Counting the size of a supposed presentation of $S_4$

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?

• Find elements $c,d\in G$ that map to $a$, respectively $b$, and evaluate the relators for $P$ in these elements $c,d$. These will be normal subgroup generators for the kernel. Jan 27 at 18:45
• Alternatively, to show that $|P| \ge 24$, you just need to find a surjective homomorphism from $P$ to $S_4$, which is not difficule. Jan 27 at 19:15
• @DerekHolt as usual, I was overthinking things. If you post this as an answer, I will accept it. Jan 27 at 19:53

To show that $$|P| \ge 24$$, you just need to find a surjective homomorphism fro $$P$$ to $$S_4$$ such as $$a \mapsto (1,2)$$, $$b \mapsto (1,2,3,4)$$.