I would like to show $$P :=\langle a,b \, | \, a^2,b^4,(ab)^3\rangle \cong S_4.$$ I'll assume $$S_4 \cong G:= \langle x,y,z \, | \, x^2,y^2,z^2, (xy)^3, (yz)^3, (xz)^2 \rangle.$$
Let $\phi: G \to P$ be defined by $x \mapsto a, y \mapsto bab^3, z \mapsto b^2ab^2$. It is easy enough to show $\phi$ is a well-defined, surjective homomorphism.
I'm struggling to show that $\phi$ is injective. One could begin, 'suppose $\phi(k)=e$, then $\phi(k)$ is a reduced word on $a^2,b^4$, and $(ab)^3$...' but that does not seem feasible (you would need to prove that the only words on $x,y,z$ mapping to words on $a^2,b^4,(ab)^3$ are actually words on the relations.)
What I suspect to be an easier proof is to show $|G| = 24 \leq |P|$. This would suffice, for $\phi$ is surjective.
Now my problem is that I'm unsure how to handle the order of $P$. Actually, I'm not even sure this is easier, since I know that in general it is hard to determine the order/structure of a group presentation. In this case, is there a nice technique for getting a hold of $|P|$, and if there is not, is there another proof method I'm overlooking?