2
$\begingroup$

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?

  • 0
    Thanks for the correction. I hope it's ok now.2012-09-09

1 Answers 1

2

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.

  • 0
    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.2012-09-09