamma_i(G) \times \gamma_{i-1}(G)/\gamma_i(G) \to \gamma_i(G)/\gamma_{i+1}(G)$ defined by $\kappa(g\gamma_i(G),h\gamma_i(G)) = [g,h]\gamma_{i+1}(G)$ is bilinear. (I think that is proved somehwere in Robinson's book.) $\endgroup$
    – Derek Holt
    Mar 17 '16 at 15:21

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