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

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?

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.

• Thanks so much @jpvee for your answer. Could you please point me in the right direction in showing Step 1? I'm sort of going in circles with commutator arithmetic. – Möbius Dickus Nov 17 '17 at 7:40
• Under the conditions of step 1, you have $[gg',nn']=[g,n][g,n'][g',n][g',n']$ for all $g,g'\in G$, $n,n'\in N$ (cf. e.g. identity #3 in en.wikipedia.org/wiki/Commutator#Identities_.28group_theory.29). From this, it follows that $[G,N]$ is generated by the set of $[g_i,n_j]$ where $G=\langle g_1,\ldots,g_m\rangle$ and $N=\langle n_1,\ldots,n_k\rangle$. It also implies that $[g,n]^{o(n)}=1$ for $g\in G$, $n\in N$, so those generators of $[G,N]$ all have finite order. – jpvee Nov 17 '17 at 8:03
• OK, this makes sense. And then in putting these three pieces together, this shows $\gamma_n(G) / \gamma_{n+1}(G)$, correct? – Möbius Dickus Nov 17 '17 at 8:05
• Yes, because in step 2 you show that $\gamma_n(G)$ is finitely generated by finite order elements, and of course the same is true for $\gamma_n(G)/\gamma_{n+1}(G)$ which is abelian, so step 3 tells you that that factor group is finite. – jpvee Nov 17 '17 at 8:10

Hint: first induction step: $G/[G,G]$ is an abelian group that inherits every element having finite order and being finitely generated. Hence this group must be finite. Now look at $\gamma_2(G)/\gamma_3(G)$. This group is also abelian and has finite index (because of the previous step) in $G/\gamma_3(G)$. Since also $G/\gamma_3(G)$ is finitely generated, $\gamma_2(G)/\gamma_3(G)$ must be finitely generated. Again $\gamma_2(G)/\gamma_3(G)$ is a finitely generated abelian group with all elements of finite order, whence finite. Can you finish?

• How does $G/\gamma_3(G)$ being finitely generated imply that $\gamma_2(G)/\gamma_3(G)$ is also finitely generated? – RaiRsXOT6L Nov 6 '18 at 8:09
• I am using the fact that a subgroup of finite index of a finitely generated group is again finitely generated (Schreier's Lemma, see for example groupprops.subwiki.org/wiki/Schreier%27s_lemma). – Nicky Hekster Nov 6 '18 at 10:51