Let $G$ be the group given by the presentation 
$$\langle x,y,z,w \ | \ xy = yx, yz = zy, zw = wz\rangle.$$


This is a right-angled Artin group (RAAG) whose graph is a path on $4$ vertices.


We can also write $G$ as $(\mathbb{Z} \times F\_2) \*\_{\mathbb{Z}} \mathbb{Z}^2$ where $F\_2$ is a free group of rank $2$ generated by $\{x, z\}$, the leftmost copy of $\mathbb{Z}$ is generated by $y$, the $\mathbb{Z}^2$ is generated by $\{z,w\}$, and the amalgamated subgroup is generated by $z$.



> 
> Are there infinitely many finite index subgroups $U$ of $G$ that can
>  be generated by at most $6$ elements?
> 
> 
> 



Finite index subgroups of a RAAG
This is not a satisfactory answer, but in the absence of anything better I will put it on record. It would be nice if someone could come up with a more illuminating answer.


Computer experiments strongly indicate that the answer is yes, and I could write down a proof by hand if forced, but it would be technical and tedious, so I would prefer not to unless this is really important!


In fact there appear to be infinitely many finite index subgroups $H$ with minimal generator number $d(H) = 4$, which is best possible since $d(G/[G,G])=4$.


Indeed, wth increasing index $n$, we find subgroups $H$ of that index with all possible values of $d(H)$ between $4$ and $m(n)$, where $m(n)$ increases with $n$. For example, for $n=8$ (which is the largest index for which I can easily compute all subgroups of that index), there are subgroups $H$ of index $8$ with $d(h) = d$ for all $d$ with $4 \le d \le 18$.


It was relatively easy to find a sequence of subgroups $H$ of increasing index with $d(H)=4$. The subgroup $$\langle xy,yz,zw,x^mw^{-m} \rangle$$ has index $2m$ whereas $$\langle xy,yz,zw,x^mw^{-m+1} \rangle$$ has index $2m-1$. The first of these has right transversal $$\{ x^k,y^l : 0 \le k \le m, 1 \le l < m \}$$ and the second has right transversal $$\{ x^k,y^l : 0 \le k < m, 1 \le l < m \}.$$ As I said, I could write down a hand proof of that, but would prefer not to.


It is perhaps worth noting that for a group $\langle X \mid R \rangle$ with $|X| - |R| \ge 2$, the answer to the question would be no, because the Reidemieister-Schreier process generates a presentation of a subgroup $H$ of index $n$ with $(n-1)|X|+1$ generators, and $n|R|$ relators, so if $|X|-|R| \ge 2$ then the deficiency of the presentation would increase with increasing index, and hence so would $d(H)$.


