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Does the following group have a name? Is it amenable?

Fix $p$ and $q$

$\langle g,h: hg^qh^{-1}=gh^pg^{-1}\rangle$

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If $q=-1, p=1$ (or vice versa) this group is the Braid group on 3 strings, so it is called $B_3$. I am not sure it has a name if $p>2, q>1$. In general, the group seems to be CAT(0) since its presentation complex is locally so, which means it is not going to be amenable unless I missed some degenerate cases. –  Misha Jan 6 '13 at 5:08
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This paper should answer your question: ams.org/mathscinet-getitem?mr=1489891 I found this reference in Misha's comment on this question: mathoverflow.net/questions/93677/… –  Ian Agol Jan 6 '13 at 7:17
    
Thanks for the comments and references! –  Kate Juschenko Jan 6 '13 at 19:47

1 Answer 1

up vote 5 down vote accepted

For 1-related groups, over a 2-letter alphabet, draw the relator on the plane grid: $g$'s are horizontal, $h$'s are vertical, starting at the point $O=(0,0)$. Connect $O$ with the endpoint $М$ of the resulting path $P$ by a vector $\vec{v}=\vec{OM}$. Consider the two support lines of the path $P$ that are parallel to $\vec{v}$. If both support lines intersect the path only once, the group is free-by-cyclic. If only one of the support lines intersects the path once, the group is an ascending HNN extension of a free group, and the rank of the free group is easily computed using Magnus rewriting. It can be all found in the old paper Kenneth S. Brown, Trees, valuations, and the Bieri-Neumann-Strebel invariant. Invent. Math., 90(3):479--504, 1987.

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Thanks, Mark. Basically, the only amenable group in this list will be when p=q=1. –  Kate Juschenko Jan 8 '13 at 3:15
    
These groups are very easy since they are ascending HNN extensions of free groups or even free-by-cyclic: residually finite, coherent, etc. But not amenable. –  Mark Sapir Jan 8 '13 at 3:38

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