$G=\langle a,b\mid aba=b^2,bab=a^2\rangle$ 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. :)

-
You can (and should) use \mid as a separator rather than | and a space. – Asaf Karagila Nov 18 '12 at 8:33
I don't see right now how to correct the mistake you were well to point out in my answer, so I have deleted it for now. – Asaf Karagila Nov 18 '12 at 12:27
I can undelete. I just didn't want people to not-think about this because they see an upvoted solution. – Asaf Karagila Nov 18 '12 at 13:27
You're welcome. Please post a solution when you finish. – Asaf Karagila Nov 18 '12 at 14:02
I don't see any particularly intelligent way to do this. The group G is isomorphic to SL(2,3) via $$a \mapsto \left(\begin{smallmatrix}-1&0\\-1&-1\end{smallmatrix}\right), b \mapsto \left(\begin{smallmatrix}-1&-1\\0&-1\end{smallmatrix}\right)$$ and SL(2,3) is clearly not metabelian. The subgroup generated by {ab,ba} is a quaternion Sylow 2-subgroup of order 8, contained in the derived subgroup of G, so that also works, but it seemed tedious to prove any basic relations amongst a and b from the presentation given. – Jack Schmidt Nov 18 '12 at 16:02

$abab=a^3=b^3$, 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.
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.