$G=\langle a,b\mid aba=b^2,bab=a^2\rangle$ 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. :)
 A: $abab=a^3=b^3$, so $Z := \langle a^3 \rangle$ is central. Modulo $Z$, we get the standard presentation $\langle a,b \mid a^3, b^3, (ab)^3 \rangle$ of $A_4$. Also, module $G'$, we have $a^2=b$, $b^2=a$, so $a^3=1$, and hence $Z \le G'$. Also, $ab,ba \in G'$ and $abba = a^2ba^2=bab^3ab=baabb^3$, so $G'$ is not abelian provided that $Z$ is nontrivial.
So to prove the group is not metabelian we need to prove that $Z$ is nontrivial, and the only sensible way of doing that, other than by coset enumeration, which is very tedious to do by hand, is to find an explicit homomorphic image of the group in which $Z$ is nontrivial. Knowing that $G$ is a nonsplit central extension of $Z$ by $A_4$, we might suspect at this stage that $G \cong {\rm SL}_2(3)$, which might help us find an explicit map, like the one described by Jack Schmidt in his comment.
A: 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.
