I am trying to apply what I have just learnt in group theory to find the group symmetries of a trigonal bipyramidal molecule. I know that the atoms forming an equilateral triangle is the dihedral group of order 6. However, how do I find the group of symmetries of the whole molcule i.e. including the three atoms that are collinear? Do I split the molecule up and count them separately and add them up?
The attached image is the molecule I wish to find the group of symmetries of.
There are two kind of symmetries in your molecule: the symmetries of the equilateral triangle (the dihedral group of order 6 = the symmetric group $S_3$) and the vertical reflexion. Since those commute, the group of symmetries is the direct product $S_3 \times \mathbb{Z}/2\mathbb{Z}$.