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.

  • $\begingroup$ 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, 2016 at 5:46

1 Answer 1


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.

  • $\begingroup$ 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, 2011 at 13:49

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