$G=\langle a,b\mid aba=b^2,bab=a^2\rangle$ is not metabelian of order $24$ This is my self-study exercise:

Let $G=\langle a,b\mid aba=b^2,bab=a^2\rangle$. Show that $G$ is not metabelian.


I know; I have to show that $G'$ is not an abelian subgroup. The index of $G'$ in $G$ is 3 and doing Todd-Coxeter Algorithm for finding any presentation of $G'$ is a long and tedious technique (honestly, I did it but not to end). Moreover GAP tells me that $|G|=24$. May I ask you if there is an emergency exit for this problem. Thanks for any hint. :)

We have a homomorphism $a\rightarrow (1,2)(3,4,8,5,6,7)$, $b\rightarrow (1,3,6,2,5,4)(7,8)$. The image has order 24 so it is an isomorphism by your T-C result.
The kernel $G'$ of the map to $Z_3$ is generated by $ab, ba$ which is quaternion of order 8; so the group is not metabelian.
