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Given a presentation $$ \langle x,y,z : x^y=x^2, y^z=y^2, z^x = z^2 \rangle, $$ where $x^y$ is just the usual conjugation (that is, $x^y$ is defined to be $y^{-1} xy$). Can we say for sure, whether this presentation defines a nontrivial group?

darij grinberg
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AlexCon
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    This is an amalgamation of three copies of the Baumslag–Solitar group $B(1,2)$, which is a special case of the Serre-Bass theory of «graphs of groups» Googling for those keywords should be of help. – Mariano Suárez-Álvarez Feb 03 '14 at 05:36
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    It appears that this group is not an amalgam of three Baumslag-Solitars but a fund group "triangle of groups" with BS vertex groups. In general, a triangle of nontrivial groups can be trivial. A sufficient condition for non triviality is a certain "nonpositive curvature property". The book by Bridson and Haefliger has the while cheaper dealing with these issues. – Moishe Kohan Feb 03 '14 at 08:22
  • I have checked with GAP and, barring mistakes, it says this is the trivial group. Could someone please check with another CAS? – Andreas Caranti Feb 03 '14 at 09:04
  • This type of presentation, where the relators form a cycle, are called cyclically presented groups, and they are notoriously hard to deal with! I think they were first introduced by john Conway, where he posed a "fun" problem asking if a certain cyclically presented group was cyclic of order 11. It took a few years to solve... The group was (a,...,e:ab-c,bc-d,...,ea-b), where - means equals, and i apologise for the rubbishness of my phone, which lacks a dollar sign and an equals sign... – user1729 Feb 03 '14 at 10:41
  • I've posted an algebraic proof that this group is trivial as an answer to this question. – Jim Belk Jul 03 '15 at 17:24

1 Answers1

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This is actually a "well-known" presentation of the trivial group. I think it might have been originally proved trivial by Graham Higman, but I am not not certain. It can be done by hand, but it is IIRC a challenging exercise to do so!

It had the interesting feature that it can be used to construct a sequence of more complicated presentations of the trivial group that defeat coset enumeration programs. Write the presentation as

$G_1 = \langle x,y,z \mid y^{-1}xyx^{-2} = z^{-1}yxy^{-2} = x^{-1}zxz^{-2}=1 \rangle$.

Now define a new group $G_2$ with generators $a,b,c$, where the three relations are derived by substituting $x=b^{-1}aba^{-2}$, $y=c^{-1}bcb^{-2}$, $z=a^{-1}cac^{-2}$ in the presentation of $G_1$. So $G_2$ has three generators and three relations each of length $25$.

Since $G_1$ is trivial, the elements $b^{-1}aba^{-2}$, $c^{-1}bcb^{-2}$ and $a^{-1}cac^{-2}$ of $G_2$ must be trivial, but then, using the triviality of $G_1$ again, we deduce that $G_2$ is trivial. But I think GAP will struggle to prove that given the presentation of $G_2$.

You can repeat this construction to give groups $G_n=1$ with three generators and three relations, each of length $5^n$.

Derek Holt
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    The sequence of examples here is due to B.H. Neumann. It is mentioned in the GAP manual. Apparently $G_2$ has recently been "cracked" by the coset enumeration program ACE written by Havas and Ramsey. – Derek Holt Feb 03 '14 at 12:25