6
$\begingroup$

Let $x$ and $y$ be two permutations of $\mathbb{Z}^2$ defined as follows. The permutation $x$ sends $(n,0)$ to $(n+1,0)$ and fixes all else while $y$ sends $(0,n)$ to $(0,n+1)$ and fixes all else. Is the group generated by $x$ and $y$ amenable?

I do know that the group does not contain a copy of the free group on two generators, so it is very likely to be amenable. I also know that if $y$ is defined, instead, by sending $(n,m)$ to $(n,m+1)$ then the group generated by $x$ and $y$ is amenable, in fact, it is a solvable extension of a locally finite group.

$\endgroup$
  • $\begingroup$ This is closely related to Houghton's groups. It's naturally a subgroup of $H_4$. I'm not sure right now if it's commensurable to Houghton's group $H_3$. $\endgroup$ – YCor Nov 4 '16 at 5:46
8
$\begingroup$

The derived subgroup of your group consists of permutations with finite support. Indeed, suppose that $w$ is a commutator word in $a$ and $b$ so the total exponent of $a$ (of $b$) is 0. Take a point $(m,n)$ where $m$ or $n$ are very large (comparing to $|w|$). Then $w(a,b)$ fixes that point. Therefore your group is an extension of a locally finite group by the Abelian group ${\mathbb Z}^2$, and is amenable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In fact, throwing away the off-axis points which are fixed by everything, the derived subgroup is the full finitary alternating group. $\endgroup$ – ndkrempel Mar 31 '11 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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