Burnsides lemma and beyond: Permutations of vertices and color

1
Burnsides lemma and beyondPermutations of
vertices and color

• by Lucas Wagner

2
A harmless problem

• How many distinguishable ways are there to color
  a 33 checkerboard with 3 colors, Red, White, and
  Blue, if two colorings are indistinguishable
  through rotations, reflections, shiftings, and
  also a cycle of colors (R?W?B?R)?
• (What does that even mean?)

3
Patriotic Examples
R?W?B?R
4
So lets solve it!

• There are 39 19,683 possible colorings.
• The distinguishable ones would be tedious to
  find, and we wouldnt even know when to stop
  looking.
• Lets develop the background we need in group
  theory to tackle this problem.

5
Groups

• A set of elements, G, along with one operation,
  , is a group if the following hold
• 1. The group is closed WRT .
• For every a, b in G, ab is also in G.
• 2. is associative.
• a(bc) (ab)c for every a, b, c in G.
• 3. There is an identity element, e, in G.
• a e e a a for every a in G.
• 4. There exist inverses for all a in G.
• a a-1 a-1 a e for some a-1 in G.

6
Subgroups

• (H, ) is called a subgroup of (G, ) if the set
  H is contained in the set G and it is also closed
  and has inverses.
• Example Z6 0, 1, 2, 3, 4, 5
• Subgroups are 0, 0, 3, 0, 2, 4, and Z6.
• Lagranges über-theorem for finite groups states
  that the size of H must divide the size of G.

7
Applying groups

• Our main problem boils down to what sorts of
  movements (e.g. rotations, reflections) that we
  allow for our checkerboard.
• These movements can be abstracted to elements in
  a subgroup of a symmetric group, which are
  permutations on the integers.

8
Permuting checkerboard vertices

• Start small consider the 22 case.
• Label the vertices, the interesting points that
  are moved around.
• The following is the identity permutation

9
Permute those vertices
10
The operation

• Every group has an operation.
• The symmetric groups operation is composition,
  ?, which is not necessarily commutative, and
  works right to left.
• E.g., (1 2)(3 4)?(1 2 3 4) (1)(2 4)(3)

11
Expanding our set

• We need a group for our theory to work. If we
  have an large group of permutations, we dont
  want to write them all out by hand.
• Start with a small set of permutations that
  represent each of the basic movements.
• e.g. Rotation by 90º and reflection
• Let a computer expand the set through
  composition. E.g. Rotation by 90º twice becomes
  rotation by 180º.
• This is a surefire way to guarantee closure,
  which for a subset of a finite group guarantees a
  subgroup.

12
Symmetries of the square

• For the 22 checkerboard, there are 8 elements
  in our group G.

13
Symmetries (ctd.)
14
Permutations acting on the colorings

• Now we can begin to talk about coloring our
  checkerboards.
• With the 22 checkerboard, lets consider an easy
  case two colors, red and white.
• Let S be the set of all possible colorings. Then
  the size of S is S 24 16.

15
Some observations

• If u, v belong to S, and u and v are
  indistinguishable WRT our group G, then there is
  some h in G such that h(u) v.
• In this example, h is the 90º rotation.

16
Orbits

• Making general this observation, the collection
  of indistinguishable objects that an element s of
  S belongs to is
• Orb(s) g(s) for all g in G
• The orbit is the set of all colorings which are
  reachable by the action of the elements of G on a
  particular coloring, s.

17
Stabilizers

• Fact The following is a subgroup of G.
• Gs is called the stabilizer of s. It is the set
  of all permutations, g, that leave a particular
  coloring, s, unchanged.

18
Orbits and stabilizers

• Since Gs is a subgroup of G, the size of Gs must
  divide the size of G by Lagranges theorem.
• Neat fact Orb(s) is exactly the divisor.
• i.e. G Gs Orb(s)
• If u and v are indistinguishable colorings, then
  they belong to the same orbit. Therefore Gu
  Gv.

19
Orbit and stabilizer example

• Let s be this particular coloring
• There are four permutations that leave the
  coloring as it looks nowthe identity, 180º
  rotation, and reflections over the diagonals.
  Therefore Gs 4.

20
Example (conclusion)

• Recall that the orbit is the set of all colorings
  which are reachable by the action of the elements
  of G.
• With a 90º rotation, one other coloring is
  reached. Therefore Orb(s) 2.
• Therefore Gs Orb(s) 4 2 8 G.

21
Preparation for the lemma

• Let N be the number of distinguishable objects,
  which it is the goal of our problem to find.
• If we can count the number of total orbits, we
  would know how many distinguishable objects there
  are. We could find N.

22
But how do we do that?

• Tell us now, Lucas!

23
Burnsides lemma

• Historically, Burnside proved the lemma in Theory
  of Groups of Finite Order in 1897.
• But Cauchy knew of it in 1845 and Frobenius in
  1887.

24
The lemma that is not Burnsides

• where
• which is the size of the set of colorings in S
  which are unchanged when acted on by g.

25
Burnside example

• How many red and white colored 22
  distinguishable checkerboards are there?
• For each of the 8 group elements, find ?(g). For
  the 90º rotation, ? 2. Why?
• Answer There are only two colorings in S that
  are unchanged when rotated by 90º.

26
Example (ctd.)

• For 180º rotation, there are four elements of S
  that are unchanged. So here, ? 4.
• 2 choices each for top two, then the bottomones
  must be whatevertheir kitty corner was

27
Example (ctd.)

• Recall S 24 16.

28
Example (ctd.)

• Notice a pattern yet?

29
Example (conclusion)

• The number of colors, raised to the number of
  cycles for a given permutation, will give that ?.
  
• The expression for N gives

30
Proof of the lemma
Make a table with g and s. Put 1s wherever
gi(sj) sj and 0s everywhere else.
? Colorings
? Elements of the group
31
Proof (ctd.)Start counting up the total number
of ones
32
Proof (ctd.)

• The seemingly uninteresting result is
• Which we will use it later. It is easier to find
  ?(g) than Gs. Typically G is much smaller than
  S.

33
Proof (ctd.)

• Now pick some coloring t, and sum up the
  stabilizer sizes, Gs, for all elements s in the
  orbit of t.
• Since the elements of an orbit have the same
  stabilizer size, Gt,
• by that starred fact.

34
Proof (ctd.)

• So for each orbit whose elements we sum Gs
  over, we will get exactly one G.
• If we sum over all colorings in S, we will get
  exactly the number of orbits times G.
• I see an N. Victory is nigh.

35
Proof (conclusion)

• Then use the uninteresting result
• to get this beauty
• quod erat demonstrandum.

36
Implementation

• Code can be written to do most of the
  tediousness.
• Think up any permutations with anything in mind
  (e.g. hexagons, tetrahedra, etc.) and get a
  group.
• The computer can do the calculations for you,
  summing up all of the ?(g), where ?(g) is just
  number of color choices to the number of
  vertex cycles of gth power.

37
Extensions to color cycling

• For every group element g, include color cycles
  in addition to the vertex cycles.
• For example, with the 22 checkerboard, allow the
  switching of colors. Then the usual 90º rotation
  that leaves the colors alone is (1 2 3 4),
  (R)(W) and (1 2 3 4), (R W) rotates 90º and
  cycles the colors
• This doubles the size of the group, to 16.

38
Finding ?(g) with color cycling

• Consider the ith vertex cycle of g, i ranging
  from 1 to the number of vertex cycles in g.
• Let mi(g) be the number of color choices for the
  ith vertex cycle of g. ?(g) will then be the
  product of all mi(g).
• To calculate mi(g), consider each cycle within
  the color permutation. Add in the length of that
  cycle (its number of colors) if it divides the
  length of the ith vertex cycle.

39
Example calculations
!!!
40
Example (ctd.)

• There is no way to color a checkerboard such that
  a diagonal reflection color switch preserves
  the original coloring.
• (1)(2)(3)(4), (R W) suffers from a worse case
  of the same problem.
• Therefore ? for these cases is zero.
• ?sad ?

41
Example (conclusion)

• Use the lemma
• There are only 4 distinguishable boards.

42
Initial problem solved

• Question
• How many distinguishable ways are there to color
  a 33 checkerboard with 3 colors, Red, White, and
  Blue, if two colorings are indistinguishable
  through rotations, reflections, shiftings, and
  also a cycle of colors (R W B)?
• Answer 150 out of the total 19,683.

Case closed.
43
Applications?
44
Sources

• Modern Algebra by John Durbin
• Modern Algebra II Notes, lectures by Dr.
  Biebighauser
• http//en.wikipedia.org/wiki/Burnside's_lemma
• http//en.wikipedia.org/wiki/William_Burnside
• http//en.wikipedia.org/wiki/Cauchy
• http//en.wikipedia.org/wiki/Ferdinand_Georg_Frobe
  nius
• Coding done in Python

45
Question Time

• (e.g., What just happened?)