Using the simulation website below, you can see the symmetry elements (axes and planes) of the molecule ethane.
http://symmetry.otterbein.edu/gallery/index.html
I am trying to construct a Cayley table given these symmetry elements to show that the point group of this molecule is, in fact, a group (obeys closure property).
This is how I wrote the group U, of the symmetry elements of the molecule:
Here is my attempt to construct a Cayley table for this. I realized this was wrong as C3*C2= C5/6, which is not part of the original set of symmetry elements, meaning the algebraic structure is not closed and therefore not a group.
I think the problem is a misinterpretation of the symmetry element. I look forward to your answer!
The group of symmetries looks like $D_3 \cong S_3$ it is of order 6. It has 2 elements of order 3 which are inveres of one another and 3 elements of order 2.
You have noted one element of order three. its square is the other one.
To be specific. Label the atoms on top as 1, 2, and 3 in clockwise order and the ones right underneath them as 1', 2', and 3'. Since we have 6 atoms I will represent the group as a subgroup of $S_6$ on the set $\{1, 2, 3, 1', 2', 3'\}$ The group has the following elements
$$\{e,(123)(1'2'3'), (132)(1'3'2'),\\(11')(23')(32'), (22')(13')(31'), (33')(12')(21') \}$$ In order to develop a Cayley Table define $a:=(11')(23')(32')$ and $b:=(123) (1'2'3')$ then the elements of the group can be written in the form $a^{\epsilon}b^i \text { for } \epsilon =0,1 \text { and } i = 0,1,2$ and multiplication can be derived from the relations $$a^2=b^3=(ab)^2=e$$