Find the Symmetry Group of : Tetrahedron and Cube.
I know that there is a duplicate question Symmetry group of Tetrahedron
but my professor suggested us to write coordinates of both cube and tetrahedron in $\mathbb R^4$ and after see what happen. My question is:
how can I write coordinates of cube and tetrahedron in this way?
I don't know where to start this problem sorry if I have not published my attempt.
The four permutations of $(1,0,0,0)$ form a tetrahedron. For a matrix $P$ to turn basis vectors into basis vectors, each column of $P$ must be a basis vector, and if $P$ is invertible they must be distinct, which means $P$ must be a permutation matrix. Conversely all $4!=24$ permutations work as isometries.
A cube in 3D may be described as the points $(\pm1,\pm1,\pm1)$. This has two inscribed tetrahedra, corresponding to the vectors with either an even or odd number of minus signs. A cube in 4D may be described as the points $(\pm1,\pm1,\pm1,\pm1)$ with an even number of minus signs - note then the subsets of points $(1,\pm1,\pm1,\pm1)$ and $(-1,\pm1,\pm1,\pm1)$ within this form two tetrahedra. The $24$ permutations still work as isometries, but now we may include multiplication by $-1$ as an isometry for a total of $48$ (which is the correct number, as we can work out using the orbit-stabilizer theorem). This makes the symmetry group $S_4\times C_2$.