# What is the simplest known finite presentation of a free nilpotent group?

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.

• 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.) – YCor Apr 29 '19 at 14:52
• "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 ?). – J. Darné Apr 29 '19 at 16:46
• 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. – Sean Eberhard May 2 '19 at 10:28
• 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? – Sean Eberhard May 3 '19 at 9:15
• 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] – grok May 3 '19 at 12:02

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: