Why do the elements of finite order in a nilpotent group form a subgroup?

I would like to prove the following statement:

Let $G$ be a nilpotent group. Then the set of elements of $G$ of finite order is a subgroup of $G$.

I have no idea but the straightforward approach by induction: We choose a central subgroup $A$ of $G$ and can assume that the elements of finite order of $G/A$ form a group. So, for $x,y\in G$ of finite order, there exists $n\in\mathbf{N}-\{0\}$ such that $(xy)^n\in A$. But what now? Is this the right track?

I've actually found a proof or two on the web, but they use a lot of theory with which I'm not familiar, e.g. tensor products and Sylow theory for infinite groups. It would be great if someone could link to or sketch a proof that avoids or minimizes the use of these tools.

• Sorry what is the definition of the nilpotent group? – user17090 Nov 6 '11 at 11:26
• @AliBleybel: see Wikipedia: a nilpotent group is a group that has a central series of finite length. – Stefan Nov 6 '11 at 11:37
• @Ali: while it is nice that questions be sensibly self-contained, googling for "nilpotent group" would have answered your question in 2 seconds... – Mariano Suárez-Álvarez Nov 6 '11 at 12:06
• Have you seen the proof in Philip Hall's lecture notes on nilpotent groups? It is very basic. – user641 Nov 6 '11 at 18:30

Recall the definition of the lower central series of a group $$G$$:

The first term of the lower central series is $$G_1 = G$$. The second term is $$G_2=[G,G]$$.

Having defined the $$n$$th term of the lower central series, the $$n+1$$st term is $$G_{n+1}=[G_n,G]$$.

A group $$G$$ is nilpotent of class at most $$c$$ if and only if $$G_{c+1}$$ is trivial, if and only if $$G_c\subseteq Z(G)$$.

Lemma 1. Some commutator identities: given a group $$G$$ and $$x,y,z\in G$$, we have:

• $$[xy,z] = [x,z]^y[y,z]$$;
• $$[x,yz] = [x,z][x,y]^z$$.

Proof. Direct computation establishes the equality. $$\Box$$

Lemma 2. Let $$G$$ be a group. If $$[x,G]\subseteq Z(G)$$, then $$[xy,z] = [x,z][y,z]$$ for all $$y,z\in G$$.

Proof. This follows from the first identity above, since $$[x,z]$$ is central. $$\Box$$

Proposition 3. Let $$G$$ be a group, and assume that $$G$$ is generated by elements of finite order. Then for every $$c\gt 0$$, $$G_c/G_{c+1}$$ is a torsion group.

Proof. We proceed by induction on $$c$$. If $$c=1$$, then $$G_1/G_2=G^{\rm ab}$$, and an abelian group generated by elements of finite order is torsion.

Assume the result is true for $$c$$, and consider $$G_{c+1}/G_{c+2}$$. This is abelian, since $$[G_{c+1},G_{c+1}]\subseteq [G_{c+1},G]=G_{c+2}$$, so it suffices to show that the generators are of finite order. $$G_{c+1}/G_{c+2}$$ is generated by elements of the form $$[x,g]$$ with $$x\in G_c$$ and $$g\in G$$. By assumption, $$x$$ is torsion modulo $$G_{c+1}$$, so there exists $$n\gt 0$$ such that $$x^n\in G_{c+1}$$. Moreover, since $$G_{c+1}/G_{c+2}$$ is abelian, by Lemma 2 we have that $$[x^n,g] \equiv [x,g]^n\pmod{G_{c+2}}$$. But since $$G_{c+1}/G_{c+2}$$ is abelian, $$[h,g]\equiv 1$$ if $$h\in G_{c+1}$$. Since $$x^n\in G_{c+1}$$, then $$[x,g]^n \equiv [x^n,g] \equiv 1 \pmod{G_{c+2}}$$, so $$[x,g]$$ is torsion modulo $$G_{c+2}$$, as desired. $$\Box$$

Theorem. Let $$G$$ be a nilpotent group. Then the set of torsion elements of $$G$$ is a subgroup of $$G$$.

Proof. It suffices to show that the product of two elements of finite order is of finite order. Let $$x,y\in G$$ be of finite order; since all computations will occur in $$\langle x,y\rangle$$, we may assume without loss of generality that $$G=\langle x,y\rangle$$. We proceed by induction on the class $$c$$ of $$G$$. If $$c=1$$, then $$G$$ is abelian, and the result is immediate: $$xy$$ is of finite order since $$x$$ and $$y$$ are of finite order.

Assume the result holds for groups of class $$c$$, and that $$G$$ is of class $$c+1$$. Then $$xy$$ is of finite order modulo $$G_{c+1}$$, so there exists $$n$$ such that $$(xy)^n\in G_{c+1}$$. By Proposition 3, $$G_{c+1}/G_{c+2}$$ is torsion, so there exists $$m\gt 0$$ such that $$(xy)^{nm}\in G_{c+2}$$. But since $$G$$ is of class $$c+1$$, $$G_{c+2}=\{1\}$$, so $$(xy)^{nm}=1$$. This shows that $$xy$$ is of finite order, as desired. $$\Box$$

Added. Alternatively, working in $$G=\langle x,y\rangle$$, we have a normal (in fact, central) series $$\{1\}=G_{c+1} \triangleleft G_{c} \triangleleft\cdots \triangleleft G_2 \triangleleft G_1 = G,$$ and each factor group $$G_{i}/G_{i+1}$$ is torsion; hence $$G$$ itself is torsion.

• Thanks for your answer. I'm afraid I won't have the time to take a closer look until weekend. – Stefan Nov 9 '11 at 19:34
• Thanks a lot. There is a small typo: The last exponent of Lemma 1 should be "z". – Stefan Nov 17 '11 at 20:39
• @StefanWalter: Thank you; fixed! – Arturo Magidin Nov 17 '11 at 20:43
• One of the hypotheses of Lemma 2 is left unjustified in the proof of Proposition 3, namely that $[x,g]$ is central. It is not immediately apparent to me why this is the case. – pre-kidney Jun 22 at 6:07
• @pre-kidney: $[x,g]G_{c+2}$ is central in $G/G_{c+2}$. Because $x\in G_c$, so $[x,g]\in G_{c+1}$, and $G_{c+1}/G_{c+2}$ is contained in $Z(G/G_{c+2})$. – Arturo Magidin Jun 22 at 6:23