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 $F$?


1 Answer 1


The following nice argument, due to Alexander Olshanskii, shows that the answer to both questions is negative.

It uses the theorem of Golod-Shafarevich (see http://arxiv.org/abs/1206.0490 for a survey on this topic). For any given $n \in \mathbb N$, choose $\tau=2/3$ and $r \in \mathbb{N}$ such that $$(1) \qquad 1-2 \tau+ n \tau^{r}=-1/3+n\left(\frac{2}{3}\right)^r <0, $$ and let $D_rF$ denote the $r$-th term of the Jennings-Zassenhaus filtration of the free group $F$ for some fixed prime $p$. (Recall that $D_rF$ is a normal subgroup of finite index in $F$.)

Now, using the notation of Question 2, suppose that we have a set of elements $\{u_{ij} \mid 1 \le i \le n, 1 \le j \le d\} \subset F$ for some $d > |F/D_rF|^n$. Then, by the pigeon hole principle, there are indices $j,k$, $1 \le j < k \le d$, such that $u_{ij}$ represents the same element of the finite group $F/D_rF$ as $u_{ik}$, for all $i=1,\dots,n$.

It follows that each of the elements $u_{1j}^{-1}u_{1k},\dots,u_{nj}^{-1}u_{nk}$ is contained in $D_rF$, so its Jennings-Zassenhaus degree is at least $r$. In view of (1), we can apply the theorem of Golod-Shafarevich to conclude that the group given by the presentation $$\langle a,b \,\|\, u_{1j}^{-1}u_{1k},\dots,u_{nj}^{-1}u_{nk} \rangle$$ must be infinite.

Therefore no natural number $n$ from Question 2 exists.


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