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Let $n\geq 1$ be an integer and $p$ a prime. Suppose that $\mathcal{F}(n,p)$ is the free prop-p-group of rank $n$.

Question: For each pair $(n,p)$, is it known a discrete free group $\mathfrak{F}$ and an explicitly described subset $\mathfrak{X}$ of $\mathfrak{F}$ such that $\mathcal{F}(n,p)\cong \frac{\mathfrak{F}}{\langle \mathfrak{X} \rangle^\mathfrak{F}}$, where $\langle \mathfrak{X} \rangle^\mathfrak{F}$ is the normal closure of the subgroup generated by $\mathfrak{X}$ in $\mathfrak{F}$?

Certainly such a free group $\mathfrak{F}$ and subset $\mathfrak{X}$ exist. My question is about giving explicitly such a pair $(\mathfrak{F},\mathfrak{X})$. The rank of such free groups $\mathfrak{F}$ must be uncountable as $\mathcal{F}(n,p)$ is uncountable.

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  • $\begingroup$ For the question itself, you have the canonical choice (as in any group), the free group over $\mathfrak{F}(n,p)$, with all the length three relations as relators. $\endgroup$
    – YCor
    Commented Jun 21, 2015 at 10:26
  • $\begingroup$ @YCor: Can you write down the presentation for $\mathbb{Z}_p$ in your way? $\endgroup$
    – Alireza Abdollahi
    Commented Jun 21, 2015 at 10:43
  • $\begingroup$ I already did: for every group $G$ you consider the free group $F_G=\{e_g:g\in G\}$, the subset $R_G=\{e_{gh}e_h^{-1}e_g^{-1}:g,h\in G\}$, and you get $G=F_G/\langle\langle R_g\rangle\rangle$, where $\langle\langle X\rangle\rangle$ means the normal subgroup generated by $X$. $\endgroup$
    – YCor
    Commented Jun 21, 2015 at 10:46
  • $\begingroup$ @YCor: Is it possible to present the free pro-p-group as a free group quotient "without referring to the free pro-p-group itself"? Or maybe better to say, find a subset $X$ in $R_G$ which is minimal with respect to the inclusion such that $\langle \langle X \rangle \rangle=\langle \langle R_G \rangle \rangle$. Anyway, this may have a more or less trivial answer as the problem is to how define "explicit". $\endgroup$
    – Alireza Abdollahi
    Commented Jun 21, 2015 at 11:06
  • $\begingroup$ I'm not very optimistic that we can get anything beyond my suggestion and easy variants. Otherwise it sounds hopelessly hard. Are you trying to compute the cohomology of $F(n,p)$ as a discrete group? $\endgroup$
    – YCor
    Commented Jun 21, 2015 at 12:45

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