An “easy” way to construct an epimorphism from S4 to S3

Question

I'm trying to construct an epimorphism φ from S4 to S3 such that:

H = ker(φ) = {(1),(12)(34),(13)(24),(14)(23)}

where H is a normal subgroup of S4, contained in A4 and isomorphic to the Klein 4-group.

I've tried to look up the question here and I've found out some similar threads like these ones:

• An epimorphism from S4 to S3 having the kernel isomorphic to Klein four-group

• Constructing a homomorphism from S4 to S3 that satisfies specific conditions

but if it's possible I'm looking for a more immediate and natural way to construct it.

Does anyone have some ideas?

Thanks in advance for your kindly help.

Editing:

Now it's clear how to construct the required morphism.

What about finding it without using that H is normal?

Answer 1

Here is a suggestion: to get such a map, what you want is a group action of $S_4$ on three things. It would be natural to look at the left-multiplication action on cosets of a subgroup of index 3/order 8 in $S_4$. The dihedral subgroup $D_8$ of $S_4$ generated by $(1, 2, 3, 4)$ and $(1, 3)$ has order 8 (it consists of $V_4$ together with the elements $(1, 2, 3, 4), (1, 3), (2, 4)$, and $(1, 4, 3, 2)$), and it will work.

You need to show that you can realise any permutation of the three cosets of $D_8$ as left-multiplication by an element of $S_4$, and you need to check that the kernel of the action is $V_4$. To get started, find a complete set of coset representatives of $D_8$ in $S_4$...