Lecture Notes for CMSC 603 March 5, 1999 Scribed by Xin Zhou

Let us first consider the following question:

Question: If the vertex of a bilateral triangle can be one of two color, how many such triangle exist? One method to solve this problem is to enumerate all such triangles to get the result. However, in this lecture we will see see a symmetric method to solve this problem.
Our first example is geometrical. Let be an equilateral triangle --- it may help to think of the triangle as a flat piece of card with its corners labeled ABC. There are six different transformations of the triangle which have the property that the triangle occupies the same position in space before and after the transformation. These transformations are known as symmetries of the triangle, and there are indicated in the following graph:

r²:

r²f:

Graph 1: Symmetries of equilateral triangle

where "e" means identity, "r" means rotation along centriod 120^o, and "f" means flip.

We have two way to get the above triangle: (1) rotation and flip (2) reflection.

Definition: A group consists of a set G, together with binary operation * defined on G which satisfies the following axioms:

(Closure) .For all x and y in G, x * y is in G
(Associativity). For all x, y, and z in G (x*y)*z = x*(y*z).
(Identity). There is an element e in G such that e*x = x*e = x for all x in G.
(Inverse). For all x in G there is an x' in G such that x * x' = x' * x = e

In the case of the triangle, let C = set of colors = {c, d} and X = C³, where X means the set of all possible of coloring of a triangle. The above group is called D3 (or D6, some people call it because it has 6 element).

When we consider the problem of different coloring, it is more convenient to consider what's going to happen. How many essentially different way to color the triangle? The first step to do this kind of problem is to identify the group.

In the above question, the group element and corresponding cycle notation are:

Group Element Cycle Notation

e (A)(B)(C)

r (ABC)

r² (ACB)

f (A)(BC)

rf (C)(AB)

r² (B)(AC)

Now, let us see some term which are useful to us.

Definition: For an element x in X, orbit_G(x) = {g(x) : g in G}
Definition: The stabilizer of x is stabilizer_G(x) = G_x= {g in G : g(x) = x}
Definition: fix_G(g) = {x in X : g(x) = x}

Now, we can draw a partition of X that breaks down by orbit of G. We have

X = {{ccc}, {ddd}, {cdd, dcd, ddc}, {dcc, cdc, ccd}}.

Next, we can decompose G by stabilizer, we get:

G = {{e, f}, {r, rf}, {r², r²f}}

Lagrange's Theorem: If G is a finite group of order n and H is a subgroup of order m, then m is a divisor of n.

Burnside's Lemma : The number of orbits of G on X is:

# of G-orbits = 1/|G|*\sum_{g \in G}|fix(g)|
By burnside's lemma, we can do the above problem as follows:
# of 2-colors of a triangle = 1/6 ( 8 + 2*2 + 3*4) = 1/6( 8 +4 +12 ) = 4

Thus, we do not need to enumerate all the possible coloring of the triangle. We enumerate group element in stead of enumerate coloring.

Triangle	Group Element	Cycle Notation	fix(g)	\|fix(g)\|
	e	(A)(B)(C)	X	2³= 8
	^r	(ABC)	ccc ddd	2¹= 2
	r²	(ACB)	ccc ddd	2¹= 2
	f	(A)(BC)	ccc ddd cdd dcc	2²= 4
	rf	(C)(AB)	ccc ddd ccd ddc	2²= 4
	r²f	(B)(AC)	ccc ddd cdc dcd	2²= 4

Proof of Busrside's Lemma:
Start with table method, we use the method of counting pairs. let E = {g(x) : g(x) = x} Then the row total r_g(E) is equal to the number of x fixed by g, that is |fix(g)|. Also the column total c_x(E) is equal to the number of g which fix x, that is |Gx|. Hence, the two methods for counting E lead to the equation
\sum_{g \in G}|fix(g)| = \sum_{x \in X}|Gx| = \sum_1^t\sum_{x in orbit(i)}|Gx| = \sum_1^t\sum_{x in orbit(i)}|G|\|orbitG(x)|

==> |Gx| = |G|/|orbitG(x)|

Remark: The mathematics notation is written by Latex grammar.