6
$\begingroup$

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 $G$.

Improve this question
$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Nov 7, 2017 at 2:59
  • $\begingroup$ @LSpice thanks, I changed the title $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Nov 7, 2017 at 6:36
  • 3
    $\begingroup$ To be length-$n$ unfree for some $n$ is equivalent to the better known notion of satisfying a group law (or "group identity") $\endgroup$
    – YCor
    Commented Nov 7, 2017 at 8:39
  • $\begingroup$ @YCor It may be obvious, but I don't see why the equivalence holds. $\endgroup$
    – Luc Guyot
    Commented Nov 10, 2017 at 23:42
  • 1
    $\begingroup$ @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 $c(a,b)=1$. $\endgroup$
    – YCor
    Commented Nov 11, 2017 at 0:12

1 Answer 1

Reset to default
12
$\begingroup$

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).

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.