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Jun 21, 2015 at 12:53 comment added Alireza Abdollahi @YCor: No. I am not trying to compute the cohomology. I am interested in mathoverflow.net/questions/209438/… It is known that the number $R_G$ defined in the latter link is the minimum number of relations needed to present the finite $p$-group $G$ as a (topological) quotient of the free pro-$p$-group of rank $d_G$ the minimum number of generators of $G$.
Jun 21, 2015 at 12:45 comment added YCor 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?
Jun 21, 2015 at 11:06 comment added Alireza Abdollahi @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".
Jun 21, 2015 at 10:46 comment added YCor 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 $X$.
Jun 21, 2015 at 10:43 comment added Alireza Abdollahi @YCor: Can you write down the presentation for $\mathbb{Z}_p$ in your way?
Jun 21, 2015 at 10:34 history edited Alireza Abdollahi CC BY-SA 3.0
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Jun 21, 2015 at 10:26 comment added YCor 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.
Jun 21, 2015 at 9:25 history asked Alireza Abdollahi CC BY-SA 3.0