I read somewhere that Maltcev proved that for any finitely generated torsion free Nilpotent group $G$ there are canonical generators, i.e.
$g\_{1},\ldots,g\_{k}$ such that any $g \in G$ can be written uniquely as:
$g\_{1}^{a\_{1}}g\_{2}^{a\_{2}} \cdots g\_{k}^{a\_{k}}$, where the $a$-s are in $Z$.
I could not find the paper online, probably since it is very old.


I have several questions (2-4 are versions of the same question):


1) If the size of the set of canonical generators is $k$ does this implies that the group has polynomial growth with exponent $k$?


2) Can we always pick one of the canonical generators to be in the center of the group?


3) If 2 fails, is there always a subgroup of finite index for which 2 is true?


4) I think this is can imply 3: can we pick canonical generators, $g\_{1},\ldots,g\_{k}$ and $z$ in the center, such that if we throw away one of the canonical generators, the rest of the canonical generators do not have z in the group they generate?


The motivation is: consider a general group G of polynomial growth with exponent d>2. I want to be to find a copy of $Z^{2}$, call it M, inside G and a group of polynomial growth with exponent d-2, call it H, such that if we look at $mH$ are disjoint for different $m-s \in M$. For some application I want to treat the $mH$-s as hyperplanes and it is important that I can pick them to be disjoint.



On Canonical generators of torsion free nilpotent group
An alternative latinization of the name is Malcev (or Mal'cev). It is less fashionable now, but most of the literature uses that spelling.


1. No. The Heisenberg group $[a,b]=c$, $c$ central is a 3-dimensional algebraic group with has canonical expression $a^nb^mc^p$, but quartic growth. Since $b^na^n=a^nb^nc^{-n^2}$, an expression of word length $N$ is expressible in the canonical coordinates bounded by $(n,m,p)\le(N,N,N^2)$, so at most quartic growth. A lower bound can be achieved by expressing the exponent of $c$ in binary. If the word growth of a group matches its algebraic dimension, it is abelian.
2. I think so. I imagine that one proves the theorem by showing that the quotient of a finitely generated torsion-free nilpotent group by its center is again torsion-free.
3. Yes. Even if the above approach fails, I think it should be possible to induct by considering the abelianization. This is not torsion-free, but it has a finite index subgroup which is. So a finitely-generated torsion-free nilpotent group contains a finite index subgroup (whose image in the abelianization is torsion-free) that is an extension of a torsion-free abelian group by a group of smaller nilpotence length.
4. I'm not sure I understand this question, but I think the example of the Heisenberg group should answer it. It is three dimensional, with generators $a,b,c$. If you throw out any of them, the remaining two generate a subgroup containing the center $c$. In particular, $a$ and $b$ alone generate the group, just with more complicated expressions for the elements. More generally, integral matrices that are upper triangular and 1s on the diagonal give accessible examples of torsion-free nilpotent groups.


