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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.