Problem

Let \(G\) be the group of rigid motions in \(\mathbb{R}^3\) of a tetrahedron. Show that \(|G| = 12\).

Solution

Label the verticies as \(v_1\), \(v_2\), \(v_3\), and \(v_4\). One rigid motion maps the vertext \(v_1\) to one of them. There are \(4\) choices for that.

Then, to complete the motion, one other vertex, say \(v_2\), needs to be mapped to a specific vertex. Given that one vertex is already taken for the previous mapping, there are \(3\) choices for that.

As a result, we end up having \(4 \times 3 = 12\) such rigid motions.

$$\tag*{$\blacksquare$}$$