1
$\begingroup$

What group is $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_3x_2=x_1, x_1x_4=x_3, x_3x_4=x_1\rangle$?

Thoughts:

The group is infinite according to GAP, $x_2=x_4$, and $x_2^2=id.$

$\endgroup$
  • $\begingroup$ Is there any context where this comes up? $\endgroup$ – Alex Provost Nov 2 '17 at 10:53
  • $\begingroup$ Yes, @AlexProvost: use Cayley's theorem to convert $C_2\times C_2$ into a group of permutations, namely $\{id., (13), (24), (13)(24)\}$, then permute the generators according to these permutations. $\endgroup$ – Shaun Nov 2 '17 at 10:56
2
$\begingroup$

Setting $x_3:=x_1x_2$ the relations become $x_1x_2^2=x_1$, so $x_2^2=1$, and $x_4=x_1^{-1}x_1x_2=x_2$. The last relation $x_3x_4=x_1$ then just says $x_1=x_1$. So, writing $x_1=a,x_2=b$ the group is given by $$ \langle a,b \mid b^2=1\rangle, $$ and now you see it.

$\endgroup$
  • $\begingroup$ So, it's the free product of $\Bbb Z$ with $C_2$? $\endgroup$ – Shaun Nov 2 '17 at 10:59
  • $\begingroup$ Right - in contrast to this group. $\endgroup$ – Dietrich Burde Nov 2 '17 at 11:58

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.