I thought that maybe it would be possible to answer this question using the concept of a group presentation.

Let $x_1,x_2,\ldots,x_k$ be k different elements of a group G and $k\geq4$. If we know that $x_i$ commutes with $x_{i+1}$ and $x_k$ commutes with $x_1$, can we say that all $x_i$ commute with each other ?

Thinking in terms of group presentations, we can ask: does the group

$$\langle x_1, \ldots, x_k \mid (\forall i<k)\ x_ix_{i+1}=x_{i+1}x_i,\ x_kx_1=x_1x_k \rangle$$

satisfy $x_ix_j$ for all $i,j$ ? In other words, does the normal subgroup generated by $[x_i, x_{i+1}]$ for each $i<k$ and $[x_k, x_1]$ contain $[x_i, x_j]$ for each $i, j$ ?

But how could we prove or disprove that? Are there any methods that can be used to work out if a certain element is in a certain generated normal subgroup?

  • 1
    $\begingroup$ I believe in this particular case the answer is 'no' and the easiest way to show it is to find a group that satisfies those relations but $x_1x_3\neq x_3x_1$ like the automorphism group of a tiling of the hyperbolic plane with orthogonal polygons. For some level of generality, this is a very difficult, in fact undecidable problem. But this particular presentation has nice properties: I think the group is hyperbolic and hence automatic. See the book 'Word processing in groups' by Epstein et. al. (coauthored by @DerekHolt btw). $\endgroup$ – Myself Jan 17 '15 at 23:24
  • $\begingroup$ @Myself Providing counterexamples was the type of answer provided on the original question - I'm specifically interested in "direct" approaches with the presentation. $\endgroup$ – Jack M Jan 17 '15 at 23:27
  • $\begingroup$ That won't work I'm afraid, at least not in general. The situation is entirely comparable to the relation between proof theory and model theory (maybe even an instance of it). You cannot know from the proof theory if no proof exists because you have not sought hard enough or because the statement is undecidable. Maybe some logician could explain you precisely why that is. But sometimes, like here, things are easier for instance because your group is automatic, or your (in model theory) the theory admits quantifier elimination etc. $\endgroup$ – Myself Jan 17 '15 at 23:39
  • $\begingroup$ @JackM: I do not really understand your question, but the group you consider is just the right-angled Artin group associated to a cycle of length $k$. $\endgroup$ – Seirios Jan 19 '15 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.