Let $F=F(a,b)$ be the free group of rank $2$.

**Question 1:** *Given any positive integer $d$, can one always find elements $u_j,v_j,w_j \in F$, $j=1,\dots,d$, such that if $1 \le j <k \le d$ then the normal closure of the three elements $u_j^{-1}u_k$, $v_j^{-1}v_k$ and $w_j^{-1}w_k$ has finite index in $F$?*

The last statement can be reformulated by saying that, for any pair of distinct indices $j,k$, the group $P_{jk}$, defined by the presentation $$\langle a,b \,\|\, u_j=u_k,v_j=v_k,w_j=w_k \rangle ,$$ is finite.

Intuitively, I would guess that the answer is negative, however, I do not see how to justify this. The pigeon hole principle can be used to show that the index of the normal closure must grow with $d$.

It is easy to ensure that the image of any given element $f \in F$ has finite order in any $P_{jk}$: just take $u_j=f^j$ for $j=1,\dots,d$. Similarly, by taking $u_j=a^j$, $v_j=b^j$ and $w_j=(ab)^j$ one can ensure that the images of $a,b$ and $ab$ all have finite orders in any $P_{jk}$.

If you think that three relators may not be enough, then I would also be happy with an answer to a more general question (Question 1 is a particular case of it for $n=3$):

**Question 2:** *Does there exist an integer $n \ge 2$ such that for every positive integer $d$ there is a set of elements $\{u_{ij}\mid 1 \le i \le n, 1 \le j \le d\} \subset F$ satisfying the following condition. For any pair of indices $1 \le j<k \le d$ the normal closure of the subset $\{ u_{1j}^{-1}u_{1k}, \dots, u_{nj}^{-1}u_{nk} \}$ has finite index in $F$?*