Which one of the following groups is isomorphic to the group $G\ $?
Hint: Try letting $a=(12),b=(23)$ and $c=(34)$. Try to prove that $S_4=\langle a,b,c\rangle$, and that they satisfy the given relations.
As I have said in my comments, the other answers show that there is a surjective homomorphism $G \to S_4$. To prove that this is an isomorphism, it is enough to prove that $|G| \le 24$, which can be done as follows.
Let $H = \langle a,b \rangle$ be the subgroup of $G$ generated by $a$ and $b$. I am assuming you know that $\langle a,b \mid a^2=b^2=1, aba=bab \rangle$ defines the dihedral group of order $6$, so $|H| \le 6$, and it is enough to prove that $|G:H| \le 4$.
To do that, we shall prove that $G = H \cup Hc \cup Hcb \cup Hcba$, and to do that it is sufficent to prove that if we multiply any of these cosets by any of the generators $a,b,c$ of $G$, then we will get another one of these cosets. So let's do that.
- Multiply coset $H$ by $a,b,c$: $Ha=H$, $Hb=H$, $Hc=Hc$.
Multiply coset $Hc$ by $a,b,c$: $Hca=Hac=Hc$, $Hcb=Hcb$, $Hcc=H$.
Multiply coset $Hcb$ by $a,b,c$: $Hcba=Hcba$, $Hcbb=Hc$, $Hcbc=Hbcb=Hcb$.
Multiply coset $Hcba$ by $a,b,c$: $Hcbaa=Hcb$, $Hcbab=Hcaba=Hacba=Hcba$, $Hcbac=Hcbca=Hbcba=Hcba$.
This proof generalizes to similar presentations of $S_n$ for all $n$.