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.

  • 3
    $\begingroup$ These groups are polycyclic, i.e. iterated extensions with $\mathbb Z$ quotients, and moreover the associated series of groups is central. Taking generators in $\mathbb Z$'s and lifting them to the group gives you "canonical generators". The very first $\mathbb Z$ is in the center, so (at least) one of the generators is central. The growth exponent is at least $k$, and equals $k$ iff the group is abelian, see Bass, "The degree of polynomial growth of finitely generated nilpotent groups", Proc. London Math. Soc. (3) 25 (1972), 603–614. $\endgroup$ Feb 22, 2013 at 22:24
  • $\begingroup$ You can find basic results on fg nilpotent groups in Kargapolov-Merzljakov "Fundamentals of the theory of groups", Graduate Texts in Mathematics, 62. $\endgroup$ Feb 22, 2013 at 22:26
  • $\begingroup$ Let me remark that the type of presentation mentioned in Igor's great comment is a polycyclic presentation, and is very amenable to computation (GAP can tell you almost anything you want to know about such a group). $\endgroup$
    – Steve D
    Feb 22, 2013 at 22:48
  • 2
    $\begingroup$ For completeness: Bass-Guivarch theorem computes growth function $f(n)$ for nilpotent groups G; it equals $n^d$, $d=\sum_i i m_i$, where $m_i$ is the rank of the abelian group $C^iG/C^{i+1}G$. In your setting $m_i$ is the number of "Mal'cev" generators of $G$ which are in $C^i G$ but not in $C^{i+1}G$. $\endgroup$
    – Misha
    Feb 22, 2013 at 23:45

1 Answer 1


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.


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