Let $G$ be the group of rigid motions in $R_3$ of a tetrahedron. Show that $|G |= 12$.($|G|$ refers to the order of $G$)
Proof: We first note that the tetrahedron has $4$ vertices. In order for it to be a rigid motion, we want to preserve the distance between every pair of points. In other words, if one face is adjacent to another, we wish to ensure that after the transformation, the vertices are still adjacent. Take any vertex of the solid. We can "send" this vertex of the solid to any of the $4$ positions. This gives us $4$ options. Take any other vertex that is adjacent to it, we have $3$ other potential options (All vertices are adjacent in a tetrahedron, and one vertex is already "occupied"). Once the position of the $2$ vertices have already been specified, the action on the other vertices is completely determined. Thus we have $4 \times 3=12$ possible rigid motions.
Is this proof too informal? How can it be improved? Thanks for your help!
Best Answer
WLOG fix one side and rotate the $3$ distinct possible ways. This can be done in the same way for each side and this can be represented as a bijection to the set of all rigid motions and so $|G| = 12$. Is this idea not rigorous enough? Although I might be misunderstanding your definition of "rigid motion".