what is the the shape of an irregular tetrahedral?

Answer 1

The isometries of an irregular tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3-dimensional point group is formed.

An equilateral triangle base and isosceles (and non-equilateral) triangle sides gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C3v, isomorphic to S3.
Four congruent isosceles (non-equilateral) triangles gives 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D2d.
Four congruent scalene triangles gives 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-group V4 ≅ Z22, present as the point group D2.
Two pairs of isomorphic isosceles (non-equilateral) triangles. This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C2v, isomorphic to V4.
Two pairs of isomorphic scalene triangles. This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C2 isomorphic to Z2.
Two unequal isosceles triangles with a common base. This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group Cs isomorphic to Z2.
No edges equal, so that the only isometry is the identity, and the symmetry group is the trivial group.