Is there a function $f$ such that for any presentation $$G=\langle x_1,\ldots,x_n \mid r_1,\ldots,r_k\rangle\quad \text{with}\quad |r_i|\leq 3$$

$k\leq f(n)$ implies that $G$ has non-abelian free subgroups.

Of course $f=0$ works trivially, I am asking for bigger functions.

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    $\begingroup$ $f=n-2$ works too. The group is large then. $\endgroup$ – Mark Sapir Apr 5 '12 at 21:57
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    $\begingroup$ The reference that Mark has in mind is: B.Baumslag and S.Pride, "Groups with two more generators than relators", J. London Math. Soc. (2) 17 (1978), no. 3, 425-426. Every group $G$ of deficiency $\ge 2$ virtually admits an epimorphism to the free group of rank 2. In particular, $G$ contains free nonabelian subgroups. $\endgroup$ – Misha Apr 6 '12 at 3:48
  • $\begingroup$ If one of your relators is a proper power, $r_i=s_i^n$ and $n>1$, then $f=n-1$ works. $\endgroup$ – ADL Apr 8 '13 at 10:04

$f(n)=n-2$ works as I said in the comment above. On the other hand, $f(n) < n-1$ for $n\ge 4$. Indeed, the metabelian Baumslag-Solitar group can be given by relations $\langle x_1,...,x_{n-1},u,v,t\mid x_1=x_2, x_2=x_3, ..., x_{n-2}=x_{n-1}, tx_1=u, ut^{-1}=v, x_1^2=v\rangle $ with $n+2$ generators and $n+1$ relations. So the answer is $f(n)=n-2$ - for every $n\ge 4$.


Another way to prove that $f(n)=n-2$ works is to recall Romanovskii's Freiheitssatz:

In any finitely presented group
$$ >G=⟨x_1,…,x_n∣r_1,…,r_k⟩, >$$ some $n-k$ of {$x_i$} freely generate a free subgroup of $G$ (if $n\ge k$).


N.S. Romanovskii, "Free subgroups of finitely presented groups" Algebra i Logika , 16 (1977) pp. 88–97 (In Russian).


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