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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:

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?

Simone Fiorio
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3 Answers3

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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 $S_4$...

Matthew Towers
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  • Sounds really interesting, I'll go through it. As a beginner in the subject I'm still wondering how I can prove that this map is actually a morphism... – Simone Fiorio Sep 15 '22 at 13:54
  • you can do it directly: once you figure out exactly what you're trying to prove, it will simply be the associativity property for $S_4$. – Matthew Towers Sep 15 '22 at 14:08
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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.

enter image description here

(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.

Mathmo123
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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 $g\in hH\cap S_3.$

Anne Bauval
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  • It definitely works but how can I prove that each class in S4/H really contains a unique element of S3? I mean, I'm quite conviced about that but it seems a bit "euristic" as an approach – Simone Fiorio Sep 15 '22 at 14:03
  • Now it's perfect, thanks a lot – Simone Fiorio Sep 15 '22 at 14:42
  • I inserted another proof that each class contains a unique element of $S_3$. I think it is better. I want to erase the first one. – Anne Bauval Sep 15 '22 at 14:46
  • What can I do if I want to deduce that H is normal by the fact that it is the kernel of a morphism of groups? I mean, is there any way to adapt this demonstration in order not to assume that H is normal and deducing it in retrospect after finding the ker of the morphism? – Simone Fiorio Sep 15 '22 at 14:59
  • This is not exactly what you ask for, but may please you. – Anne Bauval Sep 15 '22 at 15:11
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    I was able to prove that H is normal in S4 by using conjugation definition, mine was just a curiosity. If there's no easy way to construct a map from S4 to S3 without assuming H is normal it's not a problem, your previous answer is fine. – Simone Fiorio Sep 15 '22 at 15:20