# Identifying $\langle x_1, x_2, x_3, x_4\mid x_1x_2=x_3, x_3x_2=x_1, x_1x_4=x_3, x_3x_4=x_1\rangle$.

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 $x_2^2=id.$

• Is there any context where this comes up? – Alex Provost Nov 2 '17 at 10:53
• 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. – Shaun Nov 2 '17 at 10:56

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.
• So, it's the free product of $\Bbb Z$ with $C_2$? – Shaun Nov 2 '17 at 10:59