Identifying $\langle x_1, \dots, x_6\mid x_1x_2=x_3, x_2x_3=x_1, x_3x_1=x_2, x_4x_5=x_6, x_5x_6=x_4, x_6x_4=x_5\rangle.$

Question

Consider the presentation $$G:=\langle x_1, \dots, x_6\mid x_1x_2=x_3, x_2x_3=x_1, x_3x_1=x_2, x_4x_5=x_6, x_5x_6=x_4, x_6x_4=x_5\rangle.$$

What group is it?

Thoughts:

Calculations in GAP show that the group $G$ is infinite; namely, that the abelian invariants of the derived subgroup of $G$ are [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2 ].

Please help :)

Answer 1

It is visibly a free product $Q*Q$ of two copies of the group $Q = \langle a,b,c \mid ab=c, bc=a, ca=b \rangle$, and $Q$ is isomorphic to the quaternion group $Q_8$.