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.

Yes, there is. My favorite way of doing that is the following. There are exactly three ways of partitioning the set $\{1,2,3,4\}$ to two disjoint pairs, namely $$ P_1=\{\{1,2\},\{3,4\}\},\quad P_2=\{\{1,3\},\{2,4\}\},\quad\text{and}\quad P_3=\{\{1,4\},\{2,3\}\}. $$ Now given a permutation $\sigma\in S_4$ it acts naturally on the set $\{P_1,P_2,P_3\}$ of such partitions "elementwise", and thus gives us a permutation $\overline{\sigma}\in Sym(\{P_1,P_2,P_3\})$. This correspondence $\sigma\mapsto \overline{\sigma}$ is (one of) the epimorphism(s) you are looking for.

More details: $\overline{\sigma}$ takes the partition $P_1$ to the partion $\{\{\sigma(1),\sigma(2)\},\{\sigma(3),\sigma(4)\}\}$ and similarly for the others. For example, when $\sigma=(234)$ we get that $$ \begin{aligned} \overline{\sigma}(P_1)&=\{\{1,3\},\{4,2\}\}=P_2,\\ \overline{\sigma}(P_2)&=\{\{1,4\},\{3,2\}\}=P_3,\\ \overline{\sigma}(P_3)&=\{\{1,2\},\{3,4\}\}=P_1,\\ \end{aligned} $$ so the resulting permutation is the 3-cycle $P_1\mapsto P_2\mapsto P_3\mapsto P_1$.

It is tedious but straightforward to check that the mapping $\sigma\mapsto \overline{\sigma}$ is surjective. It is a bit easier to check the all the permutations of the subgroup $N$ leave all the partitions $P_j,j=1,2,3,$ invariant.

Take a tetrahedron. The symmetries form the group $S_4$ on the vertices. Consider the action of these symmetries on the three pairs of opposite edges of the tetrahedron. i.e. Pair 1 - edges 12, 34; Pair 2 edges 13, 24; pair 3 edges 14, 23.

I'll leave you to work out the details. The other platonic solids also give some geometric realisations of other relationships between groups.

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 $S_3$.