Show that a periodic (every element has finite order), finitely generated, nilpotent group has finite order.
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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 '13 at 19:21
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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. $\endgroup$ – Mikko Korhonen Aug 15 '13 at 20:14
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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 '13 at 22:36
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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.
