An epimorphism from $S_{4}$ to $S_{3}$ having the kernel isomorphic to Klein four-group Exercise $7$, page 51 from Hungerford's book Algebra.

Show that $N=\{(1),(12)(34), (13)(24),(14)(23)\}$ is a normal subgroup
  of $S_{4}$ contained in $A_{4}$ such that $S_{4}/N\cong S_{3}$ and
  $A_{4}/N\cong \mathbb{Z}_{3}$.

I solved the question after many calculations. I would like to know if is possible to define an epimorphism $\varphi$ from $S_{4}$ to $S_{3}$ such that $N=\ker(\varphi)$.
Thanks for your kindly help.

Here is an approach:
Proof Idea: $S_4/N$ is a group with 6 elements. There are only two such groups, one is cyclic and the other is $S_3$, and $S_4/N$ cannot have elements of order $6$ thus must be $S_3$.
First it is easy to show that $N$ is normal in $S_4$. It follows that $S_4/N$ is a group with $6$ elements. Let us call this factor $G$. 
Now, $G$ is a group of order $6$. As no element of $S_4$ has order $6$, it follows that $G$ has no element of order $6$.
Pick two elements $x,y \in G$ such that $\operatorname{ord}(x)=2$ and $\operatorname{ord}(y)=3$. Then, $e, y, y^2, x, xy, xy^2$ must be 6 distinct elements of $G$, and hence 
$$G= \{ e, y, y^2, x, xy, xy^2 \}$$
Now, let us look at $yx$. This cannot be $xy$, as in this situation we would have $\operatorname{ord}(xy)=6$. This cannot be $e, x, y, y^2$ either. This  means that 
$$yx=xy^2$$
Now it is trivial to construct an isomorphism from $G$ to $S_3$.
