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

The morphism $S_3\to S_4/H,\;g\mapsto gH$ is injective (since $S_3\cap H=\{1\}$) hence bijective by a counting argument.

The composition of its inverse isomorphism $S_4/H\to S_3$ by the canonical epimorphism $S_4\to S_4/H$ gives the epimorphism you looked for.

Its direct explicit definition is : send any $h\in S_4$ to the unique $g\in hH\cap S_3.$

Answer 2

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 $S_4$...

Answer 3

Here's a geometric construction.

The group of rotations of a cube is isomorphic to $S_4$. Indeed, you can view it as the action on the four diagonals of the cube.

(I took the picture from this answer)

Under this action, 180 degree rotations correspond to the non-trivial elements of the Klein 4 group.

Now consider the set of pairs of opposite faces of a cube. There are three such pairs. Labelling the pairs as $\{1 , 2, 3\}$ we get an action of $S_4$ on $\{1,2,3\}$, in other words, a homomorphism $S_4 \to S_3$.

The kernel of this homomorphism is exactly the rotations of the cube that fix each of the three pairs of opposite faces - i.e. it is exactly the Klein 4 group.