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The group $G$ generated by $x,y,z$ subject to the relations $[x,y]=y$, $[y,z]=z$, $[z,x]=x$ is trivial. This isn't the case for the corresponding group with 4 generators, which is the famous Higman group. I know several direct algebraic proofs that $G$ is trivial. But they are all quite long and fiddly. I am not looking for such proofs here, so please don't post them.

Question. Is there a geometric and good motivated proof that $G$ is trivial?

For example is it possible to realize $G$ as the fundamental group of a nice space which is simply-connected by geometric reasons? Can we use the Cayley graph of $G$, or a faithful action of $G$ on some nice space, etc.?

EDIT: What about a proof using the van Kampen diagram?

Martin Brandenburg
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    I don't understand the question. Since $G$ is in fact trivial, how can its Cayley graph help, and how can it have an interesting action? – Derek Holt Aug 30 '13 at 11:01
  • @DerekHolt One could interpret the question as "is there a nice geometric argument what the 4-generator case is infinite (non-trivial) but this fails for three-generators." (Having given this interpretation some thought over the past hour, I would be interested in seeing a solution!) – user1729 Aug 30 '13 at 11:09
  • @Derek: These were just naive ideas. – Martin Brandenburg Aug 31 '13 at 17:23
  • @user1729: Yes this is an interesting interpretation. – Martin Brandenburg Aug 31 '13 at 17:24
  • @MartinBrandenburg Have you looked up Higman's original paper? Essentially he builds up the group using free products with amalgamation and HNN-extensions, which have a geometric interpretation (Siefert-van Kampen, or they act on a tree). The construction proves that they're infinite for an even number of generators. So that might satisfy you. But...the $4$ rings a curvature/small cancellation bell... – user1729 Aug 31 '13 at 17:59
  • I know (the contents of ) Higman's original paper, but it doesn't answer my question. – Martin Brandenburg Sep 04 '13 at 08:36
  • @Derek: since you don't initially know that that the Cayley graph is trivial, you could imagine a geometric proof based on the initial assumption that this graph is not trivial. Isn't it the principle of any proof by contradiction? Anyway I don't know of such a proof. – YCor Sep 22 '13 at 19:58
  • What are you looking for in a geometric proof? For example you mention using van Kampen diagrams, and I am sure one could "van Kampenfy" a known proof, but I have a feeling it would not be any more illuminating. I have been playing with this, hoping for an to come across an illuminating idea or strategy, and in playing with van Kampen diagrams, I believe I came across a proof (will have to double check it), but it is essentially just sticking relations together, and I don't have an overarching idea that makes it illuminating –  Jun 14 '15 at 00:31
  • I did not realize I sent the above comment, I was actually going to double check it before sending it, and I did find a mistake. –  Jun 14 '15 at 00:54
  • I hope that a geometric approach offers a visible strategy for finding the proof and is therefore much easier than any random algebraic manipulation. – Martin Brandenburg Jun 14 '15 at 08:47

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