THE* SURJECTIVE MAP FROM S4 ONTO S3

* Unique up to the action of \(\text{Aut}(S_3)\).

The group \(S_4\) is the group of permutations on a \(4\)-element set. A particularly nice case is to take the \(4\)-element set to be the vertices of a regular tetrahedron. Then \(S_4\) acts on the regular tetrahedon by permuting the vertices. A visualisation of each of these permutations can be seen below.

The group \(S_3\) is the group of permutations on a \(3\)-element set. One may also see this as symmetries of a regular tetrahedon, but is it not immediately obvious how. It turns out that \(S_3\) acts on a regular tetrahedron by permuting pairs of opposite edges of the tetrehedron. Once again, these permutations can be visualised below, with each pair of opposite edges being assigned a different colour.

In short, this is 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 \(S_3\).

cycle notation v.s. one-line notation

\(S_4\) acts on the vertices of the regular tetrahedron. These vertices are labelled \(1,2,3,4\), 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 that \( S_4\) acts on the set \( \{1,2,3,4 \} \) and \(S_3\) acts on the set {a, b, c} as above. The explicit map is as follows (using chosen element notation type above):

Observe that \(4\) elements are 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 \( S_4 \big/K_4 \cong S_3 \).