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I have a torsion-free non-abelian nilpotent group $\Gamma$ of cohomological dimension $n$. Is it possible to say anything about the number of generators of $\Gamma$ in a minimal presentation?

Can I assume that the number of generators can be chosen to be less than $n$?

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  • $\begingroup$ Think about surface groups... $\endgroup$
    – Steve D
    Commented Apr 16, 2013 at 3:28
  • $\begingroup$ ...or about nonabelian free groups... $\endgroup$
    – Misha
    Commented Apr 16, 2013 at 3:36
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    $\begingroup$ Since the question has been answered in comments I have voted to close as no longer relevant. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 16, 2013 at 3:38
  • $\begingroup$ An alternative would be to make a community wiki answer that quotes Steve D and Misha's answers, and have that accepted. $\endgroup$
    – Ryan Budney
    Commented Apr 16, 2013 at 5:42
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    $\begingroup$ It seems, with the edit adding the word "nilpotent", that this became a reasonable question, so closing is perhaps unnecessary now. $\endgroup$
    – Lee Mosher
    Commented Apr 17, 2013 at 15:11

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Finitely generated torsion-free nilpotent groups are polycyclic. Therefore, their cohomological dimension equals their Hirsch length.This is a result of Gruenberg. One can find it in Gruenberg's book 'Cohomological topics in group theory' in section 8.8 or in Robert Bieri's Book on 'homological dimension of discrete groups' as Th. 7.14.

On the other hand, the minimal number of generators of a polycyclic group is smaller or equal than its Hirsch length. (This is proven by a simple induction argument on the Hirsch length)

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  • $\begingroup$ For a concrete example consider the discrete Heisenberg group in the 3-dimensional case: en.wikipedia.org/wiki/… $\endgroup$
    – Khalid Bou-Rabee
    Commented Apr 16, 2013 at 11:37
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    $\begingroup$ Could you provide a reference, especially for your second sentence? $\endgroup$
    – Earthliŋ
    Commented Apr 17, 2013 at 9:13

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