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Finitely generated nilpotent groups are always finitely presented. This is true for abelian groups, and can be shown by induction for nilpotent ones, using the classical lift of a presentation of $N$ and a presentation of $G/N$ to a presentation of $G$. As a consequence, the free $c$-nilpotent group of rank $n$, $F_n/\Gamma_{c+1}(F_n)$, is finitely presented. In fact, I do know a finite presentation : kill all iterated commutators of the generators of length between $c+1$ and $2c$. This works because all iterated commutators of length greater than $c$ can be written as a product of iterated commutators of the generators of length greater than $c$ (using basic commutator calculus), and every such commutator has a sub-commutator of length between $c+1$ and $2c$ (it can help to think of commutators as rooted planar binary trees).

But I am unable to find a simpler presentation (say, with less relations). In fact, I can think of many presentations making the lower central series to stop, and giving the right Lie algebra (using linear trees, or Lyndon trees), but I do not know how to show that the result is nilpotent, and it may well not be. And I have not been able to find a simpler presentation from the induction process described above, either. Whence my question.

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  • $\begingroup$ A "measure" of a presentation counts the number of relator, but it's also tempting to take into account the length of relators. (PS I distinguish between relator of a presentation, and relation, that is a consequence of relators.) $\endgroup$
    – YCor
    Apr 29, 2019 at 14:52
  • $\begingroup$ "Simple" does not have a precise meaning in my question. Any measure would do the trick ; in fact, I do not seek "the simplest", but anything more tractable than the one I describe in the post. Indeed, something with only iterated commutators of length $c+1$ would be great, but I do not know if this can work. Also, one could want to take into account something less measurable, like the form of the relators (e.g: only linear trees ?). $\endgroup$
    – J. Darné
    Apr 29, 2019 at 16:46
  • $\begingroup$ Isn't it true that you just have to kill all basic commutators of weight $c+1$? The group $\Gamma_{c+1} / \Gamma_{c+2}$ is free abelian with rank given by Witt's formula, so you need exactly this many relators. $\endgroup$
    – Sean Eberhard
    May 2, 2019 at 10:28
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    $\begingroup$ I revise my guess to the following: You have to kill all basic commutators of the form $[x, y]$, where $x, y$ are basic commutators of weight at most $c$ whose weights add to at least $c+1$. This is a bit more efficient than just taking all commutators of weight between $c$ and $2c$. Does this list contain any redundancy? $\endgroup$
    – Sean Eberhard
    May 3, 2019 at 9:15
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    $\begingroup$ I think this is a hard question -- in that it's connected to many other questions and to some K-theory, so an answer to it would probably answer questions already asked and studied by other (smart) people. It has been studied by computer for small $c$, see e.g. [MR2440289. Jackson, David A.(1-STL-CS), Basic commutators in weights six and seven as relators. Comm. Algebra 36 (2008), no. 8, 2905–2909] $\endgroup$
    – grok
    May 3, 2019 at 12:02

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Taking "simplest" to mean "having fewest generators", the question is how many generators you need to normally generate $\Gamma_n$ in a free group. As mentioned already in the comments, $\Gamma_n / \Gamma_{n+1}$ is free abelian with rank given by Witt's formula, so certainly you need at least this many generators. It seems to be unknown (!) whether the basic commutators of weight $n$ normally generate $\Gamma_n$. A couple of references I found:

Sims, Charles C. Verifying nilpotence. J. Symbolic Comput. 3 (1987), no. 3, 231--247. https://doi.org/10.1016/S0747-7171(87)80002-0

Jackson, David A. Basic commutators in weights six and seven as relators. Comm. Algebra 36 (2008), no. 8, 2905--2909. https://doi.org/10.1080/00927870802108148

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