Prove that the group $$ G = \langle a,b,c \mid a^2=b^2=c^2=1,\, ac=ca,\, (ab)^3=(bc)^3=1 \rangle $$ is isomorphic to $S_4$.
Obviously, take $a=(12),\, b=(23),\, c=(34)$ and use Von Dyck's Theorem to get an epimorphism $G \twoheadrightarrow S_4$.
However, I met some problem trying to proof $|G|=24$. I feel $abc$ would be an element with order $4$, but I don't know how to say it.
Let $H$ be the subgroup generated by $b$ and $c$. We will show that $G = H \cup H a \cup H ab \cup H abc$. It suffices to prove that $H \cup Ha \cup Hab \cup Habc$ is closed under right-multiplication by $a,b,c$. There are a lot of things to check but they are all easy, using the relations, especially $aba=bab$ and $bcb = cbc$.
Multiplication by $a$: $Ha = Ha$, $Ha^2 = H$, $Haba = Hbab = Hab$, $Habca = Habac = Hbabc = Habc$.
Multiplication by $b$: $Hb = Hb$, $Hab = Hab$, $Hab^2 = Ha$, $Habcb = Hacbc = Hcabc = Habc$.
Multiplication by $c$: $Hc = Hc$, $Hac = Hca = Ha$, $Habc = Habc$, $Habc^2 = Hab$.
This shows that $H$ has index at most $4$ in $G$. Under the known map $G \to S_4$, $H$ maps to the stabilizer of $1$, which has index $4$, so $[G:H] = 4$.
Now one can use a similar approach with $H = \langle b, c \rangle$, given that $b^2 = c^2 = (bc)^3 = 1$. Let $K = \langle c \rangle$. The claim is that $H = K \cup Kb \cup Kbc$. It's just like the above and I will leave it to you. It follows that $[H:K] \le 3$, and comparison with the known map $G \to S_4$ shows that $[H:K]=3$.
Finally $K = \langle c \rangle = \{1, c\}$ obviously has order at most $2$, and since the known map $G \to S_4$ maps $c$ to a nontrivial element $(3,4)$ we know that $|K| = 2$.
As mentioned in the comments, this is a special case of a general procedure called coset enumeration. There is a chapter about it in Rotman's book An introduction to the Theory of Groups.