Show that a periodic (every element has finite order), finitely generated, nilpotent group has finite order.

  • 1
    $\begingroup$ What have you tried? (Off the cuff - you're going to be inducting on the length of your central series, and wanting to show something like that every element has the form $x_1^{i_1}\ldots x_n^{i_n}$, where $x_j$ are your generators. So...start there.) $\endgroup$
    – user1729
    Aug 15, 2013 at 19:21
  • 1
    $\begingroup$ You could use the fact that finitely generated nilpotent groups are supersolvable. Or if that's too much, use the fact that in a finitely generated group factors of the lower central series are finitely generated. $\endgroup$ Aug 15, 2013 at 20:14
  • 2
    $\begingroup$ The key here is that when the abelianization of a nilpotent group is finite, the group itself is finite. This follows from a surjection from the iterated tensor powers of the abelianization to the successive quotients of the lower central series. $\endgroup$
    – user641
    Aug 15, 2013 at 22:36

1 Answer 1


let $G$ be a periodic, finitely generated and nilpotent group.

by recurrence on the nilpotency class $c$ of $G$

if $c=1$ then $G$ is abelian periodic finitely generated, so it existe $x_{1},x_{2},...,x_{r} \in G$ in which $G$ isomorphic to $\langle x_{1} \rangle \times \langle x_{2}\rangle \times ...\times \langle x_{r} \rangle$, where the order of $x_{i}$=$p_{i}^{\alpha_{i}}$, so $|G|=|\langle x_{1} \rangle|...|\langle x_{r} \rangle|=p_{1}^{\alpha_{1}}..p_{r}^{\alpha_{r}}<\infty$.

Assume $c>1$, we know that $\quad\forall x \in G$, $\langle x,G^{'} \rangle=H$ is nilpotent of class $c-1$; it's clear that $H$ is periodic. As $H$ is nilpotent finitely generated, it satisfies max, so $H$ is finitely generated, whence $H$ isi finite by the recurrence hypotesis, therefor $G^{'}$ is finite.

On the other hand $G/G^{'}$ is abelian periodicfinitely generated, so $G/G^{'}$ is finite. finally $G$ is finite.


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.