0
$\begingroup$

Let $F$ be a free group of finite rank $n > 2$, and let $S \subseteq F$ be a subset of cardinality at most $n-2$. Denote by $S^F$ the normal subgroup of $F$ generated by $S$. Must $F/S^F$ be nonabelian?

What if $F$ is free profinite instead? Is it immediate that $F/S^F$ is nontrivial?

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ Of course: it is even large by a theorem of Baumslag-Pride [Baumslag, Benjamin; Pride, Stephen J. Groups with two more generators than relators. J. London Math. Soc. (2) 17 (1978), no. 3, 425–426] $\endgroup$
    – Ashot Minasyan
    Commented May 21, 2015 at 10:27
  • $\begingroup$ @AshotMinasyan I am (sadly enough) not familiar with the terminology (what is a large group?). I would also like to know if the exists some (very elementary) argument showing that $F/S^F$ is nonabelian or merely nontrivial. I would be very glad if you could expand your comment to answer, clarifying these things for me. $\endgroup$
    – Pablo
    Commented May 21, 2015 at 10:36
  • 3
    $\begingroup$ A group is large if it has a finite index subgroup which maps onto the free group $F_2$ of rank $2$. Any large group contains a copy of $F_2$. You can read the proof of the theorem of Baumslag-Pride: it is very short. In the beginning they show that any group of deficiency at least 1 maps onto $\mathbb Z$; in particular, it is non-trivial. $\endgroup$
    – Ashot Minasyan
    Commented May 21, 2015 at 10:40
  • $\begingroup$ @AshotMinasyan I see thanks. I will edit the question to include the profinite case. $\endgroup$
    – Pablo
    Commented May 21, 2015 at 10:46

1 Answer 1

Reset to default
6
$\begingroup$

My answer to your question here included a proof of this. Indeed, such a group has a non-abelian 2-nilpotent quotient. In the discrete case, it even shows that there is a non-abelian torsion-free 2-nilpotent quotient. In the profinite case, it shows that for all $p$ (although I gave a full proof only for odd $p$) there is a non-abelian 2-nilpotent $p$-group quotient. It still works for a (profinite or discrete) group on $n$ generators with any family of relators such that at most $n-2$ of the relator do not belong to $[F,[F,F]]$, where $F$ is free (discrete or profinite) on $n$ generators.

Of course this gives no info about largeness.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .