I am given the following group: $$G = \langle x_1,x_2,x_3 | x_1^2 = x_2^2 = x_3^2 = e, \langle x_1, x_2 \rangle = \langle x_2, x_3 \rangle = e \rangle,$$ where $\langle a,b \rangle = e$ is the triple relation, meaning that $$aba = bab.$$

I want to prove that the element $x_1 x_3 x_1 x_2$ has infinite order. This seems to me rather trivial, since (intuitively speaking) the triple relations in $G$ cannot reduce the number of generators in $(x_1 x_3 x_1 x_2)^n$.

However, when trying to prove this formally, I could not finish the proof. I tried using induction, playing with the evenness of the number of the generators $x_1, x_2, x_3$ that appear in any power of this element, but without success.

  • $\begingroup$ Note that your intuitive argument would still apply even if we had the additional relations $e_1e_3=e_3e_1$. But in that case, the group is the symmetric group on four elements, so it is finite. Second point: the group you are looking at is a Coxeter group; this might be helpful. $\endgroup$ – Hugh Thomas Apr 27 '18 at 16:08

This is a Coxeter group.

Let $a = x_1x_2$ and $b = x_2x_3$. Then the subgroup $H:=\langle a,b \rangle$ has index $2$ in $G$ and has the presentation $\langle a,b \mid a^3=b^3=1 \rangle$. That is just an instance of a general result for Coxeter groups, but you could prove it by direct calculation in this example.

So $H$ is a free product of the cyclic groups $\langle a \rangle$ and $\langle b \rangle$ of order $3$. Your element is $aba$ and $(aba)^n = a(ba^2)^{n-1}ba \ne 1$.

  • $\begingroup$ What is the general result you mention? $\endgroup$ – Hugh Thomas Apr 27 '18 at 16:10
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    $\begingroup$ @HughThomas: That there is always an index 2 subgroup: the kernel given by identifying all generators. $\endgroup$ – Steve D Apr 27 '18 at 16:11
  • $\begingroup$ How do you conclude that $a(ba^2)^{n-1}ba \ne 1$? Maybe $a$ and $b$ satisfy some further relations in $G$. $\endgroup$ – Mike Apr 27 '18 at 16:30
  • $\begingroup$ It follows immediately from the fact that $H$ is a free product of two cyclic groups of order $3$. So the element has infinite order in $H$ and hence also in $G$. $\endgroup$ – Derek Holt Apr 27 '18 at 16:55
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    $\begingroup$ But $a(ba^2)^{n-1}ba = aba^2b \cdots ba^2ba$ is an alternating product of nontrivial elements from the two free factors, so it is manifestly nontrivial. $\endgroup$ – Derek Holt Apr 27 '18 at 20:38

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