most recent 30 from http://mathoverflow.net 2013-05-22T17:01:17Z http://mathoverflow.net/feeds/question/48116 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48116/does-this-group-have-a-name dan 2010-12-03T00:12:34Z 2010-12-03T01:38:55Z

<p>In my work this week I came across a group with presentation with two generators $a$ and $b$ subject to the relations $baba=1$, $a^2b=ba^2$, and $ab^{-n}ab^n=b^nab^{-n}a$. This group looks like the lamplighter group or something to me, but I couldn't get a sequence of Tietze transformations from this group to the standard presentation for the lamplighter. Does anyone know what this group is? thanks.</p>

http://mathoverflow.net/questions/48116/does-this-group-have-a-name/48119#48119 Denis Osin 2010-12-03T01:15:42Z 2010-12-03T01:26:04Z

<p>All relations of the form $ab^{-n}ab^n=b^nab^{-n}a$ follow from $baba=1$, $a^2b=ba^2$ (exercise). So the group is isomorphic to $G=\langle a,b\mid baba=1, a^2b=ba^2\rangle$. The later splits as a central extension $1\to \mathbb Z\to G \to D_{\infty }\to 1$. I do not think the group has a name.</p>

http://mathoverflow.net/questions/48116/does-this-group-have-a-name/48120#48120 ndkrempel 2010-12-03T01:38:55Z 2010-12-03T01:38:55Z

<p>The first two relations alone give a polycyclic group of Hirsch length 2 ($a^2$ is central, quotienting by it gives the infinite dihedral group $D_\infty$), which, thanks to Denis Osin's answer, is already the whole group. Even without that knowledge, it is still a quotient of this group, and so polycyclic of Hirsch length $\leq 2$. In particular, it is far too small to be the lamplighter.</p>