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Orbit and Stabilizer in Rotational Symmetry

Date: 11/11/2004 at 15:09:23
From: Bridget
Subject: Orbit and Stabilizer

Calculate the orders of the following groups of rotations:

a. of a regular tetrahedron
b. of a regular octahedron
c. of a regular dodecahedron
d. of a regular icosahedron

I'm having trouble figuring out the stabilizers.  I know that the 
order of the group of rotations is equal to the order of the orbit
times the order of the stabilizer.

a.  A regular tetrahedron is a solid with four congruent equilateral 
triangles as faces.  So the orbit would be 4.  As for the stabilizer, 
I get that for a face, there are only 3 rotations that would keep 
that face in the same position.  So I get the order is 4*3 = 12 
rotational symmetries, but I'm not sure if that's right.

b.  A regular octahedron is a solid with 8 congruent equilaterial 
triangles as faces.  Thus, the orbit would be 8, but I'm not sure what 
the stabilizer would be.

c. A regular dodecahedron would have orbit of 12.  And I think that 
the stabilizer would be 5 so the order of the group of rotations would 
be 12*5 = 60?

d.  A regular icosahedron would have orbit 20.  I believe that the 
stabilizer is 6, so it would have 20*6 = 120 rotational symmetries.

Are these correct?  Could you help me figure out the stabilizers?

Date: 11/12/2004 at 02:34:22
From: Doctor Jacques
Subject: Re: Orbit and Stabilizer

Hi Bridget,

You answers for (a) and (c) are correct.

In fact, there is a very intuitive way of seeing this. Consider, for 
example, the case (a)--the regular tetrahedron.

Imagine you want to place a regular tethrahedron on a table, on a 
predetermined (triangular) footprint.

You can place any of the 4 faces on the table (that's the orbit).  
Once a face is on the table, you can turn it in 3 different ways 
(that's the stabilizer--since the face is a triangle, there are 3 
rotations that keep it fixed).  In total, there are 4*3 = 12 ways to 
rotate the tetrahedron.

The only thing to remember is the number of faces and the number of 
vertices per face (i.e., the shape of the faces) for the regular 
polyhedra--if you don't remember that, you can find it almost anywhere 
on the Internet, for example, in our FAQ: 

Does this help?  Write back if you'd like to talk about this some 
more, or if you have any other questions.

- Doctor Jacques, The Math Forum 

Date: 11/12/2004 at 11:02:07
From: Bridget
Subject: Orbit and Stabilizer

So for an octahedron, there are 8 faces of equilateral triangles, and 
3 vertices per triangle, so would it be 8*3 = 24?

For the regular icosahedron, there are 20 equilateral triangles, and 
3 vertices per triangle, so it would be 20*3 = 60?

Date: 11/13/2004 at 03:05:47
From: Doctor Jacques
Subject: Re: Orbit and Stabilizer

Hi again Bridget,

You are right. In fact, we can say even more.

Consider, for example, the cube and the octahedron.  The sizes of the 
rotation groups are 6*4 and 3*8, i.e., they are both equal to 24.  The 
nice point is that, not only do the groups have the same size, they 
are, in fact, isomorphic.

The way to see this is to note that the centers of the 6 faces of a 
cube are the 6 vertices of a regular octahedron, and the converse is
true as well.  So, any rotation of the cube corresponds to a rotation
of a regular octahedron, and conversely.  Such polyhedra are called
dual of each other (to be precise, this is not the exact definition of
duality, but the difference is only in a scaling factor).

The same applies of the dodecahedron (12 faces, 20 vertices) and the 
icosahedron (20 faces, 12 vertices)--their rotation groups are 
isomorphic to each other, and of order 60.

If you apply the process to a regular tetrahedron (4 faces, 4 
vertices), you get nothing new--that one is called self-dual.

By the way, another interesting fact is that the rotation group of the 
cube (and octahedron) is in fact isomorphic to S4, the group of 
permutations of 4 objects.  See, for example:

  Symmetries of a Cube 

It can also be shown that the group of the dodecahedron (and 
icosahedron) is isomorphic to A5, the group of even permutations of 5 
objects, but this is a little more complicated (and practically 
requires a cardboard model to follow the proof...).

- Doctor Jacques, The Math Forum 
Associated Topics:
College Polyhedra
High School Polyhedra

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