A symmetry of a figure $\mathcal{H}$ in $\mathbb{R}^3$ is an isometry which keeps $\mathcal{H}$ invariant. The set of all symmetries of $\mathcal{H}$ obviously form a group. We know that the symmetry group of a regular tetrahedron is isomorphic to $S_4$, which consists of 24 elements.
My question is: ''What are possible symmetry groups of a tetrahedron (possibly nonregular)?''. I can show that a symmetry of a tetrahedron will send vertices to vertices, so its symmetry group is a subgroup of $S_4$. However it seems that not all subgroups of $S_4$ can be realised as the symmetry group of a tetrahedron. For instance, I tried to find tetrahedrons whose symmetry groups are of order 3, 4, 8 and 12 but I didn't succeed.
Can anyone help me? Thanks a lot!
Symmetry group of order $3$ cannot happen. Let tetrahedron have vvertices $1,2,3,4$. Subgroups of order $3$ of $S_4$ are generated by three-cycles $(a,b,c)$. WLOG we can assume it is $(1,2,3)$. Then the edge $[13]$ should go to $[21]$ under a symmetry which must fix $4$. Therefore edges $[14].[24],[34]$ have equal lengths and $[12],[23]$ have equal lengths. But then the permutation $(13)$ is a symmetry and the group of symmetries must have even order.