Is the following fact true? 



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> Let $N$ be a finitely generated nilpotent group, and denote its lower central series by $(N\_r)\_{r\ge 1}$, that is, $N\_1=N$ and $N\_{k+1}=[N,N\_k]$ is the commutator group of $N$ and $N\_k$. Then there is a finite index subgroup $H$ of $N$ which has the following property - if $H\_r$ is $H$'s lower central sequence, then all the quotients $\frac{H\_r}{H\_{r+1}}$ are torsion-free.
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Thanks in advance to anyone who is willing to help :-).


**Edit:**
I read Woess' paper wrong - he actually only claims that $H$ has to be torsion-free. Thanks for all the helpers.


***Edit 2:***
 Well, i'm still interested in an answer to the original question. I'll take off the reference request and make it a question instead.


***Edit 3*** *(YC):* in the initial version of the question the fact was attributed to Woess in "Random walks on infinite groups and graphs" during the classification of recurrent groups, but actually Woess claims the weaker and more classical fact that every finitely generated nilpotent group has a torsion-free subgroup of finite index, for which a reference was given in one answer (after the 1st edit).



Quotients of finitely generated nilpotent groups
I believe that for class $2$ and the original question (finding a subgroup $H$ of finite index such that $H\_i/H\_{i+1}$ is torsionfree for all $i$) the following works.


Suppose that $N$ is of class $2$. Write $N^{\rm ab} = C\_{k\_1}\oplus\cdots \oplus C\_{k\_r}\oplus C\_0^{s}$, where $C\_k$ is the cyclic group of order $k$ and $C\_0$ is the infinite cyclic group, with $k\_1|k\_2|\cdots|k\_r$, $1\lt k\_1$, and let $x\_1,\ldots,x\_r$ project onto generators of the finite cyclic summands, and $y\_1,\ldots,y\_s$ project onto generators of the torsionfree part.


If $s=0$, then $N^{\rm ab}$ is finite, and hence so is $N\_2$; so $N$ itself is finite and we may take $H=\{e\}$. Assume then than $s\gt 0$. 


The commutator subgroup $N\_2$ is generated by commutators of the form $[x\_j,x\_i]$, $1\leq i\lt j\leq r$; $[y\_b,y\_a]$, $1\leq a\lt b\leq s$; and $[y\_c,x\_d]$, $1\leq c\leq s$, $1\leq d\leq r$. All these commutators except perhaps for those of the form $[y\_b,y\_a]$ are torsion. The subgroup generated by the commutators $[y\_b,y\_a]$ may have torsion; let $t$ be the exponent of the torsion part of this subgroup. 


Let $H=\langle y\_1^t,\ldots,y\_s^t\rangle$. This is easily seen to be of finite index in $N$. Moreover, $H\_2 = \langle [y\_b,y\_a]^{t^2}\mid 1\leq a\lt b\leq s\rangle$, and all of these elements lie in the torsionfree part of $N\_2$, hence $H\_2$ is a subgroup of a torsionfree group and so torsionfree. On the other hand, if an element of $H^{\rm ab}$ is torsion then we can express it as $y\_1^{ta\_1}\cdots y\_s^{ta\_s}H\_2$ with $(y\_1^{ta\_1}\cdots y\_s^{ta\_s})^m\in H\_2$, $m\gt 0$. But this implies that $y\_1^{mta\_1}\cdots y\_s^{mta\_s}\in H\_2\subseteq N\_2$, which by the choice of $y\_j$ means that $mta\_j=0$ for all $j$. Since $m,t\gt 0$, this means $a\_j=0$ for all $j$, so the original element was trivial.


I would expect something similar to this to work for arbitrary class, but I can see how it might start getting a bit hairy for higher class.


