# How do you make a Cayley table for the molecule ethane from the D3h group?

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!

• There is (up to isomorphism) only one group with exactly five elements, and I doubt ethane has that group. More likely, there's a sixth element of the group that you're missing. – Gerry Myerson Dec 6 '17 at 23:09
• There are six rotational symmetries (reflections are there as well as geometric symmetries, but they revert the orientation of the entire space and would break the bonds, so you probably don't include those): 1) rotations around the axis connecting the carbon atoms by 0, 120 or 240 degrees. These form a subgroup. 2) 180 degree rotations about three axes through the midpoint between the two carbon atoms. These form a coset of the subgroup formed by the other triple. I don't know which one you are missing :-) – Jyrki Lahtonen Dec 6 '17 at 23:24
• Some of the animations in Otterbein's gallery look a bit strange. I couldn't grok the one including a pumping motion (which is how I would animate a reflection). It seems to be missing the 240 degree rotation altogether. Observe that the rotations about the axis connecting the carbon atoms are not their own inverses. The inverse of the 120 degree rotation is the 240 degree rotation (or the cw and ccw rotations are inverses of each other). Only the three 180 degree rotations are their own inverses. – Jyrki Lahtonen Dec 6 '17 at 23:32
• I could easily build animations of these myself (I have some experience and useful code ready). Unfortunately I probably won't have time for that until Friday. We have plenty of users capable of filling in that Cayley table, so that is hardly a loss :-) – Jyrki Lahtonen Dec 6 '17 at 23:35
• While it is a good idea to do this to get some experience with Cayley tables, I find it backwards to use the table to show that this is a group. The fact that these are the actual symmetries of something is precisely the reason it is a group. – Tobias Kildetoft Dec 7 '17 at 7:14

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.
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 $$a^2=b^3=(ab)^2=e$$
• Good job! ${}{}$ – Jyrki Lahtonen Dec 7 '17 at 7:52