Burnside’s lemma and beyond:
Permutations of vertices and color
by Lucas Wagner
A harmless problem
How many distinguishable ways are there
to color a 3×3 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?)
Patriotic Examples
R→W→B→R
So let’s solve it!
• There are 39 = 19,683 possible colorings.
• The distinguishable ones would be tedious
to find, and we wouldn’t even know when
to stop looking.
• Let’s develop the background we need in
group theory to tackle this problem.
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, a*b is also in G.
2. * is associative.
a*(b*c) = (a*b)*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.
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.
• Lagrange’s über-theorem for finite groups
states that the size of H must divide the
size of G.
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.
Permuting checkerboard vertices
• Start small: consider the 2×2 case.
• Label the “vertices,” the interesting points
that are moved around.
• The following is the identity permutation:
(1)(2)(3)(4)
Permute those vertices
(1 2 3 4)
(1 2)(3 4)
The operation
• Every group has an operation.
• The symmetric group’s 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)
Expanding our set
• We need a group for our theory to work. If we
have an large group of permutations, we don’t
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.
Symmetries of the square
For the 2×2 checkerboard, there are 8
elements in our group G.
g
permutation
Identity
(1)(2)(3)(4)
90º rotation
(1 2 3 4)
180º rotation
(1 3)(2 4)
270º rotation
…
(1 4 3 2)
…
Symmetries (ctd.)
g
permutation
H reflection
(1 4)(2 3)
V reflection
(1 2)(3 4)
D1 reflection
(1)(2 4)(3)
D2 reflection
(1 3)(2)(4)
Permutations
acting on the colorings
• Now we can begin to talk about coloring
our checkerboards.
• With the 2×2 checkerboard, let’s 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.
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.
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.
Stabilizers
• Fact: The following is a subgroup of G.
Gs {g G | g(s) s}
• Gs is called the stabilizer of s. It is the set
of all permutations, g, that leave a
particular coloring, s, unchanged.
Orbits and stabilizers
• Since Gs is a subgroup of G, the size of Gs
must divide the size of G by Lagrange’s
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|.
Orbit and stabilizer example
• Let s be this particular coloring:
• There are four permutations that leave the
coloring as it looks now—the identity,
180º rotation, and reflections over the
diagonals. Therefore |Gs| = 4.
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|.
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.
But how do we do that?
Tell us now, Lucas!
Burnside’s 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.
The lemma that is not Burnside’s
1
N
G
( g )
gG
where
( g ) s S | g (s) s
which is the size of the set of colorings in S
which are unchanged when acted on by g.
Burnside example
• How many red and white colored 2×2
distinguishable checkerboards are there?
• For each of the 8 group elements, find
ψ(g). For the 90º rotation, ψ = 2. Why?
• Answer: There are only two colorings in S
that are unchanged when rotated by 90º.
Example (ctd.)
• For 180º rotation, there are four elements
of S that are unchanged. So here, ψ = 4.
• 2 choices each for top
two, then the bottom
ones must be whatever
their kitty corner was:
Example (ctd.)
Recall |S| = 24 = 16.
g
permutation
ψ(g)
Identity
(1)(2)(3)(4)
24
90º rotation
(1 2 3 4)
21
180º rotation
(1 3)(2 4)
22
270º rotation
…
(1 4 3 2)
…
21
Example (ctd.)
Notice a pattern yet?
g
permutation
ψ(g)
H reflection
(1 4)(2 3)
22
V reflection
(1 2)(3 4)
22
D1 reflection
(1)(2 4)(3)
23
D2 reflection
(1 3)(2)(4)
23
Example (conclusion)
• The number of colors, raised to the
number of cycles for a given permutation,
will give that ψ.
• The expression for N gives
1
N
G
1
( g ) (16 2 4 2 4 4 8 8) 6
8
gG
Proof of the lemma
Make a table with g and s. Put 1s wherever
gi(sj) = sj and 0s everywhere else.
sS
s1
s2
…
s|S|
g1
0
0
…
1
g2
1
1
…
0
g|G|
0
1
…
1
g G
↑ Elements of the group
← Colorings
Proof (ctd.)
Start counting up the total number of ones:
s1
s2
…
s|S|
g1
0
0
…
1
( g1 )
g2
1
1
…
0
( g2 )
g|G|
0
1
…
1
( g|G| )
Gs1
Gs 2
Gs|S|
f
sS
g G
Proof (ctd.)
• The seemingly uninteresting result is:
G
sS
s
f ( g )
gG
• Which we will use it later. It is easier to
find ψ(g) than Gs. Typically |G| is much
smaller than |S|.
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|,
G
sOrb ( t )
s
Gt
G
1 G
sOrb ( t )
t
Orb(t )
by that starred fact.
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|.
G
sS
s
NG
• I see an N. Victory is nigh.
Proof (conclusion)
Then use the uninteresting result:
( g ) G
gG
sS
s
NG
to get this beauty:
1
N
G
( g )
gG
quod erat demonstrandum.
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 ψ(g), where
ψ(g) is just [number of color choices] to
the [number of vertex cycles of g]th power.
Extensions to color cycling
• For every group element g, include color
cycles in addition to the vertex cycles.
• For example, with the 2×2 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.
Finding ψ(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. ψ(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.
Example calculations
Group elements, g
m1
m2
m3
ψ(g)
[ (1 2 3 4), (R W) ]
2
n/a
n/a
2
[ (1 2)(3 4), (R W) ]
2
2
n/a
4
[ (1)(2 4)(3), (R)(W) ]
2
2
2
8
[ (1)(2 4)(3), (R W) ]
0
2
0
0
!!!
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 ψ for these cases is zero.
←sad ψ
Example (conclusion)
1
• Use the lemma: N
G
( g )
gG
1 16 2 4 2 4 4 8 8 64
N
16 0 2 4 2 4 4 0 0 16
• There are only 4 distinguishable boards.
Initial problem: solved
Question:
How many distinguishable ways are there
to color a 3×3 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.
Applications?
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_Frobenius
• Coding done in Python
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