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It is known that there are non-amenable groups not containing $F_2$, the free group on two generators. We can even have that every 2-generated subgroup is finite.

But is there a non-amenable group $G$ where for some $n$, $G$ is length-$n$ unfree in the following sense?

Definition. $G$ is length-$n$ unfree if for all $a,b\in G$, there exist words $u\ne v$ of length $n$ over the alphabet $\{a,b\}$ such that $u=v$ in $G$.

Bjørn Kjos-Hanssen
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  • I'm sure I'm being silly, but your title asks for a group that "looks like $\mathrm F_2$ up to level $n$", whereas your question asks for a group that "is level-$n$ unfree". These seem like opposite goals. – LSpice Nov 07 '17 at 02:59
  • @LSpice thanks, I changed the title – Bjørn Kjos-Hanssen Nov 07 '17 at 06:36
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    To be length-$n$ unfree for some $n$ is equivalent to the better known notion of satisfying a group law (or "group identity") – YCor Nov 07 '17 at 08:39
  • @YCor It may be obvious, but I don't see why the equivalence holds. – Luc Guyot Nov 10 '17 at 23:42
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    @LucGuyot index nontrivial elements of the 2n-ball in $F_2=\langle x,y\rangle$ as $w_0,\dots,w_m$ with $w_0=x$, $w_1=y$. Define by induction $c_0=x$, $c_n=[w_n,c_{n-1}]$ if this $\neq 1$, and otherwise $c_n=[w_n,[w_{n-1},c_{n-1}]]$. Here I use hat $c_{n-1}$ has the form $[w_{n-1},c_{n-1}]$ and is $\neq 1$ and hence does not commute with $w_{n-1}$, and $w_n$ can't commute with both $c_{n-1}$ and $[w_{n-1},c_{n-1}]$. Set $c=c_m$. Then in any $n$-unfree group, for all $a,b$ there's $i$ such that $w_i(a,b)=1$ and hence $c(a,b)=1$. – YCor Nov 11 '17 at 00:12
  • By the way I interpreted "words" as "group words". If the OP really means positive words (so "unfree" would be a misleading terminology), my above argument is still valid, but the converse probably fails (e.g., I guess there is no law of the form $uv^{-1}$ with $u,v$ positive words (possibly null), satisfied by all metabelian groups). – YCor Nov 11 '17 at 00:20
  • @YCor yes I did mean positive words, although it's of interest either way – Bjørn Kjos-Hanssen Nov 11 '17 at 03:15
  • OK, so it's indeed very confusing, especially in view of the informal definition "looks like $F_2$ up to level $n$". – YCor Nov 11 '17 at 08:49

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Burnside groups of exponent $n$ are length $n$ unfree. If $n\ge 665$, odd, then the free Burnside group of exponent $n$ of rank 2 or more is not amenable (see Adian's book "The Burnside problem" or Olshanskii's book "Geometry of defining relations" or my book "Combinatorial algebra: syntax and semantics", Chapter 5).