What are all the elements of the group of symmetries of a regular tetrahedron?

Question

I can see that why the order of the group of symmetries of a regular tetrahedron is $12$ : Roughly speaking, each time one of ${\{1,2,3,4}\}$ is on 'top' and we do to the other $3$ as we did in a regular triangle. But I need to know the elements of the group and I don't have not only a physical regular tetrahedron but also my imagination is not strong. Videos like $1$ or $2$ didn't help, because I can't understand for example what happens to the other $2$ points when the top point and one of the surface points is exchanged.

I would appreciate any simple clear explanation.

Added - For next permutations (with order $3$) now simply we consider the ‘surface’ under the other three-points, i.e. each time $4$ on top is replaced by $1$ or $2$ or $3$. And, we perform rotations as we were doing when $4$ was on top. So we will get:

‘$4$’ on top : ${\{(123),(132)}\}$

‘$1$’ on top : ${\{(234),(243)}\}$

‘$2$’ on top : ${\{(134),(143)}\}$

‘$3$’ on top : ${\{(124),(142)}\}$.

But what about $(12)(34)$, $(13)(24)$ and $(14)(23)$? How to get them?

Answer 1

One of them

Use different axes (connecting the midpoints of two opposite edges of the tetrahedron) to get the other products of two disjoint 2-cycles as 180 degree rotations.

Here is the same animation with the cube surrounding the tetrahedron shown as a wireframe. The axis of rotation joins the centers of two opposite faces of the cube

Answer 2

Consider the set of half edges of the tetrahedron. It is easy to convince yourself that the symmetry group of the solid allows you to move from one such half edge to any other in exactly one way. Since there are six edges, there are twelve half edges, and this implies that the group of symmetries has exactly 12 elements.

Now the group obviously permutes the four vertices. Since there is exactly one subgroup of order 12 in $S_4$, namely the alternating group, it must be our group.