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.

1$\begingroup$ You can fit the group presentation into the title of the question and I would suggest doing so. $\endgroup$– j.c.Dec 3, 2010 at 0:23

$\begingroup$ You've probably already observed that its abelianization is $C_\infty \times C_2$. This much certainly matches the lamplighter... $\endgroup$– ndkrempelDec 3, 2010 at 1:14
2 Answers
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.

$\begingroup$ It looks like the group is virtually cyclic. In fact it is an extension of the 4element Klein group by the infinite cyclic group that permutes the two generating involutions. Do a change of variables $c=ba$. Get $c^2=1$, $c$ commutes with all conjugates by $b^n$. So the group is a factorgroup of the lamplighter group. Now the relation $a^2b=ba^2$ means, if I am not mistaken, $cbcb=bcbc$. Which means $cc^b=c^bc^{b^2}$ or $c=c^{b^2}$. So the normal subgroup generated by $c$ is of order 4. Right? $\endgroup$– user6976Dec 3, 2010 at 1:21

$\begingroup$ It is not quite clear for me why $c$ commutes with all conjugates by $b^n$. $\endgroup$ Dec 3, 2010 at 1:38

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.

$\begingroup$ Denis has beat me to most of that with his edit! $\endgroup$ Dec 3, 2010 at 1:39