Let $G$ be a non-elementary group generated by a finite set $S$. Here, a group is called non-elementary if it is not virtually abelian. Denote $S^{\le n}:=\{g\in G: |g|\_S\le n\}$ for any $n\in \mathbb N$ and $C(H,K):=\{[h,k]: h\in H, k\in K\}$ for any two subsets $H,K\subset G$. It is easy to see that $|S^{\le n}|\ge n$ as $G$ is infinite. After some research, I know that the whole commutator subset $C(G,G)$ is still infinite. So my question is that whether there is a strictly increasing function $\kappa: \mathbb N\_{\ge 1}\to \mathbb N\_{\ge 1}$ such that $|C(S^{\le n}, S^{\le n})|\ge \kappa(n)$ for any $n\in \mathbb N\_{\ge 1}$?


Maybe a weaker question is that: suppose more that $G$ is not nilpotent, then $\kappa(n)$ can be chosen like $c\log(n)$?



The growth rate of a commutator set in a non-elementary group
You can take $\kappa(n) = n/2$ if $G$ is not virtually nilpotent of class $\le 2$.


Let $B\_n = S^{\le n}$ and $C\_n = \{[b\_1, b\_2] : b\_1, b\_2 \in B\_n\}$. Suppose $|C\_n| < n/2$. By pigeonhole there is some $m < n$ such that $C\_m = C\_{m+2}$. For $[b\_1, b\_2] \in C\_m$ and $s \in S$ we have $[b\_1,b\_2]^s = [b\_1^s, b\_2^s] \in C\_{m+2} = C\_m$, so $C\_m$ is a union of conjugacy classes. Since $G'$ is generated by the conjugates of $C\_1 \subseteq C\_m$, it follows that $G'$ is generated by $C\_m$. The conjugation action of $G$ on $C\_m$ has finite-index kernel $K = C\_G(G')$, and $[K, G'] = 1$. In particular $K$ is nilpotent of class $\le 2$.


If $G$ is nilpotent of class $2$ and not virtually abelian then there is some commutator $[s\_1, s\_2]$ of infinite order, and $[s\_1^a s\_2^b, s\_1^{a'} s\_2^{b'}] = [s\_1, s\_2]^{ab' - a'b}$. I think the set $[n]\cdot[n] - [n] \cdot[n]$ has positive density in $[-n^2,n^2]$; maybe some expert can comment. This implies that we can take $\kappa(n) = c n^2$ for some constant $c$.


If $G$ has class-2 subgroup of index $k$ then by using Schreier generators you can argue similarly that $\kappa(n) = cn^2/k$ works.


