The symmetry groups of the Platonic solids

 I love the Platonic solids. Just writing this reminds me of a joke on Pee Wee's Playhouse, where Pee-Wee said "I love fruit salad!" and all his friends yell "If you love it so much, why don't you marry it?"

And he replies "All right, I will!"

A Platonic solid is a three dimensional shape whose faces are all equal sized regular polygons. There are exactly five, the 4-sided tetrahedron, the 6-sided cube, the 8-sided octahedron, the 12-sided dodecahedron and the 20-sided icosahedron.

The cube is clearly the best known of the five shapes. I like to say the cube is the Justin Timberlake of the Platonic solids. You might not remember any other member of NSYNC, but you probably know Timberlake's name. The cube is so commonly known, it doesn't have a fancy Greek based name like hexahedron. The cube is a special case of a rectangular solid, and rectangular solids are everywhere. Most boxes and buildings are rectangular solids, and if you see a 90° angle somewhere, it's a good bet that some human being put it there.

Let's count the faces, edges and vertices of these shapes.

Tetrahedron: 4 faces, 6 edges, 4 vertices

Cube: 6 faces, 12 edges, 8 vertices

Octahedron: 8 faces, 12 edges, 6 vertices 

Dodecahedron: 12 faces, 30 edges, 20 vertices

Icosahedron: 20 faces, 30 edges, 12 vertices

 

Notice that vertices + faces = edges + 2. This is called Euler's formula and it is true for any three dimensional shape with polygons as faces that does not have a "hole" in it. To deal with holes, there is the more complex Euler-Poincaré formula, which I won't discuss here.

 

If we switch the face and vertex numbers, we will see the octahedron is related to the cube, the dodecahedron is related to the icosahedron and the tetrahedron is related to itself. This relationship is called duality. One way to think of this is to put a point in the middle of every face of a Platonic solid and connect those points with edges. If you do this to any of the shapes, you will get its dual. For our purposes, this means that the symmetries of the cube is the same group and the symmetries of the octahedron, and likewise the symmetries of the icosahedron is the same as the symmetries of the dodecahedron.

The tetrahedron stands alone, it is its own dual. I have already mentioned in passing that the group of symmetries of the tetrahedron is isomorphic to A4, the alternating group. Let's look at this in greater detail, using the permutation notation to identify the group elements.

The identity

(1)

 

Physically, this means leaving the object alone.

The double transpositions

(12)(34), (13)(24), (14)(23)

If we swap the position of any two vertices, the other two vertices must swap as well.

The 3-cycles

(123), (132), (124), (142), (134), (143), (234), (243)

In these physical movements, one vertex remains fixed and we rotate the opposite face either clockwise or counterclockwise.

If we look at A4 as a subgroup of S4, there would be three conjugacy classes, the lists I enumerated, but conjugacy in a subgroup of the permutation group is not this easy. Because we no longer have all the values of x, the form xax⁻ⁱ will no longer make every 3-cycle conjugate to every other 3-cycle. We still have the property that a 3-cycle can't be conjugate to a double transposition, but these large conjugacy classes may find themselves partitioned into more than one class.

 

In A4, here are the conjugacy classes.

 

(1)

The identity always stands alone.

(12)(34), (13)(24), (14)(23)

Every double transposition can be shown to be conjugate to every other double transposition by using a 3-cycle and its inverse as the x and x⁻ⁱ values in the conjugacy form xax⁻ⁱ.

 

(123), (134), (142), (243)

The conjugacy class that numbers 8 in S4 is split in half, and no 3-cycle is conjugate to its inverse.

 

(132), (143), (124), (234)

A small preview of group representation theory: the order of the finite group G must equal to the sum of k squares, where k is the number of conjugacy classes of G and at least one of the squares is 1², corresponding to the conjugacy class of the identity alone. In this case, k = 4, and the four squares that add up to 12 are 1² + 1² + 1² + 3².

 

Commentary

 

Since I mentioned the name of Henri Poincaré, it's only good manners to include a link to his biography. If people argued about the top ten mathematicians the way they argue about the top ten athletes in any sport, Henri Poincaré would get a lot of votes. My personal top eight are:

Archimedes

Newton

Euler

Gauss

Riemann

Von Neumann

Hilbert 

Poincaré

 

There are multiple people I might consider to fill this list to ten, but I could not make a list that didn't include these eight.

Later this week we will look at the other symmetry groups of the Platonic solids.

The number of finite abelian groups of order n

 In the last post, we learned about how to classify finite abelian groups in two ways, and how much the classification systems rely on the prime factorization of n, the order of the group. Let me present a corollary to the classification standards.

Let the prime factorization of n be written as (p1^e1)(p2^e2)...(pk^ek) where all the values of e are positive integers. The number of different abelian groups of order n is the product of the exponents e1e2...ek.

 

For example, 30 = 2¹*3¹*5¹, also known as a square free number. There is only one finite abelian group of order 30. On the other hand, we can write 60 = 2²*3¹*5¹, so there are two different abelian groups of order 60, Z60 and Z2 ✕ Z30.

 

The fields of group theory and number theory are closely intertwined. 

The classification of finite abelian groups

 Stated without proof: Every finite abelian group is the direct product of a finite number of cyclic groups of the form Zn.

 

If I make up an arbitrary finite list of cyclic groups, it's a finite abelian group. For example,

 

 Z5 ✕ Z14 ✕ Z2 ✕ Z17 ✕ Z44 ✕ Z10 

 

The order of this group is 5 ✕ 14 ✕ 2 ✕ 17 ✕ 44 ✕ 10 = 1,057,200. Because this number is not prime, there are a lot of groups of order 1,057,200, and we need a way to tell them apart and a consistent way to represent them. Mathematicians have decided on two consistent ways to write down a group like this, primary decomposition and cyclic decomposition.

 

First step: write each component subgroup as a direct product of groups whose order are powers of primes.

 

We have three components that are already powers of primes, so I will put them at the front of the list in numerical order.

 Z2 ✕ Z5 ✕ Z17 ✕ Z14 ✕ Z44 ✕ Z10

Now we split of the composite numbers into powers of primes.

Z14 = Z2 ✕ Z7  

Z44 = Z4 ✕ Z11 

Z10 = Z2 ✕ Z5

Notice that Z44 is not Z2 ✕ Z2 ✕ Z11. Z4 has a single generating element, Z2 ✕ Z2 but needs two generators. Now we replace the last three groups with their primary decompositions.

 

Z2 ✕ Z5 ✕ Z17 ✕ Z2 ✕ Z7 ✕ Z4 ✕ Z11 ✕ Z2 ✕ Z5 

 

We put them in numerical order with this proviso: powers of 2 first with highest power at the front of that list, then powers of 3, powers of 5, powers of 7, etc. There are no powers of 3 in the list, so we skip that prime. I will use parentheses to separate the lists of different primes from one another.

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

This is the primary decomposition of our arbitrary group.

 

For the cyclic group, we take the representatives of the primes in order and multiply them together. I have marked the numbers in red.

 

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

4 ✕ 5 ✕ 7 ✕ 11 ✕ 17 = 26,180

 

Next: We multiply the second group orders in each primes' list, here marked in blue.

 

(Z4 ✕ Z2 ✕ Z2 ✕ Z2) ✕ (Z5 ✕ Z5) ✕ (Z7) ✕ (Z11) ✕ (Z17)

 

2 ✕ 5 = 10

 

Now we just have the two remaining copies of Z2. The list {26180, 10, 2, 2} has the property that each number is a multiple of the number that comes after it. So the cyclic decomposition is

 

Z26180 ✕ Z10 ✕ Z2 ✕ Z2 

 

One nice thing about cyclic decomposition is we know largest subgroup generated by a single element has order 26,180. Another point in its favor is it the direct product of just four groups instead of the nine groups we needed for the primary decomposition.