Explicit isomorphism $S_4/V_4$ and $S_3$ [duplicate]

Let $S_4$ be a symmetric group on $4$ elements, $V_4$ - its subgroup, consisting of $e,(12)(34),(13)(24)$ and $(14)(23)$ (Klein four-group). $V_4$ is normal and $S_4/V_4$ if consisting of $24/4=6$ elements. Hence $S_4/V_4$ is cyclic group $C_6$ or a symmetric group $S_3$ (really, there are only two groups consisting of $6$ elements). It is easy to see, that an order of each element of $S_4$ is $1,2,3$ or $4$. So, $S_4/V_4$ is isomorphic to $S_3$.

My question: how to build the isomorphism explicitly?

Let $A = \{a = (12)(34), b = (13)(24), c = (14)(23)\}$ (it's a conjugation class from $S_4$). Let $S(A)$ be the group of permutations of $A$.

$S_4$ acts by conjugation on $A$ : if $\sigma \in S_4$ and $a \in A$, $\sigma.a = \sigma a \sigma^{-1} \in A$. This gives a group morphism $S_4 \to S(A)$.
Moreover, because $V_4$ is commutative and $A \subset V_4$, if $\sigma \in V_4$ then $\sigma.a = a$, hence $\sigma$ acts trivially, and so the kernel of that map contains $V_4$.

Next, look at the action of a transposition, say $\tau = (12)$. You should find that $\tau.a = a, \tau.b = c, \tau.c = b$. Similarly, the other transpositions of $S_4$ should acts as the other transpositions of $S(A)$.

Since $S(A)$ is generated by transpositions, this means that the map $S_4 \to S(A)$ is surjective, so its kernel has cardinality $4$, but it contains $V_4$ so it has to be $V_4$, and finally, this gives an isomorphism $S_4 / V_4 = S(A)$