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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?

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    $\begingroup$ 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. $\endgroup$
    – ahulpke
    Jan 27 at 18:45
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    $\begingroup$ Alternatively, to show that $|P| \ge 24$, you just need to find a surjective homomorphism from $P$ to $S_4$, which is not difficule. $\endgroup$
    – Derek Holt
    Jan 27 at 19:15
  • $\begingroup$ @DerekHolt as usual, I was overthinking things. If you post this as an answer, I will accept it. $\endgroup$ Jan 27 at 19:53

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

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