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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.

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You can fit the group presentation into the title of the question and I would suggest doing so. – j.c. Dec 3 '10 at 0:23
You've probably already observed that its abelianization is $C_\infty \times C_2$. This much certainly matches the lamplighter... – ndkrempel Dec 3 '10 at 1:14
up vote 9 down vote accepted

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.

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It looks like the group is virtually cyclic. In fact it is an extension of the 4-element 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 factor-group 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? – Mark Sapir Dec 3 '10 at 1:21
It is not quite clear for me why $c$ commutes with all conjugates by $b^n$. – Denis Osin Dec 3 '10 at 1:38
The normal closure of $c$ is infinite. – Steve D Mar 23 '11 at 2:42

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.

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Denis has beat me to most of that with his edit! – ndkrempel Dec 3 '10 at 1:39

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