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.

  • $\begingroup$ Sorry what is the definition of the nilpotent group? $\endgroup$
    – user17090
    Nov 6, 2011 at 11:26
  • $\begingroup$ @AliBleybel: see Wikipedia: a nilpotent group is a group that has a central series of finite length. $\endgroup$
    – Stefan
    Nov 6, 2011 at 11:37
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    $\begingroup$ @Ali: while it is nice that questions be sensibly self-contained, googling for "nilpotent group" would have answered your question in 2 seconds... $\endgroup$ Nov 6, 2011 at 12:06
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    $\begingroup$ Have you seen the proof in Philip Hall's lecture notes on nilpotent groups? It is very basic. $\endgroup$
    – user641
    Nov 6, 2011 at 18:30

1 Answer 1


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.

  • $\begingroup$ Thanks for your answer. I'm afraid I won't have the time to take a closer look until weekend. $\endgroup$
    – Stefan
    Nov 9, 2011 at 19:34
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    $\begingroup$ Thanks a lot. There is a small typo: The last exponent of Lemma 1 should be "z". $\endgroup$
    – Stefan
    Nov 17, 2011 at 20:39
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    $\begingroup$ @StefanWalter: Thank you; fixed! $\endgroup$ Nov 17, 2011 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – pre-kidney
    Jun 22, 2019 at 6:07
  • $\begingroup$ @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 $Z(G/G_{c+2})$. $\endgroup$ Jun 22, 2019 at 6:23

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