4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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 $(aba)^n = a(ba^2)^{n-1}ba \ne 1$.

$\endgroup$
  • $\begingroup$ What is the general result you mention? $\endgroup$ – Hugh Thomas Apr 27 '18 at 16:10
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.