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Let $G$ be a f.g. residually torsion-free nilpotent group. Let $x$ be a nontrivial element of G, then there is a normal subgroup $N$ of $G$ such as $G/N$ torsion-free nilpotent, $x \notin N$. Let $\gamma_{n}(K)$ be the n-term of the lower central series of K. Since $G/N$ is nilpotent, there is $n$ such as $N\gamma_{n}(G)/N = \gamma_{n}(G/N) = 1$. It follows that $\gamma_{n}(G) \le N$. Is it true that $G/ \gamma_{n}(G)$ is torsion free?

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What is $K$?$ $ – user641 Sep 9 '12 at 16:42
Thanks for the correction. I hope it's ok now. – Dennis Sep 9 '12 at 16:56

With the way you have worded the question, I think the answer is no. Let $G$ be the group defined by the presentation $\langle x,y,z \mid xz=zx, yz=zy, [x,y] = z^2 \rangle$. Then $G$ itself is torsion-free nilpotent. Choose your $x$ to be the same as my $x$. Then we could take $N = \langle z \rangle$ and $n=2$. (I am using the notation $\gamma_1(G)=G$, $\gamma_2(G)=[G,G]$.) But in the group $G/\gamma_2(G)$ the image of $z$ has order 2, so it is not torsion-free.

Of course I could have chosen $N=1$ and/or $n=3$, and then the answer would have been yes.

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Just a comment: $G$ is isomorphic to a subgroup of index 2 in the integral Heisenberg group. – YCor Sep 9 '12 at 18:32
Thanks. Is it possible to choose $n$ such as $x \notin \gamma_{n}(G)$ and $G/ \gamma_{n}(G)$ is torsion free? – Dennis Sep 9 '12 at 18:34
I would guess not. How about a group generated by $x_i$ ($i \ge 1$) with $[x_1,x_2]=x_3^2$, $[x_1,x_i]=x_{i+1}$ for $i>2$, and $[x_i,x_j]=1$ for $i,j \ge 2$. – Derek Holt Sep 9 '12 at 19:46
But this group isn't finitely generated. – Dennis Sep 9 '12 at 20:57
It is finitely generated by $x_1,x_2,x_3$. But I don't think it works, because I haven't defined the commutators of the inverses of the generators. – Derek Holt Sep 9 '12 at 23:08

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