The symmetric group \(S_4\) is the group of permutations on a 4-element set. A nice geometric realisation is to take the 4-element set to be the vertices of a regular tetrahedron. Then \(S_4\) acts on the regular tetrahedron by permuting the vertices.
The symmetric group \(S_3\) is the group of permutations on a 3-element set. This can also be realised via symmetries of a regular tetrahedron, but not in an immediately obvious way. It turns out that \(S_3\) acts on a regular tetrahedron by permuting pairs of opposite edges of the tetrahedron.
In short, these realisations describe the surjective map! The group \(S_4\) permutes the vertices of the tetrahedron. This induces a permutation on the pairs of opposite edges (of which there are three), and this induced permutation is (by definition) an element of \(S_3\).
\(S_4\) acts on the vertices of the regular tetrahedron. These vertices are labeled \(1, 2, 3, 4\). So \(S_4\) acts on the set \(\{1, 2, 3, 4\}\).
\(S_3\) acts on pairs of opposite edges. The pairs of edges are coloured red, blue, and green. They will be referred to as a, b, and c respectively. So \(S_3\) acts on the set {a, b, c}.
We keep the notation of \( S_4\) acting on the set \( \{1,2,3,4 \} \) and \(S_3\) acting on the set {a, b, c}, just as above. Explicit images of each element are as follows:
Observe that there are 4 elements in the kernel of the map from \(S_4\) to \(S_3\). These four elements form a subgroup of \(S_4\) isomorphic to \(C_2 \times C_2 \), which is often denoted by \( K_4 \). Explicitly, this subgroup is:
Therefore, by the First Isomorphism Theorem, we have \( S_4 \big/K_4 \cong S_3\).
In fact, the collection of elements in the top row of each grouping of elements (formally, each \(K_4\) coset) yields a copy of \(S_3\) inside \(S_4\). Therefore, the surjective map splits and \(S_4 \cong K_4 \rtimes S_3\).
