# Quotients of finitely generated nilpotent groups

Is the following fact true?

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.

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).

• I have asked this question in math.se, but failed to get an answer, which I do need urgently. Thanks for the help! – Miel Sharf May 28 '15 at 19:24
• Note that what you call "its central series" is usually called the lower central series. – Arturo Magidin May 28 '15 at 19:59
• Is there some reason that you cannot just take $N^k$, where $k$ is a common multiple of the exponents of the torsion subgroups of all $N_i/N_{i+1}$ ? – Arturo Magidin May 29 '15 at 3:23
• @ArturoMagidin: The solution requires more care than that. The OP's condition is satisfied for the integral Heisenberg group $H$. However, for $N = H^k$, the quotient $N/[N,N]$ contains the torsion subgroup $[H,H]^k/[H^k,H^k]$ of order $k$. Therefore, an example where your suggestion does not work is $H^2$. Your suggestion would replace this with $(H^2)^2 = H^4$, but that makes the situation worse, not better. – Dave Witte Morris May 29 '15 at 6:03
• Ah well, I shall stop thinking about the original question now, but I would still be interested to know whether it is correct! I played with a few examples, and I could not find any easily described procedure for finding the torsion-free subgroup with the calimed properties. – Derek Holt May 29 '15 at 8:25

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.

• For arbitrary class, I would try to do some induction; assuming we can do it for class $c$, with $N$ of class $c+1$, we could try finding a subgroup of finite index $H$ such that $H/N_{c+1}$ has the desired property in $N/N_{c+1}$, so that the only possible problem lies in $H_{c+1}$, and then take a finite index subgroup of $H$ given by adequate powers of the generators so that we are in the torsionfree part of $H_{c+1}$ once we get down to it. – Arturo Magidin May 29 '15 at 20:09
• For class 2 it's easier: assume that $N$ is torsion-free 2-nilpotent. Let $f$ be the projection $N\to A=N/[N,N]$ and write $A=A_1\times T$ with $T$ finite and $A_1$ torsion-free. Define $H=f^{-1}(A_1)$. Then I claim that $[H,H]=[N,N]$ (which implies that $H/[H,H]$ is torsion-free). Indeed the bracket induces an alternating bilinear map $b:A\times A\to [N,N]$, whose image generates $[N,N]$; $b$ has to be trivial on $T\times A$, so the image of $A_1\times A_1$ generates $[N,N]$, proving $[H,H]=[N,N]$. – YCor May 30 '15 at 7:44
• @YCor: How do we get to $N$ torsionfree to begin with? – Arturo Magidin May 30 '15 at 22:32
• I assumed $N$ torsion-free because this step is known (and classical). – YCor May 31 '15 at 7:46
• @YCor: Thanks! I confess to not knowing it before. – Arturo Magidin May 31 '15 at 16:55