Which one of the following groups is isomorphic to the group $G\ $? 
Let group $$G = \left \langle a,b,c\ |\ a^2 = b^2 = c^2 = 1, aba = bab, bcb = cbc, ac = ca \right \rangle.$$ Which one of the following groups is isomorphic to the group $G\ $?
(1) $D_8,$ Dihedral group of order $8.$
(2) $\Bbb Z_2 \oplus \Bbb Z_2 \oplus \Bbb Z_2.$
(3) $S_3,$ Symmetric group of order $6.$
(4) $S_4,$ Symmetric group of order $24.$

What I find is that $ab,ba,bc,cb$ are all distinct elements in $G$ of order $3.$ Moreover $ab$ and $ba,$ $bc$ and $cb$ are inverses of each other. Since $3$ doesn't divide $8,$ $G$ can't be isomorphic to $D_8$ or $(\Bbb Z_2 \oplus \Bbb Z_2 \oplus \Bbb Z_2).$ Also since $G$ has at least four distinct elements of order $3,$ it can't be isomorphic to $S_3$ as well. Hence the group $G$ has to be isomorphic to $S_4.$ So the last option is the only correct option.
But I can't able to show it explicitly that $(4)$ is indeed the correct option. Can anybody please check my reasoning above and suggest some technique to actually prove it explicitly?
Thanks in advance.
 A: 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.
A: 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 $n$.
A: That $a^2=b^2=c^2=1$ suggests that we try transpositions. $ac=ca$ suggests most easily that $a=(12),c=(34)$; because $b$ does not commute with either $a$ or $c$, we try $b=(23)$. And indeed these assignments work; because these are transpositions spanning every element of a four-element set, they generate $S_4$.
A: Hint. $S_4$ is spanned by the cycles $(12),(13)$ and $(34)$.
