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

Solution 1:

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).

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.

Solution 2:

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.