Finding a Rotational Symmetry Group

[Kalinka35]

1. The problem statement, all variables and given/known data
Glue two dodecahedra together along a pentagonal face and find the rotational symmetry group of this solid. What is its full symmetry group?


2. Relevant equations



3. The attempt at a solution
Since this clearly is a finite subgroup of SO3, it must be isomorphic to a cyclic group, a dihedral group, or the group of rotational symmetries of the tetrahedron, cube, or icosahedron.
I have only found two types of axes of rotational symmetry so far: one that runs through the centers of the faces at the ends of the figure and these have order five.
Or we can run an axis along the face where the two dodecahedra are joined and rotate about that by pi. There are 5 of this type of axis. I can't seem to find any other ones, but with just these, the rotational symmetry group is not isomorphic to any of the ones mentioned above.
And I'm not entirely sure how a full symmetry group differs from a rotational one.

Thanks.

[matt grime]

Isn't gluing an object to itself a lot like looking in a mirror? I.e. the other symmetries are reflections, not just rotations.

I'm very bad at doing the geometry in my head (and I don't have a pair of dodecahedra lying around) so can't help with the full question yet, sorry.

[Obraz35]

Well, my guess is that the number of rotational symmetries for the new object is less than the number for the original dodecahedron. I would guess it's dihedral or icosahedral (since I know of one axis with order 5) symmetry even though I do see a couple of problems with those theories, but I can't see what else it would be.
I think I understand the way the reflections work together with the rotations in the full symmetry group although I'm not sure what group they are isomorphic to.

[Billy Bob]

Kalinka35, I'm getting the same 10 rotations you are. Since this rotational group is not cyclic, we are left with D5. How can this be, if there are no reflections? Aha! It is isomorphic to D5. Here's how. Look straight on at the pasted face. (I don't have two dodecahedra either, but I do have one.) Your five rotations about the axis "that runs through the centers of the faces at the ends of the figure" are the five rotations of this pentagon. Now consider the one rotation about the axis "along the face where the two dodecahedra are joined and rotate about that by pi." When you are looking only at the pasted face, this is equivalent to a reflection in the plane about that axis. In other words, the reflection in the plane is realized by a 180 degree rotation in space, which I know we all knew, but surprise here it is!

For the full symmetry group, don't you simply take these 10 rotations and supplement them with their 10 ("true 3d") reflections across the plane of pasting? That can't be D10 though. D5 x Z2? I'd have to think about it some more.

[Billy Bob]

After further thought, the reference to "dodecahera" is a red herring. All that really matters is the pentagonal pasted face. Besides the pentagon, all you need is an additional way to tell if you are looking at the front or back, plus whether it has undergone a reflection in 3 space.

So, a pentagon with blank front, and words on the back. One symmetry (180 degree rotation in 3 space) brings the readable words to the front. Another (reflection in 3 space) brings the words to the front but unreadable.

Or, consider symmetries of a right pentagonal cylinder.

[matt grime]

After further thought, the reference to "dodecahera" is a red herring. All that really matters is the pentagonal pasted face.


Nice observation.