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Given: Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements.

Show that $G$ is finite.

Also would love to know if it's possible to show that an infinite finitely generated nilpotent group has an infinite center.

Alex Ravsky
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Ilan Aizelman WS
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    It would suffice to show $G'$ is also finitely-generated by torsion elements, since then $G/G'$ is finite (it is finitely generated and abelian) and $G'$ is finite by induction on nilpotence class. – anon Jun 13 '19 at 00:43
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    @runway44 Would love to know if you have a way to approach/solve that. – Ilan Aizelman WS Jun 15 '19 at 12:12
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    I would suggest using the lower central series $\gamma_i(G)$. If $\gamma_c(G)$ is the last nontrivial term, then it is central. By induction $G/\gamma_c(G)$ is finite, and you can then show that $\gamma_c(G) = [G,\gamma_{c-1}(G)]$ is finite. Similarly for the second problem, if $\gamma_c(G)$ is finite then by induction $G/\gamma_{c-1}(G)$ has infinite centre,$Z/\gamma_c(G)$,, and you can show that $|Z:Z \cap Z(G)|$ is fihite. – Derek Holt Jun 17 '19 at 12:21
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    You can also prove it using the results proved in detail in this post – Derek Holt Jun 17 '19 at 12:32

1 Answers1

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The following argument is incomplete, although I did not realize it until after writing it out. In fact, I now see that it is a fleshing out of the comment by runway44 above. After further thought, the missing step (torsion generation of $[G,G]$) is quite substantial and indicates this is not the right approach. Instead, the argument outlined by Derek Holt appears to be the correct path.

We prove the result by induction on the length of the lower central series. The base case consists of an abelian group, in which case the claim is immediate.

The inductive hypothesis applied to $[G,G]$ implies that it is a finite group. Let $g_1,\ldots,g_n$ be a generating set of torsion elements for $G$. Then every $g\in G$ has a representation of the form $$ g=h\prod_{i=1}^{n}g_i^{x_i},\qquad h\in [G,G], $$ where each $x_i$ ranges from $0$ to $\textrm{ord}(g_i)-1$. Indeed, we may iteratively all instances of $g_1$ to the end (keeping a commutator at the start of $g$), then $g_2$, and so on, always keeping track of the commutators incurred during the rearrangement in $h$.

From this representation, we obtain that $|G|\leq \bigl|[G,G]\bigr| \prod_{i=1}^n\textrm{ord}(g_i)$, in particular showing that $G$ is finite.

pre-kidney
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