Finitely generated nilpotent group where every element is of finite order is finite

Question

Let $G = \langle g_1,g_2,\ldots,g_m\rangle$ be a nilpotent group, where each $g_i$ has finite order. Prove $G$ is finite.

I'd like to show this by showing that the lower central series has successive quotients that are finite, i.e.,

$$|\gamma_n(G) / \gamma_{n+1}(G)| < \infty$$

I'm going to show this by induction. I'm on the base case

$$\gamma_1(G)/\gamma_2(G) = G/[G,G]$$

Any reason why this should be finite? Any idea how I can argue that $\gamma_{n+1}(G) / \gamma_{n+2}(G)$ should be finite in the inductive step, assuming $\gamma_n(G) / \gamma_{n+1}(G)$ is finite?

Answer 1

This is a nice problem because you don't use induction directly but prove it in several steps (only one of which requires induction):

Step 1: If $N\trianglelefteq G$ such that $[G,N]\le Z(G)$ and both $N$ and $G$ are generated by a finite number of elements of finite order, then $[G,N]$ is also generated by a finite number of elements of finite order.

(This can be proved by commutator arithmetic.)

Step 2: For $G$ as above, all elements of the lower central series are generated by finitely many elements of finite order.

(That's the part that requires induction, but is otherwise a trivial consequence of step 1.)

Step 3: An abelian group generated by a finite number of elements of finite order is finite.

(This is trivial as well.)

Putting the three steps together solves your problem.

Answer 2

Let $G$ be a nilpotent group of class $c$, finitely generated by torsion elements. Then $G/\gamma_2(G)$ is an abelian group, also finitely generated by torsion elements, hence it must be finite.

Consider $\overline{G}=G/\gamma_3(G)$. This is a nilpotent group (of class $\leq 2$, and again finitely generated by torsion elements) and hence $\gamma_2(\overline{G}) \subseteq \zeta(\overline{G})$. By the argument above, $\overline{G}/\gamma_2(\overline{G}) \cong G/\gamma_2(G)$ is finite. Hence $\overline{G}/\zeta(\overline{G})$ is finite.

At this point we need a lemma, mostly attributed to I. Schur; its proof relies on a simple application of the transfer map: if $X$ is a group and $X/\zeta(X)$ is finite, then $\gamma_2(X)$ is finite (see for example D.J.S. Robinson, A Course in the Theory of Groups, 10.1.4. [here][1]).

This yields $\gamma_2(\overline{G})$ is finite. But $\gamma_2(\overline{G}) \cong \gamma_2(G)/\gamma_3(G)$. It follows that $G/\gamma_3(G)$ is finite.

Continuing this argument to $\gamma_{c+1}(G)=1$, this shows that eventually $G$ is finite.

Note (added May 25th 2023) Another way to prove the statement is to apply Dietzmann’s Lemma: a finite subset of a group consisting of torsion elements and closed under conjugation generates a finite subgroup. This result can be found in the quoted book of Robinson.

[1]: https://eclass.uoa.gr/modules/document/file.php/MATH676/D.J.S.%20Robinson%20A%20Course%20in%20the%20Theory%20of%20Groups.pdf