I am trying to give an explicit description of the group $$G=\langle a,b,c\mid[a,b]=b\,,\,[b,c]=c\,,\,[c,a]=a\rangle\,.$$ Generalizing to fewer generators, one ends up with the trivial group, i.e. $$G_0=\langle\,\rangle\cong G_1=\langle a\mid [a,a]=a\rangle\cong G_2=\langle a,b\mid[a,b]=b\,,\,[b,a]=a\rangle\cong 1\,.$$ But I don't see a reason, why this should hold for $G=G_3$ or $G_n$, $n\in\mathbb{N}$.

Edit: I am sorry, I thought the symbols were standard. $[a,b]$ is defined as $aba^{-1}b^{-1}$. This makes it a little less trivial.

  • $\begingroup$ Your edit does not change the argument given in the answer. $\endgroup$ – Tobias Kildetoft Jun 22 '14 at 12:41
  • $\begingroup$ @Tobias Kildetoft: It does! $aba^{-1}b^{-1}=b$ does not imply $a=b=1$. We will need to use all three relations. $\endgroup$ – Jakob Werner Jun 22 '14 at 12:43
  • $\begingroup$ Ahh, right. I got the signs mixed up in my head. $\endgroup$ – Tobias Kildetoft Jun 22 '14 at 12:45
  • $\begingroup$ Asking for an "explicit description" is a rather vague question. Can you be more explicit about what kind of explicitness you want? $\endgroup$ – Lee Mosher Jun 22 '14 at 15:12
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    $\begingroup$ In fact the group is trivial, so it is automatic, word hyperbolic, etc. The $4$-generator version, known as the Higman group, is infinite, and is much more interesting. $\endgroup$ – Derek Holt Jun 22 '14 at 22:31

If your definition of the commutator $[g,h]=g^{-1}h^{-1}gh$ then this presentation is the trivial group.

EDIT / REMARK If the commutator is defined as $[g,h]=ghg^{-1}h^{-1}$ (and this is what is meant following all comments), then surprisingly, the group $G$ is still trivial. This is Exercise 1 in Jean-Pierre Serre's famous book Trees, pag. 10, in the chapter that deals with Amalgams.

  • $\begingroup$ Here, $[g,h]$ is defined as $ghg^{-1}h^{-1}$. $\endgroup$ – Jakob Werner Jun 22 '14 at 12:32
  • $\begingroup$ Ok that makes a big difference! $\endgroup$ – Nicky Hekster Jun 22 '14 at 13:28
  • $\begingroup$ Thank you for the reference. Do you know a proof of this result, or a place where I can find one? $\endgroup$ – Jakob Werner Jun 24 '14 at 14:36
  • $\begingroup$ @JakobWerner: matheplanet.com/matheplanet/nuke/html/… $\endgroup$ – HeinrichD Apr 5 '17 at 9:06

Since $a^{-1} b^{-1} ab = b$, by left multiplying both sides by $a$, we get $b^{-1} ab = ab$, whence $b^{-1}=1$. Thus, $b=1$. Similarly, we get that $a$ and $c$ also equal the identity.


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